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Contents

PART A. PROJECTIVE PLANE CURVES 5

1 Fundamental ideas 7

1.1 Basic definitions 71.2 Polynomials 91.3 Plane affine curves 101.4 Plane projective curves 121.5 The Hessian curve 161.6 Exercises 201.7 Notes 21

2 Elimination and the resultant 23

2.1 Elimination of one unknown 232.2 The discriminant 322.3 Elimination in a system in two unknowns 332.4 Exercises 362.5 Notes 37

3 Singular points and intersections 39

3.1 The intersection number of two curves 393.2 Bezout’s Theorem 473.3 Rational and birational transformations 513.4 Quadratic transformations 533.5 Resolution of singularities 563.6 Exercises 623.7 Notes 63

PART B. BIRATIONAL PROPERTIES OF PLANE CURVES 65

4 Branches and parametrisation 67

4.1 Formal power series 674.2 Branch representations 784.3 Branches of plane algebraic curves 844.4 Quadratic transformations 864.5 Noether’s Theorem 954.6 Analytic branches 1024.7 Exercises 1114.8 Notes 112

1

2 CONTENTS

5 The function field of a curve 113

5.1 Generic points 1135.2 Rational transformations 1155.3 Places 1215.4 Zeros and poles 1225.5 Separability and inseparability 1245.6 Frobenius rational transformations 1265.7 Derivations and differentials 1275.8 The genus of a curve 1325.9 Higher derivations in positive characteristic 1405.10 The dual and bidual of a curve 1515.11 Exercises 1555.12 Notes 155

6 Linear series and the Riemann–Roch Theorem 157

6.1 Divisors and linear series 1576.2 Linear systems of curves 1666.3 Special and non-special linear series 1736.4 Reformulation of the Riemann–Roch Theorem 1766.5 Some consequences of the Riemann–Roch Theorem 1786.6 The Weierstrass Gap Theorem 1816.7 The structure of the divisor class group 1866.8 Exercises 1876.9 Notes 188

PART C. SPACE CURVES 189

7 Algebraic curves in higher-dimensional spaces 191

7.1 Basic definitions and properties 1917.2 Rational transformations 1957.3 Hurwitz’s Theorem 2007.4 Linear series composed of an involution 2037.5 The canonical curve 2087.6 The osculating hyperplane and the ramification

divisor 2097.7 Non-classical curves with respect to a linear system 2187.8 Classical and non-classical curves with respect to lines 2197.9 Non-classical curves with respect to conics 2207.10 Dual curves of space curves 2287.11 Complete linear series of smaller degree 2327.12 Examples of curves 2447.13 The Linear General Position Principle 2477.14 Castelnuovo’s Bound 2477.15 A generalisation of Clifford’s Theorem 2507.16 The Uniform Position Principle 2507.17 Algebraic varieties and curves 2527.18 Exercises 2547.19 Notes 255

CONTENTS 3

8 Rational points and places over a finite field 257

8.1 Plane curves defined over a finite field 2578.2 Fq-rational branches of a curve 2588.3 Fq-rational places, divisors and linear series 2618.4 Space curves overFq 2668.5 The Stohr–Voloch Theorem 2718.6 Frobenius classicality with respect to lines 2848.7 Frobenius non-classicality with respect to conics 2918.8 Singular plane curves defined overFq 3038.9 The dual of a Frobenius non-classical curve 3088.10 Elliptic curves 3098.11 Classification of non-singular cubics 3158.12 Exercises 3198.13 Notes 320

9 Valuations and residues 321

9.1 Valuation rings 3219.2 Residues of differential forms 3279.3 Notes 332

10 Zeta functions and curves with many rational points 333

10.1 The zeta function of a curve over a finite field 33310.2 The Hasse–Weil Theorem 34610.3 Refinements of the Hasse–Weil Theorem 35110.4 Asymptotic bounds 35610.5 Other estimates 35910.6 Counting points on a plane curve 36110.7 Further applications of the zeta function 36810.8 The Fundamental Equation 37210.9 Exercises 37710.10 Notes 379

11 Maximal and optimal curves 381

11.1 Background on maximal curves 38211.2 The Frobenius linear series of a maximal curve 38411.3 Embedding in a Hermitian variety 39211.4 Maximal curves lying on a quadric surface 40711.5 Maximal curves with high genus 41411.6 Castelnuovo’s number 41611.7 Plane maximal curves 42511.8 Maximal curves of Hurwitz type 42711.9 Non-isomorphic maximal curves 43111.10 Optimal curves 43311.11 Exercises 43811.12 Notes 440

12 Automorphisms of an algebraic curve 443

12.1 The action ofK-automorphisms on places 44312.2 Linear series and automorphisms 449

4 CONTENTS

12.3 Automorphism groups of plane curves 45212.4 A bound on the order of aK-automorphism 45412.5 Automorphism groups and their fixed fields 45712.6 The stabiliser of a place 45912.7 Finiteness of theK-automorphism group 46412.8 Tame automorphism groups 46712.9 Non-tame automorphism groups 46912.10 K-automorphism groups of particular curves 48312.11 Fixed places of automorphisms 49112.12 Large automorphism groups of function fields 49512.13 K-automorphism groups fixing a place 51312.14 Largep-subgroups fixing a place 51912.15 Notes 522

13 Some Families of Algebraic Curves 525

13.1 Plane curves given by separated polynomials 52513.2 Curves with Suzuki automorphism group 54213.3 Curves with unitary automorphism group 55113.4 Curves with Ree automorphism group 55313.5 A curve attaining the Serre Bound 56313.6 Notes 566

14 Applications: codes and arcs 569

14.1 Algebraic geometry codes 56914.2 Maximum distance separable codes 57314.3 Arcs and Ovals 57714.4 Segre’s generalization of the theorem of Menelaus 58314.5 The connection between arcs and curves 58714.6 Arcs in ovals in planes of even order 59014.7 Arcs in ovals in planes of odd order 59114.8 The second largest complete arc 59414.9 Notes 603

15 Appendix on Field Theory and Group Theory 605

15.1 Field Theory 60515.2 Galois Theory 61115.3 Norms and Traces 61315.4 Finite fields 61415.5 Group Theory 61615.6 Notes 626

Bibliography 627

Index 647

PART A

Projective plane curves

Chapter One

Fundamental ideas

The classical idea of an algebraic curve as the set of zeros ofa polynomial is notsufficient. The classical idea works for the study of projective invariants, such asa singular point, but not for the study of birational invariants. It is the birationalinvariants, such as the genus, which are the most important.

The first step in this exposition is to examine projective invariants; for this, thenaive definition of a curve suffices. The next step is to study the effect of thestandard quadratic transformation. This enables a curve tobe transformed to onewith only ordinary singularities.

At this stage, to consider further birational invariants, adifferent definition of thecurve is required. It needs to be considered as the set of centres of its places.

The main projective invariants aredegree, inflexion,k-fold point, ordinary sin-gularity, intersection number.

The main birational invariants aregenus, place, linear series, order and dimen-sion of a linear series, automorphism.

In this chapter, the story begins with basic properties of curves, also includingsome of the peculiarities that occur for finite characteristic, such as the existenceof strange curves, that is, curves whose tangents lines at non-singular points have acommon point.

1.1 BASIC DEFINITIONS

Over the real numbers,R, consider the parabolaF given byF = Y − X2; itspoints form, as in Figure 1.1, the set

(t, t2) | t ∈ R.However, there are two other types of points associated withF , namely, (a)

those at infinity and (b) those with coordinates inC, the algebraic closure ofR.For example, regarding (b), the line with equationy + 1 = 0 meetsF in the twopoints(i,−1), (−i,−1), wherei2 = −1. Regarding (a), ifF is homogenised toF ∗ = Y Z − X2, then the line with equationZ = 0 meets the correspondingprojective curveF∗ at the point(0, 1, 0).

All these ideas need to be considered for a general curve and ageneral field. First,some notation and fundamental definitions for the spaces that appear are explained.

DEFINITION 1.1 (i) For a fieldK, letKn = (x1, x2, . . . , xn) | xi ∈ K, then-fold Cartesian product ofK.

7

8

Figure 1.1 The parabolay = x2 in the real affine plane

r r r r r r r r r r r r r r r r r r r rr

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

(ii) Let V (n,K) ben-dimensional vector space overK , which may be regardedas(Kn,+, ·), where, forxi, yi, λ ∈ K,

(x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1 + y1, x2 + y2, . . . , xn + yn),

λ(x1, x2, . . . , xn) = (λx1, λx2, . . . , λxn).

(iii) The affine planeAG(2,K) = A2(K) is a pair(P,L) where

P = P = (x, y) | x, y ∈ K, L = ℓ = aX + bY + c | a, b, c ∈ K,and apointP = (x, y) lies on aline ℓ = aX + bY + c if ax+ by + c = 0.

(iv) More generally,affine space ofn-dimensionsis AG(n,K) = An(K) withpoints x = (x1, x2, . . . , xn) and r-dimensional subspacesx + S, for r-dimensional subspacesS of V (n,K).

(v) Theprojective planePG(2,K) = P2(K) is a pair(P,L) where

P = P = (x, y, z) = (λx, λy, λz) | x, y, z, λ ∈ K,λ 6= 0,L= ℓ = aX + bY + cZ | a, b, c ∈ K,

and apointP = (x, y, z) lies on aline ℓ = aX+bY +cZ if ax+by+cz = 0.

(vi) More generally,projective space ofn-dimensionsis PG(n,K) = Pn(K)with points

x = (x0, x1, x2, . . . , xn) = (λx0, λx1, λx2, . . . , λxn)

and r-dimensional subspacesS, for (r + 1)-dimensional subspacesS ofV (n+ 1,K).

In each type of space, it is important to consider the structure-preserving trans-formations.

Fundamental ideas 9

DEFINITION 1.2 (i) A linear transformationT : V (n,K) → V (n,K), isgiven as follows:

T (x) = x′ wherex′T = AxT for a suitable non-singular matrixA,with x = (x1, x2, . . . , xn), x′ = (x′1, x

′2, . . . , x

′n).

A semi-linear transformationT : V (n,K)→ V (n,K), is given as follows:T (x) = x′ wherex′T = Aσ(x)T for a suitable non-singular matrixA

whereσ(x) = (σ(x1), σ(x2), . . . , σ(xn)) for some automorphism ofK.

(ii) An affine transformationS : AG(n,K)→ AG(n,K), is given as follows:S(x) = x′ = T (x) + b,

whereT is a linear transformation andb = (b1, b2, . . . , bn)

An affine collineationS : AG(n,K)→ AG(n,K), is given as follows:S(x) = x′ = T (σ(x)) + b.

(iii) A projectivityT : PG(n,K)→ PG(n,K) is given as follows:T (x) = x′ wherex′T = AxT for a suitable non-singular matrixA,

with x = (x0, x1, . . . , xn),x′ = (x′0, x′1, . . . , x

′n). It is also called aprojec-

tive transformationor linear collineation.

A collineationT : PG(n,K)→ PG(n,K) is given as follows:T (x) = x′ wherex′T = Aσ(x)T for a suitable non-singular matrixA.

(iv) WhenK contains the finite fieldFq, the mappingΦ :Fq → Fq

x 7→ xq

is theFrobenius automorphism.

TheFrobenius collineationassociated to the Frobenius automorphism is thecollineation ofPG(n,K) with σ = Φ; that is,

x 7→ x′, x′T = AΦ(x)T ,

with Φ(x) = (xq0, x

q1, . . . , x

qn), for some non-singular matrixA.

1.2 POLYNOMIALS

DEFINITION 1.3 (i) A polynomial f in the ringK[X1,X2, . . . ,Xn] of poly-nomials in the indeterminatesX1,X2, . . . ,Xn is reducibleif there exist non-constantf1, f2 in K[X1,X2, . . . ,Xn] with f = f1f2; otherwise,f is irre-ducible.

(ii) The degreeof a monomialXr11 Xr2

2 . . . ,Xrnn is r1 + r2 + . . .+ rn.

(iii) A polynomial is homogeneousif all its terms have the same degree.

(iv) The degreeof a polynomialf is the largest degree of all its terms; writedeg f .

10

1.3 PLANE AFFINE CURVES

In the first instance, a curve is associated both to a set of points and to a polynomial.LetK be an algebraically closed field, and letF ∈ K[X,Y ]. Then an affine curveis viewed as a set of points.

DEFINITION 1.4 (i) Theaffine curve

F = va(F ) = P = (x, y) ∈ AG(2,K) | F (x, y) = 0.

(ii) The degree ofF , writtendegF , is degF .

LetF = va(F ) be an affine curve withdegF = d, and letℓ = aX+bY +c be aline containing the pointP0 = (x0, y0) onF . Then, for any pointP = (x, y) ∈ ℓ,

ax+ by = ax0 + by0,

a(x− x0) = −b(y − y0) = abt,

x = x0 + bt, y = y0 − atfor somet ∈ K. Then

F (x, y) = F (x0 + bt, y0 − at) = G(t) =G0 +G1 t+G2 t2 + . . .+Gd t

d

=Gm tm + . . .+Gd td, (1.1)

with Gm 6= 0, Gd 6= 0.

DEFINITION 1.5 The affine curveF = v(F ) is irreduciblewhenF is irreducible.

L EMMA 1.6 The two irreducible curvesF1 = v(F1) andF2 = v(F2) are thesame if and only ifF2 = λF1 for someλ ∈ K\0.Proof. This is a consequence of Theorem 2.10. 2

DEFINITION 1.7 If F ∈ K[X,Y ] satisfies

F = Fn11 Fn1

1 . . . Fnss

with eachFi irreducible, thenF = v(F ) hascomponentsFi = v(Fi) with multi-plicity ni for i = 1, . . . , s.

DEFINITION 1.8 Let ℓ be a line which is not a component ofF .

(i) The integerm of (1.1) is theintersection number ofℓ andF at P0: write

m = I(P0, ℓ ∩ F);

(ii) if m = 1 for some lineℓ throughP0, thenP0 is a simpleor non-singularpoint ofF ;

(iii) if m ≥ 2 for all linesℓ throughP0, thenP0 is asingularor multiplepoint ofF ;

Fundamental ideas 11

(iv) if m0 = minm | ℓ a line throughP0, thenm0 is themultiplicity ofP0 onF , orP0 is anm0-fold point ofF , and write

m0 = mP0(F) = mP0

(F );

(v) if m > m0 for any lineℓ, thenℓ is atangent toF at P0.

DEFINITION 1.9 If mP (F) = 2, thenP is adoublepoint ofF . A double pointPwith two distinct tangents toF atP is anode, and with only one tangent toF atP is acusp. If mP (F) = 3, thenP is atriple point ofF .

REMARK 1.10 LetM be a subfield ofK and suppose thatF is defined overM ,that is,F = v(f(X,Y )) with f(X,Y ) ∈ M [X,Y ]. If P is a double point withtwo distinct tangents, neither of them is defined overM , thenP is anisolated pointoverM .

L EMMA 1.11 If P0 is a simple point ofF , then in (1.1)

G1 =∂F

∂X

∣∣∣∣P0

b − ∂F

∂Y

∣∣∣∣P0

a.

COROLLARY 1.12 The tangent toF at a simple pointP = (x, y) is

ℓP =∂F

∂X

∣∣∣∣P

(X − x) +∂F

∂Y

∣∣∣∣P

(Y − y).

Note the meaning of this corollary: the lineℓP has intersection multiplicity at least2 with F atP .

DEFINITION 1.13 A non-singular pointP of F is apoint of inflexionof F if

I(P, ℓP ∩ F) ≥ 3.

Here,P is also called aninflexionor, in some sources, aflex; the tangentℓP atP isthe inflexional tangent.

REMARK 1.14 The behaviour ofP = (0, 0) for a curveF = v(F ) follows simplyfrom the form ofF . Write

F (X,Y ) = Fm + Fm+1 + . . .+ Fd,

whereFi is homogeneous of degreei in X andY , andFm 6= 0. Then

(1) if m > 0, the pointP lies onF ;

(2) if m = 1, the pointP is simple andF1 is the tangent atP ;

(3) if m ≥ 2, the termFm =∏ℓi, whereℓ1, . . . , ℓm are the tangents atP ;

(4) if ℓ1, . . . , ℓm are distinct, thenP is anordinarymultiple point.

EXAMPLE 1.15 LetK be any field.

(1) If F = Y − X3, thenF = va(F ) has no singular points but(0, 0) is aninflexion.

(2) If F = Y 2 −X3, thenF = va(F ) has a singular point(0, 0) of multiplicity2 with the repeated tangentY .

12

1.4 PLANE PROJECTIVE CURVES

For any polynomialF ∈ K[X,Y ] of degreed, associate the homogeneous polyno-mial F ∗ ∈ K[X0,X1,X2], given by

X = X1/X0, Y = X2/X0, F ∗(X0,X1,X2) = Xd0F (X1/X0X2/X0).

Similarly, for any homogeneous polynomialF ∈ K[X0,X1,X2], associate thepolynomialF∗ ∈ K[X,Y ], given by

F∗(X,Y ) = F (1,X, Y ).

DEFINITION 1.16 GivenF ∈ K[X,Y ], theprojective curve

F = v(F ) = v(F ∗) = (x0, x1, x2) ∈ PG(2,K) | F ∗(x0, x1, x2) = 0.This definition implies that the projective curve consists of the affine points plusthe points at infinity ; that is,

v(F ) = (1, x, y) | (x, y) ∈ va(F ) ∪ (0, x, y) | F ∗(0, x, y) = 0

REMARK 1.17 (i) For a linear formL = a0X0 + a1X1 + a2X2, the corre-sponding line is indicated both byL andv(L).

(ii) The notions of irreducibility, components, and degreeextend in a natural wayfrom affine curves to projective curves.

DEFINITION 1.18 (i) With F homogeneous, a pointP = (x0, x1, x2) of F issingular if

∂F

∂X0=

∂F

∂X1=

∂F

∂X2= 0

atP .

(ii) Otherwise,P is simple and thetangent atP is

∂F

∂X0

∣∣∣∣P

X0 +∂F

∂X1

∣∣∣∣P

X1 +∂F

∂X2

∣∣∣∣P

X2.

Throughout, write

X∞ = (0, 1, 0), Y∞ = (0, 0, 1), O = (1, 0, 0).

Here,X∞ is thepoint at infinityon theX-axis,Y∞ is thepoint at infinityon theY -axis, andO is theorigin.

In Example 1.15 (1),Y∞ is a cusp onF and, in (2), it is an inflexion.In general, to determine the behaviour of a point, translateit to the origin, and

use Remark 1.14.Sometimes, it is convenient to use the notation:

U0 = (1, 0, 0), U1 = (0, 1, 0), U2 = (0, 0, 1).

REMARK 1.19 The properties of projective curves such as the degree, multiplic-ity of singular points, multiplicity of contact of a tangentare covariant, that is,invariant under projective transformations.


LetF = v(F (X0,X1,X2)) be a plane projective curve of degreed.

DEFINITION 1.20 For any pointP = (x0, x1, x2), if

G(X0,X1,X2) =∂F

∂X0x0 +

∂F

∂X1x1 +

∂F

∂X2x2

is not the zero polynomial, the plane curveG = v(G(X0,X1,X2)) of degreed− 1 is thepolar curve ofP with respect toF . Otherwise, the polar curve ofP isvanishing.

THEOREM 1.21 The polar curve is covariant.

Proof. If (X ′0,X

′1,X

′2) is a new homogeneous coordinate system, the link between

the old frame and the new one is given by a linear substitutionXi =∑2

j=0 aijX′j ,

i = 0, 1, 2, such that the matrix(aij) is non-singular. Under this change, the imageof F is the curveF ′ = v(F ′) with

F ′(X ′0,X

′1,X

′2) = F (

∑2j=0 a0jX

′j ,∑2

j=0 a1jX′j ,∑2

j=0 a2jX′j),

while the image ofP is the pointP ′ = (∑2

j=0 a0jxj ,∑2

j=0 a1jxj ,∑2

j=0 a2jxj).Covariance means that the diagram below is commutative, which is now shown.

F = v(F ) F ′ = v(F ′)

G = v(G) G′ = v(G′)

-(aij)

?

polar ofP

?

polar ofP ′

-(aij)

By the usual rules of the derivatives of composite functions,

∂F ′

∂X ′k

=∑2

i=0 aik∂F

∂Xi, k = 0, 1, 2.

Therefore,

∑2k=0

∂F ′

∂X ′k

x′k =∑2

i=0

∂F

∂Xixi.

2

THEOREM 1.22 Assume that the polar curveF ′ of P with respect toF is notvanishing.

(i) LetA be a non-singular point ofF . ThenA ∈ F ′ if and only if the tangenttoF atA passes throughP .

(ii) LetA be anr-fold point ofF , with r > 1.

(a) ThenA is at least an(r − 1)-fold point ofF ′.

(b) If at least one tangent toF atA has multiplicity not divisible byp, thenthis holds for all but finitely many pointsP .

14

Proof. LetA = (a0, a1, a2). The tangent line toF atA is

∂F

∂X0X0 +

∂F

∂X1X1 +

∂F

∂X2X2,

where the partial derivatives are evaluated at(a0, a1, a2). Thus, (i) is a consequenceof the definition of a polar curve.

To show (ii), note that the covariance of the polar curve allowsA to be mappedto the origin. Write

F = Φ0Xn0 + Φ1X

n−10 + . . .+ ΦiX

i0 + . . .+ Φn,

whereΦi is a homogeneous polynomial inX1 andX2 of degreei. SinceA is anr-fold point,Φ0 = . . . = Φr−1 = 0, butΦr 6= 0. Then

G = G1 +G2,

where

G1 =∂Φr

∂X1x1 +

∂Φr

∂X2x2,

and where the terms inG2 have degree at leastr in X1 andX2. SinceG1 hasdegree at leastr − 1, so (a) follows.

If there are infinitely many pointsP for whichA is ans-fold point ofF ′ withs ≥ r, then both∂Φr/∂X1 and∂Φr/∂X2 vanish. This only occurs whenΦr is ap-th power of some polynomialΨr. But then every tangent toF atA has multiplicitydivisible byp. 2

THEOREM 1.23 If F is irreducible of degreed, then there is at most one pointPwith vanishing polar curveF ′.

Proof. Assume, on the contrary, that the polar curves of two distinct points arevanishing. By the covariance of polar curves, let these points beX∞ andY∞. Then∂F/∂X1 and∂F/∂X2 are both zero. Hence the general term inF (X0,X1,X2) is

aijXd−(i+j)0 Xi

1Xj2

with both i and j divisible by p. Thenp dividesd, as otherwiseX0 dividesF ,contradicting the irreducibility ofF . So,p also dividesd − (i + j). Now, definethe coefficientsbij by bpij = aij , and the polynomialL =

∑bijX

d−(i+j)0 Xi

1Xj2 .

Then,F = Lp andF is reducible, a contradiction. 2

Points with vanishing polar curves are characterised by a purely geometric prop-erty.

THEOREM 1.24 If F is irreducible, then the polar curve ofP is vanishing if andonly if the tangents toF at non-singular points pass throughP .

Proof. By the covariance, suppose thatP is the pointY∞. With F in its inhomo-geneous formF (X,Y ) =

∑aijX

iY j , the polar curve ofP is vanishing if andonly if aij = 0 wheneverj 6≡ 0 (mod p). This occurs if and only if there is noX2-term in the form of the tangent toF at any non-singular point. 2

This condition can be weakened.


THEOREM 1.25 If F is irreducible, then the polar curve ofP is vanishing if andonly if infinitely many tangents toF pass throughP .

Proof. If the polar curveF ′ of P is vanishing, the assertion follows from Theorem1.24. IfF is not vanishing, then the weak form of Bezout’s Theorem 3.13 appliedto F andF ′ implies that there are finitely many common points. SinceF has afinite number of singular points,P lies only on finitely many tangents toF . 2

As becomes apparent later, vanishing polar curves can causeserious difficultiesin certain situations, especially when the point lies on thecurve. Here, some exam-ples are given after formalising these concepts.

DEFINITION 1.26 (i) A point is thenucleusof an irreducible curveF if it isthe common point of all tangents toF at non-singular points.

(ii) A curveF is strangeif it admits a nucleusN .

Following this definition, Theorem 1.24 states that a pointP is a nucleus ofF ifand only if the polar curve ofP with respect toF is vanishing. Theorem 1.23 hasthe following corollary

THEOREM 1.27 An irreducible curve has at most one nucleus.

The simplest example of a strange curve is an irreducible conic in characteristic2.Note that the nucleus does not lie on the conic.

EXAMPLE 1.28 An example of a strange curveF whose nucleus lies in the curveis the following. Assume thatK has characteristic2, and letq = 2r. For an integerk < q with gcd(k, q − 1) = 1, let k = 2su with 1 < s < r, u > 1 andu odd. LetF be the irreducible curvev(f(X,Y )), where

(Xk + 1)(Y + 1) + (Y k + 1)(X + 1)

= f(X,Y )(X + 1)(Y + 1)(X + Y ). (1.2)

To show thatN = (1, 1, 1) is a nucleus ofF , write (1.2) as

(Xu)2s

Y + (Y u)2s

X + (Xu + Y u)2s

= f(X,Y )(X + 1)(Y + 1)(X + Y ). (1.3)

Using this form, the tangent toF at a non-singular pointA = (a0, a1, a2) of F is

(ak1 + ak

2)X0 + ak1X1 + ak

2X2.

This shows that the tangent passes through the pointN . It should be noted thatnecessary conditions forN to lie onF are thatu is odd ands > 1.

The singular points ofF are as follows:

(i) N is an ordinary(2s − 2)- fold point with a single tangent

(m+ 1)X0 +mX1 + (m+ 1)X2

for m2s−1 = 1 butm 6= 1;

16

(ii) for eachb with bu = 1 but b 6= 1, the pointP = (1, b, 1) is a (2s − 1)-plepoint with a single tangent;

(iii) for eachc with cu = 1 but c 6= 1, the pointP = (1, 1, c) is a (2s − 1)-plepoint with a single tangent;

(iv) for all b, c with bu = cu = 1 but b 6= 1, c 6= 1, the pointP = (1, b, c) is a2s-ple point with a single tangent.

EXAMPLE 1.29 An example in odd characteristic is the following.Let p > 2, andF = v(Y m− g(X)) with m 6≡ 0 (mod p). ThenF is strange if

and only if one of the following hold:

(i) g(X) = f(Xp) + cX, wheref ∈ K[X] andc ∈ K with c = 0 for m 6= 1;

(ii) g(X) = (X + a)rf(Xp) wherea ∈ K, f ∈ K[X] and1 ≤ r < p withr ≡ m (mod p).

Strange curves have exceptional behaviour with respect to duality; this is treatedin Section 5.10. They also cause difficulties in the resolution of singularities; seethe last part of the proof of Theorem 3.27.

1.5 THE HESSIAN CURVE

LetF = v(F (X0,X1,X2)) be a plane curve of degreed. Write

Fi =∂F

∂Xi, Fij =

∂2F

∂Xi∂Xj. (1.4)

DEFINITION 1.30 If the determinant

H(X0,X1,X2) =

∣∣∣∣∣∣

F00 F01 F02

F01 F11 F12

F02 F12 F22

∣∣∣∣∣∣

is not vanishing, then the curveH = v(H(X0,X1,X2)) of degree3(d− 2) is theHessiancurve ofF . Otherwise, the Hessian curve isvanishing.

Strange curves have vanishing Hessian; see Exercise 3. Further relevant examplesof curves with vanishing Hessian are the Hermitian curves.

THEOREM 1.31 The Hessian curve is covariant.

Proof. Arguing as in the proof of Theorem 1.21, for0 ≤ i, j ≤ 2,

∂2F ′

∂X ′i∂X

′j

=∑2

k=0

∑2m=0 akiamj

∂2F

∂Xk∂Xm,

This shows that

H ′(X ′0,X

′1,X

′2) = H(X0,X1,X2) · det(aij)

2.

Sincedet(aij) 6= 0, the assertion follows. 2


THEOREM 1.32 If d ≡ 1 (mod p), then the Hessian curve is vanishing.

Proof. Let G ∈ K[X0,X1,X2] be a homogeneous polynomial of degreem. ByEuler’s formula,

∂G

∂X0X0 +

∂G

∂X1X1 +

∂G

∂X2X2 = mG.

WhenG = F , this becomes

∂F

∂X0X0 +

∂F

∂X1X1 +

∂F

∂X2X2 = dF, (1.5)

while forG = ∂F/∂Xi with i = 0, 1, 2,

∂2F

∂Xi∂X0X0 +

∂2F

∂Xi∂X1X1 +

∂2F

∂Xi∂X2X2 = (d− 1)

∂F

∂Xi. (1.6)

Then, with the notation of (1.4),

X0H(X0,X1,X2) =

∣∣∣∣∣∣

(d− 1)F0 F01 F02

(d− 1)F1 F11 F12

(d− 1)F2 F12 F22

∣∣∣∣∣∣.

Therefore,H(X0,X1,X2) = 0 whend ≡ 1 (mod p). 2

The fundamental property of the Hessian curve is stated in the following theo-rem.

THEOREM 1.33 Assume thatH is not vanishing.

(i) Letp 6= 2. Then a non-singular point ofF is an inflexion if and only if it is acommon point ofF andH.

(ii) Every singular point ofF lies onH.

Proof. Let P be a non-singular point ofF . Again by the covariance, suppose thatP is the origin and that the tangent lineℓ toF atP is theX-axis. Then

F (X0,X1,X2) = Xd−10 X2 +Xd−2

0 (a20X21 + a11X1X2 + a02X

22 ) + . . . ,

where the other terms are powers ofX0 with exponent at mostd − 3. A straight-forward calculation shows that

H(X0,X1,X2) = −2(d− 1)2a20X3(d−2)0 + . . . ,

where other the terms are powers ofX0 with exponent less than3(d − 2). Sincep 6= 2 andd 6≡ 1 (mod p) by Theorem 1.32,H passes throughP if and only ifa20 = 0; that is,P an inflexion. IfP is singular, then the above computation showsthat no term containingX0 appears inH(X0,X1,X2). HenceP ∈ H. 2

In affine form, whenF = v(f(X,Y )), write

fX =∂f

∂X, fY =

∂f

∂Y, fXX =

∂2f

∂X2, fXY =

∂2f

∂X∂Y, fY Y =

∂2f

∂Y 2.

18

L EMMA 1.34 Let d 6≡ 1 (mod p). If F = v(f(X,Y )), thenH = v(h(X,Y )),where

h(X,Y ) = (fX)2fY Y + (fY )2fXX − 2fXfY fXY

−d(d− 1)−1(fXXfY Y − (fXY )2

)f.

Proof. As in the proof of Theorem 1.32, the equations (1.5) and (1.6)are useful.Substituting inH(X0,X1,X2), the expression reduces to

X20H(X0,X1,X2) =

∣∣∣∣∣∣

(d− 1)dF F1 F2

(d− 1)2F1 F11 F12

(d− 1)2F2 F12 F22

∣∣∣∣∣∣.

In inhomogeneous coordinates, this becomes

h(X,Y ) =

∣∣∣∣∣∣

(d− 1)d f fX fY

(d− 1)2 fX fXX fXY

(d− 1)2 fY fXY fY Y

∣∣∣∣∣∣.

Expansion of the determinant and division by−(d− 1)2 give the result. 2

When studying the inflexion points ofF or, more generally, the intersection ofH andF , the last term of the polynomial in Theorem 1.34 can be omitted. So, it isalso possible to define the Hessian curve asH′ = v(h′(X,Y )), with

h′(X,Y ) = (fX)2fY Y + (fY )2fXX − 2fXfY fXY . (1.7)

An advantage is that the Hessian curve in this form is no longer necessarily van-ishing whenF has degreed ≡ 1 (mod p), see Exercise 2. Therefore, curves withsuch degrees can be considered in the study of inflexion points. With the approachabove, just two remarks are needed.

(1) Theorem 1.31 holds true ifH is given by (1.7).

(2) Rewording the proof of Theorem 1.33 in terms of inhomogeneous coordi-nates,h(X,Y ) = −2a20 + . . ., which shows that conditiond 6≡ 1 (mod p)depending on Theorem 1.32 disappears.

Therefore, the following result holds.

THEOREM 1.35 Let p 6= 2. If either (1.7) is identically zero or the HessianH ofF as in (1.7) contains all points ofF , then every non-singular point ofF is aninflexion. The converse also holds.

REMARK 1.36 The form (1.7) of the Hessian curve is inadequate to deal withtheeven characteristic case. What is actually needed is to modify the definition of thesecond partial derivatives, as suggested by Hasse. Iff =

∑aijX

iY j then theHasse second partial derivatives are

∂(2)f(X,Y )

∂X(2)=∑

aij

(i

2

)Xi−2Y j ,

∂(2)f(X,Y )

∂Y (2)=∑

aij

(j

2

)XiY j−2.


Setf = f(X,Y ) for brevity. If p 6= 2, then

∂2f

∂X2= 2

∂(2)f

∂X(2),

∂2f

∂Y 2= 2

∂(2)f

∂Y (2).

Now, the Hessian curveH = vh, with inhomogeneous form is defined to be thecurve of equation

h(X,Y ) =

(∂f

∂X

)2∂(2)f

∂X(2)+

(∂f

∂Y

)2∂(2)f

∂Y (2)− ∂f

∂X

∂f

∂Y

∂2f

∂X∂Y. (1.8)

With this definition,H is still covariant, and Theorem 1.33 holds true forp = 2.

REMARK 1.37 The vanishing of the Hessian is related to exceptional behaviour ofthe dual curve in positive characteristic. This is treated in Section 5.10. The Hassederivatives are useful in several other contexts; see Sections 5.9 and 7.6.

EXAMPLE 1.38 Let n be a positive integer which is not divisible byp. The curve

F = v(Xn0 +Xn

1 +Xn2 )

is theFermat curve of degreen.WhenK is the finite fieldFq2 with q a power ofp andn = q+1, then the Fermat

curve is theHermitian curveand denoted byHq; that is,

Hq = v(Xq+10 +Xq+1

1 +Xq+12 ).

The properties of the Hermitian curve are further developedin Section 13.3.The Fermat curve is the locus of its inflexion points if and only if n ≡ 1 (mod p)

butn 6≡ 0 (mod 4) whenp = 2. In particular, this holds for the Hermitian curveHq whenp 6= 2.

EXAMPLE 1.39 Assume thatK has characteristicp > 0. Choose a powerq of psuch thatq ≡ −1 (mod 3). LetF = v(F ) be the plane curve with homogeneousform

F (X0,X1,X2) = Xq0X2 +Xq

1X0 +Xq2X1 − 3(X0X1X2)

(q+1)/3.

It is first shown thatF is an irreducible curve with only ordinary singularities.Each of the fundamental lines meetsF in only two points. IfG is a component

of F thenG meets each fundamental line in at least one point, and so at least two ofthe vertices of the fundamental triangle are onG. If G1 is another component ofF ,then it also passes through two of the vertices of the fundamental triangle. So, oneof these points is common toG andG1, and hence is a singular point ofF . However,this is not the case, since a first partial derivative ofF is not zero at each of thesepoints. More precisely,FX0

(0, 1, 0) = FX1(0, 0, 1) = FX2

(1, 0, 0) = 1 6= 0.Next, it is shown thatF has exactly(q2−q+1)/3 singular points, each of which

is an ordinary double point. This requires some typical computations involvingpolynomials and their partial derivatives.

Since no vertex of the fundamental triangle is a singular point ofF , it suffices toconsider affine points. So, putX = X1/X0, Y = X2/X0; thenF = v(f(X,Y ))with

f(X,Y ) = Y +Xq +XY q − 3(XY )(q+1)/3.

20

LetA = (a, b) be a singular point ofF . Thenab 6= 0 andfX(a, b) = fY (a, b) = 0.A direct calculation shows that this implies

1 = a(q−2)/3b(1−2q)/3, (1.9)

1 = a(q+1)/3b(q−2)/3. (1.10)

Thus,a(q−2)/3b(1−2q)/3 = a(q+1)/3b(q−2)/3, whencea = b1−q. Now, (1.10) im-plies the following:

b(q2−q+1)/3 = 1, a = b−q2

.

Conversely, ifa, b ∈ K satisfy these conditions, thenfX(a, b) = fY (a, b) = 0,showing that every pointA = (a, b) such thatb(q

2−q+1)/3 = 1, a = b1−q = b−q2

is a singular point ofF . Also,

fXX(a, b) = 23 a

(q−5)/3b(q+1)/3,

fY Y (a, b) = 23 a

(q+1)/3b(q−5)/3,

fXY (a, b) = 13 (ab)(q−2)/3,

fXX(a, b)fY Y (a, b)− fXY (a, b)2 = 13 (ab)(2q−4)/3.

It follows that, ifp 6= 2, thenA is a double point, more precisely, a node. The sameholds true forp = 2, but the proof requires Hasse partial derivatives, as explainedin Remark 1.36.

Finally, the HessianH ofF isv((X0X1X2)(2q−4)/3), which shows thatH splits

into three linear components, each counted(2q − 4)/3 times.

EXAMPLE 1.40 For a divisork ≥ 2 of q − 1, andu, v ∈ K with uv 6= 1, theplane curveF = v(vXkY k −Xk − Y k + u) is irreducible and has the followingproperties:

(i) F has two singular points, namelyX∞ andY∞, both ordinaryk-fold points.

(ii) The HessianH = v(H) of F has

H = k3Xk−2Y k−2(Xku− 1)(Y ku− 1)(2XkY ku+ (k − 1)(Xk + Y k)).

In particular,F is not the locus of its inflexions.

(iii) For eacha ∈ K with ak = u, the pointsA1 = (1, 0, a) andA2 = (1, a, 0) ofF are inflexions. Fori = 1, 2, the tangent toFk atAi is ℓi = v(Xi − aX0).Also, I(Ai,Fk ∩ ℓi) = k.

1.6 EXERCISES

1. LetK have odd characteristicp. Put

F (X0,X1,X2) = Xq+10 +Xq+1

1 +Xq+12 +Xq−1

0 X1X2,

andC = v(F ). Show that the Hessian curveH of C in its inhomogeneousform (1.7) is non-vanishing.


2. LetK have characteristic2. Show that the Hessian curveH of the HermitiancurveH2 in its inhomogeneous form (1.8) is not vanishing.

3. LetK have odd characteristicp. Prove that the Hessian of a strange curvevanishes.

4. Prove the a plane curveF of degreek ≥ 2 is irreducible if it has both prop-erties:

(a) F contains a pointA such thatI(A,F ∩ ℓ) = k for the tangentℓ of FatA whenA is a simple point, and for at least one tangentℓ toF atAwhen isA is a singular point. In the latter case,ℓ is required to havemultiplicity 1.

(b) F contains no linear component through the point above pointA.

5. Prove that any plane curve of degree at least3 whose tangent lines at collinearpoints are concurrent is either strange or projectively equivalent to the Her-mitian curve.

1.7 NOTES

There are many works on the elements of algebraic curves: Bertini [30], Coolidge[51], Enriquez and Chisini [74], Fischer [78], Fulton [84],Lefschetz [185], Hilton[125], Salmon [232], Segre [251], Seidenberg [252], Sempleand Kneebone [254],Semple and Roth [255], Severi [260], Walker [303].

Example 1.29 comes from [138]. For Exercise 5, see [142].For strange curves, as in Section 1.4, see Hartshorne [114, Chapter IV].In [20], strange curves invariant under a cyclic linear collineation group fixing a

triangle are investigated.In the classical literature, polar curves and their generalisations, them-ic po-

lar curves, together with Hessian curves are used to establish the Plucker formu-las and their generalisations; these are equations linkingnumerical parameters ofcurves, such as the number of nodes, cusps, inflexion points,singularities of highermultilpicities, and other constants; see [51, Chapter 9]. Polar curves in positivecharacteristic are studied in [118].

Hessian curves are even important in investigating the number of points of curvesdefined over a finite field. For instance, letF = v(F ) be a projective irreduciblecurve of degreed whereF ∈ Fq[X0,X1,X2]) with Fq a finite field of odd orderq. Suppose that its Hessian curve is not vanishing. Then the number of points ofF lying in PG(2, q) is bounded above by12d(q + d− 1); see Theorem 8.40. Forqsquare andd =

√q + 1, this upper bound is12

√q(√q + 1)2.

On the other hand, the Hermitian curveH√q has vanishing Hessian and has

q√q+1 points lying inPG(2, q). Also, the polar curve of any point with respect to

H√q is a reducible curve, being a line counted

√q times. See Chapter 13 for more

examples of plane curves in positive characteristic with properties that a complexalgebraic curve cannot have.

Chapter Two

Elimination and the resultant

To obtain information on the intersection of two curves, it is natural to consider theexistence of common roots of two polynomials.

2.1 ELIMINATION OF ONE UNKNOWN

Theresultantdecides whether or not two given polynomials have a root in common.Two polynomialsf(X) andg(X) with coefficients in a fieldL, not necessarilyalgebraically closed, have a root in common inL or in its closureK if and only ifthey have a factor in common.

Consider the two polynomials:

f(X) = a0Xn + a1X

n−1 + · · ·+ an−1X + an,g(X) = b0X

s + b1Xs−1 + · · ·+ ns−1X + bs.

(2.1)

Let α1, . . . , αn be the roots off(X) and letβ1, . . . , βs be the roots ofg(X); theymay lie inK if L is not algebraically closed.

DEFINITION 2.1 Theresultantof the polynomialsf(X) andg(X) is

R(f, g) = (a0sb0

n)∏n

i=1

∏sj=1(αi − βj).

It is a symmetric polynomial in the two sets of indeterminatesαi, βj .

L EMMA 2.2 The resultantR(f, g) is zero if and only if there exists a common rootof the polynomialsf(X) andg(X).

L EMMA 2.3 The follow properties hold:

(i) R(f, g) = a0s∏n

i=1g(αi);

(ii) R(f, g) = b0n∏s

j=1f(βj);

(iii) R(f, g) = (−1)nsR(g, f);

(iv) R(f1f2, g) = R(f1, g) ·R(f2, g).

Proof. (i) By definition,

R(f, g) = a0sb0

n∏ni=1


Therefore

g(X) = b0(X − β1) · . . . · (X − βs) = b0∏s

j=1(X − βj).

23

24

Now, substitute in place ofX the valueαi:

g(αi) = b0(αi − β1) · . . . · (αi − βs) = b0∏s

j=1(αi − βj).

Then∏n

i=1g(αi) = b0(α1 − β1) · . . . · (α1 − βs)b0(α2 − β1) · . . . · (α2 − βs) · . . .b0(αn − β1) · . . . · (αn − βs)

= b0n∏n

i=1


So∏n

i=1g(αi) = bn0∏n

i=1


Multiplying both sides of this equation byas0 and using Definition 2.1 shows that

R(f, g) = as0

∏ni=1g(αi).

(ii) The proof of this is analogous.(iii) This follows immediately.(iv) Let deg g = s, deg(f1f2) = m+ n, with m = deg f1 andn = deg f2. The

leading coefficient off1f2 is a10a20, wherea10 anda20 are the leading coefficientsof f1 andf2. The roots off1f2 are given byx1, . . . , xn andy1, . . . , ym, which arejust the roots off1 andf2; write these roots asαi for i = 1, . . . ,m+ n. Then

R(f1f2, g) = (a1,0a2,0)s∏m+n

i=1 g(αi)

= fs1,0

∏ni=1g(αi) · fs

2,0

∏n+mi=n+1g(αi)

=R(f1, g) ·R(f2, g).

2

COROLLARY 2.4 The product rule for resultants is as follows:

R(∏n

i=1fi, g) =∏n

i=1R(fi, g).

The resultantR(f, g) can be expressed as a polynomial in the indeterminatesa0, . . . , an, b0 . . . , bs. To show this, the theory of symmetric polynomials is used.From Definition 2.1,

R(α1, . . . , αn, β1, . . . , βs) = (a0sb0

n)−1R(f, g) =∏n

i=1

∏sj=1(αi − βj)

is a symmetric polynomial with respect to the two setsα1, . . . , αn andβ1, . . . , βs.

Let σ1, . . . , σn be the elementary symmetric polynomials in the indeterminatesα1, . . . , αn and letτ1, . . . , τs be the polynomials inβ1, . . . , βs. By Newton’s theo-rem on symmetric functions,R(α1, . . . βs) can be expressed in terms of theσi andτj . Up to sign,σ1, . . . , σn andτ1, . . . , τs are the coefficients of the polynomialsa−10 f(X) andb−1

0 g(X). SoR(α1, . . . , βs) can be expressed as a polynomial in theindeterminates

a1

a0, . . . ,

an

a0,b1b0, . . . ,

bsb0.

The factora0sb0

n makes the denominator inR(α1, . . . , βs) disappear.Now, it is shown thatR(f, g), as given by Definition 2.1, is equal to a determinant

D of ordern+ s.

Elimination and the resultant 25

THEOREM 2.5 Given

f(X) = a0Xn + a1X

n−1 + · · ·+ an−1X + an,

g(X) = b0Xs + b1X

s−1 + · · ·+ bs−1X + bs,

thenR(f, g) = D, where

D =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a0 a1 . . . an 0 0 . . . 00 a0 a1 . . . an 0 . . . 0...

......

0 0 . . . 0 a0 a1 . . . an

b0 b1 . . . bs 0 . . . 00 b0 b1 . . . bs 0 . . . 0...

......

0 . . . 0 b0 b1 . . . bs

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

.

s rows

n rows

(2.2)

Proof. Let α1, . . . , αn be the roots off(X) andβ1, . . . , βs the roots ofg(X).Consider another determinant of ordern+ s:

M =

∣∣∣∣∣∣∣∣∣∣∣∣∣

βn+s−11 βn+s−1

2 · · · βn+s−1s αn+s−1

1 · · · αn+s−1n

βn+s−21 βn+s+−2

2 · · · βn+s−2s αn+s−2

1 · · · αn+s−2n

......

......

...β2

1 β22 · · · β2

s α21 · · · α2

n

β1 β2 · · · βs α1 · · · αn

1 1 · · · 1 1 · · · 1

∣∣∣∣∣∣∣∣∣∣∣∣∣

. (2.3)

The product(a0sb0

n)DM is now calculated in two different ways. First,M is aVandermonde determinant , whose value is given by the following formula:

M =∏

1≤ i < j ≤s(βi − βj)∏s

j=1

∏ni=1(βj − αi)

∏1≤ i < j ≤n(αi − αj).

Now,

a0sb0

nDM= Da0

sb0n∏s

j=1

∏ni=1(βj − αi) ·

∏1≤ i < j ≤s(βi − βj)

∏1≤ i < j ≤n(αi − αj)

= D ·R(f, g)∏

1≤ i < j ≤s(βi − βj)∏

1≤ i < j ≤n(αi − αj).(2.4)

The product of the matrices associated to the determinantsD andM is the matrix

N =

[B 0

0 A

],

where

B =

βs−11 f(β1) βs−1

2 f(β2) · · · βs−1s f(βs)

βs−21 f(β1) βs−2

2 f(β2) · · · βs−2s f(βs)

...... · · ·

...f(β1) f(β2) · · · f(βs)

,

A =

αn−1

1 g(α1) · · · αn−1n g(αn)

... · · ·...

g(α1) · · · g(αn).

.

26

Expanding the determinantdetN of N by Laplace’s method and taking out col-umn and row factors, and calculating the corresponding Vandermonde determinantsgives the following:

(a0sb0

n) detN

= (as0 b

n0 )∏s

j=1f(βj)∏n

i=1g(αi)∏

1≤k<l≤s(βk − βl)∏

1≤r<t≤n(αr − αt).

Regarding this as a polynomial relation between the indeterminatesα1, . . . , αn,β1, . . . , βs, and using the fact that

R(f, g) = as0

∏ni=1g(αi), R(g, f) = bn0

∏sj=1f(βj),

it follows that

(as0 b

n0 ) detN = R(f, g) ·R(g, f)

∏1≤i<j≤s(βi−βj)

∏1≤i<j≤n(αi−αj). (2.5)

Equating the expressions (2.4) and (2.5), and simplifying,gives the required resultthatR(f, g) = D. 2

PROPOSITION 2.6 The determinantD is isobaric of weightn·s in the indetermin-atesa0, . . . , an, b0, . . . , bs.

Proof. Given a general term ofD,

cai00 a

i11 · . . . · ain

n bj00 b

j11 · . . . · bjs

s ,

define the weight of such a term as the following quantity:∑n

h=1hih +∑s

k=1kjk.

Let λ be a non-zero element ofL. Consider, other than the polynomialsf(X)andg(X), two new polynomials:

fλ(X) = a0Xn + a1λX

n−1 + · · ·+ an−1λn−1X + anλ

n,

gλ(X) = b0Xs + b1λX

s−1 + · · ·+ bs−1λs−1X + bsλ

s.

If αi is a root off(X), thenλαi is a root offλ(X) for eachi = 1, . . . , n. In fact,

fλ(λαi) = a0λnαn

i + a1λλn−1αn−1

i + · · ·+ anλn

=λn(a0αni + · · ·+ an)

=λnf(αi)

= 0.

In a similar way, ifβj is a root ofg(X), thenλβj is a root ofgλ(X) for eachj = 1, . . . , s.

The resultantR(fλ, gλ) equals the determinant

Dλ =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a0 λa1 · · · λnan


......


b0 λb1 · · · λsbsb0 λb1 · · · λsbs

......

b0 λb1 · · · λsbs

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

,

s rows

n rows


whose general term is

cai00 (λa1)

i1(λ2a2)i2 · . . . · (λnan)inbj00 (λb1)

j1 · . . . · (λsbs)js .

In terms ofλ, this becomes

cλi1+2i2+···+nin+j1+2j2+···+sjs · (ai00 a

i11 · . . . · ain

n bj00 b

j11 · . . . · bjs

s ).

So,D andDλ differ solely by the factor

λi1+2i2+···+nin+j1+2j2+···+sjs .

Again, by Definition 2.1,

R(f, g) = a0sb0

n∏ni=1


So,

R(fλ, gλ) = a0sb0

n∏ni=1

∏sj=1(λαi − λβj)

= a0sb0

nλns∏ni=1

∏sj=1(αi − βj) = λnsR(f, g).

Hence,

Dλ = R(fλ, gλ) = λnsR(f, g) = λnsD,

from which it follows that

λi1+2i2+···+j1+···+sjs = λns,

as required. 2

The expansion of the determinant (2.2) gives a homogeneous polynomial in then+s+2 indeterminatesa0, . . . , an, b0, . . . , bs, whose general term has degreen+s.Now, some properties of the polynomialD and then ofR(f, g) are described.

THEOREM 2.7 As a polynomial, the determinantD(a0, . . . , an, b0, . . . , bs) is ho-mogeneous both ina0, . . . , an and inb0, . . . , bs.

Proof. The general term of the determinantD is

ai00 a

i11 · . . . · ain

n bj00 b

j11 · . . . · bjs

s ,

where the factorsah, each counted with multiplicityih, are in number∑ih = s,

since they are precisely the factors coming from the firsts rows of (2.2). SinceD(a0, . . . , bs), viewed as a polynomial only in the indeterminatesa0, . . . , an, isa sum of monomials all of the same degrees. So, such a polynomial is homoge-neous ina0, . . . , an. Analogously, the assertion is established for the indetermi-natesb0, . . . , bs. 2

THEOREM 2.8 The polynomialD(a0, . . . , an, b0, . . . , bs) is irreducible overK.

Proof. To show the theorem, the relation betweenD andR(f, g) is used.Suppose thatD is reducible; that is, there exist non-constant homogeneous poly-

nomialsP, Q ∈ K[a0, . . . , an, b0, . . . , bs] such thatD = P ·Q.Now, introduce new indeterminatesa′1, a

′2, . . . , a

′n, with a′i = (−1)iai/a0, and

b′1, b′2, . . . , b

′s, with b′j = (−1)jbj/b0. Then

D(a0, a1, . . . , an, b0, b1, . . . , bs) = D(a0, a0a′1, . . . , a0a

′n, b0, b0b

′1, . . . , b0b

′s).

28

By the homogeneity ofD,

D(a0, a0a′1, . . . , a0a

′n, b0, b0b

′1, . . . , b0b

′s)

= a0sb0

nD(1, a′1, . . . , a′n, 1, b

′1, . . . , b

′s).

Put

D(1, a′1, . . . , a′n, 1, b

′1, . . . , b

′s) = D′(a′1, . . . , a

′n, b

′1, . . . , b

′s);

the polynomialD′ is not constant, sincea0sb0

n does not divideD. Hence

D′(a′1, . . . , a′n, b

′1, . . . , b

′s)

= P ′(a′1, . . . , a′n, b

′1, . . . , b

′s) ·Q′(a′1, . . . , a

′n, b

′1, . . . , b

′s).

Consider

R′(f, g) = R(f, g)/(a0sb0

n) = D/(a0sb0

n) = D′ = P ′Q′.

Expressing thea′i andb′j as elementary symmetric functions in theαi andβj givesthat

P ′ ·Q′ =∏n

i=1


Since the factor(αi−βj) dividesP ′ ·Q′, it must divide one of the two polynomials.SupposeP ′ contains it as a factor; then, asP ′ is a symmetric polynomial in theαi and in theβj , it contains them all. Therefore,Q′ is a constant, whence theirreducibility ofD′ and so ofD follows. 2

THEOREM 2.9 Given two polynomialsf(X), g(X) with coefficients inL, thereexist polynomialsA(X), B(X) with degA(X) ≤ s − 1 anddegB(X) ≤ n − 1such that

R(f, g) = A(X)f(X) +B(X)g(X).

Proof. Write

f(X) = a0Xn + · · ·+ an−1X + an,

g(X) = b0Xs + · · ·+ bs−1X + bs,

and consider the following identities:

Xs−1f(X) = a0Xn+s−1 + · · ·+ an−1X

s + anXs−1,

Xs−2f(X) = a0Xn+s−2 + · · ·+ an−1X

s−1 + anXs−2,

...

f(X) = a0Xn + · · ·+ an−1X + an,

Xn−1g(X) = b0Xs+n−1 + · · ·+ bs−1X

n + bsXn−1,

Xn−2g(X) = b0Xs+n−2 + · · ·+ bs−1X

n−1 + bsXn−2,

...

g(X) = b0Xs + · · ·+ bs−1X + bs.


The determinant of the coefficients of the right-hand side isthe same as the deter-minant (2.2).

LetAi for i = 0, . . . , s− 1 andBj for j = 0, . . . , n− 1 be the cofactors of thedeterminant (2.2) with respect to the last column. Multiplying thei-th of the firsts rows byAi and, similarly, thei-th of the remainingn rows byBi, and summinggives

A(X)f(X) +B(X)g(X) = R(f, g),

with A(X) = A0Xs−1 + · · ·+As−1 andB(X) = B0X

n−1 + · · ·+Bn−1. 2

THEOREM 2.10 (Study)Let f andg be elements ofK[X1, . . . ,Xr], and letf beirreducible overK. If each rootξ1, . . . , ξr of f is a root ofg, thenf dividesg.

Proof. Put

f(X1, . . . ,Xr) = a0(X1, . . . ,Xr−1)Xnr + a1(X1, . . . ,Xr−1)X

n−1r

+ · · ·+ an(X1, . . . ,Xr−1),

and consider it as a polynomial in the single variableXr. Suppose that the indeter-minateXr does not occur ing; then, there existξ1, . . . , ξr−1 ∈ K such that

g(ξ1, . . . , ξr−1) · a0(ξ1, . . . , ξr−1) 6= 0,

since, otherwise,

g(X1, . . . ,Xr−1) · a0(X1, . . . ,Xr−1) = 0,

for all X1, . . . ,Xr−1. So one of the two terms of the product must be zero; but thisis impossible by hypothesis.

SinceK is algebraically closed, the polynomialf(ξ1, . . . , ξr−1,Xr) has a rootXr = ξr. From this, it follows that ther-ple (ξ1, . . . , ξr) are roots off but not ofg, contrary to hypothesis.

Therefore, writeg as a polynomial inXr :

g(X1, . . . ,Xr)

= b0(X1, . . . ,Xr−1)Xsr + b1(X1, . . . ,Xr−1)X

s−1r + · · ·+ bs(X1, . . . ,Xr−1).

Consider the indeterminatesX1, . . . ,Xr−1 as parameters so thatf andg becomepolynomials in the single variableXr; that is,f = f(Xr) andg = g(Xr). ByTheorem 2.9, there exist polynomialsA(Xr) andB(Xr) such that

A(Xr)f(Xr) +B(Xr)g(Xr) = R(f, g),

whereR(f, g) is a polynomial inX1, . . . ,Xr−1, andA = A(Xr) andB = B(Xr)depend on the parametersX1, . . . ,Xr−1.

If ξr is a root off(Xr), thenξr is also a root ofg(Xr) by hypothesis; therefore

A(X1, . . . ,Xr)f(X1, . . . ,Xr) +B(X1, . . . ,Xr)g(X1, . . . ,Xr) = 0.

Hence

A(X1, . . . ,Xr)f(X1, . . . ,Xr) = −B(X1, . . . ,Xr)g(X1, . . . ,Xr).

30

Sof(X1, . . . ,Xr) divides−B(X1, . . . ,Xr)g(X1, . . . ,Xr), and sincef is irred-ucible, it must divide one of the two polynomials. However, asdegB < deg f, sof dividesg. 2

The polynomialsf andg, as in (2.1), witha0, b0 6= 0, have a root in common ifand only ifR(f, g) = 0.

If a0 or b0 or both are zero, the expression in Definition 2.1 for the resultant isno longer utilisable. Instead,D can be used, keeping in mind thatD = R(f, g)whena0b0 6= 0. It can no longer be affirmed that the existence of roots commontof andg is equivalent to the resultant being zero. In fact, ifa0 = b0 = 0, then, ineach case,R(f, g) = 0, independently of the existence of common roots; but thisis the only case in which the resultant being zero does not guarantee that there arecommon roots.

THEOREM 2.11 If the leading coefficients of the polynomialsf and g given by(2.1)are not both zero, thenR(f, g) = 0 if and only if the two polynomials have aroot in common.

Proof. Suppose thata0 6= 0, b0 = b1 = · · · = bk−1 = 0, but bk 6= 0. Put

g(X) = bkXs−k + bk+1X

s−(k+1) + · · ·+ bs−1X + bs.

Since the leading coefficients off andg are both non-zero,R(f, g) = 0 if and onlyif f andg have a root in common.

Substituting zero for the elementsb0, b1, . . . , bk−1 in (2.2) and expanding byLaplace’s theorem, it follows thatR(f, g) = ak

0R(f, g). The polynomialsg(X)andg(X) have the same roots, whenceR(f, g) = 0 implies thatak

0R(f, g) = 0. 2

These results extend to homogeneous polynomials. As in Section 1.4, withf apolynomial of degreen,

f(X) = a0Xn + a1X

n−1 + · · ·+ an−1X + an,

let

F (X,Y ) = f∗(X,Y ) =Y nf(X/Y )

= a0Xn + a1X

n−1Y + · · ·+ an−1XYn−1 + anY

n.

Theorem 2.11 immediately gives the comparable result for homogeneous poly-nomials.

THEOREM 2.12 The homogeneous polynomials

F (X,Y ) = a0Xn + a1X

n−1Y + · · ·+ an−1XYn−1 + anY

n,

G(X,Y ) = b0Xs + b1X

n−1Y + · · ·+ bs−1XYn−1 + bsY

n

have a non-constant common factor if and only if the determinant D in (2.2) iszero.

Proof. If a0 = b0 = 0, thenD = 0 andX is a common factor ofF andG.Otherwise, withf(X) = F (X, 1) andg(X) = G(X, 1), Theorem 2.11 gives theresult. 2


THEOREM 2.13 Given the homogeneous polynomials,

F (X0, . . . ,Xr) =A0Xnr +A1X

n−1r + · · ·+An−1Xr +An,

G(X0, . . . ,Xr) =B0Xsr +B1X

s−1r + · · ·+Bs−1Xr +Bs,

whereAi andBi are homogeneous inK[X0, . . . ,Xr−1], with A0B0 6= 0, theresultantR(F,G) is a homogeneous polynomial of degreens if F andG have nocommon factor.

Proof. Let Ai = Ai(X0, . . . ,Xr−1). ThenRXr(F,G) = D(X) is given by the

determinant,∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

A0 A1 · · · · · · An 0 · · · 00 A0 A1 · · · · · · An · · · 0

......

...0 · · · 0 A0 A1 · · · · · · An

B0 B1 · · · Bs 0 · · · 00 B0 B1 · · · Bs 0 · · · 0

......

...0 · · · 0 B0 B1 · · · Bs

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

s rows

n rows

.

ReplaceX by tX. SinceAi andBj are homogenous polynomials,D(tX) is equalto the determinant,∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

A0 tA1 · · · · · · tnAn 0 · · · 00 A0 tA1 · · · · · · tnAn · · · 0

......

...0 · · · 0 A0 tA1 · · · · · · tnAn

B0 tB1 · · · tsBs 0 · · · 00 B0 tB1 · · · tsBs 0 · · · 0

......

...0 · · · 0 B0 tB1 · · · tsBs

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

s rows

n rows

.

Multiply the i-th row inA by ti for 1 ≤ i ≤ s, and thej-th row inB by tj for1 ≤ j ≤ n. This shows thattuD(tX) is equal to the determinant,

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

tA0 t2A1 · · · · · · tn+1An 0 · · · 00 t2A0 t3A1 · · · tn+1An−1 tn+2An · · · 0

......

...0 · · · 0 tsA0 ts+1A1 · · · tn+sAn

tB0 t2B1 · · · ts+1Bs 0 · · · 00 t2B0 t3B1 · · · ts+1Bs−1 ts+2Bs · · · 0

......

...0 · · · 0 tnB0 tn+1B1 · · · ts+nBs

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

,

whereu = 12s(s+ 1) + 1

2n(n+ 1).

32

The last determinant is equal totvD(X) with v = 12 (s + n)(s + n + 1). Thus

tuD(tX) = tvD(X) and henceD(tX) = tv−uD(X). Therefore,

R(tX1, . . . , tXr−1) = tnsR(X1, . . . ,Xr−1),

which shows thatR(X1, . . . ,Xr−1) is a homogeneous polynomial of degreens.2

2.2 THE DISCRIMINANT

The condition that a polynomial should have a multiple root is considered,As in (2.1), let

f(X) = a0Xn + a1X

n−1 + · · ·+ an−1X + an, a0 6= 0,

have rootsα1, . . . , αn.It is necessary to assume that the derivativef ′(X) is not identically zero, that is,

f(X) is separable.

THEOREM 2.14 The resultant off and its derivativef ′ is

R(f, f ′) = a2n−10

∏ni=1

∏j 6=i (αi − αj) (2.6)

Proof. From Lemma 2.3 (i),

R(f, f ′) = an−10

∏ni=1f

′(αi). (2.7)

Taking the derivative of

f(X) = a0

∏nk=1(X − αk)

gives

f ′(X) = a0

∑nk=1

∏j 6=k(X − αj),

whence

f ′(αi) = a0

∏j 6=i(αi − αj).

Substituting in (2.7) gives the result. 2

DEFINITION 2.15 Thediscriminantof f(X) is

D(f) = a2n−20

∏1≤j<i≤n (αi − αj)

2.

COROLLARY 2.16 The discriminant can be calculated from the formula,

R(f, f ′) = (−1)n(n−1)/2 a0D(f). (2.8)

Proof. In (2.7), for every pairi, j with i 6= j, there are two factors,αi − αj andαj − αi. As there are12n(n− 1) pairs, so

R(f, f ′) = (−1)n(n−1)/2 a2n−10

∏1≤j<i≤n(αi − αj)

2,

and the result follows. 2


PROPOSITION 2.17 The polynomialf(X) has a multiple zero if and only if itsdiscriminantD(f) is zero.

EXAMPLE 2.18 Let f(X) = aX2 + bX + c. Thenf ′(X) = 2aX + b and

R(f, f ′) =

∣∣∣∣∣∣

a b c2a b 00 2a b

∣∣∣∣∣∣= a(−b2 + 4ac).

From (2.8),R(f, f ′) = −aD(f); so

D(f) = b2 − 4ac.

2.3 ELIMINATION IN A SYSTEM IN TWO UNKNOWNS

LetL be any field, and letf, g ∈ L[X,Y ] be written as follows:

f(X,Y ) = a0(Y )Xn + a1(Y )Xn−1 + · · ·+ an−1(Y )X + an(Y ),g(X,Y ) = b0(Y )Xs + b1(Y )Xs−1 + · · ·+ bs−1(Y )X + bs(Y ).

(2.9)

Thenf, g ∈ L(Y )[X] and previous results can be applied forL(Y )[X].Now, let RY (f, g) be the resultant of the polynomialsf and g considered as

polynomials inX. By the property of the resultant,RY (f, g) is a polynomialD(Y )in Y with coefficients inK as follows:

D(Y ) = (2.10)∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a0(Y ) a1(Y ) . . . an(Y ) 0 0 . . . 00 a0(Y ) a1(Y ) . . . an(Y ) 0 . . . 0...

......

0 0 . . . 0 a0(Y ) a1(Y ) . . . an(Y )b0(Y ) b1(Y ) . . . bs(Y ) 0 . . . 0

0 b0(Y ) b1(Y ) . . . bs(Y ) 0 . . . 0...

......

0 . . . 0 b0(Y ) b1(Y ) . . . bs(Y )

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

;

here, the determinant has entriesai(Y ) in its first s rows and entriesbj(Y ) in itsnextn rows.

If the two polynomials (2.9) have a common root,(X,Y ) = (α, β), substitut-ing Y = β in (2.9) gives polynomials in the indeterminateX, namely,f(X,β),g(X,β). These polynomials have a common rootα, whence their resultant is zero.Soβ is a root ofRY (f, g).

Conversely, if the resultantRY (f, g) of the polynomials (2.9) admits a rootβ,then the resultant of the polynomialsf(X,β), g(X,β) is zero. Thus, the polyno-mialsf(X,β) andg(X,β) have a common root or their leading coefficientsa0(β)andb0(β) are both zero.

The determination of the common roots of the polynomials (2.9) is reduced todetermining the roots of the polynomial (2.10) inY ; this is expressed by sayingthat the indeterminateX has been eliminated from the system of the two equationsor that the equationD(Y ) = 0 is the result of eliminatingX from (2.9).

34

EXAMPLE 2.19 LetL = Q. To determine the common roots off(X,Y ) = X2Y + 3XY + 2Y + 3, g(X,Y ) = 2XY − 2X + 2Y + 3,

the resultant is calculated asRY (f, g) = 2Y 2 + 11Y + 12; its roots arey1 = −4,y2 = − 3

2 . The leading coefficients off(X, yi) andg(X, yi) are non-zero; hence,for eachyi, there is a common root inX. In fact,

f(X,−4) = −4X2 − 12X − 5, g(X,−4) = −10X − 5,

and they have the common rootx1 = − 12 . Similarly, x2 = 0 is the common root

of f(X, 32 ) = − 3

2X2 − 9

2X andg(X, 32 ) = −5X. Therefore, the solution of the

problem is given by the two pairs(− 12 ,−4) and(0,− 3

2 ).

THEOREM 2.20 The polynomialsf andg given by (2.9) have a non-constant fac-tor in common if and only ifRY (f, g) = D(Y ) is identically zero.

Proof. From Theorem 2.9, there exist, for each fixedy, two polynomialsAy(X),By(X) such that

RY (f, g) = Ay(X)f(X, y) +By(X)g(X, y)

with degAy(X) ≤ s− 1 anddegBy(X) ≤ n− 1. In the proof, the coefficients ofAy(X) andBy(X) are cofactors of the determinantD; it follows that there existpolynomialsA(X,Y ), B(X,Y ) such thatAy(X) = A(X, y), By(X) = B(X, y)for eachy. Hence,

Ry(f, g) = A(X,Y )f(X,Y ) +B(X,Y )g(X,Y )

with A(X,Y ) andB(X,Y ) of degree at mosts− 1 andn− 1 in Y .If RY (f, g) = 0 for everyY = y, thenA(X,Y )f(X,Y ) = −B(X,Y )g(X,Y ).

Sincef(X,Y ) cannot divideB(X,Y ), some non-constant factor off(X,Y ) mustdivide g(X,Y ), as required. Indeed, ifh(X,Y ) is common factor off(X,Y ) andg(X,Y ), for eachy there exists at least onex such thath(x, y) = 0. It follows thatRy(f, g) = 0 for eachy, whenceRY (f, g) is the zero polynomial. 2

THEOREM 2.21 If RY (f, g) is not identically zero,, thendegRY (f, g) ≤ N S,whereN = deg f(X,Y ) andS = deg g(X,Y ).

This theorem can be reformulated as follows: iff(X,Y ) andg(X,Y ) have nocommon factors, then the number of solutions of the system

f(x, y) = 0, g(x, y) = 0

is at mostN S.

Proof. Rewritef(X,Y ) andg(X,Y ) in decreasing powers ofX:f(X,Y ) = a0(Y )Xn + · · ·+ an(Y ), g(X,Y ) = b0(Y )Xs + · · ·+ bs(Y ).

This gives the following inequalities:deg a0(Y ) ≤ N − n, deg b0(Y ) ≤ S − s,deg a1(Y ) ≤ N − (n− 1), deg b1(Y ) ≤ S − (s− 1),

......

deg ah(Y ) ≤ N − (n− h), deg bk(Y ) ≤ S − (s− k),...

...deg an(Y ) ≤ N, deg bs(Y ) ≤ S.

(2.11)


The general term in the expansion of the determinant in (2.10) is

ca0(Y )i0a1(Y )i1 · . . . · an(Y )inb0(Y )j0b1(Y )j1 · . . . · bs(Y )js ,

whose degree is, by (2.11), at most

(N − n)i0 + · · ·+ (N − (n− h))ih + · · ·+Nin

+(S − s)j0 + · · ·+ (S − (s− k))jk + · · ·+ Sis.

This can be written as

(N − n)∑n

i=0ih + (S − s)∑sk=0jk +

∑nh=1hih +

∑sk=1kjk. (2.12)

Since∑n

i=0ih = s,∑s

k=0jk = n,∑n

h=1hih +∑s

k=1kjk = ns.

the sum (2.12) becomes(N − n)s+ (S − s)n+ ns. Sincen ≤ N ands ≤ S, thedegree of the general term of the determinantD(Y ) is at mostN · S. 2

THEOREM 2.22 If f(X,Y ) andg(X,Y ) are two polynomials containing no termsof degree less thanu andv respectively, theny = 0 is a root ofRY (f, g) = D(Y )of multiplicity at leastuv.

Proof. As an example, consider first the caseu = 2, v = 2,with deg f(X,Y ) = 3,deg g(X,Y ) = 2 :

f(X,Y ) = a20X2 + a11XY + a02Y

2 + a30X3 + a21X

2Y + a12XY2 + a03Y

3,

g(X,Y ) = b20X2 + b11XY + b02Y

2.

It must be shown thatRY (f, g) = D(Y ) is divisible byY 4. First,

D(Y ) =∣∣∣∣∣∣∣∣∣∣

a30 a21Y + a20 a11Y + a12Y2 a02Y

2 + a03Y3 0

0 a30 a21Y + a20 a11Y + a12Y2 a02Y

2 + a03Y3

b20 b11Y b02Y2 0 0

0 b20 b11Y b02Y2 0

0 0 b20 b11Y b02Y2

∣∣∣∣∣∣∣∣∣∣

.

In each entry of the determinant, consider only the terms of lowest degree anddelete the others. This gives a new determinantE(Y ) as follows:

E(Y ) =

∣∣∣∣∣∣∣∣∣∣

a30 a20 a11Y a02Y2 0

0 a30 a20 a11Y a02Y2

b20 b11Y b02Y2 0 0

0 b20 b11Y b02Y2 0

0 0 b20 b11Y b02Y2

∣∣∣∣∣∣∣∣∣∣

.

Then Y 4 dividesD(Y ) if and only if Y 4 dividesE(Y ). Now, multiply rows1, 2, 4, 5 by Y, Y 2, Y, Y 2 respectively, giving a new determinantF (Y ), which is

F (Y ) =

∣∣∣∣∣∣∣∣∣∣

a30Y a20Y a11Y2 a02Y

3 00 a30Y

2 a20Y2 a11Y

3 a02Y4

b20 b11Y b02Y2 0 0

0 b20Y b11Y2 b02Y

3 00 0 b20Y

2 b11Y3 b02Y

4

∣∣∣∣∣∣∣∣∣∣

.

36

Here,F (Y ) = Y 6E(Y ).So, it must be shown thatF (Y ) is divisible byY 10. In F (Y ), Columns2, 3, 4, 5

are divisible byY, Y 2, Y 3, Y 4 respectively, whenceY 10 dividesF (Y ), as required.Now, consider the general case. The two polynomials have thefollowing form:

f(X,Y ) = au,0Xu + au−1,1X

u−1Y + · · ·+ a1,u−1XYu−1 + a0,uY

u + · · · ,g(X,Y ) = bv,0X

v + bv−1,1Xv−1Y + · · ·+ b1,v−1XY

v−1 + b0,vYv + · · · .

Let a(i) be the lowest exponent in the termai(Y ), andb(i) the lowest inbi(Y ).Then

f(X,Y ) = a0(Y )Xn + a1(Y )Xn−1 + · · ·+ an−1(Y )X + an(Y ),

a(n− i) ≥ u− i;g(X,Y ) = b0(Y )Xs + b1(Y )Xs−1 + · · ·+ bs−1(Y )X + bs(Y ),

b(s− i) ≥ v − i.

SinceRY (f, g) = D(Y ), consider, as in the example, a new determinantE(Y )obtained fromD(Y ) by retaining only the terms of lowest degree in each row.

For the firsts rows, multiply the(s− i+1)-th row byY v−i+1 for 1 ≤ i ≤ v. Forthe remainingn rows multiply the(n − j + 1)-th row byY u−j+1 for 1 ≤ j ≤ u.After these operations, the(n + s − k)-th column is divisible byY u+v−k, for0 ≤ k ≤ u+ v. Therefore,RY (f, g) is divisible byY t, where

t=∑u+v

k=1k −∑u

i=1i−∑v

j=1j

= 12 (u+ v)(u+ v + 1)− 1

2u(u+ 1)− 12v(v + 1).

Hencet = uv, completing the proof. 2

2.4 EXERCISES

1. Letf(X) = a0Xn +a1X

n−1 +an andg(X) = b0Xs +b1X

n−1 + . . .+bn.Show thatR(f, g) is equal to the determinant whose entries are the coeffi-cients of the polynomials,

hk(X) = (a0Xk−1 + a1X

k−1 + . . .+ ak−1)g(X)

−(b0Xk−1 + b1X

k−1 + . . .+ bk−1)f(X),

wherek = 1, . . . , n.

2. Assume that eitherp = 0 or p ∤ n,m. Let Fn(X) be then-th cyclotomicpolynomial. Determine the resultant of the polynomialsFn(X) andXm−1.Solution: Letd = gcd(n,m) andn1 = n/d. Then

R(Fn(X),Xm − 1) =

0 when eitherm | n or n1 = 1,eϕ(n)/ϕ(n1) whenn1 = ek with e prime,

1 otherwise.


3. Assume that eitherp = 0 or p ∤ n,m. Determine the resultant of the cyclo-tomic polynomialsFn(X) andFm(X). Solution:

R(Fn(X), Fm(X)) =

0 whenm = n,eϕ(n) whenm = nek with e prime,

1 otherwise.

4. Show that the discriminant of

f(X) = X3 + aX2 + bX + c

is

D(f) = a2b2 − 4b3 − 4a3c+ 18abc− 27c2.

5. Solve the following system:

Y 2 − 7XY + 4X2 + 13X − 2Y − 3 =0,

Y 2 − 14XY + 9X2 + 28X − 4Y − 5 =0.

6. Assume that the characteristicp of K does not dividen. Show that thediscriminant off(X) = Xn + uX + v is

D(f) = (−1)n(n−1)/2nnvn−1 + (−1)(n−1)(n−2)/2(n− 1)n−1un.

2.5 NOTES

For the basic theory of resultants, see Redei [221] and Kurosh [174]. The theoryhas been expanded to any finite number of polynomials; see Gelfand, Kapranov,and Zelevinsky [100], Lang [179, Chapter 9], Segre [251], and van der Waerden[300].

Chapter Three

Singular points and intersections

The aim of this chapter is to give an account of how two curves intersect, how tocount the total number of intersections of two curves, and how to obtain equiva-lent curves with well-behaved singularities. This includes the study of quadratictransformations and the definition of the virtual genus of a curve.

3.1 THE INTERSECTION NUMBER OF TWO CURVES

In this section, acurve in A2(K) is simply an equivalence class of polynomials inK[X,Y ] under multiplication by a non-zero scalar. Given two curvesF andG, theintersection numberI(P, F ∩G) of F andG at the pointP = (a, b) is defined bythe following axioms:

(I 1) I(P, F ∩ G) is a non-negative integer whenF andG have no commoncomponent;

(I 2) I(P, F ∩G) =∞ if F andG have a common component throughP ;

(I 3) I(P, F ∩G) = 0 if P 6∈ V (F ) ∩ V (G);

(I 4) I(P, F ∩G) = 1 if F andG are two distinct lines throughP ;

(I 5) I(P, F ∩G) = I(P,G ∩ F );

(I 6) I(P, F ∩ (G+AF )) = I(P, F ∩G) for anyA ∈ K[X,Y ];

(I 7) I(P, F ∩GH) = I(P, F ∩G) + I(P, F ∩H) for anyH ∈ K[X,Y ].

The first result shows the relation between the geometry and the algebra.

THEOREM 3.1 The intersection numberI(P, F ∩ G) depends only on the ideal(F,G) generated byF andG, in the sense that, if F ′ andG′ are curves such that(F,G) = (F ′, G′), then

I(P, F ∩G) = I(P, F ′ ∩G′). (3.1)

Proof. If (F,G) = (F ′, G′), then

F ′ =AF +BG, (3.2)

G′ =CF +DG, (3.3)

39

40

with A,B,C,D ∈ K[X,Y ]. By (3.1), there existA′, B′, C ′,D′ ∈ K[X,Y ] suchthat

F =A′F ′ +B′G′, (3.4)

G=C ′F ′ +D′G′, (3.5)

If H divides bothF andG, then it also divides bothF ′ andG′, and vice versa. By(I 3), this says that, ifI(P, F ∩G) =∞, thenI(P, F ′ ∩G′) =∞.

So, it may be supposed thatF ′ andG′ have no common factor. It is first shownthat, if P is a common point of the curvesF andG, then∆(P ) 6= 0, where∆ = AD −BC. A direct calculation, using (3.2), (3.3), (3.4), (3.5), gives

F ′(1− (AA′ +BC ′)) = (AB′ +BD′)G′,

G′(1− (CB′ +DD′)) = (CA′ +DC ′)F ′.

Suppose first thatAD′ + BD′, and hence also(1− (AA′ + BC ′), is not the zeropolynomial. ThenF ′ andG′ divide both these. SinceF (P ) = G(P ) = 0 impliesthatF ′(P ) = G′(P ) = 0, it follows that, with

R1 = AB′ +BD′, R2 = AA′ +BC ′,

R1(P ) = 0, R2(P ) = 1. (3.6)

Similarly, with

R3 = CA′ +DC ′, R4 = CB′ +DD′,

R3(P ) = 0, R4(P ) = 1, (3.7)

provided thatR3, and so1−R4, is not the zero polynomial.It follows immediately that (3.6) and (3.7) remain true whenR1 andR3 are the

zero polynomials. Therefore,

R2(P )R4(P )−R1(P )R3(P ) = 1.

Now,

R2R4 −R1R3

=(AA′ +BC ′)(CB′ +DD′)− (AB′ +BD′)(CA′ +DC ′)

= (AD −BC)(A′D′ −B′C ′).

Hence∆(P ) = A(P )D(P )−B(P )C(P ) 6= 0, as required.From (I 3), it follows that

I(P,∆ ∩ F ) = 0. (3.8)

Now, the result can be deduced, using (I 5), (I 6), (I 7) and (3.7).

I(P, F ′ ∩G′)= I(P, [AF +BG] ∩ [CF +DG])= I(P, [AFD +BGD] ∩ [CF +DG])− I(P,D ∩ [CF −DG])= I(P, [AFD −BCF +BCF +BGD] ∩ [CF +DG])− I(P,D ∩ CF )= I(P, [(AD −BC)F +B(CF +DG)] ∩ [CF +DG])− I(P,D ∩ CF )= I(P,∆F ∩ [CF +DG])− I(P,D ∩ CF )= I(P,∆ ∩ [CF +DG]) + I(P, F ∩ [CF +DG])− I(P,D ∩ CF )= I(P, F ∩DG)− I(P,D ∩ CF )= I(P, F ∩D) + I(P, F ∩G)− I(P,D ∩ C)− I(P,D ∩ F )= I(P, F ∩G)− I(P,C ∩D).

Singular points and intersections 41

It remains to see thatI(P,C ∩D) = 0. If not, thenC(P ) = D(P ) = 0, and soA(P )D(P )−B(P )C(P ) = 0, contradicting that∆ 6= 0. 2

The following useful formula for reducible curves is a consequence of (I 7) and(I 5).

THEOREM 3.2 If F =∏

i Fri

i andG =∏

j Gsj

j , then

I(P, F ∩G) =∑

i,j

risjI(P, F ∩G).

Now, some propositions are established that are required later.

PROPOSITION 3.3 For everyF ∈ K[X,Y ] not havingY as a component, iff(X) = F (X, 0) = Xr(X − a1)

r1 . . . (X − am)rm , then I(O,F ∩ Y ) = r,whereO = (0, 0) is the origin.

Proof. SinceF (X,Y ) = f(X) + Y H(X,Y ), from (I 7) it follows that

I(O,F ∩ Y ) = I(O, f(X) ∩ Y ).

So, by Theorem 3.2,

I(O,F ∩ Y )

= rI(O,X ∩ Y ) + r1I(O, (X − a1) ∩ Y ) + . . .+ rmI(O, (X − am) ∩ Y ).

Then the result follows from (I 3) and (I 4). 2

Now, define the homomorphism

Φ : K[X,Y ]→ K[X], (3.9)

H(X,Y ) 7→ h(X) = H(X, 0).

Given two non-zero polynomialsF,G ∈ K[X,Y ], let f = Φ(F ), g = Φ(G),and letd ∈ K[X] be the greatest common divisor off andg. Then there existpolynomialsa, b, c, e ∈ K[X] such thatd = af + bg, f = cd, g = ed. Therefore,

ac+ be = 1. (3.10)

Now put

D = aF + bG, H = −eF + cG. (3.11)

With O the origin, calculateI(O,D,H). SinceΦ(D) = d andΦ(H) = 0, itfollows thatH = Y H ′ withH ′ ∈ K[X,Y ]. From the above, the ideals(F,G) and(D,H) are the same. Then, from (I 7) and Theorem 3.1,

I(O,F ∩G) = I(O,D ∩H) = I(O,D ∩ Y ) + I(O,D ∩H ′). (3.12)

PROPOSITION 3.4 If I(P, F ∩ G) is a function satisfying(I 1),. . . ,(I 7), then, forevery pair(F,G) with F,G ∈ K[X,Y ],

I(O,F ∩G) = I(O,F ∩G). (3.13)

42

Proof. Put i = I(O,F ∩ G); if i = 0 the proposition follows from (I 3). Now,proceed by induction oni.

Suppose thatI(O,U ∩ V ) = I(O,U ∩ V ) for all U, V ∈ K[X,Y ] under thehypothesis thatI(O,U ∩ V ) < i. First, letG be linear; that is,G = λX + µYwith λ, µ ∈ K andλ 6= 0. With the above notation,g(X) = λX. Also, f(X) isdivisible byX, sincef(0) = F (0, 0) = 0. Therefore,

f(X) = cX, a = 0, b = λ−1, d = X, e = λ, D = λ−1G = X + µλ−1Y.

From (I 4) and (I 6),

I(O,D ∩ Y ) = I(O,X + µλ−1Y ∩ Y ) = I(O,X ∩ Y ) = 1= I(O,X ∩ Y ) = I(O,X + µλ−1Y ∩ Y ) = I(O,D ∩ Y ).

Using (3.12), it follows that

I(O,D ∩H ′) = I(O,D ∩H)− I(O,D ∩ Y )

= I(O,F ∩G)− 1 = i− 1.

By the inductive hypothesis,I(O,D ∩H ′) = I(O,D ∩H ′). Therefore,

I(O,F ∩G) = I(O,D ∩ F ) + I(O,D ∩H ′)

= I(O,D ∩ Y ) + I(O,D ∩H ′)

= I(O,D ∩H ′) = I(O,F ∩G),

as required.It remains to consider the case thatG is not linear. Next, by (3.12), the inequality

I(O,D ∩ Y ) ≤ i holds. By (I 2),Y does not divideD. Also, Theorem 3.2 assertsthat

I(O,D ∩ Y ) = I(O,D ∩ Y ). (3.14)

SinceF (0, 0) = G(0, 0) = 0, alsoD(0, 0) = 0. Hence,I(O,D ∩ Y ) ≥ 1 by (I 3).Next,I(O,D ∩H ′) < i by (3.12); by the inductive hypothesis,

I(O,D ∩H ′) = I(O,D ∩H ′).

Using (3.14),

I(O,F ∩G) = I(O,D ∩ Y ) + I(O,D ∩H ′)

= I(O,D ∩ Y ) + I(O,D ∩H ′)

= I(O,D ∩H) = I(O,F ∩G).

This completes the proof of Proposition 3.4. 2

PROPOSITION 3.5 If D ∈ K[X,Y ], then

I(O,D, Y ) ≥ mO(D),

with equality if and only ifY is not tangent toD atO.

Proof. By (I 1), it may be supposed thatY is not a component ofD. Write

D(X,Y ) = d(X) + Y D′(X,Y ), (3.15)


where

d(X) = Xm(λm + λm+1X + . . .), λm 6= 0. (3.16)

With m0(D) = n,

D(X,Y ) = Dn(X,Y ) +Dn+1(X,Y ) + . . . , (3.17)

whereDi(X,Y ) is a homogeneous polynomial of degreei. From (3.15), (3.16),(3.17), it follows thatm ≥ n and

Dn(X,Y ) = λmXPn−1(X,Y ) + Y P ′n−1(X,Y ),

whereλm = 0 form > n andP ′n−1 is a homogeneous polynomial of degreen−1.

Therefore,n = m if and only if Y does not divideDn(X,Y ); that is, ifY is nottangent toD atO. From (I 5), (3.15), (I 6), (3.16), (I 7), (I 3), (I 4),

I(O,D ∩ Y ) = I(O, Y ∩ d+ Y D′)

= I(O, Y ∩ d)= I(O, Y ∩Xm(λm + λm+1X + . . .)

= I(O ∩X,Xm) + I(O ∩ Y, λm + λm+1X + . . .)

=mI(O, Y ∩X) + 0

=m.

2

The next result is important for calculating the intersection number at a point.

PROPOSITION 3.6 The intersection number satisfies

I(O,F ∩G) ≥ mO(F ) ·mO(G),

with equality if and only if the tangents atO to F are all distinct from the tangentsatO toG.

Proof. By (I 3), the proposition is true whenI(O,F ∩ G) = 0. Suppose, byinduction, that the proposition has been shown forU, V ∈ K[X,Y ] such thatI(O,U ∩ V ) < i. To show thatI(O,F ∩G) = i, with the notation of (3.4), fromProposition 3.5 it follows thatI(O,D∩Y ) > 0.Now, by (3.12),I(O,D∩H ′) < i.By the inductive hypothesis,I(O,D ∩H ′) ≥ mO(D) ·mO(H ′), with equality ifand only ifD andH ′ do not have the same tangent atO. In any case, Proposition3.5 together with (3.12) implies that

I(O,F ∩G) = I(O,D ∩ Y ) + I(O,D ∩H ′)

≥mO(D) +mO(D)mO(H ′)

=mO(D)mO(H), (3.18)

with equality if and only ifD has no tangent in common either withY orH ′, andso also not withH.

Now a relationship is found betweenmO(D)mO(H) andmO(F )mO(G). Sup-pose first thatY divides neitherFm(X,Y ) norGn(X,Y ), where

F (X,Y ) =Fm(X,Y ) + Fm+1(X,Y ) + . . . ,

G(X,Y ) =Gn(X,Y ) +Gn+1(X,Y ) + . . . ,

44

with Fi(X,Y ) andGi(X,Y ) homogeneous polynomials of degreei. MultiplyingF andG by suitable scalars ensures thatFm andGm have the following form:

Fm(X,Y ) =Xm + Y · F ′m−1(X,Y ),

Gn(X,Y ) =Xn + Y ·G′n−1(X,Y ),

whereF ′m−1 andG′

n−1 are homogeneous polynomials of degreesm−1 andn−1.With f(X) = Fm(X, 0), g(X) = Gn(X, 0)

f(X) = Xm(1 + f1(X)), g(X) = Xn(1 + g1(X)).

Without loss of generality, letm ≤ n. Thend = gcd(f, g) is divisible byXm

but not byXm+1. Therefore,

d=Xm + dmXm+1 + . . . , (3.19)

c= 1 + c1X + . . . ,

e=Xn−m + en−m+1Xn−m+1 + . . . .

Put

a=αrXr + αr+1X

r+1 + . . . ,

b=βsXs + βs+1X

s+1 + . . . .

Then, from (3.10),

1 = (αrXr + αr+1X

r+1 + . . .)(1 + c1X + . . .)

+(βsXs + βs+1X

s+1 + . . .)(Xn−m + en−m+1Xn−m+1 + . . .)

= (αrXr + . . .) + (βsX

n−m+s + . . .).

Hence

r = 0, α0 = 1, if n+ s > m;β0 = 1, if n = m, s = 0, r > 0;a+ β0 = 1, α0β0 6= 0, if n = m, r = s = 0.

(3.20)

SinceD = aF + bG andH = −eF + cG, as in (3.11),

D=(αrXr + αr+1X

r+1 + . . .)(Fm + Fm+1 + . . .)

+(βsXs + βs+1X

s+1 . . .)(Gn +Gn+1 + . . .)

= (αrXrFm + . . .) + (βsX

sGn + . . .); (3.21)

H =−(Xn−m + en−m+1Xn−m+1 + . . .)(Fm + Fm+1 + . . .)

+(1 + c1X + . . .)(Gn +Gn+1 + . . .)

= (−Xn−mFm +Gn) + . . . . (3.22)

It follows thatmO(H) ≥ n andmO(D) ≥ m.On the other hand, (3.15) and (3.19) imply that

D = d+ Y D′ = (Xm + am+1Xm+1 + . . .) + Y D′(X,Y ).

Therefore,mO(D) ≤ m; hencemO(D) = m. Now, writeD = Dm + . . . withDm a homogeneous polynomial of degreem. From (3.21) and (3.20),

Dm =

Fm, if n+ s > m;Gm, if n = m, s = 0, r > 0;α0Fm + β0Gm, if n = m, r = s = 0.

(3.23)


Now, the proof can be completed. IfmO(H) > n, then (3.22) implies thatGn = Xn−mFm, so that each tangent toF atO is also tangent toG atO. By(3.18),

I(O,F ∩G) ≥ mO(D) ·mO(H) > mn.

If mO(H) = n, thenmO(D) ·mO(H) = mn, and

Hn = −Xn−mFm +Gn 6= 0. (3.24)

It remains to show thatF andG have tangents in common if and only ifD andHhave; in other words,Hn andDm have a non-constant divisor if and only ifFm andGn have. But this follows immediately from the fact thatHn andDm are linearcombinations ofFm andGn with coefficients inK[X] and vice versa; see (3.24),(3.23), (3.20). So the result is true formO(H) = n.

It remains to show that the assumption thatY divides neitherFm nor Gn isunnecessary. Givena, b, c, d ∈ K with ad− bc 6= 0, define a new function

I(P, F ∩ G) = I(P, F ∩G),

where

F (X,Y ) =F (aX + bY, cX + dY ),

G(X,Y ) =G(aX + bY, cX + dY ).

It follows immediately thatI(F , G;P ) also satisfies the postulates (I 1),. . . ,(I 7).Now, choosea, b, c, d ∈ K so that as a sum of homogeneous polynomials,

F (X,Y ) = Fm(X,Y ) + Fm+1 + . . . ,

G(X,Y ) = Gn(X,Y ) + Gn+1 + . . . ,

the termsFm and Gn are not divisible byY . All the same,Fm and Gn have acommon divisorcX + dY . By the preceding result applied toI, it follows thatI(F , G;O) > mO(F )mO(G). SincemO(F ) = mO(F ),mO(G) = mO(G), italso follows thatI(F,G;O) > mO(F )mO(G), as required. 2

The results so far forI(P, F∩G) haveP = O, the origin. Now they are extendedto the case thatP 6= O. With P = (x0, y0), the translation

τ : (X,Y ) 7→ (X − x0, Y − y0)sendsP to the origin. The sameτ sends the curvesF andG to the curvesF ′ andG′, where

F ′(X,Y ) =F (X − x0, Y − y0),G′(X,Y ) =G(X − x0, Y − y0).

Define the functionI(P, F ∩G) by

I(P, F ∩G) = I(O,F ′ ∩G′).

Then it follows immediately thatI(P, F ∩ G) satisfies all of (I 1). . . (I 7). Con-versely, if a functionI(P, F ∩G) satisfies these properties for a fixed pointP , thenalso the functionI(O,F ′ ∩G′) = I(P, F ∩G) enjoys the same properties with re-spect to the origin. By Propositions 3.4 and 3.6, the following theorems are finallyobtained.

46

THEOREM 3.7 I(P, F ∩ G) ≥ mP (F ) ·mP (G), with equality if and only if thecurvesF andG have no common tangent atP .

THEOREM 3.8 (Uniqueness Theorem)For two curvesF andG in A2(K), thereis at most one functionI(P, F ∩G) that satisfies the seven postulates(I 1). . . (I 7).

The following result completes the theory of the intersection number.

THEOREM 3.9 (Existence Theorem)There exists a functionI(P, F ∩G) that sat-isfies the seven postulates(I 1). . . (I 7).

Proof. Suppose thatF andG have no common component. letRY (F,G) be theirresultant whenF andG are considered as polynomials inY . ThenRY (F,G) is apolynomialD(X) ∈ K[X]. Now, choose coordinates so that

(i) O = (0, 0) is not onF orG;(ii) O is not on any line through two distinct points ofF ∩G;(iii) O is not on the tangent toF orG at any point ofF ∩G.

(3.25)

It is now shown that, forP = (x0, y0) ∈ F ∩G,

I(P, F ∩G) = maxk ∈ N | (X − x0)k dividesRY (F,G).

Let

D(X) = c∏

i(X − αi)ri , c ∈ K;

then, for each pointP = (x0, y0) such thatF (P ) = G(P ) = 0, the abscissax0 isone of the rootsαi of D(X). In accord with this, put

I(P, F ∩G) =

∞, if F,G have a common factorH andH(P ) = 0;0, if F (P )G(P ) 6= 0;ri, if F (P ) = G(P ) = 0 andx0 = αi.

(3.26)The truth of postulates (I 1). . . (I 7) is a consequence of results on the resultant

and elimination theory, as can now be verified.The conditions (I 1),(I 2),(I 3),(I 5) follow from the definition. For (I 4), let

F =A1(X − x0) +B1(Y − y0),G=A2(X − x0) +B2(Y − y0).

Then

RY (F,G) = (B2A1 −B1A2)(X − x0),

with B2A1 −B1A2 6= 0. SoI(P, F ∩G) = 1.For (I 6), note that the resultantRY (F,G + AF ) in the form of (2.2) equals

RY (F,G), since the former is obtained from the latter by adding linear combina-tions of the firstm rows to the lastn rows, wheren = degF, m = degG.

It follows from Lemma 2.3 thatRY (F,GH) = RY (F,G)RY (F,H), fromwhich (I 7) follows.

Thus (I 1),. . . , (I 7) are satisfied. 2


REMARK 3.10 For another proof of this theorem in terms of the local ring atP ,see Exercise 6

These results lead to the following geometric and covariantconcept.

DEFINITION 3.11 For the affine curves,F = vα(F ), G = vα(G), and the pointP = (a, b) the intersection numberis

I(P,F ∩ G) = I((a, b), F ∩G).

3.2 BEZOUT’S THEOREM

For a projective curve, local properties are the same as for the corresponding affinecurve.

DEFINITION 3.12 For the projective curves,F = v(F ), G = v(G), and the pointY∞ = (0, 0, 1) the intersection numberis

I(Y∞,F ∩ G) = I((0, 0), F∗, G∗).

whereF∗ andG∗ are the polynomials associated toF andG as in Section 1.4.Intersection numbers ofF = v(F ) andG = v(G) at another pointP are cal-

culated, as in the affine case, by using covariant properties; that is, a projectivity isapplied to changeP to Y∞.

The covariance of the intersection number of projective curves at a point can beshown by using Theorem 3.9 and (3.26). This important property also follows fromthe proof of Bezout’s Theorem 3.14 below. Now, some global results are obtained.

THEOREM 3.13 If the plane projective curvesF andG have degreesm andn,and no common component, then they have at mostmn points in common.

Proof. Suppose, on the contrary, that the curves have more thanmn points in com-mon. Choose anymn+1 of the common points and join every pair by a line. Thereis a pointP on none of these lines nor onF orG. Take a frame for whichP = U2.Then, withF = v(F ), G = v(G),

F (X,Y,Z) = a0Zm + a1Z

m−1 + · · ·+ am,G(X,Y,Z) = b0Z

n + b1Zn−1 + · · ·+ bn,

(3.27)

whereai, bi are homogeneous inX,Y of degreei anda0, b0 6= 0.By Proposition 2.6 and Theorem 2.7, the resultantR(F,G) is either a homoge-

neous polynomialD of degreemn in X,Y or zero. Also,D(x0, y0) = 0 if andonly if there is a point(x0, y0, 1) onF andG. However, every one of themn + 1points has a different value forx0/y0, since the join of any two does not containU2. HenceD(X,Y ) = 0, soF andG have a common factor, andF andG have acommon component. 2

THEOREM 3.14 (Bezout’s Theorem)If the plane projective curvesF = v(F ) andG = v(G) defined overFq have degreesm andn, and no common component, then

∑I(P, F ∩G) = mn.

48

Proof. By Theorem 3.13,F andG have at mostnm common points. Leth ≥ 0denote the exact number of such points.

LetR = (ξ0, ξ1, ξ2) andQ = (η0, η1, η2) be two distinct points, and letℓ be theline throughR andQ. A pointP = (x0, x1, x2) lies onℓ if and only if there existλ, µ ∈ K such that

P = (λξ0 + µη0, λξ1 + µη1, λξ2 + µη2).

A necessary and sufficient condition forP to be a point ofF is

F (λξ0 + µη0, λξ1 + µη1, λξ2 + µη2) = 0.

SinceF (λξ0 +µη0, λξ1 +µη1, λξ2 +µη2) can be viewed as a homogeneous poly-nomialF ′(λ, µ) in the indeterminatesλ andµ, write

F ′(λ, µ) = a0λm + a1λ

m−1µ+ . . .+ am−1λµm−1 + amµ

m.

To calculateas, let

F (X0,X1,X2) =∑

i+j+k=m

aijkXi0X

j1X

k2 .

Then

as =∑

u+v+w=m−s

∑

i+j+k=m

(i

u

)(j

v

)(k

w

)aijk(ξu

0 ξv1ξ

w2 )(ηi−u

0 ηj−v1 ηk−w

2 ).

It may be noted thatas is a homogeneous polynomial of degreem in the indeter-minatesξ0, ξ1, ξ2, η0, η1, η2. If η0, η1, η2 are considered to be constants, thenas isa homogeneous polynomial of degreem− s in the indeterminatesξ0, ξ1, ξ2. Like-wise, if ξ0, ξ1, ξ2 are constants, thenas is a homogeneous polynomial of degreesin the indeterminatesη0, η1, η2. The non-trivial roots ofF ′(λ, µ) correspond to thecommon points of the curveF with the lineℓ.

The similar result holds for the curveG. The lineℓ throughR andQ meetsG inthe pointsP such that(λ, µ) is a non-trivial root of the polynomial,

G′(λ, µ) = b0λn + b1λ

n−1µ+ . . .+ bn−1λµn−1 + bnµ

n.

Therefore,F andG have a non-trivial common root, equivalently

R(F ′(λ, µ), G′(λ, µ)) = 0,

if and only ifF andG have a common point,

Pi = (x(i)0 , x

(i)1 , x

(i)2 ),

on the line throughR andQ. It follows thatR(F ′(λ, µ), G′(λ, µ)) = 0 if and onlyif at least one of the determinants,

D(ξ, η, x(i)) =

∣∣∣∣∣∣

ξ0 ξ1 ξ2η0 η1 η2

x(i)0 x

(i)1 x

(i)2

∣∣∣∣∣∣(3.28)

is zero.Now, fix the pointR and considerQ as a variable point; that is, fix the ho-

mogeneous triple(ξ0, ξ1, ξ2) and considerη0, η1, η2 as indeterminates. Then the


determinantD(ξ, η, x(i)) in 3.28 is a linear homogeneous polynomial while theresultantR(F ′(λ, µ), G′(λ, µ)) is a homogeneous polynomial of degreemn.

Every root ofD(ξ, η, x(i))) is also a root ofR(F ′(λ, µ), G′(λ, µ)). By Study’sTheorem 2.10,D(ξ, η, x(i)) dividesR(F ′(λ, µ), G′(λ, µ)). Conversely, every rootof R(F ′(λ, µ), G′(λ, µ)) is a root of someD(ξ, η, x(i)). Therefore, for certainpositive integersri,

R(F ′(λ, µ), G′(λ, µ)) = c∏k

i=1 D(ξ, η, x(i))ri , (3.29)

wherec is a non-zero constant andk ≤ h. This implies that

nm =∑k

i=1 ri. (3.30)

As a consequence,k ≥ 1 and henceh ≥ 1.Next, it is shown thatri is covariant. To do this, the notation in the proof of

Theorem 1.21 is used. It is a straightforward to show that

D(ξ, η, x(i)) = D(ξ′, η′, (x′(i)

)) det(aij),

which shows the covariance ofD(ξ, η, x(i)). Also, from the definition ofF ′(λ, µ)andG′(λ, µ), it follows that their resultant is a covariant.

Therefore,F andG may be assumed to have no common point on the linev(X2).Then their intersection is in the affine plane with coordinates(x, y) where, as usual,x = x0/x2 andy = x1/x2. So, take the curves in their the affine forms, withF = v(F∗(X,Y )) andG = v(G∗(X,Y )). It may also be supposed that no twosuch common points lie on the same linev(X − c).

If R = (u, 0) andQ = Y∞, then

D(ξ, η, x(i)) = X − ui,

whereui is the abscissa of the pointRi. Also,R(F ′(λ, µ), G′(λ, µ)) is the resul-tantRY (X) of F∗(X,Y ) andG∗(X,Y ), both considered as polynomials in theindeterminateY . Now, (3.29) reads:

RY (X) = c∏k

i=1 (X − ui)νi .

This shows that, ifPi = (ui, vi) is a common point of the curvesF andG, thenriis equal to the multiplicity of the factorX − ui in the resultantRY (X). Finally,

ri = I(Pi,F ∩ G), (3.31)

from Theorem 3.9 and (3.26). 2

EXAMPLE 3.15 With K = C, let

F =Y 5 −X(Y 2 −XZ)2,

G=Y 4 −X2 + Y 3.

Then

ZF − Y 2G=−XZ(Y 2 −XZ)2 − Y 2(Y 2 −XZ)(Y 2 +XZ)

= (Y 2 −XZ)(−Y 2XZ +X2Z2 − Y 4 − Y 2XZ)

= (Y 2 −XZ)(X2Z2 − 2Y 2XZ − Y 4).

50

If F = v(F ),G = v(G) andP = (x, y, z) ∈ F ∩ G, theny2 − xz = 0 ⇒ y = 0 ⇒ P = P1 or P2

with P1 = (1, 0, 0), P2 = (0, 0, 1);

y4 − x2z2 = −2y2xz ⇒ −2y2xz + y3z = 0

⇒ y = 2x ⇒ 16x4 − x2z2 + 8x3z = 0

⇒ 16x2 + 8xz − z2 = 0

⇒ P = P3 or P4

with P3 = (−1 +√

2,−2 + 2√

2, 4), P4 = (−1−√

2,−2− 2√

2, 4).

The pointsP3 andP4 are simple on both the curvesF andG, which intersecttransversally at both points, whence

I(P3, F ∩G) = I(P3, F ∩G) = 1.

ForP2 = (0, 0, 1), putf(X,Y ) = F (X,Y, 1), g(X,Y ) = G(X,Y, 1).

Thenf =Y 5 −X(Y 2 −X)2 = −X3 + 2X2Y 2 −XY 4 + Y 5

g=Y 4 −X2 + Y 3 = −X2 + Y 3 + Y 4

h= f −Xg = 2X2Y 2 −XY 3 − 2XY 4 + Y 5

= Y 2(2X − Y )(X − Y 2);

I(P2, F ∩G) = I(P2, G ∩ F ) = I(P2, g ∩ f) = I(P2, g ∩ h)= I(P2, g ∩ Y 2) + I(P2, g ∩ (2X − Y )) + I(P2, g ∩ (X − Y 2))

= 4 + 2 + I(P2, (X − Y 2) ∩ Y 3)

= 4 + 2 + 3 = 9.

ForP1 = (1, 0, 0), putf(Y,Z) = F (1, Y, Z), g(Y,Z) = G(1, Y, Z).

Thenf =−Z2 + 2Y 2Z − Y 4 + Y 5

g=−Z2 + Y 3Z + Y 4

h= g − f = −2Y 2Z + Y 3Z + 2Y 4 − Y 5

= −Y 2(2− Y )(Z − Y 2);

I(P1, F ∩G) = I(P1, f ∩ g) = I(P1, f ∩ h)= I(P1, f ∩ Y 2) + I(P1, f ∩ (2− Y )) + I(P1, f ∩ (Z − Y 2))

= 4 + 0 + I(P1, (Z − Y 2) ∩ f)

= 4 + I(P1, (Z − Y 2) ∩ (f + (Z − Y 2)2))

= 4 + I(P1, (Z − Y 2) ∩ Y 5)

= 4 + 5 = 9.

Hence ∑4i=1 I(Pi, F ∩G) = 1 + 1 + 9 + 9 = 20 = degF · deg G.


3.3 RATIONAL AND BIRATIONAL TRANSFORMATIONS

DEFINITION 3.16 Given two projective planesπ andπ′ over the same fieldK,then the mappingω : π → π′ is arational transformationif there exist independenthomogeneous polynomialsϕ0, ϕ1, ϕ2 ∈ K[X0,X1,X2] of the same degreen and,for any pointP = (a0, a1, a2) ∈ π,

ω(P ) = P ′ = (b0, b1, b2),

ρy0 = ϕ0(a0, a1, a2),ρy1 = ϕ1(a0, a1, a2),ρy2 = ϕ2(a0, a1, a2),

(3.32)

whereρ ∈ K. Also,n is theorderof the transformation.

Denote byζ the net of curvesv(µ0ϕ0 +µ1ϕ1 +µ2ϕ2). AsP ′ traverses the linev(λ0Y0 + λ1Y1 + λ2Y2) of π′, its pre-imageP describes inπ the curve

v(λ0ϕ0(X0,X1,X2) + λ1ϕ1(X0,X1,X2) + λ2ϕ2(X0,X1,X2)).

Here, to a pointP ′ = (η0, η1, η2), centre of a pencil of lines, corresponds the setof base points of a pencil of curvesϕ of the netζ. Therefore, to the pointP ′

correspond the points common to two curves:

(1) v(ϕ1, ϕ2), for (η0, η1, η2) = (c, 0, 0), c 6= 0;

(2) v(η0ϕ1 − η1ϕ0, ϕ2), for η1 6= 0, η2 = 0;

(3) v(η2ϕ0 − η0ϕ2, η2ϕ1 − η1ϕ2), for η2 6= 0.

To ensure thatω is bijective,n2 − 1 of these points must be fixed: denote themP1, . . . , Pn2−1. Then then2-th pointP = (ξ0, ξ1, ξ2) is the pre-image ofP ′ andX0,X1,X2 can be obtained as rational functions ofY0, Y1, Y2:

ω−1 : π′ → π,

ρX0 = f0(Y0, Y1, Y2),ρX1 = f1(Y0, Y1, Y2),ρX2 = f2(Y0, Y1, Y2).

(3.33)

Heref0, f1, f2 are independent homogeneous polynomials of the same degree. Inthis case, the transformationω is birational; it is also called aCremonatransforma-tion. Then through the pointsP0, P1, . . . , Pn2−1 pass all curves of the netζ. Thisis therefore ahomaloidalnet; that is, every two of its curves have a single variableintersection.

PROPOSITION 3.17 (i) The transformationω changes the curvesϕ of degreen of the homaloidal netζ passing through the pointsP0, P1, . . . , Pn2−1 in πinto the lines ofπ′.

(ii) In π, the pointsP0, P1, . . . , Pn2−1 constitute the set offundamentalpoints ofω.

52

(iii) The imageP ′ of a pointP other than a fundamental point is the image underω of the unique pencil inζ passing throughP .

(iv) When there are two curvesC1, C2 in ζ with a common componentC, everyP on C different from a fundamental point has the same imageP ′, which isa fundamental point forω−1.

Of particular interest among birational transformations are those of order two,that is, thequadratic transformations. Ifω is a quadratic transformation fromπto π′, to the setL of lines ofπ correspond inπ′ the set of conics of a homaloidalnetζ ′, that is, the set of conics passing through three fixed points, Q0, Q1, Q2, thefundamental pointsof ω in π′.

Also the inverse transformationω−1 is quadratic, since to the lines ofπ′ corre-spond inπ the conics passing through three fixed pointsP0, P1, P2, the fundamen-tal points inπ, and constitute the homaloidal netζ.

In the general case, the pointsP0, P1, P2 are distinct and not collinear, as are thepointsQ0, Q1, Q2. If the reference system is chosen so that

P0 = Q0 = U0, P1 = Q1 = U1, P2 = Q2 = U2, ω(U) = U,

then

ω : π → π′,

ρY0 = X1X2,ρY1 = X0X2,ρY2 = X0X1.

(3.34)

The transformationω is not defined at the pointsP0, P1, P2. The lines joiningpairs of these points are theexceptional lines. Every non-fundamental point onone of these linesv(Xj) is transformed to the point(δ0j , δ1j , δ2j), whereδij is theKronecker symbol. So, for example, underω, the lineP1P2 goes to the pointQ0,while, underω−1, the pointQ0 dilates toP1P2. Similarly,ω takes the linesP0P2

andP0P1 to the pointsQ1 andQ2, andω−1 reverses these.If P is any point off the exceptional lines, its imageP ′ is also off the exceptional

lines ofω−1, andω−1(P ′) = P . Therefore, with

∆ = P = (x0, x1, x2) ∈ π | x0x1x2 6= 0,∆′ = Q = (y0, y1, y2) ∈ π′ | y0y1y2 6= 0,

the restrictionω|∆ : ∆ −→ ∆′ is bijective. The inverseω−1 of ω has the followingequation:

ω−1 : π′ → π,

τX0 = Y1Y2,τX1 = Y0Y2,τX2 = Y0Y1.

(3.35)

DEFINITION 3.18 The transformationω is thestandard quadratic transformation.


3.4 QUADRATIC TRANSFORMATIONS

Given a curveF = v(F ) with F = F (X0,X1,X2), the pointsQ = (y0, y1, y2)on the image ofF underω lie on the curveG = v(G), with

G(Y0, Y1, Y2) = F (Y1Y2, Y0Y2, Y0Y1). (3.36)

The curveG given by (3.36) is thealgebraic transformof F .To better understand the geometric relation betweenF andG, consider first the

case thatF is a line withF = a0X0 + a1X1 + a2X2. There are three possibilities.(1) F does not pass through any fundamental point.In this case,a0a1a2 6= 0

and the algebraic transform ofF is the conic

C2 = v(a0Y1Y2 + a1Y0Y2 + a2Y0Y1).

This conicC2 passes through each of the fundamental points inπ′ but has no furtherintersection with any exceptional line. The fundamental pointsQ0, Q1, Q2 corre-spond to the points of intersectionS0, S1, S2 of F with the exceptional lines inπ.So,ω establishes a bijection betweenF\S0, S1, S2 andC2\Q0, Q1, Q2.

(2)F is a line throughP0 other than an exceptional line. HenceF = X1+λX2,and its algebraic transform underω splits into two distinct lines. They are theexceptional linev(Y0) and the lineF ′ = v(F ′) with F ′ = Y2 + λY1. As in (1),ωinduces a bijection betweenF andF ′.

(3) F coincides with one of the exceptional lines. Now, the algebraic transformof F splits into two exceptional lines; they arev(Y1) andv(Y2). Every point ofFfor whichω is defined has as its image the pointQ0 that can be considered as thetransform ofF .

Given a curveF = v(F (X0,X1,X2)) containing no exceptional line as a com-ponent, letG = v(G(Y0, Y1, Y2)) = v(F (X1X2,X0X2,X0X1)) be its algebraictransform. Then

G(Y0, Y1, Y2) = m(Y )F ′(Y0, Y1, Y2),

wherem(Y ) = Y0r0Y1

r1Y2r2 butF ′(Y0, Y1, Y2) is not divisible by anyYi.

DEFINITION 3.19 The curveF ′ = v(F ′) is thegeometric transformof F underω.

THEOREM 3.20 (i) If F ′ is the geometric transform ofF for ω, thenF is thegeometric transform ofF ′ for ω−1.

(ii) Apart from a finite number of points ofF andF ′, the remaining points ofFandF ′ are in bijective correspondence.

Proof. By definition,

F (Y1Y2, Y0Y2, Y0Y1) =M1(Y )F ′(Y0, Y1, Y2), (3.37)

F ′(X1X2,X0X2,X0X1) =M2(Y )F ′′(X0,X1,X2),

whereF ′′ = v(F ′′) is the geometric transform ofF ′ for ω−1. PutYi = XjXk in(3.37) fori, j, k = 1, 2, 3:

F (X20X1X2,X0X

21X2,X0X1X

22 ) =M3F

′(X1X2,X0X2,X0X1)

=M4(X)F ′′(X0,X1,X2),

54

whereM3,M4 are monomials inX0,X1,X2. SinceF has degreen, it follows that

F (X20X1X2,X0X

21X2,X0X1X

22 ) = (X0X1X2)

mF ′′(X0,X1,X2).

Since noXi dividesF or F ′′, soF = F ′′. Further, bothF andF ′′ have a finitenumber of points in common with the exceptional lines, and the bijection betweenthe components ofF andF ′′ is a consequence of the formulas that defineω andω−1. 2

Now consider the behaviour of a singular point of a plane curve under a quadratictransformation. Here, use the following correspondence between the non-except-ional lines throughP0 and the non-fundamental points of the linev(Y0):

v(X1 + λX2)←→ (0, 1,−λ).

As P = (1,−λu, u) describes the lineℓ on π, the imageP ′ = (−λu, 1,−λ)describes the linev(Y0 − uY2) on π′. As P 7→ P0, that is, the parameteru 7→ 0,thenP ′ 7→ Q = (0, 1,−λ).

THEOREM 3.21 LetF = v(F (X0,X1,X2)) be a curve of degreed such that

(a) the fundamental pointPi has multiplicitymi onF , 0 ≤ i ≤ 2;

(b) no exceptional line through a fundamental point is tangent toF at that point.

Then

(i) the algebraic transformG ofF admits the fundamental linev(Yi) as a com-ponent of multiplicitymi;

(ii) the geometric transformF ′ has degree2d−m0 −m1 −m2;

(iii) there is a multiplicity-preserving bijection between the tangents toF at Pi

and the intersections ofF ′ with the exceptional linev(Yi), apart from thetwo fundamental points ofv(Yi);

(iv) I(R,F ′ ∩ v(Yi)) = j, whereR is the point corresponding to a line countedj times among the tangents toF at Pi.

(v) the curveF ′ has a point of multiplicityd − mj − mk at the fundamentalpointQi, wherei, j, k = 1, 2, 3;

(vi) the tangents toF ′ at Qi are not exceptional lines and correspond to theintersections ofF with the linev(Xi), apart from the fundamental points.

Proof. (i), (ii) Here the discussion is limited to considering whathappens to thepointP0. The polynomialF of degreed is written in the form

F (X0,X1,X2) =∑d

j=m0Xd−j

0 Fj(X1,X2),

whereFj is a homogeneous polynomial of degreej in X1,X2 andFm0, Fd are

both non-zero. The algebraic transform ofF is G = v(G) with

G(Y0, Y1, Y2) =F (Y1Y2, Y0Y2, Y1Y2)

=∑d

j=m0(Y1Y2)

d−jFj(Y0Y2, Y0Y1)

=Y d−m01 Y d−m0

2 Y m00 Fm0

(Y2, Y1) + . . .+ Y d0 F (Y2, Y1).


SoY m00 dividesG butY m0+1

0 does not, and similarly forY m11 andY m2

2 . Therefore,G = Y m0

0 Y m11 Y m2

2 F ′, whereF ′ is the geometric transform. It follows thatF ′ hasdegree2d− (m0 +m1 +m2).

(iii) On the one hand, the tangents toF atP0 correspond to the linear factors of

Fm0(X1,X2) =

∏k=nk=1 (µkX1 + νkX2)

αk ,

with k1 + . . . + kt = d; on the other hand,I(Q,F ′ ∩ v(Y0)) at pointsQ is givenby the multiplicity of the roots of the polynomial

Y d−m0−m11 Y d−m0−m2

2 Fm0(Y2, Y1) =

∏k=nk=1 (µkY2 + νkY1)

αk ,

with k1 + . . .+ kt = m0.(iv) Note that

F ′ = Y d−m0−m11 Y d−m0−m2

2 Fm0(Y1, Y2)+ . . .+Y d−m0

0 Y −m11 Y −m2

2 Fd(Y2, Y1);

then the multiplicity ofF ′ atQ0 = (1, 0, 0) is given by the degree of

B(Y1, Y2) = Y −m11 Y −m2

2 Fd(Y1, Y2),

and this isd−m1 −m2.(v) The tangents toF ′ atQ0 are given by the factors ofB(Y1, Y2). If one of

these wereY1 or Y2, thenFd(Y2, Y1) would be divisible by a powerm > m1

of Y1 or m > m2 of Y2. But, if that happens, thenI(P1,F ∩ v(X0)) = m orI(P2,F ∩ v(X0)) = m. Therefore, one of the tangents toF atP1 or atP2 wouldbe the exceptional linev(X0), a contradiction.

(vi) Applying (v) to ω−1 gives the result. 2

THEOREM 3.22 (i) To each pointP external to the fundamental triangle ofπcorresponds a pointQ external to the fundamental triangle ofπ′.

(ii) If such a pointP is anm-ple point of the curveF , then its imageQ is anm-ple point of the geometric transformF ′ ofF .

Proof. TakeP = (1, 1, 1), and soQ = (1, 1, 1). Since the algebraic transformGand the geometric transformF ′ differ by a component not passing throughP , itsuffices to considerG instead ofF ′. In π, take a new system of coordinates:

ρZ0 = X0,ρZ1 = X1 −X0,ρZ2 = X2 −X0.

(3.38)

The pointP is now(1, 0, 0), and

F (X0,X1,X2) = F (Z0, Z0 + Z1, Z0 + Z2)

= F (Z0, Z1, Z2)

= Zd−m0 Fm(Z1, Z2) + . . .+ Fd(Z1, Z2)

= Xd−m0 Fm(X1 −X0,X2 −X0) + . . .+ Fd(X1 −X0,X2 −X0).

Therefore, the algebraic transformG = v(G(Y0, Y1, Y2)) is given by

G=(Y1Y2)d−mFm(Y0Y2 − Y1Y2, Y0Y1 − Y1Y2)

+ . . .+ Fd(Y0Y2 − Y1Y2, Y0Y1 − Y1Y2).

56

In π′, use a similar change of coordinates:

ρW0 = Y0,ρW1 = Y0 − Y1,ρW2 = Y0 − Y2.

(3.39)

This takesQ = (1, 1, 1), the image ofP underω, toQ′ = (1, 0, 0) in π′. Also,

G = G(Y0, Y1, Y2) = G(W0,W0 −W1,W0 −W2) = G(W0,W1,W2).

Now,

G(W0,W1,W2)

= [(W0 −W1)(W0 −W2)]d−mFm(W0W1 −W1W2,W2W0 −W1W2)

+[(W0 −W1)(W0 −W2)]d−m−1Fm−1(W0W1 −W1W2,W2W0 −W1W2)

+ . . .+ Fd(W0W1 −W1W2,W2W0 −W1W2)

= W 2d−m0 Fm(W1,W2) + . . .+W 2d−m−1

0 Fm+1(W1,W2)

+ . . .+W d0 Fd(W1,W2) + . . . ,

in descending powers ofW0. The highest power ofW0 appears in the first term.Therefore,Fd(W1,W2) determines the configuration of the tangents toG at Q′,and the theorem is proved. 2

3.5 RESOLUTION OF SINGULARITIES

This section is devoted to the classical result that every irreducible plane algebraiccurve is birationally equivalent to a curve with only ordinary singularities. Here, apointP of a curve isordinary if the tangents to the curve atP are distinct.

DEFINITION 3.23 If Cn is an irreducible plane algebraic curve of degreen, itsvirtual genusis

g∗(Cn) = 12 (n− 1)(n− 2)−∑1

2r(r − 1), (3.40)

where the sum is over all singular pointsP of Cn, with r the multiplicity ofP .

L EMMA 3.24 The virtual genus of a curve is non-negative.

Proof. LetP1, P2, . . . Pk be the singular points of the curveCn, with multiplicitiesr1, r2, . . . rk. First, it is shown that the integer

s = 12 (n− 1)(n+ 2)−∑k

i=112ri(ri − 1) (3.41)

is non-negative.If Cn = v(F (X,Y,Z)), then the polar curve ofCn with respect toU0 = (1, 0, 0)

is Cn0 = v(∂F/∂X). When charK = p > 0, then it is possible that∂F/∂X is the

zero polynomial. In that case, Theorem 1.23 ensures that∂F/∂Y is not the zeropolynomial. Thus it may be assumed thatCn

0 exists.A point P such thatmP (Cn) = r satisfiesmP (Cn

0 ) ≥ r − 1. By Bezout’stheorem,

∑ki=1ri(ri − 1) ≤ n(n− 1).


Sincen(n− 1) < (n− 1)(n+ 2), sos = 0 for n = 1 ands > 0 otherwise.Now, chooses pointsQ1, . . . , Qs on Cn. First, it is shown that there exists a

curveCn−1 of degreen− 1 passing through each pointPi with multiplicity at leastri − 1 and through eachQj . If G = v(G) is a general plane curve of degreen− 1,thenG has(n−1)(n+2)/2 coefficients. Its passage through an(r−1)-ple pointPrequires at most12r(r − 1) linear conditions on its coefficients. In fact, ifP = U0

andG =∑gi(Y,Z)Xn−1−i, the condition thatP is an(r − 1)-ple point is that

g0 = g1 = . . . = gr−2 = 0;

these terms contain12r(r−1) coefficients. It follows that the passage through thek

singular points requires at most∑k

i=112r(r−1) linear conditions on its coefficients.

Therefore the number of conditions onCn−1 to pass through thePi andQj is atmost

s+∑k

i=112ri(ri − 1) ≤ 1

2 (n− 1)(n+ 2),

by the definition ofs in (3.41). Hence, such aCn−1 exists.Now apply Bezout’s theorem to the curvesCn andCn−1:

n(n− 1) =∑I(P, Cn ∩ Cn−1)

≥∑I(Qj , Cn ∩ Cn−1) +∑I(Pi, Cn ∩ Cn−1)

= s+∑k

i=1ri(ri − 1)

= 12 (n− 1)(n+ 2)−∑k

i=112ri(ri − 1) +

∑ki=1ri(ri − 1)

=n(n− 1)− 12 (n− 1)(n− 2) +

∑ki=1

12ri(ri − 1)

=n(n− 1)− g∗(Cn).

Henceg∗(Cn) ≥ 0. 2

DEFINITION 3.25 The indexof the irreducible plane curveF = v(F ) is the non-negative integer

I(F) = I(F ) =∑

P

∑T (mT − 1), (3.42)

where the outer sum is over all singular pointsP of F and the inner sum is overall distinct tangentsT toF atP , withmT the multiplicity of the tangent.

It is worth noting explicitly the following property of the index.

L EMMA 3.26 The indexI(F) is zero if and only if each singular point ofF isordinary.

Now, a start is made on the proof of the famous theorem on the resolution ofsingularities.

THEOREM 3.27 Through a finite sequence of quadratic transformations, an irre-ducible plane algebraic curve can be transformed into a curve with only ordinarysingularities.

58

Proof. Given a finite sequence of quadratic transformations,ω1, . . . , ωr, if F isan irreducible plane curve, the symbolF (i) indicates the geometric transform ofF (i−1) underωi, for i = 1, . . . , r − 1 with F (0) = F .

Now, letF be an irreducible plane curve of degreed with a non-ordinary singu-larity; it suffices to exhibit a finite sequenceω1, . . . , ωr of quadratic transformationssuch thatI(F (k)) ≤ I(F (k−1)) and, if equality holds, then

g∗(F (k)) < g∗(F (k−1)).

If F (k) still has non-ordinary singularities, then, with a new sequence of quadratictransformations, another curve is obtained with strictly lower index or virtual genus.This procedure is repeated till a curve is obtained with index zero, which necessar-ily has only ordinary singularities by Lemma 3.26.

To construct such a sequence, a particular system of coordinates is used thatsatisfies the following conditions for a givenm-ple point ofF :

(1) them-ple point isP0 = (1, 0, 0);

(2) the curveF does not pass through the other two fundamental points, andFintersects each of the two exceptional lines throughP0 in d−m points otherthanP0;

(3) the exceptional linev(X0) meetsF in d distinct points.

The existence of such a system of coordinates does not alwayshappen, for in posi-tive characteristic there exist strange curves containingtheir own nucleus, for whichcondition (2) does not hold. This possibility is treated below. 2

The utility of such a system of coordinates lies in the following result.

L EMMA 3.28 If ω is the quadratic transformation relative to a triangle satisfyingconditions(1), (2), (3) and, if F ′ is the geometric transform ofF underω, then thefollowing properties hold:

(i) g∗(F ′) ≤ g∗(F), I(F ′) ≤ I(F);

(ii) if P0 is not an ordinary singularity, then

I(F ′) = I(F) =⇒ g∗(F ′) < g∗(F).

Proof. Denote byℓ1, . . . , ℓs the tangents toF at P0, and byα1, . . . , αs their re-spective multiplicities. From Theorem 3.22, every singular point ofF other thanP0 is transformed into a point ofF ′ with the same multiplicity; an ordinary sin-gularity ofF external to the fundamental triangle is transformed into anordinarysingularity ofF ′, also external to the fundamental triangle.

By part (v) of Theorem 3.21,F ′ has three “new” singular points:Q0 of mul-tiplicity d, andQ1, Q2 both of multiplicity d − m. From part (iv) of the sametheorem, it follows thatF ′ hass pointsR′

1, . . . , R′s on the linev(Y0) distinct from

the fundamental points, withr′i the multiplicity ofR′i; then

r′i ≤ I(R′i,F ′ ∩ v(Y0)) = αi


for 1 ≤ i ≤ s. Also suppose thatF ′ has degree2m− d as in (ii) of Theorem 3.21.To prove the first assertion of Lemma 3.28, it suffices to see that

g∗(F ′) = g∗(F)−∑sj=1

12r

′j(r

′j − 1). (3.43)

If F hask singular points other thanP0 with multiplicities r1, . . . , rk, then

g∗(F ′) = 12 (2d−m− 1)(2d−m− 2)− 1

2d(d− 1)− (d−m)(d−m− 1)

−∑ki=1

12ri(ri − 1)−∑s

j=112r

′j(r

′j − 1).

This expression can also be written in the form:

12 (d2 − 3d+ 2)− 1

2m(m− 1)−∑ki=1ri(ri − 1)−∑s

j=1r′j(r

′j − 1)

= g∗(F)−∑sj=1

12r

′j(r

′j − 1),

as required.For part (ii), note that, ifI(F) = I(F ′), then

∑si=1(αi − 1)−∑s

i=1

∑Ti

(m′Ti− 1) = 0,

where, in the second summation,Ti varies in the set of tangents toF ′ at R′i. It

follows that∑k

i=1[αi −∑

Tim′

Ti] + T − s = 0,

whereT is the total number of tangents toF ′ at the pointsR′1, . . . , R

′s. Since

αi = I(Ri,F ′ ∩ v(Y0)) ≥ r′i =∑

Tim′

Ti

andT ≥ s, this implies thatαi = r′i andT = s.It follows that if, as in the proof of (ii) of Lemma 3.28, someαi > 1, then also

r′i > 1. Note also thatT = s if and only if F ′ has a unique tangent at each pointR′

i. Hence the sum∑s

i=112r

′i(r

′i − 1) is positive, which proves (ii). 2

By virtue of Lemma 3.28, the procedure described immediately after the enunci-ation of Theorem 3.27 carries over to an irreducible plane curve with only ordinarysingularities once it is possible to find, for each curveF (i) of the sequence, a systemof coordinates for which conditions (1),(2),(3) hold. To finish the proof of Theorem3.27, it is necessary to construct the sequenceF (1) = F ′,F (2), . . . in such a waythat points not satisfying (2) are avoided.

To do this, use non-homogeneous coordinatesX = X1/X0, Y = X2/X0, andletF = v(f(X,Y )). To see that condition (2) can be satisfied, the intersectionofF with a lineℓ = v(Y − tX), t ∈ K∗, must be found:

f(X,Y ) = 0, Y − tX = 0. (3.44)

Write

f(X,Y ) = Fm(X,Y ) + Fm+1(X,Y ) + . . . Fd(X,Y ),

whereFi(X,Y ) is a homogeneous polynomial of degreei with Fm 6= 0, Fd 6= 0.Now, (3.44) gives

Fm(X, tX) + Fm+1(X, tX) + . . .+ Fd(X, tX) = 0,

60

which becomes

Xm(Fm(1, t) +XFm+1(1, t) + . . .+Xd−mFd(1, t)) = 0. (3.45)

If ℓ is not a tangent toF atP0, thenFm(t) 6= 0.Suppose this is the case, and putFi(t) = Fi(1, t). Then (2) is satisfied by the

line ℓ since

Pt(X) = Fm(t) +XFm+1(t) + . . .+Xd−mFd(t)

has no multiple roots. This is equivalent to saying thatPt(X) and its derivativeQt(X) have no common root. Now,

Qt(X) = Fm+1(t) + 2XFm+2(t) + . . .+ . . .+ (d−m)Xd−m−1Fd(t).

To see that this happens, distinguish two cases according asQt(X) is identicallyzero or not.

In the second case, sinceFj(t) is a polynomial int, soPt(X) can be written as apolynomial in the indeterminatesX, t asP (X, t). Similarly, defineQ(X, t). Usingthe resultant, the system,

P (X, t) = 0 Q(X, t) = 0, (3.46)

has a finite number of solutions(x, λ) unless the two polynomials have a commonfactor. This possibility can be ruled out as follows.

First, it is shown thatP (X, t) is irreducible. Otherwise,

P (X, t) = U(X, t)V (X, t),

with U, V non-constant; therefore,

P (X,Y/X) = U(X,Y/X)V (X,Y/X).

Now, if both sides are multiplied byXm, the left becomesf(X,Y ) and the rightthe product ofXiU(X,Y/X) andXjV (X,Y/X) with i + j = m. But this isa contradiction since both are non-constant in that they contain Y ; otherwise,XdividesP (X, t), which is impossible sinceFm(t) 6= 0 for all t.

To show that the two polynomials in (3.46) have no common factors, it sufficesto show thatdegP (X, t) > degQ(X, t). To see this, note that, asFj(t) has degreeat mostj, so2i + m0 − 1 is an upper limit for the degree ofFm+i(t)X

i−1. Thisnumber reaches its maximum for the largest value ofi allowed that equalsd −m.Therefore, the degree ofQ(X, t) cannot be greater than

2(d−m) +m+ 1 = 2d−m+ 1.

SincedegP (X, t) = 2d − m, the result is proved. So it has been shown that ifQt(X) is not identically zero, then (3.38) has a finite number of solutions(x, λ).Therefore there is an infinite number of values ofλ for which (x, λ) is not a solu-tion; so the lineℓ = v(Y − λX) satisfies condition (2).

Now consider the case thatQt(X) is the zero polynomial. This can only happenwhen the characteristicp of K is positive, andp | (d − m). Also p | i for eachi, with 1 ≤ i ≤ d − m, for whichFm+i(t) is not identically zero. This gives anecessary and sufficient condition forQt(X) to be zero. The following exampleshows that this can occur.


EXAMPLE 3.29 Given a fieldK of characteristicp > 0, letf(X,Y ) = Y −Xp+1.The plane curveF = v(f(X,Y )) is non-singular. The intersection ofF withℓ = v(Y − tX), apart fromP0, is the single pointP = (x, xp+1), wherex is theunique root ofXp − t. Therefore there is no line throughP0 satisfying condition(2).

DEFINITION 3.30 The pointP is a terrible point ofF if, when it is transformedto P0, it does not satisfy condition (2).

It follows that a point ofF is terrible if and onlyF is strange andP is its uniquenucleus. Summarising so far, if none of the singular points of F is terrible, Theorem3.27 is proved.

Now it is shown how to get rid of the the obstacle of terrible points. SupposePis a terriblem-ple point ofF ; then, as above,d ≡ m (mod p).

Take a triangle with sides not containingP and with vertices not lying on thecurve if d 6≡ 0 (mod p), but with a single vertex on the curve at a non-singularpoint if d ≡ 0 (mod p). Apply the quadratic transformationω relative to thistriangle as fundamental triangle; it is shown thatω takesP to a non-terrible pointQ of the geometric transformF ′. Note thatF ′ has degree2d or 2d − 1 accordingas the vertices of the triangle are all external toF or as one is a non-singular pointof F .

If Q were terrible, in the first cased 6≡ 0 (mod p) and2d−m ≡ 0 (mod p);in the second case,d ≡ 0 (mod p) and2d− 1−m ≡ 0 (mod p). But both casesare impossible in that they contradictd ≡ m (mod p). ThereforeQ is not terrible.

So it has been shown that, if there is a singular terrible point P onF , the strategyis to first apply the above quadratic transformation to give anew curveF ′, the geo-metric transform ofF , for which the corresponding pointQ is no longer terrible.Using this, Theorem 3.27 is finally proved.

EXAMPLE 3.31 Let q = 2s and letK be the algebraic closure ofFq. Let u < 2s

and letD be the absolutely irreducible curve in affine formv(g(X,Y )) such that

(Xu + 1)(Y 2s

+ 1) + (Y u + 1)(X2s

+ 1)

= g(X,Y )(X + 1)(Y + 1)(X + Y ). (3.47)

Let g1(X,Y ) = g(X + 1, Y + 1). Then

((X + 1)u + 1)Y 2s

+ ((Y + 1)u + 1)X2s

= g1(X,Y )XY (X + Y ). (3.48)

The analysis of the singular points ofD is as follows.

(i) E = (1, 1) is an ordinary(2s − 2)-ple point, with the appropriately distincttangentsv(Y +mX +m+ 1), wherem2s−1 = 1 butm 6= 1.

(ii) For eachb with bu = 1 butb 6= 1, the pointP = (b, 1) is a(2s−1)-ple pointwith a single tangent.

(iii) For eachc with cu = 1 butc 6= 1, the pointP = (1, c) is a(2s−1)-ple pointwith a single tangent.

62

(iv) For all b, c with bu = cu = 1 but b 6= 1, c 6= 1, the pointP = (b, c) is a2s-ple point with a single tangent.

Let ω be the standard quadratic transformation(X,Y,Z) 7→ (Y Z,XZ,XY ).Then the geometric transformD′ of D is the curvev(g2(X,Y )) of degree2s − 2,where

(Xu−1 + . . .+ 1)X2s−u + (Y u−1 + . . .+ 1)Y 2s−u

= g2(X,Y )(X + Y ). (3.49)

Then the curveD′ has only ordinary singularities. Also, the genus ofD′, and so ofC, is (u− 2)(2s − u− 2).

The analysis of the singular points ofD′ is as follows:

(1) O = (0, 0) is an ordinary(2s − u− 1)-ple point;

(2) E = (1, 1) is an ordinary(u− 1)-ple point.

No point at infinity is a singular point. So, ifP = (b, c) is a singular point ofD′,then∂g2/∂X = 0 atP , and so

(1 + b)u−1b2s−u−1 = 0.

Thenb = 0 or 1; similarly, c ∈ 0, 1. So neitherP = (0, 1) norQ = (1, 0)lies onD′, whenceD′ has only two singular points,O = (0, 0) andE = (1, 1).More precisely,O = (0, 0) is an ordinary(2s − u − 1)-ple point with tangentsv(Y + λX), whereλ2s−u = 1 butλ 6= 1.

To find the tangents toD′ atE = (1, 1), consider the translation

(X,Y ) 7→ (X + 1, Y + 1)

that takesE toO. ThenD′ becomesD′′ = v(g3(X,Y )), where

Xu(X + 1)2s−u +X2s

+ Y u(Y + 1)2s−u + Y 2s

= g3(X,Y )(X + Y ).

This shows thatE is a(u− 1)-ple point with tangentsv(Y + λX + λ+ 1), whereλu = 1 butλ 6= 1.

3.6 EXERCISES

1. LetF = v(X1X20 −X3

2 ) andG = v(X1X40 −X5

2 ). Show that the commonpoints ofF andG are

P1 = (1, 0, 0), P2 = (1, 1, 1), P3 = (1,−1,−1), P4 = (0, 0, 1),

where

I(Pi,F ∩ G) =

3 wheni = 1,1 wheni = 2, 3,10 wheni = 4.

2. (Pascal’s theorem) Assume that the points of intersection of the oppositesides of a hexagon inPG(2,K) are distinct. Show that these points, thediagonal points, are collinear if and only if hexagon is inscribed in a conic.


3. (Maclaurin’s Theorem) LetF be an irreducible plane cubic curve, and letP be any non-singular point ofF . If P is not an inflexion point ofF , thetangential pointP ′ of P is the unique point ofF on the tangent line toF atP other thanP . If P is an inflexion point, letP ′ = P . Let P,Q,R ∈ Fbe three non-singular collinear points ofF . Show that their tangential pointsP ′, Q′, R′ are also collinear.

4. (Lame’s Theorem) Letℓ1, ℓ2, ℓ3 andr1, r2, r3 be two triples of distinct linesin PG(2,K). Assume that no line from one triple passes through the com-mon point of two lines from the other triple. For1 ≤ j, k ≤ 3, letRjk denotethe common point of the linesℓj andrk. There are nine such common points.Show that if eight of these points lie on a cubic curve then theninth also does.This is the particular case of the Theorem of the Nine Associated Points usedin Section 8.10.

5. Generalise the preceding exercise. LetN = 12n(n + 3), and consider a set

S of N distinct common points of two projective plane curves of degreen.Then eitherS contains a pointP and subsetS ′ of N − 2 points distinct fromP such that any curve of degreen throughS ′ also passes throughP , or,for every pointP ∈ S, any curve of degreen throughS\P also passesthroughP .

6. For two polynomialsF,G ∈ K[X,Y ] and a pointP , the ideal generated byF andG in the local ringOP is (F,G) · OP . Show that the residue classring OP /((F,G) · OP ) is a finite-dimensional vector space overK whosedimensiondimK OP /((F,G) · OP ) satisfies the postulates(I 1), . . . , (I 7)studied in Section 3.1. Note that, by Theorem 3.8,

dimK OP /((F,G) · OP ) = I(P, F ∩G).

3.7 NOTES

Section 3.1 is based on Scherk [234], which in turn is a variant of Fulton [84,Chapter 3]. See also the latter for more on the local ring approach as in Remark3.10.

The proof of Bezout’s Theorem 3.14 is based on [300, Chapter 3]; see also [101,Chapter 2]. For other proofs, see [84], [163], [252], [303].

For Sections 3.3, 3.4, 3.5, see Fulton [84, Chapter 7] and Moreno [203].

PART B

Birational properties of plane curves

Chapter Four

Branches and parametrisation

4.1 FORMAL POWER SERIES

LetK be a field and letK[[X,Y ]] be the set of elements of the form

a00 + a10X + a01Y + a20X2 + · · · , aij ∈ K,

in which two operations, the sum and the product, are defined as follows:

(a00 + a10X + a01Y + · · · ) + (b00 + b10X + b01Y + · · · )= (a00 + b00) + (a10 + b10)X + (a01 + b01)Y + · · · ;

(a00 + a10X + a01Y + · · · )× (b00 + b10X + b01Y + · · · )= a00b00 + (a00b10 + a10b00)X + · · · .

Then(K[[X,Y ]],+,×) is a commutative ring whose elements are calledformalpower series. The identity is represented by the element1 + 0X + 0Y + · · · andthe zero by the element0 + 0X + 0Y + · · · . A formal power series can be writtenin the form

F = F0 + F1 + F2 + · · ·whereFi is homogeneous polynomial inX andY of degreei.With this convention,if

G = G0 +G1 +G2 + · · · ,then

F +G=(F0 +G0) + (F1 +G1) + · · · , (4.1)

F G=(F0G0) + (F0G1 + F1G0) + · · · . (4.2)

If F = Fr +Fr+1 + · · · , with Fr 6= 0, thenr is called theorderor subdegreeof F.The element0 = 0 + 0X + 0Y + · · · does not have an order since there is no firstnon-zero term; the convention is to write ord0 =∞. Then, forF,G ∈ K[[X,Y ]],

ord (F +G)≥minordF,ordG,ord (FG) = ordF + ordG.

SinceK[[X,Y ]] is an integral domain, it has a quotient field

K((X,Y )) = F/G | F,G ∈ K[[X,Y ]], G 6= 0,called thequotient field of formal power series.

67

68

GivenF1, F2, . . . ∈ K[[X,Y ]], the definition of∑n

i=0 Fi follows from the de-finition of the sum of two power series, (4.1). However, to define

∑∞i=0 Fi, the

following procedure suffices. LetFi = Fi0 + Fi1 + · · ·+ Fij + · · · , and supposethat ordFi →∞ for i→∞; then define∑∞

i=0 Fi =∑∞

i=0 Fi0 +∑∞

i=0 Fi1 + · · · .This make sense since each sum

∑∞i=0 Fij is well-defined. The following proper-

ties hold: ∑∞i=0 Fi +

∑∞i=0 Gi =

∑∞i=0 (Fi +Gi); (4.3)

F∑∞

i=0 Gi =∑∞

i=0 FGi. (4.4)

THEOREM 4.1 The units ofK[[X,Y ]] are the formal power series of order zero.

Proof. Let F be a unit inK[[X,Y ]]. Then there existsG ∈ K[[X,Y ]] such thatFG = 1; hence0 = ord (1) = ord (FG) = ordF + ordG. Since both ordF andordG are non-negative, then ordF = ordG = 0.

To show the converse, let ordF = 0; then, witha00 6= 0,

F = a00 + a10X + a01Y + · · ·= a00(1 +

a10

a00X +

a01

a00Y + · · · )

= a00(1−G(X,Y )),

with ordG(X,Y ) > 0; that is,

F = a00(1−G). (4.5)

Now, note that, if ordG = r, then ordGi = ir and so ordGi → ∞ for i → ∞.Hence

∑∞0 Gi makes sense. So, multiplying (4.5) by this expression gives

F (1 +G+G2 + · · ·+Gi + · · · )= a00(1−G)(1 +G+G2 + · · ·+Gi + · · · )= a00(1−G)

∑∞i=0 G

i

= a00

∑∞i=0 (1−G)Gi

= a00

∑∞i=0 (Gi −Gi+1)

= a00(1−G+G−G2 + · · · )= a00.

Dividing by a00 gives

F (a00−1(1 +G+G2 + · · ·+Gi + · · · )) = 1;

that is,F is invertible with inversea00−1(1 +G+G2 + · · ·+Gi + · · · ). 2

The preceding ideas and results for two indeterminates can be extended to anyfinite number of indeterminates, including the case of a single indeterminate. Somefurther properties of

K[[t]] = F (t) = artr + ar+1t

r+1 + · · · , ai ∈ Kare now given. First, if ordF = r, write

F = tr(ar + ar+1t+ · · · ) = trE(t),

with E(t) invertible.

Branches and parametrisation 69

THEOREM 4.2 The ringK[[t]] is a unique factorisation domain.

Proof. Every element has the formf = trE(t), with E(t) invertible. If it is irre-ducible, thenr = 1; further,tE(t) andt are associates. Therefore, every completefactorisation off ∈ K[[t]] has exactlyr = ord f factors of the formt or tE(t).Hence, the only complete factorisation off is t · . . . · t · tF (t), for a unitF (t). 2

DEFINITION 4.3 The field

K((t)) = F/G | F,G ∈ K[[t]], G 6= 0

=

tsE1(t)

trE2(t)= ts−rE(t) = tmE(t)

,

withE1(t), E2(t), E(t) invertible andm an integer, is thefield of rational functionsof formal power seriesin the indeterminatet.

Now, all theK-monomorphisms andK-automorphisms ofK[[t]] are determin-ed. Here, aK-homomorphismof K[[t]] is a mapping of the ring to itself pre-serving addition and multiplication, and fixing every element of K: it is a K-monomorphismif it is injective, that is, it has zero kernel. AK-automorphismis asurjectiveK-monomorphism, that is, a bijectiveK-homomorphism. If τ ∈ K[[t]],τ 6= 0 with ordt τ ≥ 1, then thesubstitutiont 7→ τ is the mapping

K[[t]]→ K[[t]],∑∞

i=0 citi 7→∑∞

i=0 ciτi.

(4.6)

Theorder of the substitution is the order ofτ . It is immediate that such a mappingis aK-monomorphism ofK[[t]]. The following theorem shows that everyK-monomorphism ofK[[t]] is of this type.

THEOREM 4.4 (i) TheK-monomorphisms ofK[[t]] are precisely the substitu-tionst 7→ τ .

(ii) AK-monomorphism ofK[[t]] is aK-automorphism if and only if the asso-ciated substitution has order1.

(iii) AnyK-monomorphism ofK((t)) induces aK-monomorphism ofK[[t]], andconversely.

Proof. It is first shown that, if the substitutiont 7→ τ has order1, then it is sur-jective. Giveng = g0 + g1t + g2t

2 + . . . in K[[t]], it is opportune to ensure theexistence of an elementf = f0 + f1t+ f2t

2 + . . . in K[[t]] such that

f0 + f1τ + f2τ2 + . . . = g0 + g1t+ g2t

2 + . . . .

Equivalently, withτ = d1t + d2t2 + d3t

3 . . ., the coefficientsf0, f1, . . . can bechosen withf0 = g0 and successively

f1d1 = g1,

f2d21 + f1d2 = g2,

f3d31 + f22d1d2 + f1d3 = g3,

... .

70

Sincefidi1 appears in thei-th row, but there is no other term in this row containing

fi, these equations determine the coefficientsf0, f1, f2, . . . , uniquely, and sof isobtained.

It remains to consider aK-monomorphismσ of K[[t]], and show that it is ob-tained by a substitutiont 7→ τ , for a suitable elementτ ∈ K[[t]] of order at least1.First, note thatσ transforms every invertible element ofK[[t]] to another invertibleelement ofK[[t]]. In fact, iff ∈ K[[t]] is invertible, there existsg such thatfg = 1;thenσ(f)σ(g) = σ(1) = 1, and soσ(f) is invertible. The invertible elements haveorder zero; soσ sends elements of order zero to elements of order zero.

Now, it is shown thatσ(t) has positive order. If it were otherwise, put

σ(t) = c0 + c1t+ . . . ;

then

σ(−c0 + t) = c1t+ . . .

with c0 6= 0, while σ(−c0 + t), being of order0, must be invertible; this is acontradiction. It follows thatσ sends non-invertible elements to non-invertible el-ements inK[[t]]; in other words,σ sends elements of positive order to elementsof positive order. In fact, iff = tnE(t) with n ≥ 1 andE(t) is invertible, thenσ(f) = τnF (t) with ordt τ ≥ 1, andF (t) = σ(E(t)) is invertible. More pre-cisely,σ sends an element of orderv to one of orderv · ordt τ , and fixes the ordersif and only if ordt τ = 1.

This last observation also shows that, ifσ is an automorphism, thenσ(t) musthave order1. Now, sinceσ is aK-monomorphism, then, for every positive integerr,

σ(∑r

i=0 citi) =

∑ri=0 ciσ(ti) =

∑ri=0 ci(σ(t))i =

∑ri=0 ci(τ)

i.

More must be shown, namely that

σ(∑∞

i=0 citi) =

∑∞i=0 ciσ(t)i =

∑∞i=0 ci(τ)

i.

For this, write∑∞

i=0 citi = (

∑ri=0 cit

i) +Rr,

with Rr =∑∞

i=r+1 citi ∈ K[[t]] and ordRr > r. Therefore,

σ(∑∞

i=0 citi) = σ(

∑ri=0 cit

i) + σ(Rr) =∑r

i=0 ci(τ)i + σ(Rr); (4.7)

Since ordσ(Rr) ≥ ordRr > r, it follows from (4.7) that

ordt

(σ(∑∞

i=0 citi)−∑r

i=0 ciτi)> r.

Sincer can be chosen arbitrarily, this implies that

σ(∑∞

i=0 citi)−∑∞

i=0 ciτi = 0,

as required.To prove that everyK-monomorphismσ ofK((t)) induces aK-monomorphism

of K[[t]], it suffices to add just a remark to the above proof, namely that σ(t) has


non-negative order. Assume on the contrary that ordt σ(t) < 0. Then, for anynon-zeroc ∈ K,

ordt(σ(t+ c)) = ordt σ(t),

and hence ordt σ(t+x) also has negative order. On the other hand,t+c is invertible.As was shown before, this implies that ordt σ(t + c) = 0, a contradiction. Theconverse follows from (i) since any substitution defines aK-monomorphism ofK((t)). 2

It is worth noting explicitly that the preceding proof also shows the followingresult.

THEOREM 4.5 Let f(t) ∈ K[[t]] and g(τ) ∈ K[[τ ]]. If f(t) = g(τ) for someτ ∈ K[[t]], then

ordt f = ordτ g · ordt τ. (4.8)

The following result relates polynomials and power series.

THEOREM 4.6 Let F (X,Y ) ∈ K[X,Y ] with F (0, 0) = 0. Let ∂F/∂Y 6= 0 at(0, 0). Then there is a uniquef(X) = c1X + c2X

2 + . . . in K[[X]] such that

F (X, c1X + c2X2 + . . .) = 0.

Proof. Without loss of generality, assume that∂F/∂Y = 1 at (0, 0), Then

F (X,Y ) =Y + a10X + . . . ∈ K[X,Y ],

F (X,Y + Z) =F (X,Y ) +∂F

∂YZ + . . . ,

where the partial derivatives are calculated with respect toX andY . Since

∂F

∂Y= 1 +G(X,Y )

with ordG(X,Y ) > 0, it follows that

F (X,Y + Z) = F (X,Y ) + [1 +G(X,Y )]Z +G2Z2 + . . . . (4.9)

First, the uniqueness off is shown. Suppose that both the formal power series

c1X + . . .+ ci−1Xi−1 + ciX

i + . . . ,

c1X + . . .+ ci−1Xi−1 + diX

i + . . .

satisfy the conditions of the theorem. Substitute

Y = c1X + . . .+ ci−1Xi−1, Z = ciX

i + . . .

in (4.9):

0 =F (X, c1X + . . .+ ci−1Xi−1 + ciX

i + . . .)

=F (X, c1X + . . .+ ci−1Xi−1) + [1 +G(X, c1X + . . .+ ci−1X

i−1)]

×(ciXi + . . .) +Xi+1H(X).

Hence it follows thatci is uniquely determined by the coefficientsc1, . . . , ci−1. Butthis implies thatci = di for all i.

72

For the existence ofF , construct inductively for eachi, a polynomial

c1X + . . .+ ciXi

such that ordF (X, c1X + . . . + ciXi) > i. For i = 1, put c1 = −a10. Suppose

then that a polynomialc1X + . . .+ ci−1Xi−1 has been determined with

ordF (X, c1X + . . .+ ci−1Xi−1) > i− 1.

Hence

F (X, c1X + . . .+ ci−1Xi−1 + ciX

i)

=F (X, c1X + . . .+ ci−1Xi−1) + [1 +G(X, c1X + . . .+ ci−1X

i−1)]

×(ciXi +Xi+1H(X))

= (dXi + . . .) + ciXi +Xi+1H(X)).

Now, chooseci = −d to ensure thatF (X, c1X1 + . . .+ ciXi + . . .) = 0. In fact,

by (4.9),

F (X, c1X1 + . . .+ ci−1Xi−1 + ciX

i + . . .)

=F (X, c1X1 + . . .+ ci−1Xi−1) +XiH(X).

By the inductive hypothesis, ordF (X, c1X1 + . . .+ ci−1Xi−1) > i− 1, whence

ordF (X, c1X1 + . . . + ciXi) > i. Since this is true for alli, the only possibility

is thatF (X, c1X + c2X2 + . . .) = 0. 2

It may be noted that the condition in this theorem can be expressed geometri-cally: the algebraic curveF = v(F (X,Y )) has a non-singular point at the originand the tangent there is not theY -axis. SinceY = c1X + c2X

2 + . . . is a solutionin K[[X]] of the equationF (X,Y ) = 0, it follows from Theorem 4.6 that

F (X,Y ) = [Y − (c1X + c2X2 + . . .)]F1(X,Y ),

with F1(X,Y ) ∈ K[[X]][Y ]. Also,F1(0, 0) 6= 0, since the origin is a simple pointof the curve; that is, ordF1 = 0. It follows that

ordF (X, 0) = ord (c1X + c2X2 + . . .).

So the following theorem has been proved.

THEOREM 4.7 LetF be an algebraic curve passing through the originO. If O isa non-singular point ofF and the tangent toF at O is different from theY -axisv(Y ), then the intersection multiplicity ofF andv(Y ) atO is

I(O,F ∩ v(Y )) = ord (c1X + c2X2 + . . .).

This theorem is generalised as follows.

THEOREM 4.8 LetF = v(F (X,Y )) andG = v(G(X,Y )) be two curves withno common components, and suppose thatF has a non-singular point at the originwith tangent different from theY -axis. If c1X + c2X

2 + . . . is the unique formalpower series such thatF (X, c1X + c2X

2 + . . .) = 0, then

I(O,F ∩ G) = ordG(X, c1X + c2X2 + . . .).


L EMMA 4.9 (Hensel’s Lemma)LetK be a field and

F (X,Y ) = Y n + a1(X)Y n−1 + · · ·+ an(X) ∈ K[[X]][Y ]

a monic polynomial of degreen > 0 in Y with coefficientsa1(X), . . . , an(X) inK[[X]]. Suppose thatF (0, Y ) = G(Y )H(Y ), where

G(Y ) =Y r + b1Yr−1 + · · ·+ br,

H(Y ) =Y s + c1Ys−1 + · · ·+ cs

are monic polynomials of degreer > 0 ands > 0 in Y with coefficients inK suchthatgcd(G(Y ),H(Y )) = 1. Then there exist two unique monic polynomials

G(X,Y ) =Y r + b1(X)Y r−1 + · · ·+ br(X),

H(X,Y ) =Y s + c1(X)Y s−1 + · · ·+ cs(X)

in K[[X]][Y ] of degreer, s in Y with coefficientsb1(X), . . . , br(X) and c1(X),. . . , cs(X) in K[[X]] such that

G(0, Y ) = G(Y ), H(0, Y ) = H(Y ),

F (X,Y ) = G(X,Y )H(X,Y ). (4.10)

Proof. Write F = F (X,Y ) as a power series inX whose coefficients are polyno-mials inY :

F = F0(Y ) + F1(Y )X + · · ·+ Fm(Y )Xm + · · · ,whereF0(Y ) = F (0, Y ) ∈ K[Y ] is monic withdegF0 = n andFm(Y ) ∈ K[Y ]with degFm < n for all m > 0.

It is necessary to find

G=G0(Y ) +G1(Y )X + · · ·+Gi(Y )Xi + · · · ,H =H0(Y ) +H1(Y )X + · · ·+Hj(Y )Xj + · · · ,

with

G0 = G0(Y ) = G(Y ) ∈ K[Y ] monic of degreer,

H0 = H0(Y ) = H(Y ) ∈ K[Y ] monic of degrees,

Gi = Gi(Y ) ∈ K[Y ] anddeg Gi(Y ) < r for eachi > 0,

Hj = Hj(Y ) ∈ K[Y ] anddeg Hj(Y ) < s for eachj > 0,

such thatF = GH. It should be noted thatF = GH means that

Fm =∑

i+j=m GiHj for eachm ≥ 0.

The proof proceeds by induction onm. Form = 0, the result is that

F0 =∑

i+j=0 GiHj ,

F0 = F0(Y ) = F (0, Y ) = G(Y )H(Y ) = G0(Y )H0(Y ).

Let m > 0. Assume that there existGi, Hj ∈ K[Y ], with degGi < r anddegHj < s for 1 ≤ i, j < m satisfying the equation (4.10) for values smaller than

74

m. Then it is necessary to findGm, Hm in K[Y ] with degGm < r, degHm < ssatisfying

Fm =∑

i+j=m GiHj .

This equation may be written as follows:

G0Hm +H0Gm = Um,

where

Um = Fm −∑

i+j=m

i,j<m

GiHj

is known, since it contains only terms already calculated bythe inductive hypothe-sis.

It may be observed thatdegUm < n; in fact,

degFm < n, degGi < r < n, degHj < s < n,

from which it follows thatdeg(GiHj) = deg(Gi) + deg(Hj) < r + s ≤ n andhencedegUm < n.

Sincegcd(G0,H0) = 1, so

G0H∗ +H0G

∗ = 1

with G∗, H∗ ∈ K[Y ]. Multiplying both sides byUm gives that

G0H∗m +H0G

∗m = Um

with G∗m = UmG

∗ andH∗m = UmH

∗. Using the division algorithm to divideH∗m

byH0, the polynomialHm is obtained:

H∗m = EmH0 +Hm,

whereEm, Hm ∈ K[Y ] with degHm < s. Substituting this expression forH∗m in

G0H∗m +H0G

∗m = Um = G0Hm +H0Gm

gives that

G0(EmH0 +Hm) +H0G∗m = G0Hm +H0Gm,

from which, simplifying and dividing byH0,

Gm = G0Em +G∗m.

Also, sincedegUm < n, it follows thatdegGm < r. By induction,Gi andHj

have been calculated so thatFm =∑

i+j=mGiHj . Thus the existence ofG andH with the desired properties has been established.

To prove the uniqueness, suppose there exist two other polynomialsgm andhm

that satisfy the hypothesis. Then

G0Hm +GmH0 = G0hm + gmH0

⇒G0(Hm − hm) = (Gm − gm)H0

⇒G0 | (Gm − gm), sinceG0 andH0 are coprime,

⇒Gm − gm = 0, asdeg(Gm − gm) < degG0,

⇒Gm = gm, Hm = hm.


The lemma has thus been proved. 2

An equivalent way of stating Hensel’s Lemma is the following.LetF (X,Y ) ∈ K[[X]][Y ] be monic of degreen in Y and such thatF (0, Y ) has

at least two roots. ThenF (X,Y ) is reducible inK[[X]][Y ]. Also, if

F (0, Y ) = G0(Y )H0(Y )

with degG0 = r, degH0 = (n− r) andgcd(G0,H0) = 1, then

F (X,Y ) = G(X,Y )H(X,Y )

withG(X,Y ),H(X,Y ) ∈ K[[X]][Y ] of degreesr, n− r in Y , where

G(0, Y ) = G0(Y ), H(0, Y ) = H0(Y ).

THEOREM 4.10 (Weierstrass Preparation Theorem)Let

F (X,Y ) =∑

aijXiY j ∈ K[[X,Y ]]

with aij ∈ K andF (0, Y ) 6= 0. If d = ordF (0, Y ). Then there exist uniqueG(X,Y ), H(X,Y ) ∈ K[[X,Y ]] such that

F (X,Y ) = G(X,Y )H(X,Y ),

whereG(0, 0) 6= 0 and

H(X,Y ) = Y d + c1(X)Y d−1 + · · ·+ cd(X)

with ci(X) ∈ K[[X]] andci(0) = 0 for 1 ≤ i ≤ d.

Note that

(1) G(X,Y ) is a unit;

(2) F (X,Y ) has no independent term;

(3) H(X,Y ) is calledspecial.

Proof. The proof is analogous to that of Hensel’s lemma. WriteF (X,Y ) as fol-lows:

F (X,Y ) = F0(Y ) + F1(Y )X + F2(Y )X2 + · · ·+ Fm(Y )Xm + · · ·with Fm(Y ) ∈ K[[Y ]]. Then it is necessary to find power series

G(X,Y ) =G0(Y ) +G1(Y )X + · · ·+Gi(Y )Xi + · · · ,H(X,Y ) =H0(Y ) +H1(Y )X + · · ·+Hj(Y )Xj + · · · ,

where

G0(Y ) ∈ K[[Y ]] with G0(0) 6= 0 andGi(Y ) ∈ K[[Y ]] for i > 0,

H0(Y ) = Y d andHj(Y ) ∈ K[Y ] with deg Hj(Y ) < d for j > 0

such thatF (X,Y ) = G(X,Y )H(X,Y ).

76

Now F (X,Y ) = G(X,Y )H(X,Y ) means that

Fm(Y ) =∑

i+j=m Gi(Y )Hj(Y ) for eachm ≥ 0. (4.11)

Proceed by induction onm. Since ordF (0, Y ) = d, with H0(Y ) = Y d, thereexists a uniqueG0(Y ) ∈ K[[Y ]] with G0(0) 6= 0 such thatF0(Y ) = F (0, Y ) =G0(Y )H0(Y ). So the case thatm = 0 has been established.

Let m > 0, and suppose that there existGi(Y ) ∈ K[[Y ]] for 1 ≤ i < m andHj(Y ) ∈ K[Y ] with deg Hj(Y ) < d for 1 ≤ j < m that satisfy (4.11). So, thetask is to findGm(Y ) ∈ K[[Y ]] andHm(Y ) ∈ K[Y ] with deg Hm(Y ) < d thatsatisfy the equation

Fm(Y ) =∑

i+j=m Gi(Y )Hj(Y ).

This equation may be written as

G0(Y )Hm(Y ) +H0(Y )Gm(Y ) = Um(Y ) = Fm(Y )−∑

i+j=m

i,j<m

Gi(Y )Hj(Y ).

SinceG0(Y ) is a unit, defineHm(Y ) to be the sum of the terms of degree less thand in Um(Y )/G0(Y ) with degHm(Y ) < d, from which it follows that

Gm(Y ) = Um(Y )/G0(Y )−Hm(Y )Y −dG0(Y ) ∈ K[[Y ]].

ThusGi(Y ), for eachi > 0, andHj(Y ), with deg Hj(Y ) < d, that satisfy (4.11)have been calculated.

To prove the uniqueness, again proceed by induction. SinceG0(Y ) is unique,suppose thatGi(Y ) andHj(Y ) are unique fori < m andj < m. Then, by theconstruction ofGm(Y ) andHm(Y ), their uniqueness follows. 2

A generalisation of Theorem 4.10 to more variables may be stated as follows.

THEOREM 4.11 If F (X1, . . . ,Xn) ∈ K[[X1, . . . ,Xn]] and

F (0, . . . , 0,Xn) = cνXνn + cν+1X

ν+1n + · · · , cν 6= 0,

then there exists a unique unitG in K[[X1, . . . ,Xn]] such that

GF = Xνn +A1(X1, . . . ,Xn−1)X

ν−1n + · · ·+Aν(X1, . . . ,Xn−1) (4.12)

with Ai(X1, . . . ,Xn−1) ∈ K[[X1, . . . ,Xn−1]] andAi(0, . . . , 0) = 0 for eachi = 1, . . . , ν.

An element ofK[[X1, . . . ,Xn]] that contains a term in whichXn appears, iscalledregular in Xn; henceF is regular. A polynomial inXn with form (4.12) iscalledspecial.

THEOREM 4.12 (Weierstrass Division Theorem)GivenF (X1, . . . ,Xn, Y ) andH(X1, . . . ,Xn, Y ) in K[[X1, . . . ,Xn, Y ]], assume thatF (0, . . . , 0, Y ) 6= 0 andd = ordY F (0, . . . , 0, Y ). Then there exist uniqueA,B such that

H = AF +B

withA ∈ K[[X1, . . . ,Xn, Y ]], B ∈ K[[X1, . . . ,Xn]][Y ], anddegY B < d.


Proof. Proceed by induction onn. The casen = 1 coincides with Theorem 4.10.Forn > 1, write

H =H0(X2, . . . ,Xn, Y ) +H1(X2, . . . ,Xn, Y )X1 + · · · ,A=A0(X2, . . . ,Xn, Y ) +A1(X2, . . . ,Xn, Y )X1 + · · · ,F =F0(X2, . . . ,Xn, Y ) + F1(X2, . . . ,Xn, Y )X1 + · · · ,B=B0(X2, . . . ,Xn, Y ) +B1(X2, . . . ,Xn, Y )X1 + · · · .

Then the required equationH = AF +B is equivalent to the following:

H0 =A0F0 +B0,

H1 =A0F1 +A1F0 +B1,

...

Hi =A0Fi + · · ·+Ai−1F1 +AiF0 +Bi,

... .

SinceF (0, · · · , 0, Y ) 6= 0, soF0(0, · · · , 0, Y ) 6= 0. Also,

ordY F (0, . . . , 0, Y ) = d

implies that ordY F0(0, . . . , 0, Y ) = d. Since

B(X1, · · · ,Xn, Y ) ∈ K[[X1, . . . ,Xn]][Y ]

anddegY B < d, then also

Bi(X1, · · · ,Xn, Y ) ∈ K[[X1, . . . ,Xn]][Y ]

anddegY Bi < d for all i ≥ 0.By induction,Ai andBi are found successively; so they exist and are unique by

construction. Therefore the same is true forA andB. 2

The proof of Theorem 4.11 follows from Theorem 4.12 by takingH = Y d.The following corollaries are a consequence of the general Weierstrass Prepara-

tion Theorem.

COROLLARY 4.13 GivenF ∈ K[[X1, . . . ,Xn]] with

F (0, . . . , 0,Xn) = cνXνn + cν+1X

ν+1n + · · · , cν 6= 0,

andA′F ∈ K[[X1, . . . ,Xn−1]][Xn] of degree < ν, withA′ ∈ K[[X1, . . . ,Xn]],thenA′ = 0.

Proof. The proof follows from Theorem 4.12. PuttingH = 0, it follows thatA′F = B′ andF = 0 by the uniqueness ofA andB. HenceA′ = 0. 2

COROLLARY 4.14 Given a special polynomialF in K[[X1, . . . ,Xn−1]][Xn] ofdegreeν andA′ ∈ K[[X1, . . . ,Xn]]. If A′F = B′ ∈ K[[X1, . . . ,Xn−1]][Xn] hasdegreeν, thenA′ ∈ K[[X1, . . . ,Xn−1]].

78

Proof. LetB′ = CXνn + · · · , with C ∈ K[[X1, . . . ,Xn−1]]. Then

CXνn −A′F = CXν

n −B′

has degree less thanν. PutH = CXνn. Then the relationH −AF = B is satisfied

not only byA = A′ andB = CXνn − B′ but also byA = C andB = H − CF.

From Theorem 4.12, sinceA andB are unique, it follows thatA′ = C. 2

THEOREM 4.15 If F is special inXn and irreducible inK[[X1, . . . ,Xn]], thenF is irreducible inK[[X1, . . . ,Xn−1]][Xn].

Proof. Suppose thatF is reducible inK[[X1, . . . ,Xn−1]][Xn] with F = G1G2,wheredegXn

Gi < degXnF. SinceF is irreducible inK[[X1, . . . ,Xn]], then one

of theGi is a unit in this ring, If it isG1, thenG−11 F = G2 and, by Corollary 4.13,

G2 = 0, which is impossible. 2

4.2 BRANCH REPRESENTATIONS

DEFINITION 4.16 A branch representationis a point of the planePG(2,K((t)))not belonging toPG(2,K).

Given a point(x0(t), x1(t), x2(t)) of PG(2,K((t))), it is therefore a branchrepresentation ifρ(t)xi(t) 6∈ K for at least onei and for allρ(t) ∈ K((t))\0.If x0(t) 6= 0, putx(t) = x1(t)/x0(t)

−1, y(t) = x2(t)/x0(t)

−1; then the branchrepresentation becomes the point(x(t), y(t)) of the affine plane

AG(2,K((t))) = PG(2,K((t)))\ℓ∞,whereℓ∞ = v(X0).

Let (x0(t), x1(t), x2(t)) be a branch representation and let

m = minordt xi(t).Then, withξi(t) = t−mxi(t), the branch representation(ξ0(t), ξ1(t), ξ2(t)) has theorder of one of its components zero and the other two non-negative; it is aspecialbranch representation and its coordinates are alsospecial. In this case, let

xk(t) = ck + ck1t+ . . .+ ckiti + . . . , k = 0, 1, 2;

the pointP = (c0, c1, c2) of PG(2,K) is thecentreof the branch representation.Now, the further notion of theorder of a branch representationis introduced.

With (x0(t), x1(t), x2(t)) already in special coordinates, theorder is the positiveinteger

minordt(a0x0(t) + a1x1(t) + a2x2(t))for all points(a0, a1, a2) ∈ PG(2,K) with a0c0+a1c1+a2c2 = 0. If x0(t), x1(t),x2(t) are linearly independent overK, then ordt(a0x0(t) + a1x1(t) + a2x2(t)) iszero or positive according asa0c0 + a1c1 + a2c2 is non-zero or zero. Further, it


takes one of two positive values,j1 andj2, wherej1 < j2; the value isj1 or j2according asa0c01 + a1c11 + a2c21 is non-zero or zero. In fact,

ordt(a0x0(t) + a1x1(t) + a2x2(t)) = j2

for a unique point(a0, a1, a2), in which case the linev(a0X0 + a1X1 + a2X2)of PG(2,K) is the tangentof the branch representation. The triple(0, j1, j2)is the order sequenceof the branch representation andj1 is its order. Whenx0(t), x1(t), x2(t) are linearly dependent overK, the same holds true providedthatj2 is defined to be∞.

The notion of branch representation is projective in the sense that every projec-tive transformation ofPG(2,K) transforms a branch representation into a branchrepresentation; also, special coordinates become specialcoordinates. Similarly, thenotions of order and tangent are projective invariants.

If one of the special coordinates is1, thenξ0(t) is invertible. Hence, every branchrepresentation can be expressed inspecial affine coordinates(x(t), y(t)) with

x(t) = u+ u1t+ . . . , y(t) = v + v1t+ . . . ,

with centre the point(u, v). This notion is preserved under a linear transformationof the affine planeAG(2,K). In what follows, the branch representations mostlyconsidered are those in special affine coordinates, as thesesimplify the exposition.

DEFINITION 4.17 (1) Two branch representations in special affine coordinates,

(x(t), y(t)) and (ξ(t), η(t)),

areequivalentif there exists aK-automorphismσ of K[[t]] such that

x(t) = σ(ξ(t)) and y(t) = σ(η(t)).

(2) Two branch representations areequivalentif they are so when put in the formof special affine coordinates.

In other words,(x(t), y(t)) and(ξ(t), η(t)) are equivalent branch representations ifthere exists a substitutiont 7→ τ of order1 such thatx(t) = ξ(τ) andy(t) = η(τ).

DEFINITION 4.18 (1) A branch representation in special affine coordinates

(x(t), y(t))

is imprimitive if there exists a branch representation in special affine coordinates

(ξ(t), η(t))

such that

x(t) = σ(ξ(t)) and y(t) = σ(η(t))

with σ aK-monomorphism ofK[[t]] that is not aK-automorphism.(2) A branch representation isimprimitive if it is so when put in the form of

special affine coordinates.

80

This definition can also be stated in terms of substitutions:(x(t), y(t)) is imprim-itive if there exists a branch representation in special affine coordinates(ξ(t), η(t))such thatx(t) = ξ(τ) andy(t) = η(τ) for a suitableτ ∈ K[[t]] with ordt τ > 1. Ifthis is the case, then(x(t), y(t)) hasramification index(at least)ν = ordt τ > 1.

The reason for this definition is thatξ(t), η(t) have, in general, simpler expres-sions thanx(t), y(t). Now, if (ξ(t), η(t)) is also imprimitive, then a further branchrepresentation(ξ1(t), η1(t)) with ramification indexν1 > 1 is obtained, and soon until a primitive branch representation is obtained. As this happens in a finitenumber of steps, the following result is reached.

THEOREM 4.19 Let (x(t), y(t)) and(ξ(t), η(t)) be two branch representations inspecial affine coordinates. If

(a) x(t) = σ(ξ(t)), y(t) = σ(η(t)),

(b) σ is aK-monomorphism ofK[[t]],

(c) ν = ordt σ(t),

then

ordt(x(t), y(t)) = ν · ordt(ξ(t), η(t)).

Proof. If (u, v) is the centre of(x(t), y(t)), then, sinceσ is aK-monomorphismof K[[t]], the same point(u, v) is also the centre of(ξ(t), η(t)). Therefore, givena non-trivial triple(a0, a1, a2), with ai ∈ K satisfyinga0 + a1u + a2v = 0,then(x(t), y(t)) has order ordt(a0 + a1x(t) + a2y(t)) and(ξ(t), η(t)) has orderordt(a0 + a1ξ(t) + a2η(t))). Now,

ordt(a0 + a1x(t) + a2y(t)) = ordt(a0 + a1σ(ξ(t)) + a2σ(η(t)))

= ordt σ(a0 + a1ξ(t) + a2η(t)).

By Theorem 4.5, this is the required result. 2

Since every product ofK-monomorphisms ofK[[t]] is also such aK-monomor-phism, for each imprimitive branch representation in special affine coordinates(x(t), y(t)) there exists aprimitive branch representation in special affine coor-dinates(ξ(t), η(t)) such that

(x(t)) = σ(ξ(t)), (y(t)) = σ(η(t)),

whereσ is aK-monomorphism ofK[[t]] that is not aK-automorphism. To showthe uniqueness, three further results are required that areinteresting in themselves.

THEOREM 4.20 Let (x(t), y(t)) be a branch representation in special affine co-ordinates.

(i) If τ is of positive minimum order in the subfieldK(x(t), y(t)) of K((t)),then the order of each elementz ∈ K(x(t), y(t)) is divisible byg = ordt τ .

(ii) The elementz can be expanded in a formal power series inτ with coefficientsin K.


(iii) In the particular case that(x(t), y(t)) is primitive, there is an element oforder1 in K(x(t), y(t)).

Proof. (i) If µ is the order ofz, the Euclidean algorithm gives integersq, r suchthatµ = qg + r and0 ≤ r < g. Hencez/τ q is an element of orderr belonging toK(x(t), y(t)). Sinceτ has minimum order, thenr = 0, as required.

(ii) Now, write z/τ q = c+z1 with c ∈ K, c 6= 0 andz1 ∈ K[[t]]; thenz1 is alsoin K(x(t), y(t)). As in the previous argument, there is a positive integerq1 suchthatz1/τ q1 = c1 + z2. Hence

z = cτ q + c1τq+q1 + τ q+q1z2.

Continuing this procedure gives the required formal power series.(iii) In particular, the same coordinatesx(t) andy(t) become formal power series

in τ , so that, if(x(t), y(t)) is primitive,τ is necessarily of order1. 2

THEOREM 4.21 Given a branch representation(x(t), y(t)) in special affine co-ordinates, let (x1(t), y1(t)) and (x2(t), y2(t)) also be branch representations inspecial affine coordinates, of which the first is primitive. If the relations,

x(t) = x1(τ1) = x2(τ2), y(t) = y1(τ1) = y2(τ2), (4.13)

with τ1, τ2 ∈ K[[t]], are satisfied, thenτ1 can be written as a formal power seriesin τ2. Further, if (x2(t), y2(t)) is also primitive, thenordt τ1 = ordt τ2.

Proof. Let τ be an element of minimum positive order inK(x1(τ1), y1(τ1)). ByTheorem 4.20 and the primitivity of(x1(t), y1(t)), this order must be1. Therefore,τ = cτ1 + . . . with c 6= 0, whenceτ = σ(τ1) for someσ that is aK-automorphismof K[[t]]. Sinceσ−1 is also such an automorphism, so alsoτ1 = dτ + . . . withd 6= 0.

Sinceτ is a rational function of the elementsx1(τ1) andy1(τ1), there exist poly-nomialsa(X,Y ), b(X,Y ) ∈ K[X,Y ] such that

τ =a(x1(τ1), y1(τ1))

b(x1(τ1), y1(τ1)).

From (4.13), it follows that

τ =a(x2(τ2), y2(τ2))

b(x2(τ2), y2(τ2)).

Therefore,τ = e τ2i + . . . with i ≥ 0. Henceτ1 = de τ2

i + . . .; that is,τ1 is aformal power series inτ2 of positive order. When(x2(t), y2(t)) is primitive, thentheir roles may be interchanged, giving thatτ2 is a formal power series inτ1. Butthis is possible if and only ifτ1 is an element of order1 in K[[τ2]]. 2

From this theorem and Theorem 4.19, the following result is obtained.

THEOREM 4.22 Let (x(t), y(t)) be an imprimitive branch representation in spe-cial affine coordinates. If(x1(t), y1(t)) and (x2(t), y2(t)) are branch representa-tions in special affine coordinates satisfying the relations,

x(t) = x1(τ1) = x2(τ2), y(t) = y1(τ1) = y2(τ2),

with τ1, τ2 ∈ K[[t]], then(x1(t), y1(t)) and(x2(t), y2(t)) have the same order.

82

Now the idea of ramification index can be introduced.

DEFINITION 4.23 (i) If (x(t), y(t)) is an imprimitive branch representation inspecial affine coordinates and if(ξ(t), η(t)) is a primitive branch representa-tion in special affine coordinates withx(t) = σ(ξ(t)), y(t) = σ(η(t)), thenν = ordt σ(t) is theramification indexof (x(t), y(t)).

(ii) The ramification index of a branch representationis its ramification index inits special affine coordinate form.

Theorem 4.22 thus has the following corollary.

COROLLARY 4.24 Equivalent imprimitive branch representations have the sameramification index. Further, the primitive representations to which they give riseare also equivalent.

A sufficient condition for imprimitivity is reducibility, where a branch represen-tation in special affine coordinates(x(t), y(t)), and any equivalent one, isreducibleif, in Definition 4.18,σ(t) = tm; that is,x(t), y(t) ∈ K[[tm]] for an integerm > 1.In fact, the adjective “reducible” signifies “imprimitivity for a substitution of typet 7→ τ with τm = t”. A useful criterion for reducibility is given by the followingtheorem.

THEOREM 4.25 Let charK = 0 or p with p 6 |n. A branch representation inspecial affine coordinates, wherea1 6= 0 and0 < n1 < n2 < . . . ,

x(t) = u+ tn,y(t) = v + a1t

n1 + a2tn2 + . . . ,

(4.14)

is reducible if and only ifgcd(n, n1, n2, . . .) > 1.

Proof. That the condition is sufficient is immediate.To show the necessity, suppose that the representation is reducible. Then there

exists a substitutiont 7→ τ of order1 such thatx(τ), y(τ) ∈ K[[tm]] with m > 1.Next it is shown thatτ/t ∈ K[[tm]]. Suppose, by way of contradiction, that

τ = t(b0 + b1tm + . . .+ bht

hm + cts + . . .)

with c 6= 0 but withm not dividings. It follows that

x(τ) =u+ tn(b0 + b1tm + . . .+ bht

hm + cts + . . .)n

=u+ tn(b0 + . . .+ bhyhm)n−1 + . . . .

Sincex(τ) ∈ K[[tm]], it is immediate thatm | n. Further,m | (n+ s) sincen 6≡ 0(mod p). Hencem | s, a contradiction.

Therefore, writeτ = zt with z = u0 + u1tm + . . . ∈ K[[tm]] andu0 6= 0. Now,

suppose that at least one of the exponentsn1, n2, . . . is not divisible bym, and letnk+1 be the smallest such exponent. Then

y(t)− (v + a1tn1zn1 + . . .+ akt

nkznk)

= ank+1tnk+1(u0 + u1t

m + . . .)nk+1 + . . .

= ank+1unk+10 tnk+1 + . . . .


This shows that the first expression, but not the second, belongs toK[[tm]]. This isa contradiction, which shows thatm must divide not onlyn but also each exponentn1, n2, . . .; this finishes the proof. 2

THEOREM 4.26 LetK be of characteristic0 or of characteristicp > 0 with p6 |n.Every branch representation of ordern has a special affine coordinate form of thefollowing type:

x(t) = u+ tn, y = v + η(t), (4.15)

with ordt η(t) ≥ n.

Proof. Let

x(t) =u+ untn + ukt

k + . . . ,

y(t) = v + y1(t),

with un 6= 0, ordt y1(t) ≥ n, be a branch representation. First, the equation

τn = untn + ukt

k−n + . . . (4.16)

is solved forτ ∈ K[[t]]. The casen = 1 already provides a solution; so, supposethatn > 1. Dividing both sides of (4.16) bytn gives the equation

zn = un + uktk−n + . . . (4.17)

in the unknownz ∈ K[[t]]. Sincep 6 |n, the polynomialZn − un certainly doesnot have multiple roots, and therefore has at least two distinct roots. By Hensel’sLemma 4.9,Zn − (un + ukt

k−n + . . .) is reducible inK[[t]][Z]; that is,

Zn − (un + uktk−n + . . .) = (Zr + ρ(Z))(Zs + σ(Z)),

whereρ(Z), σ(Z) ∈ K[[t]][Z]. Write

Zr + ρ(Z) = Zr + ρ1(t)Zr−1 + . . .+ ρr(t), ρi ∈ K[[t]].

The polynomialg(Z) = Zr + ρ1(0)Zr−1 + . . . + ρr(0), with coefficients inK,dividesZn − un. Henceg(Z) has no multiple roots and so at least two distinctroots. Applying Hensel’s Lemma iteratively shows that the polynomial

Zn − (un + uktk−n + . . .)

splits into linear factors inK[[t]][Z]. LetZ− ζ(t) be one of these factors; note thatordt ζ(t) = 0. Then

ζ(t)n − (un + uktk−n + . . .) = 0,

whence(ζ(t)t)n = untn + ukt

k . . .. Puttingτ = ζ(t)t givesx = u + τn. Sinceordt ζ(t)t = 1, the relationτ = ζ(t)t gives t; that is, there exists an elementf(τ) ∈ K[[τ ]] such thatt = f(τ). Finally, puttingη(τ) = y1(f(τ)) gives

x = u+ τn, y = v + η(τ),

with ordτ η(τ) ≥ n, as required. 2

THEOREM 4.27 LetK be of characteristic0 or of characteristicp > 0 with p6 |n.A branch representation of ordern is imprimitive if and only if it is reducible.

84

Proof. It suffices to show that the imprimitivity implies the reducibility. By The-orem 4.26, the branch representation in special affine coordinates(x(t), y(t)) maybe taken in the form (4.15). If(x(t), y(t)) imprimitive, there exists a represen-tation (ξ(τ), η(τ)) with ξ(τ), η(τ) ∈ K[[τ ]] such that, for a suitable elementτ = cit

i + . . . , i > 1,

x(t) = ξ(τ), y(t) = η(τ).

By Theorem 4.19, the order of the primitive representation(ξ(τ), η(τ)) is n/i,which is coprime top; writem = n/i. Therefore, still by Theorem 4.26, it may besupposed that(ξ, η) is of type (4.15), that is, that

ξ(τ) = u+ τm, η(τ) = v + b1τm1 + b2τ

m2 + . . . .

Hence

tn = τm, (4.18)

a1tn1 + a2t

n2 + . . .= b1τm1 + b2τ

m2 + . . . . (4.19)

From (4.18) it follows that(ti/τ)m = 1, whenceτ = cti sincec is anm-th rootof unity; in particularc ∈ K\0. Now, (4.19) implies thatnj = imj for everyj.This shows thatgcd(n, n1, n2, . . .) > 1; hence(x(t), y(t)) is reducible. 2

Finally, in this section, the definition of abranchis given; it is one of the centralconcepts in the following sections.

DEFINITION 4.28 A branchis an equivalence class of primitive branch represen-tations. Thecentre, theorder, and theorder sequenceof a branch are the centre,order, and the order sequence of any branch representation.

4.3 BRANCHES OF PLANE ALGEBRAIC CURVES

The next definition is the foundation for developing the study of algebraic curvesbased on the idea of branch; in fact, branches constitute an important tool for thelocal study of algebraic curves.

DEFINITION 4.29 A branch of an algebraic curveis a branch whose representa-tions(x0(t), x1(t), x2(t)) satisfy the equationF (X0,X1,X2) = 0 of the curve.

THEOREM 4.30 The centre of a branch of a curve is a point of the curve.

Proof. LetF = v(F ) be a curve and let(ξ0(t), ξ1(t), ξ2(t)) be any special branchrepresentation; then

F (ξ0(t), ξ1(t), ξ2(t)) = 0.

Puttingt = 0 givesF (ξ0(0), ξ1(0), ξ2(0)) = 0. Therefore, the point

P (ξ0(0), ξ1(0), ξ2(0)),

which is the centre of the branch representation, is a point of the curveF . 2


THEOREM 4.31 There exists a unique branch of a curveF centred at a simplepoint ofF .Proof. Without loss of generality, let the point be the origin and let the tangentthere be any line other thanv(X). If F = v(F (X,Y )), then, by Theorem 4.25,the representation

x = t, y = c1t+ c2t2 + · · ·

is primitive and represents a branch ofF centred at the origin.To show the uniqueness, let(ξ(t), η(t)) be a primitive branch representation of

F centred at the origin; then ordξ > 0, ord η > 0 andF (ξ, η) = 0. If the originOis a simple point ofF and the tangentO toF is notv(X), by Theorem 4.6, thereexists a unique formal power series

c1X + c2X2 + · · ·

such that

F (X, c1X + c2X2 + · · · ) = 0.

ThenF (X,Y ) factorises inK[[X]][Y ]:

F (X,Y ) = (Y − (c1X + c2X2 + · · · ))H(X,Y ),

with ordH(X,Y ) = 0. In K[[t]],

F (ξ, η) = (η − (c1ξ + c2ξ2 + · · · ))H(ξ, η) = 0.

Henceη = c1ξ + c2ξ2 + · · · .

Since(ξ, η) is a primitive representation,ξ must be of order1. The substitutionσ : t → ξ is of order 1, and so, by Theorem 4.4, is an automorphism ofK[[t]];its inverseσ−1 is also an automorphism. Also,σ−1 sendsξ to t andc1ξ + . . . to(t, c1t + c2t

2 + · · · ). It follows that(ξ, c1ξ + . . .) is equivalent to the branch rep-resentation(x(t), y(t)) introduced at the beginning of the proof; so the uniquenessis established. 2

DEFINITION 4.32 Let G = v(G) be a curve andγ a branch centred at the pointP . If (ξ0(t), ξ1(t), ξ2(t)) is a representation ofγ in special coordinates, then theintersection multiplicityis

I(P,G ∩ γ) =

ordt G(ξ0(t), ξ1(t), ξ2(t)) if γ /∈ G,∞ if γ ∈ G.

Note thatI(P,G ∩ γ) depends only onγ, and not on the representation chosen.

THEOREM 4.33 The intersection numberI(P,G ∩ γ) is covariant, that is, inde-pendent of the choice of plane coordinate system.

Proof. For i = 0, 1, 2, letXi =∑2

j=0 ai,jX′j be the equation of the change of the

coordinate system(X0,X1,X2) into (X ′0,X

′1,X

′2). If

G′(X0,X1,X2) = G(∑2

j=0 a0jX′j ,∑2

j=0 a1jX′j ,∑2

j=0 a2jX′j)

86

andxi(t) =∑2

j=0 aijx′j(t) for i = 0, 1, 2, then

ordt G(x0(t), x1(t), x2(t)) = ordt G′(x′0(t), x

′1(t), x

′2(t)).

If G(x0(t), x1(t), x2(t)) = 0, thenI(P,G ∩ γ) =∞. 2

The following result shows how Definition 4.32 appears naturally in the study ofthe intersection multiplicity at a common point of two curves.

THEOREM 4.34 (i) If the branchγ of the curveF is centred at a simple pointP ofF , andG is any other algebraic curve, then

I(P,G ∩ γ) = I(P,G ∩ F).

(ii) More generally, if P is a singular point of the irreducible curveF andG isa curve not containingF as a component, then

I(P,G ∩ F) =∑

γ

I(P,G ∩ γ),

whereγ runs over all branches ofF centred atP.

(iii) If P is anmP -fold point ofF thenmP is a bound on the number of branchesofF with centreP .

Before dealing with Theorem 4.34 it needs to be shown that every point ofF isthe centre of at most a finite number of branches. This ensuresthat the definition∑

γ I(P,G ∩γ) is meaningful. The proof requires some more results on the behav-iour of branches under local quadratic transformations. These results are stated inthe forthcoming section after proving the following usefultheorem.

THEOREM 4.35 Two distinct irreducible algebraic curves have no branchesincommon.

Proof. Suppose, by way of contradiction, that the two curvesCn andCm have abranchγ in common. Let(ξ(t), η(t)) be a branch representation; it is a point of theplane overK((t)) and so is a common point of the two curves regarded as curvesof the plane overK((t)). Applying Bezout’s Theorem over the fieldK, the twocurves being irreducible, gives that

∑I(P, Cn ∩ Cm) = nm. Similarly, Bezout’s

Theorem overK((t)) gives the same result. Hence, the two curves have all theirintersections in the smaller field, contradicting that, over K((t)), they have thepoint (ξ(t), η(t)) in common. 2

COROLLARY 4.36 Every branch(ξ, η) with ξ = 0 is a branch of the linev(X),but of no other irreducible curve.

4.4 QUADRATIC TRANSFORMATIONS

The study of branches centred at a singular point of an algebraic curve is helped bylocal quadratic transformations, that is, transformations of the form,

σ(X,Y ) = (X ′, Y ′),X ′ = X,Y ′ = Y/X.

(4.20)


It is not defined for points on the linev(X), but the pointY∞ = (0, 1, 0) is takento be the image of these points, other than the origin. Let(ξ(t), η(t)) be a branchrepresentation with centre the origin andξ(t) 6= 0. Then the branch representation

(ξ(t), η(t)/ξ(t))

is the image of the branch(ξ(t), η(t)) underσ, and so a branch representationcorresponds to another branch representation. This gives the following result.

THEOREM 4.37 The local quadratic transformationσ given by (4.20) defines abijective correspondence between branches centred at the origin with tangent notv(X) and those centred at any affine point onv(X).

Taking a pointP = (a, b), with a 6= 0, on an algebraic curveF = v(F (X,Y )),then the image ofP underσ belongs to the algebraic curvev(F (X,Y )), whereF (X,Y ) = F (X,XY ). This gives the following idea.

DEFINITION 4.38 The algebraic transformof a curveF = v(F (X,Y )) is thecurveF = v(F (X,Y )), whereF (X,Y ) = F (X,XY ).

A useful property is the following relation: given a branchγ centred atP and acurveG, let γ′ andG be their images under a local quadratic transformation. IfPis the centre ofγ′, then

I(P,G ∩ γ) = I(P , G ∩ γ′). (4.21)

To show (4.21), setG = v(G(X,Y ), G = v(G(X,Y )), and let(x′(t), y′(t)) be aprimitive representation ofγ′. Then

ordt G(x(t), y(t)) = ordt G(x(t), y(t)/x(t)) = ordt G(x′(t), y′(t)),

showing (4.21).Every polynomialF (X,Y ) ∈ K[X,Y ] can be written in the form

F (X,Y ) = Fr(X,Y ) + Fr+1(X,Y ) + . . . , Fr 6= 0,

with Fi(X,Y ) homogeneous of degreei. The curveF = v(F (X,Y )) passesthrough the originO if and only if r = 0; whenr ≥ 1, thenO is a point ofF ofmultiplicity r. Since

F (X,Y ) =F (X,XY )

=Fr(X,XY ) + Fr+1(X,XY ) + . . .

=Xr(Fr(1, Y ) +XFr+1(1, Y ) + . . .),

so the algebraic transform ofF hasXr as a component. For this reason, it isappropriate to eliminateXr and introduce the notion of the geometric transform.

DEFINITION 4.39 Thegeometric transformof a curvev(F (X,Y )) with anr-plepoint, withr ≥ 0, at the origin is the curvev(F ′(X,Y )) given by

F ′(X,Y ) = F (X,Y )/Xr = F (X,XY )/Xr.

Now the first, important property of the geometric transformis established.

88

THEOREM 4.40 If F is irreducible and is not the linev(X), then the geometrictransformF ′ is also irreducible.

Proof. Let F (X,Y ) be irreducible and different fromcX, c ∈ K. It must beshown thatF ′(X,Y ) = F (X,XY )/Xr is irreducible.

Suppose thatF ′(X,Y ) = G(X,Y )H(X,Y ) with G(X,Y ),H(X,Y ) non-constant polynomials. First, consider the possibility that H(X,Y ) = h(X). ThenF (X,XY ) = Xrh(X)G(X,Y ). PuttingY = Z/X gives

F (X,Z) = Xrh(X)G(X,Z/X).

Let d be the minimum value for whichXdG(X,Z/X) ∈ K[X,Z]; then

Xr−dh(X) ∈ K[X],

whence

h(X) = h1Xd−r + h2X

d−r+1 + . . . .

Therefore,

F (X,Z) = (h1 + h2X + . . .)(XdG(X,Z)).

SinceF (X,Y ) containsY , thenF (X,Z) must containZ and consequently is nota constant. The irreducibility ofF (X,Z) therefore implies thath(X) = h1X

d−r.This means thatF (X,XY ) = XdG(X,Y ). Then, from the definition ofr, neces-sarilyd = r; that is,h(X) is a constant, as required.

Finally, the possibility thatH(X,Y ) orG(X,Y ) containsY must be excluded.If it were so, then

F (X,Z) = XrG(X,Z/X)H(X,Z/X) = [XmG(X,Z)][(Xr−mH(X,Z)],

where both factors in square brackets are polynomials. But they are not constantpolynomials since they containZ and that contradicts the irreducibility of bothF (X,Z) andF (X,Y ). 2

As a corollary, there is the following result.

THEOREM 4.41 LetF1 = v(F1(X,Y )), . . . ,Fs = v(Fs(X,Y )) be irreduciblecurves, none of themv(X). Then the geometric transform of

F = v(Fm11 · · ·Fms

s )

is

F ′ = v(F ′1m1 · · ·F ′

sms).

Now, Theorem 4.37 may be made more precise.

THEOREM 4.42 If F is a curve such that the linev(X) is not tangent toF atOandF ′ is the geometric transform ofF by the local quadratic transformationσ asin (4.20), then there exists a bijection between the branches ofF centred atO andthe branches ofF ′ centred at an affine point ofv(X).


Proof. Considering Corollary 4.36, it follows from Theorem 4.35 that the branchesof F and those ofF ′ admit a representation(ξ(t), η(t)) with ξ(t) 6= 0. Further, if(ξ(t), η(t)) satisfiesF (X,Y ) = 0, then(ξ(t), η(t)/ξ(t)) satisfiesF (X,XY ) = 0and alsoF (X,XY )/Xr = 0; and conversely. This gives a bijective correspon-dence between the branches ofF andF ′.

To show that the branches ofF ′ obtained in this way are centred at affine pointsof v(X), it suffices to see that ordξ(t) ≤ ord η(t); that is,

ξ(t) = aiti + . . . , η(t) = bjt

j + . . .

with i ≤ j. Suppose instead thatj < i. The hypothesis thatv(X) is not tangent toF atO implies thatY r is present inF (X,Y ), whence, withcr 6= 0,

Fr(X,Y ) = c0Xr + c1X

r−1Y + . . .+ crYr.

So ordFr(ξ(t), η(t)) = jr, but ordFr+k(ξ(t), η(t)) ≥ (r+ 1)j for k > 0. There-fore, ordF (ξ(t), η(t)) = rj, which contradicts thatF (ξ(t), η(t)) = 0. This fin-ishes the proof. 2

THEOREM 4.43 If O is an ordinary singular point on the curveF = v(F (X,Y ))of multiplicity r, then

(i) it is the centre of exactly r branches ofF ;

(ii) the branches are all linear, and their tangents are precisely the tangents toF atO.

Proof. With O at the origin,F (X,Y ) =∏r

i=1(Y −miX) + · · · . Then, the geo-metric transformF ′ = v(F ′(X,Y )) of F under the local quadratic transformation(4.20) is given by

F ′(X,Y ) =r∏

i=1

(Y −mi) +XH(X).

The linev(X) meetsF ′ in the pointsQi = (0,mi) for i = 1, . . . , r.Now, it is shown that these points are all simple onF ′. ChooseQ1, say, and take

it to (0, 0) via the translation(X,Y ) 7→ (X,Y −m1). The transformed curve isG = v(G(X,Y )) with

G(X,Y ) =F ′(X,Y +m1)

=∏r

j=1((Y +m1)−mj) +XH1(X)

=Y (Y − (m2 −m1)) · · · (Y − (mr −m1)) +XH2(X)

= (−1)r−1Y∏r

i=2(mi −m1) + · · · .Since all themi −m1 are non-zero, it follows thatG(X,Y ) contains a single termof degree1 given by(−1)r−1Y

∏ri=2(mi −m1); this shows thatQ1 is simple for

v(G), and so also forF ′. From Theorem 4.31 it now follows that there are preciselyr branches ofF ′ with centre at an affine point onv(X). By the previous theorem,F has exactlyr branches centred at the origin. By Theorem 4.31 the branch ofF ′

centred atQi has a representation(t,mi + . . .). The corresponding branch ofFtherefore has a representation(t,mit+ . . .). Since the tangent of such a branch isthe linev(Y −miX), the theorem is proved. 2

90

THEOREM 4.44 Each pointP ofF is the centre of at least one and at most a finitenumber of branches ofF .Proof. Since each branch ofF = v(F ) must annul at least one irreducible factorFi, the polynomialF may be supposed irreducible. In fact, let

F = F1 ∪ F2 ∪ ... ∪ Fl

with eachFi = v(Fi) an irreducible component ofF ; then

F (X,Y ) = F1(X,Y ) · · ·Fl(X,Y ).

Let γ = (ξ(t), η(t)) be a branch ofF ; soF (ξ(t), η(t)) = 0, or

F1(ξ(t), η(t)) · · ·Fl(ξ(t), η(t)) = 0.

HenceFi(ξ(t), η(t)) = 0 for at least onei and soγ is a branchFi for thati.It may also be supposed thatP , of multiplicity r, is at the origin and that the line

v(X) is not tangent toF atP. Then,F =∏r

i=1(Y −miX) + · · · with themi notnecessarily distinct. Applyingσ toF givesF ′ = v((F ′(X,Y ′)/Xr) with

F ′(X,Y ′)/Xr =∏r

i=1(Y′ −mi) +XH(X) = 0.

Therefore, the linev(X) meetsF ′ in the pointsQ′i(0,mi), and the sum of the

intersection multiplicities at these points isr, that is∑s

i=1 I(Q′i,v(X) ∩ F ′) = r. (4.22)

If at least two of themi are distinct, each point(0,mj) of F ′ has multiplicity lessthanr. By an induction argument, with the assumption of the existence of at leastone and at most a finite number of branches centred at the point(0,mj), thenF ′

has at least one and at most a finite number of branches centredat an affine point onv(X). By Theorem 4.42,F has at least one and at most a finite number of branchescentred atO.

This leaves the case that all themi are equal:

F (X,Y ) = (Y −m1X)r + · · · , r > 1.

In this case,F ′ = v(F ′(X,Y ′)/Xr) with

F ′(X,Y ′)/Xr = (Y ′ −m1)r +XH(X).

Now, the translationX = X, Y = Y ′ − m1, which takes the point(0,m1) to(0, 0), transformsF ′ to v(Y r +XG(X,Y )). Hence,F ′ may be assumed to havethe singular point of multiplicityr at the origin.

With a similar argument to the above, by induction onr, it may be supposed thatthe curveF ′ = v(F ′(X,Y )) with

F ′(X,Y ) = (Y −m2)r +H(X,Y ).

Then, applyingσ gives a curveF (2) = v(F (2)(X,Y )), which may be taken tohave a point of multiplicityr at (0,m2), whence

F (2)(X,Y ) = (Y −m2)r +XH(X,Y ).

Is it possible that an infinite sequence of points of multiplicity r is obtained?The answer is, in fact, in the negative, as is shown below. Therefore, this proceduregives rise to a point that is not of multiplicityr. Induction now finishes the proof.


To show that the answer is negative, introducem3,m4, . . . through the curvesF (2), F (3), . . . exactly asm1 andm2 were obtained in relation toF andF ′. Foreachi ≥ 1, write F (X,Y ) in the following form:

F (X,Y ) =∑

b>0 cabXa(Y −m1X−m2X

2−· · ·−miXi)b+XhiPi(X) (4.23)

wherecab ∈ K, hi ∈ K, Pi(X) ∈ K[X], Pi(0) 6= 0; this can be done uniquely.With Y = η + (m1X +m2X

2 + · · ·+miXi), write

F (X,Y ) =F (X, η + (m1X +m2X2 + · · ·+miX

i))

=G(X, η) =∑

cabXaηb

=∑

b>0 cabXaηb +XhiPi(X)

=∑

b>0 cabXa(Y −m1X −m2X

2 − · · · −miXi)b +XhiPi(X).

In the expansion of the sum, the termcabXaY b is not eliminated by other terms.

SinceF has a point of multiplicityr at the origin, it follows thata + b ≥ r. Also,hi ≥ r. Using the form (4.23) forF , the algebraic transformF = v(F (X,Y ′)) isas follows:

F (X,Y ′)

=F (X,XY ′)

=∑

cabXa(XY ′ −m1X −m2X

2 − · · · −miXi)b +XhiPi(X)

=∑

cabXa+b(Y ′ −m1 −m2X − · · · −miX

i−1)b +XhiPi(X) = 0.

Since, by assumption, the geometric transformF ′ of F has a point of multiplicityr at the origin, soF ′ = v(F ′) has

F ′(X,Y ′) = F (X,Y ′)/Xr

=∑

cabXa+b−r(Y ′ −m1 −m2X − · · · −miX

i−1)b

+Xhi−rPi(X) = 0.

Sincehi(F) = hi(F ′)+ r = (hi(F ′′)+ r)+ r, repeating the processi times giveshi(F) = hi(F i) + ir, from whichhi(F) ≥ ir. It follows thathi →∞ asi→∞.

SubstitutingY = m1X +m2X2 + · · · in (4.23) gives

F (X,m1X +m1X2 + · · · ) =

∑cabX

a(mi+1Xi+1 + · · · )b +XhiPi(X).

Hence

ordx F (X,m1X +m2X2 + · · · ) ≥ mini+ 1, hi

for eachi. Therefore ordx F (X,m1X +m2X2 + · · · ) =∞ or 0; in other words,

there is a branchγ = (ξ = t, η = m1t+m2t2+ · · · ) centred atO, which is the first

part of the theorem. The next proposition completes the proof of Theorem 4.44.2

PROPOSITION 4.45 The sequence of quadratic transformations appearing in theabove proof does not give rise, for r > 1, to an infinite sequence of points ofmultiplicity r,

Proof. Assume at first that the polynomial∂F/∂Y does not vanish. Then the polarcurveG of Y∞ = (0, 0, 1) with respect toF exists. SinceF is irreducible, the

92

branchγ = (ξ = t, η = m1t +m2t2 + · · · ) cannot be a branch ofG by Theorem

4.35. LetG′ be the polar curve ofY∞ with respect to the geometric transformF ′

of F under the local quadratic transformation. ThenG′ = v(∂F ′/∂Y ′), and(∂F (X,Y )

∂Y

/Xr−1

)

Y =XY ′

=

(∂F (X,XY ′)

∂Y ′∂Y ′

∂Y

)/Xr−1

=∂

∂Y ′

(1

XrF (X,XY ′)

),

whence

1

Xr−1

∂F

∂Y=∂F ′

∂Y ′ .

With O = (0, 0) the centre ofγ andO′ = (0,m1) the centre of

γ′ : (ξ′ = t, η′ = η/ξ = m1 +m2t+ · · · ),then

I(O,G ∩ γ) = ordt

(∂F

∂Y

)

X=t,Y =m1t+···

= ordt

(∂F ′

∂Y ′

)

X′=t,Y ′=m1+···+ ordt(X

r−1)X=t

= I(O′,G′ ∩ γ′) + ordt(tr−1)

= I(O′,G′ ∩ γ′) + r − 1.

Therefore, ifr > 1,

I(O,G ∩ γ) > I(O′,G′ ∩ γ′) > 0.

Continuing this argument gives an infinite sequence of decreasing positive integers,a contradiction which proves the assertion under the initial hypothesis.

If this hypothesis does not hold, then choose another non-tangent line toF tobe theY -axis. Then the polynomial∂F/∂Y does not vanish any more. Now, thepolar curve ofF for Y∞ exists. By (4.24), the polar curve ofF ′ for Y∞ also exists,and hence this proof still works. 2

Now Theorem 4.34 can be proved. To do this, define

J(P,G ∩ F) =∑

γ I(P,G ∩ γ),whereγ ranges over all branches ofF centred atP . Now, check thatJ(P,G ∩ F)satisfies the seven postulates in Chapter 3.1. By Theorem 4.33, J(P,G ∩ F) iscovariant; so we may assume thatP coincides with the origin and that theY -axisis is not tangent toF atP .

If F andG have no common component throughP , then the branches ofFcentred atP are distinct from those ofG centred atP . Hence,J(P,G ∩ F) is anon-negative integer for every branchγ of F . This shows the validity of (I 1).

If F andG have a common component, sayC, throughP , then any branchγ of Ccentred atP is a common branch ofF andG. This implies thatJ(P,G ∩ F) =∞showing (I 2).


If P is not a point ofF , no branch ofF is centred atP ; soJ(P,G ∩ F) = 0.If P is a point ofF andγ is a branch ofF centred atP given by the primitiverepresentation(x(t), y(t)), then ordt x(t) > 0, ordt y(t) > 0. Thus

G(x(t), y(t)) = G(0, 0) + tH(t).

If P is not a point ofG, thenG(0, 0) 6= 0. Hence

J(P,G ∩ F) = ordt G(x(t), y(t)) = 0,

and this shows (I 3). Also, (I 4) follows from Theorem 4.34 (i).To show (I 5), the following technical result is required.

THEOREM 4.46 Let the originP be anr-fold point ofF = v(F ) and ans-foldpoint ofG = v(G) such thatv(X) is not a tangent atP toF or G. LetF ′ andG′be the geometric transforms ofF andG under a local quadratic transformation. IfP ′

1, . . . , P′n are the common affine points ofF ′ andG′ onv(X), then

J(P,F ∩∆) = rs+∑n

i=1 J(P ′j ,F ′ ∩ G′).

Proof. Let γ1, . . . , γm be the branches ofG centred atP , and(xi(t), yi(t)) a prim-itive representation ofγi for i = 1, . . . ,m. Let the branches ofG′ centred at affinepoints ofv(X) areγ′1, . . . , γ

′m, and(x′i(t) = xi(t), y

′i(t) = yi(t)/xi(t)) is a prim-

itive representation ofγ′i, for i = 1, . . . ,m. Hence,

J(P,F ∩ G) =∑m

i=1 ordt F (xi(t), yi(t))

=∑m

i=1 ordt xi(t)rF ′(x′i(t), y

′i(t))

= r∑m

i=1 ordt xi(t) +∑m

i=1 ordt F′(x′i(t), y

′i(t))

= rs+∑n

i=1 J(P ′i ,F ′ ∩ G′).

2

Now, (I 5) is shown by induction onk = J(P,F ∩ G). SinceJ(P,F ∩ G) = 0if and only if P is not a common point ofF andG, the assertion is true fork = 0.Now, assume it is true for every integer less thank, and letJ(P,F ∩ G) = k. ByTheorem 4.46,

J(P,F ∩ G) = rs+∑n

i=1 J(P ′i ,F ′ ∩ G′).

As rs ≥ 1, soJ(P ′i ,F ′∩G) < k. Hence,J(P ′

i ,F ′∩G′) = J(P ′i ,G ∩F ′). Hence,

J(P,F ∩ G) = J(P,G ∩ F).The validity of (I 6) and (I 7) follows from the next lemma.

L EMMA 4.47 Let γ be a branch ofF = v(F ) centred at the pointP, and letG = v(G). If Φ = v(G+ FH) withH ∈ K[X,Y ], then

J(P,G ∩ γ) = J(P,Φ ∩ γ).If Λ is a reducible curve that splits into the componentsG andΨ, then

I(P,Λ ∩ γ) = I(P,G ∩ γ) + I(P,Ψ ∩ γ).

94

Proof. Let (x(t), y(t)) be a primitive representation ofγ. Then

J(P,Φ ∩ γ) = ordt(G(x(t), y(t)) + F (x(t), y(t))H(x(t), y(t)))

= ordt G(x(t), y(t)),

whence the first assertion follows.Let Ψ = v(M) andΛ = v(GM). Then

I(P,Λ ∩ γ) = ordt GM(x(t), y(t))

= ordt G(x(t), y(t))M(x(t), y(t))

= ordt G(x(t), y(t)) + ordt M(x(t), y(t))

= I(P,G ∩ γ) + I(P,Ψ ∩ γ),whence the second assertion follows. 2

The uniqueness of the intersection number, as in Theorem 3.8, gives the follow-ing important result.

THEOREM 4.48 LetF andG be curves. Then, for any pointP,

I(P,F ∩ G) = J(P,F ∩ G).

It is possible to show directly thatJ(P,F ∩ G) = J(P,G ∩ F) whenG is a line.To do this, fix the coordinate system in such a way thatP = (0, 0), G = v(Y )and that no tangents toF at P is v(X). By Theorem 4.34, ifF = v(F ), thenJ(P,F ∩ v(Y )) = ordx F (x, 0). So, the following result is required:

J(P,v(Y ) ∩ F) = ordx F (x, 0). (4.24)

To prove this, supposeP to be anr-fold point ofF with r > 1, as the assertion istrue forr = 1 by Theorem 4.34. Then

F (X,Y ) =∏r

i=1(Y −miX) +G(X,Y ),

with ordG > r.Let γi, for i = 1, . . . , k, be the branches ofF centred atP and let(xi(t), yi(t))

be a primitive representation ofγi. Under a local quadratic transformation,γi istransformed into a branchγ′i of the geometric quadratic transformF ′ of F .

With γ′i centred atP ′i = (0,mi), let (x′i(t), y

′i(t)) be a primitive representation

of γ′i. Since the algebraic quadratic transform ofG = v(Y ) is the conicv(XY ),from (4.21),

J(P,v(Y ) ∩ F) =∑k

i=1 ordt y(t) =∑k

i=1 ordt x′i(t) +

∑ki=1 ordt y

′i(t).

By (4.22), this yields

J(P,v(Y ) ∩ F) = r +∑k

i=1 ordt y′i(t).

In fact, asy′i(t) = yi(t)/xi(t), the branches giving a positive contribution to thesum are those centred atP . If their number iss, relabel the indicesi so that thebranches ofF ′ centred atP have indices1, . . . , s. Then

J(P,v(Y ) ∩ F) = r +∑s

i=1 ordt y′i(t) = r + J(P,v(Y ) ∩ F ′).


Now, if a least two ofP ′i = (0,mi) are distinct, then each pointP ′

i = (0,mi) is anr′i-fold point ofF ′ with r′i < r, and induction can be applied onr. Hence,

∑ki=1 ordt y

′i(t)r + ordx F

′(x, 0).

On the other hand,r = ordx xr. Thus

r + ordx F (x, 0) = ordx (xrF ′(x, 0)) = ordx F (x, 0).

Even if there is only one pointP ′i , this can be done whenr′i < r. Otherwise,

there is only oneP ′i , sayP ′, andr′i = r. In this case, takeP ′ to the origin by a

translation, and apply the local quadratic transformation. Perhapsr is reduced. Ifnot, repeat the procedure. As in the proof of Theorem 4.44, this procedure ends ina finite number of steps. This completes the proof.

4.5 NOETHER’S THEOREM

Let F = v(F (X,Y )) andG = v(G(X,Y )) be two curves with no commoncomponents. By Bezout’s Theorem,F ∩ G = P1, . . . , Ps. Given a polynomialH(X,Y ), it is proposed to find conditions for it to belong to the ideal generatedby F andG; that is,H = AF + BG for some polynomialsA andB. A neces-sary condition is thatH vanishes at the pointsP1, . . . , Ps. This condition is alsosufficient if the multiplicity ofF andG at each pointPi is equal to1.

The result is not proved directly, but follows from other results.

DEFINITION 4.49 For a pointP = (x0, y0), the local ring OP at the pointP isthe ring of all rational functions defined atP ; that is,

OP = A/d | A, d ∈ K[X,Y ], d(P ) 6= 0.

DEFINITION 4.50 Given any pointP of the plane, put

qP = H ∈ K[X,Y ] | H = (AF +BG)/d; d(P ) 6= 0; d,A,B ∈ K[X,Y ].

PROPOSITION 4.51 The setqP is an ideal.

Proof. It must be shown that

(i) if H1 andH2 are inqP , then(H1 +H2) ∈ qP ;

(ii) if H ∈ qP andD ∈ K[X,Y ], thenHD ∈ qP .

LetH1 = (A1F +B1G)/d1, H2 = (A2F +B2G)/d2 with d1(P ), d2(P ) 6= 0.Then

H1 +H2 =(A1F +B1G)

d1+

(A2F +B2G)

d2=CF +BG

h

with C = d1A1 +d2A2, D = d1B1 +d2B2 and0 6= h(P ) = d1(P )d2(P ). HenceH1 +H2 ∈ qP . Also,

DH = D

(AF +BG

d

)=

(DA)F + (DB)G

d,

96

and soDH ∈ qP . 2

The proof of the following lemma is omitted, although the lemma is importantlater.

L EMMA 4.52 Letdi(X,Y ) be a set of polynomials with no common zeros. Thenthe ideal generated by thedi(X,Y ) is (1); that is, 1 =

∑Aidi for some polyno-

mialsAi.

DEFINITION 4.53 The polynomialH satisfiesNoether’s conditionsfor F, G andP if H ∈ qP .

THEOREM 4.54 If H satisfies Noether’s conditions forF, G and for every pointP, thenH ∈ (F,G).

Proof. For every pointP , there is a polynomialdP such thatdP (P ) 6= 0 anddP ·H ∈ (F,G). From the previous lemma, it follows that the ideal generatedbythedP is equal to(1), whence1 =

∑AP dP .

Since(F,G) is an ideal anddP ·H ∈ (F,G), so alsoAP · (dPH) ∈ (F,G) andhenceH = 1H = (

∑AP dP )H ∈ (F,G). 2

THEOREM 4.55 The idealqP is (1) except forP = P1, . . . , Ps.

Proof. Let P 6= Pi for i = 1, . . . , s. Then one of the polynomialsF,G, sayF, isnot zero atP ; that is,F (P ) 6= 0. Since1 = F/F , so1 ∈ qP , whereA = 1 andB = 0 in (AF +BG)/F. But, if 1 ∈ qP , thenqP = (1).

Now, letP = Pi and suppose that1 ∈ qP = qPi. By the definition ofqPi

,

1 = (AF +BG)/d

with d(Pi) 6= 0. From this,d = AF +BG. However, asF andG pass throughPi,sod(Pi) = 0, a contradiction. So,qPi

6= (1). 2

COROLLARY 4.56 (Lasker–Noether)The ideal generated byF andG can be de-composed as follows:

(F,G) = qP1∩ · · · ∩ qPs

.

Proof. If H ∈ (F,G) thenH = AF + BG = (AF + BG)/1. Taked = 1; thend(Pi) = 1 6= 0 for all Pi. SoH ∈ qPi

for eachi; that is,H ∈ ⋂si=1 qPi

. Hence

(F,G) ⊂ qP1∩ · · · ∩ qPs

.

Conversely, ifH ∈ qP1∩ · · · ∩ qPs

, then there exists a polynomiald such thatd(Pi) 6= 0 for eachi = 1, . . . , s andH = (AF +BG)/d. SoH satisfies Noether’sconditions by Theorem 4.54, giving thatH ∈ (F,G); that is,

qP1∩ · · · ∩ qPs

⊂ (F,G).

From the two inclusions, it follows that(F,G) =⋂s

i=1 qPi2

COROLLARY 4.57 If P1 = (0, 0), then all powers of sufficiently high degree ofX,Y and the ideal(X,Y ) are inqP1

.


Proof. Write F andG in descending powers ofY , and calculate the resultant withrespect toY , that is, takingX as parameter. This gives a polynomial inX, namely

RY (F,G) = AF +BG = Xρ(c0 + c1X + · · ·+ cmXm),

with ci ∈ K, c0 6= 0 andA,B ∈ K[X,Y ]. SinceP1 = (0, 0) belongs to bothFandG, soρ > 0. Then

Xρ = (AF +BG)/(c0 + c1X + · · ·+ cmXm) ∈ qP1

;

so also all powers greater thanρ belong toqP1.

If, instead of consideringY as unknown andX as parameter, it is the other wayround, then there existsσ > 0 such thatY σ ∈ qP1

. Let τ = ρ+ σ − 1. It is shownthat(X,Y )τ ⊂ qP1

.Now, (X,Y )τ = (X,Y ) · . . . · (X,Y ); if u ∈ (X,Y )τ thenu = u1 · . . . · uτ

with ui = AiX +BiY ∈ (X,Y ) for i = 1, . . . , τ . So

u= (A1X +B1Y ) · . . . · (AτX +BτY )

=CτXτ +Xτ−1Y Cτ−1 + · · ·+ C1XY

τ−1 + C0Yτ .

It remains to see thatXiY j ∈ qP1. Then, sinceqP1

is an ideal, it follows thatu ∈ qP1

and so(X,Y )τ ⊂ qP1.

Since, in the powers of the productXiY j with i+ j = τ , eitheri ≥ ρ or j ≥ σ,since, if i ≤ (ρ − 1) andj ≤ (σ − 1), thenτ = i + j ≤ ρ + σ − 2; but this isimpossible by the definition ofτ . HenceXiY j ∈ qP1

. In fact, if i ≥ ρ, then it hasbeen shown above thatXi ∈ qP1

. SinceqP1is an ideal, soXiY j ∈ qP1

. 2

THEOREM 4.58 Let P1 = (0, 0). ThenH is in the idealqP1if and only if there

existA,B ∈ K[[X,Y ]] such thatH = AF +BG.

Proof. LetH ∈ qP1. ThenH = (A1F +B1G)/d with A1, B1, d ∈ K[X,Y ] and

d = d0,0 + d1,0X + d0,1Y + · · · ,with d0,0 6= 0 becaused(P1) 6= 0. Henced is a unit inK[[X,Y ]] and alsoA = A1d

−1, B = B1d−1 are the required elements inK[[X,Y ]].

Conversely, suppose thatH = AF + BG with A, B ∈ K[[X,Y ]]. Now, writeA = Ar + A′

r andB = Br + B′r, whereAr andBr are the sums of the terms of

degree less than or equal tor in A andB. So

H = ArF +BrG+A′rF +B′

r = ArF +BrG+ Tr+1,

whereTr+1 contains only terms of degree greater thanr. SinceH andArF +BrGare polynomials, their differenceTr+1 is also a polynomial. By Corollary 4.57,Tr+1 ∈ qP1

for r sufficiently large. Further,ArF + BrG ∈ qP1, by takingd = 1,

and soH ∈ qP1. 2

Consider the ringK[[X]][Y ], and letG, H, f be three of its elements, where

f =∏r

i=1(Y −miX − · · · ) with mi 6= mj for i 6= j.

THEOREM 4.59 If, for i = 1, . . . , r,

ordX H(X,miX + · · · ) ≥ ordX G(X,miX + · · · ) + r − 1,

then

H = Af +BG withA, B ∈ K[[X]][Y ].

Also,A andB can be chosen so thatordX B ≥ r − 1.

98

Proof. The theorem is proved by induction onr. For r = 1,

ordX H(X,miX + · · · ) ≥ ordX G(X,miX + · · · ).Then

H(X,m1X + · · · ) = B(X)G(X,m1X + · · · );soY = m1X+· · · is a root of the polynomialH(X,Y )−B(X)G(X,Y ),whence

H(X,Y )−B(X)G(X,Y ) = A(X,Y )(Y −m1X − · · · ).So the caser = 1 has been proved.

Suppose the theorem is true in the following case:

f1 =∏r

i=2(Y −miX − · · · ), G1 = XG.

Then

ordX H(X,miX + · · · )≥ordX G1(X,miX + · · · ) + (r − 1)− 1

= ordX (XG(X,miX + · · · )) + r − 2

= ordX X + ordX G(X,miX + · · · ) + r − 2

= ordX G(X,miX + · · · ) + r − 1.

By the inductive hypothesis,H = A1f1 + B1G with ordX B1 ≥ r − 2. Sincemi 6= mj for i 6= j,

ordX A1(X,m1X + · · · ) + r − 1

= ordX (H(X,m1X + · · · )−B1(X,m1X + · · · )XG(X,m1X + · · · ))≥ordX G(X,m1X + · · · ) + r − 1.

Here, use has been made of the following two inequalities:

ordX H(X,m1X + · · · ≥ ordX G(X,m1X + · · · ) + r − 1;

ordX (B1(X,m1X + · · · )XG(X,m1X + · · · ))≥ ordX G(X,m1X + · · · ) + r − 1.

It has therefore been shown that

ordX A1(X,m1X + · · · ) ≥ ordX G(X,m1X + · · · );so

A1(X,Y ) = A2(X,Y (Y −m1X − · · · ) +B2(X,Y )G(X,Y ).

Substituting the expression forA1 in H = A1f1 +B1XG gives that

H(X,Y ) = [A3(Y −m1X − · · · ) +B3G(X,Y )]f1 +B1XG(X,Y )

=A4f + [A5f1 +B1X]G(X,Y )

=Af +BG,

whereA3, A4, A5, B1, B3 ∈ K[X,Y ] and

ordX B = ordX B1 + ordX X ≥ r − 2 + 1 = r − 1.2


Let F = v(F (X,Y )) be a curve for whichO = (0, 0) is an ordinary singularpoint with multiplicity r. Suppose thatv(X) is not a tangent toF atO. Then

F (X,Y ) =∏r

i=1(Y −miX) + · · · ,with mi 6= mj for i 6= j.

From the previous chapter, the curveF hasr branches atO that are of the fol-lowing type:

x = t, y = mit+ · · · .Then, by the definition of a branch, the point(X,miX+ · · · ) is a root ofF (X,Y );that is,F (X,miX+ · · · ) = 0. HenceF (X,Y ) is divisible by(Y −miX−· · · ) inK[[X]][Y ] and, ifmi 6= mj for i 6= j, then(Y −miX−· · · ) and(Y −mjX−· · · )are not associated. This proves the following result.

COROLLARY 4.60 The polynomial

F (X,Y ) =∏r

i=1(Y −miX − · · · ) · g(X,Y )

with g ∈ K[[X]][Y ] andg(0, 0) 6= 0, whereordX F = r.

COROLLARY 4.61 LetF = v(F (X,Y )) be a curve for whichP1 is an ordinarysingular point with multiplicityr, and letγ1, . . . , γr be the branches ofF centredat P1. If G = v(G(X,Y )) andH = v(H(X,Y )) are two curves such that

I(P1,H ∩ γj) ≥ I(P1,G ∩ γj) + r − 1 for j = 1, . . . , r,

then

H ∈ qP1.

Proof. By the previous corollary,

F (X,Y ) =∏r

i=1(Y −miX − · · · )g(X,Y )

with g(0, 0) 6= 0. With∏r

i=1(Y −miX − · · · ) = f , thenF = fg. Sinceg is aunit, f = g−1F. Let γj = (x, y) with

x = t,y = mit+ · · · .

be a branch ofF . Then

I(P1,G ∩ γj) = ordt G(t,mjt+ · · · ) = ordX G(X,mjX + · · · );analogously,

I(P1,H ∩ γj) = ordt H(t,mjt+ · · · ) = ordX H(X,mjX + · · · ).Since ordX H ≥ ordX G+ r − 1 by Theorem 4.59,

H = Af +BG with A,B ∈ K[[X,Y ]].

Sincef = g−1F , it follows thatH = (Ag−1)F + BG and, by Theorem 4.58,H ∈ qP1

. 2

Let γ be the set of all branches of an irreducible curveF .

100

DEFINITION 4.62 A formal sum∑nγγ, where eachnγ is an integer andnγ 6= 0

only for a finite number ofγ, is adivisorof F .

The set of divisors is an abelian group under addition:∑nγ γ +

∑mγγ =

∑(nγ +mγ)γ.

Its identity is thezero divisor, that is, the divisor for which everynγ is zero.

DEFINITION 4.63 (i) A divisor∑nγγ is effectiveif nγ ≥ 0 for each branch

γ; write∑nγγ ≻ 0.

(ii) A non-effective divisor isvirtual.

(iii) Given two divisorsD andD′, defineD ≻ D′ if D −D′ ≻ 0.

Let F be a curve with only ordinary singularitiesP1, . . . , Ps having respectivemultiplicities r1, . . . , rs. Also, let γi,j , for j = 1, . . . , ri, be the branches ofFcentred atPi. Then the divisor

D =∑

i,j (ri − 1)γi,j

is thedouble-point divisorof F .For two curvesF andG, denote byG · F the following divisor:

G · F =∑

γ I(P,G ∩ γ)γwhereγ varies over all branches ofF . ThenG · F represents the divisor cut out onF by G.

To formulate Noether’s Theorem, letF = v(F (X,Y )) be a curve with onlyordinary singularities and suppose thatG = v(G(X,Y )) andH = v(H(X,Y ))are two other curves, whereF andG have no common components.

THEOREM 4.64 (Noether)If H · F ≻ D + G · F , then

H = AF +BG,

withA, B ∈ K[X,Y ].

Proof. Let γ be a branch ofF with centre a pointP of multiplicity r ≥ 1 for F .By hypothesis,

I(P,H ∩ γ) ≥ I(P,G ∩ γ) + r − 1.

By Corollary 4.61,H ∈ qP and soH satisfies Noether’s conditions. Hence, byTheorem 4.54,H belongs to the ideal(F,G). 2

Now, the projective formulation of this theorem is given. Let

F = v(F ∗(X0,X1,X2), G = v(G∗(X0,X1,X2)), H = v(H∗(X0,X1,X2))

be the corresponding projective curves.

THEOREM 4.65 (Noether)If H · F ≻ D + G · F , then

H∗ = A∗F ∗ +B∗G∗,

withA∗, B∗ homogeneous polynomials inK[X0,X1,X2].


Proof. Applying a projectivity if necessary, assume thatG andF do not meet onv(X0) and thatv(X0) is a component of neitherG norF ; therefore,

G∗(0,X1,X2) 6= 0, F ∗(0,X1,X2) 6= 0.

Consider the affine polynomials of the curves:

F (X,Y ) = f0,0 + f1,0X + f0,1Y + · · ·+ fn,0Xn + f0,nY

n, degF = n;

G(X,Y ) = g0,0 + g1,0X + f0,1Y + · · ·+ gm,0Xm + g0,mY

m, degG = m;

H(X,Y ) =h0,0 + h1,0X + h0,1Y + · · ·+ hs,0Xs + h0,sY

s, degH = s.

By the preceding theorem,

H∗(1,X, Y ) = A(X,Y )F ∗(1,X, Y ) +B(X,Y )G∗(1,X, Y ) (4.25)

with degA = n0, degB = m0. PuttingX = X1/X0, Y = X2/X0 in (4.25)gives

1

Xs0

H∗(X0,X1,X2) =1

Xn0+n0

A∗(X0,X1,X2)F∗(X0,X1,X2)

+1

Xm0+m0

B∗(X0,X1,X2)G∗(X0,X1,X2)

with A∗ andB∗ homogeneous. Putt = maxs, n0 + n,m0 +m; then

Xt−s0 H∗ = X

t−(n0+n)0 A∗F ∗ +X

t−(m0+m)0 B∗G∗,

which can be re-written as follows:

Xρ0H

∗(X0,X1,X2) =A∗1(X0,X1,X2)F

∗(X0,X1,X2)

+B∗1(X0,X1,X2)G

∗(X0,X1,X2) (4.26)

with A∗1 andB∗

1 homogeneous. Suppose thatρ > 0, and putX0 = 0:

A∗1(0,X1,X2)F

∗(0,X1,X2) +B∗1(0,X1,X2)G

∗(0,X1,X2) = 0.

Noting thatF ∗(0,X1,X2) is a homogeneous polynomial in the indeterminatesX1, X2, it may be factorised as

F ∗(0,X1,X2) =∏n

i=1(αiX1 − βiX2)

with (0, αi, βi) ∈ F . Analogously,

G∗(0,X1,X2) =∏m

j=1(γjX1 − GjX2)

with (0, γj ,Gj) ∈ G.Since the curvesF andG have no common component, so points onv(X0),

namely of type(0, βi, αi), are inv(B∗1). Hence the terms of the product

∏ni=1(αiX1 − βiX2)

divideB∗1 , whence

A∗1(0,X1,X2) =−B

∗1(0,X1,X2)

F ∗(0,X1,X2)G∗(0,X1,X2)

=A∗2(X1,X2)G

∗(0,X1,X2).

102

Therefore

A∗1(X0,X1,X2) = A∗

2(X1,X2)G∗(X0,X1,X2) +X0A

∗3(X0,X1,X2)

with A∗2, A

∗3 homogeneous. Substituting this expression forA∗

1 in (4.26) gives

Xρ0H

∗(X0,X1,X2) =X0A∗3(X0,X1,X2)F

∗(X0,X1,X2)

+B∗2(X0,X1,X2)G

∗(X0,X1,X2).

Now putX0 = 0; henceB∗2(0,X1,X2) = 0 and so

B∗2(X0,X1,X2) = X0B

∗3(X0,X1,X2).

EliminatingX0 gives

Xρ−10 H∗(X0,X1,X2) =A∗

3(X0,X1,X2)F∗(X0,X1,X2)

+B∗3(X0,X1,X2)G

∗(X0,X1,X2).

Repeating this procedure reducesρ to 0, which proves the theorem. 2

4.6 ANALYTIC BRANCHES

In this section, another definition of the intersection multiplicity of two algebraiccurves is given, and it is shown that the new definition coincides with the previousone.

For an algebraic curveF = v(F ) with F (X,Y ) ∈ K[X,Y ], the idea of abranch was introduced. Now, this idea is extended to the casethatF (X,Y ) is aformal power series; that is,F (X,Y ) ∈ K[[X,Y ]].

DEFINITION 4.66 An analytic branch centred atO is a branch centred atO in thesense of Definition 4.28.

The branchbelongstoF (X,Y ) ∈ K[[X,Y ]] if, for any representation(x(t), y(t)),thenF (x(t), y(t)) = 0.

It is shown below that there is a unique branch that belongs toa given irreduciblepower series other than 1 and that, givenF1, F2 ∈ K[[X,Y ]], both irreducible anddifferent from 1, if a branch belongs to both of them, then they are associates. So,it is possible to introduce the idea of analytic branch as an irreducible formal powerseries different from 1 or, more precisely, as an equivalence class of irreduciblepower series other than 1 with respect to the equivalence relation: F1 ∼ F2 ifF1 = EF2 with E a unit.

GivenF (X,Y ) ∈ K[[X,Y ]] with F = Fr +Fr+1 + · · · , perform a transforma-tion of the type

X = aX ′ + bY ′, Y = cX ′ + dY ′

with ad− bc 6= 0 in such a way thatFr is not divisible byX.Now, sinceFr is not divisible byX, then

F (0, Y ) = arYr + ar+1Y

r+1 + · · · , with ar 6= 0.


The Weierstrass Preparation Theorem applied toF (X,Y ) determines a polyno-mial in Y with coefficients inK[[X]] which is an associate ofF (X,Y ); that is,there exists a unitE(X,Y ) ∈ K[[X,Y ]] such that

F (X,Y ) = E(X,Y )(Y r +A1(X)Y r−1 + · · ·+Ar(X))

withAi(X) ∈ K[[X]] andAi(0) = 0 for i = 1, . . . , r. This allows the substitution,for each formal power seriesF (X,Y ), of its associated polynomial inK[[X]][Y ].

DEFINITION 4.67 (i) An analytic curveis an elementF (X,Y ) ∈ K[[X,Y ]]that is not a unit and that has no repeated factors.

(ii) If F (X,Y ) is a formal power series withF 6= 0, thenF is ananalytic cycleandF (X,Y ) = 0 is its equation.

DEFINITION 4.68 An analytic cycleF is centred atO = (0, 0) if F (0, 0) = 0.More generally,F (X,Y ) is centred at(0,m) if F (0,m) = 0.

DEFINITION 4.69 The centre(0, 0) is r-ple for F (X,Y ) if

F = Fr + Fr+1 + · · · , Fr 6= 0,

and issimpleif r = 1.

The factors that appear inFr represent the tangents ofF (X,Y ) at its centre.

REMARK 4.70 BothF andEF,whereE is a unit, refer to the same analytic curve.

THEOREM 4.71 LetF (X,Y ) ∈ K[[X,Y ]] with ordF = 1. Then there is one andonly one branch(x(t), y(t)) such thatF (x(t), y(t)) = 0.

Proof. The proof is similar to that of Theorem 4.31, considering a formal power se-ries instead of a polynomial. The existence of the branch is obtained as in Theorem4.6 and, ifF = Y − c1X − · · · , then the branch is(t, c1t+ · · · ). 2

DEFINITION 4.72 If (x(t), y(t)) is a primitive representation of the branchγ andG is an analytic cycle with equationG(X,Y ) = 0, then put

I(P,G ∩ γ) = ordt G(x(t), y(t)).

The idea of multiplicity of a branch at its centre is defined asin the case of abranch of a plane algebraic curve.

THEOREM 4.73 Two distinct irreducible analytic curvesF = v(F (X,Y )) andG = v(G(X,Y )) cannot have a branch in common.

Proof. The transformation

X = aX ′ + bY ′, Y = cX ′ + dY ′,

with ad− bc 6= 0, can makeF andG regular inY and substitute the correspondingspecial polynomialsF1 andG1 obtained from the Weierstrass Preparation Theo-rem.

104

The special polynomials are irreducible inK[[X]][Y ] and also inK((X))[Y ].By hypothesis,F andG are not associates inK[[X,Y ]]; soF1 andG1 are also as-sociates neither inK[[X]][Y ] nor inK((X))[Y ]. Therefore, their greatest commondivisor inK((X))[Y ] is 1; so1 = aF1 + bG1 with a, b ∈ K((X))[Y ].

Note that

a(Y ) = an(X)Y n + · · ·+ ai(X)Y i + · · ·+ a0(X)

with ai(X) ∈ K((X)); that is,

ai(X) =si(X)

ti(X)= XkiEi(X),

whereEi(X) is a unit fori = 0, . . . , n. Also,

b(Y ) = bm(X)Y m + · · ·+ bj(X)Y j + · · ·+ b0(X),

with bj(X) ∈ K((X)); that is,

bj(X) =rj(X)

uj(X)= XhjHj(X),

with Hj(X) a unit forj = 0, . . . ,m.It may happen that someki andhj are negative; writeρ = max|ki|, |hj | for

i = 0, . . . , n andj = 0, . . . ,m. Hence, multiplying byXρ gives

Xρ = AF1 +BG1, (4.27)

with A, B ∈ K[[X]][Y ].Suppose that there is a branch(x(t), y(t)) common to the two analytic curves.

Substituting in (4.27) gives thatxρ = 0, whencex = 0. HenceF1 is special; thatis, F1 = Y r + c1(X)Y r−1 + · · · + cr(X) and(x(t), y(t)) is a branch ofF1, asit is an associate ofF . Therefore,0 = F1(0, y(t)) = y(t), whencey = 0. Thiscontradicts the definition of a branch; so they cannot have branches in common.2

COROLLARY 4.74 Let (ξ(t), η(t)) be a branch withξ(t) = 0; then it is a branchof the linev(X) and of no other irreducible analytic curve.

Proof. The branch considered satisfies the equationX = 0 and no other irreducibleanalytic curve can have such a branch. 2

THEOREM 4.75 LetF be an analytic cycle centred atO = (0, 0) not containingv(X) as a component. ThenF has a unique special equationF = 0, whereF isreducible inK[[X]][Y ] if and only if it is reducible inK[[X,Y ]].

Proof. Suppose thatF is regular inY ; otherwise, it can be made so by an affinetransformation. Then, by the Weierstrass Preparation Theorem, there exist uniqueG andH such thatF = GH with H a special polynomial which is an associate ofF. By Theorem 4.15 the second part follows. 2

DEFINITION 4.76 An analytic cycle alongv(X) is a finite number of irreducibleanalytic cycles centred onv(X) together with their associated multiplicity. Thenotation ism1F1 + · · ·+msFs.


If F ∈ K[[X]][Y ] is monic and irreducible, Hensel’s Lemma 4.9 shows that thepolynomialF (0, Y ) has only one rootλ andF defines an irreducible analytic cyclecentred at(0, λ) and different fromv(X). If λ = 0, thenF is special; in general,F is special relative toY1 = Y − λ.

Conversely, every irreducible analytic cycle centred onv(X), but not this line,is represented by a unique monic irreducible elementF ∈ K[[X]][Y ].

If F1, . . . ,Fs are such cycles with monic irreducible seriesF1, . . . , Fs, then bydefinitionFm1

1 · · ·Fmss is the monic seriesm1F1 + · · ·+msFs. Conversely every

monic elementF ∈ K[[X]][Y ] has a unique factorisation into monic irreduciblefactors:F = Fm1

1 · · · · · Fmss . It defines the cyclem1F1 + · · ·+msFs.

Also, in the theory of analytic branches, it is useful to consider the local quadratictransformation

X ′ = X, Y ′ = Y/X.

Let F = v(F (X,Y )) be an analytic cycle centred atO = (0, 0), and define thetransform ofF only in the case thatv(X) is not tangent toF . DefiningF ′ as thespecial series ofF , it may be assumed directly thatF is special.

DEFINITION 4.77 The curve given byF ′ = v(F (X,XY ′)/Xr), where(0, 0) isr-ple forF , is thetransformF ′ of F .

Theorems 4.40, 4.41, 4.42, 4.43 that connectF andF ′ in the context of branchescontinue to hold for analytic curvesF centred at(0, 0) and that do not have the linev(X) as a tangent. The proofs are the same, apart from taking the special seriesF (X,Y ) of F .

Corresponding to Theorem 4.44 is the following result.

THEOREM 4.78 LetF = v(F (X,Y )) be an analytic curve centred atO = (0, 0).Then there is at least one and at most a finite number of branches (x(t), y(t)) thatsatisfyF.

Proof. The proof is similar to that of Theorem 4.44 with some small changes. LetF = Fr(X,Y ) + Fr+1(X,Y ) + · · · ; then, by a transformation

X ′ = aX + bY, Y ′ = cX + dY,

it can be ensured thatFr is not divisible byX. Applying Weierstrass’ Theoremagain gives the series

Yr + a1(X)Y r−1 + · · ·+ ar(X),

which is the special series ofF associated toF (X,Y ). Suppose further thatF is ir-reducible; then the transformF ′ ofF having polynomialF ′(X,Y ′) = F (X,XY ′)is still irreducible. By Hensel’s Lemma,F ′(0, Y ′) has only one rootλ. By a trans-lationY ′′ = Y ′ − λ, it may be assumed thatλ = 0; then

F (X,Y ′) = Y ′r + b1(X)Y ′r−1+ · · · .

If r does not decrease, thenF ′ is still special and the axisv(X) is not tangentto F ′. The difference from Theorem 4.44 is the need to make transformations

106

apart from translations on the analytic curves before applying the local quadratictransformations.

2

It remains to prove, as in Theorem 4.44, thatr decreases. This then gives thefollowing result.

COROLLARY 4.79 The sequence of quadratic transformations and translationsthat occur in the preceding proof cannot produce a sequence of r-ple points forr > 1.

Proof. LetF be irreducible withF (X,Y ) = 0 as its special equation, and assumethatF is not tangent to the linev(X). As in the proof of Theorem 4.45, if thesequence of quadratic transformations does not produce a reduction inr, then

F (X,Y ) = (Y −m1X −m2X2 − · · · )G(X,Y ).

SinceF is irreducible and monic, it follows that

F (X,Y ) = Y −m1X −m2X2 − · · · ,

whencer = 1. 2

THEOREM 4.80 An irreducible analytic cycleF has only one branch.

Proof. The proof is by induction onr. Let O = (0, 0) be r-ple forF . If r = 1,then, by Theorem 4.71, there is one and only one branch ofF .

Suppose therefore thatr > 1 and ensure by a suitable transformation thatv(X)is not tangent toF . Consider the special equationF (X,Y ) = 0 and apply the localquadratic transformation. If there is a reduction inr, the result is obtained by theinductive hypothesis. Ifr does not reduce, first make a translation and then anotherquadratic transformation; by Corollary 4.79,r reduces and the result follows. 2

Let F (X,Y ) = Y e + a1(X)Y e−1 + · · · + ae(X) ∈ K[[X]][Y ] be monic andirreducible. By Hensel’s Lemma,F (0, Y ) has only one rootλ,which may be takenasλ = 0. Let

x = ctd + · · · , y = y(t), c 6= 0,

be a branch ofv(F (X,Y )). It is shown thatd = e.

L EMMA 4.81 If (x(t), y(t)) is a primitive branch representation, then the fieldK(x(t), y(t)) contains an element of order1.

Proof. The proof is given in Theorem 4.20. 2

DEFINITION 4.82 An element of order1 is auniformizing parameter.

L EMMA 4.83 Letx = ctd + · · · , c 6= 0. Then

K[[x]][t] = K[[t]] and [K((x))(t) : K((x))] = d.


Proof. Let f ∈ K[[x]][t]. Thenf is a polynomial int with coefficients inK[[x]];that is,

f = ah(x)th + ah−1(x)th−1 + · · ·+ a0(x),

with ai(x) = a0,i + a1,ix + a2,ix2 + · · · for i = 0, . . . , h. Substituting inai(x)

the value ofx, in f only powers oft are obtained. Sof ∈ K[[t]]; this shows thatK[[x]][t] ⊂ K[[t]].

Conversely, consider

1, t, . . . , td−1, x, tx, . . . , td−1x, x2, tx2, . . . ;

they have degrees0, 1, 2, . . . in t and constitute a basis forK[[t]].Take an elementh ∈ K[[t]]; it can be expressed as a linear combination of the

basis elements:

h = h0 + h1t+ h2t2 + · · ·+ hd−1t

d−1 + hdx+ · · · .Soh can be written as a polynomial int of degreed−1 with coefficients inK[[x]].In fact,

h = h′0 + h′1t+ · · ·+ h′d−1td−1,

with

h′0 =h0 + hdx+ h2dx2 + · · ·+ hidx

i + · · · ,h′1 =h1 + hd+1x+ h2d+1x

2 + · · ·+ hid+1xi + · · · ,

......

h′j =hj + hd+jx+ h2d+jx2 + · · ·+ hid+jx

i + · · · .Henceh ∈ K[[x]][t] and soK[[x]][t] = K[[t]].

For the second part of the theorem, consider an elementF ∈ K[[X,T ]] withF = X − cT d − · · · andc 6= 0. Note thatF is irreducible with subdegree1. SinceF (0, T ) = −cT d + · · · with c 6= 0, the Weierstrass Preparation Theorem 4.10shows the existence of a unitE(X,T ) such that

E(X,T ) = T d + a1(X)T d−1 + · · · .The polynomial associated toF is irreducible inK[[X]][T ]. As in the proof ofTheorem 4.15, it is also irreducible inK((X))[T ]. By the choice ofF , the point(x(t), t) with F (x(t), t) = 0 also satisfies

T d + a1(X)T d−1 + · · · = 0.

Hencet is a root of

T d + a1(x(t))Td−1 + · · · = 0,

which is a polynomial of degreed with coefficients inK((x(t))). So, it has beenshown that[K((x(t)))(t) : K((x(t)))] = d. 2

COROLLARY 4.84 If (x = ctd + . . . , y = etf + . . .) is a primitive branch repre-sentation centred at(0, 0) andx 6= 0, theny is algebraic overK((x)).

108

Proof. As in Lemma 4.81, choose an elementτ = at + . . . , a 6= 0 of order1 inK(x, y). Thenx = cτd + . . . , y = eτf + . . . and, further,τ = a(x, y)/b(x, y) witha(X,Y ), b(X,Y ) ∈ K[X,Y ]. If τ ∈ K(x), then,a fortiori, τ ∈ K((x)), whencey ∈ K(x). If, instead,τ 6∈ K(x), write τb(x, y) = a(x, y) and then re-order it inpowers ofy, giving

a1(x, τ)ym + a2(x, τ)y

m−1 + . . . = 0,

with m ≥ 1 andai(x, τ) ∈ K(x, τ). Therefore,y is algebraic overK((x))(τ). ByLemma 4.83, the result follows. 2

THEOREM 4.85 Given a monic, irreducible polynomial

F = Y e + a1(X)Y e−1 + · · · ∈ K[[X]][Y ],

if (x = ctd + · · · , y = y(t)), with c 6= 0, is a branch representation ofF, thend = e.

Proof. By Lemma 4.83,

y(t) ∈ K[[t]] = K[[x(t)]][t] and [K((x(t)))(t) : K((x(t)))] = d.

It follows that y is a root of a polynomialF of degreeℓ ≤ d with coefficientsin K((x)). ThenF is a multiple ofY e + a1(X)Y e−1 + · · · . This shows thate ≤ ℓ ≤ d. To see that alsod ≤ e, choose an elementτ of order1 in K(x, y).Thenτ ∈ K((x))(y) and[K((x))(y) : K((x))] = e, from which it follows thatτis a root of a polynomial of degree at moste. But, thend ≤ e, as required. 2

THEOREM 4.86 Given a field of characteristic0, consider a monic, irreduciblepolynomialF = Y d + a1(X)Y d−1 + · · · ∈ K[[X]][Y ] and one of its branchesγ = (x(t), y(t)). ThenF (x(t), Y ) splits completely into linear factors inK((t)).

Proof. Let (x = ctd + · · · , y = y(t)) with c 6= 0 be the branch representationγ.By applying Theorem 4.26 it is possible to transform such a pair into an equivalentbranch representation(x = td, y = η(t)). Hence, it may be assumed thatx = td.

Let

y =∑

cijtij with cij

6= 0. (4.28)

Write only the terms that have non-zero coefficients. Then, with

e = gcd(d, . . . , ij , . . .),

e = 1. For, if e > 1, then(x(t), y(t)) can be re-written in terms ofte, but thiscontradicts the primitive character of(x(t), y(t)).

Before continuing with the proof, some algebraic argumentsneed to be made.Suppose thatK = C; then thed-th roots of unity are thed complex numbers,

ζd,0, . . . , ζd,d−1,

where

ζd,k = cos(2kπ/d) + i sin(2kπ/d),

0 ≤ k < d.


A d-th root of unityζ is primitive if ζk 6= 1 for 1 ≤ k ≤ d− 1 andζd = 1. Letζ = cos(2π/d) + i sin(2π/d) be primitive. Similarly, every algebraically closedfield of characteristic0 has a primitived-th root of1.

Consider thed substitutions,

t 7→ ζit, i = 0, . . . , d− 1. (4.29)

Such a substitution sendstd to td andy(t) to yi(t) for i = 0, . . . , d − 1. Sincesuch substitutions areK-automorphisms ofK[[t]], theyi(t) are roots ofF (td, Y ).Hence,(x(t), y(t)) is a branch ofF ; that is,F (x(t), y(t)) = 0, whencey(t) is aroot ofF (x(t), Y ) = 0. By (4.29),y(t) 7→ yi(t) by an automorphism; ify(t) is aroot, so also isyi(t).

If it is shown that theyi are distinct, then they are thed roots ofF (td, Y ) andthe proof is complete. The expression foryλ is

yλ =∑

cijζλij tij , (4.30)

where thecijare the same as in (4.28). Ifyλ = yµ, thenζλij = ζµij , whence

ζ(λ−µ)ij = 1 and(λ− µ)ij ≡ 0 (mod d). Pute = gcd(. . . , ij , . . .); then

(λ− µ)e ≡ 0 (mod d).

Since gcd(d, e) = 1, it follows thatλ−µ ≡ 0 (mod d), soλ = µ, as0 ≤ λ ≤ d−1and0 ≤ µ ≤ d− 1. Since theyi are distinct, the result is proved. 2

There is a more general result valid in any characteristic.

THEOREM 4.87 LetK be a field of characteristicp and let

F = Y d + a1(X)Y d−1 + · · · ∈ K[[X]][Y ]

be a monic polynomial. Then there existsx(v) ∈ K[[v]] such thatF (x(v), Y ) splitscompletely into linear factors inK[[v]][Y ].

Proof. If (x(t), y(t)) is a branch representation ofF , theny(t) is a root of thepolynomialF (x(t), Y ) ∈ K[[t]][Y ]. Therefore,F (x(t), Y ) = (Y −y(t))F1(t, Y ),with

F1(t, Y ) = Y d−1 + b1(t)Yd−2 + . . . ∈ K[[t]][Y ].

If d > 1, there is a branch representation(t(t1), y1(t1)) of F1 = v(F1) and theargument can be repeated to obtain that

F1(t1, Y ) = (Y − t(t1))(Y d−2 + c1(t1)Yd−3 + . . .) ∈ K[[t1]][Y ].

After this procedure has been repeatedd− 1 times,

F (x(t), Y ) = (Y − y(t))(Y − y1(t1)) · (Y − yd−1(td−1)).

Putv = td−1, and expresstd−2 in terms ofv; successively, do this for eachti withi = d− 3, . . . , 2, 1, whence the result follows. 2

THEOREM 4.88 Given a monic, irreducible polynomial

F = Y d + a1(X)Y d−1 + · · · ∈ K[[X]][Y ],

110

let F be an analytic cycle with special equationF = 0 centred atO = (0, 0) andlet γ = (x(t), y(t)) be one of its branches. IfG = v(G) is another analytic cyclealso centred atO. Then

I(P,G ∩ γ) = ordX RY (F (X,Y ), G(X,Y ));

that is,

ordt G(x(t), y(t)) = ordX RY (F (X,Y ), G(X,Y )).

Proof. First, a proof is given in the case of characteristic zero; the proof in thegeneral case is similar.

Recall thatRY (F,G) is a polynomial inX. By a preceding theorem,

F (x(t), Y ) =∏d

i=1(Y − yi(t))

and there areK-automorphisms ofK[[t]] that sendx to x andy to yi. Then

ordt G(x(t), yi(t)) = ordt G(x(t), y(t)).

From Section 2.1,

RY (∏d

i=1(Y − yi), G) =∏d

i=1G(x(t), yi(t)),

whence

ordt RY (F (x, Y ), G(x, Y )) = dordt G(x, y).

Recall that ordt x = d; so it follows that

ordt RY (F (x, Y ), G(x, Y )) = dordx RY (F (x, Y ), G(x, Y )),

and the theorem is proved. 2

In the general case, write

F (x(u), Y ) =∏d

i=1(Y − y1(u)),where(x(u), y(u)) and(x(u), yi(u)) are not-necessarily-primitive representations.By a substitution of the typet = ciu

ρi + · · · , the pair(x(u), yi(u)) is transformedto a primitive representation(x′i(ti), y

′i(ti)). By Theorem 4.85, ordti

x′i(ti) = d;so ordu x(u) = ρid. Thus all the redundanciesρi are equal: writeρi = ρ.

Now, (x′i(ti), y′i(ti)) andx′j(tj), y

′j(tj)) are primitive branch representations of

F . The substitutionti 7→ tj transforms(x′i(ti), y′i(ti)) to x′j(tj), y

′j(tj)), which

still represents a branch ofF . Since such a substitution is an automorphism, then

ordtiG(x′i(ti), y

′i(ti)) = ordtj

G(x′j(tj), y′j(tj)),

whence

ordu G(x(u), yi(u)) = ordu G(x(u), yj(u)).

The remainder of the proof proceeds as before.Now, an alternative proof can be given for the following:

J(P,G ∩ F) = I(P,G ∩ F).


COROLLARY 4.89 LetF andG be plane algebraic curves with no common comp-onents, and letP be a point of the plane. Then the intersection multiplicities definedby the resultant coincides with the intersection defined using the branches ofFcentred atP ; that is, I(P,G ∩ F) = J(P,G ∩ F).

Proof. If P does not belong to both curves, then, from the definitions, both inter-section multiplicities are zero. Therefore, suppose that the point belongs to bothcurves and that it is the origin; suppose also that(0, 1, 0) does not belong to bothcurves and that the curves meetv(X) only in P = (0, 0) = (0, 0, 1).

LetF = v(F (X,Y )) andG = (G(X,Y )); if F does not pass through(0, 1, 0),thenF (X,Y ) is monic inY. Let F = F1F2 · · ·Fs be the complete factorisationof F in K[[X]][Y ] with Fi monic and irreducible for eachi. Let F1, . . . , Fk thepolynomials centred at(0, 0),. The polynomialFi defines the branchγi of F cen-tred at the origin. Note thatRY (Fi, G) = Pi(X) is a unit inK[[X]] for i > k,sinceG(0, Y ) andFi(0, Y ) could have a root in common only if the two curvesmetv(X) at a point other than the origin. Then

I(P,G ∩ F)= I(P,F ∩ G)= ordX RY (F,G)

= ordX RY (∏s

i=1Fi, G) = ordX

∏si=1RY (Fi, G)

=∑s

i=1 ordX RY (Fi, G) =∑k

i=1 ordX RY (Fi, G)

=∑k

i=1 I(P,G ∩ F) = J(P,G ∩ F).

2

4.7 EXERCISES

1. If K is a field of characteristicp > 0, show that the branch representation

x(t) = u+ tp + tp+1, y(t) = v + tp + tp+2

has orderp, and that it is not equivalent to a branch representation of type

x(t) = u+ tp, y(t) = v + η(t), with ordt η(t) ≥ p.

2. IfR denotes the set of all branch representations in special affine coordinates,for eachK-monomorphismσ of K[[t]], define the mappingσ⋆ : R → R asfollows:

σ⋆ : (x(t), y(t)) 7→ (σ(x(t), σ(y(t)).

Let T ⊂ R be the union of the imagesσ⋆(T ) asσ varies in the set ofK-monomorphisms ofK[[t]] that are not automorphisms. Show that a branchrepresentation(x(t), y(t)) in special affine coordinates is primitive if andonly if it is in R\T .

3. Show that aK-automorphism ofK((t)) induces an automorphism ofK[[t]].

112

4. Foru ∈ K((t)) with ordt u = m ≥ 1, the fieldK((u)) can be viewed as asubfield ofK((t)). Show that[K((t)) : K((u))] = m.

5. Letp = 3. Show that the irreducible plane curve

F =v(X3Y 3 + 2X2Y 4 +X4Y +X3Y 2 + 2XY 4

+2X2Y 2 + 2XY 3 + 2X3 +X2Y +XY + Y 2)

has five singular points, namely

O = (1, 0, 0), X∞ = (0, 1, 0), Y∞ = (0, 0, 1),

E = (1, 1, 1), P = (1, 2, 1),

each the centre of exactly two branches.

6. Give a direct proof of Noether’s Theorem 4.64 in its simplest case: AssumethatF = v(F ) andG = v(G) intersect inmn distinct points wherem =degF, n = degG. Show that anyH of degreer which passes through allcommon points ofF andG has equation of the formH = AF + BG = 0whereA andB are polynomials of degreesr −m andr − n respectively.

7. Show that the assertion in Exercise 6 holds true if each common pointP ofF andG is a simple point ofF andI(P,H ∩ F) ≥ I(P,G ∩ F).

8. LetF = v(F ) andG = v(G) be plane curves of degreem andn respec-tively. Assume that at every common pointP the tangents toF are distinctfrom the tangents toG. LetH = v(H) be a curve of degreer which has anh+ k− 1-fold point atP wheneverP is h-fold onF andk-fold onG. ThenH = AF +BG whereA andB are polynomials of degreesr−m andr−nrespectively such that the curvesA = v(A) andB = v(B) have points ofmultiplicity k − 1 andh− 1 at least atP .

9. Assume thatF = v(F ) andG = v(G) intersect inmn distinct points wherem = degF, n = degG. Show that ifnr of these points lie on a curve ofdegreer, then remaining(m − r)n lie on a curve of degreem − r. DeducePascal’s theorem from this result.

10. (Cayley–Bacharach theorem). Let∆ be a set ofmn points of intersectionof two curves of degreesm andn. Show that that every curve of degreem+ n− 3 containing all but one point of∆ must contain∆.

4.8 NOTES

Chapter 4 is based on Seidenberg [Chapters 11,12,15][252].For generalisations ofthe Cayley–Bacharach theorem, see [Chapter 5][255] and [68].

Chapter Five

The function field of a curve

5.1 GENERIC POINTS

An (affine)pointnow means any pointP = (x, y) with coordinatesx, y belongingto some extension of the fieldK. As in Section 4.2, branch representations arepoints in this sense. A point isconstantif both its coordinatesx, y belong toK; inthe other case, the point isvariable.

DEFINITION 5.1 A point P = (x, y) is genericon the curveF = v(F (X,Y ))if, for everyG(X,Y ) in K[X,Y ] with G(x, y) = 0, it follows thatG(a, b) = 0for all pointsQ = (a, b) of F . In other words, the pointP = (x, y) is generic ifG(x, y) = 0 implies thatG ≡ 0 (mod F ).

REMARK 5.2 It follows from this definition that two distinct irreducible curvescannot have a common generic point.

THEOREM 5.3 A reducible curve has no generic points.

Proof. Let C be a curve with two componentsF = v(F ) andG = v(G), withF 6⊂ G, G 6⊂ F . ThenC = v(FG). In fact, if P = (a, b) is a point, constant orvariable, such thatF (a, b)G(a, b) = 0; thenF (a, b) = 0 orG(a, b) = 0.

Suppose thatC has a generic pointP = (x, y); thenF (x, y) = 0 orG(x, y) = 0.Suppose the former holds; that is,F (x, y) = 0. ThenF is zero at all points ofCand so also at all points ofG. SoG must be a component ofF , a contradiction. 2

THEOREM 5.4 A variable pointP = (x, y) of an irreducible curveF is a genericpoint.

Proof. Let F = v(F (X,Y )) and suppose thatG(x, y) = 0 but G(X,Y ) 6≡ 0(mod F (X,Y )),withP = (x, y) a variable point ofF . TakingF to be irreducible,it follows from Bezout’s Theorem that the common points of the curvesF andG = v(G(X,Y )) are finite in number and so belong to the planeAG(2,K). Inother words, each of the common points ofF andG is constant. In particular,sinceP = (x, y) is a variable point ofF , it cannot belong to the curveG, and soG(x, y) 6= 0, as required. 2

From now on, the term “point” of an irreducible curveF only indicates any ofits constant points, while “generic point” ofF stands for any of its variable points.

The above discussion extends to the projective plane. LetF = v(F ) with ho-mogeneousF = F (X0,X1,X2) ∈ K[X0,X1,X2]. A generic pointof F is

113

114

defined to be a pointP = (x0, x1, x2) with F (x0, x1, x2) = 0, whose coordi-natesx0, x1, x2 come from an extension field ofK such that ifG is a homoge-neous polynomial inK[X0,X1,X2] that vanishes atP thenG vanishes at everypoint of F . This amounts to saying thatG is divisible byF . As in the affinecase, a curve has a generic point if and only if it is irreducible. Using homo-geneous polynomials, the function field ofF can be described as the subfield inK(x0, x1, x2) consisting of all elementsξ = u(x0, x1, x2)/v(x0, x1, x2) whereu(X0,X1,X2), v(X0,X1,X2) ∈ K[X0,X1,X2] are homogeneous of the samedegree, andv(x0, x1, x2) 6= 0. In fact, if x0 6= 0, this subfield isK-isomorphic toK(1, x1/x0, x2/x0) which is justΣ = K(x, y). Also, Σi = Σj whenxi, xj 6= 0,and0 ≤ i, j ≤ 2.

If all coordinatesx0, x1, x2 are constant, thenP is aconstantpoint; otherwise, itis avariablepoint. As before, a point of an irreducible curve is generic if and onlyif it is variable.

REMARK 5.5 With the notion of universal domainΩ overK, all points are in theplane overΩ. SinceΩ is algebraically closed field, one can think about a curveFoverK as a curve overΩ, meaning that the coefficients in the equation ofF arein Ω. All theorems previously proved remain true as stated, though may have adifferent content because of the change of meaning for point.

DEFINITION 5.6 The generic pointsP = (x, y) andP1 = (x1, y1) of an irreduc-ible curveF areK-isomorphicif there exists an isomorphism

σ : K[x, y]→ K[x1, y1]

such thatσ(x) = x1 andσ(y) = y1.

THEOREM 5.7 Any two generic points of an irreducible curve areK-isomorphic.

Proof. LetP = (x, y) andP1 = (x1, y1) be generic points of the irreducible curveF . Now, letσ : K[x, y] → K[x1, y1] be a map such thatσ(G(x, y)) = G(x1, y1)for everyG(X,Y ) ∈ K[X,Y ]. It must be shown thatG(x, y) = H(x, y) im-pliesG(x1, y1) = H(x1, y1) for everyH(X,Y ) ∈ K[X,Y ]. In other words, ifG(x, y) − H(x, y) = 0, then alsoG(x1, y1) − H(x1, y1) = 0. But this followsimmediately from the notion of generic point. Therefore themapσ is well-defined.Both the following relations hold:

σ(G1(x, y) +G2(x, y)) =σ(G1(x, y)) + σ(G2(x, y));

σ(G1(x, y)G2(x, y)) =σ(G1(x, y))σ(G2(x, y)).

This shows thatσ is a homomorphism fromK[x, y] to K[x1, y1]. Further, thekernel ofσ is zero sinceG(x1, y1) = 0 impliesG(x, y) = 0 by the hypothesis thatP = (x1, y1) is a generic point ofF . 2

COROLLARY 5.8 An irreducible curveF has infinitely many generic points.

Proof. To prove the assertion, it is enough to show thatF has infinitely many pointsP = (x, y) such thatx, y ∈ K((t)). Given a constant pointQ of F , let (x(t), y(t))

The function field of a curve 115

be any representation of a branch centred atQ. ThenP = (x(t), y(t)) is a variablepoint ofF . By Theorem 5.4,P = (x(t), y(t)) is a generic point ofF . For anyK-monomorphismσ of K[[t]], the pointP ′ = (x′(t), y′(t)) with x′(t) = σ(x(t))andy′(t) = σ(y(t)) is also a variable point ofF . By Theorem 4.4, eachτ ∈ K[[t]]defines aK-monomorphism ofK[[t]]. SinceK[[t]] has infinitely many suchK-monomorphisms, each of them defines a generic point ofF . 2

DEFINITION 5.9 In the case thatP = (x, y) is a generic point of an irreduciblecurveF = v(F (X,Y )), thenΣ = K(x, y) is thefield of rational functionsor,more briefly, thefunction fieldof F .Note that, by Theorem 5.7, this field is uniquely determined up toK-isomorphism.Also, its transcendency degree overK is 1. In other words, the following theoremholds.

THEOREM 5.10 For any two elementsξ, η ∈ Σ, there exists an irreducible poly-nomial g(X,Y ) ∈ K[X,Y ] such thatg(ξ, η) = 0; that is, there is a polynomialrelation between any two elements inΣ.

An immediate corollary is the following result.

COROLLARY 5.11 Every pointP = (ξ, η) of AG(2,Σ) not in AG(2,K) is ageneric point of an irreducible curveF .

5.2 RATIONAL TRANSFORMATIONS

In classical literature on projective geometry, a rationaltransformation is a mapping

X ′ =f(X,Y )

h(X,Y ), Y ′ =

g(X,Y )

h(X,Y ),

with f, g, h ∈ K[X,Y ].Sinceh(X,Y ) could vanish at some points, these formulas may not allow the

images of such points to be calculated. Further, if the domain of the transformationis to be non-empty and be contained in a given algebraic curveF = v(F (X,Y )),then it is necessary thath(X,Y ) 6≡ 0 (mod F (X,Y )). In this case,h(a, b) iszero only at a finite number of points ofF . Then, for anyP = (a, b) ∈ F andh(a, b) 6= 0, the following transformation can be defined:

T(a, b) =

(f(a, b)

h(a, b),g(a, b)

h(a, b)

).

In particular,T is defined at every generic point ofF and transforms such a genericpointP = (x, y) into a pointP ′ = (x′, y′) of AG(2,Σ). Now the proper definitioncan be given.

DEFINITION 5.12 A rational transformationT of an algebraic curveF is givenby a generic pointP = (x, y) of F and a pointP ′ = (x′, y′), x′, y′ ∈ K(x, y).Then,

x′ =f(x, y)

h(x, y), y′ =

g(x, y)

h(x, y),

116

with f, g, h ∈ K[X,Y ] andh(x, y) not vanishing identically onF . From this, foreachP = (a, b) ∈ F such thath(a, b) 6= 0,

T(a, b) =

(f(a, b)

h(a, b),g(a, b)

h(a, b)

).

REMARK 5.13 The expressions forx′, y′ as rational functions ofx, y are notuniquely determined; butT(a, b) does not depend on the particular functions cho-sen.

Proof. Suppose that

x′ =f1(x, y)

h1(x, y), y′ =

g1(x, y)

h1(x, y),

and

T′(a, b) =

(f1(a, b)

h1(a, b),g1(a, b)

h1(a, b)

),

with h1(a, b) 6= 0. Then it must be shown thatT′(a, b) = T(a, b); that is, if

f(x, y)

h(x, y)=f1(x, y)

h1(x, y),

thenf(a, b)

h(a, b)=f1(a, b)

h1(a, b).

Thus, the mapping is well-defined and does not depend on the representative func-tions. From the hypothesis that

f(x, y)

h(x, y)=f1(x, y)

h1(x, y),

it follows that

f(x, y)h1(x, y)− f1(x, y)h(x, y) = 0.

So, the polynomial

G(X,Y ) = f(X,Y )h1(X,Y )− f1(X,Y )h(X,Y )

vanishes at the generic pointP = (x, y) of F . Hence,G(a, b) = 0 for every pointQ = (a, b) ∈ F , whence

f(a, b)h1(a, b)− f1(a, b)h(a, b) = 0.

Sinceh(a, b)h1(a, b) 6= 0, it may be concluded that

f(a, b)/h(a, b) = f1(a, b)/h1(a, b).

2

The mappingT is defined atP = (a, b) if, for some expression inh(x′, y′),the valueh(a, b) 6= 0. If there exists no such expression, thenT is not definedatP = (a, b).


EXAMPLE 5.14 Let C = v(Y −X2) be a conic andP = (x, y) one of its genericpoints. LetT be defined by the generic pointP ′ = (x′, y′) where

x′ = x, y′ = y/x.

ThenT((0, 0)) is defined as, although0/0 is not defined,y/x = x. In fact,x′, y′

can be written in the formx′ = y, y′ = x, whenceT((0, 0)) = 0. On the otherhand, consider the transformationT determined byx′ = y, y′ = x/y. The expres-sionx/y cannot be rewritten in the formg(x, y)/h(x, y) with h(0, 0) 6= 0. In fact,sinceh(x, y)x− g(x, y)y = 0, so

h(X,Y )X − g(X,Y )Y = d(X,Y )(Y −X2)

with d(X,Y ) ∈ K[X,Y ], which is impossible, as the terms of least degree inh(X,Y )X − g(X,Y )Y contain a termcX with c ∈ K, c 6= 0, which cannot bethe case in the right-hand side.

In terms of homogeneous coordinates, arational transformationT of F is de-fined by a generic pointP = (x0, x1, x2) of F and a pointP ′ = (x′0, x

′1, x

′2) such

thatK(P ) ⊃ K(P ′). Then

x′0 : x′1 : x′2 = f0(x0, x1, x2) : f1(x0, x1, x2) : f2(x0, x1, x2),

with f0, f1, f2 homogeneous polynomials of the same degree. IfQ = (a0, a1, a2)is a point ofF , and f0, f1, f2 do not vanish atQ, thenT(Q) is defined to be(f0(Q), f1(Q), f2(Q)). The triple(f0, f1, f2) is not unique. But, if

f0(x0, x1, x2) : f1(x0, x1, x2) : f2(x0, x1, x2)

= g0(x0, x1, x2) : g1(x0, x1, x2) : g2(x0, x1, x2),

where thegi are also homogeneous of the same degree, and ifg0, g1, g2 do notall vanish atQ, then(f0(Q), f1(Q), f2(Q)) and(g0(Q), g1(Q), g2(Q)) define thesame pointQ′ = T(Q). In this way,T is defined at all but a finite number of pointsof F .

EXAMPLE 5.15 In Example 5.14 in whichC = v(Y −X2) andT is given by theequationsx′ = y, y′ = x/y, the mapping is not defined at the pointP = (0, 0). InPG(2,K), however,

(1, x′, y′) = (y, y2, x) = (x2, yx2, x) = (x, yx, 1)

= (f0(1, x, y), f1(1, x, y), f2(1, x, y)),

where

f0 = X0X1, f1 = X1X2, f2 = X20 .

Then, withP = (1, 0, 0),

T(P ) = P ′ = (0, 0, 1).

In other words,T is not defined atP viewed as a point ofAG(2,K), but it isdefined atP when that point is regarded as a point inPG(2,K). For an example ofa rational transformation in the projective plane not everywhere defined on a givencurve, consider the curveC′ = v(Y 2 −X2 −X3) and the transformationT givenby x′ = x, y′ = y/x. It can be shown thatT is not defined atP = (1, 0, 0).

118

Returning to the affine case, two cases occur forx′, y′ ∈ K(x, y).

(1) If x′, y′ ∈ K, then all points ofF are transformed into the same point,namelyP ′ = (x′, y′). This is thetrivial case.

(2) If at least one ofx′, y′ is not inK, then the field extensionK(x′, y′)/Khas still transcendency degree1. HenceP ′ = (x′, y′) is a generic point ofan irreducible curveF ′, theimage curveof F underT. This is thenon-trivial case, and it is usually tacitly assumed that this is the case.

These definitions give the following two results.

THEOREM 5.16 All transformed points ofF underT are on the image curve.

THEOREM 5.17 A rational transformationS of F is always defined on genericpoints ofF and the image of a generic point ofF is a generic point of the imagecurveF ′.

As a special case, take a branch representation ofF as a generic point.

THEOREM 5.18 The image of a branch representation of a curveF under a rat-ional transformation is a branch representation of the image curveF ′.

The image of a primitive branch representation need not be primitive. Still, equiva-lent branch representations yield equivalent branch representations, so that a givenbranchγ of F gives rise (underT) to a definite branchγ′ of F ′. The branchγ′ isthe imageof γ underT.

Let T be given by a generic pointP ′ = (x′, y′) of F , and define a mapT∗ asfollows: if P = (a, b) is a point ofF andγ is a branch ofF centred atP (a, b), setT∗(P ) = P ′ whereP ′ is the centre of the imageγ′ of γ underT. HereT∗ neednot be aunivalentmap, asF may have several branches centred atP . ThusT∗(P )should be defined to be the set of all centres of the branches ofF ′ arising from thebranches ofF centred atP . It will also not be univalent, in general. The centre ofthe transformed branchγ′ of γ need not be at finite distance. However, from thediscussion on the projective plane,T∗ will be defined at every point ofF , sincethere is at least one branch ofF centred at any point ofF .

Let P = (a0, a1, a2) be a point ofF and assume thatT is defined atP . Then

1 : x′ : y′ = f0(1, x, y) : f2(1, x, y) : f2(1, x, y),

wheref1, f2, f3 are homogeneous of the same degree andfi(a0, a1, a2) 6= 0 forsomei. Let

x0(t) = a0 + ty0(t), x1(t) = a1 + ty1(t), x2(t) = a2 + ty2(t)

be a special representation of a branch ofF . Putt = 0 in fi(x0(t), x1(t), x2(t));this shows thatγ′ is centred at the point

P ′ = (f0(a0, a1, a2), f1(a0, a1, a2), f2(a0, a1, a2)).

In other words,T∗ extendsT, and it has the advantage of being defined everywhereonF . Nevertheless,T cannot be defined at a point whereT∗ is multivalued.


DEFINITION 5.19 A rational transformationS : F → F ′, given by the genericpointsP = (x, y) ∈ F andP ′ = (x′, y′) ∈ F ′, is birational if

K(x′, y′) = K(x, y).

In this case, the generic pointP ′ = (x′, y′) of F ′ together with the pair of elements(x, y) yield a rational transformationS : F ′ → F .

THEOREM 5.20 If a birational transformationS is defined at the pointP andS

is defined at the pointS(P ), thenS(S(P )) = P.

Proof. Let P = (a, b) and letr, s, t ∈ K[X,Y ] such that

S : x′ =r(x, y)

t(x, y), y′ =

s(x, y)

t(x, y),

with t(a, b) 6= 0. ForP ′ = (a′, b′) = S(P ), let r′, s′, t′ ∈ K[X,Y ] be such that

S : x =r′(x′, y′)

t′(x′, y′), y =

s′(x′, y′)

t′(x′, y′),

with t′(a′, b′) 6= 0. Now,

x=r′(r(x, y)/t(x, y), s(x, y)/t(x, y))

t′(r(x, y)/t(x, y), s(x, y)/t(x, y)),

y=s′(r(x, y)/t(x, y), s(x, y)/t(x, y))

t′(r(x, y)/t(x, y), s(x, y)/t(x, y)).

There exists an integerh ≥ 1 such that

w(X,Y ) = (t(X,Y ))h

[Xt′

(r(X,Y )

t(X,Y ),s(X,Y )

t(X,Y )

)− r′

(r(X,Y )

t(X,Y ),s(X,Y )

t(X,Y )

)]

is a polynomial with coefficients inK. It follows simply thatw(x, y) = 0, andso the pointP = (x, y) is generic point ofF . Hencew(a, b) = 0 for each pointQ = (a, b) ∈ F . Sincet(a, b) 6= 0, t′(a′, b′) 6= 0,

a =r′(r(a, b)/t(a, b), s(a, b)/t(a, b))

t′(r(a, b)/t(a, b), s(a, b)/t(a, b))

and similarly

b =s′(r(a, b)/t(a, b), s(a, b)/t(a, b))

t′(r(a, b)/t(a, b), s(a, b)/t(a, b)).

Therefore

a =r′(a′, b′)

t′(a′, b′), b =

s′(a′, b′)

t′(a′, b′)

andS(a′, b′) = (a, b). However,(a′, b′) = S(P ) and(a, b) = P , from which itfollows thatS(S(P )) = P, as required. 2

The rational transformationS is theinverseof S and so is writtenS = S−1.SinceS is defined at every generic pointP ofF and sinceS(P ) is generic point

of F ′, the following corollary is immediate.

120

COROLLARY 5.21 A birational transformationS of a curveF defines a bijectivecorrespondence between the generic points ofF and those of the image curveF ′.

DEFINITION 5.22 Let Σ be the function field of a curve; equivalently, letK havean extensionΣ = K(x, y) of transcendence degree 1 overK. A model ofΣ isgiven by a pointP = (ξ, η) such thatΣ = K(ξ, η) and by the curveF havingPas generic point; in symbols, write(F ; (ξ, η)).

It should be noted that two models(F ; (x, y)) and(F ′; (x′, y′)) are distinct if(x, y) 6= (x′, y′). It may, however, happen thatF = F ′ but (x, y) 6= (x′, y′). So,a given curveF may give rise to more than one model, consistent with the factthat a curve can have non-trivial automorphisms, that is, can be birational imagesof themselves under a non-identical birational transformation. Automorphisms areinvestigated in Section . Here two examples illustrate thissituation.

EXAMPLE 5.23 Let C andP = (x, y) be as in Example 5.14. IfS is the map

x′ = x+ c,y′ = 2cx+ y + c2,

with c ∈ K, thenS−1 is as follows:

x = x′ − c,y = y′ − 2cx′ − c2.

Hence

y − x2 = (y′ − 2cx′ − c2)− (x′ − c)2 = y′ − x′2.This shows thatS transformsC into itself, asy−x2 = 0 implies thaty′−x′2 = 0.Therefore,(C; (x, y)) is transformed into the model(C; (x′, y′)) by S.

EXAMPLE 5.24 Let q0 = 2e andq = 2q20 . The irreducible plane curve

D = v(X2q0(Xq +X) + yq + Y )

has only one singular point, namelyY∞, which is a2q0-fold point. For a genericpoint P = (x, y) let K(D) = K(x, y) be the function field ofD. Now, puth = xy + x2q0+2 + y2q0 . A straightforward computation shows that the rationaltransformation

S : x′ =y

h, y′ =

x

hleavesD invariant. In particular, the associated mapT onD is defined on all pointsof D apart fromY∞ andO. Further, the branchγ centred atO is transformed toa branch centred atY∞. Also, S has order two as it coincides with its inverse.Therefore,Y∞ is the centre of only one branch, sayγ′, which is the image ofγunderT. Since(x(t), y(t)), with

x(t) = t, y(t) = t2q0 + tq+2q0 + t2qq0 + tq(q+2q0) + . . . ,

is a primitive representation ofγ, so(ξ(t), η(t)), with

ξ(t) =1 + tq + t2q0(q−1) + . . .

t(1 + t+ t2q−2q0−1 + . . .), η(t) =

1

t2q0(1 + t+ t2q−2q0−1 + . . .),

is a primitive representation ofγ′.


If (F ; (ξ, η)) and (F ′; (ξ′, η′)) are two models ofΣ, then Σ = K(ξ, η) andΣ = K(ξ′, η′). HenceK(ξ, η) = K(ξ′, η′) and so, by Definition 5.19, the pointsP = (ξ, η) andP ′ = (ξ′, η′) determine a birational transformation fromF toF ′.

THEOREM 5.25 Any two models of a given fieldΣ are birationally equivalent.

A propertyof the fieldΣ is a birationally invariant property of any one of its models.

Now, consider the irreducible curvesF = v(F (X,Y )) andG = v(G(X,Y )),and take generic pointsP = (x, y) of F andQ = (ξ, η) of G. Assume that theirfunction fieldsΣ = K(x, y) andΣ′ = K(ξ, η) are isomorphic overK. Let

ϕ : Σ→ Σ′

be aK-isomorphism, and putx′ = ϕ(ξ), y′ = ϕ(η). ThenΣ′ = K(x′, y′) andϕF (x, y) = 0 implies F (x′, y′) = 0. Hence(F ; (x′, y′)) is a model ofΣ′, inparticular a birationally equivalent model to the model(G; (ξ, η)).

This discussion motivates the following definition.

DEFINITION 5.26 Two irreducible algebraic curves inPG(2,K) arebirationallyisomorphicif their function fields are isomorphic overK.

5.3 PLACES

Let (F , (x, y)) be a model of a fieldΣ of transcendency degree1. If (ξ(t), η(t)) is anot-necessarily-primitive branch representation ofF , then, By Theorem 5.7, thereis aK-isomorphismσ of Σ intoK((t)) such thatσ(x) = ξ(t), σ(y) = η(t). For anautomorphismρ ofK[[t]], extended toK((t)) in a natural way, putσ′ = ρσ. Thenσ′ is also aK-isomorphism ofΣ intoK((t)) such that the branch representations(σ′(x) = ξ′(t), σ′(y) = η′(t)) and(ξ, η) are equivalent. In a diagram,

K(ξ(t), η(t))

Σ = K(x, y)

K(ξ′(τ), η′(τ)).?

ρ

*σ

HHHHHHjσ′=ρσ

This leads to the following fundamental concepts.

DEFINITION 5.27 (i) A K-isomorphismσ : Σ → K((t)) is aplace represen-tation.

(ii) A place representationσ is primitive if (σ(x), σ(y)) for a pair, and hence forany pair, of generatorsx, y of Σ is a primitive branch representation ofF .

122

(iii) Two place representationsσ andσ′ areequivalentif there is aK-automorph-ismρ of Σ such thatσ′ = ρ σ.

(iv) A placeis an equivalence class of primitive place representations.

(v) The set of all places ofΣ is denoted byP(Σ).

A place is merely the field-theoretic or birational counterpart of the geometric con-cept of a branch. The following two theorems explain this point.

THEOREM 5.28 The places ofΣ and the branches of any model ofΣ are in one-to-one correspondence, in a natural way.

Proof. Let (F ; (x, y)) be a model ofΣ. It has been shown how a primitive branchrepresentation(ξ(t), η(t)) leads to a primitive place representationσ in a naturalway, and also how to go back to a primitive branch representation (ξ(t), η(t)) froma primitive place representationσ by σ(ξ) = ξ(t), σ(η) = η(t). In this way,a branch representation and a place representation give rise to one other. Also,equivalent branch representations correspond to equivalent place representations;so branches and places are in one-to-one correspondence. 2

THEOREM 5.29 Let (F ; (x, y)) and (F ; (x′, y′)) be two models ofΣ, and letSbe the birational transformation sendingP = (x, y) into P ′ = (x′, y′). ThenS

induces a one-to-one correspondence between the branches of F and those ofF ′,two branches corresponding if and only if they correspond tothe same place ofΣ.

Proof. Let the placeP, that is, one of its primitive representations, send(x, y) into(x(t), y(t)) and(x′, y′) into (x′(t), y′(t)). In a diagram,

(x, y) (x(t), y(t))

(x′, y′) (x′(t), y′(t)).

-P

?τ

ppppppppp?

-P

There existr(X,Y ), s(X,Y ) ∈ K[X,Y ] such thatx′ = r(x, y)/s(x, y). Apply-ing P givesx′(t) = r(x(t), y(t))/s(x(t), y(t)), and similarly fory′. This showsthatS(P (x(t), y(t))) = (x′(t), y′(t)) justifying the fourth arrow in the diagram,and proving one part of the theorem. A similar argument givesthe converse. 2

5.4 ZEROS AND POLES

Let F = v(F (X,Y )) be an irreducible curve with a generic pointP = (x, y).For a placeP of the field Σ = K(x, y), let τ : Σ → K((t)) be a primitive

representation ofP. If ξ is an element ofΣ, then ordt τ(ξ) does not depend on thechoice ofτ . To show this, take another primitive representationτ ′ : Σ→ K((t)) ofP. Thenτ ′ = ρτ with an automorphismρ ofK[[t]] naturally extended toK((t)).


Hence ordt τ ′(ξ) = ordt ρ(τ(ξ)) = ordt τ(xi) by Theorem 4.5. This leads to thefollowing concept.

DEFINITION 5.30 Let τ be any primitive representation of a placeP of Σ.

(i) Theorder ofξ atP is ordP ξ = ordt τ(ξ).

(ii) If ordP ξ = 1, thenξ is auniformizing elementor local parameteratP.

Let γ be the branch ofF associated to the placeP. Thecentreof P is the centreof γ. Since, by Theorem 5.28, there is a bijection between the branches ofF andthe places ofΣ, and by Theorem 4.44 there is a only a finite number of branchescentred at the same point, there is a only a finite number of places ofΣ with thesame centre. Put

ordγ ξ = ordP ξ.

In particular, ordγ x = ordt x(t), ord γ y = ordt y(t) which can be thought of asalternative notation for ordt x(t) and ordt y(t). From part (iii) of Theorem 4.20 thefollowing result appears.

L EMMA 5.31 The function fieldΣ has a uniformizing element at every place.

DEFINITION 5.32 (i) The placeP is azeroof ξ if ordP ξ > 0; if P is a zero ofξ, it is a zero ofmultiplicity ordP ξ.

(ii) The placeP is apoleof ξ if ordP ξ < 0; if P is a pole ofξ, it is a pole ofmultiplicity −ordP ξ.

(iii) If the placeP is not a pole ofξ, thenξ is regularatP.

Henceforth,ζ denotes a function ofΣ, that is, an element ofΣ\K.

THEOREM 5.33 The functionζ has only finitely many zeros. The number of zeroscounted with their multiplicity is[Σ : K(ζ)].

Proof. Sinceζ is transcendental overK, thenΣ is a finite algebraic extension ofK(ζ) and[Σ : K(ζ)] is an integer. IfL is a finite algebraic extension of a fieldM ,thenL is a simple extension ofM . By the Theorem of Primitive Element, thereexistsη ∈ Σ such thatΣ = K(ζ, η). Write ζ = x, η = y; in other words, considerζ as the first coordinate of a generic point of a modelF of Σ.

Let

F (X,Y ) = c0(X)Y m + c1(X)Y m−1 + · · ·+ cm(X)

be an irreducible polynomial with coefficients inK of which (x, y) is a zero. Then

c0(X)m−1F (X,Y ) = (c0(X)Y )m + c1(X)(c0(X)Y )m−1 +

· · ·+ cm(X)c0(X)m−1

=Zm + c1(X)Zm−1 + · · · ,

124

having putc0(X)Y = Z. For this polynomial inK[X,Z], the pair(x, c0(x)y) is azero. Sincec0(x)y is a generator ofΣ overK(x), thenc0(x)y may be substitutedfor y; in other words, it may be supposed thatc0(X) = 1. Then[Σ : K(x)] = m.

Let (F ; (x, y)) be a model ofΣ = K(x, y) with (x, y) as a generic point. Thenthe branchesγ of F such that ordγ x > 0 are to be counted each with its order.These branches have centres on theY -axis, buta priori possibly atY∞. Thiscan happen but not whenc0(X) = 1. To check it, assumeγ is a branch centredat Y∞. Note that if (x(t), y(t)) is a primitive branch representation ofγ, thenordγ y = ordy(t) < 0. So

ordγ (ym + c1(x)ym−1 + · · · ) = ordγ y

m < 0,

which contradicts that

ym + c1(x)ym−1 + · · · = 0, ordγ 0 =∞.

Let γ1, γ2, . . . , γs be the branches for which ordγ x > 0; they are precisely thebranches centred at points(0, b, 1) on they-axis. Consider now such a pointPj onF . Let γj1, . . . , γjk be the branches ofF centred atPj . Then

I(Pj ,v(X) ∩ F) = ordγj1x+ . . .+ ordγjk

x.

Summing for all such pointsPj , it follows that∑I(Pj ,v(X) ∩ F) =

∑ordt x(t);

the right-hand side counts the zeros ofx with appropriate multiplicity. To finishthe proof, it suffices to note that, ifPj = (0, yj , 1) andyj is anmj-ple root ofthe polynomialY m + c0(0)Y m−1 + . . ., thenI(Pj ,v(X) ∩ F) = mj ; hence∑

jI(Pj ,v(X) ∩ F) = m. 2

THEOREM 5.34 The functionζ has finitely many poles. The number of poles eachcounted with its multiplicity is[Σ : K(ζ)].

Proof. Poles ofζ are zeros of1/ζ,. Hence, there are only a finite number of them,and this number, counting with multiplicity, is[Σ : K(1/ζ)] = [Σ : K(ζ)]. 2

COROLLARY 5.35 The number of zeros ofζ equals the number of its poles. Equiv-alently, ∑

P∈P(Σ)ordP ζ = 0,

where the summation extends over all placesP.

5.5 SEPARABILITY AND INSEPARABILITY

Any field of transcendency degree1 overK can be obtained by a transcenden-tal extensionK(x)/K followed by an algebraic extensionΣ/K(x). In positivecharacteristicp, the latter extensionΣ/K(x) can be split into a separable stepΣs/K(x) and a purely inseparable stepΣ/Σs. Separable extensions have sim-ilar properties to algebraic extensions in characteristiczero. Purely inseparableextensions are related to certain subfields ofΣ containingK, namely, the subfieldsΣq = ξq | ξ ∈ Σ, whereq denotes a power ofp.


THEOREM 5.36 For everyq, the extensionΣ/Σq is purely inseparable of degreeq. Conversely, if K ⊂ Σ′ ⊂ Σ andΣ/Σ′ is purely inseparable of degreeq, thenΣ′ = Σq.

Proof. SinceK is perfect, it follows thatΣq is a field that containsK. The exten-sionΣ/Σq is inseparable, asξq ∈ Σq for everyξ ∈ Σ. To calculate the degree ofΣ/Σq, choose a local parameterx ∈ Σ at some placeP of Σ.

By the Theorem of the Primitive Element, there existsy ∈ Σ such thatΣ =K(x, y). Actually Σ = K(x, yq) holds. In fact,Σ = K(x, yq)(y)/K(x, yq) isa purely inseparable extension sincey satisfies the equationXq − yq = 0 overK(x, yq). On the other hand,K(x) ⊂ K(x, yq) ⊂ Σ, and thereforeΣ/K(x, yq)is also a separable extension. This is only possible whenΣ = K(x, yq).

Now, Σq = K(xq, yq) andΣ = K(x, yq) imply thatΣ = Σq(x). Sincex is aroot of the polynomialXq−xq with coefficients inΣq, it follows that[Σ : Σq] ≤ q.

To prove the reverse inequality, letaq0X

m + . . .+ aqm be a minimal polynomial

of the extensionΣ = Σq(x)/Σq. It is enough to show thatm ≥ q. Assume thatordP aq

ixi is minimal; that is, ordP aq

ixi ≤ ordP aq

jxj for everyj = 0, . . . ,m,

Thenaq0x

m + . . . + aqm = 0 yields that ordP aq

ixi = ordP aq

kxk for somek, with

0 ≤ k ≤ m and i 6= k. Since ordP x = 1, it follows that i ≡ k (mod q).Therefore,m ≥ q.

For the converse, assume that the extensionΣ/Σ′ is a purely separable of degreeq. Thenξq ∈ Σ′ for everyξ ∈ Σ. HenceΣq ⊂ Σ′ ⊂ Σ. Since the degree ofΣ/Σq

is q, so it follows thatΣq = Σ′. 2

Let Σ′ be a subfield ofΣ which containsK and has transcendency degree1 overK. The extensionΣ/Σ′ is finite. Also, there is a unique intermediate fieldΣ′′,with Σ′ ⊂ Σ′′ ⊂ Σ such thatΣ′′/Σ′ is separable andΣ/Σ′′ is purely inseparable.Then,[Σ : Σ′] = [Σ : Σ′′][Σ′′ : Σ′], where[Σ′′ : Σ′] is theseparable degreeand[Σ : Σ′′] is theinseparable degreeof the extension[Σ : Σ′]. An elementξ ∈ Σ isaseparableor inseparable variableof Σ according as the finite extensionΣ/K(ξ)is separable or inseparable.

A separable variableζ of Σ is characterised in the following manner. For anyelementξ ∈ Σ\K, the irreducible polynomialf(X,Y ) ∈ K[X,Y ] linking ξ andζ, that isf(ξ, ζ) = 0, has the property that the polynomialf(X, ζ) ∈ K(ζ)[X]is separable; that is, its derivativedf(X, ζ)/dX is not identically zero. Note thatdf(X, ζ)/dX = ∂f(X,Y )/∂X with Y = ζ.

A useful criterion for separability inΣ is given in the following lemma.

L EMMA 5.37 An elementζ ∈ Σ is a separable variable if and only ifζ 6∈ Σp, thatis, ζ is not ap-th power of an element inΣ.

Proof. If ζ is a separable variable,K(ζ) 6⊂ Σp, becauseΣ/Σp is purely inseparableof degree greater than1. Conversely, ifζ ∈ Σ\K is not separable, there is anintermediate fieldΣ′ with K(ζ) ⊂ Σ′ ⊂ Σ such thatΣ/Σ′ is purely inseparable ofdegreep. By Theorem 5.36,Σ′ = Σp, whence it follows thatζ ∈ Σp. 2

126

5.6 FROBENIUS RATIONAL TRANSFORMATIONS

In this section,K has positive characteristicp, andq denotes a power ofp. Also,if g(X,Y ) ∈ K[X,Y ] andg(X,Y ) =

∑cijX

iY j , thengq(X,Y ) denotes thepolynomial

∑cqijX

iY j . The previous notation is maintained:F = v(f(X,Y ))is an irreducible curve;P = (x, y) is a generic point ofF ; the fieldΣ = K(x, y)is the associated function field;ω is a rational transformation ofΣ; the subfieldω(Σ) = ω(ξ) | ξ ∈ Σ is the image ofΣ underω.

DEFINITION 5.38 The rational transformation,

ωq : x′ = xq, y′ = yq,

of Σ is theq-th Frobeniustransformation.

Let f(X,Y ) =∑aijX

iY j . Note thatfq(X,Y ) is also irreducible. Also, theimage curve ofF underωq is F ′ = v(fq(X,Y )). In fact, asx′ = xq, y′ = yq,sofq(x′, y′) =

∑aq

ijx′iy′j = (

∑aijx

iyj)q = 0. In particular,P ′ = (x′, y′) isa generic point ofF ′, and the subfieldΣ′ = K(x′, y′) of Σ can be viewed as thefunction field associated withF ′.

THEOREM 5.39 The mappingωq is not birational. However, Σ ∼= Σ′.

Proof. If ωq were birational, then bothx andy, and hence every element inΣ,would be aq-th power. But this is impossible. Now, it is shown that the mapping

ρ : Σ −→ Σ′,

U(x, y)

V (x, y)7−→ Uq(x′, y′)

V q(x′, y′),

with U(X,Y ), V (X,Y ) ∈ K[X,Y ], V (x, y) 6= 0, is well-defined and does notdepend on the representative functions. To do this, chooseU1, V1 ∈ K[X,Y ] suchthat

U1(x, y)

V1(x, y)=U(x, y)

V (x, y).

ThenU(x, y)V1(x, y)− U1(x, y)V (x, y) = 0. AsP = (x, y) is a generic point ofF , there existsh(X,Y ) ∈ K[X,Y ] such that

f(X,Y )h(X,Y ) = U(X,Y )V1(X,Y )− U1(X,Y )V (X,Y ).

It follows that

fq(X,Y )hq(X,Y ) = Uq(X,Y )V q1 (X,Y )− Uq

1 (X,Y )V q(X,Y ).

As fq(x,′ y′) = 0, this shows that

U(x′, y′)V1(x′, y′)− U1(x

′, y′)V (x′, y′) = 0,

whenceUq

1 (x′, y′)

V q1 (x′, y′)

=Uq(x′, y′)

V q(x′, y′).

Since the mappingρ induces an automorphism ofK, so ρ is a homomorphismfrom Σ to Σ′. Actually, ρ is injective, asξq = 0 impliesξ = 0, and surjective as∑bijx

′iy′j is the image of∑aijx

iyj whenaqij = bij for everyi andj. Therefore,

ρ is an isomorphism. 2


DEFINITION 5.40 A rational transformationω of Σ is eitherseparable, or insep-arableaccording as the field extensionΣ/ω(Σ) is separable or inseparable.

THEOREM 5.41 (i) Every inseparable rational transformationω of Σ is theproduct of a separable rational transformation by aq-th Frobenius transfor-mation.

(ii) The inseparability degree ofω is equal toq.

Proof. Let ω be given by the generic pointsP = (x, y) of F andP ′ = (x′, y′)of F ′. If eitherx or y is not ap-th power,Σ/ω(Σ) is separable. Otherwise, letqdenote the highest power ofp such thatx′ = ξq andy′ = ηq. By Theorem 5.36,ω(Σ) is purely inseparable and its inseparability degree is equal to q. Further, therational transformationωs given by the generic point(ξ, η) is separable. Henceωis the product ofωs and theq-th Frobenius transformation. 2

5.7 DERIVATIONS AND DIFFERENTIALS

DEFINITION 5.42 A derivationD in a fieldL is a mappingD : L → L given bya 7→ a′ = D(a) such that

(i) (a+ b)′ = a′ + b′;

(ii) (ab)′ = a′b+ ab′,

for all a, b ∈ L.

L EMMA 5.43 (i) 0′ = 0;

(ii) 1′ = 0;

(iii) (a/b)′ = (a′b− ab′)/b2 for all a, b ∈ L with b 6= 0.

Proof. (i) Here,0′ = (0 + 0)′ = 0′ + 0′, whence0′ = 0′ − 0′ = 0.(ii) In this case,1′ = (1 · 1)′ = 1′ · 1+1 · 1′ = 1′ +1′, whence1′ = 1′− 1′ = 0.(iii) Finally, a′ = (b · a/b)′ = b′ · a/b+ b · (a/b)′. So

(a/b)′ = (a′ − b′ · a/b)/b = a′/b− b′ · a/b2 = (a′b− ab′)/b2. 2

DEFINITION 5.44 Given a subfieldM of L, a derivationD of L is said to beoverM if D(c) = 0 for eachc ∈M.

LetD be a derivation ofL overM and letθ, φ ∈ L; then

D(θiφj) = iθi−1φjDθ + jθiφj−1Dφ.

So

D(F (θ, φ)) =∂F

∂θDθ +

∂F

∂φDφ,

128

whereF ∈M [X,Y ]. Here,∂F/∂θ means∂F/∂X evaluated forX = θ, Y = φ.Consider now the particular case in whichL = K((t)). InK((t)) a derivation is

given by the following rule:d

dt

∑cit

i =∑icit

i−1,

with ci ∈ K. Takedt as an indeterminate and consider the fieldK((t))(dt). Ifθ ∈ K((t)), define thedifferentialof θ in the following way:

dθ =

(dθ

dt

)dt.

Let φ ∈ K((t))\K and, if the characteristicp of K is positive, suppose also thatdφ/dt 6= 0; that is,φ 6∈ K((tp)). Define, in the fieldK((t))(dt),

dθ

dφ

asdθ divided bydφ. From the rule thatd

dt(θ1θ2) = θ1

dθ2dt

+ θ2dθ1dt,

multiplication bydt gives

d(θ1θ2) = θ1dθ2 + θ2dθ1

and similarly

d(θ1 + θ2) = dθ1 + dθ2.

Dividing by dφ givesd(θ1θ2)

dφ= θ1

dθ2dφ

+ θ2dθ1dφ

,

d(θ1 + θ2)

dφ=dθ1dφ

+dθ2dφ

.

Therefore

θ 7→ dθ

dφ

is a derivation ofK((t)), which coincides withd/dt for φ = t. Such a derivationis denoted byd/dφ.

Asd

dtF (θ, φ) =

∂F

∂θ

dθ

dt+∂F

∂φ

dφ

dt,

multiplying bydt gives[d

dtF (θ, φ)

]dt =

∂F

∂θ

dθ

dtdt+

∂F

∂φ

dφ

dtdt,

and hence

dF (θ, φ) =∂F

∂θdθ +

∂F

∂φdφ,

whereF is a rational functionf/g with f, g ∈ K[X,Y ] andg(θ, φ) 6= 0.To deal withΣ, a characterisation for separability in terms of derivatives is

needed.


THEOREM 5.45 The elementζ ∈ Σ is a separable variable if and only ifdζ 6= 0for each placeP of Σ.

Proof. By the Theorem of the Primitive Element, there existsξ ∈ Σ such thatΣ = K(ξ, ζ). Note that for any placeP, eitherdξ or dζ does not vanish. In fact,for a primitive place representationτ of P, the equalitydτ(ξ) = 0 occurs if andonly if τ(ξ) ∈ K((t)) has no termciti with i 6≡ 0 (mod p). If dτ(ζ) = 0 alsohappens then neitherτ(ζ) ∈ K((t)) has a termciti with i 6≡ 0 (mod p). Then(τ(ξ), τ(ζ)) is an imprimitive representation of the branch of the model(F ; (ξ, ζ)).But this would mean thatτ is an imprimitive place representation.

Let U(X,Y ) ∈ K[X,Y ] be an irreducible polynomial for whichU(ξ, ζ) = 0,andF = v(U) the corresponding irreducible curve havingP = (ξ, ζ) as a genericpoint. For any placeP of Σ, the equationU(ξ, ζ) = 0 holds; hence, by derivation,

(∂U/∂ξ)dξ + (∂U/∂ζ)dζ = 0.

Assume the existence of a placeP ∈ P(Σ) such thatdζ = 0. Thenζ = τ(ζ)for a primitive representationτ of P, and so

(∂U/∂ξ)dξ = 0.

Sincedξ 6= 0, it follows that

∂U/∂ξ = 0.

Therefore, either∂U(X,Y )/∂X = 0 andζ is inseparable or∂U(X,Y )∂X 6= 0andP = (ξ, ζ) is also a generic point of the curvev(∂U(X,Y )∂X). The lattercase cannot actually occur, asdeg ∂U(X,Y )∂X < degU(X,Y ), whileU(X,Y )is irreducible.

Conversely, supposeζ is inseparable; then there existξ in Σ andU(X,Y ) inK[X,Y ] such that

U(ξ, ζ) = 0, ∂U(X, ζ)/∂X = 0.

Therefore, for each placeP of Σ,

U(ξ, ζ) = 0, ∂U(ξ, ζ)/∂ξ = 0.

It follows that also

(∂U(ξ, ζ)/∂ζ)dζ = 0.

Here,

∂U(ξ, ζ)/∂ζ 6= 0

by the irreducibility ofU(X,Y ). Sodζ = 0, as required. 2

Given a placeP of Σ, let τ : Σ → K((t)) be a primitive representation ofP.Having fixed a separable variableζ of Σ, let ξ be any element ofΣ. Then thereexists a polynomialf(X,Y ) ∈ K[X,Y ] such thatf(ξ, ζ) = 0. it follows thatf(ξ, ζ) = 0, where, with the usual conventions,ξ(t) = τ(ξ) and ζ(t) = τ(ζ).Thereforedf(ξ, ζ) = 0, and so

∂f

∂ξdξ +

∂f

∂ζdζ = 0.

130

Note that∂f/∂ξ 6= 0. If this were not true, then∂f/∂X = 0 is satisfied by(ξ, ζ);but this implies that the polynomial∂f/∂X is identically zero, simply from the factthat such a polynomial, if not zero, would have degree less than that off(X,Y ),contradicting the irreducibility off(X,Y ). Since∂f/∂X = 0, so∂f/∂ξ = 0;but this is impossible withζ a separable variable ofΣ. By the previous theorem,dζ 6= 0.

Therefore it makes sense to write

dξ

dζ= −∂f/∂ζ

∂f/∂ξ∈ K((t)), (5.1)

and to make the following definition.

DEFINITION 5.46dξ

dζ= −∂f/∂ζ

∂f/∂ξ∈ Σ. (5.2)

In a diagram,

ξ dξ/dζ

ξ dξ/dζ

-d/dζ

?τ

?τ

-d/dζ

The mappingξ 7→ dξ/dζ is a derivation ofΣ, sinceξ 7→ dξ/dζ ∈ K((t)) and themappingτ : Σ→ K((t)) is aK-isomorphism.

Now, the idea of the differential of a separable variable ofΣ is introduced in asimilar way to the analogous idea forK((t)).

DEFINITION 5.47 (i) Let ζ be a separable variable ofΣ. A differential is anyelementxdζ, with x ∈ Σ, of the fieldΣ(dζ), a transcendental extension ofΣ by the symboldζ.

(ii) The differentialdξ of ξ ∈ Σ is defined by

dξ =dξ

dζdζ. (5.3)

The differentials with respect to the fixed elementζ form a vector space of dimen-sion1 overΣ, called thevector space of exact differentials.

It should be noted that the differential does not depend on the choice ofζ in thefollowing sense.

If ζ1 is another separable variable ofΣ, thendζ1/dζ 6= 0, and if one identifiesthe “new” indeterminatedζ1 with (dζ1/dζ)dζ, thendξ as defined throughdζ1 canbe considered as the same asdξ defined throughdζ. If this is done, then

dξ

dζdζ =

dξ

dζ1dζ1,

by the following theorem.


THEOREM 5.48 If ζ, ζ1 ∈ Σ are both separable, then, for eachξ ∈ Σ,

dξ

dζ1· dζ1dζ

=dξ

dζ. (5.4)

Proof. LetP be a place ofΣ. If σ is a primitive representation ofP, then

σ

(dξ

dζ1· dζ1dζ

)= σ

(dξ

dζ1

)· σ(dζ1dζ

)=

dξ

dζ1· dζ1dζ

=dξ

dζ.

Sinceσ is aK-isomorphism fromΣ intoK((t)), the result is proved. 2

The above definition of differential is functional but awkward in that it singlesout a separable variableζ for special role. An alternative introduction of the con-cept of the vector space of exact differentials which avoidssuch an inconvenienceis possible. To do this, one introduces for everyζ ∈ Σ, including inseparablevariables and elements ofK, a symboldζ and considers the (infinite-dimensional)vector space overΣ whose vectors are the formal finite sumsα1dζ1 + α2dζ1 + . . .with ζ1, ζ2, . . . ∈ Σ andα1, α2, . . . ∈ K; then one subjects these symbols to theequations

∂F

∂ζ1dζ1 +

∂F

∂ζ2dζ2 + . . . = 0,

and no others, whereF (X1,X2, . . .) is a rational function overK satisfied byζ1, ζ2, . . .. This (one-dimensional) vector space overΣ is the desired vector spaceof exact differentials.

Familiar results on derivatives, such as the first assertionin the following theo-rem, hold true in zero characteristic. Extensions to positive characteristic are alsopossible but inseparability needs to be considered carefully.

THEOREM 5.49 (i) If K has zero characteristic0, then dξ = 0 is only forξ ∈ K.

(ii) If K has positive characteristic, thenξ is a separable variable ofΣ if andonly if dξ 6= 0.

Proof. Let f(X,Y ) ∈ K[X,Y ] be irreducible and such thatf(ξ, ζ) = 0. Iff(X,Y ) is written in the usual canonical form,

f(X,Y ) = a00 + a10X + a01Y + . . . aijXiY j + . . . ,

then, withg(X,Y ) = ∂f(X,Y )/∂Y ,

g(X,Y ) = a01 + a11X + 2a02Y + . . .+ jaijXiY j−1 + . . . .

Suppose thatdξ/dζ = 0. From (5.2) it follows that(ξ, ζ) satisfies the equationg(X,Y ) = 0. SinceP = (ξ, ζ) is a generic point of the curvev(f(X,Y )), thatcan happen only wheng(X,Y ) is the zero polynomial.

Then, in zero characteristic,

f(X,Y ) = f(X, 0) = a00 + a10X + . . .+ an0Xn,

132

whencef(ξ, 0) = 0, which implies thatξ ∈ K. Instead, in positive characteristicp,

f(X,Y ) =∑aikX

iY pk.

Since∂f(ξ, Y )/∂Y = 0, soξ is inseparable. Finally, asdx = 0 for any inseparablevariable inΣ, so (ii) holds.

2

By (5.2),dξ/dζ is an element inΣ. Hence, it makes sense to consider the secondderivative ofξ, that is,

d2ξ

dζ2=d(dξ/dζ)

dζ,

and iteratively the higher derivatives ofξ:

di+1ξ

dζi+1=d(diξ/dζi)

dζ.

Hence

dn(ξη)

dζn=∑n

j=0

(n

j

)djξ

dζj· d

n−jη

dζn−j, (5.5)

which is essentially a consequence of the product rule:

d(ξη)

dζ=dξ

dζη +

dη

dζξ.

5.8 THE GENUS OF A CURVE

Given a placeP of Σ and a primitive representationτ, if ξ is a separable variableof Σ andτ(ξ) = ξ, the order ofdξ atP is defined as follows.

DEFINITION 5.50

ordP dξ = ordtdξ(t)

dt. (5.6)

This concept is well defined by Theorem 5.45 and does not depend on the choiceof the representativeτ .

DEFINITION 5.51 Let ξ be a separable variable ofΣ.

(i) A placeP with ordP dξ > 0 is azeroof dξ; if P is a zero, then it is azero oforderordP dξ.

(ii) A placeP with ordP dξ < 0 is apole; if P is a pole, then it is apole of order−ordP dξ.

L EMMA 5.52 If x ∈ Σ is separable, thenordP dx = 0 except for a finite numberof places; that is, dx has only a finite number of zeros and poles.


Proof. Let y be an element ofΣ such thatΣ = K(x, y). Consider the irreduciblecurveF = v(F (X,Y )) for whichP = (x, y) is a generic point. Now, it is shownthat the affine simple points ofF with tangentv(X) are finite in number. In fact,such a pointP = (a, b) of F also belongs to the polar curveF ′ of F at the pointQ = (0, 1, 0). Note that, sincex is a separable variable ofΣ, so∂F (X,Y )/∂Y isnot identically zero; hence the polar curveF ′ is v(∂F (X,Y )/∂Y ).

Then the number in question is at mostdegF (degF −1) by Bezout’s Theorem.It remains to calculate ordP dx at the placesP of Σ centred at affine simple pointsP with tangent distinct fromv(X). As before, this includes all places except per-haps for a finite number. IfP = (a, b) is such a point, thenx = a+ct+ · · · , wherec 6= 0; otherwise,F either hasv(X) as tangent atP or P is a singular point. Itfollows thatdx/dt = c, whence ordP dx = 0. This proves the result. 2

Lemma 5.52 shows that the sum∑

P∈P(Σ) ordP dζ is meaningful, since all buta finite number of terms are zero.

A fundamental result that enables the genus of a curve to be defined is the fol-lowing.

THEOREM 5.53 For every two separable variableξ, η of Σ,∑

P∈P(Σ)ordP dξ =∑

P∈P(Σ)ordP dη,

where the summation is over all places ofΣ.

Proof. Sincedξ = (dξ/dζ)dζ, so∑

P∈P(Σ)ordP dξ =∑

P∈P(Σ)ordP(dξ/dζ) +∑

P∈P(Σ)ordP dζ.

Sincedξ/dζ ∈ Σ, it follows from Corollary 5.35 that∑

P∈P(Σ) ordP(dξ/dζ) = 0,and so

∑P∈P(Σ) ordP dξ =

∑P∈P(Σ) ordP dζ, as required. 2

DEFINITION 5.54 Let g be the integer for which∑

P∈P(Σ)ordP dζ = 2g − 2.

Theng, necessarily non-negative, is thegenus ofΣ.

It is shown in the next section that the genus ofΣ is greater than or equal to thevirtual genus of each curveF which is a model ofΣ, and equality holds if and onlyif F has only ordinary singularities . The following theorem provides a formula forordP for a place centred at a simple point.

THEOREM 5.55 LetP be a simple point of an irreducible curveF = v(F (X,Y ))such that the vertical lineℓ throughP is the tangent toF at P, and letP be theunique place ofF centred atP . Given a generic point(x, y) of F such thatx isseparable, then

ordP dx = ordP

(∂F (x, y)

∂y

).

134

Proof. Without loss of generality, takeP to be the origin. Then

F (X,Y ) = X + Φ2(X,Y ) + . . .+ Φn(X,Y ).

Suppose thatY r, with r ≥ 2, is the term of least degree not containingX. Write

F (X,Y ) = X + cY r + . . . ,

where all other terms are divisible byX2,XY orY r+1. ThenI(P,F ∩v(X)) = rand a primitive representation of the branch ofF centred atP is

x(t) = −ctr + . . . ,y(t) = t.

Hencedx(t)/dt = −crtr−1 + . . .; that is, ordP dx = r − 1. Also,

∂F (X,Y )

∂Y= cr Y r−1 + . . . ,

where all other terms are divisible byX or Y r. Hence,

ordP(∂F (x, y)/∂y) = r − 1,

as required.This proof fails when the characteristicp of K is positive, andr = np with n a

positive integer.To prove the result in this case, take the derivative ofF (x(t), y(t)) = 0. Then

(∂F (x(t), y(t))

∂X

)x′(t) +

(∂F (x(t), y(t))

∂Y

)y′(t) = 0.

Hence

∂F (x(t), y(t))

∂Xx′(t) = −∂F (x(t), y(t))

∂Y.

Since ordt y′(t) = 0, so

ordt(∂F (x(t), y(t))/∂Y ) = ordt(∂F (x(t), y(t))/∂X) + ordt x′(t).

To obtain the result, it is necessary to show that ordt(∂F (x(t), y(t))/∂X) = 0.Suppose, therefore, that ordt(∂F (x(t), y(t))/∂X) > 0 and consider the polar

curveF ′ = v(∂F (X,Y )/∂X) at (1, 0, 0). Hence∂F (x(0), y(0))/∂X = 0, fromwhich it follows thatF ′ passes throughP . So the tangent toF atP passes through(1, 0, 0). This gives a contradiction since the tangent toF atP is v(X). Therefore

ordt(∂F (x(t), y(t))/∂X) = 0,

as required. 2

THEOREM 5.56 If P1, . . . ,Pk are the singular points of an irreducible curveF ,each of them ordinary, then the virtual genus ofF is equal tog; that is,

g = 12 (d− 1)(d− 2)− 1

2

∑ki=1mi(mi − 1)

whered is the degree ofF andmi is the multiplicity of the pointPi.


Proof. Choose a reference system(X0,X1,X2) in the plane so thatv(X0) is theline at infinity and so that the following conditions are satisfied:

(1) the points at infinity ofF ared distinct, simple points, none of which isY∞ = (0, 0, 1);

(2) no tangent toF at a singular point is theY -axis.

Let P = (x, y) be a generic point ofF . Now, calculate∑

P ordP dx that, bydefinition, equals2g − 2.

Given these assumptions, the following properties are demonstrated:

(a) dx has poles only at places centred on the line at infinity and ordP dx = −2for each of these places;

(b) dx has zeros only at places centred at simple points ofF with vertical tangentand ordP dx = ordP ∂F (X,Y )/∂Y ;

(c) if P is a point of multiplicitym of F andF ′ = v(∂F (X,Y )/∂Y ), thenI(P,F ∩ F ′) = m(m− 1).

To show (a), a primitive representation of a branchγ of F centred at the simplepointP = (0, 1, a), with a 6= 0, must be obtained. InterchangingX0 andX1, thepointP becomes the affine point(1, 0, a) and the branchγ is centred at(0, a) withtangentv(X2 − aX0 − bX1), with b ∈ K; in non-homogeneous coordinates, thisbecomesv(Y − bX − a), which, it should be noted, is notv(X) for (0, a). Aprimitive representation of the branchγ has the following form:

x(t) = t,

y(t) = a+ ctr + . . . ,

with r ≥ 1. In homogeneous coordinates,

x0(t) = 1,

x1(t) = t,

x2(t) = a+ ctr + . . . .

It follows that a representation ofγ centred at(0, 1, a) has the following form:

x0(t) = t,

x1(t) = 1,

x2(t) = a+ ctr + . . . ;

this, in non-homogeneous coordinates, becomes

x(t) = t−1,

y(t) = (a+ ctr + . . .)/t.

Hence,dx(t)/dt = −1/t2 and ordP dx = −2.

136

To prove (b), consider a simple affine pointP = (a, b) ofF . If the linev(X−a)throughP is not the tangent toF atP, a branch representation centred atP is givenby

x(t) = a+ t,

y(t) = b+ ctr + . . . ,

whence ordP dx = 0.If, instead, the tangent toF atP is v(X − a), a branch representation is given

by

x(t) = a+ dtr + . . . ,

y(t)= b+ t,

with ≥ 2. By Theorem 5.55,

ordP dx = ordP

(∂F (x(t), y(t))

∂Y

).

This means that ordP dx = I(P,F ∩ F ′), whereF ′ = v(∂F/∂Y ).To finish the proof, it remains to show (c); that is, it is necessary to calculate

I(P,F ∩ F ′), whereP is anm-fold singular point ofF . Note thatF ′ has an(m− 1)-fold singularity atP .

Now it is shown that the tangents atP for F are all distinct from those forG,and hence thatI(P,F ∩ F ′) = m(m− 1).

Without loss of generality, letP = (0, 0). Then

F (X,Y ) = Φm(X,Y ) + . . .+ Φd(X,Y ),

with

Φm(X,Y ) = am,0Xm + . . .+ ai,m−iX

iY m−i + . . .+ a0,mYm.

Note thata0,m 6= 0, as otherwisev(X) is a tangent.Let t1, . . . , tm be the roots of the polynomial

Φm(Z) = Φm(1, Z) = a0,m + . . .+ ai,m−iZi + . . .+ a0,mZ

m.

SinceP is an ordinary singular point ofF , the roots ofΦm(Z) are distinct; soΦm(Z) and its derivativeΦ′

m(Z) have no common roots.To determine the tangents toF ′, let

∂F (X,Y )

∂Y= Ψi(X,Y ) + . . .+ Ψd−1(X,Y ),

where

Ψi(X,Y ) =∂Φm(X,Y )

∂Y;

also∂Φm(X,Y )/∂Y cannot be identically zero, as otherwiseΦm(Z) would bethep-th power of a polynomial and its roots would have multiplicity p. Hence, thetangents toF ′ atP are given by the roots ofΨi(Z) = Ψi(1, Z), except possiblyfor tangents equal tov(X).


The curvesF andF ′ have a tangent atP in common if and only ifΦm(Z) andΨi(Z) have a root in common; but that cannot happen sinceΨi(Z) = ∂Φm(Z)/∂ZandΦm(Z) has no multiple roots. This proves (c).

Now, if γ is a branch centred at a singular point ofF , it must be linear sinceP is an ordinary singularity. Hence a primitive representation of γ has the formx(t) = a+ t, y(t) = b+ ctr + . . . . It follows that ordγ dx = 0.

As a consequence,∑

P∈P(Σ)ordP dx =∑I(P,F ∩ F ′)− 2d,

where the second summation is over all simple points ofF with tangentv(X − a).Applying Bezout’s Theorem, the intersection ofF andF ′ can be determined. Withd = degF andd− 1 = degF ′,

d(d− 1) =∑k

i=1mi(mi − 1) +∑

P∈P(Σ)ordP

(∂F (x, y)

∂y

),

where the second summation is over all places centred at affine non-singular pointsof F with vertical tangent. From (b),

d(d− 1) =∑k

i=1mi(mi − 1) +∑

P∈P(Σ)ordP dx,

and so

∑P∈P(Σ)ordP

(∂F (x, y)

∂y

)= d(d− 1)−∑k

i=1mi(mi − 1).

Therefore,∑

P∈P(Σ)ordP dx = d(d− 1)−∑ki=1mi(mi − 1)− 2d.

Finally,

d(d− 3)−∑ki=1mi(mi − 1) = 2g − 2,

whence

g = 12 (d− 1)(d− 2)− 1

2

∑ki=1mi(mi − 1).

2

It has been shown that the genus of an irreducible algebraic curve with onlyordinary singularities coincides with its virtual genus. This leads to the followingdefinition.

DEFINITION 5.57 Thegenusof an irreducible algebraic curve is the genus of itsfunction fieldΣ, whence it is the genus of each model(F ; (x, y)) of Σ, and so isthe virtual genus of a plane model with only ordinary singular points.

However, the genus of an irreducible curve with some non-ordinary singularities isin general smaller than its virtual genus. This follows fromTheorem 3.27 and (i)of Lemma 3.28. The following example illustrates this case.

EXAMPLE 5.58 For a positive integerh ≥ 4, let

F (X,Y ) = Y 2 − f(X), with f(X) =∏h

i=1(X − αi),

138

whereαi 6= αj for 1 ≤ i < j ≤ h. LetF be the irreducible curvev(F (X,Y )) andlet P = (x, y) be a generic point ofF . ,

First, all the affine points ofF are simple, since all the roots off(X) are distinct,andY∞ is a singular point of multiplicityh − 2. So, the virtual genusgv of F isequal toh− 2.

To determine the genusg of F , the sum∑

P∈P(Σ) ordP dx is calculated. Forthis, the simple points ofF with vertical tangent and the unique point at infinitymust be considered. Among the simple points ofF the only ones with verticaltangent are the pointsQi = (αi, 0) for i = 1, . . . , h. If P = (x0, y0) is a simplepoint ofF , its tangent isv(G) with

G = FX(x0, y0) (X − x0) + FY (x0, y0) (Y − y0);this is vertical whenFY (x0, y0) = 0, that is, if and only ify0 = 0.

A primitive branch representation centred atQi is

x(t)=αi + ct2 + . . . , c 6= 0,

y(t)= t.

Then,dx(t)/dt = 2ct + . . . and so ordt(dx(t)/dt) = 1. Hence, ifP is the placecorresponding toQi, then ordP dx = 1.

Actually, F has only one point at infinity, which isY∞ = (0, 0, 1). Now, it isshown that are one or two branches centred atY∞, according ash is odd or even.The corresponding places are poles whose order is3 in the former case, and2 forboth places in the latter case.

TakeY∞ to the originO = (0, 0) by interchanging theX-axis and the line atinfinity. Then the transformed curve isF ′, given by

Y h−2 −Xh + f(X,Y ), with f(X,Y ) = Xh −∏hi=1(X − αiY ),

which shows thatX∞ is an(h− 2)-ple point ofF with only the one tangent, givenby ℓ∞. A primitive representation(x′(t), y′(t)) of a branchγ′ of F ′ centred atP = (0, 0) has the form,

x′(t) = ti + . . . ,y′(t) = a tj + . . . .

(5.7)

with i < j. From the equationy(t)h−2 − x(t)h + y(t)h = 0, it follows that(h− 2)j = hi, whence one of the following occurs:

(1) i = h− 2, j = h;

(2) h is even andi = 12 (h− 2), j = 1

2h.

The two cases are investigated separately.In case (1),F ′ has only one branch centred atP = (0, 0). In fact,

I(P,v(X) ∩ F ′) = h− 2,

while ordt x′(t) = h− 2 for every branchγ′. By Theorem 4.48,

h− 2 =∑

ordt x′(t),


where the summation is for all branchesγ′ of F ′ centred atP (0, 0), whence theassertion follows. Now, (5.7) reads,

x′(t) = th−2 + . . . ,y′(t) = a th + . . . .

(5.8)

Going back toF , the corresponding branchγ of F centred atY∞ has a primitiverepresentation,

x(t) = t−2E1(t),y(t) = t−hE2(t)),

(5.9)

with invertibleE1(t), E2(t) ∈ K[[t]]. So ordt dx(t)/dt = −3, since charK 6= 2.Hence, ordP∞

dx = −3, whereP∞ is the place coming fromγ. This implies that2g − 2 = h− 3, whenceg = 1

2 (h− 1) andh is odd.In case (2),h is even. Now, Theorem 4.48 together withx′(t) = 1

2 (h−2) showsthatF ′ has exactly two branches centred atO = (0, 0), each having a primitiverepresentation,

x′(t) = t(h−2)/2 + . . . ,y′(t) = a th/2 + . . . .

(5.10)

Each of the two corresponding branches ofF centred atY∞ has primitive repre-sentation,

x(t) = t−1F1(t)),y(t) = t−h/2F2(t),

(5.11)

with invertibleF1(t), F2(t) ∈ K[[t]]. It follows that ordt dx(t)/dt = −2. If P(i)∞

with i = 1, 2 are the corresponding places, ordP(i)∞

dx = −2. Therefore,

2g − 2 =∑

PordP dx = h− 4,

and so the genus ofF is g = 12 (h− 2) for h even.

THEOREM 5.59 Let m be the smallest non-gap ofΣ at a placeP ∈ PK(X ).Suppose thatξ ∈ K(X ) has a pole of orderm at P but it is regular elsewhere. IfF is a subfield ofΣ containingK(ξ) such that[F : K(ξ)] > 1, thenF has positivegenus.

Proof. SupposeF is rational, and letF = K(η). Thenξ = f(η)/g(η), wheref(X), g(X) ∈ K[X] and the roots off(X) are distinct from those ofg(X). As-sume first thatg(X) is constant. Thenf(X) has positive degreek. Actually,k > 1as [F : K(ξ)] > 1. Since any pole ofη is also a pole ofξ, the hypothesis im-plies thatη has a unique pole, that is, div(η)∞ = −nP. Now, n = mk; but thiscontradicts the definition ofξ.

Assume now thatg(X) is divisible byX−a. ThenP is a zero ofη−a. Actually,P is the unique zero ofη − a, becauseP is the unique pole ofξ. Thus(η − a)−1

has a pole atP but is regular elsewhere; letn = div ((η−a)−1)∞. The fact thatPis the unique pole ofξ also implies thatg(X) = c(X − a)r with r ≥ 1. AsP is apole of ordern = mr of (η− a)−l, the definition ofm implies thatr = 1. If f(X)

140

is constant, thenξ = c(η − a)−1 with c ∈ K, c 6= 0, which yields thatF = K(ξ),a contradiction. Iff(X) has degreek > 0, then every pole ofη is a pole of orderk of ξ. Again, by the definition ofξ, this can only happen whenk = 1. But thenc(η−b)(η−a)−1 = ξ with a, b, c ∈ K, c 6= 0, whenceK(ξ) = F , a contradiction.2

THEOREM 5.60 Suppose thatΣ = K(x, y). If Σ has genusg, then

g ≤ ([Σ : K(x)]− 1)([Σ : K(y)]− 1).

Proof. Put n = [Σ : K(x)] andm = [Σ : K(y)]. Let F = v(f(X,Y )) bethe irreducible plane curve withP = (x, y) as a generic point. Letd denote thedegree ofF . If Y∞ ∈ F , choose a non-tangent lineℓ = v(X − a) to F at Y∞and transformX to X − a, and similarly forX∞. ThenF = v(g(X,Y )) withg(X,Y ) = f(X+a, Y + b), andP = (ξ, η) with ξ = x−a, η = y− b is a genericpoint ofF . Note thatK(ξ) = K(x), K(η) = K(y). Also, no place arising from abranch ofF centred atY∞ is a zero ofξ. By Theorem 5.33,Y∞ is a(d − n)–foldpoint ofF . Similarly,X∞ is a(d−m)–fold point ofF . Hence, the virtual genus

g∗ ≤ 12 (d2 − 3d− 2− (d− n)(d− n− 1)− (d−m)(d−m− 1).

Sinceg ≤ g∗, it suffices to check that12 (d2 − 3d− 2− (d− n)(d− n− 1)− (d−m)(d−m− 1)) ≤ (n− 1)(m− 1).

A straightforward calculation also shows that equality occurs in two cases, namely,for d = n+m andd = n+m− 1. Sinceg∗ = g only happens whenF has onlyordinary singularities, it follows thatg = (n− 1)(m− 1) if and only if each of thefollowing occurs:

(i) eitherd = n+m or d = n+m− 1;

(ii) F has at most two singular points, namelyX∞ andY∞;

(iii) F has only ordinary singularities.

2

5.9 HIGHER DERIVATIONS IN POSITIVE CHARACTERISTIC

Formula (5.5) shows thatdnζn/dζn = n!. Unfortunately, this implies thatdnζn/dζn = 0 for any separable variableζ ∈ Σ when the characteristicp ofK is less than or equal ton. To avoid all the negative consequences of this unusualbehaviour, a slight modification on the higher derivations is made by replacing fac-torials with binomial coefficients.

The binomial coefficient(

nm

)is usually defined for non-negative integersn and

m with n ≥ m. Here, it is convenient to extend this definition to all integersn andall non-negative integersm by putting(

n

m

)= n(n− 1) · · · (n− (m− 1))/[m(m− 1) · · · 1],

(n

0

)= 1.


Note that(

nm

)= 0 when0 ≤ n < m.

Higher derivatives inK((t)) are considered first.

DEFINITION 5.61 For i = 0, 1, 2, . . ., thei-th Hasse derivativeof∑akt

k is

D(i)t (∑akt

k) =∑(

ki

)akt

k−i.

Note thatD(i)t for i ≥ 1 is overK, that isD(i)

t c = 0 for c ∈ K andi ≥ 1. Also,

D(i)t f = (1/i!)dif/dti

for f ∈ K((t)) provided that the characteristicp of K is either0, or i < p. Inparticular,D(0)

t f = f for everyf ∈ K((t)), whileD(1)t f coincides with the usual

derivativedf/dt of f . Further,

D(i)t tj =

1 for i = j,0 for i > j ≥ 0.

(5.12)

Now, some fundamental equations for Hasse derivatives are shown.

L EMMA 5.62 Letf, g ∈ K((t)) ands ∈ K. Then

D(i)t (f + g)=D

(i)t f +D

(i)t g, D

(i)t (sf) = sD

(i)t f ; (5.13)

D(i)t (∑∞

k=0ckfk) =

∑∞k=0D

(i)t (ckf

k) for ordt f > 0; (5.14)

D(i)t (fg) =

∑ij=0D

(j)t fD

(i−j)t g; (5.15)

D(i)t D

(j)t f =

(i+ j

i

)D

(i+j)t f ; (5.16)

D(i)t (f1 · · · fr) =

∑D

(k1)t f1 · · ·D(kr)

t fr, (5.17)

where the last sum is over allr-ples of non-negative integers(k1, . . . , kr) withk1 + . . .+ kr = i.

Proof. Both equations in (5.13) are immediate. As a consequence, for every posi-tive integerm,

D(i)t (∑m

k=0ckfk) =

∑mk=0D

(i)t (ckf

k).

If ordt f > 0, then all terms of a given orderm in D(i)t (∑∞

k=0ckfk) are al-

ready present inD(i)t (∑m+i

k=0 ckfk), and the same holds for

∑∞k=0D

(i)t (ckf

k) and∑m+ik=0D

(i)t (ckf

k). Thus (5.14) follows from a finite number of repeated applica-tions of (5.13).

To show (5.15), letf =∑cvt

v, g =∑dvt

v. Then the coefficient oftv in fgis∑ckdv−k. Therefore,D(i)

t (fg) =∑evt

v−i with ev =(vi

)∑ckdv−k. On the

other hand, fromD(j)t f =

∑(kj

)ckt

k−j andD(i−j)t g =

∑(w

i−j

)dwt

w−(i−j),

D(j)t fD

(i−j)t g =

∑k

∑w

(k

j

)(w

i− j

)ckdwt

k+w−i.

Hence, the coefficient oftv−i in∑i

j=0D(j)t fD

(i−j)t g is

∑ij=0

∑k

(k

j

)(v − ki− j

)ckdv−k.

142

Since∑i

j=0

(k

j

)(v − ki− j

)=

(v

i

),

which may be established by comparing the coefficient ofXi on both sides of(1 +X)v = (1 +X)k(1 +X)v−k, so (5.15) follows.

Now,

D(i)t D

(j)t f =

∑cv

(v

j

)(v − ji

)tv−(i+j).

This, together with the identity(v

j

)(v − ji

)=

(i+ j

i

)(v

i+ j

),

proves (5.16). Finally, (5.17) follows from (5.15) by induction onr. 2

For anyb ∈ K and non-negative integersi,m, since

D(j)t (b+ t) =

b+ t if j = 0,1 if j = 1,0 if j ≥ 2,

so (5.17) implies that

D(i)t (b+ t)m =

(m

i

)(b+ t)m−i. (5.18)

Another fundamental equation is the chain rule.

L EMMA 5.63 Letf, g ∈ K[[t]] andordt g > 0.

D(i)t f(g(t)) =

[(D

(1)t g(t))iD

(i)T f(T ) +

∑i−1j=1gjD

(j)T f(T )

]∣∣∣T=g(t)

, (5.19)

where

gj =∑

D(r1)t g(t) · · ·D(rj)

t g(t),

and the sum is over allj-ples of positive integers(r1, . . . , rj) with r1+. . .+rj = i.

Proof. By (5.14), it is enough to prove (5.19) forf(t) = tn. From (5.17),

D(i)t g(t)n =

∑k1+...+kn=iD

(k1)t g(t) · · ·D(kn)

t g(t).

In each summand, among theki some positive integers occur; let(s1, . . . , sm) bethem-ple arising from(k1, . . . , kn) by omitting the zeros. Then1 ≤ m ≤ i,s1 + . . .+ sm = i, and

D(k1)t g(t) · · ·D(kn)

t g(t) = g(t)n−m(D(s1)t g(t) · · ·D(sm)

t g(t)).

Note that each termg(t)n−m(D(s1)t g(t) · · ·D(sm)

t g(t)) comes from exactly(

nm

)

termsD(k1)t g(t) · · ·D(kn)

t g(t). Hence

D(i)t g(t)n =

i∑

m=1

∑

s1+...+sm=i

(n

m

)g(t)n−mD

(s1)t g(t) · · ·D(sm)

t g(t). (5.20)


Since (n

m

)g(t)n−m = D

(m)T Tn

∣∣∣∣T=g(t)

,

the assertion follows. 2

Another useful equation is given in Lemma 5.65. The proof requires the number-theoretical result given in the following Lemma.

L EMMA 5.64 Let p be a prime and let the positive integersn,m have thep-adicexpansions

n =∑nip

i, m =∑mip

i.

with 0 ≤ ni ≤ p− 1, 0 ≤ mi ≤ p− 1. Then(n

m

)6≡ 0 (mod p)

if and onlyni ≥ mi for all i.

Proof. For an indeterminateX, consider the following expansion overFp:

(1 +X)n = (1 +X)n0+n1p+...+nrpr

= (1 +X)n0(1 +Xp)n1 . . . (1 +Xpr

)nr .

Now, pick out the coefficient ofXm from both sides:(n

m

)=

(n0

m0

)(n1

m1

)· · ·(nr

mr

).

This gives the result. 2

L EMMA 5.65 Let q = ph, wherecharK = p. For f ∈ K((t)) with ordt f > 0,and0 ≤ j, k < q,

D(q)t f jq+k = f jqD

(q)t fk + fk(Dtf

j)q. (5.21)

Proof. From (5.15),

D(q)t f jq+k =

∑qm=0D

(m)t f jqD

(q−m)t fk.

By Lemma 5.64, this sum consists of two terms only, namely thefirst and the last.It remains to show thatD(q)

t f jq = (D(1)t f j)q.

Putf j =∑cnt

n. Then

D(1)f j =∑ncnt

n−1,

(D(1)f j)q =∑nqcqnt

nq−q;

f jq = (f j)q

=∑cqnt

qn,

D(q)t f jq =

∑cqn

(nq

q

)tnq−q.

As(nqq

)≡ n (mod p), by Lemma 5.64, so the result follows. 2

Using similar arguments, the following lemma is established.

144

L EMMA 5.66 Let i = rpm + s with 1 ≤ r < p and0 ≤ s < pm. Then, for allf ∈ K((t)),

D(i)t f =

1

r!

(i

s

)−1

D(s)t

(D

(pm)t · · ·

(D

(pm)t f

)

︸︷︷︸r times

).

Proof. By (5.16),

D(pm)t · · ·

(D

(pm)t f

)=

(i

s

)(rpm

pm

)· · ·(

2pm

pm

)D

(i)t f,

where, as in Lemma 5.64,(npm

pm

)≡ n (mod p) for all 1 ≤ n ≤ r. Sincer < p and

s < i, sor! 6≡ 0 (mod p) and(

is

)6= 0. Hence the result follows. 2

Now, Hasse derivatives inΣ are introduced using a similar idea that led fromd/dt to d/dζ via any place ofΣ; see Section 5.7. This time, however, not all placesare suitable, or admissible, as it may happen at some places that the idea does notwork. In fact, computational and some theoretical complications arise when eitherordP ζ < 0, or ordP ζ ≥ 0 but ordP(ζ − b) > 1 with σ(ζ) = b + . . .. If thissituation does not occur forP, thenζ−b is alocal parameter; that is,τ(P) = b+ tfor some primitive representation ofP equivalent toσ.

By Lemma 5.52, all but finitely many places ofΣ are neither zeros nor poles ofdζ. Choose one of these infinite places, sayP. Let σ be a primitive representationof P, and putσ(ζ) = ζ. Thenζ = b+ φ(t) with φ(t) ∈ K[[t]] and ordt φ(t) = 1.Sincet 7→ φ(t) is an automorphism ofK((t)), it may be assumed thatζ = b + t.In other words,ζ is a local parameter atP.

For an elementξ ∈ Σ, let f(X,Y ) ∈ K[X,Y ] be an irreducible polynomialsuch thatf(ξ, ζ) = 0. Let F = v(f(X,Y )) be the corresponding irreduciblecurve. All but finitely many branches ofF are centred at simple affine points suchthat the corresponding places are neither zeros nor poles ofdζ. Hence,P may beassumed to come from a branchγ of F with a primitive representation in the form(ξ, ζ) with ξ = a+ψ(t), ζ = b+t, andψ(t) ∈ K[[t]]. Such a placeP is admissiblewith respect toζ andξ.

Now, Hasse derivatives are generalised to more than pne variable. As pointedout in Remark 1.36, in positive characteristic the usual partial derivatives of a poly-nomialf(X,Y ) =

∑am,kX

mY k, that is,

∂α+βf(X,Y )

∂Xα∂Xβ=∑am,kα!

(m

α

)β!

(k

β

)Xm−αY k−β

are changed to Hasse partial derivatives

∂(α+β)f(X,Y )

∂X(α)∂Y (β)=∑am,k

(m

α

)(k

β

)Xm−αY k−β .

L EMMA 5.67 The i-th Hasse derivative ofξ with respect tot can be defined as


follows:

D(i)t ξ =− 1

∂f(ξ, ζ)/∂ξ

∂

(i)f(ξ, ζ)

∂ζ(i)

+

i−1∑

j=1

∂(i−j+1)f(ξ, ζ)

∂ξ∂ζ(i−j)

D(j)t ξ+

i∑

n=2

i∑

j=n

∑

r1+...+rn=j

∂(i−j+n)f(ξ, ζ)

∂ξ(n)∂ζ

(i−j)D

(r1)t ξ · · ·D(rn)

t ξ

,

(5.22)

where


∂ξ(n)∂ζ

(i−j)=∂(n)f(ξ, ζ)

∂ξ(n)

whenj = i.

Proof. Let f(X,Y ) =∑akmX

kY m. Then∑akmξ

kζ

m= 0. From (5.13),∑

akmD(i)t (ξ

kζ

m) = 0. By (5.18),D(j)

t ζm

=(mj

)ζ

m−j. From (5.15),

D(i)t (ξ

kζ

m) =∑i

j=0D(j)t ξ

kD

(i−j)t ζ

m

=

(m

i

)ζ

m−iξ

k+∑i

j=1

(m

i− j

)ζ

m−(i−j)D

(j)t ξ

k.

By (5.20), for everyj with 1 ≤ j ≤ i,

D(j)t ξ

k= kξ

k−1D

(j)t ξ +

j∑

n=2

∑

r1+...+rn=j

(k

n

)ξ

k−nD

(r1)t ξ · · ·D(rn)

t ξ.

Hence∑

akmD(i)t (ξ

kζ

m) =

∑amk

[(m

i

)ζ(m−i)

ξk+

i−1∑

j=1

(m

i− j

)kζ

m−(i−j)ξ

k−1D

(j)t ξ + kζ

mξ

k−1D

(i)t ξ·

i∑

n=2

i∑

j=n

∑

r1+...+rn=j

(m

i− j

)(k

n

)ζ

m−i+jξ

k−nD

(r1)t ξ · · ·D(rn)

t ξ

= 0.

(5.23)

Since∂f(ξ, ζ)/∂ξ 6= 0, the assertion now follows by dividing the last equation by∂f(ξ, ζ)/∂ξ. 2

Although the above proof is constructive, for higher valuesof i the computationof the coefficients ofc(i)(X) is usually long, since it depends on the number ofpartitions of a natural number. Forj = n, the condition thatr1 + . . . + rn = jimplies thatr1 = . . . = rn = 1 and the corresponding term is

∂(i)f(ξ, ζ)

∂ξ(n)∂ζ

(i−n)(D

(1)t ξ)n;

146

but, forn = 2 andj > n, there are already⌊j/2⌋ summands with coefficient

∂(i−j+2)f(ξ, ζ)

∂ξ(2)∂ζ

(i−j).

For small values ofi,

D(1)t ξ = Dtξ = −∂f(ξ, ζ)/∂ζ

∂f(ξ, ζ)/∂ξ; (5.24)

D(2)t ξ =

− 1

∂f(ξ, ζ)/∂ξ

(∂(2)f(ξ, ζ)

∂ξ(2)

(Dtξ)2 +

∂(2)f(ξ, ζ)

∂ξ∂ζDtξ +

∂(2)f(ξ, ζ)

∂ζ(2)

).

(5.25)

To check (5.25), the procedure in the proof of Lemma 5.22 is used. It suffices to

calculateD(2)t (ξ

kζ

m), and repeat the final argument:

D(2)t (ξ

kζ

m) = ζ

mD

(2)t ξ

k+

(m

2

)ζ

m−2ξ

k+mζ

m−1Dtξ

k

= kξk−1

ζmD

(2)t ξ +

(k

2

)ζ

mξ

k−2(Dtξ)

2 +

(m

2

)ζ

m−2ξ

k

+mkζm−1

ξk−1

Dtξ.

Similarly it is possible to check the third and fourth derivatives:

D(3)t ξ = − 1

∂f(ξ, ζ)/∂ξ

(∂(2)f(ξ, ζ)

∂ξ(2)

(D(2)t ξ)(Dtξ) +

∂(3)f(ξ, ζ)

∂ξ(3)

(Dtξ)3

+∂(2)f(ξ, ζ)

∂ξ∂ζ(D

(2)t ξ) +

∂(3)f(ξ, ζ)

∂ξ(2)∂ζ

(Dtξ)2

+∂(3)f(ξ, ζ)

∂ξ∂ζ(2)

(Dtξ) +∂(3)f(ξ, ζ)

∂ζ(3)

),

(5.26)

D(4)t ξ = − 1

∂f(ξ, ζ)/∂ξ

(∂(2)f(ξ, ζ)

∂ξ(2)

(D(3)t ξ)(Dtξ) +

∂(2)f(ξ, ζ)

∂ξ(D

(2)t ξ)2

+∂(3)f(ξ, ζ)

∂ξ(3)

(D(2)t ξ)(Dtξ)

2 +∂(4)f(ξ, ζ)

∂ξ(4)

(Dtξ)4 +

∂(2)f(ξ, ζ)

∂ξ∂ζ(D

(3)t ξ)

+∂(3)f(ξ, gz)

∂ξ(2)∂ζ

(D(2)t ξ)(Dtξ) +

∂(4)f(ξ, ζ)

∂ξ(3)∂ζ

(Dtξ)3 +

∂(3)f(ξ, ζ)

∂ξ∂ζ(2)

(D(2)t ξ)

+∂(4)f(ξ, ζ)

∂ξ(2)∂ζ

(2)(Dtξ)

2 +∂(4)f(ξ, ζ)

∂ξ∂ζ(3)

(Dtξ) +∂(4)f(ξ, ζ)

∂ζ(4)

).

(5.27)

As in Lemma 5.67, higher derivatives can be defined using the same approachthat producedDζ fromDt via a place ofΣ.


DEFINITION 5.68 Let ζ be a separable variable ofΣ. For i = 0, 1, 2, . . ., thei-thHasse derivativeD(i)

ζ ξ of ξ ∈ Σ is defined iteratively: D(0)ζ ξ = ξ, and

D(i)ζ ξ =− 1

∂f(ξ, ζ)/∂ξ

∂

(i)f(ξ, ζ)

∂ζi+

i−1∑

j=1

∂(i−j+1)f(ξ, ζ)

∂ξ∂ζ(i−j)D

(j)t ξ+

i∑

n=2

i∑

j=n

∑

r1+...+rn=j


∂ξn∂ζ(i−j)D

(r1)t ξ · · ·D(rn)

t ξ

.

(5.28)

Diagrammatically,

ξ D(i)ζ ξ

ξ D(i)t ξ

-D

(i)ζ

?τ

?τ

-D(i)t

Some previous remarks also hold true forD(i)ζ . For i ≥ 1, the Hasse derivative

D(i)ζ is overK, and

i!D(i)ζ ξ = diξ/dζi

for i < p,

REMARK 5.69 By definition, σ(D(i)ζ ξ) = D

(i)t σ(ξ) for every admissible place

P of Σ, whereσ is a primitive representation ofP such thatσ(ζ) = b + t withb ∈ K. Sinceσ is aK-isomorphism fromΣ ontoK((t)), the previous results onD

(i)t extend toD(i)

ζ .

L EMMA 5.70 All equations but (5.17) in Lemma 5.62 as well as (5.12) and Lemma5.66 remain true whenK((t)) andt are replaced byΣ andζ.

EXAMPLE 5.71 Suppose thatp = 2. Let f(X,Y ) = Y + 1 + Y 5X23 andΣ = K(x, y) with y + 1 = y5x23. Thenx is a separable variable ofΣ. Bystraightforward calculations,

∂f(x, y)

∂x= x22y5;

∂f(x, y)

∂y=

1

y;

∂(2)f(x, y)

∂x(2)= x21y5;

∂(2)f(x, y)

∂x∂y= x22y4;

∂(2)f(x, y)

∂y(2)= 0;

∂(3)f(x, y)

∂x(3)= x20y5;

∂(3)f(x, y)

∂x(2)∂y= x21y4;

∂(3)f(x, y)

∂x∂y(2)=∂(3)f(x, y)

∂y(3)= 0;

∂(4)f(x, y)

∂x(4)= x11y5;

∂(4)f(x, y)

∂x(3)∂y= x20y4;

∂(4)f(x, y)

∂x(4)= x11y5;

∂(4)f(x, y)

∂x(2)∂y(2)=∂(4)f(x, y)

∂x∂y(3)= 0;

∂(4)f(x, y)

∂y(4)= x20y4.

148

From (5.24),

Dxy =x22y5

1 + x23y4= x22y6.

From (5.25), (5.26) and (5.27),

D(2)x y = x21y7; D(3)

x y = x20y8; D(4)x y = y6(u4 + u3 + 1)x19.

THEOREM 5.72 Letζ andη be separable variables ofΣ. Then there are elementsd1, . . . , di−1 in Σ that are polynomials inD(k)

η (ζ) for 1 ≤ k ≤ i such that, for anyξ ∈ Σ,

D(i)η ξ = (dζ/dη)iD

(i)ζ ξ +

∑i−1k−1diD

(k)ζ ξ. (5.29)

Proof. LetP be an admissible place ofΣ with respect to the pair(ξ, ζ). To compareDη andDζ , admissibility with respect to the pair(ξ, η) also enters into considera-tion. Since there remain infinitely many places with the required properties,P maybe assumed admissible with respect to both pairs(ξ, ζ) and(ξ, η).

Choose a primitive representationσ of P such thatσ(ξ) = ξ = a + ψ(t),Thenσ(D

(i)ζ ξ) = D

(i)t ξ, andσ(ζ) = ζ = b + t. Also, σ(η) = η = c + λ(t)

and ordt λ(t) = 1. By the latter equation,K((t)) has an automorphismh withh(t) = λ(t). Letm be its inverse, and putm(t) = τ . Then

m(η) = c+ t, m(ξ) = a+m(ψ(t)).

Sincem operates onK((t)) by substitution,µ(ξ) = a+ ψ(τ) holds. Put

ξ∗ = a+ ψ(τ), η∗ = c+ t, ζ∗ = b+ τ.

Also, mσ(D(i)η ζ) = D

(i)t ζ∗ = D

(i)t τ . Theorem 5.72 follows from Lemma 5.63

applied tof(t) = a+ ψ(t) andg(t) = τ . 2

Assume thatK has positive characteristicp. Again, letζ be a separable variableof Σ. For any positive integerm, put

Σm = ξ ∈ Σ | D(i)ζ ξ = 0, 1 ≤ i < pm,

Σ∞ = ξ ∈ Σ | D(i)ζ ξ = 0, i = 1, 2, . . ..

ThenΣm ⊃ K andΣ∞ ⊃ K.

L EMMA 5.73 For any positive integerm,

(i) Σm is a subfield ofΣ containingK;

(ii) D(m)ζ is a derivation ofΣm;

(iii) Σm+1 = ξ ∈ Σm | D(pm)ζ ξ = 0.


Proof. (i) For eachξ, η ∈ Σm, c, λ ∈ K and1 ≤ i < pm,

D(i)ζ (ξ) +D

(i)ζ (η) = D

(i)ζ (ξ + η) = 0;

λD(i)ζ (ξ) = D

(i)ζ (λξ) = 0;D

(i)ζ (c) = 0.

HenceK ⊂ Σm ⊂ Σ.(ii) From (5.15),

D(pm)ζ (ξη) = ηD

(pm)ζ ξ + ξD

(pm)ζ η

for ξ, η ∈ Σm. This, together with (5.13), shows thatD(pm)ζ is a derivation ofΣm.

(iii) Put T = ξ ∈ Σm | D(pm)ζ ξ = 0. If it is shown thatT = Σm+1, then the

lemma follows by induction. By definition, it suffices to showthatD(i)ζ ξ = 0 for

ξ ∈ T andpm < i < pm+1. Now, write i = rpm + s with 1 ≤ r ≤ p − 1 and0 ≤ s ≤ pm − 1. From Lemmas 5.66 and 5.70,

D(s)ζ

(D

(pm)ζ

)r

ξ = r!D(s)ζ D

(rpm)ζ ξ = r!

(i

s

)D

(i)ζ ξ.

From Lemma 5.64,(

is

)6≡ 0 (mod p). Hence, forξ ∈ T ,

D(i)ζ ξ =

1

r!

(i

s

)−1

D(s)ζ

(D

(pm)ζ

)r

ξ = 0. 2

L EMMA 5.74 For any positive integerm,

Σm = Σpm

= upm | u ∈ Σ.Proof. Let η ∈ Σ. Choose an admissible placeP with respect toζ andη. For aprimitive representationτ of P, τ(η) = η =

∑ckt

k. As τ(ηpm

) =∑cp

m

k tkpm

, itfollows that

D(i)t ηpm

=∑cp

m

k

(kpm

i

)tkpm−i, (5.30)

which equals zero for1 ≤ i < pm. Sinceτ is an isomorphism fromΣ intoK((t)),this only occurs whenD(i)

ζ ηpm

= 0. Thusηpm ∈ Σm for everyη ∈ Σ.To show the converse, note that everyξ ∈ Σ1 is either a constant or an insepa-

rable variable ofΣ, by Lemma 5.37. The assertion form = 1 follows from (iii) ofthe same theorem. Now, the proof is by induction onm.

Assume that every element inΣm is apm-th power of an element inΣ. Thenξis in Σm+1 whenξ = ηpm

andD(i)ζ ξ = 0 for pm ≤ i < pm+1. Again, choose an

admissible placeP with respect toζ andη, and writeτ(η) = η =∑ckt

k with τa primitive representation ofP. Arguing as before, (5.30) follows, and this is zerosincepm < i < pm+1 is assumed. By thep-adic criterion of Lemma 5.64, thisimplies that everyk in the expansion ofη is divisible byp. ThusDtη = 0. ByLemma 5.37,η is an inseparable variable ofΣ. Hence, (iii) of the same theoremshows thatη = up for someu ∈ Σ. Therefore,ξ = ηpm

= upm+1

. 2

L EMMA 5.75 Σ∞ = K.

150

Proof. Let ξ be any non-zero element in∈ Σ∞. For a placeP of Σ, letk = ordP ξ.From the definition,Σ∞ ⊂ Σm for any positive integerm. Choosem such thatpm > k. By Lemma 5.74,ξ = ηpm

for someη ∈ Σ. Hence ordP ξ = pmordP η.But this only occurs when ordP ξ = 0 and ordP η = 0. Therefore, ordP ξ has nozeros. By Theorem 5.33,ξ ∈ K. 2

THEOREM 5.76 Let x0, . . . , xr ∈ Σ. Thenx0, . . . , xr are linearly independentoverΣm if and only if there exist integersǫ0, . . . , ǫr, with 0 = ǫ0 < ǫ1 < . . . <

ǫr < pm, such thatdet(D(ǫi)ζ xj) 6= 0.

Proof. Suppose thatx0, . . . , xr ∈ Σ are linearly dependent overΣm; that is, thereexistz0, . . . , zr ∈ Σm not all zero for which

∑rj=0 zjxj = 0. By (5.15) and the

definition ofΣm,∑r

j=0zjD(i)xj = 0.

This proves the “if” part of the theorem.The converse is proved by induction onr. The caser = 0 is trivial. Let

x0, . . . , xr be linearly independent overΣm, and assume that the assertion of thetheorem does not hold; that is, ther + 1 vectors

uj = (xj ,D(1)ζ xj ,D

(2)ζ xj , . . . ,D

(pm−1)ζ xj)

are linearly dependent overΣ. Then there existz0, . . . , zr ∈ Σ such that∑r

j=0zjD(i)ζ xj = 0 for 0 ≤ i ≤ pm − 1. (5.31)

Sincex0, . . . , xr−1 are linearly independent overΣm, from the inductive hypothe-sis it follows, for0 ≤ j ≤ r − 1, that the abover vectors are linearly independentoverΣ. Hencezr 6= 0. Therefore,zr = 1 can be assumed. It suffices to show thatzj ∈ Σm for j = 0, . . . , r − 1. In fact, this together with (5.31) fori = 0, gives acontradiction that completes the proof.

To prove thatzj ∈ Σm, it is enough to show by Lemma 5.73 thatD(n)ζ zj = 0

for 1 ≤ n ≤ pm−1. This is done by induction onn.Forn = 1, from (5.31),

D(1)ζ

∑rj=0zjD

(i)ζ xj = (i+ 1)

∑rj=0zjD

(i+1)xj +∑r−1

j=0D(1)ζ zjD

(i)ζ xj = 0.

From (5.31),(i + 1)∑r

j=0 zjD(i)ζ xj = 0 for i < pm − 1. This holds true for

i = pm − 1 since in this casei+ 1 ≡ 0 (mod p). Hence∑r−1

j=0D(1)ζ zjD

(i)ζ xj = 0

for 0 ≤ i ≤ pm − 1. Since the abover vectors for0 ≤ j ≤ r − 1 are linearlyindependent overΣ, the conclusion is thatD(1)

ζ zj = 0 for j = 0, . . . , r − 1.

Assume thatD(s)ζ zj = 0 for 1 ≤ s < n ≤ pm−1 and for0 ≤ j ≤ r − 1. From

(5.31) by derivation,(i+ n

i

)∑rj=0zjD

(i+n)ζ xj +

∑r−1j=0D

(n)ζ zjD

(i)ζ zj = 0.


As in the casen = 1, to prove thatD(n)ζ zj = 0, it suffices to show that

(i+ n

i

)∑rj=0zjD

(i+n)ζ xj = 0 for 0 ≤ i ≤ pm − 1.

For i+ n < pm, this follows from (5.31). Fori+ n ≥ pm,

pm ≤ i+ n < pm + pm−1,

pm > i ≥ pm − n ≥ (p− 1)pm−1.

Hence the coefficient ofpm−1 in thep-adic expansion ofi+ n is zero and that ofiis p− 1. From Lemma 5.64,

(i+n

i

)≡ 0 (mod p), and this completes the proof.2

The following result is a consequence of Theorem 5.76.

THEOREM 5.77 Let x0, . . . , xr ∈ Σ. Thenx0, . . . , xr are linearly independentoverΣ∞ if and only if there exist integersǫ0, . . . , ǫr, with 0 = ǫ0 < ǫ1 < . . . < ǫr,

such thatdet(D(ǫi)ζ xj) 6= 0.

5.10 THE DUAL AND BIDUAL OF A CURVE

Let F be an irreducible plane curve of degreen > 1. To each non-singular pointof F there corresponds a unique tangent line toF at the point. In the dual plane,tangent lines determine points and, at least intuitively, the point set obtained is acurve. A rigorous proof is the main purpose of the present section, in which someunexpected behaviour of the dual curves in positive characteristic is also shown.

DEFINITION 5.78 Thedual curveofF is an irreducible plane curveF ′ containingall but finitely many pointsP ′ = (b0, b1, b2) such that the lineb0X0+b1X1+b2X2

in PG(2,K) is tangent toF .

Assume thatF = v(F ) for a homogeneousF ∈ K[X0,X1,X2] of degreen.With the notation of (1.4), Euler’s formula gives thatnF = X0F0+X1F1+X2F2.SinceF is irreducible, one of the partial derivatives, sayF0, is not vanishing. Then,for the generic pointP with coordinatesx = (x0, x1, x2),

x0F0(x) + x1F1(x) + x2F2(x) = 0. (5.32)

One of the coordinates ofP , sayx0, is distinct from0. Without loss of generality,ξ = x1/x0 is a separable variable of the function fieldΣ of F .

DEFINITION 5.79 The rational transformation,

ωF : Σ→ Σ, (x′0, x′1, x

′2) = (F0(x), F1(x), F2(x)), (5.33)

is therational Gauss mapassociated toF .

First, it is shown that the rational Gauss map is non-trivial. Without loss ofgenerality, suppose thatF0(x) 6= 0. In the trivial case, both polynomialsF1 − sF0

152

andF2−bF0 are divisible byF . Since they have both degree less thann, the trivialcase only occurs whenF1 = sF0 andF2 = tF0. Then, Euler’s formula gives that

nF = X0F0 +X1F1 +X2F2 = (X0 + sX1 + tX2)F0.

SinceF is irreducible andF0 6= 0, this only happens forn = 1 andF0 is constant,contradicting the hypothesisn > 1. So, the transform ofF under the rationalGauss map is an irreducible curveF ′ which is actually the dual curve ofF .

Now, some important properties of the dual curve are established. Choose anyseparable variableζ of Σ, and deriveF (x) = 0. Then

dx0

dζF0(x) +

dx1

dζF1(x) +

dx2

dζF2(x) = 0. (5.34)

So the triple(F0(x), F1(x), F2(x)) provides a non-trivial solution of the followingsystem of homogeneous linear equations in the unknownsu0, u1, u2:

x0u0 + x1u1 + x2u2 = 0, (5.35)dx0

dζu0 +

dx1

dζu1 +

dx2

dζu2 = 0. (5.36)

There is a unique non-trivial solution, up to a non-zero factor, since the matrix ofthe system has rank2. To show this, assume thatx0 = 1. Thendx0/dζ = 0, andthe rank is1 if and only if dx1/dζ = 0 anddx2/dζ = 0. But x1/x0 is a separablevariable, and thusdx1/dζ 6= 0 by Theorem 5.49(ii).

Taking derivatives and subtracting, a further equation is obtained:

x0dF0(x)

dζ+ x1

dF1(x)

dζ+ x2

dF2(x)

dζ= 0. (5.37)

Now consider the dual curveF ′′ of F ′, that is, thebidual of F . In the realand complex planes,F ′′ coincides withF . But this is not true for every curve inpositive characteristic, and this behaviour is now investigated.

Putyi = Fi(x) for i = 0, 1, 2, and writey = (y0, y1, y2). If F ′ = v(F ′) for anirreducible homogeneous polynomialF ′ with coefficients inK, thenF ′(y) = 0;that is,Q = (y0, y1, y2) is a generic point ofF ′. The rational Gauss map associatedtoF ′ is

ωF ′ : Σ→ Σ, (y′0, y′1, y

′2) = (F ′

0(y), F′1(y), F

′2(y)), (5.38)

whereF ′i is the partial derivative ofF ′ with respect toXi. If F ′ is not a line,

that is,F is not a strange curve, thenF ′ is transformed into the curveF ′′, andR = (F ′

0(y), F′1(y), F

′2(y)) is a generic point ofF ′′. SinceF ′′ arises fromF ′ in

the same way asF ′ from F , the triple(F ′0(y), F

′1(y), F

′2(y)) is a solution of the

system of homogeneous linear equations in the unknownsv0, v1, v2:

y0v0 + y1v1 + y2v2 =0, (5.39)dy0dζ

v0 +dy1dζ

v1 +dy2dζ

v2 =0. (5.40)

Another solution is the triple(x0, x1, x2). This follows from (5.37) and (5.32). Inthis system, two cases occur according to whether the matrix

y0 y1 y2

dy0dζ

dy1dζ

dy2dζ

=

F0(x) F1(x) F2(x)

F0(x)

dζ

F1(x)

dζ

F2(x)

dζ

(5.41)


has maximal rank or not. If the rank is2, then there is a unique solution up toa non-zero factor, andF andE share a generic point. Since bothF andF ′′ areirreducible, this is only possible when they coincide. Therefore,ωF ′ωF is a bira-tional transformation leavingF and alsoF ′ invariant. In terms of separability, asin Section 5.5, this means that both Gauss mapsωF andωF ′ , and hence also theirproduct, are separable; that is,F is areflexivecurve.

Now, the possibility that the matrix in (5.41) has rank1 is examined. For thispurpose, letx0 = 1 andζ = x1. Note thatF2(x) 6= 0, sincex1 is a separablevariable. If the rank is1, then

F1(x)dF2(x)

dx1= F2(x)

dF1(x)

dx1. (5.42)

Also, from (5.34),

F1(x) +dx2

dx1F2(x) = 0. (5.43)

Taking derivatives:

dF1(x)

dx1= −d

2x2

dx21

F2(x)−dx2

dx1

dF2(x)

dx1.

This, together with (5.42) and (5.43), gives

d2x2

dx21

= 0. (5.44)

Note that, ifp = 2, then (5.44) holds trivially. Forp 6= 2, (5.44) implies that allnon-singular point ofF are inflexions. To show this, derive (5.43). In terms ofF ,

∂2F (x)

∂x21

+ 2dx2

dx1

∂2F (x)

∂x1∂x2+

(dx2

dx1

)2∂2F (x)

∂2x2+dx2

2

dx21

∂F (x)

∂x2= 0.

Writing f(X,Y ) = F (1,X, Y ), this, together with (5.43) and (5.44), yields(∂f

∂x1

)2∂2f

∂x22

+

(∂f

∂x2

)2∂2f

∂x21

− 2∂f

∂x1

∂f

∂x2

∂2f

∂x1∂x2= 0.

Sincef(x1, x2) = 0, by Theorem 1.35 the assertion holds. Another consequenceof (5.44) forp > 2 is that the Gauss mapωF is inseparable; that is,F is a non-reflexive curve. In fact, bothF0(x)/F2(x) andF1(x)/F2(x) are inΣp. Thisdepends on the above equations: from (5.44),

d(F1(x)/F2(x))

dx1= 0,

and this, together with (5.32), implies that

d(F0(x)/F2(x))

dx1= 0.

Let pm be the largest power ofp for which

F0(x)/F2(x) = zpm

0 , F1(x)/F2(x) = zpm

1 ,

for somez0, z1 ∈ Σ. Then (5.32) becomeszpm

0 + zpm

1 x1 + x2 = 0. Therefore,

H(X,Y )f(X,Y ) = Z0(X,Y )pm

+ Z1(X,Y )pm

X + Z2(X,Y )pm

Y, (5.45)

with Z0, Z1, Z2,H ∈ K[X,Y ]. In summary, the following theorem is established.

154

THEOREM 5.80 Letp 6= 2.

(i) A sufficient condition for an irreducible plane curveC to coincide with itsbidual curve is thatC is reflexive.

(ii) The curveF = v(f(X,Y )) is non-reflexive if and only iff(X,Y ) satisfies(5.45) withpm the inseparability degree of the Gauss map ofC.

(iii) F is non-reflexive if and only if it is the locus of its inflexion and singularpoints.

Forp = 2, the situation is different.

THEOREM 5.81 Letp = 2.

(i) Every irreducible plane curveF = v(f(X,Y ) is non-reflexive, andf(X,Y )satisfies (5.45) withpm the inseparability degree of the Gauss map ofF .

(ii) If pm > 2, thenF is the locus of its inflexion and singular points.

Proof. The proof of Theorem 5.81 requires a different argument starting from(5.42). Forp = 2, (5.42) becomesd(F1(x)F2(x))/dx1 = 0. From Theorem 5.49and Lemma 5.37,F1(x)F2(x) is Σ2, that is, it is a square inΣ. SinceF2(x) 6= 0,this yieldsF1(x)/F2(x) ∈ Σ2; similarly, F0(x)/F2(x) ∈ Σ2. In particular, theGauss map is inseparable. By (5.32), this shows that (5.45) holds forp = 2, where2m is the inseparability degree of the Gauss map.

Conversely, if the Gauss map is inseparable, thenF0(x)F2(x) andF1(x)F2(x)are both squares. Hence their derivatives are zero, and thusthe above matrix hasrank1.

It remains to prove the last assertion in Theorem 5.80. From (5.45),

z0(x1, x2)2m

+ z1(x1, x2)2m

x1 + z2(x1, x2)2m

x2 = 0.

If m > 1, this yieldsD(2)x1 x2 = 0. Therefore, the assertion follows from (5.25) and

(1.8). 2

REMARK 5.82 For pm = 2, Theorem 5.81 (iii) does not hold true. A counterex-ample is the Hermitian curveH2 = v(X3

0 + X31 + X3

2 ) which is non-reflexive,but has only nine inflexion points. A necessary and sufficientcondition for a non-reflexive curveF to have only finitely many inflexion points is that the matrix[

F0(x) F1(x) F2(x)

D(2)x1F0(x) D(2)

x1F1(x) D(2)

x1F0(x)

]

has rank2. This is shown in Section 7.8.

REMARK 5.83 It is possible for a non-reflexive curve to coincide with its dualcurve, up to a Frobenius collineation. An example of this is the Hermitian curveHq = v(Xq+1

0 +Xq+11 +Xq+1

2 ) for any powerq of p. In fact, the Gauss map is

ωHq: Hq → Hq, (x0, x1, x2) 7→ (xq

0, xq1, x

q2),

and hence is the product of the identity rational transformation and theq-th Frobe-nius collineation.


REMARK 5.84 The dual curve of a strange curve is a line, and hence strangecurves have no bidual curve.

5.11 EXERCISES

1. LetΣ = K(x, y). It follows from Lemma 5.37 that eitherx, or y is a sepa-rable variable ofΣ. Give another proof here by using Theorem 1.23.

2. With Σ a field of transcendence degree1, give a new proof for Bezout’stheorem using Theorem 5.35.

3. Show that a uniformizing element at any place is a separable variable ofΣ.

4. Show that, if charK = 0, thendξ/dζ = c if and only if ξ = cξ + d, withc, d ∈ K. This result extends to positive characteristic provided that ξ isseparable.

5. Let γ be a branch ofF with order sequence(0, j1, j2). If K has positivecharacteristicp, assume thatp ≥ 3 but thatp does not divide eitherj1 orj2 − j1. Show that the corresponding branchγ′ of the dual curveF ′ of Fhas orderj2 − j1.

6. Prove the following generalisation of Theorem 5.60. LetΣ1 andΣ2 be twosubfields ofΣ both of transcendency degree1. Fori = 1, 2, putni = [Σ : Σi]and denote bygi the genus ofΣi. If Σ = Σ1Σ2 is the compositum ofΣ1 andΣ2, that is, the smallest subfield ofΣ containing bothΣ1 andΣ2, then

g ≤ n1g1 + n2g2 + (n1 − 1)(n2 − 1).

5.12 NOTES

Chapter Six

Linear series and the Riemann–Roch Theorem

6.1 DIVISORS AND LINEAR SERIES

As in the previous chapter, the fieldΣ = K(x, y) is associated to an irreduciblealgebraic curveF equipped with a generic pointP = (x, y). From an algebraicpoint of view,Σ is a function field of transcendency degree1 whose places are in aone-to-one correspondence with branches ofF .

A natural generalisation of a place is a divisor.

DEFINITION 6.1 (i) A divisor of Σ is a formal sum of places

D =∑

P∈P(Σ)nPP,with nP integers such thatnP = 0 except for a finite number of places.

(ii) The coefficientnP is themultiplicity or weightof P in D.

(iii) The supportof D, denoted by SuppD, is the set of all places appearing withnon-zero coefficients inD.

L EMMA 6.2 The set of all divisors ofΣ is an abelian group with respect to theaddition, thedivisor groupDiv(Σ) of Σ.

Proof. ThesumD + E of two of divisorsD =∑nPP andE =

∑mPP is the

divisor∑

(nP +mP)P. The identity element(0) is thezero divisorwhich has nonon-zero weight. 2

DEFINITION 6.3 (i) If nP ≥ 0 for every placeP, the divisorD =∑nPP is

effective, or positive.

(ii) A non-effective divisor isvirtual.

(iii) Given two divisorsD =∑nPP andE =

∑mPP, the notationD ≻ E is

used whenD − E is effective.

The most important divisors arise directly from elements ofΣ, due to Theorem5.33. Letξ ∈ Σ\0.

(i) div(ξ) =∑

P∈P(Σ)nPP with nP = ordPξ is theprincipal divisor ofξ.

(ii) div(ξ)0 =∑

P∈P(Σ)nPP with nP = ordPξ andnP ≥ 0 is thedivisor ofzerosof ξ.

157

158

(iii) div (ξ)∞ =∑

P∈P(Σ)(−nP)P with nP = ordPξ andnP ≤ 0 is thedivisorof polesof ξ.

Note that div(ξ) = div(ξ)0 − div(ξ)∞.

THEOREM 6.4 Let ξ ∈ Σ\0. Thendiv(ξ) = (0) if and only ifξ ∈ K\0.Proof. If ξ ∈ K\0, thenξ has neither zeros no poles. Assumex 6∈ K. Fromthe proof of Theorem 5.33,Σ has a model(F ; (ξ, η)) such that some branch ofFis centred at an affine point on theX-axis. The corresponding place ofΣ is a zeroof div(ξ). Hence div(ξ)0 6= (0). Since zeros ofζ ∈ Σ\0 are poles of its inverseζ−1, so div(ξ)∞ 6= (0) also holds. Hence div(ξ) 6= (0). 2

L EMMA 6.5 The elements inDiv(Σ) which are principal divisors constitute a sub-groupPrin(Σ).

Proof. If f, g ∈ Σ\0, then div(fg) = div(f) + div(g). 2

DEFINITION 6.6 (i) The factor groupPic(Σ) = div(Σ)/Prin(Σ) is thedivisorclass groupof Σ.

(ii) For a divisorD of Σ, the corresponding element in the factor groupPic(Σ)is denoted by[D], thedivisor classof D.

DEFINITION 6.7 Two divisorsA andB of Σ areequivalent, writtenA ≡ B if[A] = [B], that is if there existsξ ∈ Σ\0 for whichA−B = div(ξ).

L EMMA 6.8 (i) The relation of Definition 6.7 is an equivalence relation.

(ii) LetA,B,C,D be divisors ofΣ.

(a) If A ≡ B, thenA+ C ≡ B + C.

(b) If A ≡ B, then−A ≡ −B.

(c) If A ≡ B, C ≡ D, thenA+ C ≡ B +D.

Proof. (ii)(a) SinceA ≡ B, there existsξ ∈ Σ\0 for whichA − B = div(ξ).SinceA−B = (A+ C)− (B + C), the assertion follows.

(ii)(b) If A ≡ B, thenA − B = div(ξ) with ξ ∈ Σ\0, whence it follows that−A− (−B) = div(ξ−1).

(ii)(c) If C ≡ D also holds, that is,C − D = div(η) with η ∈ Σ\0, thenA+ C − (B +D) = A−B + (C −D) = div(ξ) + div(η) = div(ξη). 2

DEFINITION 6.9 Thedegreeof a divisorD =∑

P∈P(Σ)nPP is degD =∑nP .

Note thatϕ : A 7→ degA is an additive map. In other words,ϕ is a homomor-phism from the divisor group to the additive group of integers.

THEOREM 6.10 If A ≡ B, thendegA = degB.

Linear series and the Riemann–Roch Theorem 159

Proof. LetA− B = div(ξ) with ∈ Σ\0. If ξ ∈ K, thenA = B. Otherwise, asdiv(ξ) = 0 by Corollary 5.35, deg(A−B) = 0, whence the assertion follows.2

Theorem 6.10 shows that deg[A] is meaningful. Hence the mapϕ : [A] 7→ deg[A]is a homomorphism from the divisor class group to the additive group of integers.

DEFINITION 6.11 The subgroup of divisors of degree zeroDiv0(Σ) of Div(Σ)consists of all divisors ofΣ of degree zero.

Under the natural homomorphismDiv(Σ) 7→ Pic(Σ), the subgroupDiv0(Σ) cor-responds to the subgroupPic0(Σ) of divisor classes of degree zero; that is,

Pic0(Σ) = [A] ∈ Pic(Σ) | degA = 0.Certain subsets of divisors, not necessarily subgroups, play an important role in thestudy ofΣ. The most relevant are the linear series.

Let x0, x1, . . . , xr be elements ofΣ, not all zero, and letB be a fixed divisorof Σ. An orderedr-ple c = (c0, c1, . . . , cr) of elements inK is permissiblewhenc0x0 + c1x1 + . . .+ crxr 6= 0. For such anr-plec, consider the divisor

Ac = div(c0x0 + c1 + . . .+ crxr) +B. (6.1)

DEFINITION 6.12 The setL of divisorsAc for all permissibler-plesc is a linearseries.

Note that there may be a finite number of places occurring in every divisor ofL.Such places are thefixedplaces ofL. A sum of fixed places is afixeddivisor ofL.

Generally, it is supposed thatx0, x1, . . . , xr are linearly independent overK.Then, the condition of permissibility onc is thatc 6= 0 = (0, 0, . . . , 0). The nexttwo results follow readily.

THEOREM 6.13 All divisors of a linear seriesL are equivalent toB and henceare equivalent to each other.

DEFINITION 6.14 The common degree of the divisors of a linear series is theor-der of the linear series.

THEOREM 6.15 For everyζ 6= 0,

Ac = div(c0x0ζ + c1x1ζ + · · ·+ crxrζ)− div(ζ) +B.

COROLLARY 6.16 If B ≡ B′, then divisors inL can be represented in the form

div(c0x0 + c1x1 + · · ·+ crxr) +B′

for somex0, x1, . . . , xr of Σ linearly independent overK.

Proof. If B ≡ B′, thenB−B′ = div(ζ) for an elementζ ∈ Σ\0. Then Theorem6.15 applies. 2

160

THEOREM 6.17 Lety0, . . . , ys ∈ Σ be linearly independent elements overK suchthat the divisorsA′

d= div(d0y0+d1y1+. . . dsys)+B

′,with a fixed divisorB′, areexactly the divisors ofL arising fromx0, x1, . . . , xr, linearly independent elementsoverK. Thenr = s and, for some non-zero elementζ ∈ Σ,

yi = ζ∑cijxj

with det(cij) 6= 0.

Proof. Since div(yi) +B′ ∈ L, there existsci such that div(yi) +B′ = Aci. Thus

Aci≡ B′, whenceB ≡ B′. ThereforeB − B′ = div(ζ) for a non-zeroζ ∈ Σ.

Hence

div(yi) = div(∑r

j=0cijxj) +B −B′ = div(∑r

j=0cijxj) + div(ζ)

= div(∑r

j=0(cijxj)ζ) = div(∑r

j=0cijxjζ).

Thus div(∑r

j=0(cijxj)/yi) = 0. By Theorem 6.4, there existsk ∈ K\0 suchthat yi = k

∑rj=0cijxjζ. Absorbing the constantk into the coefficientscij , this

readsyj =∑r

j=0cijxjζ. As a consequence,s ≤ r. Similarly, r ≤ s, whencer = s follows. 2

COROLLARY 6.18 The integerr depends only on the seriesL and not on its rep-resentations.

DEFINITION 6.19 (i) The integerr is thedimensionof the linear seriesL.

(ii) A linear series of dimensionr and ordern is denotedgrn.

It should be noted that the linear seriesL is a projective space overK whosepoints are the divisorsL. The dimension ofL is equal to the dimension of theprojective space. HencePG(r,K) can be regarded as arepresentativeof L.

If d = kc, with k ∈ K\0, thenAd = Ac. Conversely, if

div(∑r

i=0cixi) +B = div(∑r

i=0dixi) +B,

then div(∑r

i=0cixi) = div(∑r

i=0dixi), whence, by Theorem 6.4,∑r

i=0cixi = k∑r

i=0dixi,

with k ∈ K\0. Assuming the elementsxi to be linearly independent overK,this only occurs whend = kc. Thus the set of divisorsAc of the linear seriesL is in one-to-one correspondence with the points of ther-dimensional projectivespacePG(r,K) overK, the divisorAc corresponding to the point with coordinates(c0, c1, . . . , cr).

A change of representation (6.1) of divisors inL changesc by a projectivity ofPG(r,K). To check this, use the same notation as in the proof of Theorem 6.17. IfA′

d = Ac, that is,

div(∑r

i=0diyi) +B = div(∑r

i=0cixi) +B′,

then div(∑r

i=0diyi) = div(∑r

i=0ciyi). Sinceyi = ζ∑r

j=0cijxj , so

div(∑r

i=0

∑rj=0dicijxj) = div(

∑ri=0cixi).


By Corollary 5.35,cj =∑r

i=0cijdj . Thusc = (c0, . . . , cr) andd = (d0, . . . , dr)are related in the way stated. PutC = (cij) and letCT be the transpose of the

matrixC; then the relation can also be written asc = λCT d andd = µCT −1c

with λ, µ ∈ K\0.

DEFINITION 6.20 A subseries of a linear seriesL is the set of all divisors corre-sponding to the point of a projective subspace ofPG(r,K), the projective spacerepresentative ofL.

The following result shows thatPG(r,K) is a good representative ofL.

THEOREM 6.21 Every subseries of a linear series is a linear series.

Proof. Let PG(s,K) be ans-dimensional subspace ofPG(r,K) whose points arein one-to-one correspondence with the divisors of a linear seriesL. Givens + 1linearly independent points inPG(s,K), say,

c0 = (c00, . . . , c0r), . . . , cr = (cs0, . . . , csr),

the subseries corresponding toPG(s,K) consists of the divisors

div((∑s

i=0λici0)x0 + · · ·+ (∑s

i=0λicir)xr) +B (6.2)

= div(λ0(c00x0 + · · ·+ c0rxr) + · · ·+ λs(cs0x0 + · · ·+ csrxr)) +B.

Put yi = ci0x0 + · · · + cirxr for i = 0, 1 . . . , r. Then the divisor (6.2) can berewritten as div(

∑ri=0yi)+B, that is, as the divisor of a linear seriesL′. A straight-

forward calculation shows thaty0, . . . , yr are linearly independent if and only if thematrixC = (cij) has maximum rankr + 1. This condition is satisfied here, sincec0, . . . , cr have been chosen to represent linearly independent points in PG(r,K).2

Let L be a linear series consisting of all divisors (6.1). If thereexistsc such that∑ri=0cixi = 1, thenL is normalised.

THEOREM 6.22 A linear seriesL is normalised if and only if its fixed divisorBbelongs toL.Proof. If L is normalised, take the(r + 1)-tuple c so that

∑ri=0cixi = 1; then

Ac = div(1) +B = B, showing thatB ∈ L.Conversely, ifB ∈ L, then there existsc 6= 0 such thatB = Ac. For such

an (r + 1)-ple c, the divisor div(∑r

i=0cixi) is equal to0. Therefore,∑r

i=0cixi

has no zeros or poles. By Corollary 5.35,∑r

i=0cixi is equal to a nonzero constantk ∈ K. Puttingdi = ci/k, this sum can be rewritten in the form

∑ri=0dixi = 1,

as required. 2

Actually, it may be supposed thatL is given in normalised form. In fact, as inTheorem 6.17, in (6.1) the elementsxi may be replaced byxiζ with an arbitraryζ ∈ Σ\0 andB by a suitableB′. Now, putζ =

∑ri=0cixi, andyi = xi/ζi,

bearing in mind thatζ 6= 0 asx0, . . . , xr are linearly independent overK. Thisgives thatAc = div(

∑ri=0ciyi) +B′, with

∑ri=0ciyi = 1.

162

THEOREM 6.23 If L andL′ are two linear series and an element ofL is equivalentto one ofL, thenL andL′ are both contained in a linear series.

Proof. Let

L= Ac | Ac = div(∑r

i=0cixi) +B, c 6= 0;L′ = A′

d | A′d = div(

∑sj=0diyi) +B′, d 6= 0.

In L′, the divisorB′ may be replaced by any divisor equivalent to it, in particularB. HenceB = B′ may assumed. Now, the linear series

L0 = div(∑r

i=0cixi +∑s

j=0djyj) +B

contains bothL andL′. 2

In this proof, x0, . . . , xr, y0, . . . , ys are not necessarily linearly independent.HencedimL0 ≤ 1 + dimL + dimL′. Another remark is made in the follow-ing theorem.

THEOREM 6.24 An arbitrary divisorC can be added to or subtracted from thedivisors of a linear series to obtain another linear series of the same dimension.

THEOREM 6.25 The multiplicities with which a placeP appears in the divisorsof a linear seriesL is bounded above and below.

Proof. The divisorB in L does not affect the question of boundedness; so assumethatB = (0). LetP be a place ofΣ. For i = 0, . . . , r, let−νi = ordP xi. SinceordP(x+ y) ≥ minordP x,ordP y, so ordP(

∑ri=0cixi) ≥ nP , where

nP =

0 if P is not a pole of anyxi;

min−νi if P is a pole of somexi.

which gives a lower bound. This also shows that

−nQ ≥ −ordQ(∑r

i=0cixi)

for each placeQ of Σ. Hence∑−nQQ ≻ div(

∑ri=0cixi)∞,

with the summation over all placesQ of Σ. Since

div(∑r

i=0cixi) = div(∑r

i=0cixi)0 − div(∑r

i=0cixi)∞,

so∑−nQQ+ div(

∑ri=0cixi) ≻ div(

∑ri=0cixi)0.

Hence∑−nQ +

∑ordQ(

∑ri=0cixi) ≥

∑mQ,

wheremQ is thezero number of∑r

i=0cixi; that is,mQ = ordQ(∑r

i=0cixi)0. Onthe other hand,

∑ordQ(div

∑ri=0cixi) = 0, by Corollary 5.35. This means that∑

nQ ≥∑mQ, whence

∑nQ ≥ mP . Since

∑nQ depends only onL but is

independent of the particular choice of the coefficientsc0, . . . , cr, it follows thatordP(

∑ri=0cixi) is bounded above by a number only depending onL. 2


THEOREM 6.26 LetP be a place ofΣ andν the smallest multiplicity with whichP occurs in the divisors ofL. Then the divisors ofL in which P occurs withmultiplicity at leastν + 1 make up a subseries ofL of dimensiondimL − 1.

Proof. Let (6.1) be a divisor ofL. ChooseB ∈ L in which P occurs inB withmultiplicity ν; that is,nP(B) = ν. ThenP is not a pole of anyxi, for otherwisePwould occur with multiplicity less thanν in some div(x) +B. In fact,

nP((xi) +B) = ordP(xi) + nP(B) = ordP(xi) + ν,

and, if ordP(xi) < 0 were true, thennP((xi) + B) < ν would also be true,contradicting the minimality ofν. Also, P is not a zero for everyxi. If this werenot true, then

nP((∑r

i=0cixi) +B) = ordP(∑r

i=0cixi) + ν > ν,

since

ordP(∑r

i=0cixi) ≥ min ordP(xi) ≥ νand ordP(xi) > 0 for 0 ≤ i ≤ r.

Let σ : Σ→ K((t)) be a primitive representation of the placeP. Then

σ(xi) = ai0 + ai1t+ · · · .Since ordP(xi) = 0 for somei, the coefficientai0 does not vanish for everyi.Therefore,

Ac = div(c0x0 + · · ·+ crxr) +B,

(c0(a00 + a01t+ · · · ) + · · ·+ cr(ar0 + ar1t+ · · · )) +B

= ((c0a00 + · · ·+ crar0) + tM1 + · · · ) +B,

whence

nP(Ac) =nP((c0a00 + · · ·+ crar0) + tL1 + · · · ) + nP(B)

=nP((c0a00 + · · ·+ crar0) + tL1 + · · · ) + ν.

Now,nP(Ac) ≥ ν + 1 if and only if

nP(div(c0a00 + · · ·+ crar0) + tL1 + · · · ) ≥ 1,

whencec0a00 + · · ·+ crar0 = 0.This means that, in the projective spacePG(r,K) representingL, the hyperplane

H = v(a00X0 + . . . , a0rXr) passes through the pointA = (c0, . . . , cr). Hence,the divisors ofL in whichP appears with multiplicity at leastν + 1 correspond tothose pointsA lying in H, and hence they constitute a linear series of dimensiondimL − 1. 2

COROLLARY 6.27 The divisors ofL in which a given placeP appears with mul-tiplicity at leastµ constitute a subseries ofL.

Proof. If µ ≤ ν, thenL itself is the subseries. Ifµ = ν + 1, then the corollary isjust Theorem 6.26. Ifµ = ν + 2, apply Theorem 6.26 to the subseries consistingof all divisors in whichP occurs with multiplicity at leastν + 1; continue thisprocess. 2

164

REMARK 6.28 With the notation of Theorem 6.13, ifν < µ, the subseries isempty. Also, the divisors ofL in whichP appears with multiplicity at leastν+2 isa subseriesL′ of L whose dimension is at leastdimL − 2. In fact,L may happennot to contain divisors in whichP appears with multiplicityν + 1. If this is thecase, Theorem 6.13 does not apply toL′ andν + 1.

COROLLARY 6.29 If P1, . . . ,Ps is a set of places ofS andµ1, . . . , µs are integ-ers, then the divisors ofL in which eachPi appears with multiplicity at leastµi

constitute a subseries ofL.

THEOREM 6.30 The effective divisors ofL constitute a subseries.

Proof. There are only finitely many places ofΣ in which the divisors (6.1) can havenegative degree, namely, the places occurring inB and the poles of the elementsxi.The theorem now follows from the previous corollary whenP1, . . . ,Ps are definedto be these places andµi = 0 for i = 1, . . . , s. 2

A linear series of effective divisors is denoted bygrn, wheren is its order andr

is its dimension.

THEOREM 6.31 In a linear seriesgrn, the order and dimension satisfyr ≤ n.

Proof. If r = 0, thenr ≤ n. Let r > 0, and take a placeP that does not occurin every divisor ofgr

n. Then the divisors ofgrn in which P occurs constitute a

subseriesgr−1n . SubtractingP from each of these divisors, a subseriesgr−1

n−1 isobtained. Applying induction onr shows thatr− 1 ≤ n− 1, whence the assertionfollows. 2

DEFINITION 6.32 Let gsm andgr

n be two linear series ofΣ. Thengsm is contained

in grn if every divisorD ∈ gs

m is contained in some divisorE ∈ grn; that is,E ≻ D.

Here,

gsm = div(c0x0, . . . , cmxm) +B | c = (c0, . . . , cm) ∈ PG(m,K)

is contained ingrn if and only if there existxs+1, . . . , xr ∈ Σ, andF ≻ B such that

grn = div(c0x0, . . . , crxr) + F | c = (c0, . . . , cr) ∈ PG(r,K).

THEOREM 6.33 If grn andgr′

n′ are linear series ofΣ such that a divisor of one ofthem is equivalent to a divisor of the other, thenn = n′ and both are contained ina linear series of effective divisors.

Proof. Theorem 6.25 together with Theorem 6.13 yieldn = n′. From Theorem6.23, gr

n and gr′

n are both contained in linear seriesL. Let L′ be the set of alleffective divisors inL. By Theorem 6.30,L′ is a linear series, and bothgr

n andgr′

n′

are subseries ofL′. 2

THEOREM 6.34 The effective divisors equivalent to a divisorC constitute a linearseries.


Proof. It may be that no effective divisor is equivalent toC, for instance whenordP C < 0. If this is the case then such effective divisors constitutethe emptylinear series. Otherwise there is at least one effective divisorB equivalent toC, andC may be replaced byB since the divisors equivalent toB are the same as thoseequivalent toC.

Let ordP B = n. The linear seriesL0 = div(x0) + B consists of effectivedivisors, more precisely,L0 = g0

n. Every divisorE equivalent toB can be writtenasE = div(f) + B for somef ∈ Σ. Let Γ the set of such elementsf ∈ Σ withE ranging over the set of all effective divisors equivalent toB. Note thatΓ is asubspace ofΣ regarded as a vector space overK. To show this, take a placeP ofΣ and two elementsf1, f2 ∈ Γ together with two constantsc1, c2 ∈ K. Since

ordP(c1f1 + c2f2) ≥ minordP f1,ordP f2,it follows that

div(c1f1 + c2f2) ≻∑

P∈ΣPnPP,

with nP defined to beminordPf1, ordPf2. Sincef1, f2 ∈ Γ impliesnP ≥ 0,the assertion follows.

To show thatΓ has dimension at mostn, suppose on the contrary thatΓ containsaK-independent setx0, . . . , xr with r > n. ThenL = Ac | c 6= 0 is a linearseries with dimensionr > n; but this contradicts Theorem 6.31. 2

DEFINITION 6.35 The linear series consisting of all effective divisors equivalentto a divisorC is acompletelinear series and it is denoted by|C|.

L EMMA 6.36 Let |A|, |B| be two complete linear series, withA andB not neces-sarily effective. IfA1 ≡ A andB1 ≡ B, then

|A1 +B1| = |A+B|, |A1 −B1| = |A−B|.Proof. By Lemma 6.8 (iii),A1 +B1 ≡ A+B, whence|A1 +B1| = |A+B|. Thesame holds when “+” is replaced by “-”. 2

DEFINITION 6.37 (i) The sumof two complete linear series|A| and|B| con-sists of all effective divisors equivalent toA+B, is denoted by||A|+ |B||.

(ii) The differenceof |A| and |B| is the linear series of all effective divisorsequivalent toA−B.

THEOREM 6.38 LetA andB be two effective divisors. Then||A| − |B|| consistsof all divisorsD −B, whereD ∈ |A| andD ≻ B.

Proof. SupposeC ∈ ||A|−|B|| = |A−B|. ThenC ≡ A−B. HenceC+B ≡ A,whenceC + B ∈ |A|. ThereforeC + B = A1 with A1 ∈ |A|, which yieldsC = A1 −B. Conversely, ifC = A1 −B with A1 ∈ |A| andA1 ≥ B, then

C ∈ |A1 −B| = ||A1| − |B|| = ||A| − |B||. 2

166

In Section 5.7 the concept of a differential of the formdx for a separating variablex ∈ Σ is defined and investigated. In the context of divisors, associate withdx thedivisor of the differentialdx defined by

(dx) =∑

ordPdxP.Every differential is of the formξdx with ξ ∈ Σ. Hence(ξdx) = div(ξ) + (dx),which shows that all divisorsξdx belong to the same equivalence class of divisors,thecanonical class. A canonical divisor has degree2g − 2, by Definition 5.54.

DEFINITION 6.39 The complete linear series|dx| is thecanonical series.

The canonical series has order2g − 2. In the next section, an explicit descriptionof the canonical series is given which demonstrates that itsdimension isg − 1.

6.2 LINEAR SYSTEMS OF CURVES

The aim of the present section is to describe the construction of a linear series froma linear system of curves by means of intersection divisors.LetΣ be a function fieldof transcendency degree1 and(F ; (x, y)) a model ofΣ, whereF is an irreduciblecurve andP = (x, y) is a generic point ofF . A divisor of Σ arises from everycurveG that containsF as a component. Such a divisor is theintersection divisorcut out onF by Φ and defined to be

F · G =∑I(P,G ∩ γ)P,

where the summation is over all places ofΣ, andγ is the branch ofF correspondingtoP.

THEOREM 6.40 Suppose that

(a) F = v(F (X0,X1,X2)) is an irreducible curve with generic point(1, x, y);

(b) G = v(G(X0,X1,X2) is any curve of degreem;

(c) F 6= ℓ∞ = v(X0), the line at infinity;

(d) F does not divideG.

Then

div G(1, x, y) = G · F −m(ℓ∞ · F).

Proof. For a placeP of Σ, let (x0(t), x1(t), x2(t)) be a primitive representationin special coordinates of the branch ofF corresponding toP. Note thatℓ∞ 6= Fimplies x0(t) 6= 0 . As usual, letx(t) = x1(t)/x0(t) andy(t) = x2(t)/x0(t).Then

ordP G(x0(t), x1(t), x2(t)) = ordP x0(t)m + ordP G(1, x, y).

Multiplying by P and summing over all placesP, it follows that∑

ordP G(x0(t), x1(t), x2(t)) = m∑

ordP x0(t)P +∑

ordP G(1, x, y)P.


By (4.32) the left-hand side isD · F , while∑

ordP x0(t)P = ℓ∞ · F . Therefore,

G · F = m(ℓ∞ · F) + div G(1, x, y),

whence the assertion follows. 2

COROLLARY 6.41 The curveG cuts out onF the divisorG · F equivalent tom(l∞ · F).

COROLLARY 6.42 If G1 = v(G1(X0,X1,X2)) is another curve of degreem thatdoes not containF as a component, thenG1 andG cut out equivalent divisors onF .

DEFINITION 6.43 Let Φ0 = v(ϕ0(X0,X1,X2)), . . . ,Φr = (ϕr(X0,X1,X2))be curves of the same degreem. Assume thatϕ0, . . . , ϕr are linearly independentoverK.

1. The set of curvesF = v(∑r

i=0ciϕi) with c = (c0, . . . , cr) ranging over allnon-trivial (r + 1)-ples of elements ofK is a linear system of dimensionr,and isthe linear system generated byΦ0, . . . ,Φr;

2. whenr = 1, the system is apencil;

3. whenr = 2, the system is anet;

There is a one-to-one correspondence between the curves of alinear system ofdimensionr and the points ofPG(r,K). It is worth noting that, if the polynomialsare written in their inhomogeneous form, that is,ϕ(X,Y ) = ϕ(1,X, Y ), thensome of them could have degree less thanm.

THEOREM 6.44 The curves in a linear system of degreem that do not containF as a component cut out onF the divisors of a linear series. The converse alsoholds, up to a fixed divisor.

Proof. For the converse, letL be a linear series consisting of all divisors (6.1).SinceΣ = K(x, y) with P = (x, y) a generic point ofF , there are polynomialsϕi(X,Y ), d(X,Y ) ∈ K[X,Y ] such thatxi = ϕ(x, y)/d(x, y) for i = 0, . . . , r.Hence

∑ri=0cixi =

∑ri=0ciϕi(x, y)/d(x, y).

Also,

Ac = div (∑r

i=0ciϕi(x, y)/d(x, y)) +B

= div (∑r

i=0ciϕi(x, y))− div(d(x, y)) +B.

If m is the maximum degree of the polynomialsϕi(X,Y ), define the homoge-neous polynomialϕi(X0,X1,X2) to beXm

0 ϕ(X1/X0,X2/X0), for i = 0, . . . , r.Let Fi = v(ϕ(X0,X1,X2)) be the corresponding curve, and letD = v(D) withD(X0,X1,X2) a homogeneous polynomial such thatD(1,X, Y ) = d(X,Y ).

168

Now, the linear systemΓ generated by the curvesFi is well defined. LetΦc bethe curvev(

∑ri=0ciϕi(X0,X1,X2)). From Theorem 6.40,

div (∑r

i=0ciϕ(x, y)) = Φc · F −m(ℓ∞ · F).

Therefore,

div

(∑ri=0

ciϕi(x, y)

d(x, y)

)+B = Φc · F −m(ℓ∞ · F)− div d(x, y) +B.

SinceC = −m(ℓ∞ · F)− (d(x, y)) + B does not depend onc, the linear systemΓ cuts out onF the linear series consisting of the divisors (6.1), up to thefixeddivisorC.

The direct part of Theorem 6.44 can be proved by reversing these steps. 2

REMARK 6.45 Theorem 6.44 establishes a close relationship between linear seriesand linear systems. However, the dimensiondim Γ of a linear systemΓ differs fromthe dimension of the linear seriesL cut out byΓ onF . In fact,dim Γ ≥ dimL andequality only holds when no two distinct curves inΓ cut out the same divisor inL.

To eliminate this difference, linear dependency moduloF is needed.

DEFINITION 6.46 If F = v(F (X0,X1,X2)) and

F (X0,X1,X2) dividesφ(X0,X1,X2)− ψ(X0,X1,X2),

thenφ ≡ ψ (mod F ).

Given a linear systemv(∑r

i=0ciϕi(X0,X1,X2)), if one of theϕi, sayϕ0, is lin-early dependent on the others moduloF , thenϕ can be eliminated; that is, thelinear systemv(

∑ri=1ciϕi) cuts out the same linear series onF asv(

∑ri=0ciϕi)

does. So, theϕi may be supposed linearly independent moduloF . Note that, ifm < degF , then linear independence means linear independence modulo F .

To achieve deeper results on linear series and especially onthe canonical series, amodel(F ; (x, y)) of Σ is chosen such that the irreducible curveF has only ordinarysingularities.

LetP1, . . . , Pk denote the singular points ofF and letri be the multiplicity ofPi

for i = 1, . . . , k. Let γij | j = 1, . . . , ri be the set of all branches ofF centredatPi.

DEFINITION 6.47 (i) Thedouble-point divisorof F , more precisely ofΣ withrespect to the model(F ; (x, y)), is

D =∑k

i=1

∑ri

j=1(ri − 1)Pij ,

wherePij is the place ofΣ corresponding toγij .

(ii) A curve G is anadjointof F if G · F ≻ D.

In this case,G · F = D + E with E an effective divisor , andE is thedivisor cutout byG onF , up toD. Also, G passes throughE′ for any effective divisorE′

such thatE ≻ E′.


THEOREM 6.48 A curveG is an adjoint ofF if and only if G has at least an(r − 1)-fold point at everyr-fold point ofF .

Proof. If P is an(r − 1)-fold point ofG andγ is a branch ofF centred atP , thenI(P,G ∩ γ) ≥ r− 1. Suppose now thatG has only ans-fold point, withs ≤ r− 2,at a point of multiplicityr of F . SinceP is ordinary singularity,G has at moststangents atP . Sinces < r, some tangent toF atP , sayℓ, is not tangent toG atP . Let γ be the branch whose tangent isℓ. Sinceγ is linear,I(P,G ∩ γ) = s. Ass ≤ r − 2, soG is not an adjoint. 2

THEOREM 6.49 The adjoints of a given degree cut out a complete linear series onthe curveF .

Proof. Let Φ be an adjoint of degreem. ThenΦ · F = D + E, with E ≻ (0). IfE′ is another effective divisor withE′ ≡ E, the existence of an adjointΦ′ of F forwhichΦ′ · F = D + E′ needs to be established.

First it is shown thatE ≡ E′ implies the existence of two curvesΨ andΨ′ suchthat

Ψ · F = E + F, Ψ′ · F = E′ + F.

SinceE′ = E+ div(u) for an elementu = a(x, y)/b(x, y) with a(X,Y ), b(X,Y )both inK[X,Y ], so

E + div b(x, y) = E′ + div a(x, y).

Applying Lemma 6.40 to the curvesC1 = v(a(X,Y )) andC2 = v(b(X,Y )) gives

div a(x, y) = C1 · F − (deg C1)(ℓ∞ · F),div b(x, y) = C2 · F − (deg C2)(ℓ∞ · F).

(6.3)

Hence,

E + C2 · F − (deg C2)(ℓ∞ · F) = E′ + C1 · F − (deg C1)(ℓ∞ · F).

Now, suppose that the reference system is chosen with the following two properties:

(a) ℓ∞ ∩ F is disjoint from bothC1 ∩ F andC2 ∩ F ;

(b) if P appears either inE or inE′, then the centre ofP is not a point ofℓ∞.

Then (6.3) implies thatdeg C1 = deg C2. Therefore,

E + C2 · F = E′ + C1 · F ,whenceC1 · F −E = C2 · F −E′. PuttingF = C1 · F −E, the assertion follows.

Note thatF need not to be effective even if this happens in some cases. Forinstance, ifE andE′ do not have places in common. Nevertheless, a slight changeis enough to ensure thatF is effective. In fact, writeF = F0 − F∞ with effectivedivisorsF0 andF∞. If F∞ = n0P0+. . .+ntPt, choose a lineℓi through the centreof Pi and define the totally reducible curveC whose components are the linesℓi fori = 1, . . . t, each countedni times. PutU = C · F . ThenU is an effective divisor

170

such thatU ≻ F∞. Then alsoU + F ≻ (0). ReplacingΨ by CΨ andΨ′ by CΨ′

ensures that

(CΨ) · F = C · F + Ψ · F = U + E + F = U + F,

(CΨ′) · F = C · F + Ψ′ · F = U + E′ + F = E′ + U + F.

This shows thatF ≻ (0) can be assumed. Then

(ΦΨ′) · F = D + E + E′ + F ≻ D + Ψ · F .

Put

Φ = v(ϕ), Ψ = v(ψ)), Ψ′ = v(ψ′).

By Theorem 4.64,

ϕψ′ = AF + ϕ′ψ,

whereA andϕ′ are homogeneous polynomials, andϕ′ 6≡ 0 (modF ). Thus

D + E + E′ + F = Φ · F + Ψ′ · F= (ΦΨ′) · F= (AF + ϕ′ψ) · F= (ϕ′ψ) · F= Φ′ · F + Ψ · F= Φ′ · F + E + F.

Hence,Φ′ · F = D + E′. 2

THEOREM 6.50 Every complete linear series is cut out, up to a fixed divisor, bythe adjoint of some degreem.

Proof. Let |E| be a complete linear series. Apart from the trivial case when|E| isempty, it may be assumed thatE is effective. Choose an adjoint passing throughE.Such adjoints certainly exist; for example, take lines passing through the centres ofthe branches corresponding to the places occurring inE, as well as, for everyr-foldpoint ofF , somer − 1 lines passing through it. The totally reducible curve whosecomponents are these lines is an adjoint ofF passing throughE. Any adjointΦ ofF throughE satisfiesΦ · F = D + E + E′ with E′ ≻ (0).

Let m denote the degree ofΦ, By Theorem 6.49, the adjoints of degreem cutout the complete linear series|E + E′| onF . The divisors of|E + E′| containingE′ are cut out by the adjoints of degreem which pass throughE′. By Theorem6.38, the linear series||E+E′| − |E′|| consists of all divisorsH −E′, whereH iscut out by the adjoints of degreem passing throughE′. To complete the proof it isenough to note that|E| = ||E + E′| − |E′||. 2

THEOREM 6.51 An irreducible plane curveF has genus0 if and only if F isbirationally equivalent to a line.


Proof. From Theorem 5.56, lines have genus0. For the converse,F = v(f(X,Y ))is assumed to have only ordinary singularities. Also,deg f(X,Y ) ≥ 2 may besupposed. From the proof of Theorem 3.24, the complete linear series cut out onF by the adjoint curves ofF of degreen− 1 has dimension

r ≥ 12 (n− 1)(n+ 2)− 1

2

∑ki=1ri(ri − 1)− (2n− 3) = 1.

Supposer > 1. Then there is an adjoint curveG of degree1 containing two non-singular points ofF . By Bezout’s theorem applied toF andG, it turns out that theymust have a common component, since

∑ki=1ri(ri − 1) + (2n− 3) + 2 > n(n− 1).

This is impossible sinceF is irreducible andG has lower degree thanF . Thereforer = 1. So, adjoint curves ofF degreen− 1, with one exception, can be expressedin the formGλ = v(gλ(X,Y )) with λ ∈ K andgλ(X,Y ) = g1(X,Y )λg2(X,Y ),whereG1 = v(g1(X,Y ) andG2 = v(g2(X,Y )) are two fixed, distinct adjoints ofdegreen− 1.

Choose a coordinate system so that

f(X,Y ) = uY n + . . . , g1(X,Y ) = v1Yn−1 + . . . , g2(X,Y ) = v2Y

n−1 + . . . ,

with u, v1, v2 6= 0, and so that no common point ofF andG1 is at infinity. LetR(X,λ) be the resultant with respect toY of F andG. Then

R(X,λ) = b0(λ) + b1(λ)X + . . .+ bs(λ)Xs,

wheres = n(n− 1). Now,R(X, 0) is the resultant ofF andG1, and it has degreen(n − 1). Hencebs(0) 6= 0, and sobs(λ0) = 0 for at most a finite set of valuesof λ0. Dismissing this finite set,F andGλ have no common point at infinity. ThepolynomialR(X,λ) ∈ K[X] has a root ofai of multiplicity at leastri(ri − 1)at eachri-fold pointPi = (ai, bi), and has a simple rootci at each of the2n − 3chosen simple pointsQi = (ci, di) of F . Since

k∑

i=1

ri(ri − 1) + 2n− 3)n(n− 1)− 1,

all but one of the roots ofR(X,λ) = 0 have been accounted. The remaining rootis, therefore,

ϕ(λ) =bs−1(λ)

bs(λ)−∑k

i=1ri(ri − 1)ai −∑k

i=1ci.2

Note thatϕ(λ) is a rational function. Similarly, from the resultant with respect toX, another rational functionψ(λ) is obtained.

For each valueλ which has not been dismissed, the pointQ = (ϕ(λ), ψ(λ))is onF . If Q = (x0, y0) is a point ofF not onG2, there is a uniqueλ0 suchthatQ ∈ Gλ0

, namelyλ0 = −g1(x0, y0)/g2(x0, y0). Hencex0 is a root of thepolynomialR(λ0,X) ∈ K[X], and sox0 = ϕ(λ0). Similarly,y0 = ψ(λ0). Thereis only a finite set of points ofF which are onG2. LetΣ be the function field of theline ℓ of equationY = 0. ThenΣ = K(ξ).

172

Let ω be the rational transformation

x′ = ϕ(ξ), y′ = ψ(ξ)

of Σ. Then the image curveF ′ of ℓ by ω is an irreducible plane curve whichhas infinitely many common points withF . By Bezout’s theorem, they coincide.HenceF is birationally equivalent to the lineℓ.

DEFINITION 6.52 For a curveF of degreen > 3, the adjoints of degreen− 3 arethecanonicalor special adjoints ofF .

THEOREM 6.53 If F has degreen > 3 and genusg, there areg linearly indepen-dent canonical adjoints ofF .

Proof. The curves of degreen − 3 form a linear system of dimension12n(n − 3).If F is non-singular, all these curves are adjoints and the result follows since

g = 12 (n− 1)(n− 2) = 1

2n(n− 3) + 1.

SupposeP is anr-fold point ofF . If G is an adjoint,P is an(r − 1)-fold pointof G by Theorem 6.48. To find a canonical adjointG of F , take any polynomialG(X,Y ) ∈ K[X,Y ] of degreen− 3. Arguing as in the proof of Theorem 3.24, acurveG = v(G(X,Y )) with

G(X,Y ) = a00 + a10X + . . .+ a0,n−3Yn−3

is an adjoint ofF if and only if (a00, a10, . . . , a0,n−3) is a solution of a certainsystem of linear equations. The number of the equations is the sum of the numberof singular points ofF , each counted1 + 2 + . . . + (r − 1) =

∑ki=1

12ri(ri − 1)

times, whereP1, . . . , Pk are the singular points ofF andri is the multiplicity ofPi. These equations can be interpreted as hyperplane equations inPG(N,K) withN = 1

2n(n−3). In this way,12∑r(r−1) is the total number of hyperplanes arising

from singular points ofF . The intersection of these hyperplanes is a subspaceΠM

of dimensionM ≥ N − 12

∑r(r − 1). ToM + 1 linearly independent points in

ΠM there correspond the same number of linearly independent canonical adjoints.Now the assertion follows from the fact thatM ≥ g−1, a consequence of Theorem5.56. 2

THEOREM 6.54 If F has degreen > 3, then the canonical series|(dx)| is cut outby the canonical adjoints.

Proof. With a change of coordinates, takeF as in the proof of Theorem 5.56. Thenthe polar curveF ′ = v(∂(F (X,Y )/∂Y ) of F at Y∞ exists and, ifP = (x, y)denotes a generic point ofF , thenx is a separable variable ofΣ. From the proof ofTheorem 5.56,

F ′ · F = D + div(dx)0, div(dx)∞ = 2(ℓ∞ · F).

If E is a canonical adjoint, thenE+2(ℓ∞ ·F) ≡ F ′ ·F , whenceE ≡ D+div(dx).Therefore, every canonical adjoint cuts out a divisor equivalent to div(dx) onF .By Theorem 6.49, the result is established provided that there exists at least onecanonical divisor. However, this follows from Theorem 6.53for g ≥ 1. Forg = 0,the canonical series is empty, and the assertion holds. 2


REMARK 6.55 The above approach to the canonical series does not work whenthe condition thatF has only ordinary singularities is dropped. However, in somecases, the canonical linear series may have a nice explicit description in terms oflinear systems of curves, as the following example shows.

EXAMPLE 6.56 Assumeh to be odd in Example 5.58, and consider the linearsystem∆ consisting of all plane curves of degree1

2 (n − 3) for which Y∞ is an12 (n − 1)-fold point. In other words,∆ consists of all completely reducible planecurves split into1

2 (n − 1) lines throughY∞. SinceY∞ is an(n − 2)-fold pointof F , ∆ cuts out onF an effective linear seriesL for which Y∞ is a fixed placeof multiplicity 1

2 (n − 2)(n − 3). It is easy to see thatL has dimension12 (n − 3).Discarding1

2 (n−2)(n−3)Y∞ from each divisor ofL, the resulting linear series isagd

2d with d = 12 (n− 3) = g− 1. Actually, this is the canonical series by Theorem

6.70 (i).

6.3 SPECIAL AND NON-SPECIAL LINEAR SERIES

Now, the Riemann–Roch Theorem can be proved; it is the main result concerningcomplete linear series. Again, a model(F ; (x, y)) of Σ is used with the propertythatF is an irreducible algebraic curve with only ordinary singularities. The degreeof F is denoted bym, andW stands for a canonical divisor.

If m ≤ 3, take a triangle meetingF in 3m distinct points. Then the imagecurve ofF under a quadratic transformation associated to the triangle is birationallyequivalent toF , and it has degree3m, which is larger than3 for m > 1. If m = 1,repeating this process, an irreducible curve of degree greater than3 is obtained.Thus, it may be assumed thatm > 3.

DEFINITION 6.57 For a complete linear seriesgrn,

(i) the index of specialityi, is the number of linearly independent canonicaldivisors through a given divisor;

(ii) the series isspecialif i > 0 andnon-specialif i = 0.

THEOREM 6.58 (Riemann)If grn is a complete linear series on the curveF of

genusg, thenr ≥ n− g.

THEOREM 6.59 (Riemann–Roch)If grn is a complete linear series|D| on the

curveF of genusg, withD an effective divisor andW a canonical divisor, then

(i) r = n− g + i;

(ii) r = g − 1 andn = 2g − 2 whenD = W ;

(iii) r = n− g whenn > 2g − 2.

174

EXAMPLE 6.60 Let F be as in Example 5.58. SetPi = (ai, 0) for i = 1, . . . , h,and denote byPi the place associated to the unique branch ofF centred atPi. LetD =

∑ni=1 ǫiPi with ǫi = 0 or ǫi = 1. By Example 6.56, every canonical divisor

is cut out onF by a completely reducible curve split into12 (h − 1) lines throughY∞ minus the fixed divisor12 (n− 2)(n− 3)P∞ whereP∞ is the place centred atY∞. From this, the index of specialityi of D is calculated:

i =

12 (h− 1)− degD for degD ≤ 1

2 (h− 1),

0 for degD > 12 (h− 1).

(6.4)

The Riemann–Roch Theorem yields thatdim |D| = 0 for degD ≤ 12 (h− 1).

In the proof of Theorem 3.24 it is shown that the maximum number of linearlyindependent adjoints of degreem−2 of F is 1

2m(m−1)−∑ 12ri(ri−1)+e with

e ≥ 0. These cut out divisors of degree

m(m− 2)−∑ri(ri − 1) = 2g − 2 +m

on F . Thus the adjoints of degreem − 2 cut out out a complete linear seriesgg−2+m+e2g−2+m . A priori, e may be positive, as the

∑ri(ri − 1) conditions imposed on

a curve of degreem−2 to be adjoint may be linearly independent. A posteriori, bythe Riemann–Roch Theorem,e = 0. For the present it may only be asserted thate ≥ 0. However, the assertion thate = 0 can be formulated as a consequence ofthe Riemann–Roch Theorem.

COROLLARY 6.61 The maximum number of linearly independent adjoints of de-greem− 2 ofF is 1

2m(m− 1)−∑12ri(ri − 1).

Similarly, the maximum number of linearly independent adjoints of degreem − 2of F is 1

2m(m − 1) −∑ 12ri(ri − 1) + e with e ≥ 0. Here, again,e = 0 follows

from the Riemann–Roch Theorem.To prove Theorem 6.59, a particular case is first examined.

L EMMA 6.62 If a complete linear seriesgrn has ordern = g + 1, thenr ≥ 1.

Proof. Sincem ≥ 2, so2g+m− 2 ≥ g+ 1. The adjoints of degreem− 2 cut outonF a seriesg2g−2+m

g−2+m+e. Given a divisorC ∈ grn, then it is first shown that there

existsD0 ∈ L with D0 ≻ C. By Theorem 6.38, the divisorsH − C, with H ∈ LandH ≻ C, is agr′

n′ with

r′ = g − 2 +m+ e− (g + 1) + e′ = m− 3 + e+ e′,

n′ =2g − 2 +m− (g + 1) = g − 3 +m,

with e′ ≥ 0. Sincer′ ≥ 0, soF exists inL with F ≻ C.Consider now the divisors inL which containF − C; they form a series of

dimension

s = (g − 2 +m)− (2g − 2 +m− (g + 1)) + e′ = 1 + e′ ≥ 1.

This series containsC and has dimension at least one. 2


COROLLARY 6.63 If a complete linear seriesgrn hasn > g + 1, thenr ≥ 1.

Proof. LetC+E ∈ grn, withC,E effective divisors anddegC = g+1. By Lemma

6.62,dim |C| ≥ 1; hence there exists an effective divisorC ′ such thatC ′ ≡ C butC 6= C ′. SoC ′ +E ≡ C +E butC +E 6= C ′ +E. Thus the dimensionr ≥ 1. 2

THEOREM 6.64 (Noether’s Reduction Theorem)Given a placeP centred at thesimple pointP ofF , if there exists a special adjoint through the divisorC but notthroughC + P, thenP is a fixed place of|C + P|.

Proof. The original proof of Noether which is reproduced here does not work inpositive characteristic whenP is a terrible point. To deal with terrible points, thesame argument as after Definition 3.30 does work; a quadratictransformation isapplied to give a curve, the geometric transform ofF , whose singular points areordinary, for which the corresponding pointQ of P is no longer terrible. Therefore,P may be assumed not to be terrible. Then there is a lineℓ through the centrePof the branch corresponding toP that meetsF in m distinct points, viewed as thecentres of the branches corresponding to the placesP = P1, . . . ,Pm. Then

(Fm−3 ℓ) · F = D + C + E + P1 + . . .+ Pm.

The adjoint curvesFm−2 cut out onF the complete linear series

|C + E + P1 + . . .+ Pm|.As above, the adjoint curvesFm−2 throughE + P2 + . . . + Pm form a linearsubsystemΓ′ which cuts out onF the complete series|C + P|; that is,

|C + P| = ||C + E + P1 + . . .+ Pm| − |E + P2 + . . .+ Pm|| .The curves ofΓ′ meetℓ in m − 1 points, and therefore containℓ as a component.It follows that each curve ofΓ′ passes through the centre ofP. SinceP is not inE + P1 + . . .+ Pm, soP is a fixed place of|C + P|. 2

REMARK 6.65 In the theorem, the caseC ≻ P is allowed, since places arecounted with multiplicity.

Proof of Theorem 6.59.The proof works by induction oni.Suppose first thati = 0; in this case, the proof is by induction onr. The first

case isr = 0, and it must be shown thatn = g. From Lemma 6.62 and Corollary6.63 it follows thatn ≤ g. As above, the canonical curves cut out agg−1+e

2g−2 onF ;therefore, through anyg − 1 places there is at least one canonical divisor. Sincei = 0, son ≥ g. Thereforen = g and the theorem is proved fori = r = 0.

Suppose, still withi = 0, thatr > 0. Given a placeP centred at a simple pointthat is not a fixed place forgr

n, Theorem 6.13 shows that the divisors ofgrn, with P

removed, form agr−1n−1. Note that the index of speciality for the latter series is zero.

In fact, if it were otherwise, there would exist a special adjoint through a divisorCof gr−1

n−1, which cannot pass throughP sincei = 0. By Theorem 6.64,P wouldbe a fixed place ofgr

n, a contradiction. By induction,r − 1 = n − 1 − g, whencer = n− g.

176

Finally, suppose thati > 0, and choose a divisorC in grn and a canonical curve

C′ throughC. LetP be a simple point ofF not lying onC′; then the placeQ withcentreQ is fixed for|C +P|, by Theorem 6.64. Therefore, this series|C +P| is agr

n+1, and its index of speciality is one less than that of|C|. Hence

r = n+ 1− g + (i− 1) = n− g + 1.

6.4 REFORMULATION OF THE RIEMANN–ROCH THEOREM

Theorem 6.59 can be written in the following form:

dim |C| = degC − g + 1 + dim |W − C|; (6.5)

here,C is an effective divisor andW a canonical divisor. This has been establishedin the case that|C| contains at least one effective divisor; that is,r ≥ 0. Now,it is shown that (6.5) is still valid when|C| contains no effective divisor; that is,r = −1.

If dim |W − C| ≥ 0, then, withW − C for C, equation (6.5) becomes thefollowing:

dim |W − C|=deg(W − C)− g + 1 + dim |W − (W − C)| (6.6)

=2g − 2− degC − g + 1 + dim |C|,sincedegW = 2g − 2. But this implies (6.5); so, consideringC orW − C givesequivalent results.

If dim |W −C| = −1, thenW −C contains no effective divisors. Since|C| alsocontains no effective divisors, it must be shown thatdegC = g−1 or, equivalently,that deg(W − C) = g − 1.

Suppose thatdegC < g − 1 and letC = A − B with bothA andB effectivedivisors. Then

dim |C +B| ≥deg(C +B)− g=degC + degB − g≥degB.

SodegB conditions can be imposed on the divisors of|C + B|; therefore, thereexistsA′ ∈ |C + B| such thatA′ ≻ B. But thenA′ − B ≡ C, andA′ − B iseffective, contradicting thatdim |C| = −1.

Now, suppose thatdegC > g − 1. Thendeg(W − C) > g − 1, and the aboveargument holds, withW − C in place ofC. Thus, (6.5) is valid for any divisorConF .

The Riemann–Roch Theorem can be reformulated in terms of functions onF .

DEFINITION 6.66 For any divisorC =∑nPP, effective or not,

L(C) = f ∈ Σ | ordP(f) ≥ −nP for all P ∈ P(Σ)= f ∈ Σ | div(f) + C ≻ 0 ∪ 0.


L EMMA 6.67 The setL(C) is a vector space overK with the following properties:

(i) if C ≻ D, thenL(D) is a subspace of the vector spaceL(C);

(ii) L(0) = K, andL(C) = 0 whendegC < 0;

(iii) dimL(C) is finite, anddimL(C) ≤ degC + 1 whendegC ≥ 0;

(iv) if C ≡ D, thendimL(C) = dimL(D).

Proof. (i) Let C = D + E with E ≻ 0. If f ∈ L(D), then div(f) + D ≻ 0,whence div(f) +D + E ≻ 0. ThusL(C) ⊃ L(D) and (i) follows.

(ii) If f ∈ Σ is a function, then div(f) = div(f)0 − div(f)∞ with div(f)∞ 6= 0.Therefore div(f) 6≻ 0, whenceL(0) = K. WhendegC < 0, then there is nofunctionf such that div(f) + C ≻ 0. HenceL(C) = 0.

(iii) A more general assertion is proved. LetA,B be divisors ofΣ. If B ≻ A,then

dimL(B)/L(A) ≤ degB − degA.

It is enough to prove it forB = A+P, as the general case follows from the specialcase by induction. It may be assumed thatL(B) 6= L(A). Choose an elementξ inL(B) that does not belong toL(A). If A =

∑nQQ, then ordP ξ = nP + 1 = m,

and hence ordP ξ−1 = −m. Now, ordP ξ−1η ≥ 0 for any elementη ∈ L(B).Given a primitive representationτ of P, let

τ(ξ−1) = ct−m + . . . , c 6= 0,

τ(η) = dtm + . . . .

Let V1(K) be one-dimensional vector space. Then the mapping

ϕ : L(B)→ V1(K),

η 7→ cd,

is K-linear, and its kernel isL(A). Thus,dimL(B)/L(A) ≤ 1. It follows thatdimL(B)/L(A) = 1.

(iv) Let D = C + div(g). The mapping

ϕ : L(C)→ L(D),

f 7→ fg,

isK-linear, and its kernel is the zero divisor0. Soϕ is an isomorphism. 2

THEOREM 6.68 For any divisorC on the curveF ,dimL(C) = degC − g + 1 + dimL(W − C). (6.7)

Proof. Let r = dimL(C)− 1 and suppose thatf0, f1, . . . , fr is a basis forL(C).Let AC = ∑cifi | ci ∈ K; thenAC is a linear series, and

∑cifi ∈ L(C) if

and only if the divisor div(f) + C is effective; that is, div(f) + C ∈ AC . Hence|C| is a linear series of effective divisors and has dimensionr. Now, Theorem 6.5gives the required result. 2

178

EXAMPLE 6.69 ConsiderF = X3 + Y 3 + Z3 defined over

F4 = 0, 1, ω, ω = ω2 = ω + 1and letF = v(F ). ThenF(F4) consists of the following points:

P0 = (0, 1, 1), P3 = (1, 0, 1), P6 = (1, 1, 0),P1 = (0, 1, ω), P4 = (1, 0, ω), P7 = (1, ω, 0),P2 = (0, 1, ω), P5 = (1, 0, ω), P8 = (1, ω, 0).

LetC = 3P0. Considerf1 = X/(Y + Z). Then div(f1) = P1 + P2 − 2P0. Notethat

Y + Z

X=

1

Y 2 + Y Z + Z2X2 = ut2,

with u = 1/(Y 2 +Y Z+Z2) invertible atP0 andt = X a uniformizing parameter;sof1 has a pole of order 2 atP0. Similarly,f2 = Y/(Y + Z) has a pole of order 3atP0, since

Y + Z

Y=

1

Y (Y 2 + Y Z + Z2)X3.

Here, div(f1) = P3 + P4 + P5 − 3P0. Hencef1 andf2 both are inL(C). Sinceg = 1, degC = 3, L(W − C) = 0, sodimL(C) = 3. Therefore, a basis forL(C) is 1, f1, f2.

6.5 SOME CONSEQUENCES OF THE RIEMANN–ROCH THEOREM

THEOREM 6.70 (i) The canonical series is the only series on a curve of genusg that has order2g − 2 and dimensiong − 1.

(ii) The canonical series has no fixed place.

(iii) A divisor of a complete specialgrn putsn − r conditions on a canonical

divisor to contain it.

Proof. (i) A gg−12g−2 with dimensionr = g − 1 andr > n − g is special; so it is

totally contained in the canonical series, and hence coincides with it, having thesame dimension.

(ii) If the canonical series had a fixed pointP , the residual series would be agg−12g−3. The addition of another placeQ to the divisors of this series would give a

gg−12g−2 different from the canonical series.(iii) If i is the index of speciality of thegr

n, then i = g − (n − r), whencei− 1 = g − 1− (n− r). 2

EXAMPLE 6.71 Let C6 be a sextic curve with one singular pointP0 that is anordinary quadruple point; for example, if charK 6= 2, let

F = XY (X − Y )(X + Y )Z2 +X6 + Y 6.


Then a special adjoint is a cubic curve with a triple point atP0, and is therefore asetS of three lines throughP0. The canonical series therefore has order 6, sinceeach line ofS meetsC6 in two points apart fromP0, and dimension 3, since thepencil of lines throughP0 has dimension 1. SinceC6 has genus

g = 12 (6− 1)(6− 2)− 1

24(4− 1) = 4,

this is consistent with the canonical series being ag36 .

Now, the converse of Theorem 6.64 is established.

THEOREM 6.72 Given a linear seriesgrn onF , if there exists a placeP such that

the complete linear seriesgrn + P = D + P | D ∈ gr

n hasP as a fixed place,then thegr

n is complete and special, and there exists a canonical divisor containinga divisor ofgr

n but not containingP.

Proof. If grn + P has the fixed placeP, since between the order and dimension

of this series there is the inequalityr ≥ (n + 1) − g, so r > n − g; hence theseriesgr

n is special. Further,grn is complete, since otherwisegr

n + P would not becomplete. Now, if each canonical divisor containing a divisor of gr

n also containedP, thengr

n +P would also be special, with the same index of speciality asgrn; that

is, a divisor ofgrn and one ofgr

n + P impose the same number of conditions on acanonical divisor. This contradicts Theorem 6.70. 2

COROLLARY 6.73 The complete seriesgrn +P hasP as a fixed place if and only

if the seriesgrn is special and there exists a canonical divisor containing adivisor

of grn but not containingP. In other words, for any effective divisorA and any

placeP, one and only one of the following equations holds:

(i) dim |A+ P| = dim |A|;

(ii) i(|A+ P|) = i(|A|).

DEFINITION 6.74 Thesumof two linear seriesgrn, g

r′

n′ is

grn + gr′

n′ = D +D′ | D ∈ grn, D

′ ∈ gr′

n′.

COROLLARY 6.75 A complete seriesgrn is non-special ifn > 2g−2 or r > g−1,

whereas it is special ifn < g.

THEOREM 6.76 (Reciprocity Theorem of Brill and Noether). If grn and gr′

n′ areresidual with respect to the canonical series|W |, that is, |gr

n + gr′

n′ | = |W |, thenn− 2r = n′ − 2r′.

Proof. The speciality indicesi andi′ of grn andgr′

n′ satisfyi = r′+1 andi′ = r+1.Thusr = n−g+ i = n−g+r′+1 and, similarly,r′ = n′−g+r+1: Subtractingthe two equations gives the result. 2

THEOREM 6.77 (Clifford’s Theorem)If a linear seriesgrn is special, thenn ≥ 2r.

180

Proof. It may be supposed that the seriesgrn is complete. IfH is the residual of a

divisorG of grn with respect to a canonical divisor, thenG imposesr conditions on

the canonical divisors throughH, while it imposesn− r conditions on an arbitrarycanonical divisor containing it, by Theorem 6.70(iii). As it cannot happen thatindependent conditions in the general case become dependent for a subseries, son− r ≥ r, whencen ≥ 2r. Equality can only occur when the series is complete.

An alternative view is as follows. LetΦ be the canonical linear system ofF . Ifthegr

n is complete, then it is cut out by a subsystemΦ′ of dimensionr of Φ. Moreprecisely, for a given divisorG of gr

n and a canonical adjointG throughG, letHbe the residual ofG in the canonical divisorC cut out byG; that is,C = G +H.ThenΦ′ consists of all special adjoints throughH.

Now, consider the subsystemΦ′′ of all the canonical adjoints throughG. Ifi− 1 is the dimension ofΦ′′, thenr = n− g + i by the Riemann–Roch Theorem.Also, Φ′ andΦ′′ haveG in common but not a pencil of curves, as otherwiseFwould be a component of a curve in such a pencil. Hence the system containingboth Φ′ and Φ′′ has dimension at mostr + i − 1. SincedimC = g − 1, sor + i− 1 ≤ g − 1 = n− r + i− 1, whencen ≥ 2r. 2

EXAMPLE 6.78 A non-singular quartic curveC4 has genus 3. The lines of theplane are the special adjoints and cut out the canonical seriesg2

4 . The lines througha pointQ off C4 cut out a specialg1

4 .

A complement to Theorems 6.31 and 6.77 is the following result.

THEOREM 6.79 (i) A necessary and sufficient condition for a curve to be ra-tional is that it possesses at least one linear seriesgn

n .

(ii) The complete linear series of a rational curve aregnn for n ∈ N.

Proof. A seriesgnn is complete as its dimension is the largest possible for its order;

see Theorem 6.31.Suppose the curve is rational; that is, it is the line at infinity ℓ∞ = v(X0) up

to a birational transformation. The linear system of all plane curves of degreencuts out onℓ∞ a linear series of ordern. Since then + 1 monomialsXiY n−i

for i = 0, . . . , n are linearly independent overK, the linear series has dimensionn, and hence it is agn

n . Since there is a curve of degreen cutting out any pre-assigned divisor of degreen, sogn

n consists of all divisors ofℓ∞ of degreen. Thus,everything apart from the sufficiency condition is proved.

Suppose now that the curve possesses a seriesgnn , with n > 0. By Theorem 6.31,

gnn is complete. Choose a simple pointP on the curve such that the corresponding

placeP is not a fixed place ofgnn . The divisorsC − P, with C ∈ gn

n andCcontainingP, constitute a linear subseriesgn−1

n−1 . By repeating this argument, theexistence of ag1

1 follows. Let

g11 = div(c0 + c1x) +B | c0, c1 ∈ K, (c0, c1) 6= (0, 0).

By Theorem 7.35,g11 is not composed of an involution. Thus,[Σ : K(x)] = 1,

whenceΣ = K(x). 2


6.6 THE WEIERSTRASS GAP THEOREM

In Section 5.4, it is shown that every functionz ∈ Σ has a finite number of poles,giving rise to the effective divisor of poles div(z)∞. The converse of this theoremis generally false, as not every effective divisor is the pole divisor of a function inΣ. This is explained in the following theorem.

THEOREM 6.80 An effective divisorA is the divisor of polesdiv(z)∞ of somefunction ofΣ if and only if|A| has no fixed place.

Proof. If A is the divisor of poles ofz, thenA ≡ div(z)0, that is, |A| containsdiv(z)0. SinceA and div(z)0 do not share any place,|A| has no fixed place.

Conversely, let|A| have no fixed places. It is shown first thatB ≡ A for somedivisor B such thatA andB have no common place. WriteA =

∑ki=1nPi

Pi.By assumption, for everyj with 1 ≤ j ≤ k, there is a divisorAj ≡ A suchthatAj does not containPj . If Fj = v(ϕj(X0,X1,X2)) cuts outAj , then thecurves of the linear systemv(

∑ki=1λjϕj) cut out divisors in|A|. Such a divisor

containsPj if and only if thek-ple (λ1, . . . , λk) is a solution of a certain system oflinear equations. Choosing coefficientsλj that satisfy none of the equations in thatsystem, a divisorB of the desired type is obtained. Now, the argument in the firstpart can be reversed. 2

As a corollary, the following theorem is obtained.

THEOREM 6.81 Every effective divisor is the divisor of poles of some function ofΣ if and only ifΣ is a rational function field.

REMARK 6.82 Theorem 6.81 remains true if the condition is only required fordivisors consisting of one place.

For the rest of this section, the genusg of Σ is assumed to be positive.

THEOREM 6.83 Let As = P1 + . . . + Ps be an effective, non-special divisorwith placesPi not necessarily distinct, and letA0 = 0, An = P1 + . . .Pn, for1 ≤ n ≤ s. Then

(i) there exist at leastg values ofn for whichAn is not the divisor of poles ofany function inΣ;

(ii) the indices of thePi can be relabelled such that this occurs precisely forgvalues ofn.

Proof. Putin = i(|An|); in particular,i0 = g, is = 0. By Theorem 6.73,

in = in−1 if dim |An−1| < dim |An|,in = in−1 − 1 if dim |An−1| = dim |An|.

This implies the existence of exactlyg values ofn for which the relationshipdim |An−1| = dim |An| holds, that is, for whichPn is a fixed place in|An|. FromTheorem 6.80, there arek values ofn, with k ≥ g, for whichAn is not the divisorof poles of a function inΣ.

182

To show (ii), merely relabel the indices of thePi as follows. Letn1 be thelargest values ofn for which |An| has a fixed place. Relabel the places ofAn1

sothatPn1

is such a fixed place. Proceeding similarly for the new labelling, letn2 bethe largest value ofn less thann1 for which |An| has a fixed place, and relabel theplaces ofAn2

so thatPn2is fixed. By continuing this procedure, an arrangement

of the placesPi is eventually obtained so that wheneverAn has a fixed place,Pn

is fixed. From the preceding argument it then follows thatk = g. 2

Now, the caseP1 = P2 = . . . = Ps is considered.

DEFINITION 6.84 For any placeP, a non-negative integern is apole numberofP if there is a functionz such that div(z)∞ = nP; otherwise, n is agap numberof P.

L EMMA 6.85 The set of pole numbers ofP is a sub-semigroupH(P), Weierstrasssemigroup atP, of the additive semigroup of the non-negative integers.

Proof. If div(z1)∞ = n1P and div(z2)∞ = n2P, then div(z1z2)∞ = (n1 +n2)P.2

The integern is a pole number ofP if and only if

dim(|nP|) > dim |(n− 1)P|.By Theorem 6.83, this occurs if and only ifi(nP) = i((n − 1)P). In particular,everyn > 2g − 1 is a pole number.

On the other hand,1 is a gap number. Suppose, on the contrary, that1 is apole number ofP. ThenH(P) is the whole semigroup of non-negative integersshowing that there are no gaps at all, contradicting Theorem6.83. Therefore, forP1 = . . . = Pn, Theorem 6.83 reads as follows.

THEOREM 6.86 (Weierstrass Gap Theorem)For any placeP, there exist exactlyg gap numbersn1 < n2 < . . . < ng, with n1 = 1, ng ≤ 2g − 1.

It is shown in Section 7.6, see Corollary 7.54, that all but finitely many places havethe same sequence of gap numbers. These are theordinary placesof Σ. If thegap sequence ofΣ at an ordinary point is(1, 2, . . . , g) thenΣ hasclassical gapsequence at a general point.The non-ordinary places are theWeierstrass points,and thei-th gap of an ordinary point is greater than or equal to thei-th gap of anyWeierstrass point. Furthermore, ifg ≥ 2, thenΣ contains at least one Weierstrasspoint. ordinary point

Two illustrative examples are now given.

EXAMPLE 6.87 Let F = v(f(X,Y )) be the irreducible plane curve in Example5.58. Assume thath is odd. ThenF has genusg = 1

2 (h− 1). It is now shown thatthe gap numbers ofK(F) at ordinary places are1, 2 . . . , g, and thatF has exactly2g + 2 Weierstrass points where the gap numbers are the odd integers less than orequal to2g − 1.


With same notation employed in Example 5.58,

div(x− ai)−1 =

−2Qi + 2P∞ for h odd,

−2Qi + P(1)∞ + P

(2)∞ for h even.

This shows that2 is a pole number atQi. Sinceg 6= 0, it follows that the semigroupH(Qi) consists of all even numbers. Therefore, the odd integers less than or equalto 2g − 1 are gap numbers. This holds true forP∞. These2g + 2 places are theonly Weierstrass points ofF .

To show the latter assertion, it suffices to prove that1, 2, . . . , g are gap numbersat any placeP corresponding to a branch ofF centred at a pointP = (a, b) withb 6= 0. This occurs if and only ifi(nP) = i((n−1)P)−1 for n = 1, . . . , g. To dothis, the canonical series ofK(F) is first determined. LetQ = (x, y) be a genericpoint of F . The conditionp ∤ h = deg f(X,Y ) ensures thatx is a separatingvariable. From the computations carried out in Example 5.58,

(dx) = Q1 + . . .+Qh − 3P∞,

div y = Q1 + . . .+Qh − hP∞.

Therefore,(y−1dx) = (h − 3)P∞. This shows that the canonical series ofK(F)is |(h− 3)P∞|. Also,

div (xi) = i(R1 +R2)− 2iP∞,

whereR1 andR2 are the places corresponding to the branches centred at the pointsof F lying at finite distance on theY -axis. The caseR1 = R2 is possible, and thisoccurs whenf(0) = 0, that is, theY -axis meetsF at finite distance only in theorigin. Hence, ifi = 0, 1, . . . , (h− 3)/2, then div(xi)+ (h− 3)P∞ is a canonicaldivisor. Since1, x, . . . , x(h−3)/2 are linearly independent overK, it follows interms ofg that canonical series ofK(F) consists of all divisors

div (c0 + c1x1 + . . .+ cg−1xg−1) + (2g − 2)P∞

asc = (c0, . . . , cg−1) ranges over the points ofPG(g − 1,K).Forn = 1, . . . , g− 1, this implies that a canonical divisorW containsnP if and

only if

W = div (c0 + c1x1 + . . .+ cg−1xg−1) + (2g − 2)P∞

such that the polynomialc0 + c1X + . . . + cg−1Xg−1 has ann-fold root in a.

Therefore,i(nP) = i((n− 1)P)− 1, for n = 1, . . . , g.

EXAMPLE 6.88 Takep = 3, and letH3 = v(Y 3 +Y −X4), the Hermitian curve.SinceH3 has genus3, it is the canonical curve of its function fieldK(H3). Hence,the canonical divisors are cut out onH3 by lines.

LetP be the place arising from the pointP = (a, b) of H3. Assume at first thatbotha, b are in the prime fieldF3 ofK. There are27 such points. Ifℓ is the tangenttoH3 atP , thenI(P, ℓ ∩H3) = 4. Thus

i(P) = 2, i(2P) = i(3P) = i(4P) = 1, i(5P) = 0.

This shows that the gap numbers ofK(H3) atP are1, 2, 5. The same holds for theplaceP∞ centred at the infinite point of theY -axis. Now, assume that eithera or bdoes not belong toF3. ThenI(P, ℓ ∩H3) = 3, and hence

i(P) = 2, i(2P) = i(3P) = 1, i(4P) = 0.

184

Therefore, the gap numbers ofK(H3) at P are1, 2, 4. To sum up, the gap se-quence at ordinary places is(1, 2, 4), while there are28 Weierstrass points withgap sequence(1, 2, 5).

LetP be any place ofΣ, and letm denote the smallest pole number ofP. Choosean elementx ∈ Σ such that div(x)∞ = mP . Every integern in the Weierstrasssemi-groupH(P ) can be written in the formn = am + i wherea andi are non-negative integers such thata > 0 andi < m. Since there are finitely many gapsat P, all the valuesi = 0, 1, . . . ,m − 1 arise. Letaim + i be the least polenumber congruent toi modulom. Choose elementsz0, z1, . . . , zm−1 ∈ Σ suchthat div(zi)∞ = (aim+ i)P. Note thatz0 may be taken to bex.

It is shown that everyy ∈ Σ which has a pole atP but is regular elsewhere canbe written in the form,

y = A(x) +B(x)z1 + . . .+ C(x)zm−1, (6.8)

with A(X), B(X), . . . , C(X) ∈ K[X].Let n be the pole number arising fromy; that is, div(y)∞ = nP . If n ≡ i

(mod m), write n = am + i. Sincen cannot be less thenaim + i, there is aconstantc so thatz′ = y− cxa−aizi is either zero or div(z′)∞ = n′P with n′ < n.Proceeding then similarly withz′, an equation (6.8) is obtained.

Applying this equation to the firstm− 1 powers ofy gives the following:

B1(x)z1 + . . .+ C1(x)zm−1 = y −A1(x),B2(x)z1 + . . .+ C2(x)zm−1 = y2 −A2(x),

......

...Bm−1(x)z1 + . . .+ Cm−1(x)zm−1 = ym−1 −Am−1(x).

(6.9)

Note that(z1, . . . , zm−1) can be viewed as the solution of a system of linear equa-tions overK(x).

If these equations are linearly independent, that is, the(Bi(x), . . . , Ci(x)) arem− 1 linear independent vectors over the fieldK(x), then Cramer’s rule gives theresult,

zi = Qi,1(x)(y −A1(x)) +Qi,2(x)(y2 −A2(x)) + . . .

+Qi,m−1(x)(ym−1 −Am−1(x)),

(6.10)

with Qi,1(X), Qi,2(X), . . . , Qi,m−1(X) ∈ K(X).Otherwise,

P (x) + P1(x)y + . . .+ Pm−1(x)ym−1 = 0, (6.11)

with P (X), P1(X), . . . , Pm−1(X) ∈ K[X]. Since

ordP(Pk(x)yk) = mdegPk(X) + kn

from (6.11), so

m degPk(X) + kn = m deg(Pk′(X)) + k′n,

for somek, k′ with 1 ≤ k < k′ ≤ m− 1. This implies that

n

m=

degPk(X)− degPk′(X)

k − k′ .


Now k andk′ are both less thanm. Therefore,m andn have a common divisor.Now taken prime tom. Then the equations in (6.9) must be linearly indepen-

dent, and (6.10) holds. This together with (6.8) applied toym gives an irreducibleequation of the form,

U0(x)ym + U1(x)y

m−1 + . . .+ Um−1(x)y + Um(x) = 0, (6.12)

with Ui(X) ∈ K[X], andU0(X) 6= 0. More precisely,U0(X) is a non-zeroconstant polynomial as bothx andy have a unique poleP. Further, ifi < j, then

mdegUi(X) + (m− i)n= ordP(Ui(x)ym−i)

= ordP(Uj(x)ym−j)

=mdegUj(X) + (m− j)nonly for j ≡ i (mod m). As this congruence is only valid fori = 0, j = m, itfollows thatdegUm(X) = n and that

degUi(X) < in/m, for i = 1, . . . ,m− 1. (6.13)

Sincegcd(m,n) = 1 implies thatΣ = K(x, y), the following result is obtained.

THEOREM 6.89 (Weierstrass normal form)Let F an irreducible curve of genusg ≥ 1. For a placeP of K(F), let m be the first non-gap atP and letn be theleast non-gap which is prime tom. Then

(i) F = v(f(X,Y )) with

f(X,Y ) = Y m + U1(X)Y m−1 + . . .+ Um−1(X)Y + Um(X), (6.14)

whereUi(X) ∈ K[X], degUm(X) = n and (6.13) holds;

(ii) P = (x, y) is a generic point ofF = v(f(X,Y )), if div(x)∞ = m anddiv(y)∞ = n;

(iii) the branch ofF associated toP is the unique branch ofF with centre atY∞.

A consequence is the following result.

PROPOSITION 6.90 LetF = v(f(X,Y )) be given in a Weierstrass normal form(6.14). Then the genusg has the following properties:

(i) g ≤ 12 (n− 1)(m− 1);

(ii) if no point ofF other thanY∞ is singular, theng = 12 (n− 1)(m− 1).

Proof. (i) The idea is to apply the Riemann–Roch Theorem 6.59 to the completelinear series|(mn)P|. For each ordered pair(i, j) of non-negative integersi, j forwhich im + jn < mn, the divisor div(xiyj) + (mn)P belongs to|(mn)P|. Thisalso holds for div(xn) + (mn)P. As in the discussion on (6.9), the hypothesisgcd(m,n) = 1 implies that1, y, . . . , ym−1 are linearly independent overK(x).Therefore,xn and the functionsxiyj form a linearly independent set in the vectorspaceL(P). So,

dim |(mn)P|≥n+m− 1 +∑m−1

j=1 ⌊n(m− j)/m⌋=n+m− 1 + 1

2 (m− 1)(n− 1). (6.15)

186

In particular,dim |(mn)P| > 12nm = 1

2 deg(mn)P0. This and Clifford’s Theorem6.77 imply that|(mn)P| is not special. Then, from (6.15) and Theorem 6.59,g ≤ 1

2 (n− 1)(m− 1). 2

(ii) Both m andn cannot be divisible byp. Suppose first thatm 6≡ 0 (mod p).Thenx is a separable variable ofK(F) and

g(X,Y ) = ∂f/∂Y = mY m−1 + . . .+ (m− 1)Ui(X)Y m−1 + . . .+ Um−1(X).

A primitive representation of the branch ofF associated toP is

x(t) = t−m, y(t) = t−n + . . . .

From (6.13),

ordt(m− 1)y(t)m−1 = −(m− 1)n

< −⌊in/m⌋m− (m− i− 1)n

≤ ordt(m− i)Ui(x(t))y(t)m−i−1.

Hence ordtg(x(t), y(t)) = −n(m− 1). From this and Corollary 5.35,∑

Q ordQg(x, y) = n(m− 1),

where the summation is over all places ofK(F) other thanP. On the other hand,since the branch ofF associated to such a placeQ is centred at a non-singular pointof F , from Theorem 5.55,

ordQ g(x, y) = ordQ dx.

Also,

ordP dx = −(m+ 1).

Now, by Definition 5.54,2g − 2 = n(m− 1)− (m+ 1), whence (ii) follows.If n 6≡ 0 (mod p), a similar argument can be used, this time with

g(X,Y ) = ∂f/∂X = nXn−1 + . . .

anddy.

6.7 THE STRUCTURE OF THE DIVISOR CLASS GROUP

In Section 6.1, basic facts on divisors of a function fieldΣ of transcendency degree1 are discussed. Now, a fundamental result on the divisor class groupPic(Σ) isstated. As the groupPic(Σ) is written additively, that is,[A+ B] = [A] + [B] for[A], [B] ∈ Pic(Σ), the meaning ofm · [A] is just [mA], for any positive integermand divisorA of Σ. So, the set[A] | m · [A] = 0, [A] ∈ Pic(Σ) is a subgroupPic(Σ,m) of Pic(Σ). Therefore,

Pic0(Σ,m) = Pic(Σ,m) ∩ Pic0(Σ)

is a subgroup ofPic0(Σ), which plays an important role in the study of the structureof the divisor class group, as its order only depends ong.


THEOREM 6.91 For any positive integerm not divisible byp,

Pic0(Σ,m) ∼= (Zm)2g,

whereg is the genus ofΣ and (Zm)2g is the direct product of2g copies of theadditive group of integers modulom.

Theorem 6.91 is a special case of a more general result on abelian varieties. Theidea is to prove the theorem first forK = C and then use reduction theory to extendit from C to anyK. No more detail is given here. A proof of Theorem 6.91. validin the case thatK is the algebraic closure of a finite field, is given in Section 10.7.The following example illustrates how to determinePic0(Σ, 2) in a particular case.

EXAMPLE 6.92 Let F andΣ be as in Example 5.58, and supposeh to be odd.With the notation of Example 6.60, letD1 be the set of all divisorsD for whichǫ1 = 0. ThendegD < h− 1.

For two distinct divisorsD1 andD2 in D1, writeE1 − E2 = D1 −D2 with E1

andE2 effective divisors. Note that bothE1 andE2 are inD1. Thus

degE1 = degE2 <12 (h− 1).

Thendim |E1| = 0 by (6.4), and henceD1 6≡ D2. Now, for anyD ∈ D1, letF = mP1 − D with m = degD; then[F ] 6= [0]. Since2P1 ≡ Pi, as shown inExample 6.60, so2mP1 ≡ 2D; that is,[2F ] = [0]. Since divisorsD in D1 can bechosen in2h−1 different ways, the total number of inequivalent divisorsF amountsto 2h−1. Sinceg = 1

2 (h− 1), Theorem 6.91 shows that such divisorsF constitutea full representative system of cosets in the divisor class groupPic(Σ, 2).

The hypothesisp6 |m cannot be dropped in Theorem 6.91, even thoughPic0(Σ,m)is finite. In particular,Pic0(Σ, p) ∼= (Zp)

γ for a non-negative integerγ, thep-rankof Σ. This important particular case appears in Theorem 12.58. Here, the mainresult is stated, which can be viewed as a weaker form of Theorem 6.91.

THEOREM 6.93 Thep-rankγ is less than or equal to the genusg.

A proof of Theorem 6.93 valid in the case whereK is the algebraic closure of afinite field is given in Section 10.7.

6.8 EXERCISES

1. For any two effective divisorsA, B,

dim |A|+ dim |B| ≤ dim |A+B|.

2. LetP be a place ofΣ. Show that the following conditions are equivalent:

(i) P is a Weierstrass point;

(ii) gP is a special divisor;

(iii) mP is a special divisor for somem ≥ g.

188

3. Show that Example 6.88 is the only irreducible plane curveof genus3 withnon-classical gap sequence.

4. Letp = 3. For function fields of genus3 show that ordinary places have theclassical gap sequence(1, 2, 3, 4).

5. For q odd, letF = v(X(q+1)/2 + Y (q+1)/2 + 1) be the Fermat curve ofdegree1

2 (q+1). Show that the Weierstrass semigroup ofF at each inflexionpoint ofF is 〈 12 (q − 1), 1

2 (q + 1)〉.

6. Forq ≡ 3 (mod 4), letF = v(Y q + Y −X(q+1)/4). Show that the Weier-strass semigroup ofF at the placeP∞ arising from the unique branch centredatY∞ is 〈q, 1

4 (q + 1)〉.

7. LetF an irreducible curve of genusg ≥ 0. For a placeP of K(F), letm bethe first non-gap atP, andn another non-gap atP. Choosex, y ∈ Σ suchthat div(x)∞ = m and div(y)∞ = m. Show that there exists an irreduciblepolynomialg(X,Y ) ∈ K[X,Y ] with the following properties:

(i) g(x, y) = 0;

(ii) g(X,Y ) = Y ℓ + U1(X)Y ℓ−1 + . . . + Uℓ−1(X)Y + Uℓ(X), whereUi(X) ∈ K[X];

(iii) If n/m = n′/m′ with gcd(n,m) = 1, thenm′ | ℓ | m;

(iv) degUℓ(X) = (nℓ)/m;

(v) degUi(X) ≤ ⌊(ni)/m⌋ for 1 ≤ i ≤ ℓ− 1;

(vi) the subfieldK(x, y) of K(F) is a separable extension ofK(x);

(vii) if m does not dividen, theng > 0;

(viii) if gcd(m,n) = 1, thenK(F) = K(x, y).

6.9 NOTES

The proof of Theorem 6.89 is an adaptation of Baker’s proof from [18, Chapter 5];see also [123], [158], [159] and [306]. For a different proofof Proposition 6.90 (i),see [154].

For direct proofs of Theorem 6.91, see [204] and [80].Exercise 7 can be viewed as a generalisation of Theorem 6.89;see [235].

PART C

Space curves

Chapter Seven

Algebraic curves in higher-dimensional spaces

Algebraic curves in higher-dimensional spaces are usuallyintroduced in the moregeneral context of arbitrary algebraic varieties. However, an equivalentad hocdefinition which fits in with the set-up of present book is alsopossible. Essentially,an irreducible algebraic curve ofPG(n,K) is a set of points which can be put ina birational correspondence with the points of an irreducible algebraic plane curve.The idea of deriving the concept of an algebraic curve in a higher dimensionalspace directly from that of an algebraic plane curve has at least two advantages. Onthe one hand, it allows the development of a theory of algebraic curves dependingonly on the basic concepts that have been investigated so far. On the other hand,it makes it possible to go directly to important results on linear series and rationaltransformations.

7.1 BASIC DEFINITIONS AND PROPERTIES

A point in then-dimensional projective spacePG(n,K) is a homogeneous(n+1)-tuple (a0, a1, . . . , an) with ai ∈ K. As before, define abranch representationtobe a point inPG(n,K(t)) but not inPG(n,K). All the basic definitions fromChapter 5 extend to projective spaces. For example, if

x0(t) = a0 + b0ti0 + . . . , b0 6= 0,

......

xn(t) = an + bntin + . . . , bn 6= 0,

is a special representation of a branch, then its centre is the pointA = (a0, . . . , an),and its order ismini0, . . . , in.

Let Q = (x0, . . . , xn) be a point in ann-dimensional projective space whosecoordinatesxk belong to an extension field ofK. Then,xi 6= 0 for some0 ≤ i ≤ n.The fieldK(x0/xi, . . . , xn/xi) is the field of the pointQ = (x0, . . . , xn) anddenoted byK(x0, . . . , xn) orK(Q). Here,

K(x0/xj , . . . , xn/xj) = K(x0/xi, . . . , xn/xi)

provided that alsoxj 6= 0. In fact, xk/xj = (xk/xi)(xi/xj) holds for everyk = 0, . . . , n. HenceK(Q) is independent of the choice of the non-zero coordinatexi of P .

As before,Σ = K(x, y) stands for a finite extension ofK of transcendencedegree1. So,x is transcendental overK andy is related tox by an irreducible

191

192

polynomial equation overK:

F (x, y) = 0

with F (X,Y ) ∈ K[X,Y ]. As usual, considerΣ as the function fieldK(F) of anirreducible plane curveF = v(F ) arising from its generic pointP = (x, y). Withthe terminology of Chapter 5,(F ; (x, y)) is a birational model ofΣ.

From now on, assume thatK(F) = K(Q). In particular,x0, . . . , xn ∈ K(F)and, ifxi 6= 0, then somexj/xi 6∈ K.

An immediate consequence of the hypothesis is that some coordinatexj/xi mustbe a separable variable inK(F), otherwise bothx and y would be inseparablewhich is impossible; see Exercise 1 in Section 7.18.

Let γ be a branch ofF andP the corresponding place ofK(F). If

τ : K(F)→ K((t))

is a primitive representation ofP, then thecorresponding branch inPG(n,K) withrespect to the pointQ = (x0, . . . , xn) is defined, namely the branch with primitiverepresentation(τ(x0), . . . , τ(xn)). The centres of all such branches form a pointset inPG(n,K) which gives the idea of a curve. The aim here is to develop thetheory of space curves based on this idea.

DEFINITION 7.1 An irreducible (algebraic) curveΓ in PG(n,K) consists of allpoints which are centres of branches corresponding to the places ofK(F) withrespect to a pointQ = (x0, . . . , xn). These branches are thebranches ofΓ, and

(Γ, (x0, . . . , xn)) = (Γ, Q)

is amodelofK(F) in PG(n,K). Thexj are thecoordinate functionsof Γ.

An advantage of Definition 7.1 is that it is a straightforwardextension of that ofan irreducible plane curve. Several results, especially the following theorems, arehigher-dimensional versions of the analogous results for plane curves.

THEOREM 7.2 Each point of an irreducible curveΓ is the centre of finitely manybranches ofΓ.

Proof. LetA = (a0, . . . , an) be a point ofΓ, andγ a branch centred atA. Choose aspecial representation(x0(t), . . . , xn(t)) of γ. Suppose thatai 6= 0. Thenxi 6= 0.Setyk = xk/xi for k = 0, . . . , n. One of them, sayyj , is not a constant. Thenyj − aj 6∈ K. HenceP is a zero ofyj − aj . By Theorem 5.33, any function inK(F) has only finitely many zeros. Hence there are only finitely many branchescentred atA. 2

As a finite number of points can be the centres for only a finite number ofbranches, Theorem 7.2 has the following corollary.

THEOREM 7.3 An irreducible curveΓ has infinitely many points.

DEFINITION 7.4 (i) A hypersurfaceΓ in PG(n,K) is the set of all points sat-isfying a polynomial equation,

F (X0, . . . ,Xn) = 0,

whereF ∈ K[X0, . . . ,Xn] is a homogeneous non-constant polynomial;write Γ = v(F ).

Algebraic curves in higher-dimensional spaces 193

(ii) The degreeof the hypersurfaceΓ is degF .

(iii) A hyperplaneH is a hypersurface with a linear equation, that is,

H = v(a0X0 + . . .+ anXn).

Most of the material in Chapter 6.2 extends from curves to hypersurfaces. Inparticular, two hypersurfacesΦ1 = v(F1) andΦ2 = v(F2) have the same set ofpoints if and only if the polynomialsF1 andF2 have the same irreducible factors.

If γ is a branch with centreA and a primitive representation(x0(t), . . . , xn(t)) inspecial coordinates, and∆ is a hypersurfacev(G), then theintersection multiplicityis

I(A,∆ ∩ γ) = ordt G(x0(t), . . . , xn(t)).

The branchγ is linear if I(A,H ∩ γ) = 1 for some hyperplaneH.

THEOREM 7.5 Let Γ be the irreducible curve ofPG(n,K) given by the pointQ = (x0, . . . , xn).

(i) A hypersurface∆ containsΓ if and only if G(x0, . . . , xn) = 0.

(ii) If ∆ does not containΓ, then they have only finitely many common points.

Proof. (i) If G(x0, . . . , xn) = 0, then it is immediate that∆ containsΓ.For the converse, supposeG(x0, . . . , xn) 6= 0 and show thatΓ is not contained

in ∆. By relabelling the indices, assume thatx0 6= 0 and putyi = xi/x0. Then

G(1, y1, . . . , yn) 6= 0,

and henceG(1, y1, . . . , yn) has a finite number of zeros. IfA = (1, a1, . . . , an) isa point ofΓ and

(1, y1(t) = a1 + . . . , . . . , yn(t) = an + . . .)

is a primitive representation of a branchγ of Γ centred atA, then the correspondingplaceP is a zero ofG(1, y1, . . . , yn). Hence∆ can contain only a finite number ofaffine points ofΓ. On the other hand,Γ has infinitely many affine points, as in theproof of Theorem 7.3. HenceΓ cannot be contained in∆.

(ii) It is only necessary to show that the hyperplanev(X0) meetsΓ in a finitenumber of points. But this follows sincex0 has only a finite number of zeros.2

THEOREM 7.6 Let γ be a branch ofΓ centred atA, and assume that the hyper-surface∆ does not containΓ. ThenI(A,∆ ∩ γ) is finite; that is,

G(x0(t), . . . , xn(t)) 6= 0,

where(x0(t), . . . , xn(t)) is a primitive representation ofγ and ∆ has equationG(X0, . . . ,Xn) = 0.

Proof. If σ is a primitive representation of the place associated toγ, thenσ mapsK(Q) isomorphically ontoK(x0(t), . . . , xn(t)) with xi mapped toxi(t). Thusthexi(t) satisfy the same polynomial relations overK as do thexi. It follows thatG(x0, . . . , xn) 6= 0 impliesG(x0(t), . . . , xn(t)) 6= 0. 2

194

This theorem, together with Theorem 7.2, provides the extension of the conceptof intersection divisor: if ∆ does not containΓ, then

∆ · Γ =∑

I(A,∆ ∩ γ)P,

whereP is the place associated toγ andγ ranges over all branches ofΓ.Some slight changes in the wording and proof of Theorem 6.40 are sufficient to

obtain the following result.

THEOREM 7.7 Let the curveΓ be given by the pointP = (x0, . . . , xn), and let∆ = v(G(X0, . . . ,Xn)) be a hypersurface of degreem. If ∆ does not containΓ,andH∞ is the hyperplanev(X0), then

div (G(x0, . . . , xn)) = ∆ · Γ−m(H∞ · Γ).

As in Section 6.2,deg(∆ · Γ) stands for the degree of the intersection divisor of∆andΓ.

COROLLARY 7.8 For any hyperplane not containingΓ,

deg(∆ · Γ) = deg ∆ · deg(H∞ · Γ), deg(H · Γ) = deg(H∞ · Γ).

The positive integerdeg(H · Γ) is the degreeof Γ. Then the first equation inCorollary 7.8 can be viewed as the extension of Bezout’s theorem to space curves.

COROLLARY 7.9 (Bezout)If a hypersurface∆ does not containΓ, then

deg(∆ · Γ) = deg ∆ · deg Γ.

Also, Theorem 6.41 and Corollaries 6.42 and 6.44 extend to space curves withsimilar proofs.

THEOREM 7.10 Let ∆ be a hypersurface of degreem not containingΓ. Then∆cuts out onΓ a divisor equivalent tom(H∞ · Γ).

THEOREM 7.11 If ∆1 is another surface of degreem not containingΓ, then∆1 ·Γis equivalent to∆ · Γ.

THEOREM 7.12 Every linear system of hypersurfaces of a given degree cuts outon Γ a linear series. Conversely, every linear series ofK(F), except for a fixeddivisor, is cut out onK(F) by a linear system of hypersurfaces of the same degree.

Let ∆0 = v(ϕ0), . . . ,∆s = v(ϕn) be hypersurfaces of degreem, where

ϕ0, . . . , ϕs ∈ K[X0, . . . ,Xn].

It is possible that some hypersurface in the linear systemv(∑s

i=0 ciϕi) containsΓeven though none of the∆i does. In the case of plane curves, there was the conceptof linear independence moduloF , whereΓ = v(F ); see Remark 6.45. Here, asimilar concept is required.


DEFINITION 7.13 The hypersurfaces∆0, . . . ,∆s are linearly dependent moduloΓ if some hypersurface in the linear system

∑si=0 ciϕi vanishes overΓ; equiva-

lently, if∑s

i=0 ciϕi(x0, . . . , xn) = 0,

whereΓ is given by the pointP = (x0, . . . , xn).

If this happens, one or more of the∆i can be eliminated and in this way the linearsystem becomes linearly independent moduloΓ.

DEFINITION 7.14 A singular point of an curveΓ is the centre of either at leasttwo branches or just one non-linear branch ofΓ. A non-singular point is also calledsimple.

THEOREM 7.15 Every irreducible curve has only finitely many singular points.

Proof. The notation of the proof of Theorem 7.2 is used. There are finitely manybranches centred onH∞, and these are put aside.

Let γ be one of the branches centred atA = (1, a1, . . . , an). One of thexi, sayx1, is separable. Let

x1 = a1 + ctv + . . . , c 6= 0.

If v > 1, then the place corresponding toγ is a zero ofdx1. By Lemma 5.52, thisonly happens for a finite number of places. Hence all but finitely many branchesof Γ are linear. Eliminating the non-linear branches, it may be supposed that thebranches centred atA are linear. SinceK(F) = K(Q),

x =f(x1, . . . , xn)

d(x1, . . . , xn), y =

g(x1, . . . , xn)

d(x1, . . . , xn).

with

d, f, g ∈ K[X1, . . . ,XN ].

Sinced(x1, . . . , xn) 6= 0, the hypersurfacev(d) does not containΓ. By Theorem7.5, this hypersurface meetsΓ in only a finite number of points. All branchescentred at these points are eliminated. Any branchγ centred atA = (1, a1, . . . , an)goes into a branch of the plane curveC centred at the point

A′ =

(f(a1, . . . , an)

d(a1, . . . , an),g(a1, . . . , an)

d(a1, . . . , an)

).

If A′ is a singular point ofC, eliminate the branchγ. In this way, finitely manybranches are eliminated. Then there remains at most one branch ofC centred atA′,and hence with at most one branch ofΓ centred atA. 2

7.2 RATIONAL TRANSFORMATIONS

Rational transformationsare defined similarly to Section 5.2 when consideringpointsQ = (x0, x1, . . . , xn), Q′ = (x′0, x

′1 . . . , x

′n′) satisfyingK(Q′) ⊆ K(Q).

196

Up to relabelling indices, neitherx0 norx′0 is zero; so it is possible to assume thatx0 = x′0 = 1. Let Γ andΓ′ denote the irreducible curves inPG(n,K) associatedwith Q andQ′.

A rational transformationω : K(Q)→ K(Q′) is given by

x′i = fi(x1, . . . , xn)/d(x1, . . . , xn) for i = 1, . . . n′,

wherefi, d ∈ K[X1, . . . ,Xn], and the hypersurfaceH = v(d) does not containΓ.

Basic facts on rational transformations stated in Section 5.2 for n = 2 hold truein any dimensionn ≥ 3. The image of a branchγ of Γ underω is a branch ofΓ′.The imageA′ of a pointA ∈ Γ exists andA′ ∈ Γ′ provided thatA is an affine pointand thatA does not lie in the hypersurfaceH = v(d). Actually, both conditionscan be dropped whenA is the centre of only one branch, in particular whenA is anon-singular point ofΓ. In fact, computing the coordinates of the imageA′ of sucha pointA ∈ Γ is possible using the fact that the coordinate functions aredeterminedup to a non-zero factor.

Now, it is shown how to carry out the calculation after replacing (1, . . . , x′n′) by(z, . . . , zx′n′) for a suitably chosenz ∈ K(Q). Putting

mP = minordPf1/d . . . ,ordPfn/d,take an indexk for which mP = ordPfk/d. In the definition ofω, replace(1, f1/d, . . . , fn/d) by (1/fk, f1/fk, . . . , fn/fk). Note thatx′k = 1. Now, allf ′j(A) = f ′j(a0, . . . , an) exist. Therefore,

A′ = φ(A) = (f ′0(A), . . . , f ′n′(A))

provided thatA is the centre of a unique branch ofΓ. In particular, ifΓ is an irre-ducible non-singular curve, then the rational transformationω defines themorphismφ : Γ −→ Γ′.

If K(Q) = K(Q′), thenΓ is birationally equivalentor birationally isomorphicto the curveΓ′ given byQ′, and branches correspond to each other when derivedfrom the same place ofK(Q). A non-singular point ofΓ and a non-singular pointof Γ′ correspondif they are the centres of corresponding branches. This definitionextends to singular points provided that they are the centreof only one branch. IfbothΓ andΓ′ are non-singular curves, the morphismφ is anisomorphism.

Since every field of degree of transcendency1 overK has a pair of generators,the following result is obtained.

THEOREM 7.16 Any irreducible curve is birationally equivalent to a planecurve.

EXAMPLE 7.17 The Klein quartic is the non-singular plane curve

F = v(X + Y 3 +X3Y )

of genus3. , LetP = (x, y) be a generic point ofF . ThenK(F) = K(x, y), with

x+ y3 + x3y = 0,

is the function field ofF . Put

x0 = x, x1 = x2y, x2 = y2, x3 = −3xy.


Let Γ be the irreducible curve inPG(3,K) given byP = (x0, x1, x2, x3). Theassociated rational transformationω defines a morphismφ = F −→ Γ such that,if A = (a, b), then

A′ = φ(A) = (a, a2b, b2,−3ab),

provided thatA is not a vertex of the triangle of reference inPG(2,K). So, as inSection 7.2, the image of such a vertex is also uniquely determined.

Assume thatp = 5. First,A = (1, 0, 0) is considered. To carry out the com-putation, choose a primitive representationσ of the unique place ofK(F) whosecorresponding branch ofF centred atA. If σ(x) = x(t), σ(y) = y(t), then

x(t) = a1ti + a2t

j + . . . , a1a2 6= 0,

y(t) = t.

Fromf(a1ti + a2t

j + . . . , t) = 0,

x(t) = −t3 + t10 + . . . , y(t) = t.

Let σ(xi) = xi(t) for i = 0, 1, 2, 3. Then

x0(t) = −t3 + t10 + . . . , x1(t) = (−t3 + t10 + . . .)2t,x2(t) = t2, x3(t) = −3(−t3 + t10 + . . .)t.

So eP = t−2 andφ(A) = B2 with B2 = (0, 0, 1, 0). The placeP correspondsto the branch ofΓ which is centred atB2 and has the following primitive branchrepresentation:

y0(t) = −t+ t8 + . . . , y1(t) = t5 − t12 + . . . ,y2(t) = 1, y3(t) = 3t2 − 3t9 + . . . .

Similar calculations show that

φ(X∞) = B0 whereB0 = (1, 0, 0, 0),φ(Y∞) = B1 whereB1 = (0, 1, 0, 0).

The intersection divisor of the planeH = v(X0) with Γ is H · Γ = 5P1 + P2,where the placesP1 andP2 correspond to the branches ofC centred at the pointsY∞ andO.

It is opportune to investigate rational transformations that are not birational. Inthe non-trivial case, the transcendency degree ofK(Q′) is 1; in the trivial case,K(Q′) = K, andΓ is transformed into a point. Unless specifically mentioned,thenon-trivial case is considered.

If P is a place ofK(Q), given by an isomorphismτ : K(Q) → K((t)), thenτinduces an isomorphismτ ′ : K(Q′) → K((t)), giving a placeP ′ of K(Q′); thenP lies overP ′. In terms of curves, ifQ, Q′ define the curvesΓ, Γ′, then a branchof Γ gives rise to a branch ofΓ′. The pointA of Γ, which is the centre of the firstbranch, is taken to the pointA′ of Γ′, which is the centre of the second. As before,A′ is well defined if just one branch is centred atA. This is the case whenA isa non-singular point, and hence for all but finitely many points by Theorem 7.15.From a geometric point of view,Γ coversΓ′, as the images of the non-singularpoints ofΓ cover all but finitely many points ofΓ′.

198

To pursue the discussion on rational transformations and morphisms, a purelyfield-theoretic question is first considered.

LetP ′ be a place ofK(Q′) andP a place ofK(Q) overP ′. LetP be given byτ and letτ induceτ ′. Then, even ifτ is primitive, which is assumed,τ ′ need notbe.

L EMMA 7.18 The redundancy ofτ ′ is independent of the choice ofτ .

Proof. As usual,Σ = K(x, y), x(t) = τ(x), y(t) = τ(y), andf ∈ K[X,Y ]is an irreducible polynomial for whichf(x, y) = 0. Chooseξ ∈ Σ, and writeξ = u(x, y)/v(x, y) with u, v ∈ K[X,Y ], andf ∤ v. Then

ordPξ = ordt u(x(t), y(t))/v(x(t), y(t)).

Let Σ′ = K(x′, y′), andf ′(x′, y′) = 0 for an irreduciblef ′ ∈ K[X,Y ]. The pair(x′(t), y′(t)) with x′(t) = τ ′(x′), y′(t) = τ ′(y′) can be viewed as a representationof the branch corresponding toP ′. If τ ′ has redundancyν ≥ 1, then there existss(t) ∈ K[[t]] such that

s(t) = cνtν + cν+1t

ν+1 + . . . , cν 6= 0

andx′(t) = x(s(t)), y′(t) = y(s(t)) for a primitive representation(x(s), y(s)) ofP ′ in K[[s]].

Suppose thatξ ∈ Σ′. Let ξ = u′(x′, y′)/v(x′, y′) with u′, v′ ∈ K[X,Y ], andf ′ ∤ v′. Then

τ(ξ) = u(τ(x′), τ(y′))/v(τ(x′), τ(y′)).

Now, since

u(τ(x), τ(y))/v(τ(x), τ(y)) = u′(τ ′(x′), τ ′(y′))/v′(τ ′(x′), τ ′(y′)),

it follows that ordPξ = ν · ordP′ξ, which proves the assertion. 2

From Lemma 7.18, the notationeP = ordP ξ/ordP′ξ is meaningful. Hence, theproof of Lemma 7.18 gives the following formula:

ordPξ = eP · ordP′ ξ. (7.1)

Now, eP is theramification index ofP. Also, if eP > 1, say thatω ramifies atP.

L EMMA 7.19 Over every placeP ′ of Σ′, there lies at least one and at most a finitenumber of places ofΣ.

Proof. Consider the linear series|mP ′|. By the Riemann–Roch theorem,

dim |mP ′| ≥ m− g′,whereg′ denotes the genus ofΣ′. In other words,|mP ′| = gr

m. Thenr ≥ 1for sufficiently largem. So there exists an effective divisorB ≡ mP ′ such thatB 6= mP ′. If B ≻ P ′, cancelP ′ as often as possible; in the end,B 6≻ P ′. Hencethere exists an elementx such that div(x) = mP ′ −B.

Now, check thatP ′ is a zero ofx. SinceB 6≻ P ′, so div(x)0 = mP anddiv(x)∞ = B. It follows that ordP′ = m. Actually, it has been shown thatx has


P ′ but no other place as a zero. By (7.1), the placesP of Σ lying overP ′ arealso zeros ofx. As the number of zeros ofx is equal to[Σ : K(x)], this numbergives not only un upper bound for the number of places ofΣ lying overP ′, but alsoensures the existence of at least one placeP of Σ at whichx vanishes. Such a placeP lies over a placeQ′ of Σ′. By (7.1)Q′ is a zero ofx; but thenQ′ = P ′. Thiscompletes the proof. 2

Let P1, . . . ,Pj be the places ofΣ lying overP ′, and leteP1, . . . , ePj

be theirramification indices, so that ordPi

x = ePi· ordP′

ix. ThenP ′ corresponds to

D(P ′) = eP1P1 + · · ·+ ePj

Pj . (7.2)

Also,

[Σ : K(x)] =∑j

i=1 ordPix =

∑ji=1 ePi

ordPi′ x =

(∑ji=1 ePi

)· [Σ′ : K(x)],

whence∑j

i=1 ePi= [Σ : Σ′] . (7.3)

As shown in Lemma 7.27,eP1= . . . = ePj

= 1 in general and, if this occurs, theneach of the[Σ : Σ′] places overP ′ is anunramified place. On the other hand,P1

ramifies completelywhenj = 1 or, equivalently, wheneP1= [Σ : Σ′].

An interpretation of (7.3) is given in the following theorem.

THEOREM 7.20 To each placeP ′ of Σ′ there corresponds[Σ : Σ′] places ofΣ,counted with multiplicity.

Geometrically speaking,Γ 7→ Γ′ is a [Σ : Σ′]-fold covering.In a rational transfor-mation, multiplicity is counted in a similar way. Hence the following criterion fora rational transformation to be birational is obtained.

COROLLARY 7.21 A rational transformation fromΓ to Γ′ is birational if and onlyif to each branch ofΓ′ there corresponds just one branch ofΓ; equivalently, if andonly if to some branch ofΓ′ there corresponds just one branch ofΓ.

THEOREM 7.22 Let Γ be given byP = (1, x1, . . . , xn). Then, for some pair ofelements

y1 = a01 + a11x1 + · · ·+ an1xn, y2 = a02 + a12x1 + · · ·+ an2xn,

with aij ∈ K, the curveΓ′ havingQ = (1, y1, y2) as a generic point is birationallyequivalent toΓ.

Proof. LetP be a place ofK(Q) centred at a non-singular pointA of Γ. To avoidpoles of the coordinate functionsxi, suppose thatA is an affine point; that is,Adoes not lie in the hyperplanev(X0). Then, for someaj1 ∈ K, the hyperplane

H1 = v(a01 + a11X1 + · · ·+ an1Xn)

passes throughA and ordP(a01 +a11x1 + · · ·+an1xn) = 1 at the placeP centredatA. Take

y1 = a01 + a11x1 + · · ·+ an1xn;

200

this elementy1 has only a finite number of zeros. Choose a hyperplane

H2 = v(a02 + a12X1 + · · ·+ an2Xn)

throughA but not through the other zeros ofy1. Take

y2 = a02 + a12x1 + · · ·+ an2xn,

and letΣ′ = K(y1, y2). SinceΣ′ ⊂ Σ, there is a placeP ′ lying underP. Then,P ′

is a common zero ofy1 andP is the only place lying overP ′. Also, the multiplicityis 1. This follows from ordPy1 = 1 together with (7.1). Corollary 7.21 completesthe proof. 2

REMARK 7.23 A geometric interpretation of Theorem 7.22 can be given in termsof projection. WithH1 andH2 as above and(aij), j = 1, . . . n + 1, i = 0, . . . , na matrix such thatdet(aij) 6= 0, apply toPG(n,K) the projectivity associated tothe matrix(aij). This can be done so thaty1 = x1, y2 = x2 in Theorem 7.22. Themapping,

π : AG(n,K) −→ AG(2,K),(a1, a2, . . . , an) 7→ (a1, a2).

(7.4)

is a projection. Thus therational transformation given in Theorem 7.22 determinedby (P,Q) is induced by a projection.

Using projective coordinates,π can be viewed as induced by the transformation(a0, a1, . . . , an) → (a0, a1, a2), which is defined at points outside the subspaceΠn−3 = v(X0,X1,X2), and which in fact is the projection fromΠn−3 as vertexto the planeΠ2 = v(X3, . . . ,Xn). The join of (a0, a1, . . . , an) andΠn−3 is asubspaceΠn−2 intersectingΠ2 in the point(a0, a1, a2, 0, . . . , 0) or, omitting thelast coordinates as superfluous, in(a0, a1, a2). It is to be noted thatΓ may intersectthe vertexΠn−3 and, if so, the transform, or transforms, of the points of intersectionare defined via the branches centred at those points.

In summary,every irreducible curve can be transformed by means of a projectioninto a birationally equivalent plane curve.

7.3 HURWITZ’S THEOREM

This section establishes a basic theorem of Hurwitz concerning the genera of sub-fields of a function field. For a finite separable extensionS/S′ of degreen, Hur-witz’s Theorem states the numerical relation

2g − 2 = n(2g′ − 2) + d,

whereg andg′ are the genera ofS andS′ andd is a non-negative integer; it alsogives an interpretation ford together with a method for computingd. A key lemmais a formula similar to (7.1).

L EMMA 7.24 If ξ ∈ S′ is a separable variable ofS,

ordPdξ = ordtds

dt+ ordP′ dξ, (7.5)


Proof. From the proof of (7.1),ξ(t) = ξ′(s(t)), whencedξ/dt = ds/dt · dξ′/dsby derivation. Note thatds/dt 6= 0, since otherwisedξ/dt = 0 contradicting theseparability ofξ. This gives the result. 2

By (7.5),

∑ordt

ds

dt=∑

ordPdξ −∑

ordP′ dξ,

where the summation is over the places ofS. From (7.1),∑

ordP′ dξ = [S : S′] ·∑′ ordP′ dξ,

the latter summation being over the places ofS′. Sincedξ has a finite number of ze-ros and poles, it follows that

∑ordt (ds/dt) is a finite sum. Also,

∑ordt (ds/dt)

is independent of the choice ofξ in S′. This leads to the following definition.

DEFINITION 7.25 Thedifferentof a finite separable extensionS/S′ is the divisorD(S/S′) =

∑dPP with dP = ordt (ds/dt), where the summation is over all

placesP of S.

THEOREM 7.26 (Hurwitz) Let S be a finite separable extension ofS′, and letgandg′ be the genera ofS andS′. Then

2g − 2 = [S : S′] · (2g′ − 2) + degD(S/S′). (7.6)

Proof. By Definition 5.54,∑

ordPdξ = 2g − 2 and∑′ordP′dξ′ = 2g′ − 2. Now,

the result follows from Lemma 7.24. 2

If eP is prime top, then (7.5) becomes

ordPdξ = (eP − 1) + ordP′dξ.

Otherwise, it can only be asserted that ordPdξ eP + ordP′dξ. In any case,

ordPdξ ≥ eP − 1 + ordP′ dξ.

Hence∑

(eP − 1) ≤∑ ordt (ds/dt) = degD(S/S′). (7.7)

where the summation is for all places ofS at whichω ramifies. In particular,∑(eP − 1) is a finite sum. This gives the following.

L EMMA 7.27 Let S be a finite separable extension ofS′. Thenω : S → S′

ramifies at a finite number of places.

Let S/S1 andS1/S2 be finite separable extensions. ThenS/S2 is also a finiteseparable extension, and it is straightforward to prove thefollowing equation.

dP(S/S2) = dP(S/S1) + [S : S1] dP′(S1/S2), (7.8)

whereP is any place ofS andP ′ is the place ofS1 lying underP.The following example shows how this machinery works.

202

EXAMPLE 7.28 Let Hq = v(Y + Xq + XY q) be the Hermitian curve. For ageneric pointP = (x, y) of Hq, let Σ = K(x, y) the function field ofHq. Sincedx 6= 0, sox is a separable variable ofΣ by Theorem 5.49. Now, fix a divisormof q2 − q + 1, and consider the following rational transformation:

ω : ξ = xm, η = ym.

Let Γ′ be the curve with generic pointQ = (ξ, η), and setΣ = K(ξ, η). Sincep ∤ m, soξ is a separable variable ofΣ, by Lemma 5.37. The differentD(Σ/Σ′) iscalculated explicitly.

Take the pointP1 = (1, 0, 0) of Γ. There is a unique branchγ of Γ centred atP1.Since the tangent toΓ atP1 is theX-axis, there is a primitive branch representationof γ of type(x = t, y = y(t)), with y(t) ∈ K[[t]]. Note that

f(t, y(t)) = y(t) + tq + ty(t)q = 0

shows that

y(t) = −(tq + tq+1 + · · ·+ ajtq2+αj(q

2−q+1) + · · · ).LetP be the place ofΣ corresponding to the branchγ, given by the primitive placerepresentationτ : Σ→ K((t)). Let τ ′ : Σ′ → K((t)) be the isomorphism inducedby τ on Σ′, and letP ′ be the place ofΣ′ arising fromτ ′. From the definition ofω, P is the unique place ofΣ lying overP ′. Let τ ′(ξ) = ξ(t), τ ′(η) = η(t). Then(ξ(t), η(t)) is a representation of the branchγ′ of Γ′ corresponding toP ′. By thedefinition ofω,

ξ(t) = tm

η(t) = (−1)m(tm)q[1 + (tm)(q

2−q+1)/m + · · ·]m

.

Therefore, it is a reducible representation with redundancy degreem. Replacementof tm by s provides a primitive representation ofγ′:

ξ(s) = s

η(s) = (−1)msq[1 + s(q2−q+1)/m + · · · ].

This shows that ordP′η = q. Since ordPη = mq, and no other place ofΣ lies overP ′, soeP = m; that is,[Σ : Σ′] = m. Also, ass(t) = tm yieldsds/dt = mtm−1,the placeP contributes(m− 1)P to the different divisorD(Σ/Σ′).

Similar arguments can be used in doing computations for the other two funda-mental pointsP2 = (0, 1, 0) andP3 = (0, 0, 1). The above results forP1 hold truefor bothP2 andP3. This can be shown directly. The change of the coordinate sys-tem by(X0,X1,X2) → (X1,X2,X0) leavesΓ invariant and takesP1 to P2 andP2 to P3, Therefore, the contribution of the three fundamental points toD(Σ/Σ′)is (m− 1)P1 + (m− 1)P2 + (m− 1)P3.

Each of the remaining pointsP of Γ is affine:P = (a, b) with a, b 6= 0. The tan-gent toΓ atP is not the horizontal linev(Y −a). Hence, a primitive representationof the unique branchγ of Γ centred atP is of type

x = a+ t+ · · · , y = b+ t.

Therefore,

ξ(t) = am + . . . , η = (b+ t)m.


Now, chooseη′ ∈ K[[s]] ands(t) ∈ K[[t]] such thatη(t) = η′(s(t)). Then

m · η(t)/dt = m(b+ t)m−1 = m · η′/dt · ds(t)/dt,whence ordt ds/dt = 0. Therefore,eP = 1 for the placeP corresponding toγ.This shows thatP does not contribute toD(Σ/Σ′). Summing up, the differentdivisorD(Σ/Σ′) of the rational transformationω is equal to

(m− 1)P1 + (m− 1)P2 + (m− 1)P3,

and its degree is equal to3(m− 1). By Hurwitz’s Theorem 7.6,

(2g − 2) = m(2g′ − 2) + 3(m− 1),

whereg = 12 (q2 − q) is the genus ofΓ. Hence the genusg′ of Γ′ is equal to

g′ = 12

(q2 − q + 1

m− 1

).

7.4 LINEAR SERIES COMPOSED OF AN INVOLUTION

Let Γ′ be a curve given by the pointQ = (y0, . . . , yr) with yi ∈ Σ andK(Q) ⊆ Σ.There are two basic ideas related toQ: the rational transformation

τ : (1, x, y) 7→ (y0, . . . , yr)

and the linear seriesL consisting of all divisors

Ac = div(∑

ciyi

)+B, c = (c0, . . . , cr) ∈ PG(r,K) (7.9)

whereB is a fixed divisor. The aim now is to give a geometric representation ofLon Γ′. The simplest case occurs whenτ is birational and, in this case, the divisorsof L are the hyperplane sections onΓ′.

The curveΓ′ lies in PG(r,K). However, if y0, . . . , yr are linearly dependentoverK, that is, if c0y0 + · · · + cryr = 0 with someci ∈ K not all zero, thehyperplaneH = v(c0Y0 + · · ·+ crYr) containsΓ′ by Theorem 7.5. Conversely, ifΓ′ lies inH, then the same theorem yields thatc0y0 + · · ·+ cryr = 0.

The significance of the usual assumption when discussing linear series, thaty0, . . . , yr are linearly independent, is now clear: this property holdsif and onlyif Γ′ is not contained in a proper subspace ofPG(r,K).

If a fixed divisorC is added to the divisors ofL, a new linear series is obtained,but bothΓ′ and the rational transformationτ remain the same. Hence it is assumedthatthe seriesL in question contains only effective divisors and has no fixedplace.Also, it is helpful to make the following purely field-theoretic definition.

LetP ′ be a place ofΣ′ and let

D(P ′) = eP1P1 + . . .+ ePs

Ps

be the divisor as in (7.2). The set of all divisorsD(P ′) obtained asP ′ ranges overthe places ofΣ′ is aninvolutionof Σ of order

[Σ : Σ′] = eP1+ . . .+ ePs

.

Also, writeν = [Σ : Σ′]; thenγν indicates an involution of orderν. Also, by abuseof notation, writeτ(P) = P ′ whenP lies overP ′, and extend this to divisors bywriting τ(

∑µiPi) =

∑µiτ(Pi). In other words,τ is a homomorphism from the

divisor group ofΣ into the divisor group ofΣ′.

204

THEOREM 7.29 For x ∈ Σ′, let div(x)0 anddiv(x)′0 denote the divisor of zeros ofx for Σ andΣ′. Then

τ(div(x)0) = [Σ : Σ′] · div(x)′0.

Proof. Let

div(x)′0 = µ1P ′1 + · · ·+ µsP ′

s,

with P ′i 6= P ′

j , for i 6= j. If∑

j ePijPij is the divisor corresponding toP ′

i, then

τ(∑

j ePijPij) = [Σ : Σ′] · P ′

i.

Also, ordPijx = ePij

ordP′

j= ePij

µj . Hence,

div(x)0 =∑

i,j

(ordPijx)Pij ,

and

τ(div(x)0) = τ(∑

ij(ordPijx) Pij) =

∑ij (ordPij

x)τ(Pij)

=∑

i

(∑j ordPij

x)P ′

i =∑

i

(∑j ePij

)µiP ′

i

= [Σ : Σ′]∑

i µiP ′i = [Σ : Σ′] div(x)′0.

2

COROLLARY 7.30 Let div(x)′0 = P ′1 + . . . + P ′

t. Thendiv(x)0 is the sum ofthe divisors corresponding to theP ′

i. Hence, it is made up of the divisors of aninvolution.

PROPOSITION 7.31 The divisors of a linear seriesL are made up of divisors ofan involution.

Proof. If the representation (7.9) ofL is not normalised, change it to a normalisedone; see Theorem 6.22. This involves replacingyi by a multiplezyi, z ∈ Σ, andhence it does not changeΓ′. By Theorem 6.22,B is then a divisor ofL.

Let P1, . . . ,Ps be the places occurring inB. The condition that the divisorAc ∈ L containsPi is a linear condition

∑λijcj = 0 on thecj ; see Theo-

rem 6.26. This condition is not trivial; that is, someλij 6= 0, sincePi is nota fixed divisor ofL. Thus the polynomial

∑λijXj is not identically zero, and

henceH = v(∑λijXj) is a hyperplane ofPG(r,K). Since finitely many hyper-

planes cannot cover all points ofPG(r,K), there are points off the hypersurface∆ = v(

∏si=1(

∑λijXj)). Choose such a pointP = (c0, . . . , cm). Then the cor-

responding divisorAc contains none of thePi. Thus,Ac is the set of zeros andBis the set of poles of

∑ri=0 ciyi. Hence,B is the set of zeros of an element inΣ.

By Corollary 7.30,B is made up of elements of an involution. 2

Thus, the linear seriesL, or the rational transformationτ , defines an associatedinvolution, γ(L) or γ(τ). If γ(L) = γν with ν > 1, thenAc is composed ofthe involutionγν ; otherwiseAc is simple. Then,τ is birational if and only if the


corresponding series is simple. In fact,Σ = Σ′ if and only if ν = 1. Applying τ to(7.9) gives

τ(Ac) = ν · div(∑ciyi) + τ(B).

Each place inτ(B) occurs as many times as a multiple ofν, and the same holds forτ(Ac). Dividing by ν gives

1

ντ(Ac) = div(

∑ciyi)

′ +1

ντ(B), ν = [Σ : Σ′]. (7.10)

The divisors on the right-hand side cut out a linear seriesL′ onΓ′. If P ′1 + . . .+P ′

t

is a divisor inL′, then the sum of the divisors corresponding to theP ′i is anAc ∈ L.

Actually, eachAc ∈ L is obtained in this way.The linear seriesL′ has no fixed place. To prove this, suppose otherwise and that

P ′ is a fixed point. LetP1, . . . ,Ps be the places overP ′. As seen before, there isa divisorAc containing none of thePi and such thatτ(Ac) does not containP ′, acontradiction. Also,L′ is effective. The hyperplanes ofPG(r,K) cut out onΓ′ thelinear series

div(∑ciyi)

′ +D.

As this has no fixed place and is effective, the multiplicityλ with which any placeP entersD is −minc(ordP(

∑ciyi). Since(1/ν)τ(B) can be described in the

same way, this also shows thatD = (1/ν)τ(B). Hence, (7.10)is the linear seriescut out onΓ′ by hyperplanes.From this the following result is obtained.

THEOREM 7.32 Let Γ′ be a curve arising from a rational transformationτ ofΣ. If τ is birational, then the corresponding linear seriesL is cut out onΓ′ byhyperplanes.

REMARK 7.33 Definition 6.32 can be interpreted geometrically. With a slightchange in the wording in Remark 7.23, the projectionπs from PG(r,K) to itss-dimensional subspaceΠs = v(Xs+1, . . . ,Xr), for 2 ≤ s < r, is defined, withthe vertex ofπs beingΠr−(s+1) = v(X0, . . . ,Xs).

Let gsm andgr

n be two simple linear series ofΣ such thatm ≤ n, s ≤ r − 1.Suppose thatgs

m is contained ingrn. Then,

gsm = div (c0x0 + . . .+ csxs) +B | c = (c0, . . . , cs) ∈ PG(s,K),gr

n = div (c0x0 + . . .+ crxr) + F | c = (c0, . . . , cr) ∈ PG(r,K),with F ≻ B.

Let Q = (x0, . . . , xs) andQ′ = (x0, . . . , xr). If Γ and∆ are the irreduciblecurves determined byK(Q) andK(Q′), then∆ is the projection ofΓ with vertexΠr−(s+1). The converse also holds. An irreducible curveΓ is normal if it cannotbe obtained in this way by a projection of another curve. Equivalently, if Γ isdetermined byK(Q) with Q = (x0, . . . , xr) then the associated simple, fixed-place-free linear series of effective divisors,

div (c0x0 + . . . , crxr) +B | c = (c0, . . . , cr) ∈ PG(r,K)is complete.

206

EXAMPLE 7.34 Let Σ = K(x), andQ = (1, x, . . . , xr). The algebraic curveΓof PG(r,K) determined byQ is normal, and it is therational normal curve ofPG(r,K).

Another fundamental result is the following.

THEOREM 7.35 Letgrn be a linear series without fixed points but composed of an

involution. IfP is a place ofΣ, then every divisor ofgrn containingP also contains

(at least) another placeQ, whereQ depends only onP, and conversely. The caseP = Q can occur when a divisor ofgr

n containsP with multiplicity at least2.

Proof. Let grn be composed of the involutionγν with ν > 1. Take any divisor

P1 + P2 + . . . + Ps of γν . Then every divisor containingP1 contains each of theremaining placesP2, . . . ,Pν . This proves the direct part.

For the converse, assume thatgrn is simple, so that it defines a birational transfor-

mation, and hencegrn is cut out onΓ′ by hyperplanes. LetP be any place centred at

a simple point ofΓ′. If Q another place ofΣ, then there is a divisor ingrn contain-

ing P butQ. In fact, if the centreQ of Q is distinct from the centreP of P, thenthere is a hyperplane throughP but not throughQ, and hence the divisor cut outhas the required property. ForP = Q, such a divisor arises from any hyperplanecontainingP but not2P, or geometrically, from any non-tangent hyperplane toΓ′

atP . 2

A characterisation for a point of space curve to be non-singular is given in thefollowing theorem.

THEOREM 7.36 Let grn be the linear series cut out onΓ by hyperplanes. Then a

pointP of Γ is non-singular if and only if the subseries ofgrn containing a placeP

centred atP, minusP, has no fixed place.

Proof. Let γ1, . . . , γk be the branches ofΓ centred atP , andP1, . . .Pk the cor-responding places ofΣ. For i = 1, . . . k, put nPi

= ordPγi. Then the divisorE =

∑ki=1 nPi

Pi is the fixed divisor of the linear subseriesL′ of grn cut out onΓ

by the hyperplanes throughP . Also, degE ≥ 1 and equality holds if and only ifP is a non-singular point ofΓ. On the other hand,degE = 1 occurs if and only ifL minusP1 has no fixed divisor. 2

DEFINITION 7.37 A singular point ofΓ is ans-fold point if degE = s.

REMARK 7.38 A point P of Γ is ans-fold point if and only if the fixed divisor ofthe linear series cut out onΓ by the hyperplanes containingP has degrees.

As a consequence of Theorem 7.36, there is the following result.

THEOREM 7.39 Let grn be a complete linear series ofΣ. If n > 2g, then the

curve Γ, arising from grn via the associated rational transformation, is free of

singularities; that is, it is a non-singular model ofΣ.

Proof. By the Riemann–Roch Theorem,r = n − g becausen > 2g > 2g − 2implies thatgr

n is not special. Also,grn has no fixed point; otherwise, a complete


non-special linear seriesgrn−1 would be obtained by subtracting a fixed point. How-

ever, thenr = n− 1− g, again by the Riemann–Roch Theorem, contradicting theprevious equationr = n− g.

Now, letP be any point ofΓ, and letP be a place centred atP . The subseries ofall divisors ofgr

n containingP minusP is a linear seriesgr−1n−1 which has no fixed

point. The latter property can be shown as before. The hypothesis thatgr−1n−1 has a

fixed point would imply the existence of a complete non-special linear seriesgr−1n−2,

contradicting the Riemann–Roch theorem. 2

For g > 1, the most important linear series is thecanonical seriesgg−12g−2. It is

possible thatgg−12g−2 may be composed of an involution. However, the following

theorem shows that this only happens in a very special situation.

THEOREM 7.40 Let g > 1. The canonical series is composed of an involution ifand only ifΣ has a completeg1

2 . If this is the case, thenΣ has a subfieldΣ′ suchthat [Σ : Σ′] = 2.

Proof. Assume first thatΣ admits a completeg12 , and take one of the divisors in

g12 , namelyC = P1 + P2. Since the canonical series|W | has no fixed point, its

subseries of divisors containingP has dimension1. If |W | were not composed ofan involution, then requiring aW ∈ |W | to containC would impose two linearconditions. So, this would omply that|W − C| = (g − 1)− 2 = g − 3 and hencei(C) = g − 2, contradicting the Riemann–Roch Theorem applied tog1

2 . Actually,it has been shown thatgg−1

2g−2 is made up of divisors ofg12 .

Conversely, let|W | be composed of an involutionγν . By Theorem 7.35, therational transformation associated to the canonical series gives rise to a curveΓ′ inPG(g−1,K) of degree(2g−2)/ν such that the hyperplanes cut out onΓ′ a linearseriesgg−1

(2g−2)/ν . Sincer ≤ n for anygrn, this is only possible forν ≤ 2. As |W |

is supposed to be composed,ν = 2 follows, showing that|W | is composed of aninvolutionγ2 of order2.

It remains to show thatγ2 is a linear series, that is, its divisors are the divisors ofa g1

2 . As ν = 2, the linear series cut out onΓ′ by hyperplanes isgg−1g−1 . HenceΓ′

has genus0 by Theorem 6.79. Therefore the fieldΣ′ of Γ′ isK(ξ) for someξ ∈ Σ,and[Σ : K(ξ)] = 2. LetP be a place ofΣ which is not a pole ofx. For a primitiverepresentationσ of P,

σ(ξ) = a+ bti + . . . .

ThenP is a zero ofξ − a with multiplicity i. LetP ′ be the place ofΣ′ underP.Hence, (7.1) implies thatP ′ is a zero ofξ − a regarded as an element inΣ′. Ifi = 1, just one more place ofΣ , sayQ, lies overP ′. By (7.1),Q is also a zeroof ξ − a. Therefore, div(ξ − a)0 = P +Q. Now, the casei > 1 is examined. By(7.1), i = ePordP′(ξ − a). SinceΓ′ is rational, ordP′(x − a) = 1. Thus,i = eP .SinceeP ≤ ν = 2, soi = eP = 2. This shows thatP is the only place ofΣ lyingoverP ′. Hence2P ∈ γ2, and div(ξ − a)0 = 2P.

Thus, it has been shown thatγ2 consists of all divisors

div(c0 + c1ξ) + div(ξ)∞. (7.11)

208

The completeness ofg12 depends on the hypothesis thatg > 0. 2

THEOREM 7.41 If Σ has more than oneg12 , then its genus is0 or 1; that is, Σ is

either rational or elliptic.

Proof. Assume thatΣ admits ag12 . From the proof of Theorem 7.40, the degree

[Σ : K(ξ)] = 2 for someξ, andg12 consists of all divisors (7.11). IfΣ has another

g12 , then likewise[Σ : K(η)] = 2 with someη 6∈ K(ξ). Therefore,Σ = K(ξ, η).

From Theorem 5.60, the assertiong ≤ 1 follows. 2

DEFINITION 7.42 A function field of genusg > 1 that has a (unique) completelinear seriesg1

2 ishyperelliptic. An irreducible curveΓ ishyperellipticif its functionfield ishyperelliptic.

THEOREM 7.43 Let K(x) be the unique subfield of a hyperellipticΣ such that[Σ : K(x)] = 2. If [Σ : K(y)] ≤ g, with g the genus ofΣ, thenK(y) is containedin K(x).

Proof. Suppose that[Σ : K(z)] ≤ g but z 6∈ K(x). ThenΣ = K(x, z). FromTheorem 5.60,

g ≤ ([Σ : K(x)]− 1)([Σ : K(y)]− 1) ≤ g − 1,

a contradiction. 2

Curves of genus2 are examples of hyperelliptic curves. For a more detailed accountof hyperelliptic curves, see Section 7.11.

For non-hyperelliptic curves, the relevant results are stated in the following sec-tion.

7.5 THE CANONICAL CURVE

In this section,C denotes a non-hyperelliptic curve of genusg. Since, the canonicalseries onC is a simplegg−1

2g−2 with no fixed points, the projective image of such aseries is a curveΓ of degree2g − 2 in PG(g − 1,K), birationally equivalent toCand unique up to projectivities. It is thecanonical curve of genusg. Conversely, ifΓ is a curve of degree2g − 2 in PG(g − 1,K) which is birationally equivalent toC, then the seriesgg−1

2g−2 cut out by the hyperplanes is the canonical series, and soΓis the canonical curve.

THEOREM 7.44 (i) The canonical curve is non-singular.

(ii) Two canonical curves that are birationally equivalent are projectively equiv-alent.

(iii) A divisorD = P1 + . . . + Pn defines a complete special series|D| = grn

if and only if the centres of then branches corresponding to the placesPi,i = 1, . . . , n span an(n− r − 1)-dimensional subspace ofPG(g − 1,K).


(iv) The canonical curve is normal.

Proof. (i) Assume on the contrary thatM is a singular point ofΓ, that is, a(k+2)-fold point with k non-negative. By Theorem 7.36, the hyperplanes throughMcut out onΓ a linear seriesgg−2

2g−4−k without fixed places. Ifγ1, . . . , γs are thebranches ofΓ centred atM andP1, . . . ,Ps are the corresponding places ofΣ, thehyperplanes cut out in addition a fixed divisorD =

∑si=1 nPP, where the degree∑s

i=1 nP = k + 2. Hence a fixed divisorD′ =∑s

i=1 n′PP, with 0 ≤ n′P ≤ nP

and∑s

i=1 n′P = k, can be added togg−2

2g−4−k, so as to produce a special series

gg2g−4 = |D′ + gg−2

2g−4−k|.The dimension of this series is smaller thang− 1, but also at leastg− 2, and henceit is a special seriesgg−2

2g−1. The residual divisors relative to the canonical seriesconstitute ags

2. By the Reciprocity Theorem 6.76 of Brill and Noether,

2g − 6 = 2(g − 2− s),whences = 1, and the series is ag1

2 . This is ruled out, however, sinceΓ is supposednot to be hyperelliptic.

(ii) This follows from Theorem 6.17 together with Theorem 6.70 (ii).(iii) If G is a special divisor ofn points on the canonical curveΓ that defines

a completegrn, the linear system of hyperplanes ofPG(g − 1,K) throughG has

dimensiong − 1− n+ r; soG is in a subspaceΠn−r−1 . Conversely, if a divisorG of n points is in a subspaceΠn−r−1 and not in any lower dimension, the indexof speciality ofG is g − n+ r, and so defines a completegr

n.(iv) This follows from Remark 7.33, since the canonical series is complete. 2

7.6 THE OSCULATING HYPERPLANE AND THE RAMIFICATION

DIVISOR

The idea is to consider for any pointP of a non-singular space curve the differentintersection multiplicities of hyperplanes with the curveatP . There is only a finitenumber of these intersection multiplicities, this number being equal to the dimen-sion of the space. There is a unique hyperplane, called theosculating hyperplane,with the maximum intersection multiplicity. The concept ofosculating hyperplanewill also be defined for singular curves.

Let Γ be an irreducible curve of degreen in PG(r,K). The hyperplanes cut outonΓ a simple, fixed-point-free, not-necessarily-complete, linear seriesL. Note thatL has degreen and dimensionr, and hence it is agr

n. Let Γ be given by the pointP = (x0, . . . , xn). Without loss of generality, suppose thatx0 6= 0; then

K(Γ) = K(x1/x0, . . . , xr/x0)

is the function field ofΓ. Further,L consists of all divisors

Ac = div (c0x0 + c1x1 + . . .+ crxr) +B

210

with c = (c0, . . . , cr) ∈ PG(r,K), andB =∑ePP, where

eP = −minordP x0, . . . ,ordPxr.Here, ordPB = n.

Geometrically speaking,Ac is cut out onΓ by the hyperplane

H = v(c0X0 + . . .+ crXr);

that is,Ac is the intersection divisorH · Γ =∑I(P,H ∩ γ)P, where

I(P,H ∩ γ) = ordt (∑r

i=0cixi(t)) + nP .

Herexi(t) = τ(xi), i = 0, . . . , r, is a primitive representationτ of P, and the(r + 1)-tuple(x0(t), . . . , xr(t)) is a primitive representation of the branchγ of Γcorresponding toP, with P the centre ofγ.

To develop local properties ofΓ, an integerj is aHermitianP-invariant or an(L,P)-order if there exists a hyperplaneH such thatI(P,H ∩γ) = j. In Example7.17, ifP = (0, 0, 1, 0) then the(L,P)-orders are0, 1, 2, 5.

In the case thatL is the canonical series, it follows from the Riemann–RochTheorem thatj is a (L,P)-order if and only ifj + 1 is a Weierstrass gap; that is,there exists no rational function inΣ which has a pole of orderj + 1 atP but isregular at all places different fromP; see Section 6.6.

For any non-negative integeri, consider the set of all hyperplanesH of PG(r,K)for whichI(P,H∩γ) ≥ i. Such hyperplanes correspond to the points of a subspaceΠi in the dual space ofPG(r,K). Then,

L = Π0 ⊃ Π1 ⊃ Π2 ⊃ . . . .An integerj is an(L,P)-order if and only ifΠj 6= Πj+1, in which caseΠj+1

has codimension 1 inΠj . SincedegL = n, soΠi is empty as soon asi > n. Thenumber of(L,P)-orders is exactlyr+1; they arej0, j1, . . . , jr in increasing order.SinceP is not a base point ofL, soj0 = 0. Note also thatj1 = 1 if and only if thebranchγ is linear.

In the case that the linear seriesL is complete, from the Riemann–Roch Theo-rem,

dim Πji= n− ji − g

for ji < n− 2g + 2; in particular,ji = i wheni ≤ n− 2g.Consider the intersectionΠi of hyperplanesH of PG(r,K), for which

I(P,H ∩ γ) ≥ ji+1.

ThenΠ0 is simplyP , the centre of the branchγ, andΠ1 is the tangent line to thebranchγ atP . Also, Πi is thei-th osculating spaceor theosculatingi-space; inparticular,Πr−1 is theosculating hyperplane. The flag

Π0 ⊂ Π1 ⊂ . . . ⊂ Πr−1 ⊂ PG(r,K)

can be viewed as the algebraic analogue of the Frenet frame indifferential geome-try.

Now, an equation of the osculating hyperplane is given in terms of the coordinatefunctionsxi. The essential tool is a generalisation of the Wronskian determinantwhose entries are Hasse higher derivatives of the coordinate functions.


Choose a local parameterζ atP, and dividexk by ζeP for k = 0, 1, . . . , r. ThenordPD

(i)ζ xk ≥ 0; that is,

τ(D(i)ζ xk) = c

(i)0k + . . . ,

with c(i)0k ∈ K. Writing the Wronskian determinant it is essential to use less

messy and more expressive notation. For this purpose, in therest of this sec-tion, points ofΓ are thought of asbranch points, that is, the term of a pointPof Γ is used to indicate the branch ofΓ associated to the placeP. In this spirit,D

(i)ζ xk(P ) is an appropriate notation forc(i)0k . In particular, the centre ofP is the

point (x0(P ), . . . , xr(P )) of Γ.

THEOREM 7.45 The (L,P)-orders can be computed iteratively. If the(L,P)-ordersj0, . . . , ji−1 are known, then the successive orderji is the smallest integersuch that thei+ 1 points inPG(r,K),

Pm = (D(jm)ζ x0)(P ), (Dζt

(jm)x1)(P ), . . . , (D(jm)ζ xr)(P )),

for m = 0, 1, . . . , i, are linearly independent. Thei-th osculating space atP isspanned by these points.

Proof. Up to a projectivity ofPG(r,K), for everyi = 0, . . . r the subspaceLi isthe intersection of ther − i hyperplanesv(Xi+1), . . .v(Xr). Then

ji+1 = minordP(ai+1xi+1 + . . .+ arxr) | ai+1, . . . , ar ∈ K.Hence, considering the suffixes in descending order,

jr = ordPxr, jr−1 = ordPxr−1, . . . , j0 = ordPx0.

Now, define the(r + 1)× (r + 1) matrixA = (ami), where

ami = (D(jm)t xi)(P ),

for 0 ≤ r ≤ n, 0 ≤ i ≤ n. Since, by Lemma 5.70 and (5.12),D(j)ζ ζj = 1

butD(k)ζ ζj = 0 for k > j, soA is lower-triangular with every diagonal element

non-zero. HenceLi is spanned by the points,

Pm = (D(jm)ζ x0(P ), . . . ,D

(jm)ζ xr(P )),

for m = 0, 1, . . . , i. 2

The minimality of theji holds in an even stronger sense.

COROLLARY 7.46 If m0,m1, . . . ,mk are integers with0 ≤ m0 < . . . < mk

such that the pointsPi = (D(mi)ζ x0(P ), . . . ,D

(mi)ζ xr(P )), for i = 0, 1, . . . , k,

are linearly independent, thenji ≤ mi for i = 0, . . . , k.

Proof. Since the pointsPs = (D(s)t x0(P ), . . . ,D

(s)t xr(P )),with s = 0, . . . , ji−1

span a subspace of dimensioni, while thei + 1 points withs = m0, . . . ,mi arelinearly independent, soji − 1 < mi; that is,ji ≤ mi. 2

212

COROLLARY 7.47 The osculating hyperplane atP has equation∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

X0 X1 . . . Xr

D(j0)ζ x0(P ) D

(j0)ζ x1(P ) . . . D

(j0)ζ xr(P )

D(j1)ζ x0(P ) D

(j1)σ x1(P ) . . . D

(j1)ζ xr(P )

...... . . .

...

D(jr−1)ζ x0(P ) D

(jr−1)ζ x1(P ) . . . D

(jr−1)ζ xr(P )

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

= 0.

The pointP is an osculation pointor, more precisely, anL-osculation pointifjr > r, that is, if there is a hyperplane intersectingΓ atP with multiplicity greaterthanr.

EXAMPLE 7.48 Let Σ = K(x) be the rational function field. Choose theX-axisC = v(Y ) as a model(C; (x, 0)) of Σ. Let s ≥ 1 be any integer. Then the elements1, x, xps

, xps+1 are linearly independent overK, and letΓ be the associated curveof PG(3,K). Note that the divisorsAc +B with

Ac = div (c0 + c1x+ c2xps

+ c3xps+1)

andB =∑ePP constitute a simple, fixed-point-free linear seriesL. Then

(j0, j1, j2, j3) = (0, 1, ps, ps+1)

for anyP of Γ, and the osculating plane has equation∣∣∣∣∣∣∣∣

X0 X1 X2 X3

1 t tps

tps+1

0 1 0 tps

0 0 1 t

∣∣∣∣∣∣∣∣= 0;

that is,

X0 − tX1 + tps

X2 − tps+1X3 = 0.

To investigate global properties ofΓ, the condition onζ to be a local parameteris weakened to be a separable variable. However, the idea of considering a determi-nant such as that in Theorem 7.45 is still useful. This depends on the existence ofa strictly increasing sequence of non-negative integersǫ0, . . . , ǫn, theorders, suchthat the generalised Wronskian

W = det(D(ǫi)ζ xj), (7.12)

with 0 ≤ i, j ≤ r, does not vanish. To determine such a sequence the idea isto chooseǫ0, . . . , ǫr in strictly increasing order such thatǫ0 = 0 and defineǫiinductively so that, ifǫ0, . . . , ǫi−1 are known, thenǫi is the smallest integer suchthat the points

Pk = (D(ǫk)ζ x0(P ), . . . ,D

(ǫk)ζ xr(P ))

k = 0, . . . , i are linearly independent inPG(r,Σ). As in Corollary 7.46, theǫi areminimal in a stronger sense: ifm0, . . . ,mi are integers with0 ≤ m0 < . . . < mi

such that the pointsQk = (D(mk)ζ x0(P ), . . . ,D

(mk)ζ xr(P )), k = 0, 1, . . . , i, are

linearly independent inPG(r,Σ), thenǫk ≤ mk for k = 0, . . . , i. To show thatthis inductive procedure works properly, the following lemma is needed.


L EMMA 7.49 (i) If yi =∑aijxj with (aij) ∈ GL(r + 1,K), then

det(D(ǫi)ζ yj) = det(aij) det(D

(ǫi)ζ xj).

(ii) If h ∈ Σ, then

det(D(ǫi)ζ (hxj)) = hr+1 det(D

(ǫi)ζ xj).

(iii) If η is another separable variable, then

det(D(ǫi)η xj) = (dζ/dη)ǫ1+...+ǫr det(D

(ǫi)ζ xj).

Proof. (i) This comes from projective geometry, and does not dependon the mini-mality of theǫi.

(ii) By the Leibniz rule, as in Lemma 5.70 and (5.15),

D(ǫi)ζ (hxj) = h ·D(ǫi)

ζ xj +∑ǫi

s=1(D(s)ζ h) · (D(ǫi−s)

ζ xj).

In the first row ofW , every element hash as a factor. The second row is

h ·D(ǫ1)ζ (x0) + . . . , . . . , h ·D(ǫ1)

ζ (xk) + . . . ,

where the dots in each entry indicate terms that are the components of a vectorw which is a linear combination of vectors(D(k)

ζ x0(P ), . . . ,D(k)ζ xr(P )) with

0 ≤ k < ǫ1. This vectorw is a multiple of(D(ǫ0)ζ x0(P ), . . . ,D

(ǫ0)ζ xrP )), by

the minimality of ǫ1. Hence, in the determinant,w can be omitted, and soh isagain a factor. Proceeding inductively on the other rows ofW gives the result.

(iii) The same argument can be employed if the chain rule (5.67) is used in placeof the Leibniz rule. Here,D(ǫi)

η xj in W may be replaced by

D(ǫi)t xj · (dζ/dη)ǫi .

2

From Lemma 7.49, it follows thatǫ0, . . . , ǫn depend only on the linear seriesL,and therefore can be called theL-orders, or simply theordersof the curveΓ.

To show the actual existence of the orders, choose a placeP and letζ be a localparameter ofΣ at P, and suppose thateP = 0. By Theorem 7.45, there existj0, . . . , jr, the(L,P)-orders, for which

det(D(ji)ζ xj(P )) 6= 0.

This implies thatdet(D(ji)ζ xj)) 6= 0. Hence there exists theorder sequence, that is,

the strictly increasing minimal sequenceǫ0, . . . , ǫr for which det(D(ǫi)ζ xj)) 6= 0

andǫi ≤ ji for i = 0, . . . , r.

DEFINITION 7.50 Theramification divisorof L is

R = div (det(D(ǫi)ζ xj)) + (ǫ0 + . . .+ ǫr)div (dζ) + (r + 1)E,

whereE =∑ePP andeP = −minordP x0, . . . ,ordPxr.

L EMMA 7.51 The divisorR depends only on the linear seriesL.

214

Proof. From Theorem 6.17, suppose that the divisors inL can also be written asAd +B whereAd = div (

∑rk=0 dkyk) +B with d = (d0, . . . , dr) andP = d in

PG(r,K). From Lemma 7.49 (i),

ordP(div (det(D(ǫi)ζ xj)) = ordP(div (det(D

(ǫi)ζ yj)),

for every placeP of Σ. Given any non zero elementh ∈ Σ, let

e∗P = −minordP(hxi), E∗ =∑e∗PP.

ThenE∗ + div (h) = E and

div (det(D(ǫi)ζ (hxj))) + (ǫ0 + . . .+ ǫr)div (dζ) + (r + 1)E∗

= (r + 1)div (h) + div (det(D(ǫi)ζ xj))

+(ǫ0 + . . .+ ǫr)div (dζ) + (r + 1)E∗

= div (det(D(ǫi)ζ xj)) + (ǫ0 + . . .+ ǫr)div (dζ) + (r + 1)E.

Also, if η is another separable variable inΣ, then

div (det(D(ǫi)η xj)) + (ǫ0 + . . .+ ǫr)div (dη) + (r + 1)E

= (ǫ0 + . . .+ ǫr)(div (dζ/dη) + div (det(D(ǫi)ζ xj))

+(ǫ0 + . . .+ ǫr)(div (dη) + (r + 1)E

= div (det(D(ǫi)ζ xj)) + (ǫ0 + . . .+ ǫr)div [(dη) · (dζ/dη)] + (r + 1)E

= div (det(D(ǫi)ζ xj)) + (ǫ0 + . . .+ ǫr)div (dζ) + (r + 1)E.

This completes the proof. 2

From Corollary 5.35 and the definition of the genus (5.54),

degR = (ǫ1 + . . .+ ǫr)(2g − 2) + (r + 1)n. (7.13)

In almost all cases,degR can be easily computed from the(L,P)-orders.

THEOREM 7.52 If j0, . . . , jr are the(L,P)-orders, then

vP(R) ≥∑ri=0(ji − ǫi),

with equality if and only if

det

((jiǫk

))6≡ 0 (modp).

Proof. Suppose again thateP = 0. For a local parameterζ atP,

vP(R) = ordP(det(Dǫk

t xi)).

Arguing as in Theorem 7.45, it is possible to takeyi ∈ Σ such that

τ(yi) = citji + . . . for i = 0, . . . , n,

where, as usual, dots indicate terms of higher orders. Then

det(D(ǫk)t τ(yi)) =det

((jiǫr

)tji−ǫr + . . .

)

=det

((jiǫr

)tji + . . .

)t−ǫ0−...−ǫn

=det

((jiǫr

))tj0+...+jn−ǫ0−...−ǫn + . . . .


Thus, ordP(det(D(ǫk)ζ yi) ≥

∑(ji − ǫi), with equality for

det

((jiǫk

))6= 0 (mod p).

2

By Theorem 7.52 the ramification divisorR is effective, andnP = 0 in thedivisor R =

∑nPP if and only if ji = ǫi all i. Henceǫ0, ǫ1, . . . , ǫr are the

(L,P)-orders for almost all placesP of Σ; so these places areL-ordinary. Theother places, finite in number, satisfy

(j0, . . . , jr) 6= (ǫ0, . . . , ǫr)

and areL-Weierstrass points; also,nP is theweightof P, and is denoted byvP(R).Note that, ifζ is a local parameter atP, then

vP(R) =

∣∣∣∣∣∣∣∣∣∣

x0 x1 . . . xr

D(ǫ1)ζ x0 D

(ǫ1)ζ x1 . . . D

(ǫ1)ζ xr

...... . . .

...

D(ǫr)ζ x0 D

(ǫr)ζ x1 . . . D

(ǫr)ζ xr

∣∣∣∣∣∣∣∣∣∣

. (7.14)

In this terminology, the number ofL-Weierstrasspoints, each counted according toits weight, is

∑vP(R) = (ǫ0 + . . .+ ǫr)(2g − 2) + (r + 1)n.

REMARK 7.53 In the particular case thatL is the canonical linear series ofΣ, thenotion ofL-Weierstrasspoint is the same as in Section 6.6.

This has the following important corollary.

COROLLARY 7.54 Every curve which is neither rational, nor elliptic, nor hyper-elliptic has finitely many Weierstrass points.

Note that this corollary holds true for hyperelliptic curves, see Theorem 7.97.

DEFINITION 7.55 The linear seriesL, or equivalently the curveΓ, is classicalifthe order sequence(ǫ0, ǫ1, . . . , ǫr) is (0, 1, . . . , r).

Now, some criteria are given that are useful in deciding whether or not a givenlinear series is classical. First, a proposition follows that refines the precedinginequalityǫi ≤ ji.

PROPOSITION 7.56 Let j0, . . . , jr be the(L,P)-orders. Ifm0, . . . ,mr are inte-gers such that0 ≤ m0 < . . . , < mr and

det

((jimk

))6≡ 0 (modp),

thenǫi ≤ mi for eachi = 0, . . . , r.

216

Proof. As in the proof of Theorem 7.52,

det(D(mk)t τ(yi)) = det

((jimk

))tj0−m0+...+jn−mn + . . . .

Therefore,ǫi ≤ mi for all i by the minimality of theǫi. 2

COROLLARY 7.57 Let j0, . . . , jr be the(L,P)-orders. If∏

i>s(ji − js)/(i − s)is not divisible by the characteristicp ofK, thenL is classical and

vP(R) =∑r

i=0(ji − i).Proof.

det

((jik

))= det(ji

k/i!+. . .) = det(jik)/(1!2! . . . n!) =

∏i>s(ji−js)/(i−s).

Henceǫi = i by Proposition 7.56 and the weight ofP in R equals∑r

i=0(ji− i) byTheorem 7.52. 2

COROLLARY 7.58 If the characteristicp of K is 0, or larger thann, thenL isclassical andvP(R) =

∑ri=0(ji − i).

Proof. The condition in Corollary 7.57 is satisfied ifji 6≡ js (modp) for all i 6= s;in particular, this holds whenjn ≤ n. 2

L EMMA 7.59 If ǫ is anL-order andµ is such that(

ǫµ

)6≡ 0 (modp), thenµ is also

anL-order. In particular ifǫ < p, then0, 1, . . . , ǫ− 1 are alsoL-orders.

Proof. Since(

ǫµ

)6= 0, so0 ≤ µ ≤ ǫ. It may be supposed thatµ > 0. Let k be the

largest integer such thatǫk < µ. The matrix

(ǫ0ǫ0

). . .

(ǫk

ǫ0

) (ǫǫ0

)

......

...(

ǫ0ǫk

). . .

(ǫk

ǫk

) (ǫǫr

)(ǫ0µ

). . .

(ǫk

µ

) (ǫµ

)

is triangular with1, . . . , 1,(

ǫµ

)on the main diagonal. Hence the rows are linearly

independent overK. Thusǫk+1 ≤ µ by Proposition 7.56, and soµ = ǫk+1 by thedefinition ofk. 2

L EMMA 7.60 Assume thatp ≥ r, and that0, 1, . . . , r − 1 are the firstr L-orders.If L is non-classical, thenǫr is a power ofp.

Proof. From Lemma 7.59,ǫr ≥ p. Assume on the contrary thatǫr is not a power ofp. By Lemma 5.64, there is an integerk such thatp ≤ k < ǫr and

(ǫr

k

)6≡ 0 (modp).

Thenk is also anL-order. But this contradicts thatǫr−1 < k < ǫr. 2

In the classical case, theL−Weierstrass points are precisely theL-osculatingpoints. In zero characteristic, every curveΓ is classical.


THEOREM 7.61 LetN be the number of Weierstrass points of a function fieldΣwhich is neither rational, nor elliptic, nor hyperelliptic. If eitherp = 0 or p ≥ g,then

2g + 2 ≤ N ≤ (g − 1)g(g + 1). (7.15)

Proof. By Corollary 7.58, the canonical curveΓ of Σ is classical, and∑vP(R) = (g − 1)g(g + 1).

For a pointP ∈ Γ, let j0 = 0, j1 = 1, . . . , jg−1 be the orders ofΓ at P . ByCorollary 7.52,

vP(R) =∑g−1

i=0 (ji − i).Sincejg−1 ≤ 2g − 2, and hencejg−1−k ≤ 2g − 2 − k, a lower bound forji isji ≤ g − 1 + i. So,ji − i ≤ g − 1. Therefore,g(g + 1) is an upper bound forN .On the other hand, an upper bound forji is

ji ≤ 2i. (7.16)

This comes from Clifford’s theorem applied to the fixed pointfree special linearseries cut out onΓ by the hyperplanesH throughP with intersection multiplicityI(P,H ∩ Γ) at leastji. So,ji − i ≥ i for i = 2, . . . g − 1, and hence

vP (R) ≥ 2 + . . . g − 1 = 12 (g + 1)(g − 2).

Therefore,2g + 2 is a lower bound forN . 2

If L is non-classical, every point isL-osculating. However, it should be notedthat non-classical linear series are rare.

THEOREM 7.62 Let ǫ0 < ǫ1 < . . . < ǫr be theL-orders.

(i) If ǫr ≥ pm, then there existz0, . . . , zr ∈ Σ such that

zpm

0 x0 + zpm

1 x1 + . . .+ zpm

r xr = 0. (7.17)

(ii) Let ǫr = pm and assume that (7.17) holds. Then the equation of the osculat-ing hyperplane at all but finitely many pointsP is

zpm

0 (P )X0 + zpm

1 (P )X1 + . . .+ zpm

r (P )Xr = 0. (7.18)

(iii) If ǫr−1 < pm, and (7.17) holds, thenǫr = pk with k ≥ m. Equality holds ifand only if somezi is a separable variable ofΣ.

Proof. Sincedet(D(ǫi)ζ xj) 6= 0, (i) follows from Theorem 5.76 and Lemma 5.74.

Choose a pointP ∈ Γ or, more precisely, a branch ofΓ centred atP such thatP isneither a pole of anyzk, nor a zero of all thezk. Then

τ(zk) = ak0 + ak1t+ . . .

for k = 0, . . . , r, and at least one termau0 is distinct from0. Now, (7.17) can bealso written as

apm

k0 x0 + . . .+ ar0zpm

r xr + (z0 − ak0)pm

x0 + . . .+ (zr − akr)pm

xr = 0.

218

Then

ordt (apm

k0 x0(t) + . . .+ ar0zpm

r xr(t)) ≥ pm.

As ǫr = pm, this shows that

apm

k0 X0 + . . .+ apm

r0 Xr = 0

is the equation of the osculating hyperplane atP . Since, by definition,zk(P ) = ak0

for 0 ≤ k ≤ r, the assertion (ii) holds.The previous argument also shows that ifǫr−1 < pm−1, thenǫr ≥ pm. From

Lemma 7.59,ǫr = pk with k ≥ m. If zi is inseparable, then (iii) of Lemma 5.49providesui such thatzi = up

i . Hence, if nozi is separable, then

upm+1

0 x0 . . .+ upm+1

r xr = 0,

and henceǫr ≥ pm+1.Conversely, ifǫr = pk with k > m andǫr−1 < pm, thenx0, . . . , xr are linearly

dependent overΣk but independent overΣm. Hence, if (7.17) holds, then eachzpm

i is Σk. By Lemma 5.74,zi = upi with ui ∈ Σ for 0 ≤ i ≤ r. Hence the last

assertion of Theorem 7.62 follows from Lemma 5.49 (iii). 2

REMARK 7.63 If ǫr = pm holds in Theorem 7.62, then the(r + 1)-ple is deter-mined up to a non-zero factoru ∈ Σ. This follows from (ii) by the uniqueness ofthe osculating hyperplane.

7.7 NON-CLASSICAL CURVES WITH RESPECT TO A LINEAR SYSTEM

Let F = v(f(X,Y )) be an irreducible plane curve. Choose any positive integers smaller than the degreen of f(X,Y ). The linear system of all plane curves ofdegrees cuts out onF agr

d that has no fixed place. Hered = ns andr = 12s(s+3).

Theorem 7.62 together with Lemma 7.60 has the following corollary.

THEOREM 7.64 Supposens > p ≥ 12s(s + 3). ThenF is non-classical with

respect to the linear series cut out onF by all curves of a given degrees if andonly if there existz0(X,Y ), . . . , zr(X,Y ), h(X,Y ) ∈ K[X,Y ] and an integerm ≥ 1 such that

h(X,Y )f(X,Y ) = z0(X,Y )pm

+ . . . zr(X,Y )pm

Y s. (7.19)

REMARK 7.65 Rewording Theorem 7.64 in its homogeneous form, (7.19) reads

h(X0,X1,X2)f(X0,X1,X2) =∑zj(X0,X1,X2)

pm

Xu0X

v1X

w2

with homogeneous polynomialszj(X0,X1,X2) of the same degree. This showsthat interchangingX0 with X1 or X2 produces a similar equation. Therefore, inthe local study ofF , it is enough to consider branches ofF centred at an affinepoint. For this purpose, the inhomogeneous equation (7.19)is used, as it is moremanageable than the homogenous one.


7.8 CLASSICAL AND NON-CLASSICAL CURVES WITH RESPECT TO

LINES

The simplest case occurs when the linear system consists of all lines, that is, thelinear seriesL1 cut out onF = v(f(X,Y )) consists of all divisors

Ac = div (c0 + c1x+ c2y) +B, c = (c0, c1, c2) ∈ PG(2,K).

Without loss of generality, takex to be a separable variable. So,∂f(X,Y )/∂Ydoes not vanish. LetR =

∑nPP be the ramification divisor.

First, the classical case is considered; that is, the order sequence ofL is assumedto be(0, 1, 2). Then

degR = 3(2g − 2) + 3n. (7.20)

Assumep 6= 2. LetP be a a non-singular point ofF . ThenP is anL-Weierstrasspoint if and only if it is inflexion point ofF . If this is the case,j2 = I(P, ℓ ∩ F)with ℓ the (inflexional) tangent toF atP , andnP = j2 − 2 whereP is the placearising from the unique branch centred atP . Now, letP be a singular point ofF , andγ a branch ofΓ centred atP . Thenγ, namely the branch pointP ofF is anL-Weierstrass point if and only ifP is non-linear, and when this occurs,nP = j1 + j2 − 3 for the place arising fromP .

All these results are straightforward consequences of the previous general results,especially Lemmas 7.59 and 7.58. A direct, elementary argument is also possible.Again letp 6= 2. Sinceǫ2 = 2 6= p, the Hessian curve ofF does not vanish; writeits equation in inhomogeneous form:

∂f2

∂2X

(∂f

∂Y

)2

− 2∂f2

∂X∂Y

∂f

∂X

∂f

∂Y+∂f2

∂2Y

(∂f

∂X

)2

= 0.

Note that

d2y

dx2=

(∂f

∂y

)∂f2

∂2x

(∂f

∂y

)2

− 2∂f2

∂x∂y

∂f

∂x

∂f

∂y+∂f2

∂2y

(∂f

∂x

)2

.

Let P be a branch point ofF . If (x, y) is a primitive representation ofP , setx′ = dx/dt andx′′(t) = d2x/dt2. Likewise fory. Then

d2y

dx2=

(x′y′′ − x′′y′)x′3

.

By (i) of Theorem 7.49,x′y′′ − x′′y′ is covariant. Therefore,

ordt

(∂f2

∂2x

(∂f

∂y

)2

− 2∂f2

∂x∂y

∂f

∂x

∂f

∂y+∂f2

∂2y

(∂f

∂x

)2)

= ordt (x′y′′ − x′′y′) + 3(ordt∂f

∂y− ordt x

′).

Since the Hessian has degree3(n− 2), Bezout’s theorem gives

3n(n− 2) = degR+ 3

(ordt

∂f

∂y− ordt x

′).

220

Bezout’s theorem applied to the polar curve atY∞ shows that

∑ordt

∂f

∂y= n(n− 1),

while∑

ordt x′ = 2g − 2 by definition. Therefore, (7.20) holds.

In the non-classical case, the order sequence is(0, 1, pm) with m ≥ 1 for podd, andm ≥ 2 for p = 2. In particular,F is a non-reflexive curve whose Gaussmap has inseparability degreepm; this is theGeneral Order of Contact Theorem.Theorem 7.62 together with Lemma 7.60 shows that

zpm

0 + zpm

1 x+ zpm

2 y = 0,

wherez0, z1, z2 ∈ K(F) and at least one of them is separable. By Remark 7.63,the triple(z0, z1, z2) is determined up to a non-zero factor inK(F). After clearingdenominators,

h(X,Y )f(X,Y ) = z0(X,Y )pm

+ z1(X,Y )pm

X + z2(X,Y )pm

Y (7.21)

with z0, z1, z2, h ∈ K[X,Y ].

PROPOSITION 7.66 The factorH(X,Y ) is not needed in (7.21) for non-singularcurves; that is,h(X,Y ) = 1 can be assumed.

THEOREM 7.67 If degF = q, thenF is a rational strange curve. Up to a changeof the coordinate system,F has homogeneous equation

Xq2 = Xq−1

1 X0 + b2Xq−21 X2

0 + . . .+ bq−2X21X

q−20 +X1X

q−10 ,

with b2, . . . , bq−2 ∈ K.

THEOREM 7.68 Assume thatdegF = q+1. Up to a change of the coordinate sys-tem, eitherF is the Hermitian curveHq, or the rational nodal curve, or a strangecurve of equation

X0Xq2 +Xq−1

0 + a1Xq0X1 + a2X

q−10 X2

2 + . . .+ aqX0Xq1 +Xq+1

1 = 0,

with suitablea1, . . . , aq ∈ K.

EXAMPLE 7.69 The irreducible plane curveF given in Example 5.24 is a non-classical curve with order sequence(0, 1, 2q0). This follows from Theorem 7.62for

z0(X,Y ) = Xq0+1 + Y q0 , z1(X,Y ) = X, z2(X,Y ) = 1.

7.9 NON-CLASSICAL CURVES WITH RESPECT TO CONICS

Next, the cases = 2 in Theorem 7.64 is examined. The linear system of all conicscuts out onF a simple, fixed-point-free linear seriesL2 = g5

2n. This gives rise to


the irreducible curveΓ of PG(5,K) of degree2n which is the image curve ofFunder theVeronese map, that is, the birational transformation

ϕ : (1, x, y) 7→ (1, x, y, x2, xy, y2).

Some or allL2-orders may be calculated from theL1-orders, whereL1 = g2n is

cut out onF by all lines.To do this, two possibilities are considered separately according asL1 is classical

or non-classical. Take any non-singular pointP ∈ Γ which is not anL1-Weierstrasspoint ofF , and letP be the corresponding place. IfF is L1 non-classical, thenthe(L,P)-orders are(j0 = 0, j1 = 1, j2 = pm) with pm > 2. Choose three linesℓ0, ℓ1, ℓ2 such thatI(P, ℓi ∩ Γ) = ǫi, for 0 ≤ i ≤ 2. There are six (degenerate)conics whose components are two of these three lines. They provide six pairwisedifferent intersection multiplicities withF atP , namely0, 1, 2, pm, pm + 1, 2pm.Therefore, these integers are the(L2,P)-orders, and also theL2-orders due to thegeneral choice ofP .

From now on,F is supposed to be classical with respect toL1. The aboveargument shows that0, 1, 2, 3, 4 are (L2,P)-orders, and henceL2-orders. Sincethe hypotheses of Theorem 7.64 are satisfied forp ≥ 5, soǫ5 is a power ofp. Theconditionp ≥ 5 cannot be omitted; see Exercise 2. Forp ≥ 5, Theorem 7.62 showsthatǫ5 ≥ pm if and only if there existz0, . . . , z5, h ∈ K[X,Y ] such that

h(X,Y )f(X,Y )= z0(X,Y )pm

+ z1(X,Y )pm

X + z2(X,Y )pm

Y+z3(X,Y )pm

X2 + z4(X,Y )pm

XY + z5(X,Y )pm

Y 2.(7.22)

Unlike the caser = 2, the non-singularity condition onF is not enough to ensurethath(X,Y ) can be assumed in this equation. An example is given in Exercise 4.

Non-classicality ofL2 gives rise to several interesting properties of the curve.Here, the case thatp > 2, F is classical with respect toL1, and theL2-orders ofF are0, 1, 2, 3, 4, pm is considered. PutF = v(f(X,Y )), and letQ = (x, y) be ageneric point ofF . Then (7.22) holds, and at least one of the elementszi(x, y) inK(F) is a separable variable. Here and in the rest of the present section, i rangesover0, 1, . . . , 5. LetP be a place ofK(F) with (L2,P)-orders0, j1, j2. Choosean indexj such that ordPzj(x, y) ≤ ordPzi(x, y), and setvi = zi(x, y)/zj(x, y).From (7.22)

vpm

0 + vpm

1 x+ vpm

2 y + vpm

3 x2 + vpm

4 xy + vpm

5 y2 = 0. (7.23)Let γ be the branch ofF corresponding toP. Without loss of generality, assumethatγ is centred at an affine point, say at the pointP = (a, b). Given a primitiverepresentation(x(t), y(t)) of P , write the formal power seriesvi(t) explicitly inthe following form:

vi(t) = µ(0)i + µ

(1)i t+ . . .+ µ

(k)i tk + . . . (7.24)

So(µ

(0)0 )pm

+ (µ(0)1 )pm

x(t) + (µ(0)2 )pm

y(t) + (µ(0)3 )pm

x(t)2+

(µ(0)4 )pm

x(t)y(t) + (µ(0)5 )pm

y(t)2 + tpm

[(µ(1)0 )pm

+ (µ(1)1 )pm

x(t)+

(µ(1)2 )pm

y(t) + (µ(1)3 )pm

x(t)2 + (µ(1)4 )pm

x(t)y(t) + (µ(1)5 )pm

y(t)2] + . . .+

tkpm

[(µ(k)0 )pm

+ (µ(k)1 )pm

x(t) + (µ(k)2 )pm

y(t) + (µ(k)3 )pm

x(t)2+

(µ(k)4 )pm

x(t)y(t) + (µ(k)5 )pm

y(t)2] + . . . = 0.

222

Fork = 0, 1, . . ., put

sk(X,Y ) = (µ(k)0 )pm

+ (µ(k)1 )pm

X + (µ(k)2 )pm

Y+

(µ(k)3 )pm

X2 + (µ(k)4 )pm

XY + (µ(k)5 )pm

Y 2.

Then, withsk(t) = sk(x(t), y(t)), it follows that

s0(t) + tpm

s1(t) + . . .+ tkpm

sk(t) + . . . = 0. (7.25)

As s0(x, y) contains a non-zero coefficient,s0(X,Y ) = 0 is the equation of aconic, possibly reducible. Puttingµ(i)

0 = µi, the following result is obtained.

L EMMA 7.70 (i) The conicC0 = v(s0(X,Y )), with

s0(X,Y ) = µpm

0 + µpm

1 X + µpm

2 Y + µpm

3 X2 + µpm

4 XY + µpm

5 Y 2,

intersects the branchγ ofF with multiplicity at leastpm.

(ii) This multiplicity is greater thanpm if and only ifs1(a, b) = 0, where

s1(a, b) = (µ(1)0 )pm

+ (µ(1)1 )pm

a+ (µ(1)2 )pm

b+ (µ(1)3 )pm

a2

+(µ(1)4 )pm

ab+ (µ(1)5 )pm

b2.

(iii) If (a, b) is not a common zero of the six polynomialszi(X,Y ), then

C0 = v(z0(a, b)pm

+ z1(a, b)pm

X + z2(a, b)pm

Y+z3(a, b)

pm

X2 + z4(a, b)pm

XY + z5(a, b)pm

Y 2).(7.26)

In particular, for all but finitely many pointsP = (a, b) ofF , the osculatingconic ofF at P is C0.

REMARK 7.71 It may happen that a particular branch does not haveC0 as osculat-ing conic, as the following example shows. The Fermat curveF = v(F ), with

F = Xp−1 + Y p−1 + 1,

satisfies (i), since

XY (Xp−1 + Y p−1 + 1) = Y pX +XpY +XY.

Now, take a pointP = (0, c) with cp−1 = −1 of F on theY -axis. ThenP is aninflexion ofF ; thus the osculating conic ofF atP is degenerate and isv((Y −c)2),consisting of the horizontal line throughP counted twice. On the other hand, theconic C is also degenerate but isv(X(Y − c)) consisting of the horizontal andvertical lines through P, sincecpX +XY = X(Y − c).

Looking back through this section, or alternatively looking forward to Section8.7, the conicC0 plays a central role in the study of non-classical curves with respectto conics. When the symbolC0 is used it should be understood thatC0 denotes theconic with equation (7.26).

Introduce a new system of reference taking the centre(a, b) of the branchγ tothe origin and the tangentℓ of γ to theX−axis.


This change of coordinates from(X,Y ) to (X, Y ) is given by

X = v11X + v12Y + a,Y = v21X + v22Y + b,

(7.27)

with v11v22 − v12v21 6= 0. Correspondingly, letK(F) = K(x, y) such that

x = v11x+ v12y + a,y = v21x+ v22y + b,

(7.28)

Then ordP x = j1, ordP y = j2, Equation (7.22) is invariant under this transforma-tion. To see this, put

zi(X,Y ) = zi(v11X + v12Y + a, v21X + v22Y + b) = zi(X, Y ),f(X,Y ) = f(v11X + v12Y + a, v21X + v22Y + b) = F (X, Y ),h(X,Y ) = h(v11X + v12Y + a, v21X + v22Y + b) = H(X, Y ),a = cp

m

, b = dpm

, vij = nijpm

, i, j = 1, 2,

and writezi = zi(X, Y ). Then, with

Z0(X, Y ) = z0 + cz1 + dz2 + c2z3 + cdz4 + d2z5,Z1(X, Y ) = n11z1 + n21z2 + 2cn11z3 + (cn11 + dn21)z4 + 2dn21z5,Z2(X, Y ) = n12z1 + n22z2 + 2cn12z3 + (cn22 + dn12)z4 + 2dn22z5,Z3(X, Y ) = n11

2z3 + n11n21z4 + n212z5,

Z4(X, Y ) = 2n11n12z3 + (n12n21 + n11n22)z4 + 2n21n22z5,Z5(X, Y ) = n12

2z3 + n12n22z4 + n222z5,

equation (7.22) becomes

H(X, Y )F (X, Y )

=Z0(X, Y )pm

+ Z1(X, Y )pm

X + Z2(X, Y )pm

Y

+Z3(X, Y )pm

X2 + Z4(X, Y )pm

XY + Z5(X, Y )pm

Y 2. (7.29)

Since the relationship betweenZi andzi is linear and invertible,

minordPzi(x, y)) = minordPZi(x, y). (7.30)

Also, if w0, . . . , w5 ∈ K[X,Y ], set

∆(w0, . . . , w5) =

w012w1

12w2

12w1 w3

12w4

12w2

12w4 w5

.

With this notation,

ordP∆(z0(x, y), . . . , z5(x, y)) = ordP∆(Z0(x, y), . . . , Z5(x, y)). (7.31)

SetWi = Zi(X, Y )/Zj(X, Y ), andVi = Wi(x, y) wherej is chosen such that

ordPZj(x, y) = minordPZi(x, y).Then, in the new reference system(X, Y ), the conicC0 has equation

W0(a, b)pm

+W1(a, b)pm

X + V2(a, b)pm

Y+W3(a, b)

pm

X2 +W4(a, b)pm

XY +W5(a, b)pm

Y 2 = 0,(7.32)

224

which proves the covariance of the conicC0. From (7.32),

V pm

0 + V pm

1 x+ V pm

2 y + V pm

3 x2 + V pm

4 xy + V pm

5 y2 = 0.

Let ki = ordPVi. Then the left-hand side is the sum of six elements formK(F)whose orders are as follows:

ordPVpm

0 = k0pm, ordPV

pm

1 x = k1pm + j1,

ordPVpm

2 y = k2pm + j2, ordPV

pm

3 x2 = k3pm + 2j1,

ordPVpm

4 xy = k4pm + j1 + j2, ordPV

pm

5 y2 = k5pm + 2j2.

At least two of these orders must be equal, and they are less than or equal to theremaining four. Hence one of the following relations must hold:

(a) (k0 − k1)pm = j1, (b) (k1 − k3)p

m = j1,(c) (k2 − k4)p

m = j2, (d) (k0 − k3)pm = 2j1,

(e) (k0 − k2)pm = 2j2, (f) (k2 − k5)p

m = j2,(g) (k0 − k4)p

m = j1 + j2, (h) (k0 − k5)pv = 2j2,

(i) (k1 − k2)pm = j2 − j1, (j) (k3 − k4)p

m = j2 − j1,(k) (k1 − k4)p

m = j2, (l) (k4 − k5)pm = j2 − j1,

(m) (k1 − k5)pm = 2j2 − j1, (n) (k2 − k3)p

v = 2j1 − j2,(o) (k3 − k5)p

m = 2(j2 − j1).If n < pm, thenj2 < pm. This leaves just three possibilities; namely, (n)k2 = k3;(m) k1 = k5 + 1; (g) k0 = k4 + 1. In particular, the following result is obtained.

THEOREM 7.72 LetF = v(f(X,Y )) be an irreducible algebraic curve of degree3 ≤ n < pm which satisfies (7.22). Letγ be a branch ofF whose centre is theorigin and whose tangent is theX− axis. If the(L2,P) orders of a branch pointP ofF are0, j1, j2, then one of the following holds:

j2 =2j1, andC0 has equationY + cX2 = 0, c 6= 0; (7.33)

j2 = 12 (j1 + pm), andC0 has equationY 2 = 0; (7.34)

j2 = pm − j1, andC0 has equationXY = 0. (7.35)

LetG(X, Y ) = H(X, Y )F (X, Y )). Then

GX = HXF +HFX =Zpm

1 + 2Zpm

3 X + zpm

4 Y , (7.36)

GY = HY F + FFY =Zpm

2 + Zpm

4 X + 2Zpm

5 Y . (7.37)

Note that bothFX andFY are not zero polynomials, otherwiseF would be non-classical with order sequence(0, 1, pv) wherev ≤ m. The Hessianv(G∗(X,Y ))of the plane curveG = v(G(X,Y )), which may be reducible, is given by

G∗ =GXX(GY )2 − 2GXY GXGY +GXY (GX)2

= 2Zpm

3 (Zpm

2 + Zpm

4 X + 2Zpm

5 Y )2 − 2Zpm

4 (Zpm

2 + Zpm

4 X + 2Zpm

5 Y )

(Zpm

1 + 2Zpm

3 X + zpm

4 Y ) + 2Zpm

5 (Zpm

1 + 2Zpm

3 X + Zpm

4 Y )2

= 2(Z3Z22 − Z1Z2Z4 + Z5Z

21 )pm

+(4Z3Z5 − Z24 )pm

(Zpm

1 X + Zpm

2 Y + Zpm

3 X2 + Zpm

4 XY + Zpm

5 Y 2)≡ 2(Z3Z

22 − Z1Z2Z4 + Z5Z

21 )pm

+ (4Z3Z5 − Z24 )pm

Zpm

0 (mod F (X, Y ))

≡−∆(Z0, . . . , Z5)pm

(mod F (X, Y )).


On the other hand,

GXX(GY )2 − 2GXY GXGY +GY Y (GX)2 ≡H3[FXX(FY )2 − 2FXY FXFY + FY Y (FX)2] (modF (X, Y )).

It follows that

H3[FXX(FY )2 − 2FXY FXFY + FY Y (FX)2]≡ −∆(Z0, . . . , Z5)

pm

(modF (X, Y )).(7.38)

It may be noted that the left-hand side, apart fromH3, gives the Hessian ofF .SinceFY is not the zero polynomial,x is a separable variable ofK(F). Hence,

d2y

dx2= −FXXF

2Y− 2FXY FXFY + FY Y F

2X

F 3Y

, (7.39)

where the polynomials on the right-hand side are to evaluateat (x, y). Therefore,

d2y

dx2= ∆(Z0(x, y), . . . , Z5(x, y))

pm

/(H(x, y)FY (x.y))3 (7.40)

AsF is supposed to be classical with respect toL1, this shows that

∆(Z0(x, y), . . . , Z5(x, y)) 6= 0.

If R is the ramification divisor with respect to the linear seriescut out onF bylines, the following result is obtained.

L EMMA 7.73

vP(R)= pmordP∆(Z0(x, y), . . . , Z5(x, y)) + 3 ordPdx

−3[ordP(Z2(x, y)pm

+ Z3(x, y)pm

x+ 2Z5(x, y)pm

y)]. (7.41)

Now, a geometric interpretation of (7.41) is given. For thispurpose, the follow-ing technical hypotheses are assumed:

(i) z0(x, y), . . . , z5(x, y) are linearly independent overK;

(ii) The linear system

λ0z0(X,Y ) + . . . λ5z5(X,Y )

cuts out a simple linear seriesL′2.

It is worth mentioning here that both hypotheses are valid whenn < pm, but theproof is postponed to the end of the section. As usual, it is also assumed thatL′

2 isfixed point free.

LetZ be the image curve ofF under the rational transformation

ϕ′ : (1, x, y) 7→ (z0(x, y), z1(x, y), z2(x, y), z3(x, y), z4(x.y), z5(x, y)).

By hypothesis(ii), Z is an irreducible curve ofPG(5,K), birationally equivalenttoF overK. It may be noted thatZ can be viewed as the dual curve ofΓ. In fact,Z is birationally equivalent to the dual curve ofΓ as the Gauss map associated toΓis the product ofϕ′ by the Frobenius transformation ofPG(5,K). For more detailson dual curves, see Section 7.10.

226

Apart fromZ, it is appropriate to consider the cubic hypersurface

Φ = v(det(∆(X0, . . . ,X5))).

This depends on the fact that

I(P,Z ∩ Φ) = ordP det ∆(z0(x, y), . . . , z5(x, y)−min ordPzi(x, y).

In particular,Z is not contained inΦ, and Bezout’s Theorem gives that

3 degZ =∑

P∈Z∩ΦI(P,Z ∩ Φ) (7.42)

According to (7.30) and (7.31), the computation of the orders on the right-hand sidecan be carried out in the new reference system. Let

ordPZj(x, y) = min ordPzi(x, y),

then

vP(R) = pm I(P,Z ∩ Φ) + 3 ordPdx− 3ordPB(x, y) (7.43)

with

B(x, y) =

(Z2(x, y)

Zj(x, y)

)pm

+

(Z3(x, y)

Zj(x, y)

)pm

x+ 2

(Z5(x, y)

Zj(x, y)

)pm

y.

Orders on the right-hand side can be directly computed for each of the15 possibil-ities listed above. As in Theorem 7.72, only the casen < pm is treated.

Summing up (7.43) gives

pm degZ = c− n+∑

degB (7.44)

wherec denotes the class ofF .If (7.33) holds, then ordPZ2(x, y) = ordPZ3(x, y) is strictly less than the orders

of the four otherZi(x, y). Therefore,

B(x, y) = 1 + [Z4(x, y)/Z2(x, y)]pm

+ [Z5(x, y)/Z2(x, y)]pm

x.

Hence, ordPB(x, y) = 0.Assume that (7.34) holds. This time, ordPZ5(x, y) is strictly less than the orders

of the other ordPZi(x, y). Therefore,

B(x, y) = [Z2(x, y)/Z5(x, y)]pm

+ [Z4(x, y)/Z5(x, y)]pm

x+ 2y).

Since ordP y < pm, so ordPB(x, y) = j2.Finally, if (7.35) holds, then ordPZ4(x, y) is strictly less than the orders of the

other ordPZi(x, y). Therefore,

B(x, y) = [Z2(x, y)/Z4(x, y)]pm

+ X + 2[Z5(x, y)/Z4(x, y)]pm

y.

Since ordP x < pm, so ordPB(x, y) = j1.As ordPdx ≥ j1 − 1, the following result is obtained.

L EMMA 7.74 Letn < pm. Then

vP(R) ≥ pm I(P,Z ∩ Φ) +

3(j1 − 1) when (7.33) holds;−3(j2 − j1 + 1) when (7.34) holds;−3 when (7.35) holds.


SincevP(R) ≥ j1 + j2 − 3, this yields the following inequality:

4vP(R) ≥ pm I(P,Z ∩ Φ)−

6 whenj1 = 1 and (7.34) holds;0 otherwise.

If j1 < p, then ordPdx = j1−1, and(j1−0)(j2−0)(j2−j1) 6≡ (mod p) in eachof the three cases in Theorem 7.72. Hence, by Corollary 7.57,vP = j1 + j2 − 3,and equality holds in Lemma 7.74. Therefore, the following result holds.

L EMMA 7.75 If j1 < p, then

I(P,Z ∩ Φ) =

0 when (7.33) holds;2 when (7.34) holds;1 when (7.35) holds

So far, places corresponding to branches centred at affine points have been consid-ered. However, these results hold true when the centre is an infinite point, as inRemark 7.65.

Now, summing the above inequality asP ranges over all branch points ofFcentred at an affine point,

12(2g − 2) + 12n ≥ 3pm degZ − 6N,

whereN denotes the number of linear branchesP satisfying (7.34). Since

vP(R) ≥ j1 + j2 − 3 = 12 (pm − 3)

for anyP counted inN , an upper bound onN is

N ≤ 23(2g − 2) + 3n

pm − 3.

Therefore, the following theorem is established.

THEOREM 7.76 Let3 ≤ n < pm. Then

degZ ≤ 4

pm − 3[(2g − 2) + n]. (7.45)

Equality holds if and only ifF is non-singular. In this case, (7.45) reads

degZ =4n(n− 2)

pm − 3. (7.46)

EXAMPLE 7.77 Let q be an even power of the characteristicp of K. When thedegreen = 1

2 (√q + 1), the Fermat curveF = v(Xn + Y n + 1) is a non-classical

curve with respect toL2, as theL2-orders are(0, 1, 2, 3, 4,√q). In fact, (7.22)

holds for

h = (Xn + Y n − 1)(Xn − Y n + 1)(Xn + Y n − 1),

z0 = 1, z1 = −2X, z2 = 2Y, z3 = X2, z4 = −2XY, z5 = Y 2.

Since the six curvesv(Zi(X,Y )) have no common points,degZ = 2n. On theother hand,F has exactly3n inflexion points, and its branch points satisfying (7.34)are precisely its inflexion points.

228

As mentioned before, hypotheses (i) and (ii) are valid whenn < pm. Now, aproof is given by using some of the above results. To prove thefirst one, assume onthe contrary that a non-trivial linear combination ofzi(x, y) is zero, that is,

λ0z0(x, y) + . . .+ λ5z5(x, y) = 0 (7.47)

with λi ∈ K not all zeros. From Bezout’stTheorem,Φ andZ have non-trivialintersection. Choose a branch pointP of Z whose centre(z0(P ), . . . , z5(P )) is apoint of Φ. Suppose, without loss of generality, thatP arises from the branchγunderϕ.

From (7.47) it follows that neither (7.34) nor (7.35) can actually occur in theTheorem 7.72. To show this, note first that since the relationship betweenZi(x, y)andzi(x, y) are linear and invertible, (7.47) implies

ρ0Z0(x, y) + . . .+ ρ5Z5(x, y) = 0 (7.48)

with ρi ∈ K not all zeros. Therefore, at least two of the six numbers ordPZi(x, y)are equal and less than of the remaining four. This means thatjust one case, namely(n), can occur from the previous long list(a), . . . , (o). Hence, (7.33) is the onlypossibility in Theorem 7.72. In particular,C0 is a non-degenerate conic,j = 2or j = 3, and ordP det ∆(Z0(x, y, . . . , Z5(x, y)) − ordPZj(x, y) = 0. But thenI(P,Z ∩ Φ) = 0, a contradiction.

If L′2 is composed of an involution of orderr, thenn > rpm must hold which is

impossible forn < pm. Thus,(ii) is valid whenb < pm.Hence, Theorem 7.45 holds for any non-classical curveF with order sequence

(0, 1, 2, 3, 4, pm) with respect to the linear seriesL2 cut out onF by all conics.

7.10 DUAL CURVES OF SPACE CURVES

In Section 5.10 the dual curve of a plane curve is investigated. Now, duality forspace curves is considered.

Let Γ be a non-degenerate irreducible curve ofPG(r,K).

DEFINITION 7.78 The dual curve ofΓ is the irreducible curveΓ′ of PG(r,K)containing all but finitely many pointsP ′ = (b0, . . . , br) such that the hyperplanev(b0X0 + . . .+ brXr) in PG(r,K) is an osculating hyperplane ofΓ.

Corollary 7.47 shows how to obtain the dual curve of a space curve as its imageunder a rational transformation. For everyi = 0, . . . , r, cross out the last rowand thei-th column in the Wronskian determinant (7.12), and denote byFi theresulting(r − 1) × (r − 1) determinant multiplied by(−1)r+i. Then the rationaltransformationωC given by

ωC((x′0, x′1, . . . , x

′r)) = (F0, F1, . . . , Fr)

is theGauss mapassociated toΓ, and the curveΓ′ given by the point,

P ′ = (F0, F1, . . . , Fr),


is thedual curveof Γ. Note that, from the definition ofFi,

F0x0 + . . . Frxr = 0

F0D(1)ζ x0 + . . . FrD

(1)ζ xr = 0.

The geometric meaning of these equations is that the osculating hyperplane con-tains the tangent line. As for plane curves, thebidual Γ′′ of Γ is the dual ofΓ′.Also,ω′ stands for the Gauss map associated toΓ′.

The argument used in Section 5.10 can be adapted to the study of duality of spacecurves. However, the present situation is much more involved in positive character-istic and no details are given here. Here, the main results are stated together withillustrative examples. Letǫ0 < ǫ1 < . . . , ǫr be the orders of a non-degenerateirreducible curveΓ of PG(r,K). Denote bydegi ω the inseparability degree of theGauss map.

THEOREM 7.79 The highest power ofp dividing ǫr is equal todegi ω.

THEOREM 7.80 Suppose that the dual curveΓ′ of Γ is also non-degenerate andlet ǫ′0 < ǫ′1 < . . . ǫ′r be its orders. For a rational transformationδ : Γ→ Γ′ whoseinseparability degree is equal toq degi ω with q ≥ 1, the following conditions areequivalent:

(i) Γ = Γ′′ andδ = ω′ ω;

(ii) q dividesǫ′r, and for a general pointP ∈ Γ,

L(d)ω(P ) ⊆ L

(r−1)P

holds where the integerd is defined byǫd < ǫ′r/q < ǫd+1.

Also, if this is the case, thenǫd+1 = ǫ′r/q, andǫd+1 6≡ 0 (mod p).

EXAMPLE 7.81 The irreducible plane curveF = v(f(X,Y )) is assumed to benon-classical with orders0, 1, q′ whereq′ = pm > 2. Then (7.21) holds, and hencethe tangentℓQ toF at a general pointQ has equation

z0(Q)q′

+ z1(Q)q′

X + z2(Q)q′

Y = 0. (7.49)

Let grd be the linear series introduced before Theorem 7.64. The corresponding

rational transformationτ is given byx′ij = xiyj with 0 ≤ i, j ≤ s, and it definesthe irreducible curveΓ = τ(F) in PG(r,K). The orders ofΓ are all integers inthe setu + vq′ | u + v ≤ s. In particular,ǫr = sq′, and the equation of theosculating hyperplane ofΓ at the general pointδ(Q) is obtained formally raising(7.49) to thes-th power and replacingXiY j with Xij . HereXij with 0 ≤ i, j ≤ sdenote the non-homogeneous coordinates inPG(r,K).

Let δ be theqq′-th Frobenius transformation. IfℓQ containsδ(Q) = Qqq′

for ageneral pointQ, then

L(s−1)τ(P ) ⊆ L

(r−1)P , L

(s)τ(P ) 6⊂ L

(r−1)P

for a generic pointP . Note thatΓ′ is not degenerate becausep > s. Also, Γ′

coincides withΓ up to a linear collineation associated to a diagonal matrix.Inparticular,Γ andΓ′ have the same degrees.

230

THEOREM 7.82 Let

(i) ǫ0 < ǫ1 < . . . , ǫr be the degrees of a non-degenerate irreducible curveΓ ofPG(r,K);

(ii) the dual curveΓ′ of Γ be non-degenerate;

(iii) ǫ′0 < ǫ′1 < . . . ǫ′r be the orders ofΓ′;

(iv) ǫr · ǫ′r 6≡ 0 (mod p).

ThenΓ′′ = Γ if and onlyǫ′r ≤ ǫr. If this is the case, thenω′ ω is the identity mapandǫr = ǫ′r.

EXAMPLE 7.83 Let q > 3. The rational curveΓ given by the point,

Q = (1, x, x2, x3, xq + x2q+3, xq+1 + x2q+2),

is a non-degenerate curve inPG(5,K) with order sequence(0, 1, 2, 3, q, q+1). Thedual curveΓ′ is also non-degenerate and its orders are(0, 1, q, q + 1, 2q, 2q + 1).This gives an example of a curve withǫrǫ′r 6≡ 0 (mod p). Thus, by Theorem 7.82,Γ 6= Γ′′.

Another generalisation of the concept of dual curve of a plane curve is the Gaussiandual, which is an irreducible curve∆ in PG(r′,K) with r′ =

(r+12

)− 1, obtained

from Γ by the following Gauss map, via the classical Plucker embedding of thelines ofPG(r,K) in PG(r′,K).

Let Γ be given by the pointQ = (1, x1, . . . , xr), wherex = x1 is a separablevariable ofK(Γ). The2× 2 minors of the matrix,

[1 x x2 . . . xr

0 1 D(1)x x2 . . . D

(1)x xr

],

can be arranged in a sequence ofr′ elements, say,y0 = 1, y1, . . . , yr′ . Then theirreducible curve∆ given by the point

R = (1, y1, . . . , yr′)

is theGaussian dualof Γ. Also, the pointsQ = (1, x, . . . , xr) andR define arational transformationω of K(Γ). PutΣ = K(Q) = K(Γ) andΣ′ = K(R).

The first of the above minors is equal to1, while the others are

αi =D(i)x xi, 2 ≤ i ≤ r;

βi =xi − xαi, 2 ≤ i ≤ r;γij =xiαj − xjαi, 2 ≤ i < j ≤ r.

Since

γij = xiαj − xjαi = (βi + xαi)αj − (β + xaj)αi = βiαj − βjαi,

it follows that

Σ′ = K(α2, . . . , αr, β2, . . . , βr).


On the other hand, sinceΣ = K(x, x2, . . . , xr) andxi = xαi + βi for every2 ≤ i ≤ r, so

Σ = Σ′(x) = K(α2, . . . , αr, β2, . . . βr).

The following result is a generalisation of Theorems 5.80 and 5.81 to spacecurves.

THEOREM 7.84 Let ǫ0 = 0, ǫ1 = 1, ǫ2, . . . , ǫr be the orders ofΓ. Thenǫ2 = q′ isa power ofp if and onlyΣ/Σ′ is inseparable with inseparability degreeq′.

Proof. Let q′ be a power ofp, and letd be any power ofp such that1 ≤ d < q′. Itsuffices to show that the following conditions are equivalent:

(i) for all 2 ≤ i ≤ r andi < j ≤ r,

D(d)x αi = D(d)

x βi = D(d)x γij = 0; (7.50)

(ii) for all 2 ≤ i ≤ r,

D(1)x D(1)

x xi = D(d)x xi = 0. (7.51)

Now,

D(d)x βi = D(d)

x xi −d∑

k=0

D(d−k)x αiD

(k)x x =

−xD(1)

x αi if d = 1,

D(d)x xi − xD(d)

x αi if d > 1.

(7.52)On the other hand,

D(d)x γij =

d∑

k=0

(D(k)x xiD

(d−k)x αj −D(k)

x xjD(d−k)x αi). (7.53)

From (7.52) and (7.53), ifD(d)x αi = 0 for 2 ≤ i ≤ r and1 ≤ d ≤ q′, then

D(d)x γij = αjD

(d)x xi − αiD

(d)x xj , (7.54)

and

D(d)x βi =

0 if d = 1

D(d)x xi if d > 1

. (7.55)

By (7.54)and (7.55), it follows that (7.50) is equivalent tothe condition,

D(d)x αi = D(d)

x D(1)x xj = 0

for all 2 ≤ i ≤ r andd ≥ 1, while

D(d)x = xj = 0

whend > 1. This condition is equivalent to (7.51). 2

232

7.11 COMPLETE LINEAR SERIES OF SMALLER DEGREE

It is a difficult problem to determine the smallestn such that a curve has a completelinear series of ordern and positive dimension, that is, agr

n with r > 0. In thissection, those curves are investigated for which this minimum degree is either1 or2. By Theorem 6.79 the former possibility only occurs for rational curves. Also,by Theorems 6.79 and 7.41, if a curve has more than oneg1

2 , it is either rational orelliptic, and in the latter but not in the former case,g1

2 is complete. Also, if a curvehas genus2 then it is hyperelliptic since its canonical series is ag1

2 .

THEOREM 7.85 An irreducible plane curveF = v(f(X,Y )) has genus1 if andonly ifF is birationally equivalent to a non-singular cubic curve. Acanonical formof f(X,Y ) is one of the following:

(I) Y 2 −X3 − uX − v with 4u3 + 27v2 6= 0, for p 6= 2, 3;

(II) Y 2 −X3 − uX2 − vX − w with v3 = u3w, for p 6= 3;

(III) Y 2 + uY +X3 + vX + w with u 6= 0, for p = 2;

(IV) Y 2 +XY +X3 + uX + v with v 6= 0, for p = 2.

Proof. By Theorem 5.56, any non-singular cubic curve has genus1. Conversely, letΣ be an elliptic curve, and take a placeP of Σ. By Theorem 7.39, the cubic curvederived from the rational transformation associated to thecomplete linear series|3P| is a non-singular plane curveF . Take the coordinate system in the plane suchthat the line at infinity is a tangent toF at the pointY∞. ThenF = v(f(X,Y ))with f(X,Y ) containing only one term of degree3, namelyX3. ReplacingX bycX andY by dX for suitable non-zero elementsc, d ∈ K, the polynomialf(X,Y )becomes

Y 2 + a1XY + a3Y −X3 + a2X2 + a4X + a5.

If p 6= 2, it is possible to get rid of both termsXY andY by a replacingYwith bY + c. So, without loss of generality, takea1 = a3 = 0. For p = 3, thisleads to (II), the conditionv3 = u3w being necessary and sufficient forF to benon-singular.

If p 6= 2, 3 theX2-term can also be removed by replacingX byX − 12a2 giving

the canonical form (I). The condition4u3 + 27v2 6= 0 says thatF is non-singular.Supposep = 2. If a1 6= 0, then linear transformations ofX andY can be made

to give a3 = a4 = 0 anda1 = 1. If a1 = 0, thena3 6= 0 anda2 = 0 can beobtained by a linear transformation ofX. Also, a linear transformation inY givesa3 = 1. The resulting canonical forms are those in (III) and (IV); the conditionsare necessary and sufficient forF to be non-singular. 2

Now, hyperelliptic curves are investigated.

THEOREM 7.86 Every hyperelliptic curve of genusg is birationally equivalent toa curve∆ of degreeg + 2 with only one singular point. This singular point is ag-fold point, and

∆ = v(a0(X)Y 2 + a1(X)Y + a2(X)).


The uniqueg12 is cut out by the vertical lines, apart from a fixed divisor.

Proof. Let Γ be irreducible of genusg. There exist distinct placesP1, . . . ,Pg

of Σ = K(Γ) which arise from branches ofΓ centred at non-singular pointsP1, . . . , Pg such that the linear series|E|, with E =

∑gi=1 Pi, is non-special.

Now, assumeΓ to be hyperelliptic, and letD = Q1 + Q2 be a divisor in theuniqueg1

2 of Σ such that the branches ofΓ corresponding toQ1 andQ2 havedifferent centresQ1 andQ2. Since the linear series|D + E| is non-special, fromthe Riemann–Roch Theorem,dim |D + E| = 2. Then, sincei(|E|) = −1, soi(|E−Pi|) = 0 for eachi = 1, . . . g. Therefore, there is a unique canonical divisorCi containingE − Pi.

If |D + E| has a fixed placeP, thenP is one of the places inE becauseg12 has

no fixed place. Also,i(|D + E − Pi|) = 1; that is, there is a unique canonicaldivisorCi containingD + E − Pi. Therefore, to ensure thatD + E has no fixedplace, it suffices to chooseQ1 different from the finitely many places appearing inthe canonical divisorsCi. So, suppose thatD + E = g2

g+2 has no fixed place. Let

g12 = div (c0 + c1x) +D | c = (c0, c1) ∈ PG(1,K).

Then there existsy ∈ K(Γ) such that

g2g+2 = div (c0 + c1x+ c2y) +D +E | c = (c0, c1, c2) ∈ PG(2,K).

Sinceg2g+2 has no fixed place, from Theorem 6.80 it follows that div(z)∞ = D+E

for somez ∈ Σ. Hence,

z = c0 + c1x+ c2y

with [Σ : K(z)] = g + 2. In particular,z cannot be written asd0 + d1x withd0, d1 ∈ K. Therefore,y may be replaced byz in the above representation of|D + E|. By Theorem 7.43,z 6∈ K(x), whence it follows thatΣ = K(x, z). Inother words,|D + E| is simple, that is, not composed of an involution. Hence, upto birationally equivalence,Γ may be supposed to be an irreducible plane curve ofdegreeg + 2, andg2

g+2 is cut out onΓ by lines.If D′ = Q′

1 +Q′2 andD′ = D + div (ξ) for someξ ∈ Σ, then

D′ + E = D + E′ + div (ξ).

In particular, if the branches ofΓ corresponding toQ′1 andQ′

2 have different centresQ′

1 andQ′2, then the common point of the linesQ1Q2 andQ′

1Q′2 is the common

centreU of the branches corresponding toP1, . . . ,Pg. This is only possible whenU is ag-fold point of Γ. This implies thatΓ has no more singular points. In fact,its virtual genusg∗ is equal tog, since

g∗ ≤ 12 ((g + 1)g − g(g − 1)) = g,

while g∗ ≥ g by Lemma 3.28 (i). Without loss of generality,U may be assumed tobeY∞. ThenΓ = v(g(X,Y )) with

g(X,Y ) = a0(X)Y 2 + a1(X)Y + a2(X),

wheredeg g(X,Y ) ≤ 2g + 2. The uniqueg12 is cut out onΓ by the vertical lines,

minus the fixed divisorE. 2

234

COROLLARY 7.87 A hyperelliptic curve of genusg is birationally equivalent to acurve

v(Y 2 + h(X)Y + g(X)), with h(X), g(X) ∈ K[X], (7.56)

anddeg h(X) ≤ g + 1, deg g(X) ≤ 2g + 2.

Proof. Let Γ = v(a0(X)Y 2 + a1(X)Y + a2(X)). For a generic pointP = (x, y)of Γ,

τ : x′ = x, y′ = a0(x)

is a rational transformation of the function fieldK(Γ) = K(x, y) of Γ. Then,τ isbirational, and the image curveΓ′ = v(g′(X,Y )) with

g′(X,Y ) = Y 2 + h(X)Y + g(X),

whereh(X) = a1(X) andg(X) = a0(X)a1(X). This proves the corollary. 2

Two cases are distinguished according asK has even characteristic or not.

THEOREM 7.88 LetK to be of zero or odd characteristic. Then

(i) a curve of genusg is hyperelliptic if and only if it is birationally equivalentto a curveX = v(Y 2 − f(X)), where the polynomialf(X) ∈ K[X] hasdegree2g + 1 and no square factors;

(ii) the uniqueg12 is cut out by the vertical lines apart from a fixed divisor;

(iii) there are finitely many placesP ofK(X ) such that2P ∈ g12 .

Proof. In zero or odd characteristic, the equation given in Theorem7.86 can besimplified further by means of the birational transformationx′ = x, y′ = y+ 1

2h(x)to the form

Y 2 = f(X), f(X) ∈ K[X], deg f(X) ≤ 2g + 2. (7.57)

Suppose thatf(X) = g(X)2k(X) with g(X), k(X) ∈ K[X] wherek(X) has nosquare factor. Then the birational transformationx′ = x, y′ = g(x)−1y reducesthe equation to the same form (7.57) but so thatf(X) has no square factors.

Supposeh = 2m, and letα be a rootα of f(X). The birational transformationx′ = (x − α)−1, y′ = yxm reduces the equation to the form (7.57) withm odd.Example 5.58 shows thath = 2g + 1.

An alternative proof depending on the Hurwitz formula is also possible. For ageneric pointP = (x, y) of Γ, letΣ = K(x, y) be the associated function field andΣ′ = K(x) its rational subfield. The field extensionΣ/Σ′ has degree2, and henceit is separable, as the characteristic ofK is distinct from2. From the calculation inExample 5.58,ω : Σ 7→ Σ′ only ramifies at the places corresponding to branchescentred at the pointsP = (a, 0) with f(a) = 0 and atY∞. The ramification indexis equal to2 for each of these places. Hence, the degree of the different divisorD(Σ/Σ′) is equal to2g + 2. If g(Σ) denotes the genus ofΣ, then (7.6) becomes

2g(Σ)− 2 = 2(0− 2) + 2g + 2,

whenceg(Σ) = g. The placesP such that2P ∈ g12 are those arising from the

branches centred atY∞ and at the points(a, 0) with f(a) = 0. So, they are finitelymany. 2


THEOREM 7.89 Let

(i) K be of zero or odd characteristic;

(ii) the polynomialsg(X), h(X) ∈ K[X] have both degree2g+1 and no squarefactors;

(iii) α0, . . . , α2g andβ0, . . . , β2g be the roots ofg(X) andh(X).

Then the hyperelliptic curves

Γ = v(Y 2 − f(X)), ∆ = v(Y 2 − g(X))

are birationally equivalent if and only if there is rationalfunction

w(X) = (aX + b)/(cX + d), ad− bc 6= 0,

which maps the setα0, . . . , α2g,∞ onto the setβ0, . . . , β2g,∞.Proof. For a generic pointP = (x, y) of Γ, let Σ = K(x, y) be the function fieldof Γ. The uniqueg1

2 of Σ consists of all divisors,

div (c0 + c1x) +B, c0, c1 ∈ K, (c0, c1) 6= 0,

whereB = 2P∞ andP∞ is the place corresponding to the unique branch ofΓcentred atY∞; see Example 5.58.

Now, assume that∆ is a birationally equivalent toΓ. ThenΣ = K(ξ, η) for ageneric pointQ = (ξ, η) of ∆. Taking into account the uniqueness ofg1

2 , Theorem6.17 ensures the existence ofζ ∈ Σ, for which

1 = ζ(cx+ d), ξ = ζ(ax+ b)

with ad− bc 6= 0. Eliminatingζ givesξ = (ax+ b)/(cx+ d).For 0 ≤ i ≤ 2g, let γi be the branch ofΓ centred at the pointPi = (αi, 0).

The corresponding placePi of Σ is such that the divisor2Pi belongs tog12 . In the

model(∆; (ξ, η)) of Σ, the place2Pi comes from a branchγ of ∆, centred at apointP , possible coincident withY∞. If σ is a primitive representation ofPi, then

σ(ξ) = (aσ(x) + b)/(cσ(x) + d),

whence eitherP is affine and itsX-coordinate is(aαi + b)/(cαi + d), orP = Y∞andcαi + d = 0. Hencew(αi) ∈ β0, . . . , β2g,∞. The previous argument stillworks if γi is replaced by the branch ofΓ centred atY∞, showing that

w(∞) = b/d ∈ β0, . . . , β2g,∞.

To show the converse, put

Σ = K(x, y) with y2 − f(x) = 0,

Σ′ = K(x′, y′) with y′2 − g(x′) = 0.

Assume first thatw(∞) = ∞. Thenw(X) = aX + b. Relabel the roots ofg(X)such thatw(αi) = βi for i = 0, . . . , 2g, and write

f(X) =u∏2g

i=0(X − αi),

g(X) = v∏2g

i=0(X − βi).

236

Then the birational transformation

ω : ξ = ax+ b, η = ǫy with ǫ2 = a2g+1v/u,

mapsΓ onto∆. In fact, since

g(ξ) = v∏2g

i=0(ξ − βi)

= v∏2g

i=0(ax+ b− βi)

= v∏2g

i=0(ax+ b− (aαi + b)

= va2g+1∏2gi=0(x− ai)

= a2g+1(v/u)f(x),

so

η2 − g(ξ) = a2g+1(v/u)(y2 − f(x)),

whenceη2 − g(ξ) = 0. Therefore, not onlyP = (x′, y′) but alsoQ = (ξ, η) is ageneric point of the curvev(Y 2 − g(X)). By Theorem 5.7,K(ξ, η) andK(x′, y′)are isomorphic. On the other hand, asω is birational,K(x, y) andK(ξ, η) areisomorphic. The proof in the casew(∞) =∞ is complete.

If w(∞) 6=∞, relabel the indices of the roots off(X) andg(X) such that

w(∞) = β0, u(α0) =∞, w(αi) = βi,

for i = 1, . . . , 2g. Then,

w(X) = (β0X + b)/(X − α0), β0αi + b = αiβi − α0βi,

for 1 ≤ i ≤ 2g. It follows that

ω : ξ =β0x+ b

x− α0, η = ǫ

y

(x− α0)g+1

with

ǫ2 = u−1(b+ α0β0)∏2g

i=1(β0 − βi),

is a birational transformation ofΓ onto∆. In fact,

g(ξ) = v∏2g

i=0(x− βi)

= v∏2g

i=0

β0x+ b

x− α0− βi

= v(b+ α0β0)1

x− α0

∏2gi=1(β0 − βi)

x− αi

x− α0

=v

u(b+ α0β0)

(∏2gi=1(β0 − βi)

) 1

(x− α0)2g+2f(x).

This shows thaty2−f(x) = 0 impliesη2−g(ξ) = 0. Now, the previous argumentdepending on Theorem 5.7 can be repeated to finish the proof. 2

Analogous results hold in the case of even characteristic.

THEOREM 7.90 Whenp = 2, a hyperelliptic curve is birationally equivalent to acurve

v(Y 2 + h(X)Y + g(X)),

as in (7.56), with no affine singular points and the leading coefficient ofg(X) equalto 1.


Proof. From Theorem 7.86, consider a hyperelliptic curve

Γ = v(Y 2 + h(X)Y + g(X)),

with h(X), g(X) ∈ K[X], deg h(X) ≤ g + 1, deg g(X) ≤ 2g + 2. If P = (a, b)is an affine singular point ofΓ, then, by a transformation(X,Y ) 7→ (X,Y + b),the pointP = (a, 0); henceg(a) = 0. SinceP is a singular point ofΓ, so

∂(Y 2 + h(X)Y + g(X))/∂Y = 0

at (a, 0); henceh(a) = 0. Also,

∂(Y 2 + h(X)Y + g(X))/∂X = 0

at (a, 0). Puttingg′(X) = dg(X)/dX, it follows thatg′(a) = 0. Thus

h(X) = (X − a)h1(X), g(X) = (X − a)2g1(X),

with h1(X), g1(X) ∈ K[X], deg h1(X) < deg h(X), deg g1(X) < deg g(X).The birational transformationx′ = x, y′ = y/(x − a) mapsΓ onto the curve∆ = v(Y 2 + h1(X)Y + g1(X)). By repeated application of this argument, ahyperelliptic curve with no affine singular point is obtained. 2

L EMMA 7.91 For p = 2, let Γ = v(Y 2+h(X)Y +g(X)) be a hyperelliptic curvewith no affine singular point. Then, under the birational transformation,

x′ = x, y′ = y + r(X)

with r(X) ∈ K[X], the image curveΓ′ of Γ has no affine singular points.

Proof. It is shown that, ifΓ′ had an affine singular point, thenΓ would also havean affine singular point. Then

Γ′ = v(Y 2 + h(X)Y + f(X))

with f(X) = r(X)2 + h(X)r(X) + g(X). If P = (a, b) is a singular point ofΓ′,then both partial derivatives are zero atP . Putting

r′(X) = dr(X)/dX, h′(X) = dh(X)/dX, g′(X) = dg(X)/dX,

gives

b2 + h(a)b+ r(a)2 + h(a)r(a) + g(a) = 0;

h(a) = 0;

bh′(a) + r′(a)h(a) + r(a)h′(a) + g′(a) = 0.

Note that the third equation can also be written as(r(a) + b)h′(a) + g′(a) = 0.Since

∂(Y 2 + h(X)Y + g(X))/∂Y =h(X),

∂(Y 2 + h(X)Y + g(X))/∂X =Y h′(X) + g′(X),

it follows that the pointQ = (a, c) with c = r(a) + b is a singular point ofΓ. 2

238

L EMMA 7.92 In Theorem 7.90, the polynomialsh(X), g(X) can be chosen suchthat

deg g(X) ≥ 2 deg h(X) + 1. (7.58)

Proof. Let Γ be as in Theorem 7.90. Putk = deg h(X), n = deg g(X). If n iseven, then the birational transformationx′ = x, y′ = y + xn/2 reduces it to thesame form (7.56) withdeg h(X) = k, deg g(X) < n. Lemma 7.91 ensures thatno affine point becomes singular. Ifdeg g(X) is still even, the argument can berepeated. Thusdeg g(X) can be supposed to have odd degree2m + 1. If k > m,then the birational transformationx′ = x, y′ = y + xk−1 mapsΓ onto the curve(7.56) withdeg g(X) ≥ 2k − 1. Hencem ≥ k; that is, (7.58) holds. 2

L EMMA 7.93 In Lemma 7.90, the polynomialsh(X), g(X) can be chosen so that

deg g(X) = 2 deg h(X)− 1, (7.59)

Proof. Assume thath(X) andg(X) satisfy (7.58). Choose an elementa ∈ K forwhich h(a) 6= 0 6= g(a). ReplacingX by X + a ensures thath(0) 6= 0 6= g(a).Now, x′ = x−1, y′ = yx−(m+1) is a birational transformation ofΓ onto the curveΓ1 = v(Y 2 + h1(X)Y + g1(X)) with h1(X), g1(X) ∈ K[X] defined as follows:

h1(X) = h

(1

X

)Xm+1, g1(X) = g

(1

X

)X2m+2.

Note thatdeg h1(X) = m + 1, anddeg g1(X) = 2m + 2. If hm+1 is the leadingcoefficient ofh1(X), letu ∈ K be a root of the polynomial

X2 + hm+1X + 1.

Then the birational transformationx′ = x, y′ = y+uxm+1 mapsΓ1 onto a hyper-elliptic curveΓ2 as in (7.56) satisfying (7.59). IfΓ2 has some affine singular points,use the argument employed in the proof of Theorem 7.90 to obtain a curve with noaffine singular point. Such a curve is still of the form (7.56)with h(X), g(X)satisfying (7.59). 2

THEOREM 7.94 For p = 2, a hyperelliptic curve of genusg is birationally equiv-alent to a curve,

Γ = v(Y 2 + h(X)Y + g(X)), deg g(X) = 2g + 1, deg h(X) ≤ g,with the following properties:

(i) Γ has no affine singular point;

(ii) Γ has only one branch centred at its unique point at infinityY∞;

(iii) if P∞ is the place corresponding toY∞, then theg12 is cut out onΓ by the

vertical lines minus the fixed divisor(2g − 1)P∞;

(iv) there are finitely many placesP of Σ such that2P ∈ g12 .


Proof. With the notation used in the preceding two proofs,k + 1 < n = 2m + 1holds and henceΓ has a unique point at infinity, namelyY∞. More precisely,Y∞is a (2m− 1)-fold point ofΓ.

Now, it is shown thatΓ has exactly one branch centred atY∞. Write

h(X) = h0 + . . .+ hkXk

with hk 6= 0, and leth(X,Y ) be the associated homogeneous polynomial; that is,

h(X,Y ) = h0Yk + . . .+ hkX

k.

Similarly, write

g(X) = f0 + . . .+X2m+1,

and let

g(X,Y ) = f0Y2m+1 + . . .+X2m+1.

Then, in homogeneous coordinates,

Γ = v(Y 2Z2m−1 + Y Z2m−kh(X,Z) + g(X,Z)).

TakeY∞ to the originO by the linear transformation(X,Y,Z) 7→ (X,Z, Y ). ThecurveΓ is mapped to the curve

∆ = v(Y 2m−1Z2 + Y 2m−kh(X,Y )Z + g(X,Y ))

or, in inhomogeneous coordinates,

∆ = v(Y 2m−1 + Y 2m−kh(X,Y ) + g(X,Y ))

The above transformation takes the line at infinity to theX-axis which is then atangent to∆ at the origin.

Now, it is shown that there is only one branch of∆ centred atO. Let (x(t), y(t))be a primitive representation of a branch of∆ centred at the origin and tangent totheX-axis. Write

x(t) = ati + · · · , y(t) = btj + · · · , with i < j.

Then

y(t)2m−1 + y(t)2m−kh(x(t), y(t)) + g(x(t), y(t)) = 0. (7.60)

Substitution gives

(btj + · · · )2m−1 + (btj + · · · )2m−kh(ati + · · · , btj + · · · )+g(ati + · · · , btj + · · · ),

which shows that the terms of possible minimum degree int are

(2m− 1)j, (2m− k)j + ik, (2m+ 1)i.

Therefore, one of the following cases may occur:

(I) (2m− 1)j = (2m− k)j + ki < (2m+ 1)i;

(II) (2m+ 1)i = (2m− k)j + ki < (2m− 1)j;

240

(III) (2m− 1)j = (2m− k)j + ki = (2m+ 1)i;

(IV) (2m− 1)j = (2m+ 1)i < (2m− k)j + ki.

In fact, none of the first three cases occurs. To show that (I) is impossible, notethat(2m−1)j = (2m−k)j+ki implies that(k−1)j = ki; hencei = k−1, j = k,since becausek − 1 andk are relatively prime. Then

(2m− 1)j < (2m+ 1)i,

(2m− 1)k < (2m+ 1)(k − 1),

m<k,

contradicting thatk ≤ m.Similarly for (II), (2m+1)i = (2m−k)j+ki impliesi = 2m−k, j = 2m−k+1.

Thus,

(2m− k)j + ki< (2m− 1)j,

(2m− k)(2m− k + 1) + k(2m− k)< (2m− 1)(2m− k + 1),

m<k,

again a contradiction.To rule out (III),(2m−1)j = (2m+1)i implies thati = 2m−1, j = 2m+1 as

in (I), while (2m+1)i = (2m−k)j+ki implies thati = 2m−k, j = 2m−k+1as in (II).

This leaves (IV). The only positive integersi, j which can occur in (IV) arei = 2m − 1 andj = 2m + 1. Hence,∆ has a branchγ centred at the origin suchthat I(O,v(Y ) ∩ γ) = 2m + 1. But then no other branch of∆ is centred at theorigin. This follows from Theorem 4.48, sinceI(O,∆ ∩ v(Y )) = 2m+ 1.

It remains to show thatg = m. As i = 2m − 1 is odd, Theorem 4.26 gives asubstitution of typet 7→ t = t+ · · · , for which

x(t) = x(t) = t(2m−1)

, y(t) = y(t) = αt(2m+1)

+ · · · ,with α ∈ K\0. Replacet by t and omit the bar overx andy. If y(t) containssome terms of odd degree larger than2m + 1, thenβt2m+1+n denotes the lowestdegree for which this happens. Otherwise, letβ = 0. Write

y(t) = αt2m+1 + · · ·+ βt2m+1+n + · · ·with β ∈ K. Now, the left-hand side of (7.60) can be written ast(2m−1)(2m+1)u(t)with

u(t) = (α+ · · ·+ βtn + · · · )2m−1 + t2m−2k+1(α+ · · ·+ βtn + · · · )2m−k

[hk + · · ·+ hk−rt2r(α+ · · ·+ βtn + · · · )r + · · ·

+ h0t2k(α+ · · ·+ βtn + · · · )k] + 1 + g2mt

2(α+ · · ·+ βtn + · · · ) + · · ·+ g0t

2(2m+1)(α+ · · ·+ βtn + · · · )2m+1,

wheregr 6= 0 butgi = 0 for i = r+1, . . . , 2m. Also, if g(X) contains some termsof even degree, thens denotes the smallest degree for which this happens.


Otherwise, letgs = 0. From the definition ofn,

du/dt =

(α+ · · ·+ βtn + · · · )2m−2(βtn−1 + tn+1G1(t))

+t2m−2k(α+ · · ·+ βtn + · · · )2m−k(hk + t2G2(t))

+t2m−2k+1(α+ · · ·+ βtn + · · · )2m−k−1(βtn−1 + . . .)(hk + t2G3(t))

+t2m−2k+1(α+ · · ·+ βtn + · · · )2m−2k+1(t2G4(t))

+(gst2(2m−s)−1(α+ · · ·βtn + · · · )2m−s(βtn−1 + tn+1G5(t)).

It should be noted that, ifβ = 0, then the termβtn−1 + tn−2G6(t) vanishesby definition. In this case,du/dt has only one term with lowest degree, namelyα2m−khkt

2m−2k, but for β 6= 0 there is one more, namelyα2m−2βtn−1. By(7.60),u(t) = 0, and hencedu(t)/dt = 0. This rules out the former possibility.Thus, the latter possibility occurs, and hence2m− 2k = n− 1 holds.

Going back toΓ, its unique branch centred atY∞ has the following primitiverepresentation:

ξ(t) =t2m−1

αt2m+1 + · · ·+ βt2m−2k+1 + · · · = t−2 1

α+ · · ·+ βt2m−2k+1 + · · ·η(t) =

1

αt2m+1 + · · ·βt2m−2k+1 + · · · ,

with α, β 6= 0.SinceK has even characteristic,

dξ(t)

dt= t−2 (2m− 2k + 1)βt2m−2k + · · ·

(α+ · · ·+ βt2m−2k+1 + · · · )2.

Hence, ifP is the corresponding place ofΣ, then

ordP(dξ) = ordt (dξ(t)/dt) = 2m− 2k − 2. (7.61)

Note thatdξ(t)/dt 6= 0, and this shows thatξ is a separable variable ofK(Γ).To calculate the genusg of Γ, take all branchesγ of Γ with ordt dx(t)/dt 6= 0,

where(x(t), y(t)) is a primitive representation ofγ. Apart from the branch centredatY∞, all branches are centred at affine points, and hence ordt dx(t)/dt ≥ 0. So,attention is focused on branches with ordt dx(t)/dt > 0. SinceΓ has no affinesingular point, a necessary and sufficient condition for ordt dx(t)/dt to be positiveis that the tangent line toΓ at the centreP = (a, b) of γ is the vertical linev(X−a).If this happens, then

ordt dx(t)/dt = ordt

(h(t)

dy(t)

dt

), with h(t) = h(x(t)),

by Theorem 5.55. SinceK has even characteristic,

ordt

(dx(t)

dt

)= ordt

(h(t)

dy(t)

dt

)= ordt

(dy(t)

dt

)+ ordt h(t).

Let j denote the multiplicity of a roota of h(X); that is

h(X) = (X − a)jh1(X)

242

with h1(X) ∈ K[X] andh1(a) 6= 0. A primitive representation of the uniquebranch centred atP = (a, b) is

x(t) = a+ ti + · · · , y(t) = b+ t,

wherei ≥ 2. Actually, i = 2 sinceΓ has degree2m+1 andY∞ is a(2m−1)-foldpoint ofΓ; then ordt h(t) = 2j.

Let a1, . . . , as with multiplicities j1, . . . , js be the roots ofh(X) . Then

j1 + . . .+ js = k.

This, together with (7.61), shows that∑

ordt dx(t)/dt = 2k + 2m− 2k − 2 = 2m− 2.

By (5.54), this sum defines the genus ofΓ, and hencem = g. The placesP suchthat2P ∈ g1

2 are those arising from the branches centred atY∞ and at the points(a,√g(a)) with h(a) = 0. So, they are finitely many. 2

THEOREM 7.95 For p = 2, let Γ and ∆ be two hyperelliptic curves of genusggiven in their canonical form as in Theorem 7.94; that is,

Γ=v(Y 2 + h(X)Y + g(X)), deg g(X) = 2g + 1, n = deg h(X) ≤ g;∆ =v(Y 2 + h1(X)Y + g1(X)), deg g1(X) = 2g + 1, m = deg h1(X) ≤ g.

Let α0, . . . , αn and β0, . . . , βm be the roots ofh(X) and h1(X), each countedwith multiplicity. ThenΓ and∆ are birationally equivalent if and only if each ofthe following conditions are satisfied:

(i) n = m, and there is rational functionw(X) = (aX + b)/(cX + d), withad − bc 6= 0, which maps the multisetα1, . . . , αn,∞ onto the multisetβ1, . . . , βn,∞;

(ii) there is a rational functionv(X) ∈ K(X) such that

v(X)2 + v(X) =g(X)

h(X)2+

g1(w(X))

h1(w(X))2.

Proof. For a generic pointP = (x, y) of Γ, let K(Γ) = K(x, y). Assume thatboth (i) and (ii) hold. Then the birational transformation

ω : ξ = w(x), η = v(x)h1(w(x)) +h1(w(x))

h(x)y,

mapsΓ onto∆. In fact, a direct computation shows thatη2 + h1(ξ)η + g1(ξ) = 0follows fromy2 + h(x)y + g(x) = 0.

To prove the converse, assume thatΣ = K(ξ, η) with η2 + h1(ξ)η+ g1(ξ) = 0.Let γi be a branch ofΓ centred at the pointPi = (αi,

√g(αi)) Let Pi be the

corresponding place ofK(Γ).As in the proof of Theorem 7.89, the uniqueness ofg1

2 implies thatξ = w(x)with w(X) = (aX + b)/(cX + d) andad + bc 6= 0. The argument used in thatproof also shows the setβ1, . . . , βm,∞ containsw(αi), for i = 1, . . . , n, and


w(∞). Interchanging the roles ofΓ and∆, it follows that the setα1, . . . , αn,∞containsw(βi), for i = 1, . . . ,m, andw(∞). Therefore,n = m and (i) holds.

To show (ii), note that

η = a(x) + b(x)y

with a(X), b(X) ∈ K(X) since[K(Γ) : K(x)] = 2. Sincey2 = h(x)y + g(x),soη2 + h1(ξ)η + g1(ξ) = 0 can be written as

(b(x)2h(x)+b(x)h1(w(x))y+a(x)2+b(x)2g(x)+a(x)h1(w(x))+g1(w(x)) = 0.

Sincey 6∈ K(x), this is only possible whenb(x) = h1(w(x))/h(x) and(

a(x)

h1(w(x))

)2

+a(x)

h1(w(x))=

g(x)

h(x)2+

g1(w(x))

h1(w(x))2.

Puttingv(X) = a(X)/h1(w(X)), (ii) follows. 2

THEOREM 7.96 Letp = 2. For u, u1 ∈ K(X), put

Σ =K(x, y) with y2 + y + u(x) = 0,

Σ1 =K(ξ, η) with η2 + η + u1(ξ) = 0.

Suppose that bothΣ andΣ1 are hyperelliptic, and thatΣ1 ⊂ Σ. ThenΣ = Σ1 ifand only if there existw(X) = (aX+b)/(cX+d), ad−bc 6= 0, andv(X) ∈ K(X)such that

ξ = w(x), v(x)2 + v(x) = u(x) + u1(w(x)). (7.62)

Proof. The hypothesis thatΣ is hyperelliptic means thaty 6∈ K(x); that is, noelementf(x) ∈ K(x) exists for whichu(x) = f(x)2 +f(x), and similarly forΣ1.A direct calculation shows that, if (7.62) holds, theny2 +y+u(x) = 0 implies that

(y + v(x))2 + y + v(x) + u1(ξ) = 0.

Hence, eitherη = y + v(x), or η = y + v(x) + 1. In both cases,y ∈ K(ξ, η).SinceK(x) = K(w(x)) = K(ξ), soΣ = Σ1.

Conversely, ifΣ = Σ1, the uniqueness ofg12 in any hyperelliptic function field

implies thatξ = w(x), wherew(x) = (ax + b)/(cx + d), ad − bc 6= 0 anda, b, c, d ∈ K. Also, η = a(x) + b(x)y with a(x), b(x) ∈ K(x) andb(x) 6= 0.Thenη2 + η + u1(w(x)) = 0 implies

(b(x)2 + b(x))y + a(x)2 + b(x)2u(x) + a(x) + u1(w(x)) = 0.

Sincey 6∈ K(x), this is only possible whenb(x) = 1 and

(a(x))2 + a(x) = u(x) + u1(w(x)).

Lettingv(x) = a(x), the second equation in (7.62) follows. 2

THEOREM 7.97 LetF be a hyperelliptic curve of genusg. If w is the number ofdistinct Weierstrass points ofF , then

w = 2g + 2 when eitherp = 0 or p > 2,1 ≤ w ≤ g + 1 whenp = 2.

Proof. Forp 6= 2, the assertion follows from Theorem 7.88 and Example 6.87. Forp = 2, if the canonical form in Theorem 7.94 is adopted to represent F , then thearguments in Example 6.87 can be used to prove thatw − 1 is equal to the numberof distinct roots ofh(X). Hence,0 ≤ w ≤ g. 2

244

7.12 EXAMPLES OF CURVES

In this section,K has positive characteristicp, andq denotes a power ofp such thatq ≡ −1 (mod 3). Two examples are worked out to illustrate some points of thetheory of space curves.

In the first example,F is the irreducible plane curve considered in Example 1.39.ThenC = v(g(X,Y )) with

g(X,Y ) = Y +Xq +XY q − 3(XY )(q+1)/3.

Also,Hq = v(Y + Xq + Y qX) is the Hermitian curve,P = (x, y) is a genericpoint ofHq, andΣ = K(x, y) is the associated function field. Note that

Y 3 +Xq3 + (XY q)3 − 3(XY )q+1 =

(Y +Xq + Y qX)(Y + ǫXq + ǫ2Y qX)(Y + ǫ2Xq + ǫY qX),

whereǫ3 = 1 but ǫ 6= 1. Puttingξ = x3, η = y3, it follows that

η + ξq + ξηq − 3(ξη)(q+1)/3 = 0.

Therefore,Q = (ξ, η) is a generic point ofC, and the function fieldΣ′ = K(ξ, η)of C is a subfield ofΣ.

From Example 1.39,F has exactly13 (q2− q+ 1) singular points, each of whichis an ordinary double point. By Theorem 5.56, the genus ofF is

12

[13 (q2 − q + 1)− 1

].

Let L denote the linear series cut out onF by the linear system of plane cubics∆ = v(c0 + c1X

3 + c2Y3 +XY ). ThenL has no fixed place andL = g3

3(q+1).The associated subfieldΣ′′ = K(ξ, η, xy) of Σ containsΣ′. It is shown below thatΣ′ = Σ′′; see Lemma 7.100.

L EMMA 7.98 The seriesg33(q+1) is composed of an involution of order3.

Proof. Let P1 be the place ofΣ arising from the unique branchγ of Hq centredat the fundamental pointP1 = (1, 0, 0). To compute the ramification indexeP1

,a representation of the placeP ′

1 of Σ′′ lying underP1 is needed. As shown inExample 7.28,P1 has a primitive representationσ such thatσ(x) = t and

σ(y) = −(tq + tq+1 + · · ·+ ajtq2+αj(q

2−q+1) + · · · ).Note that

−σ(y)3 =(tq + tq2+1 + · · · )3 =

[tq(1 + tq

2−q+1 + · · · )]3

=

=(t3)q(1 + tq2−q+1 + · · · )3 = (t3)q

[1 + (t3)(q

2−q+1)/3 + · · ·]3,

−σ(xy)= t(tq + tq2+1 + · · · ) = t

[tq(1 + tq

2−q+1 + · · · )]

= tq+1(1 + tq2−q+1 + · · · ) = (t3)(q+1)/3

[1 + (t3)(q

2−q+1)/3 + · · ·].


So, the substitutions = t3 provides a primitive representation ofP ′1, and hence

eP1= 3.

The pointP = (1, x3, y3, xy) determines an irreducible curveΓ′ in PG(3,K).The branchγ′1 of Γ′ associated withP ′

1 has a primitive representation,

(x0(s) = 1, x1(s) = s, x2(s) = squ(s)3, x3(s) = s(q+1)/3u(s)), (7.63)

with

u(s) = 1 + s(q2−q+1)/3 + · · ·+ ajs

αj(q2−q+1)/3 + · · · .

Its centre is the pointP ′1 = (1, 0, 0, 0).

If P2 is the place ofΣ arising from the branch with centre atP2 = (0, 1, 0), theargument may proceed as above or use the fact thatΓ is invariant under the cyclicchange of the coordinate system:

(X0,X1,X2) → (X2,X0,X1).

The ramification indexeP2is 3. Also,

x0(s) = (squ(s)3, x1(s) = 1, x2(s) = s, x3(s) = s(q+1)/3u(s))

is a primitive representation of the branchγ′2 of Γ′ associated withP2. Its centre isthe pointP2 = (0, 1, 0, 0).

Similarly, the placeP3 of Σ arising from the branch centred atP3 = (0, 0, 1)has ramification index3, and

(x0(s) = s, x1(s) = squ(s)3, x2(s) = 1, x3(s) = sq+1)/3u(s))

is a primitive representation of the branchγ′3 associated with the placeP ′3 of Σ′

lying underP3. The centre ofg′3 is the pointP ′3 = (0, 0, 1, 0).

Every other pointP of F is affine:P = (1, a, b). LetP be the unique place ofΣ centred atP . The corresponding branchγ centred atP has a primitive represen-tation

(x(t) = a+ a1t+ a2tq + . . . , y(t) = b+ t),

with a1 = −b−q anda2 = b−(q2+q) − ab−q. Hence the branchγ′ correspondingtoP ′ has a representation(x0(t) = 1, x1(t), x2(t), x3(t)) with

x1(t) = a3 + 3a2a1t+ 3aa21t

2 + a31t

3 + 3a2a2tq + 6aa1a2t

q+1 + · · · ,x2(t) = b3 + 3b2t+ 3bt2 + t3,

x3(t) = ab+ (a+ a1b)t+ a1t2 + a2bt

q + a2tq+1 + · · · .

Its centre is the pointP ′ = (1, a3, b3, ab). For a primitive cube root of unityǫ, putPǫ(ǫa, ǫ

2b). ThenP = (1, ǫa, ǫ2b) ∈ Hq if and only ifP = (1, a, b) ∈ Hq.Let Pǫ be the place ofΣ arising from the branchγǫ of Hq centred atPǫ. Since

q ≡ −1 (mod 3) implies thatǫq+1 = 1, it follows that the branchγ′ǫ is centred atthe pointP ′ = (1, a3, b3, ab). Therefore, the places lying overP ′ areP,Pǫ andP ′

ǫ2 . In particular, the ramification indexeP is 1. 2

As a corollary, the following lemma is obtained.

L EMMA 7.99 The curveΓ′ has degreeq + 1.

246

From the results above, the differentD(Σ/Σ′) is 2(P ′1 +P ′

2 +P ′3). By Hurwitz’s

Theorem,2g − 2 = 3(2g′ − 2) + 6, whereg = 12 (q2 − q) is the genus ofΓ.

Therefore,Γ′ andC have the same genus. SinceΣ′ is a subfield ofΣ′′, this givesthe following result.

L EMMA 7.100 Σ′ = Σ′′.

In particular,Γ′ is birationally equivalent toC. A geometric explanation of this isthatC can be viewed as the projection ofΓ′ from the pointP4 = (0, 0, 0, 1) to theplanev(X3), whereP4 6∈ Γ′.

Now, theL-orders are calculated. First, the branch pointP ′1 is considered. For

i = 0, 1, 2, 3, letHi = v(Xi). From (7.63),

I(P ′1,Hi ∩ Γ′) =

0 for i = 0,1 for i = 1,

13 (q + 1) for i = 2,

q for i = 3.

Hence the(L,P ′1)-order sequence is(0, 1, 1

3 (q+1), q). This holds true for bothP ′2

andP ′3.

If P = (1, a3, b3, ab) is another point ofΓ′, the first three(L,P )-orders are0, 1, 2, and the last one is eitherq or q+1. To decide this,I(P,H∩Γ′) is calculatedfor any plane

H = v(u0X0 + u1X1 + u2X2 + u3X3)

with

u0 = −(u1a3 + u2b

3 + u3ab.

ThenI(P,H ∩ Γ′) takes both values1 and2 for some choice ofu1, u2, u3. How-ever,I(P,H ∩ Γ′) = 3 cannot occur because the matrix

3a2a1 3b2 a+ a1b3aa2

1 3b a1

a31 1 0

has zero determinant. Also,I(P,H ∩ Γ′) > 3 implies thatI(P,H ∩ Γ′) ≥ q. Onother hand, from Lemma 7.63,I(P,H ∩ Γ′) ≤ q + 1. Since

(q+1

q

)6≡ 0 (mod p),

Lemma 7.59 shows that the fourthL-order isq. In particular, there are finitelymany Weierstrass points withL-orders(0, 1, 2, q + 1).

The precise numberM of the Weierstrass points is calculated from the ramifica-tion divisorR. Note that

degR = (1 + 2 + q)( 13 (q2 − q − 8)) + 4(q + 1) = 1

3 (q3 + 2q2 + q − 12).

The contribution of each of the three fundamental pointsP ′1, P

′2, P

′3 is 1

3 (q+1)−2.Therefore,

M = degR− (q − 5) = 13 (q3 + 2q2 − 2q + 3).

It may be noted thatM = q2 + 1 + 2g′q, whereg′ is the genus ofΓ′,


7.13 THE LINEAR GENERAL POSITION PRINCIPLE

The Riemann–Roch Theorem shows how the dimension and the degree of a com-plete linear series are related. Some more results in this direction are stated in thissection using the linear general position principle.

DEFINITION 7.101 Let Γ be a non-degenerate irreducible curve inPG(r,K). AhyperplaneH of PG(r,K) is in linear general positionif there is no subspace ofPG(r,K) of dimension less thanr− 1 that containsr common points ofΓ andH.

The main result on the Linear General Position Principle following.

THEOREM 7.102 Let Γ be a non-degenerate irreducible curve inPG(r,K). If Γis not strange, then some hyperplane is in linear general position.

In particular, ifK has zero characteristic, then the Linear General Position Prin-ciple is valid. Since the irreducible conic in characteristic 2 is the only non-singularstrange curve, see Section 7.19, the Linear General Position Principle is valid forevery non-singular curve.

7.14 CASTELNUOVO’S BOUND

If F is an irreducible plane curve of degreen and genusg, theng ≤ 12 (n−1)(n−2)

by Lemma 3.28. Castelnuovo’s bound is a generalisation of this for space curves.The following two lemmas are the main ingredients in the proof of Castelnuovo’sBound.

L EMMA 7.103 LetA andB be two positive divisors such that bothdim |A| anddim |B| are positive. Then

dim |A|+ dim |B| ≤ dim |A+B| ≤ degA+ dim |B|.Proof. It may be that|A| has a fixed divisorF ; if it is the case, replaceA byD = A−F . Sincedim |A| = dim |D| anddim |D+B| ≤ dim |A+B|, if sufficesto prove the assertion forD in place ofA. Putr = dimA = dimD. Since|D| hasno fixed divisor,dim (D−P) = r−1 for every place. Geometrically speaking, thethe divisors of|D| correspond to the points ofPG(r,K), those in|D − P| to thepoints of a hyperplaneHP of PG(r,K). Since finitely many hyperplanes cannotcover the wholePG(r,K), there is a divisorD0 ∈ |D| which does not contain anyplaceP fromB. ReplaceD0 byD, and put

|D|= div(c0 +∑r

i=1cixi) +D | c = (c0, . . . cr) ∈ PG(r,K),|B|= div(c0 +

∑si=1cr+ixr+i) +B | c = (c0, cr+1, . . . cr+s) ∈ PG(s,K),

L= div(c0 +∑r+s

i=1 cixi) +D0 +B | c = (c0, . . . cr+s) ∈ PG(r + s,K).First, div(xi) ≻ −(D + B) for every1 ≤ i ≤ r + s. Therefore,L consists ofeffective divisors. Also,dimL = r + s. To show the latter assertion, it is enoughto show that1, x1, . . . , xr, . . . , xr+s are linearly independent overK. Taking into

248

account that both sets1, x1, . . . , xr and1, xr+1, . . . , xr+s are linearly inde-pendent overK, suppose by contradiction that

ξ = u0 + u1x1 + . . .+ urxr = v1xr+1 + . . .+ vsxr+s = η,

with coefficients not all zero. LetP be a pole ofξ; thenP is also a pole ofη.It follows, on the one hand thatP is in D and, on the other hand, thatP is inB. But this contradicts the choice ofD0. SinceD + B is a common divisor ofL and |D + B|, Theorem 6.23 showss thatL is a subseries of|D + B|. Hencer + s = dimL ≤ dim |D +B|.

To show the other bound, putC = D +B. ThenC ≻ B. By Theorem 6.38,

|B| = |C −D| = ||C| − |D||is the subseries of|C| consisting of all the divisors containingD, minus its fixeddivisorD. From Theorem 6.26,

dim |B| ≥ dim |C| − ordD,

whence it follows thatdim |D +B| ≤ degD + dimB. 2

L EMMA 7.104 Let Γ be a non-degenerate, irreducible curve inPG(r,K) of de-green that is not strange. IfD is the divisor cut out onΓ by a general hyperplane,then

(i)

dim |kD| − dim |(k − 1)D| ≥ k(r − 1) + 1, (7.64)

for every positive integerk ≤ m, wherem = ⌊n−1r−1 ⌋.

(ii)

dim |kD| − dim |(k − 1)D| = n, (7.65)

for k > m.

Proof. Let D be the divisor cut out onΓ by a general hyperplane. ThenD canbe written as the sum of places corresponding to branches ofΓ centred at distinctpoints. Assume thatk ≤ m, and choosek(n− 1) + 1 places fromD. Let Ω be theset of the centres of the corresponding branches ofΓ. For each of thek(n− 1) + 1pointsQj ∈ Ω, a hypersurface of degreek exists which passes throughΩ\QjbutQj . To show this, partition the remainingk(n− 1) points ofΩ into k sets:

S1 = P1,1, P1,2 . . . , P1,n−1 , . . . , Sk = Pk,1, Pk,2, . . . , Pk,n−1of n − 1 points each. By the linear general position principle, eachis linearlyindependent overK. Therefore, for every1 ≤ u ≤ k, there is a hyperplaneHu

through every point inSu that does not containQj . The reducible hypersurface∆j of degreek whose components are the hyperplanesHu, u = 1, . . . , k is therequired hypersurface of degreek. The resultingk(n − 1) + 1 hypersurfaces∆j

with Qj ranging overΩ are linearly independent overK. From this (7.64) follows.


In fact, letΓ be the curve arising fromQ with Q = (1, x1, . . . , xn), and|(k − 1)D| = div(c0y0 + c1y1 + . . .+ cryr) + (k − 1)D

with r = dim |(k − 1)D|. If the hyperplaneH = v(F (X0,X1 . . . ,Xn)) cutsoutD, put ξ = F (1, x1, . . . , xn). Similarly, if ∆j = v(Fj(X0,X1, . . . ,Xn), putηj = Fj(1, x1, . . . , xn). LetL be the linear series

div (∑r

i=0ciyiξ +∑k(n−1)+1

j=1 djηj) + kD |(c0, . . . , cr, d0, . . . , dk(n−1)+1) ∈ PG(r + k(n− 1) + 1,K).

Note thatL entirely consists of effective divisors. Also, ther + 1 elementsyiξtogether with thek(n − 1) + 1 elementsηj form a linearly independent set overK. Hence, (7.64) holds. Fork > m, the same argument shows how to findkhyperplanes inPG(n,K) containing all but any one of the points ofΩ. From this,(7.65) follows. 2

THEOREM 7.105 (Castelnuovo’s Bound)Let Γ be a non-degenerate irreduciblecurve inPG(n,K) of degreed and genusg. If Γ is not strange, then

g ≤ 12m(m− 1)(n− 1) +mǫ,

wherem = ⌊ d−1n−1⌋ andd− 1 = m(n− 1) + ǫ.

Proof. Let L be the linear series cut out onΓ by hyperplanes. SinceL may notbe complete, consider the complete linear seriesL′ containingL. ThenL′ = |D|,whereD is a divisor cut out onΓ by a hyperplane. Asdim |D| ≥ n, from Lemma7.104, it follows by induction that

dim |kD| ≥ 12k(k + 1)(n− 1) + k, k = 1, . . . ,m.

dim |(m+ j)D| ≥ 12k(k + 1)(n− 1) + k + jd, j ≥ 1.

However, for sufficiently largej, the divisor|(m+ j)D| is non–special. So, by theRiemann–Roch Theorem,dim |(m+ j)D| = (m+ j)d− g. Hence

g≤ (j +m)d− 1

2m(m+ 1)(n− 1)−m− jd

= 12m(m− 1)(n− 1) +m(d−m(n− 1)− 1),


A slightly different wording of Castelnuovo’s Bound is stated in the followingtheorem.

THEOREM 7.106 Let Γ be as in Theorem 7.105. Ifǫ is the unique integer with0 ≤ ǫ ≤ r − 2 andd− 1 ≡ ǫ (mod (r − 1)), then

g ≤ c0(d, r) =d− 1− ǫ2(r − 1)

(d− r + ǫ) (7.66)

L EMMA 7.107 Let D be as in Lemma 7.104, and letǫ′ = d − m(r − 1) forǫ′ ∈ 2, . . . , r. Assume thatg = c0(d, r). If m ≥ 2, then

(i) the dimension of the linear series|2D| is equal to3r − 1;

(ii) there exists a base-point-free(ǫ′ − 2)-dimensional complete linear seriesD′

of degree(ǫ′− 2)(m+ 1) such that(m− 1)|D|+D′ is the canonical linearseries.

250

7.15 A GENERALISATION OF CLIFFORD’S THEOREM

Clifford’s Theorem 6.77 states thatn ≥ 2r for any specialgnr . Here, an alternative

proof depending on the linear general position principle isgiven, which also showsthatn = 2r with n > 0 is only possible for canonical or hyperelliptic curves.

THEOREM 7.108 A non-canonical specialgrn, with n > 0, on a non-hyperelliptic

curveC satisfiesn > 2r.

Proof. Supposegrn is not complete. Choose an effective special divisorD such

that |D| = grn. Then|W −D| is agr′

n′ with r′ ≥ 0. Note thatn′ > 0, since|D| isnot the canonical series. From the reciprocity Theorem 6.76of Brill and Noether,n − 2r = n′ − 2r′. Sincen + n′ = 2g − 2, this givesr − r′ = n − g + 1. Also,r+ r′ ≤ g− 1 by Lemma 7.103. Therefore,n ≥ 2r, and equality holds if and onlyif n′ = 2r′.

Sincen+ n′ = 2g − 2, eithern or n′ is less than or equal tog − 1, and supposethat this occurs forn. So,0 < n ≤ g − 1. Note as well thatn = 2r only occurswhenr + r′ = g − 1; that is, every canonical divisor is the sum of a divisor form|D| and a divisor from|W −D|.

Now, assume thatn = 2r for a non-hyperelliptic curve and consider the canon-ical curveΓ. LetH be a hyperplane inPG(g − 1,K) such thatΓ ∩H consists of2g−2 pairwise distinct points. ThenΓ∩H containsn points, sayP1, . . . , Pn suchthat the corresponding places define a divisorP1 + . . . + Pn in |D|. By Theorem7.44 (iii), these points span a12n−1-dimensional subspace ofPG(g−1,K). Sincen ≤ g − 1, either 1

2n − 1 = g − 2, orH is not in a linear general position. ButTheorem 7.102 together with (i) of Theorem 7.44 ensures the existence of a lineargeneral hyperplane section, and this completes the proof. 2

This generalisation of Clifford’s Theorem gives a refinement of the lower boundin Theorem 7.61.

THEOREM 7.109 LetN be the number of Weierstrass points of a function fieldΣwhich is neither rational, nor elliptic, nor hyperelliptic. If its canonical series isclassical, thenN ≥ 2g + 6.

Proof. It suffices to note in the second part of the proof of Theorem 7.61 that thebound (7.16) may be replaced byji ≤ i− 2, by Theorem 7.108. 2

7.16 THE UNIFORM POSITION PRINCIPLE

Another position principle on hyperplane sections of an algebraic curve, which isvalid in zero characteristic but not for all curves in positive characteristic, is treatedbriefly.

DEFINITION 7.110 A finite setS of points inPG(n,K) is in uniform positionif,for any two positive integersk,m, all subsets ofS of a sizek impose the samenumber of conditions of hypersurfaces of degreem.


With this definition, theuniform position principlestates that the points of a generalhyperplane section of an algebraic curve are in uniform position.

Like the General Linear Position Principle, the Uniform Position Principle hasbeen an important tool in investigating relationships between the genusg and thedegreed of a space curves ofPG(n,K). Here a few results are quoted withoutproofs.

A classical theorem of Halphen improving Castelnuovo’s genus bound for cer-tain curves inPG(3,C) states that an irreducible curveΓ in PG(3,C) either lieson a quadric surface or its genusg satisfies the following:

g ≤⌊

16d

2 − 12d+ 1

⌋.

THEOREM 7.111 (Halphen)In PG(3,K) let X be an irreducible non-singularcurve. If either

d ≥

7, for p = 0,7, for p > 2 andX reflexive,

25, for p > 0 andX non-reflexive,

or d = 17 for p > 0, thenX lies on a quadric surface provided that

g > c1(d, 3) = ⌊ 16 (d2 − 3d+ 6)⌋ .

For the definition of reflexivity ofΓ, see Section 7.19.

THEOREM 7.112 Letp 6= 2. ThenΓ is not reflexive if and only ifǫ2 > 2.

Halphen’s theorem extends to certain curves inPG(r,K) for r ≥ 4, and it appearsto be very useful when one looks for a boundcα(d, r) for the genus of a curveof degreed in PG(r,K) not lying on any irreducible surface of degree less thanr + α − 1. The result for the smallest caseα = 1 is stated below since it plays arole in Section 11.5.

L EMMA 7.113 Let Γ be an irreducible algebraic curve ofPG(r,K) of degreedand genusg, and let

d ≥

36r, for r ≤ 6,288, for r = 7,2r+1, for r ≥ 8.

ThenX lies on a surface of degree less than or equal tor − 1 provided that

g > c1(d, r) =d− 1− ǫ1

2r(d− r + ǫ1 + 1) +

0 if ǫ1 ≤ r − 2

1 if ǫ1 = r − 1, (7.67)

whereǫ1 is the unique integer such that0 ≤ ǫ1 ≤ r − 1 andd− 1 ≡ ǫ1 (mod r).

The above numbersc1(d, r) areHalphen’s numbers. Note that (7.67) forr = 3coincides with the formula in Theorem 7.111.

,

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7.17 ALGEBRAIC VARIETIES AND CURVES

It is shown that the concept of an irreducible variety of dimension1 is equivalentto the concept of an irreducible space curve as defined and investigated in thisbook. LetF be an irreducible plane curve, and letK(F) be its function field.Given a pointQ = (x0, x1, . . . , xr) in PG(r,K(F)) such thatK(Q) = K(F),the associated irreducible space curveΓ of PG(r,K) consists of the centres of thebranches(σ(x0, σ(x1), . . . , σ(xr)) withP ranging over all places ofK(F) andσ aprimitive representation ofP. Since not all coordinates ofQ are zero, takex0 6= 0,and writeyi = xi/x0. Here, and in the rest of the section,i ranges from1 to r.Let AG(r,K) be ther-dimensional affine space whose hyperplane at infinity isv(X0). Then the affine points ofΓ are centres of the branches ofΓ that arise fromthe placesP for which ordP(yi) ≥ 0. The set of these points may be considered asan affine part, or as an affine curve,∆ coming fromΓ.

It is shown in this section that∆ is an affine irreducible variety of dimension1and conversely. From this, the equivalence of the concepts of an irreducible curveand of a projective algebraic variety of dimension1 can be deduced by standardarguments.

For the present purpose, the most convenient wording of the definition of anaffine variety depends on the concept of a specialisation.

DEFINITION 7.114 Let x1, . . . , xr be elements of an extension fieldΣ of K. ApointA = (a1, . . . , ar) of AG(r,K) is an(algebraic) specialisationof the pointQ = (x1, . . . , xr) of AG(r,Σ) when everyf(X1, . . . ,Xr) ∈ K[X1, . . . ,Xr] thatvanishes atQ = (x1, . . . , xr) also vanishes atA = (a1, . . . , ar). If this occurs,

writeQ = (x1, . . . , xr)−→K A = (a1, . . . , ar), or brieflyQ−→

KA.

DEFINITION 7.115 A non empty set points inAG(r,K) is anaffine irreduciblevariety if it consists of all specialisations inAG(r,K) of a pointQ = (x1, . . . , xr)of AG(r,Σ), whereΣ is a field extension ofK. The pointQ is ageneric pointofthe variety. If the degree of transcendency[Σ : K] is equal tod, then the varietyhasdimension d.

This can be carried out over to projective space by using homogeneous polynomialsin such a way that the following holds. LetV be a projective irreducible variety ofPG(r,K). If H is any hyperplane inPG(r,K), thanV ∩H is either empty, or itis an affine irreducible variety.

Let Γ be an irreducible curve ofPG(r,K) given by a pointQ = (x1, . . . , xr).It is first shown that the affine points ofΓ are points of an one-dimensional affinevarietyV1 associated to the same pointQ. LetP be a place ofK(Γ) = K(Q) suchthat ordPxi ≥ 0. Choose a primitive representation ofP and write

σ(xi) = ai + bit+ . . . .

Let f ∈ K[X1, . . . ,Xr] a polynomial that vanishes at(x1, . . . , xr). Sinceσ is aK-isomorphism fromK(Γ) toK, this implies thatf also vanishes at

(a1 + . . . , . . . , ar + . . .).


Puttingt = 0 givesf(a1, . . . , ar) = 0. Therefore,(x1, . . . , xr) 7→ (a1, . . . , ar),and the assertion follows.

To show the converse, the following result must be proved.

L EMMA 7.116 Every point ofV1 is the centre of a branch ofΓ.

Proof. At least one of the coordinatesxi ofQ, sayx1, is a separable variable. Thus,the other coordinates are separable variables ofK(x1, . . . , xr)/K(x1). Theorem15.1, that is, the Theorem of Primitive Element in its strongform ensures the ex-istence of a primitive element ofK(x1, . . . , xr)/K(x1), which can be written as alinear combinationc2x2 + . . .+ crxr with cj ∈ K. Here and in the rest of the sec-tion, j ranges from2 to r. Also, replacingα+Λβ byΛ2x2+. . .+Λrxr in the proofof Theorem 15.1, a polynomialg(Λ2, . . . ,Λr) exists such thatc2x2 + . . . + crxr

is a primitive element ofK(x1, . . . , xr)/K(x1) provided thatg(c2, . . . , cr) 6= 0.Now,P = (x1, c2x2 + . . .+ crxr) is viewed as the generic point of an irreducibleplane curveF = v(f). By Definition 5.1,F is the locus of the specialisationsof its generic point. Since(x1, . . . , xr) −→ (0, . . . , 0), every polynomial in twoindeterminates which vanishes at(x1, c2x2 + . . . + crxr) also vanishes at(0, 0).Therefore,

(x1, c2x2 + . . .+ crxt) −→ (0, 0).

In particular,f(0, 0) = 0; that is,F contains the pointO = (0, 0). Hence, at leastone of the branches ofF arising from the placesPi is centred atO. Such a placeis a zero ofc2x2 + . . .+ crxr.

It may supposed thatx2, . . . , xr are linearly independent overK. Otherwise,one of them, sayxr, is a linear combination ofx2, . . . , xr−1. Then an inductiveargument onr shows the existence of a placeP of

K(x1, . . . , xr−1) = K(x1, . . . , xr)

such thatx2, . . . , xr−1 are zeros ofP. Thenxr, being a linear combination ofx2, . . . , xr−1, is also a zero ofP. Hence the branch ofΓ associated toP is centredatP = (0, . . . , 0), and the proof of Lemma 7.116 is done.

Choose elementscjk ∈ K with k ranging over1, . . . , r − 1 such that the matrix

c2,1 . . . cr,1...

......

c2,r−1 . . . cr,r−1

has non-zero determinant. In this way,r−1 elements of the formc2x2 + . . .+crxr

are obtained which are linearly independent overK.Generalising this, for any positive integerN ≥ r − 1, there areN − 1 elements

of the same formc2x2 + . . . + crxr, no n − 1 of them linearly dependent; it isenough to choose the entriescjk by requiring certain determinants not to vanish.It is also possible to require the conditionsg(c2,k, . . . , cr−1,k) 6= 0, so that allc2,kx2 + . . .+ cr−1,kxr are primitive elements ofK(x1, . . . , xr)/K(x1).

Now, assumeN = (k+1)(r−1). Then, for somer−1 values of the indexk, theelementsc2,kx2 + . . .+ cr,kxr are zeros of the same placeP. Solving these linear

254

homogeneous equations shows thatx2, . . . , xr, asx1, all are zeros ofP. Thus, thecorresponding branch ofΓ is centred atP = (0, . . . , 0). 2

REMARK 7.117 In the literature, an irreducible projective variety of dimension1is usually called aprojective, geometrically irreducible, algebraic curve. It hasbeen shown here that this term is equivalent, even if formally different, to the termused here of an irreducible algebraic curve.

7.18 EXERCISES

1. Forp > 2, letF = v(f(X,Y ) withf(X,Y ) = V p

0 + V p1 X + V p

2 Y + V p3 X

2 + V p4 XY + V p

5 Y2),

whereV0, V1, . . . , V5 ∈ K[X,Y ] are polynomials satisfying the equationV0(4V3V5 − V 2

4 ) = V 21 V5 + V 2

2 V3 − V1V2V4.

Show thatF is non-classical with respect to lines, as its Hessian curveisv(2(4V2V5 − V 2

4 )pf(X,Y )).

2. Forp = 2, let Γ = v(Y +X3 + Y 8). Show that the orders with respect tothe linear system of conics are0, 1, 2, 3, 4, 6.

3. Forp > 5, let Γ = v(G(X,Y )) be the non-singular curve withG(X,Y ) = 1 +Xp +XY p + Y p+1 +Xp−1.

Show thatΓ is non-classical with respect to the linear system of conics, andthat (7.22) holds withH(X,Y ) = X but not withH(X,Y ) = 1.

4. Let F be the curve in Example 1.40. Show that, ifp ≥ 5 and k2 6≡ 1(mod p), thenF is classical with respect to the linear system of conics.

5. Giveng + 1 distinct elementsai ∈ K, let Σ = K(x, y) with

y2 + y =

g+1∑

i=1

1

x+ ai.

Show that ifp = 2, thenΣ hasg + 1 distinct Weierstrass points.

6. Show that the Klein quartic, see Example 7.17, is not hyperelliptic.

7. LetF be a hyperelliptic curve. In characteristicp 6= 2, deduce from Exam-ple 6.87 and Theorem 7.88 that the smallest non-gap numberm1(P) at anordinary place ofK(F) is equal tog + 1. Extend the result to characteristicp = 2.

8. Two distinct irreducible curves ofPG(r,K) have only finitely many com-mon points.

9. Show thatǫi + ji(P ) ≤ ji+j(P ) , i+ j ≤ r .

10. Letp ≥ 5. For a non-hyperelliptic curveF of genus4, show the existence ofa placeP of K(F) such that4 is the first non-gap atP.


7.19 NOTES

For rational strange curves ofdegF = q+ 1, there are three types; see Ballico andHefez [21].

For the General Order of Contact Theorem, see Hefez and Kleiman [119].In Proposition 7.66, if singular points are allowed,H(X,Y ) = 1 can still be

ensured under some natural conditions; see [117]. However,it is not possible ingeneral to get rid ofh(X,Y ), the curve in Exercise 1 being such an example. Here,degF ≤ q with q = pm. Hefez [117] and Homma [137] show that equality holdsfor just one type of curve.

Fermat curves which are not classical with respect toL2 are investigated byGarcia and Voloch, see [96]. They showed that, ifp > 5, then the Fermat curve,

F = v(Xn + Y n + 1),

is non-classical with respect toL2 if and only if

p | (n− 2)(n− 1)(n+ 1)(2n− 1).

Theorem 7.66 is due to Pardini [213] form = 1, and to Homma [140] for arbi-trarym.

Curves hitting Castelnuovo’s Bound, that is, those with genus equal toCastel-nuovo’s numberc0(d, r) are extremaland they have several remarkable proper-ties; see [8], [113, Chapter 3], [12, Chapter 3, Section 2]. One of these proper-ties is stated in Lemma since it is used in Section 11.5 where certain curves withg = c0(d, r) are considered. The proof can be deduced from that of Theorem7.105.For more information, see [8, p. 361 and Lemma 3.5].

The main reference for the Gaussian duality of space curves is [98] in which theauthors determine whenΓ coincides with its gaussian dualΓ′, up to a linear trans-formation. A third generalisation of the concept of the dualcurve of an irreducibleplane curveC is the dual hypersurface of an irreducible curveΓ of PG(r,K). Theidea is to consider all hyperplanes ofPG(r,K) containing at least one tangent lineto Γ and regard them as points in the dual projective spacePG(r,K)∗ ∼= PG(r,K)of PG(r,K). Such points, but finitely many of them, lie in an irreduciblehyper-surface ofΠ which is thedual hypersurface∆ of Γ. To find a generalisation of thenotation of a reflexive curve, more are needed, namely the concept of theconormalvarietyC(Z) of an irreducible algebraic varietyZ defined overK, see [117].

Recall thatC(Z) is defined to be the closure inPG(r,K) × PG(r,K)∗ the setof all pairs(P,H) whereP is a non-singular point ofZ andH is a hyperplaneof PG(r,K) containing the tangent space ofZ at P . The dimension ofC(Z) isr − 1. Thedual varietyof Z ′ of Z is the projection ofC(Z) on PG(r,K)∗. Avariety isreflexivewhenC(Z) = C(Z ′). From the definition, ifZ is reflexive then(Z ′)′ = Z andZ ′ is also reflexive.

A characteristic-free proof of Theorem 7.102 is found in [219].A proof of the uniform position principle in characteristiczero, due to Eisenbud

and Harris, is found in [113]. Rathmann [219] extended the result to positive char-acteristic provided thatn ≥ 4. He also showed the validity of the uniform positionprinciple for a large family of curves ofPG(3,K) including all reflexive curves. It

256

should be noted that the validity of both position principles depends on the mon-odromy group of the curve, which is a highly transitive permutation group. Incharacteristicp, the classification of highly transitive permutation groups dependson the classification of all finite simple groups. For more results, see [113].

The general theory of algebraic varieties is found in several textbooks. A goodreference for Section 7.17 is [185].

The result that the irreducible conic in characteristic2 is the only non-singularstrange curve is due to Samuel; see [114, Theorem IV.3.9].

The characterisation of reflexive curves in Theorem 7.112 isdue to Hefez andKleiman [117]; see also [119].

For a proof of Lemma 7.113, see ([113, Theorem 3.22] and [219,Corollary 2.8].A full account of results related to Halphen’s theorem is found in the books,

[113], [108], as well as in the papers, [49], [48].For a historical account of Halphen’s Theorem, see [114, p. 349], [108] and

[112]. Halphen’s theorem remains essentially valid in any characteristic, see [113,Theorem 3.13] forp = 0, and [19] forp > 0.

For Exercise 10 and more results on Weierstrass points of curves with low genus,see [159].

Chapter Eight

Rational points and places over a finite field

In this chapter,K is the algebraic closure of a finite fieldFq of orderq. For everypositive integern, K contains a unique finite fieldFqn of orderqn. These finitefields coverK, that is, every elementa ∈ K is contained in anFqn for somen.

DEFINITION 8.1 (i) TheFrobenius automorphism ofK is the mapΦ that takesx ∈ K to xq ∈ K.

(ii) The n-th Frobenius automorphism ofK is the mapΦn that takesx ∈ K toxqn ∈ K.

Then,Fqn consists of all elements inK which are fixed byΦn.

8.1 PLANE CURVES DEFINED OVER A FINITE FIELD

L EMMA 8.2 An irreducible plane curveF is defined overFq if and only if forevery pointP = (a0, a1, a2) ∈ F its image pointP q = (aq

0, aq1, a

q2) under the

Frobenius collineationΦ is also inF .

Proof. Let F = v(F ) whereF ∈ K[X0,X1,X2] is a homogeneous polynomialof degreed. If

F =∑

aijXi0X

j1X

d−(i+j)2 ,

then the homogeneous polynomial

G =∑

aqijX

i0X

j1X

d−(i+j)2

defines the irreducible plane curveG = v(G). Thus,P = (a0, a1, a2) ∈ F if andonly if P q = (aq

0, aq1, a

q2) ∈ G. Now, if F is defined overFq, thenF = G and

henceP q ∈ F .Conversely, ifP q ∈ F thenP q is a common point ofF andG. If this occurs for

every pointP ∈ F , the irreducible curvesF andG have infinitely many commonpoints. By Bezout’s theorem,F = G and henceF is defined overFq. 2

L EMMA 8.3 The irreducible plane curveF = v(F (X,Y )) is defined overFq ifand only if, for every generic pointP = (x, y) ofF , the image pointP q = (xq, yq)of P = (x, y) under the Frobenius collineation is also a generic point ofF .

257

258

Proof. If F (X,Y ) ∈ Fq[X,Y ], thenF (x, y) = 0 impliesF (xq, yq) = 0. There-fore,P q = (xq, yq) is a point ofF , and hence a generic point of it. To show theconverse, assume without loss of generality that some coefficient ofF (X,Y ) isequal to1. Now,F (x, y) = 0 implies thatG(xq, yq) = 0, where

F =∑

aijXiY j , G =

∑aq

ijXiY j .

If P q = (xq, yq) ∈ F , then it is a common generic point ofF and the irreducibleplane curveG = v(G). By Remark 5.2,F (X,Y ) = cG(X,Y ). Actually, c = 1by the assumption. Therefore,F (X,Y ) ∈ Fq[X,Y ]. 2

LetF = v(f(X,Y )) be an irreducible plane curve defined overFq.

DEFINITION 8.4 For a generic pointP = (x, y) ofF , the set of all elements of theform u(x, y)/v(x, y) with U(X,Y ), V (X,Y ) ∈ Fq[X,Y ] form a subfieldFq(F)of K(F), theFq-rational function field ofF .

By Theorem 5.7, this definition is independent of the choice of the generic point.

DEFINITION 8.5 A rational transformationω of Σ = K(F) preservingFq(F)is anFq-rational transformation, andΣ′ = ω(Σ) is anFq-rational function field,as well. IfF ′ is the image curve ofF under anFq-rational transformation, thenF 7→ F ′ is anFq-rational covering. This term is also used ifF andF ′ are replacedby Fq-rational non-singular modelsX andX ′ of Σ andΣ′ = K(F ′).

8.2 Fq-RATIONAL BRANCHES OF A CURVE

Let f =∑ait

i ∈ K((t)). For any integern, then-th-conjugate off is

f (n) =∑

aqn

i ti.

For anyf, g ∈ K((t)) and integersn, m, the following properties hold:

(f + g)(n) = f (n) + g(n); (fg)(n) = f (n)g(n); (f (n))(m) = f (n+m). (8.1)

Further, everyf ∈ K((t)) can be written asf = g(n) with g ∈ K((t)). Hence, then-th-conjugate mapκ : f 7→ f (n) is an automorphism ofK((t)). Note thatκ isnot aK-automorphism asκ(a) 6= a for any constanta 6∈ Fqn . Now, f = f (n) ifand only iff ∈ Fqn((t)). Also, if f2 = α(f1) for aK-automorphismα of K((t)),

then f (n)2 = κακ−1(f

(n)1 ). Note thatκακ−1 is aK-automorphism ofK((t)).

Thus, givenξ, ξ1, η, η1 ∈ K[[t]], if the branch representations(ξ, η) and(ξ1, η1)

areK-equivalent, then(ξ(n), η(n)) and(ξ(n)1 , η

(n)1 ) areK-equivalent branch rep-

resentation as well. From this, if(ξ, η) is a primitive branch representation, then(ξ(n), η(n)) is also a primitive branch representation.

Let γ be a branch with a primitive representation(ξ, η). The branchΦn(γ) givenby the primitive representation(ξ(n), y(n)) is then-th Frobenius image ofγ.

DEFINITION 8.6 A branchγ is Fq-rational if γ has a primitive representation(ξ, η) with ξ, η ∈ Fq[[t]], that is, ifξ(1) = ξ, η(1) = η.

Rational points and places over a finite field 259

It should be observed that, ifγ is Fq-rational having a primitive representation(ξ, η) with ξ, η ∈ Fq[[t]], then other primitive representations(ξ1, η1) of γ does

not have the property thatξ(1)1 , η(1)1 ∈ Fq[[t]] unless there existsα ∈ AutFq

K[[t]]such thatξ1 = α(ξ), η1 = α(η). Here, AutFq

K[[t]], theFq-automorphism groupof K[[t]], consists of allK-automorphismsα such thatα(f) ∈ Fq[[t]] for everyf ∈ Fq[[t]]. The observation follows from the following two lemmas. Thefirst is arefinement of Theorem 4.20.

L EMMA 8.7 For anFq-rational branch(x(t), y(t)), the subfieldFq(x(t), y(t)) ofK(x(t), y(t)) contains an elementξ of order1.

Proof. Let τ be an element inFq(x(t), y(t)) of minimal positive order. Arguing asin the proof of Theorem 4.20, bothx(t) andy(t) can be expanded into power seriesof τ . Since they are the components of a primitive branch representation, this isonly possible for ordt τ = 1. 2

L EMMA 8.8 Let ξ ∈ Fq[[t]] with ordt ξ = 1. Assume thatα ∈ Aut(K[[t]]) withα(ξ) ∈ Fq[[t]]. Thenα ∈ AutFq

K[[t]].

Proof. It suffices to prove thatα(t) ∈ Fq[[t]]. Write

ξ = u1t+ . . .+ uktk + . . . ,

with u1, . . . .uk ∈ Fq, andα(t) = a1t + . . . + amtm + . . . where bothu1 anda1

are non-zero. Then

α(ξ) = u1α(t) + . . .+ ukα(t)k + . . . = a1u1t+ t2G(t).

Sinceα(ξ) ∈ Fq[[t]], soa1u1 ∈ Fq, whencea1 ∈ Fq.To prove that this holds true for everyam, induction onm is used. Suppose that

a1, . . . , am−1 ∈ Fq. Then

α(ξ) = b1t+ . . .+ bm−1tm + (am + cm)tm + . . .

with b1, . . . , bm−1, cm ∈ Fq. Sinceα(ξ) ∈ Fq[[t]], soam + cm ∈ Fq, implyingthatam ∈ Fq. 2

Now, branches of an irreducible plane curveF defined overFq are considered.From the definition, it may appear that a branch ofF is Fq-rational if and onlyif its centre is a point inPG(2, q). This is false in general, even if it holds fornon-singular points.

THEOREM 8.9 Let P be a non-singular point of an irreducible plane curveFdefined overFq. If P ∈ PG(2, q), then the unique branch ofF centred atP isFq-rational.

Proof. After a change of coordinates overFq, suppose thatP = (1, 0, 0), and thatthe tangent line toF at P is not theY -axis. By Theorem 4.6, the branch ofFcentred atP has a primitive representation(x(t), y(t)) with

x(t) = t, y(t) = c1t+ . . . cktk + . . . .

From the proof of that theorem, each coefficientci is in the subfield ofK gener-ated by the coefficients ofF (X,Y ). SinceF is defined overFq, this subfield iscontained inFq. 2

260

THEOREM 8.10 Let P ∈ PG(2, q) be a singular point of an irreducible planecurveF defined overFq. Suppose thatγ is a linear branch ofF with centre atP,and letℓ be the common tangent line toF andγ atP . Thenγ is Fq-rational if andonly if ℓ is defined overFq.

Proof. Again, assume thatP = (1, 0, 0), and thatℓ is not theY -axis. The proof ofTheorem 8.9 still works for linear branches, and gives the “if” part of the assertion.

Conversely, any primitive representation(x(t), y(t)) of γ is of type

x(t) = at+ · · · , a 6= 0, y(t) = bt+ · · ·such thatb/a is the slope ofℓ. If γ isFq-rational, then there is a representation witha, b ∈ Fq. Henceℓ is defined overFq, as it passes through a point ofPG(2, q) andits slope is inFq. 2

The notation for the set of allFq-rational branches ofF is F(Fq), andSq for|F(Fq)|. Theorems 8.9 and 8.10 show thatF(Fq) is a contained in the setF(Fq)

∗

consisting of all branches ofF centred at points ofPG(2, q). HenceSq ≤ Bq

whereBq stands for|F(Fq)∗|, and, ifF is non-singular, then equality holds. How-

ever, it may happen for singular curves thatSq < Bq. This possibility is investi-gated in Section 8.8. Here an example is given.

EXAMPLE 8.11 Let q be odd, and choose a non-square elementc ∈ Fq. Theirreducible cubic curveF = v(cX2 − Y 2 + X3) defined overFq contains thepoint P = (1, 0, 0), which is a double point. Two branchesγ+ andγ− of F arecentred atP , namely those with primitive representation(x(t), y(t)), where

x(t) = t, y(t) = ±mt+ . . . ,

with m2 = c. Here,m ∈ Fq2\Fq and hence neither branch isFq-rational.

On the other hand, the following result holds.

L EMMA 8.12 Letγ be a branch of an irreducible plane curveF defined overFq.Then there is an overfieldFqn of Fq depending onγ such thatγ has a primitiverepresentation(x(t), y(t)) with x(t), y(t) ∈ Fqn [[t]].

Proof. Let P ′ = (x′, y′) be a generic point ofF . By Theorem 3.27, there is anirreducible plane curveG = v(g(X,Y )) with only ordinary singularities which isbirationally equivalent toF . Without loss of generality, suppose that no point ofGat infinity is singular and that no vertical line is tangent toG at a singular point. Let

ω : x′ = u(x, y)/v(x, y), y′ = w(x, y)/z(x, y)

be a birational transformation that takesG to F , whereP = (x, y) is a genericpoint ofG. The subfieldFqk of K generated byFq, together with the coefficientsof the polynomialsu, v, w, z ∈ K[X,Y ], contains all coefficients ofg(X,Y ). Apossible larger subfieldFqs of K also contains the (finitely many) slopes of thetangents toG at singular points.

Now, let γ be any branch ofF . By Theorems 8.9 and 8.10,ω−1(γ) is Fqn-rational ifFqn is the subfield generated byFqs and the coordinates of the centre ofω−1(γ). Sinceω(ω−1(γ)) = γ, the subfieldFqn of K has the desired property.2

A corollary to Lemma 8.12 is the following result.


THEOREM 8.13 Let γ be a branch of an irreducible plane curveF defined overFq. Then there is a positive integern depending onγ such thatΦn(γ) = γ.

8.3 Fq-RATIONAL PLACES, DIVISORS AND LINEAR SERIES

Let f =∑aijX

iY j be any polynomial inK[X,Y ]. For any integern, then-th-conjugate off is the polynomial

f (n) =∑aqn

ij XiY j .

Then-th-conjugate mapf 7→ f (n) is an automorphism ofK[X,Y ]. Here,f is ir-reducible if and only iff (n) is irreducible; also (8.1) holds for anyf, g ∈ K[X,Y ].Then,f = f (n) only whenf ∈ Fqn [X,Y ], and

f · f (1) · . . . · f (n−1) ∈ Fq[X,Y ] if and only if f ∈ Fqn [X,Y ]; (8.2)

f + f (1) + · · ·+ f (n−1) ∈ Fq[X,Y ] if and only if f ∈ Fqn [X,Y ]. (8.3)

Now, these results together with those obtained in Section 8.2 are applied to thefunction fieldK(F) of an irreducible plane curveF = v(F (X,Y )) defined overFqn . Let P = (x, y) be a generic point ofF . Those elements inK(F) havinga representationu(x, y)/v(x, y), with u, v ∈ Fqn [X,Y ] constitute a subfield, theFqn-rational function fieldFqn(F) of F . Since any elementf ∈ K(F) can bewritten as

f =u(x, y)

v(x, y), with u[X,Y ], v[X,Y ] ∈ K[X,Y ], F (X,Y ) ∤ V (X,Y ) (8.4)

then-th conjugate off can be defined:

f (n) =u(n)(x, y)

v(n)(x, y).

Note thatv(n)(x, y) 6= 0 becauseF (X,U) ∤ v(n)(X,Y ).

L EMMA 8.14 The definition off (n) does not depend on the representation.

Proof. It must be shown that

u(x, y)

v(x, y)=w(x, y)

z(x, y)=⇒ u(n)(x, y)

v(n)(x, y)=w(n)(x, y)

z(n)(x, y).

To do this, write the the first equation as

u(x, y)z(x, y)− v(x, y)w(x, y) = 0.

Raising this to theqn-th power, this gives

u(n)(xqn

, yqn

)z(n)(xqn

, yqn

)− v(n)(xqn

, yqn

)w(n)(xqn

, yqn

) = 0.

By Lemma 8.3,P = (x(n), y(n)) is a generic point ofF . Thus

F (X,Y )G(X,Y ) = u(n)(X,Y )z(n)(X,Y )− v(n)(X,Y )w(n)(X,Y )

for some polynomialG ∈ K[X,Y ], whence it follows that

u(n)(x, y)z(n)(x, y)− v(n)(x, y)w(n)(x, y) = 0,

as required. 2

262

L EMMA 8.15 If f = f (1) thenf ∈ Fq(F).

Proof. Assume first thatf ∈ K(x) and writef = c · u(x)/v(x) with c ∈ K andu(X), v(X) ∈ K[X] monic polynomials for whichgcd(u(X), v(X)) = 1. ByLemma 8.14,f (1) = cq · u(1)(x)/v(1)(x). Fromf = f (1),

u(X)v(1)(X) = u(1)(X)v(X)

andc ∈ Fq. Sincegcd(u(X), v(X)) = 1 anddeg u(X) = deg u(1)(X), this im-plies thatu(X) = u(1)(X) whenceu(X) ∈ Fq[X]. Likewise forv(X). Therefore,the assertion holds whenf ∈ K(x).

Let n = [K(x, y) : K(x)]. SinceK(F) = K(x, y), every elementf ∈ K(F)can be uniquely written in the form

f = c0(x) + c1(x)y + . . .+ cn−1(x)yn−1

with c0(x), . . . , cn−1(x) ∈ K(x). Then

f (1) = c(1)0 (x) + c

(1)1 (x)y + . . .+ c

(1)n−1(x)y

n−1.

From f = f (1) it follows that ci(x) = c(1)i (x) for i = 0, . . . , n − 1. By the

preceding assertion, this implies thatci(x) ∈ Fq(x). Thenci(X) ∈ Fq[X], andthe assertion follows.

2

Note that (8.1) holds for anyf, g ∈ K(F). Also, for everyf ∈ K(F) there is ag ∈ K(F) such thatf = g(n). Hence then-th conjugate map is an automorphismι of K(F) that fixes every element inFqn(F). The constants fixed byι are thosein Fqn .

Let F = v(F (X,Y )) be an irreducible plane curve defined overFq. For ageneric pointP = (x, y) of F , let K(F) = K(x, y) be the associated functionfield of F . Given a placeP of K(F), let σ be any primitive place representationσ of P. As usual, putx(t) = σ(x), y(t) = σ(y). ThenF (x(t), y(t)) = 0 and, byLemma 8.3,

F (x(n)(t), y(n)(t)) = 0.

This means that the pair(x(n)(t), y(n)(t)) is a primitive representation of a branchof F = v(F (X,Y )). The corresponding placeΦn(P) is then-th Frobenius imageofP. This definition is independent of the choice ofσ.

Now, a primitive representation ofΦn(P) is given. Next,σι−1 is an isomor-phism fromK(F) into K((t)). This does not fix every constant, but the isomor-phism

τ = κσι−1 (8.5)

does. To check thatτ is a primitive place representation of theΦn(P) of P, letι−1(x) = u(x, y)/v(x, y), that is,x = u(n)(x, y)/v(n)(x, y). Then

τ(x) = κ

(u(x(t), y(t))

v(x(t), y(t))

)=u(n)(x(n)(t), y(n)(t))

v(n)(x(n)(t), y(n)(t))= x(n)(t).

Similarly, τ(y) = y(n)(t), and the result is established.


DEFINITION 8.16 For be an irreducible plane curveF defined overFq, a placePof the function fieldK(F) is anFqn-rational placeif Φn(P) = P.

Note thatΦn(P) = P if and only if, for any primitive representationσ, also (8.5)is a primitive representation ofP. Further, from Theorem 8.13, for every placePof K(F), there is a positive integern depending onP such thatΦn(P) = P.

L EMMA 8.17 LetF be an irreducible plane curve defined overFq. If the branchγ ofF is Fq-rational, then the corresponding placeP is alsoFq-rational.

Proof. With the notation above applied to the casen = 1, if (x(t), y(t)) is anFq-rational branch representation ofγ, thenx(1)(t) = x(t), y(1)(t) = y(t). Therefore,τ = σ, showing thatP is Fq-rational. 2

It is shown later that the converse of Lemma 8.17 also holds; see Theorem 8.31.

THEOREM 8.18 Let P be a non-singular point of an irreducible plane curveFdefined overFq. ThenP ∈ PG(2, q) if and only if the placeP of K(F) corre-sponding to the unique branchγ ofF centred atP is Fq-rational.

Proof. Suppose thatP = (1, a, b) is the centre ofγ, and that the tangent toFat P is not the vertical line throughP . Assume first thatP ∈ PG(2, q). ByTheorem 8.9, the branchγ has a primitive representation(x(t), y(t)) such thatx(t) = x(1)(t), y(t) = y(1)(t). Therefore,P is anFq-rational place.

To show the converse, letσ be a primitive representation ofP such that

σ(x) = a+ t, σ(y) = b+ ϕ,

with ordt ϕ ≥ 1. SinceΦ(1)(P) = P means thatσ = λτ for aK-automorphismλof K((t)), it follows that

a+ t = aq + λ(t), b+ ϕ = bq + λ(ϕ(1)).

Since both ordt λ and ordt λ(ϕ) are positive, this shows thatP ∈ PG(2, q). 2

REMARK 8.19 Let F andG be irreducible curves defined overFq, and supposethatF andG are birationally equivalent overK. It is not true generally that theymust be birationally equivalent overFq, as well.

Consider the following counter-example. For a squareq, letH√q = v(H) be

the non-singular Hermitian curve defined overFq with

H = X√

q+10 +X

√q+1

1 +X√

q+12 .

For a generic pointP = (x0, x1, x2) of H√q, letK(Hq) the associated function

field. Now, the non-singular plane curveF = v(F (X0,X1,X2)) is defined to bethe image ofH√

q under the projectivityω of PG(3,K) associated to the matrix

M =

a 1 aq+1

aq+1 a 11 aq+1 a

,

264

wheret is a suitable element of orderq2 + q + 1 in the cubic extensionFq3 of Fq.More precisely, the elementt is chosen such that

tq√

q+√

q + tq+√

q+1 + t= 0,

tq√

q+q+√

q+1 + t√

q+1 + 1 = 0,

tq√

q+√

q+1 + tq+1 + t√

q 6= 0,

anddetM 6= 0. Then

F (X0,X1,X2) = X√

q0 X1 +X

√q

1 X2 +X√

q2 X0.

HenceF can also be regarded as a curve overFq. If H√q andF were birationally

equivalent overFq, then they would have the same number ofFq-rational points.But, as shown in Section 8.5, the number ofFq-rational points ofHq is q

√q + 1,

whileF has at most12 (q√q + 2q +

√q) such points.

The mapΦn can be extended to divisors. IfD =∑nPP, then

Φn(D) =∑nPΦn(P).

DEFINITION 8.20 Let D be a divisor ofK(F), whereF is an irreducible planecurve defined overFq.

(i) If Φn(D) = D, thenD is anFqn-rational divisor.

(ii) The Fqn-rational divisors form a subgroup of div(Σ), theFqn-rational divi-sor groupdiv(Fqn(F)).

(iii) The subgroup of div(Fqn(F)) consisting of all divisors of degree zero is thezero degreeFqn-rational divisor groupdiv0(Fqn(F)).

If D =∑nPP where allP areFqn-rational places, thenD is anFqn-rational

divisor. The converse does not hold in general. To understand why this can happen,for instance whenn = 1, choose any placeP of K(F), and definek to be thesmallest positive integer for whichΦk(P) = P occurs. Then the divisor

D = P + Φ(P) + . . .Φk−1(P)

is Fq-rational even ifP is not, that is, ifk > 1. The divisorD is theclosed placeof P, andN(P) = degD is thedegree of the closed placeP.

Analogous to the trace of an element in a finite field, this observation leads to theintroduction of a similar concept for divisors.

DEFINITION 8.21 For anyFqn-rational divisorD, the divisor

T(D) = D + Φ(D) + . . .Φn−1(D)

is anFq-rational divisor, thetrace of theFqn-rational divisorD overFq.

Thetrace mapdefines a homomorphism

T : Div(Fqn(F))→ Div(Fq(F)),

D 7→ T(D)

It may be noted that, for anyξ ∈ Fqn(F),

T(div ξ) = div(T(ξ))

which justifies the terminology; see Section 15.4.


PROPOSITION 8.22 If n is not divisible byp, then the trace map is surjective.

Proof. LetD′ be anyFq-rational divisor ofK(F). ThenΦk(D′) = D′ for everyk. PutD = 1

n D′. ThenD is anFq-rational divisor, and hence anFqn-rational

divisor ofK(F) as well. Thus

T(D) =

n−1∑

k=0

Φk(D) =

n−1∑

k=0

1

nΦk(D′) =

n−1∑

k=0

1

nD′ = D′.

This proves the assertion. 2

L EMMA 8.23 LetF be an irreducible plane curve defined overFq. If ξ ∈ Fq(F),then bothdiv(ξ) anddξ areFq-rational divisor.

Proof. With the notation above, ifσ(ξ) =∑ait

i, thenτ(ξ) =∑aq

i ti. Also,

dσ(ξ)/dt =∑iait

i−1, dτ(ξ)/dt =∑iaq

i ti−1.

Thus,

ordPξ = ordΦ(P) ξ, ordPdξ = ordΦ(P) dξ. 2

L EMMA 8.24 LetF be an irreducible plane curve defined overFq. If ξ is a non-zero element ofK(F) anddiv ξ is anFq-rational divisor, thenξ ∈ Fq(F) up to aconstant factor.

Proof. If div ξ is anFq-rational divisor, then divξ = div ξ(1). By Corollary 5.35,there is a a non-zero constanta ∈ K such thataξ(1) = ξ. Chooseb ∈ K such thatbq−1 = a, and replaceξ by η = bξ. Thenη(1) = η, as

η(1) = bqξ(1) = bqa−1ξ = bqa−1b−1η = η.

Henceη ∈ Fq(F), and the assertion follows from Lemma 8.15. 2

DEFINITION 8.25 A linear seriesL is Fq-rational if there existx0, . . . , xr inFq(F) and anFq-rational divisorB such thatL consists of all divisors

Ac = div (∑r

i=0cixi) +B, (c0, . . . , cr) ∈ PG(r,K).

THEOREM 8.26 LetF be anFq-rational curve andB an Fq-rational divisor ofFq(F). Then|B| is anFq-rational linear series.

Proof. As usual,P = (x, y) is a generic point ofF , andK(F) is representedby K(x, y). Chooseg, f0, . . . , fr in K[X,Y ] for which xi = fi(x, y)/g(x, y),for i = 0, . . . , r. It may be assumed thatg(X,Y ) ∈ Fq[X,Y ] by the followingargument.

The coefficients ofg(X,Y ) generate a finite subfieldFqn of K. Put

h(X,Y ) = g(X,Y )(1) · · · g(X,Y )(n−1).

From (8.2),g(X,Y )h(X,Y ) ∈ Fq[X,Y ]. For brevity, write

h = h(x, y), hi = hfi(x, y),

266

for i = 0, . . . r, and putD = B−div(h). By Lemma 8.23,D is still anFq-rationaldivisor. Then|B| = |D| and|B| consists of all divisors

Ac = div(∑r

i=0cihi) +D, (c0, . . . , cr) ∈ PG(r,K).

The coefficients of the polynomialshi(X,Y ), 0 ≤ i ≤ r, together withFq gener-ate a finite subfieldFqk ofK. The simple field extensionFqk |Fq has degreek. So,Fqk = Fq(θ), with θ a root of a separable polynomials(X) ∈ Fq[X] of degreek,

irreducible overFq. The roots ofs(X) areθj = θqj

wherej = 0, 1, . . . , k − 1.Let d ∈ Fqk [X,Y ]. Forj = 0, 1, . . . , k − 1, put

dj(X,Y ) = θjd(X,Y ) + θjqd(X,Y )(1) + . . .+ θj

qk−1

d(X,Y )(k−1).

Then,dj(X,Y ) = d(1)j (X,Y ), and hencedj ∈ Fq(X,Y ). Also,

d0(x, y) = θd(x, y) + θqd(x, y)(1) + . . .+ θqk−1

d(x, y)(k−1),

d1(x, y) = θ1d(x, y) + θ1qd(x, y)(1) + . . .+ θ1

qk−1

d(x, y)(k−1),

...

dk−1(x, y) = θk−1d(x, y) + θqk−1d(x, y)

(1) + . . .+ θqk−1

k−1 d(x, y)(k−1).

Note that the determinant of the matrix,

θ θq . . . θqk−1

θ1 θq1 . . . θqk−1

1...

... . . ....

θk−1 θqk−1 . . . θqk−1

k−1

,

does not vanish asϑ, ϑ1, . . . , ϑk−1 is a basis ofFqk overFq. Therefore, thereexiste0, e1, . . . , ek−1 ∈ Fqk such that

d(x, y) = e0d0(x, y) + e1d1(x, y) + . . .+ ek−1dk−1(x, y).

Applying this tod(X,Y ) = hi(X,Y ) for every i = 0, 1, . . . r, it turns out that∑cihi can be written as aK-linear combination of(r+1)k elements fromFq(x, y).

Since these elements are chosen independently ofc = (c0, . . . , cr), the assertionfollows. 2

REMARK 8.27 Theorem 8.26 together with Lemma 8.23 ensure the existence ofFq-rational linear series. In particular, the canonical series ofFq(F) is anFq-rational linear series. Also, IfL is anFq-rational linear series, andP is anFq-rational place then the sub-series ofL consisting of all divisors containingP isalso anFq-rational linear series.

8.4 SPACE CURVES OVERFq

In this section,F is an irreducible plane curve defined overFq. Also, for a genericpointP = (x, y) of F , the fieldFq(F) is theFq-rational subfield of the function


fieldK(F) of F . Consistent with the concept of anFq-rational linear series, somebut not all space curves arising fromF are considered to beFq-rational curves. Themain goal is to define and characterise theFq-rational branches of anFq-rationalspace curve in terms ofFq-rational places ofK(F).

DEFINITION 8.28 (i) An irreducible curveΓ of PG(r,K) is defined overFq

if the curve is given by a pointQ = (x0, x1, . . . , xr), with xi ∈ Fq(F) fori = 0, 1, . . . r.

(ii) If Γ is defined overFq, then a branchγ of Γ is Fq-rational if it has a branchrepresentation

(x0(t), x1(t), . . . , xr(t)),

with xi(t) ∈ Fq[[t]] for i = 0, 1, . . . , r.

THEOREM 8.29 Let Γ be an irreducible curveΓ of PG(r,K) defined overFq.For a non-singular point ofΓ the following conditions are equivalent:

(i) P ∈ PG(r, q);

(ii) the unique branchγ of Γ with centre atP is Fq-rational.

(iii) the placeP ofK(F) associated toγ is Fq-rational.

Proof. As usual, letx0 = 1 and letσ be a primitive representation ofP. Putσ(xi) = xi(t) for i = 1, . . . , r. Then(1, x1(t), . . . , xr(t)) is a primitive represen-tation ofγ. SinceΓ is defined overFq, so (1, x

(1)1 (t), . . . , x

(1)r (t)) is a primitive

representation of the branchγ′ of Γ associated toΦ(P).(i)⇒ (ii) As P is a non-singular point ofF , takex1(t) = a1 + t. Write

xi(t) = ai + ϕi(t)

with ϕi ∈ K[[t]] and ordt ϕi(t) ≥ 1. Then

x(1)i (t) = aq

i + ϕ(1)i (t)

for i = 1, . . . , r. AsP ∈ PG(r, q), soP is the centre ofγ′ as well. SinceP is thecentre of exactly one branch, this implies thatγ = γ′. So, there existsτ ∈ K[[t]]

such thatxi(t) = x(1)i (τ). Fori = 1, this givest = τ . But thenxi(t) = x

(1)i (t) for

everyi = 1, . . . , r; that is,γ is anFq-rational branch ofΓ.

(ii) ⇒ (iii) As in (8.5), letτ = κσι−1. Thenτ(xi) = x(1)i (t), for 1 ≤ i ≤ r.

By hypothesis,γ has a primitive representation(1, x1(t), . . . , xr(t)) such that, fori = 1, . . . , r, thenxi(t) = x

(i)i (t). This means thatτ = σ, and henceΦ(P) = P.

(iii) ⇒ (i) If τ = σ, thenxi(t) = x(1)i (t), implying thatai = aq

i , for eachi = 1, . . . , r. 2

REMARK 8.30 Theorem 8.26 shows how to obtain non-singular models ofF de-fined overFq; in particular, the canonical curve is such a curve providedthatK(F)is not hyperelliptic. For each non-singular model, the above three conditions areequivalent.

268

THEOREM 8.31 If Γ is an irreducible curve ofPG(r,K) defined overFq, then abranchγ ofΓ isFq-rational if and only if the corresponding placeP isFq-rational.

Proof. By Theorem 8.26,Γ is birationallyFq-equivalent to a non-singular modelX of K(F) defined overFq. So, ifΓ = K(Q) andX = K(R) with

R = (1, y1, . . . , yr), yi ∈ Fq(F),

then the rational transformation

ω : yi =ui(x1, . . . , xr)

vi(x1, . . . , xr), i = 1, . . . , r,

is Fq-rational; that is,ui, vi ∈ Fq[X1, . . . ,Xr]. Note thatω−1 is alsoFq-rational.

Now, assume thatγ is Fq-rational. Then the imageγ′ of γ underω is also anFq-rational branch ofX . By Remark 8.30,P isFq-rational. To prove the converse,choose a primitive representation ofγ′,

(1, y1(t) = a1 + ϕ1(t), . . . , yr(t) = ar + ϕr(t)), ordt ϕi(t) ≥ 1, i = 1, . . . , r,

such thatτ(yi(t)) = σ(yi(α(t)) with an automorphismα of K[[t]]. Thenaqi = ai,

that is,P ∈ PG(r, q), Again, Theorem 8.29 applies, whenceγ′ is Fq-rational.Sinceγ = ω−1(γ′), the assertion follows. 2

The notation for the set of allFq-rational places of a function field of transcen-dence degreeΣ defined overFq is Σ(Fq). By Theorem 8.31, there is a one-to-onecorrespondence betweenΣ(Fq) andΓ(Fq) for each birational modelΓ of Σ; inparticular,|Σ(Fq)| = Sq.

REMARK 8.32 If a non-singular modelX of F is defined overFq, and places ofF are identified with points ofX , any closed place ofK(F) is viewed as aclosedpoint. For a pointA = (a0, . . . , ar) ∈ PG(r,K) belonging toX , let Fqk be thesubfield ofK generated byFq and the coordinates ofA. Theorem 8.31 shows thatthe closed point ofA is the set consisting of thek points,

A = A0, A1, . . . , Ak−1,whereAj = (aqj

0 , . . . , aqj

r ) for j = 0, . . . , k − 1.

L EMMA 8.33 For everyn, the fieldΣ has only finitely many effectiveFq-rationaldivisors of degreen.

Proof. With the notation in Remark 8.32, ifD = P1 + . . . + Pn is anFq-rationaldivisor, andP is a point inD, then eitherP ∈ X (Fq) or P ∈ X (Fqi) \ X (Fq)for somei > 1. If the latter case occurs then the Frobenius imageΦ(P ) of P isanother point inD, and this holds true forΦj(P ) with j = 2, . . . i− 1. Therefore,every point appearing inD with positive weight belongs toPG(r, qn). If D rangesover allFq-rational divisors of degreen, such points form a finite setP1, . . . , PsasPG(r, qn) has only finitely many points. PutdegPi = vi. It suffices to showthat the equation

∑si=1 nivi = n


has finitely many solutions(n1, . . . , ns) with non-negative integers. Actually, thenumber of the solutions is bounded above by(n+ 1)s. 2

Using Lemmas 8.24 and 8.15, it is straightforward to show that theFq-rationalprincipal divisors form a subgroupPrin(Fq(F)) of Prin(Σ). A divisor class[A],that is, an element[A] of the divisor class groupPic(Σ), is anFq-rational divisorclass if [A] contains at least oneFq-rational divisor. By Theorem 8.26, if[A]is anFq-rational divisor class, then the linear series|A| is Fq-rational. The setof all Fq-rational divisors classes form a subgroupPic(Fq(F)) of Pic(Σ), thegroup of theFq-rational divisor classes of theFq-rational function fieldFq(F).Its subgroupPic0(Fq(F)) consisting of allFq-rational divisor classes of degreezero is thegroup of the zero degreeFq-rational divisor classes of theFq-rationalfunction fieldFq(F).

THEOREM 8.34 The groupPic(Fq(F)) is finite.

Proof. Choose a non-constant elementf ∈ Fq(F) and letg = fm for a sufficientlylarge positive integerm. From Theorem 5.33,deg div(f)0 > 0. So,deg div(g)0 =mdeg div(f)0 = n > 0, andm is chosen such thatm ≥ 2g. By Lemma 8.33,it suffices to show that anyFq-rational divisor is the difference of the effectiveFq-rational divisors of degreen. LetD ∈ Div0(Fq(F)). By the Riemann–RochTheorem 6.59,

dim(div(g)0 −D) ≥ deg(div(g)0 − g ≥ g + 1 ≥ 1.

Since div(g)0 − D is anFq-rational divisor, there exists a non-zero elementh inFq(F) such that div(h) + div(g)0 −D ≻ 0. Put

D1 = div(g)0, D2 = div(h) + div(g)0 −D.Both divisorsD1 andD2 areFq-rational of the same degreen, and fromD =D1 −D2 + div(h) the equivalenceD ≡ D1 −D2 follows. 2

This theorem gives rise to following definition.

DEFINITION 8.35 Let F be an irreducible algebraic curve defined over a finitefieldFq. The order of the groupPic0(Fq(F)) is theclass number.

EXAMPLE 8.36 Assume thatF is rational. LetA andB be any twoFq-rationaldivisors of the same degreen. From Theorem 6.79, the linear series|B| has dimen-sionn, and henceA ∈ |B|. Since|B| is alsoFq-rational by Theorem 8.26, there isf ∈ Fq(F) such thatA = B + divf . Therefore, the class number is equal to1.

Let e denote the smallest degree of all positive degreeFq-rational divisors ofFq(F). It is easily seen that the degree of everyFq-rational divisorD is divisibleby e, that is,degD = m · e with m ∈ Z. Actually, e = 1, but this is shown later;see Proposition 10.4. Meanwhile no use is made of it.

PROPOSITION 8.37 Let [C1], . . . , [Ch] be all zero degreeFq-rational divisors ofFq(F), and let[C0] be a fixed degreeeFq-rational divisor ofFq(F). Then anyFq-rational divisor class[C] can be written as[C] = [vC0]+ [Ci] wheredegC = v ·eand1 ≤ i ≤ h. In particular, for anyv ≥ 0, the number ofFq-rational divisorclasses of degreev · e is equal to the class number.

270

Proof. Let [C] be anFq-rational divisor class ofFq(F). ThendegC = v · e. Theclassv[C0] = [vC0] has the same degree, and hence[C − vC0] is a degree zeroFq-rational divisor class ofFq(F). Therefore[C − vC0] = [Ci] for somei with1 ≤ i ≤ h. 2

THEOREM 8.38 The numbernq(|C|) of distinct effectiveFq-rational divisors inanFq-rational divisor class[C] is equal to

nq(|C|) =qdim |C|+1 − 1

q − 1.

Proof. Let r = dim |C|. By Theorem 8.26 there are linearly independent elementsxi ∈ Fq(F) such that the complete linear series|C| consists of the divisors

Ac = div(c0x0 + c1 + . . .+ crxr) + C

wherec = (c0, . . . , cr) runs over all a point ofPG(r,K). The divisorAc is Fq-rational if and only if the corresponding pointc belongs toPG(r, q). The assertionfollows from the fact that|PG(r, q)| = (qr+1 − 1)/(q − 1). 2

An infinite sequence associated to Pic(Fq(F)) isA0, A1, . . . , An, . . . , where

An = |A ∈ Div(Fq(F)) | A ≻ 0; degA = n|.Note thatA0 = 1, andA1 = N1, the number of allFq-rational places ofFq(F).Forn > 2g−2 the numberAn can be computed from the class numberh of Fq(F).

PROPOSITION 8.39 For any integern > 2g − 2 divisible bye,

An =h

q − 1(qn+1−g − 1).

Proof. There areh divisor classes of degreen, say [C1], . . . , [Ch]. By Theorem8.38

|A ∈ [Cj ] ∩Div(Fq(F));A ≻ 0| = 1

q − 1(qdim |Cj |+1 − 1).

This together with the Riemann–Roch Theorem 6.59 imply that

|A ∈ [Cj ] ∩Div(Fq(F));A ≻ 0| = 1

q − 1(qn+1+g − 1).

Any Fq-rational divisor of degreen of Fq(F) lies in exactly one of the abovedivisor classes[C1], . . . , [Ch]. Therefore

An =

h∑

j=1

|A ∈ [Cj ] ∩Div(Fq(F));A ≻ 0| = h

q − 1(qn+1−g − 1).

2

The trace map also defines a homomorphism

T0 : Pic0(Fqn(F))→ Pic0(Fq(F)), (8.6)

[D] 7→ [T(D)].

Like the trace map, such a homomorphism is surjective provided thatn is not di-visible byp. Therefore, Proposition 8.22 holds true forT0.


8.5 THE STOHR–VOLOCH THEOREM

In this section, the fundamental theorem of Stohr and Voloch on the numberSq

of Fq-rational points on a non-singular curveF is obtained. The essential ideais to count those pointsP on a non-singular modelX for which the osculatinghyperplane atP contains the imageP q of P under the Frobenius collineationΦ.Since everyFq-rational point has this property, an upper bound forN1 is obtained.

If an explicit presentation of a non-singular modelX is not available, or toocomplicated, it is convenient to use some simpler singular model Γ defined overFq and count theFq-rational branches ofΓ. This can be done by using the sameidea but points on the curve must be regarded as branch points, in the same way asSection 7.6.

An elementary proof of a simple case of the Stohr–Voloch theorem is first given.

THEOREM 8.40 Assume thatK has odd characteristic. LetF be an irreducibleplane curve of degreed defined overFq. If F has only finitely many points ofinflexion, then the numberSq of Fq-rational points ofF satisfies the inequality,

2Sq +N ′ ≤ d(q + d− 1), (8.7)

whereN ′ counts the non-Fq-rational pointsQ ∈ F such that the tangent line atQcontains the imageQq ofQ under the Frobenius collineation.

Proof. If F = v(F (X,Y )), the tangent toF at a non-singular pointP = (1, a, b)is

v((X − a)FX(a, b) + (Y − b)FY (a, b)).

Let

G(X,Y ) = (X −Xq)FX + (Y − Y q)FY . (8.8)

and putG = v(G). Suppose thatF (X,Y ) does not divideG(X,Y ). Then thecurvesG andF have no common components. Since

GX =FX + (X −Xq)FXX + (Y − Y q)FXY ,

GY =FY + (X −Xq)FXY + (Y − Y q)FY Y ,

the tangent toF at P ∈ PG(2, q) is also tangent toG at the same point. Hencethe intersection numberI(A,F ∩ G) ≥ 2; this remains valid whenP is a singularpoint ofF . Also, a direct calculation shows thatI(P,F ∩ G) ≥ 2 holds true for apointP = (0, a, b) of F with a, b ∈ Fq.

For any other common pointQ of F andG, the tangent toF atQ contains theFrobenius imageQq. Hence, if there areN ′ points of the latter type, then Bezout’stheorem implies (8.7).

In the case thatF (X,Y ) dividesG(X,Y ), writeG(X,Y ) = H(X,Y )F (X,Y ).Then

HXF +HFX = GX = FX + (X −Xq)FXX + (Y − Y q)FXY ,

HY F +HFY = GY = FY + (X −Xq)FXY + (Y − Y q)FY Y ,

272

Let P = (x, y) be a generic point ofF . Eitherx or y is a separable variable ofK(F). Without loss of generality, suppose that this isx. ThenFY (x, y) 6= 0, and

GX(x, y)

GY (x, y)=FX(x, y) + (x− xq)FXX(x, y) + (y − yq)FXY (x, y)

FY (x, y) + (x− xq)FXY (x, y) + (y − yq)FY Y (x, y). (8.9)

Also, sinceG(x, y) = 0, from (8.8)

FX(x, y)

FY (x, y)= − y

q − yxq − x. (8.10)

Using the expressions in (8.8), (8.9), (8.10), and eliminating x − xq andy − yq

from

G(x, y) = 0, GX(x, y) = 0, GY (x, y) = 0

shows that(x, y) is a zero of the polynomial

(FY )2FXX − 2FXFY FXY + (FX)2FY Y . (8.11)

SinceP = (x, y) is a generic point ofF , it follows thatF (X,Y ) divides (8.11).By Theorem 1.35, all non-singular points ofF are inflexions. 2

Now, the general case is considered. The notation and terminology from Section7.6 are used. In addition,Γ is defined overFq, and the simple base-point-freelinear seriesL is assumed to beFq-rational. The symbolΓ(Fq) stands for the setof all Fq-rational branches ofΓ. As in Section 7.6, the term ‘pointP ’ is adoptedto denote the branch ofΓ associated to the placeP of K(Γ). In doing so,Φ(P )denotes the Frobenius image ofP , and the point(x0(P )q, . . . , xr(P )q) is the centreof Φ(P ). Further,P ∈ Γ(Fq) means that the pointP , viewed as a branch ofΓ, isFq-rational. Bearing this in mind,P ∈ Γ(Fq) does not merely mean that the centreof P is in PG(r, q), even if this is true when the centre is a non-singular point of Γ.

Let ζ be a local parameter atP. The osculating hyperplane toΓ at P containsthe centre ofΦ(P ) if and only if ∆ = 0, where

∆ =

∣∣∣∣∣∣∣∣∣∣∣∣

x0(P )q x1(P )q . . . xr(P )q

D(j0)ζ x0(P ) D

(j0)ζ x1(P ) . . . D

(j0)ζ xr(P )

D(j1)ζ x0(P ) D

(j1)ζ x1(P ) . . . D

(j1)ζ xr(P )

...... . . .

...

D(jr−1)ζ x0(P ) D

(jr−1)ζ x1(P ) . . . D

(jr−1)ζ xr(P )

∣∣∣∣∣∣∣∣∣∣∣∣

. (8.12)

As in Section 7.6, this motivates the study of the following Wronskian-type deter-minant:

Wν0,...,νr−1

ζ (x0, . . . , xr) = detM(ν0, . . . , νr−1),

whereζ denotes a separable variable,ν0, . . . , νr−1 are non-negative integers, and

M(ν0, . . . , νr−1) =

xq0 xq

1 . . . xqr

D(ν0)ζ x0 D

(ν0)ζ x1 . . . D

(ν0)ζ xr

D(ν1)ζ x0 D

(ν1)ζ x1 . . . D

(ν1)ζ xr

...... . . .

...

D(νr−1)ζ x0 D

(νr−1)ζ x1 . . . D

(νr−1)ζ xr

.


PROPOSITION 8.41 There exists a sequence of increasing non-negative integersν0, . . . , νr−1 with ν0 ≥ 0 such that

Wν0,...,νr−1

ζ (x0, . . . , xr) 6= 0.

If theνi are chosen minimally in lexicographical order, then there exists an integerI with 0 < I ≤ r such that

νi =

ǫi for i < I,ǫi+1 for i ≥ I.

Proof. Let I be the smallest integer such that the row(xq0, x

q1, . . . , x

qr) is a lin-

ear combination of the vectors(D(ǫi)ζ x0,D

(ǫi)ζ x1, . . . ,D

(ǫi)ζ xr) with i = 0, . . . , I.

Thenν0, . . . , νr−1 = ǫ0, . . . , ǫn\ǫI. Sinceν0 = 0, soI > 0. 2

The minimal choice of theνi has a stronger significance.

COROLLARY 8.42 If m0, . . . ,mk are any increasing integers withm0 ≥ 0 suchthat the row vectors of the matrixM(m0, . . . ,mk) are linearly independent, thenνi ≤ mi for i = 0, . . . , k.

Proof. The matrixM(ν0, . . . , νk − 1) has rankk + 1, whenceνk − 1 < mk; thatis, νk ≤ mk. 2

DEFINITION 8.43 TheStohr–Voloch divisor ofL is

S = div (Wν0,...,νr−1

ζ (x0, . . . , xr)) + (ν1 + . . .+ νr−1)div (dζ) + (q + r)E.

whereE =∑ePP andeP = −minordP x0, . . . ,ordPxr.

By the lemma below, the integersν0, . . . , νr−1 and the Stohr–Voloch divisordepend only on the linear seriesL.

DEFINITION 8.44 The integersν0, . . . , νr−1 are theFrobenius ordersof the linearseriesL, or, equivalently, of the curveΓ, and(ν0, . . . , νr−1) is theFrobenius ordersequence.

Since Frobenius orders ofL are alsoL-orders,νi ≤ deg Γ for i = 0, 1, . . . , r − 1.

L EMMA 8.45 (i) If yi =∑aijxj with (aij) ∈ GL(r + 1, q), then

Wν0,...,νk−1

ζ (y0, . . . , yr) = det(aij)Wν0,...,νk−1

ζ (x0, . . . , xr).

(ii) If h ∈ Fq(Γ), then

Wν0,...,νr−1

ζ (hx0, . . . , hxr) = hq+rWν0,...,νr−1

ζ (x0, . . . , xr).

(iii) If η is another separable variable, then

W ν0,...,νr−1η (x0, . . . , xr) = (dζ/dη)ν1+...+νr−1W

ν0,...,νr−1

ζ (x0, . . . , xr).

Proof. This is similar to that of Lemma 7.49. 2

Corollary 5.35 together with the definition of the genus (5.54) shows that

ordS = (ν1 + . . .+ νr−1)(2g − 2) + (q + r)n. (8.13)

274

PROPOSITION 8.46 The coordinate functionsx0, . . . , xr may be chosen so that

Wν0,...,νr−1

ζ (x0, . . . , xr) =

∣∣∣∣∣∣∣∣∣∣∣∣

x1 − xq1 . . . xr − xq

r

D(ν1)ζ x1 . . . D

(ν1)ζ xr

D(ν2)ζ x1 . . . D

(ν2)ζ xr

... . . ....

D(νr−1)ζ x1 . . . D

(νr−1)ζ xr

∣∣∣∣∣∣∣∣∣∣∣∣

. (8.14)

Proof. Dividing x0, . . . , xr by x0 is equivalent to takingx0 = 1. Now, the resultis obtained, puttingν0 = 0. 2

REMARK 8.47 By (5.12) and Remark 5.70,D(νi)ζ xq

j = 0 if νi is not a multiple ofq. Hence, ifνi < q, thenν0, . . . , νi are the firsti+1 orders of the irreducible curveof PG(r,K) given by the pointR = (1, x1 − xq

1, . . . , xr − xqr). Such a curve is

defined overFq.

PROPOSITION 8.48 (i) If ν is a Frobenius order ofL less thanq, then everyintegerp-adically less thanν is also a Frobenius order ofL.

(ii) If νi < p, then(ν0, . . . , νi) = (0, . . . , i).

Proof. This follows from Corollary 7.59. 2

Now, the weightvP(S) of the placeP in the Stohr–Voloch divisorS is examined.Given a local parameterζ ∈ K(Γ) atP, if the componentsx0, . . . , xr are dividedby ζeP , theneP becomes zero. Therefore,

vP(S) = vP(Wν0,...,νr−1

ζ (x0, . . . , xr)) ≥ 0;

in particular,S is an effective divisor. Further,P is in Supp(S) if and only if∣∣∣∣∣∣∣∣∣∣∣∣

xq0(P ) xq

1(P ) . . . xqr(P )

x0(P ) x1(P ) . . . xr(P )

D(ν1)ζ x0(P ) D

(ν1)ζ x1(P ) . . . D

(ν1)ζ xr(P )

...... . . .

...

D(νr−1)ζ x0(P ) D

(νr−1)ζ x1(P ) . . . D

(νr−1)ζ xr(P )

∣∣∣∣∣∣∣∣∣∣∣∣

= 0. (8.15)

Let (j0, . . . , jr−1) be the(L,P)-order sequence. In the case that

(ν0, . . . , νr−1) = (j0, . . . , jr−1),

P ∈ Supp(S) if and only if Φ(P ) is centred at a point lying in the osculatinghyperplane ofΓ atP . If νi < ji for somei, thenP ∈ Supp(S).

Also, if the centre ofP is in PG(r, q), the determinant in (8.15)vanishes as itsfirst two rows coincide. Hence, Supp(S) contains every place that corresponds to abranch centred at a point inPG(r, q). In particular, allFq-rational places ofK(Γ)are in Supp(S).

DEFINITION 8.49 The linear seriesL or, equivalently, the curveΓ is Frobeniusclassicalif the Frobenius order sequence(ν0, . . . , νr−1) is (0, . . . , r − 1).


REMARK 8.50 If Γ is Frobenius non-classical, thenΓ is also non-classical in gen-eral. Exceptions only occur when the integerI in Definition 8.41 satisfies bothconditions:I + 1 ≡ 0 (mod p), andI < r. This follows from Propositions 8.41and 8.48. An exception forν1 = ǫ2 = q = 2 is the linear seriesg2

4 cut out by lineson the irreducible plane quartic

F = v((X2 +X)(Y 2 + Y ) + 1).

defined overF2.

THEOREM 8.51 The curveΓ is Frobenius non-classical if and only if, for infinitelymany placesP of K(Γ), the osculating hyperplane atP contains the centre ofΦ(P ).

Proof. Let

ξ =

∣∣∣∣∣∣∣∣∣∣∣∣

xq0 xq

1 . . . xqr

x0 x1 . . . xr

D(ǫ1)ζ x0 D

(ǫ1)ζ x1 . . . D

(ǫ1)ζ xr

...... . . .

...

D(ǫr−1)ζ x0 D

(ǫr−1)ζ x1 . . . D

(ǫr−1)ζ xr

∣∣∣∣∣∣∣∣∣∣∣∣

.

If Γ is Frobenius classical, thenξ = 0. Hence,∆ = 0, with ∆ as in (8.12), foreveryP of Γ. Conversely,∆ = 0 implies thatξ has infinitely many zeros. Thenξitself is zero, and hence(ν0, . . . , νr−1) 6= (0, . . . , r − 1). 2

Frobenius non-classical curves are not classical, and hence they have propertiesestablished in Theorem 7.62. The following result shows howthat theorem can bestrengthened for Frobenius non-classical curves.

THEOREM 8.52 LetΓ be a non-classical curve with(ǫ0, ǫ1, . . . , ǫr) its order seq-uence. Assume thatǫr ≤ pm ≤ q = ph and that (7.17) holds for somez0, . . . , zr ∈K(Γ). ThenΓ is Frobenius non-classical if and only if

z0xph−m

0 + z1xph−m

1 + . . .+ zrxph−m

r = 0. (8.16)

Proof. For a primitive representationσ of P, let

σ(xi) =xi(P ) + ait+ . . . ,

σ(zi) = zi(P ) + bit+ . . . ,

for all i. Also, let

η = zpm

0 xq0 + zpm

1 xq1 + . . .+ zpm

r xqr.

Then

η = z0(P )pm

X0(P )q + . . .+ zr(P )pm

Xr(P )q + tG(t).

This shows that the osculating plane atP contains the centre ofΦ(P ) if and onlyif P is a zero ofη.

Suppose thatΓ is Frobenius non-classical. Thenη = 0 as it has infinitely manyzeros. Therefore (8.16) holds, becauseη is thepm-th power of the left-hand side in

276

(8.16). Conversely, (8.16) impliesη = 0. Hence, for every pointP of Γ, the oscu-lating plane atP contains the centre ofΦ(P ). By Theorem 8.51,Γ is a Frobeniusnon-classical curve. 2

Now, some numerical results follow. ForP ∈ Γ, let (j0, . . . , jr) be the(L,P)-order sequence. Remember that, by convention within the present chapter andSection 7.6,P is the place ofK(Γ) corresponding to the branch pointP ∈ Γ.

PROPOSITION 8.53 (i) LetP ∈ Γ(Fq). Then

vP(S) ≥∑rj=1 (ji − νi−1),

with equality if and only if

det

((jiνk

))6≡ 0 (mod p),

0 ≤ k ≤ r − 1, 1 ≤ i < r.

(ii) LetP 6∈ Γ(Fq). Then

vP(S) ≥∑r−1j=1 (ji − νi).

If

det

((jiνk

))6≡ 0 (mod p),

0 ≤ k ≤ r − 1, 0 ≤ i < r − 1, then strict equality holds.

Proof. The argument is similar to the proof of Theorem 7.52 withτ a primitiverepresentation ofP.

(i) SinceP is Fq-rational, the osculating plane atP is overFq; that is, it is

HP = v(a0X0 + . . .+ arXr),

with a0, . . . , ar ∈ Fq. As in the proof of Theorem 7.45, it is possible to replacexi

by yi ∈ Fq(Γ) such that

y0 =1,

τ(yi) = tji + . . . ,

with ji ≥ 1 for i = 1, . . . r. Then ordPWν0,...,νr−1

ζ (y0, . . . , yr), as given in (8.14),may be computed as follows. Sinceν0 = 0 andji > 0 for i = 1, . . . , r,

ordt (Wν0,...,νr−1

t (1, σ(y1), . . . , σ(yr))

= ordt det

((jiνk

)tji−νk + . . .

)

= ordt det

((jiνk

))tj1+...+jr−ν0−...−νr−1 + . . . .

The result follows.(ii) There is a non-singular(r+ 1)× (r+ 1) matrix (aij) with entries inK, but

not necessarily inFq, such that

yi =∑r

j=0 aijxj , σ(yi) = tji + . . . , i = 0, . . . r.


Let hi =∑r

j=0 aijxqj . In contrast to (i), it may not be true thathi = yq

i . Neverthe-less, ordPhi ≥ 0 for eachi. As

Wν0,...,νr−1

ζ (x0, . . . , xr) det(aij) =

∣∣∣∣∣∣∣∣∣∣∣∣

h0 . . . hr

D(ν0)t y0 . . . D

(ν0)t yr

D(ν1)t g0 . . . D

(ν1)t gn

... . . ....

D(νn−1)t g0 . . . D

(νn−1)t gn

∣∣∣∣∣∣∣∣∣∣∣∣

=∑n

0 (−1)ihidi,

wheredi is the minor ofhi in the matrix, eachi = 0, . . . , n, so

vP(S) ≥ minordt τ(d0), . . . ,ordt τ(dr).Now, by the same computations as above,

ordt τ(di) ≥ j0 + . . .+ jr − ji − ν0 − . . .− νr−1.

If

det

((jiνk

))

i,k=0,...,r−1

≡ 0 (mod p),

then

ordt τ(dr) > j0 + . . .+ jr−1 − ν0 − . . .− νr−1.

This proves (ii). 2

Now, numerical relations between the Frobenius ordersν0, . . . , νn−1 and theHermitian ordersj0, . . . , jn of anFq-rational placeP are considered.

PROPOSITION 8.54 LetP ∈ Γ(Fq). If m0, . . . ,mr−1 are integers satisfying bothconditions0 ≤ m0 < . . . < mr−1 and

det

((ji − j1mk

))

i=1,...,r; k=0,...,r−1

6≡ 0 (mod p),

thenνi ≤ mi for eachi.

Proof. The best choices for the integersmi are the orders of the rational mapPG(1,K) −→ PG(r − 1,K) defined by

(1, x) 7→ (1, xj2−j1 , . . . , xjr−j1) = (xj1 , xj2 , . . . , xjr ).

It may therefore be supposed thatm0 = 0 and that

det

((jimk

))

i=1,...,r; k=0,...,r−1

6≡ 0 (mod p).

Assume again thatx0 = 1 andσ(xi) = tji + . . . for eachi. Thus, as in the proofof Proposition 8.53(i),

Wm0,...,mr−1

ζ (x0, . . . , xr) = det

((jimk

))tj1+...+jr−m0−...−mr−1 + . . . 6= 0.

Henceνi ≤ mi for all i, by the minimality of theνi. 2

Some useful corollaries to Propositions 8.53 and 8.54 are now stated.

278

COROLLARY 8.55 If P ∈ Γ(Fq), then

νi≤ ji+1 − j1 for eachi,

vP(S)≥ r j1.

In the case thatmi = i, this gives the next result.

COROLLARY 8.56 LetP ∈ Γ(Fq). If the integer∏

r≥k>s≥0(jk − js)/(k − s) isnot divisible byp, thenνi = i for everyi, and

vP(S) = n+∑r

i=0 (ji − i).Whenmi = ǫi, Proposition 8.54 reads as follows.

COROLLARY 8.57 SupposeP ∈ Γ(Fq). If

det

((ji − j1ǫk

))

i=1,...,r; k=0,...,r−1

6≡ 0 (mod p),

thenνi ≤ ǫi for eachi.

COROLLARY 8.58 If the Frobenius order sequence(ν0, . . . , νr−1) of Γ is notclassical, then everyP ∈ Γ(Fq) is anL-osculating point.

Proof. If there wereP ∈ Γ(Fq) with (L,P)-order sequence(0, . . . , r), then Corol-lary 8.55 would imply thatνi = i all i. 2

COROLLARY 8.59 If the Frobenius order sequence(ν0, . . . , νr−1) is other thanthe sequence(ǫ0, . . . , ǫr−1), then everyP ∈ Γ(Fq) is anL-Weierstrass point.

Proof. If there is anL-ordinary point, thenνi ≤ ǫi+1−ǫ1 by Corollary 8.55; henceνi = ǫi by Proposition 8.41. 2

COROLLARY 8.60 If L is complete and some branch pointP is Fq-rational, thenνi = i for eachi < n− 2g.

Proof. Let P ∈ Γ(Fq). SinceL is complete, soji = i for i ≤ d − 2g. Hence, byCorollary 8.55,νi = i for eachi < n− 2g. 2

COROLLARY 8.61 If some branch pointP is Fq-rational, thenνi ≤ i+ n− r.Proof. Sincejr ≤ n, soji ≤ i + n − r for eachi, and Corollary 8.55 gives theresult. 2

THEOREM 8.62 (Stohr–Voloch)Let

(i) Γ be an irreducible curve inPG(r,K) defined overFq;

(ii) n be the degree andg be the genus ofΓ;

(iii) Sq be the number ofFq-rational places ofΓ;

(iv) (ν0, . . . , νr−1) be the Frobenius order sequence of the linear seriesL cut outby hyperplanes.


Then

Sq ≤ 1r(ν1 + . . .+ νr−1)(2g − 2) + (q + r)n. (8.17)

In stronger form,

Sq ≤ 1r(ν1 + . . .+ νr−1)(2g − 2) + (q + r)n−∑A(P ), (8.18)

with

A(P ) =

∑rk=1 (jk − νk−1)− r, for P ∈ Γ(Fq);

∑r−1k=0 (jk − νk), otherwise.

Proof. By Corollary 8.55,vP(S) ≥ r for eachP ∈ Γ(Fq). SinceS is effec-tive, soSq ≤ deg(S)/r. Now, from (8.13), the first result (8.17) follows. UsingProposition 8.53, this gives (8.18). 2

COROLLARY 8.63 If Sq >1r(r− 1)(g − 1) + (q + r)n, then eachFq-rational

branch point is anL-osculating point.

EXAMPLE 8.64 (1) LetC be a plane curve of degreen. The genus

g ≤ 12 (n− 1)(n− 2).

(a) If γmd = γ2

n is the linear system of all lines of the plane, then

N1 ≤ 12 [v1(2g − 2) + n(q + 2)]. (8.19)

Here either(v0, v1) = (0, ǫ1) with ǫ = 1 or (v0, v1) = (0, ǫ2), whereǫ2 is theorder of contact ofC with the tangent at a generic point; soǫ2 ≤ n. Also, ǫ2 = 2or ǫ2 = pv for some integerv.

If not every point ofC is an inflexion andp 6= 2, then(v0, v1) = (0, 1) and

N1 ≤ 12n(n− 3) + (q + 2)n = 1

2n(n− 1 + q). (8.20)

If, under the same condition thatC is classical andp = 2, then

N1 ≤ 122n(n− 3) + (q + 2)n = 1

2n(2n− 4 + q). (8.21)

(b) Letγmd = γ5

2n be the linear system of all quadrics in the plane. Then

N1 ≤ 15 [(v1 + v2 + v3 + v4)(2g − 2) + 2n(q + 5)]. (8.22)

Suppose thatC is classical with respect toγ2n; then, with respect toγ5

2n, the ordersequence is(0, 1, 2, 3, 4, ǫ5) for someǫ5 ≤ 2n. If p 6= 2, 3, thenǫ5 = 5 or ǫ5 = pv.

If C is classical with respect toγ52n, then

(v0, v1, v2, v3, v4) = (0, 1, 2, 3, 4)

and

N1 ≤ 1510n(n− 3) + (q + 5)2n = 2

5n5(n− 2) + q. (8.23)

(c) Now, letC be defined overFp and suppose thatn ≥ 3. If n ≤ 12p andC has

s double points, then

N1 ≤ 25n5(n− 2) + p − 4s. (8.24)

280

This is obtained by considering the same system as (1)(b). ByLemma 2.32, theorder sequence is classical asd = 2n ≤ p. Also the genus of the curve is

g ≤ 12 (n− 1)(n− 2)− s.

Then Theorem 8.62 gives (8.24).(2) Now take the Hermitian curve

H√q = v(X

√q+1 + Y

√q+1 + Z

√q+1)

and the complete linear seriesγ2√q+1 cut out by the lines of the plane. The tangent

atP0 = (x0, y0, z0) toH√q is

v(x√

q0 X + y

√q

0 Y + z√

q0 Z).

This meetsH√q at (x, y, z) where

x√

q+1 + y√

q+1 + z√

q+1 = 0,

with

x√

q+10 + y

√q+1

0 + z√

q+10 =0,

x√

q0 x+ y

√q

0 y + z√

q0 z=0.

Substitution forz gives

(y√

q0 x

√q − x

√q

0 y√

q)(yq0x− xq

0y) = 0.

Thus the tangent atP0 meetsH√q with multiplicity

√q atP0 and with multiplicity

one at(xq0, y

q0, z

q0). So, with respect toΣ1, at a generic point,j2 =

√q whereas at

a rational pointj2 =√q + 1. Therefore, the order sequence forH√

q is

(ǫ0, ǫ1, ǫ2) = (0, 1,√q)

and the Frobenius order sequence is(v0, v1) = (0,√q). Hence, for a non-singular

curveC of degree√q + 1 with a linear series giving thesevi, the genus

g = 12

√q(√q − 1)

and Theorem 8.62 shows the following:

N1≤ 1m(2g − 2)(v0 + v1) + (q +m)d

= 12(q −

√q − 2)

√q + (q + 2)(

√q + 1)

= q√q + 1.

The upper bound is achieved byH√q, as the following argument shows.

First,N1 > 0, as(0, 0, 1) lies on the projectively equivalent curve,

v(X√

q+1 + Y√

qZ + Y Z√

q)

Given a rational pointP onH√q, each line ofPG(2, q) throughP meetsH√

q inthe points of a Hermitian variety, singular for the tangent at P and non-singular forthe otherq lines throughP ; in the latter case, each intersection is

√q + 1 points,

includingP . So,

N1 = 1 + q√q.


There are only a few curves overFq whose number ofFq-rational places canbe calculated by using a simple formula. However, certain Frobenius non-classicalcurves are of this kind.

THEOREM 8.65 LetX be an irreducible non-singular curve of degreen definedoverFq. If ν1 ≥ 2, thenSq = n(q − n+ 2).

Proof. The idea is to computevP(S)−vP(R) for any placeP of Γ. Up to a changeof the coordinate functions withinFq(X ), the centre ofP is an affine point, and thehyperplanev(X1) does not contain the tangent line toX atP . Thenζ = z1−x1(P )is a local parameter atP. Sinceν1 > 1,

vP(S) =

∣∣∣∣∣∣∣∣∣∣∣

1 ζq xq2 . . . xq

r

1 ζ x2 . . . xr

0 0...

... V0 0

∣∣∣∣∣∣∣∣∣∣∣

= ordP((ζ − ζq) detV ),

whereV denotes the(r − 2) × (r − 2) matrix with general entryD(i)ζ xj . On the

other hand, from Proposition 8.41,νi = ǫi+1 for i = 1, . . . , r − 1. Thus, (7.14)reads:

vP(R) =

∣∣∣∣∣∣∣∣∣∣∣

1 ζ x2 . . . xr

0 1 D(1)ζ x2 . . . D

(1)ζ xr

0 0...

... V0 0

∣∣∣∣∣∣∣∣∣∣∣

= ordP(detV ).

Hence

vP(S)− vP(R) = ordP(detV ). (8.25)

Now, it is shown that

ordP(ζ − ζq) =

1 for P ∈ Fq(X ),0 for P 6∈ Fq(X ).

(8.26)

Sinceν1 > 1, the matrix

1 ζq xq2 . . . xq

r

1 ζ x2 . . . xr

0 1 Dζx2 . . . Dζxr

has rank2. In particular,

(ζ − ζq)Dζxi = (xi − xqi )

for i = 2, . . . , r. So, if ordP(ζ − ζq) > 0, then ordP(xi − xqi ) > 0 for i = 2, . . . r,

and henceP ∈ Fq(X ). If Dζ(ζ − ζq) = 1, thenP 6∈ Fq(X ). Therefore, (8.26)holds. From (8.25) and (8.26),Sq = ord(S) − ord(R). By (7.13) and (8.13), theassertion follows. 2

282

REMARK 8.66 LetF be the irreducible plane curve considered in Examples 5.24and 7.69; thenF is defined overFq. More precisely,F is a Frobenius non-classicalplane curve withǫ = ν = 2q0. This follows from Theorem 7.17 for

z0(X,Y ) = Xq0+1 + Y q0 , z1(X,Y ) = X, z2(X,Y ) = 1.

As noted before,Y∞ is the unique singular point ofF , and it is the centre of onlyone branch ofF . Theorem 8.65 does not hold true forF , sinceF has exactlyq2+1Fq-rational branch points, but

n(q − n+ 2) = (q + 2q0)(−2q0 + 2) 6= q2 + 1.

EXAMPLE 8.67 Let q = ph be a power ofp > 3. For a powerd = pm withm < hand fora, b ∈ K with b 6= a2, the plane curveF = v(f(X,Y )), with

f(X,Y ) = XdY d − (Y + a)Xd − (X + a)Y d +XY + a(X + Y ) + b,

has the following properties:

(i) F has two singular points, namelyX∞ andY∞, both ordinaryd-fold points.

(ii) the tangent lines toF atX∞ areℓ = v(Y −λ) and atY∞ areℓ′ = v(X−λ),with λ ranging over the solutions ofλd − λ− a = 0.

(iii) I(X∞, ℓ ∩ F) = I(Y∞, ℓ′ ∩ F) = 2d;

(iv) F has no linear components at eitherX∞ or Y∞;

(v) the genus ofF is (d− 1)2.

By (ii), (iii), (iv), Exercise 4 in Chapter 1 implies the irreducibility ofF . For ageneric pointP = (x, y) of F , the birational transformation,

ω : x1 = x, x2 = y, x3 = xy,

defines an irreducible curveΓ of PG(3,K). SincefY = −(Xd−X+a) is not thezero polynomial,x is a separable variable in the function fieldK(F) of F . Also,

D(1)x y = − y

d − y − axd − x− a, D

(2)x y =

D(1)x y

xd − x− a, D(3)x y =

D(2)x y

xd − x− a. (8.27)

The generalised WronskianW = det(D(xi)x ) arising fromω is

W =

∣∣∣∣∣∣∣∣∣

1 x y xy

0 1 D(1)x y y + xD

(1)x y

0 0 D(2)x y D

(1)x y + xD

(2)x y

0 0 D(3)x y D

(2)x y + xD

(3)x y

∣∣∣∣∣∣∣∣∣,

which is equal to∣∣∣∣∣∣∣∣∣

1 x y 0

0 1 D(1)x y y

0 0 D(2)x y D

(1)x y

0 0 D(3)x y D

(2)x y

∣∣∣∣∣∣∣∣∣.


From (8.27),W = 0; henceΓ is a non-classical curve ofPG(3,K). The ordersare0, 1, 2, ǫ3, with ǫ3 = pv for somev ≥ 1 by Theorem 7.62(i). Note thatf(X,Y )can be written in the form:

f(X,Y ) = z0(X,Y )d + z1(X,Y )dX + z2(X,Y )dY + z3(X,Y )dXY, (8.28)

where

z0(X,Y ) = XY − c(X + Y ) + e, z1(X,Y ) = −Y + c,z2(X,Y ) = −X + c, z3(X,Y ) = 1,

with cd = a anded = b. Sincez2(x, y) = −x+ c is a separable variable ofK(F),soǫ3 = d by Theorem 7.62(i).

Now, takea, b ∈ Fq, andF is viewed as a curve defined overFq. It is shownthatΓ is Frobenius non-classical if and only if

q = d2, aq + a = 0, b = bd.

For this, the determinant (8.46) withνi = i, i = 0, 1, 2, 3 is calculated. By (8.27),this is

W ′ =

∣∣∣∣∣∣∣

xq − x yq − y xqyq − xyq

1 D(1)x y y

0 D(2)x y D

(1)x y

∣∣∣∣∣∣∣.

Since

xdyd − (y + a)xd − (x+ a)d + xy + a(x+ y) + b = 0,

so

W ′ =(xy)q − (yd − a)xq − (xd − a)yq + xdyd − a(xd + yd) + b

= [(xy)q/d − (y − c)xq/d − (x− c)yq/d + xy − a(x+ y) + e]d.

Thus, a necessary and sufficient condition forΓ to be Frobenius non-classical isg(x, y) = 0, where

g(X,Y ) = (XY )q/d − (Y − c)Xq/d − (X − c)Y q/d +XY − a(X + Y ) + e.

Let G = v(g(X,Y )). Note that, replacingq/d by d and(c, e) by (a, b) as well,g(X,Y ) becomesf(X,Y ). Hence,G is also irreducible. From this discussion,F is Frobenius non-classical if and only ifP = (x, y) is a common generic pointof F andG; that is,f(X,Y ) = kg(X,Y ) with k ∈ K andk 6= 0. Finally, astraightforward calculation shows that this occurs if and only if

q = d2, b = bd, aq + a = 0.

The Stohr–Voloch theorem applied toω gives

Sq ≤

2(d+ 1)((d− 1)2 − 1)/3, for q = d2, b = bd, aq + a = 0;2(d− 1)2 − 1 + 2d(q + 3)/3, otherwise.

(8.29)

If F is defined overFq, then the following result can be added to Theorem 7.64.

284

THEOREM 8.68 F is Frobenius non-classical if and only if there existst(X,Y ) inK[X,Y ] such that

t(X,Y )f(X,Y ) = z0(X,Y )pm

+ z1(X,Y )pm

Xq + . . . zr(X,Y )pm

Y qs. (8.30)

Also, Remark 7.65 remains valid. Despite the characterisation given in Theorems7.64 and 8.68 it is hard to find Frobenius non-classical curves. What it emergesis that they are rare but important curves which can have manyFq-rational pointsin some cases. In the following two sections, the two simplest cases,s = 1, 2 areconsidered. The cases = 1 refers to lines and the cases = 2 to conics.

8.6 FROBENIUS CLASSICALITY WITH RESPECT TO LINES

In this section, classicality and Frobenius classicality are intended with respect tothe linear seriesL1 = g2

n cut out by lines on an irreducible curveF of degreen defined overFq. Also, if a branch ofF centred at the pointP andP is theassociated place ofΣ = K(F), thenj0(P) = 0 < j1(P) < j2(P) denote theL1-orders ofP.

Let F be an irreducible plane curve of degreen defined overFq. For s = 1,Theorem 8.68 and (8.17) read as follows.

THEOREM 8.69 F is Frobenius non-classical if and only if there existst(X,Y ) inK[X,Y ] such that

t(X,Y )f(X,Y ) = z0(X,Y )pm

+ z1(X,Y )pm

Xq + . . . zr(X,Y )pm

Y qs. (8.31)

REMARK 8.70 For every squareq, the Hermitian curve of equationX√

q+1 +Y

√q+1 + 1 = 0 is Frobenius non-classical. By Theorem 8.65, the number of its

Fq2-rational points is equal toq√q + 1.

THEOREM 8.71 If Sq is the number ofFq-rational points ofF , then

Sq ≤ 12 [q′(2g − 2) + (q + 2)n] ≤ 1

2 n[(n− 3)q′ + q + 2]. (8.32)

Non-classical plane curves with respect to lines are investigated in Section 7.8.Now, some more material is added on Frobenius non-classicalplane curves.

PROPOSITION 8.72 LetF be an irreducible plane curve defined overFq which isFrobenius non-classical. If a branch ofF centred at a pointP ∈ PG(2, q), thenj2(P) ≥ 2j1(P) + 1.

Proof. As usual, suppose thatP is an affine point, and that the tangent toF atP isnot a vertical line. Letr = j1(P). Then a primitive representation of the branch is(x(t), y(t)) with

x(t) = a+ artr + ar+1t

r+1 + ar+2tr+2 . . . , ar 6= 0,

y(t) = b+ brtr + br+1t

r+1 + br+2tr+2 . . . .


Assume thatx is a separable variable. Thendx(t)/dt 6= 0, and hence there existsan integerh > r such that

dx(t)/dt = hahth−1 + . . . ,

with hah 6= 0. Similarly, if y is a separable variable, then

dy(t)/dt = kaktk−1 + . . . ,

with kak 6= 0. Bearing in mind that bothx andy cannot be inseparable variables,otherwise the above branch representation is not primitive, let

w =

minh, k whenx, y are separable variables,h wheny is an inseparable variable,k whenx is an inseparable variable.

(8.33)

The hypothesisν ≥ 2 means that the formal power series,

(x(t)q − x(t)) y(t)dt− (y(t)q − y(t)) x(t)

dt,

is equal to zero. In particular, the coefficient of the termtr+w−1 must be zero. Acalculation shows that this coefficient isw(arbw − awbr), sinceaq = a, bq = b.Hencearbw = awbr. Thus the intersection multiplicity of the branch with the linev(bw(X − a)− aw(Y − b)) is at leastr + s. Sincer ≥ 1 andw > r, this impliesthatj2(P) ≥ 2j1(P) + 1. 2

THEOREM 8.73 LetF be a non-singular plane curve of degreen defined overFq.Assume thatF is Frobenius non-classical withν = q′.

(i) If q′ > 2, then

√q + 1 ≤ n ≤ q − 1

q′ − 1, and q′ ≤ √q.

(ii) If q′ =√q, thenF is projectively equivalent to the Hermitian curve

Hq = v(X√

q+10 +X

√q+1

1 +X√

q+12 ).

Proof. SinceF is non-singular, its genusg = (n− 1)(n− 2)/2 by Theorem 5.56.From (10.47), which is the Hasse–Weil Bound for non-singular plane curves, seesection 10.6

Sq ≤ q + 1 + (n− 1)(n− 2)√q.

On the other hand, from Theorem 8.65,Sq = n(q − n + 2), whence the lowerbound onn in (i) follows.

Putq = ph. By Theorem 8.110,F is non-classical and1 < ν1 ≤ ǫ2 = pm ≤ ph.Then Theorem 7.62 together with Remark 7.66 shows that

F (X0,X1,X2) = Zpm

0 X0 + Zpm

1 X1 + Zpm

2 X2

with homogeneous polynomialsZ0, Z1, Z2 ∈ Fq[X0,X1,X2] of the same degreed andn = dpm + 1. On the other hand, asF is Frobenius non-classical, from the

286

proof of Theorem 8.52 there is a homogeneous polynomialG ∈ K[X0,X1,X2]such that

GF = Z0Xph−m

0 + Z1Xph−m

1 + Z2Xph−m

2 .

So,degF ≤ deg(GF ). Therefore,n ≤ d + ph−m whence the upper bound onnin (i) follows.

Eliminatingn from the first inequality in (i) givesq′ ≤ √q. If equality holds,thend = 1, and a direct computation shows that this implies(ii). 2

Now, it is shown that (ii) remains true for singular curves oflow degree.

THEOREM 8.74 If F be an irreducible plane curve of degreen defined overFq,with q square such that

(i) F is Frobenius non-classical withν =√q,

(ii) n < 2√q andF is not rational,

then(ii) in Theorem 8.73 holds.

Proof. The idea is to consider the dual curveD of F . First it is shown thatD andF are birationally equivalent overFq. The second step consists in proving thatDis a non-singular plane curve which is still Frobenius non-classical with the sameν1 = q′. After that, the assertion follows from (ii) of Theorem 8.73.

From Remark 8.50, eitherq = 4 andF is classical, orF is not only Frobeniusnon-classical but also non-classical with order sequence(0, 1,

√q). In the former

case,F is a non-singular plane cubic curve, and the assertion follows from (ii) ofTheorem 8.73.

Therefore,F may be assumed non-classical with order sequence(0, 1,√q). If

Q = (x, y) is a generic point ofF , from (7.17) and (8.16),

z√

q0 + z

√q

1 x+ z√

q2 y=0, (8.34)

z0 + z1x√

q + z2y√

q =0, (8.35)

where at least one of the elementszi is a separable variable. Letγ be a branch ofF centred at the pointA = (a, b) ∈ F , corresponding to the placeP of K(F). IfordPzj = min ordPzi with i ranging over1, 2, 3, then, by (8.34), the tangent toγis

v(m0(a, b)X0 +m1(a, b)X1 +m2(a, b)X2)

Using both hypothesesn < 2√q andg > 0, it is seen thatF is not a strange curve.

Thereforez0, z1, z2 are linearly independent overK.The dual curveD ofF is the image curve ofF under the rational transformation,

ω : x′0 = z0, x′1 = x1, x

′2 = z2,

defined overFq. Let n′ denote the degree ofD. Suppose thatω is not birational.Then the linear seriesgn′

2 consisting of all divisors,

Ac = div (c0z0 + c1z1 + c2z2) +B,


whereB =∑ePP andc = (c0, c1, c2) ∈ PG(2,K), is composed of an involu-

tion. Therefore, for infinitely many pointsA ∈ F , the tangentℓ toF atA is tangentto F at another pointA′, as well. Since bothI(A,F ∩ ℓ) andI(A,F ∩ ℓ) are atleast

√q, this implies thatn ≥ 2

√q, a contradiction.

Now, it is shown that every branchγ′ of D is linear. Take a primitive representa-tion σ of P and letσ(x) = x(t), σ(y) = y(t) wherex(t), y(t) ∈ K[[t]]. Then thebranchγ has the primitive representation:

x(t) = a+ a1t+ . . .+ aktk + . . . ,

y(t) = b+ b1t+ . . .+ bktk + . . . .

(8.36)

Put zi(t) = σ(zi) andσ(mi) = mi(t). Then a homogeneous primitive repre-sentation of the branchγ′ of D associated toP is (m0(t),m1(t),m2(t)). From(8.34),

z0(t)√

q + z1(t)√

qx(t) + z2(t)√

qy(t) = 0,

whence

m0(t)√

q +m1(t)√

qx(t) +m2(t)√

qy(t) = 0. (8.37)

When the formal power seriesmi(t) is written in the form,

mi(t) = µ(0)i + µ

(1)i t+ . . .+ µ

(k)i tk + . . . ,

for i = 0, 1, 2, then (8.37) gives the following:

[(µ(0)0 )

√q + (µ

(0)1 )

√qx(t) + (µ

(0)2 )

√qy(t)]

+ t√

q[(µ(1)0 )

√q + (µ

(1)1 )

√qx(t) + (µ

(1)2 )

√qy(t)]

+ . . .

+ tk√

q[(µ(k)0 )

√q + (µ

(k)1 )

√qx(t) + (µ

(k)2 )

√qy(t)]

+ . . . = 0;

(8.38)

here((µ

(0)0 )

√q, (µ

(0)1 )

√q, (µ

(0)2 )

√q)

is the centre of the branchγ′. This shows that

the lineℓ, where

ℓ = v((µ(0)0 )

√qX0 + (µ

(0)1 )

√qX1 + (µ

(0)2 )

√qX2), (8.39)

meetsF in P with multiplicity at least√q; that is,I(P, ℓ ∩ F) ≥ √q.

If there were another branch ofD with the same centre, thenI(R, ℓ ∩ F) ≥ √qwould hold for another branchR of F , contradicting the hypothesis thatn < 2

√q.

Now suppose thatγ′ is a non-linear branch ofD. Thenµ(1)i = 0 for i = 0, 1, 2,

and equation (8.38) implies that

[(µ(0)0 )

√q + (µ

(0)1 )

√qx(t) + (µ

(0)2 )

√qy(t)]

+ t2√

q[(µ(2)0 )

√q + (µ

(2)1 )

√qx(t) + (µ

(2)2 )

√qy(t)]

+ . . .

+ tk√

q[(µ(k)0 )

√q + (µ

(k)1 )

√qx(t) + (µ

(k)2 )

√qy(t)]

+ . . . = 0.

(8.40)

This shows that the lineℓ given by (8.39) meetsF atP with multiplicity at least2√q. But this is impossible, again by the hypothesisn < 2

√q. Hence,D has

288

only linear branches and just one is centred at each of point.Therefore,D is anon-singular plane curve.

To prove thatD is non-classical, (8.35) is used. From (8.35),

m0(t) +m1(t)x(t)√

q +m2(t)y(t)√

q = 0, (8.41)

whence

m0(t) +m1(t)x√

q0 +m2(t)y

√q

0 + t√

qG(t) = 0. (8.42)

This shows that the lineℓ = v(X0 + a√

qX1 + b√

qX2) is the tangent toD at thecentre ofγ′, and thatj2(P) ≥ √q. ThusD is non-classical. Since eitherx or y isa separable variable ofK(F), from Theorem 7.62(iii) it follows that the orders ofD are0, 1,

√q. Also, ℓ passes through the point which is the image of the centre of

γ′ under the Frobenius collineation. This follows from (8.34).By (ii) of Theorem 8.73,D is the Hermitian curveHq as in (ii), up to a change

of the coordinate system overFq. Since the dual curve of the Hermitian curvecoincides with itself,F itself is birationally equivalent to the Hermitian curve, bymeans of a birational transformation defined overFq.

To finish the proof, it suffices to show thatF is a non-singular plane curve. From(8.42) it follows thatF has only linear branches. In fact, if the branchγ were notlinear, then (8.42) would imply that

ordP [m0(t) + ξ0m1(t) + η0m2(t)] ≥ 2√q;

that is, the lineℓ = v(X0 + ξ0X1 + η0X2) would have intersection multiplicityat least2

√q with the branchγ′ of D, contradicting thatdegD =

√q + 1. If the

point (1, aq, bq) were also the centre of another branch ofF , thenℓ would be abitangent ofD, which would therefore have degree at least2

√q, a contradiction

with degD =√q + 1. Therefore, each point ofF is the centre of a single linear

branch. ThusF is non-singular, and this completes the proof. 2

Since a non-singular pointP of a classical curveF is an inflexion if and only ifj2(P) 6= 2j1(P) whereP is the place associated to the unique branch centred atP , the classical idea to distinguish inflexions from other non-singular points can beextended to branches. For branches centred at points ofPG(2, q), this produces apartition ofF(Fq)

∗ into two subsets

S1 = γ ∈ F(Fq)∗ | j2(P) = 2j1(P),

S2 = γ ∈ F(Fq)∗ | j2(P) 6= 2j1(P), (8.43)

whereP denotes the place arising from the branchγ. This gives rise to importantnumbers,Mq andM ′

q:

DEFINITION 8.75 (i) Mq is the number of branchesγ ∈ S1 each one countedj1(P) times;

(ii) M ′q is the number of branchesγ ∈ S2, each one countedj1(P) times.

Here,Bq = Mq + M ′q. As stated out in Theorems 8.9, 8.10 and Remark 8.11,F

may have branches centred at points inPG(2, q) that are notFq-rational branches


of F . In other words,Bq ≥ Sq but it is not necessary thatBq = Sq. Estimates forBq are obtained in Section 8.8.

For any Frobenius classical curve, Proposition 8.72 implies thatMq = 0, andhenceSq = M ′

q. On the other hand, ifF is classical and has degreen thenM ′q ≤

3n(n− 2).As shown in Section 8.7, for Frobenius non-classical curveswith respect to con-

ics, it is easier to estimate2Mq+M ′q rather thanSq. ThenSq can be estimated from

that bound and some bound onM ′q. Also, if F is classical, thenM ′

q ≤ 3n(n − 2)can be used. Actually, the latter case occurs whenq ≥ 25, n ≤ √q and

2Mq +M ′q ≥ n(q −√q + 1). (8.44)

This is a consequence of the following result.

THEOREM 8.76 LetF be an irreducible plane curve of degreen and defined overFq with q ≥ 25. If n ≤ √q + 1 and (8.44) holds, thenF is either classical orprojectively equivalent to the Hermitian curve

Hq = v(X√

q+10 +X

√q+1

1 +X√

q+12 ).

Proof. Suppose on the contrary thatF is non-classical. Then Corollary 7.57 im-plies that

j1(P) j2(P)(j2(P)− j1(P)) ≡ 0 (mod p), (8.45)

for any branchγ of F centred at a pointP of PG(2,K).First, the case whereF is Frobenius classical is investigated. From Theorem

8.53,

vP(S) ≥ (j2(P)− 1) + j1(P).

If γ ∈ S2, this implies thatvP(S) ≥ 2j1(P). If γ ∈ S1, it follows from (8.45) thatj1(P) ≡ 0 (mod p), which together with Theorem 8.53(ii) gives

vP(S) > j2P + j1(P)− 1 ≥ 3vP(S).

Therefore,

(2g − 2) + (q + 2)n = degS ≥ 3Mq + 2M ′q ≥ 3

2 (2Mq +M ′q) . (8.46)

Sincen(n− 3) ≥ 2g − 2, it follows from (8.46) and 8.44) that

(n− 3) + (q + 2) ≥ 32 (q −√q + 1);

that is,2n ≥ (q− 3√q+5). But this cannot actually occur when bothn ≤ √q+1

andq ≥ 25 hold. 2

THEOREM 8.77 For a polynomialf(X) ∈ Fq[X] having some simple root, letF = v(Y n − f(X)) with p 6 |n be an irreducible plane curve. Ifp > 2, then thefollowing hold.

(i) F is non-classical if and only ifn ≡ 1 (mod p) and there exist polynomialsg(X) andh(X) in K[X] with deg g(X) ≥ deg h(X) such that

f(X) = g(X)pX + h(X)p. (8.47)

290

(ii) F is Frobenius non-classical if and only if

g(X)pXq + h(X)p = (g(X)pX + h(X)p)psm′

, (8.48)

wherem′ is given by the system,n = psm+ 1, m 6≡ 0 (mod p),q − 1 = (psm′ − 1)(ps +m), m′ 6≡ 0 (mod p).

(8.49)

(iii) If F is Frobenius non-classical, thendeg f(X) = n. If, in addition, themultiplicities of the roots off(X) are relatively prime top, then all roots off(X) belong toFq.

Proof. First,K(F) = K(x, y) with yn−f(x) = 0. Here,x is a separable variableof K(F) by the hypothesisp 6 |n. The non-classicality ofF is expressed by theconditionD(2)

x y = 0. By (5.16) and Lemma 5.70, differentiatingyn − f(x) twicewith respect tox gives

2nyn−1D(2)x y +

(n− 1)f ′(x)2

nf(x)= f ′′(x),

wheref ′(x) is f ′(X) evaluated atx, and similarly forf ′′(x). So,D(2)x y = 0 is

equivalent to that(n− 1)f ′(x)2 = nf ′′(x)f(x). Calculating both sides at a simpleroot of f(X), this equation holds if and only ifn ≡ 1 (mod p) andf ′′(x) = 0.Sincex is not a constant,f ′′(x) = 0 impliesf ′′(X) = 0. Write f(X) =

∑aiX

i.

Now, f ′′(X) = 0 occurs if and only if everyai with i 6≡ 0, 1 (mod p) vanishes.Hence,f(X) = g(X)pX + h(X)p with g(X), h(X) ∈ Fq[X]. Further,deg f(X)is not divisible byp if and only if deg g(X) ≥ deg h(X). Therefore, (i) is proved.

To prove (ii), the following hypotheses may be assumed:

Y n = g(X)pX + h(X)p;n ≡ 1 (mod p);deg g(X) ≥ deg h(X).

(8.50)

Using the first equation, the Frobenius classicality ofF expressed by

yq − y = (xq − x)D(1)x y

reads

yq−1+n = g(x)px+ h(x)p, (8.51)

So, if F is Frobenius non-classical, thenyq−1+n ∈ K(x). No power ofy withexponent smaller thann is in K(x), since[K(F) : K(x)] = n. Hencen dividesq− 1 + n, and soq− 1 = λn for a positive integerλ. Sincen ≡ 1 (mod p), thereis a positive integerm such thatn = psm+1 with s a positive integer not divisibleby p. Thus,λ ≡ −1 (mod p). So,λ = prm′ − 1 with m′ a positive integer notdivisible by p. Hence,q = psrmm′ + prm′ − psm. From this, it follows thatr = s, otherwise eitherm orm′ would be divisible byp. Thenm ≡ m′ (mod p).Therefore,

yq−1+n = (yn)psm′

= (g(x)px+ h(x)p)psm′

, (8.52)


whence

g(x)pxq + h(x)p = (g(x)px+ h(x)p)psm′

.

Sincex is not a constant, this shows (8.48).Conversely, if (8.48) holds, thenF is Frobenius non-classical. This completes

the proof of (ii).Comparison of degrees in (8.48) gives

p deg g(X) + q = (p deg g(X) + 1)psm′.

Takingq − 1 = (psm′ − 1)(ps + 1) into account, it followsdeg g(X) = ps−1m.Thus,deg f(X) = p deg g(X) + 1 = psm+ 1 = n, which proves (iii). 2

Note that, writingf(X) = a0Xn + a1X

n−1 + . . . + an, a change of the in-determinateX of type X ′ = aX + b can be made witha, b ∈ Fq such thatf ′(X) = f(aX+b) has no term of degreen−1. Therefore,deg g(X) > deg h(X)is assumed without loss of generality in Theorem 8.77.

Now, the casesq = p2 or q = p3 in Theorem 8.77 are considered more closely.Assume that both (8.47) and (8.48) hold. Ifq = p2, thens = m = m′ = 1 canonly occur in (8.49). Also,deg g(X) = ps−1m = 1. Henceh(X) is a constantpolynomial. It turns out thatF is projectively equivalent overFq to the HermitiancurveHsqq.

If q = p3, there are examples which are not Fermat curves. Since

p3 − 1 = (p− 1) = (p(p+ 1) + 1),

from (8.49),s = m′ = 1, m = p+ 1, anddeg g(X) = p+ 1. Since it is assumedthatdeg g(X) > deg h(X), so (8.48) gives

g(X)Xp2

+ h(X) = g(X)p + h(X)p.

By comparison of coefficients,

g(X) = αXp+1 + βp2

, h(X) = βpXp + βX + γ, (8.53)

with α, β ∈ Fp, α 6= 0, andγ ∈ Fp2 . For instance,α = β = 1 andγ = 0 is agood choice, giving rise to the Frobenius non-classical curveF with equation

Y p2+p+1 = (Xp+1 + 1)pX + (Xp +X)p. (8.54)

It may be noted that this curve is non-singular. By Theorem 8.65, the numberSq

of its Fq-rational places is equal to(p2 + p+ 1)(p3 − p2 − p+ 1).

REMARK 8.78 Let q = p3. Up to birational equivalence overFq, the curveFfrom (8.54) is the only Frobenius non-classical curve of theform v(Y n − f(X))overFq. This follows from the fact that (8.49) is only possible whens = m′ = 1andm = p+ 1.

8.7 FROBENIUS NON-CLASSICALITY WITH RESPECT TO CONICS

This section continues the study of irreducible curvesF which are non-classicalwith respect to the linear seriesL2 cut out by conics. TheL2-orders are assumed to

292

be0, 1, 2, 3, 4, pm with p > 2. In particular,F is classical with respect toL1. Thenotation is that of Section 7.9. In addition,F is defined overFq with pm < q = ph.

The hypothesis onF to be Frobenius non-classical is expressed by the existenceof a polynomialt(X,Y ) ∈ K[X,Y ] such that

t(X,Y )f(X,Y ) =z0(X,Y )pm

+ z1(X,Y )pm

Xq + z2(X,Y )pm

Y q

+z3(X,Y )pm

X2q + z4(X,Y )pm

XqY q + z5(X,Y )pm

Y 2q.(8.55)

Equivalently,

vpm

0 + vpm

1 xq + vpm

2 yq + vpm

3 x2q + vpm

4 xqyq + vpm

5 y2q = 0. (8.56)

As in the proof of Theorem 8.73, in (8.56)h ≥ m. Then

v0 +v1xph−m

+v2yph−m

+v3x2(ph−m) +v4(xy)

ph−m

+v5y2(ph−m) = 0. (8.57)

In terms of osculating conics, as in Lemma 7.70, Frobenius non-classicality isexpressed in the following theorem.

THEOREM 8.79 If L2 is Frobenius classical, then the conicC0 passes throughthe imageAq = (1, aq, bq) of the centreA = (1, a, b) of γ under the Frobeniuscollineation. In addition, ifA ∈ PG(2, q), thenC0 meetsγ with multiplicity greaterthanpm.

Proof. Takeξ, η in K[[t]] such thatξ(tq) = x(t)q andη(tq) = y(t)q. From (8.56),

[(µ(0)0 )pm

+ (µ(0)1 )pm

ξ(tq) + (µ(0)2 )pm

η(tq) + (µ(0)3 )pm

ξ(tq)2

+(µ(0)4 )pm

ξ(tq)η(tq) + (µ(0)5 )pm

η(tq)2]+

tpm

[(µ(1)0 )pm

+ (µ(1)1 )pm

ξ(tq) + (µ(1)2 )pm

η(tq) + (µ(1)3 )pm

ξ(tq)2+

+(µ(1)4 )pm

ξ(tq)η(tq) + (µ(1)5 )pm

η(tq)2] + . . .

+ tkpv

[(µ(k)0 )pm

+ (µ(k)1 )pm

ξ(tq) + (µ(k)2 )pm

η(tq) + (µ(k)3 )pm

ξ(tq)2

+(µ(k)4 )pm

ξ(tq)η(tq) + (µ(k)5 )pm

η(tq)2] + . . . = 0.

Sincex(t) = a+ tG(t) andy(t) = b+ tH(t), it follows that

ξ(tq) = aq + tqG1(t), η(tq) = bq + tqH1(t).

Hence

(µ(0)0 )pm

+ (µ(0)1 )pm

aq + (µ(0)2 )pm

bq+

(µ(0)3 )pm

(aq)2

+ (µ(0)4 )pm

aqbq + (µ(0)5 )pm

(bq)2

= 0.

This shows that thatC0 passes through the pointAq = (1, aq, bq). Also, sincepm < q,

(µ(1)0 )pm

+ (µ(1)1 )pm

aq + (µ(1)2 )pm

bq)+

(µ(1)3 )pm

(aq)2

+ (µ(1)4 )pm

aqbq + (µ(1)5 )pm

(bq)2

= 0.

If aq = a, bq = b, then this equation iss1(a, b) = 0. By Theorem 7.70(ii), thesecond assertion follows. 2


For the rest of this section,F is a Frobenius non-classical curve with Frobeniusorders0, 1, 2, 3, pm.

The Stohr–Voloch theorem applied toL2 reads

Sq ≤ 15(6 + pm)(2g − 2) + (q + 5)2n. (8.58)

The aim is to find other good upper bounds on the numberSq of Fq-rationalbranches of such a curve. The idea is to exploit Frobenius classicality with respectto conics to compute the possible(L1,P)-orders, rather than countingFq-rationalbranches only. If there are only a few possibilities for the(L1,P)-orders, and theweightsvP(S) can be calculated using the previous numerical formulas, this ideacan work effectively, and the Stohr–Voloch divisor arising from the linear seriesL1

cut out by lines can bring about an improvement to (8.58).By Theorem 7.72, the hypothesisn < pm ensures that only three cases can occur

for (L1,P)-orders, and in each casej2 is calculated fromj1. Now, these three casesare investigated further.

PROPOSITION 8.80 Let n < pm. If (7.33) holds and the centreA = (1, a, b) ofγ is not inPG(2, q), then the Frobenius imageAq = (1, aq, bq) does not lie on thetangent toγ; also, (a− aq)v21 − (b− bq)v11 6= 0.

Proof. By Theorem 8.79,C0 passes throughAq = (1, aq, bq). On the other hand,when (7.33) holds,C0 is a non-degenerate conic whose tangent line atA = (1, a, b)coincides with the tangent ofγ. 2

PROPOSITION 8.81 Letn < pm. If (7.34) holds and the centreA = (1, a, b) of γis not inPG(2, q), then the Frobenius imageAq = (1, aq, bq) lies on the tangent toγ; also, (a− aq)v21 − (b− bq)v11 = 0.

Proof. If (7.34) holds, thenC0 is a degenerate conic and consists of the tangent lineof γ counted twice. Thus, the assertion is an immediate consequence of Theorem8.79. 2

PROPOSITION 8.82 If (7.35) holds, then the centre ofγ is not inPG(2, q).

Proof. In this case,C0 splits into two distinct lines, one of which coincides with thetangent ofγ. This shows thatC0 meetsγ with multiplicity j1 + j2 = pm. Hence,the result follows from the second assertion in Theorem 8.79. 2

Now, it is shown that Frobenius non-classicality sharply limits the possible val-ues ofj1. In particular, ifn < pm < q, thenj1 does not exceed3.

In the following theorem,Z is the curve introduced in Section 7.9,P is thebranch ofZ associated to a branchγ of F , andP is the corresponding place.

THEOREM 8.83 There is a hyperplaneH in PG(5,K) such that

I(P,Z ∩H) ≥ j1ph−m.

Proof. Without loss of generality, letγ be centred at an affine pointA = (1, a, b).Let α, β ∈ K such thata = αph−m

andb = βph−m

. The intersection multiplicityof Z with the hyperplane

H = v(X0 + αX1 + βX2 + α2X3 + αβX4 + β2X5) (8.59)

294

atP is

I(P,Z ∩H) = ordP(v0 + αv1 + βv2 + α2v3 + αβv4 + β2v5).

Puttingξ = x− a, η = y − b givesj1 = minordPξ, ordPη. From (8.57),

v0 + αv1 + βv2 + α2v3 + αβv4 + β2v5+ξph−m

v1 + ηph−m

v2 + ξ2(ph−m)v3 + (ξη)ph−m

v4 + η2(ph−m)v5 = 0.

whence

ordP(v0 + αv1 + βv2 + α2v3 + αβv4 + β2v5) ≥ j1ph−m,

which proves the result. 2

It is worth mentioning the following result that is valid forsingular curves.

PROPOSITION 8.84 If F is singular, thendegZ ≥ 2ph−m.

Proof. Let A = (1, a, b) be a singular point ofF , and letH be the hyperplane(8.59). IfA is the centre of only one branch ofF , it must have order greater than1, and the assertion follows from Proposition 8.83.

Suppose thatA is the centre of two branchesγ1 andγ2 of F . The correspondingbranchesγ′1 andγ′2 of Z are distinct. Therefore,

I(P,Z ∩H) ≥ I(P, γ′1 ∩H) + I(P, γ′2 ∩H).

By Proposition 8.83,I(P,Z ∩H) ≥ 2ph−m. 2

If n < ph−m also holds, then Proposition 8.83 together with Theorem 7.72 givesthe following sharp bound forj1.

PROPOSITION 8.85 If n ≤ minpm, ph−m, thenj1 ≤ 3.

From this, an upper bound onSq is obtained after giving an explicit formula forvP(S) that actually holds without any limitation onnwith respect topm andph−m.

SincevP(S) remains invariant under the change of coordinates (7.27) only whenall coefficients are fromFq, it is necessary to know howvP(S) changes in generalunder (7.27). For this, take a local parameterζ atP, and adopt the usual simplestnotationu′ in place ofDζ(u) for u ∈ K(F). Since

(x− xq)y′ − (y − yq)x′

= [a− aq + v11x+ v12y − (v11x+ v12y)q][v21x

′ + v22y′]

−[b− bq + v21x+ v22y − (v21x+ v22y)q][v11x

′ + v12y′]

= [(a− aq)v21 − (b− bq)v11]x′ + [(a− aq)v22 − (b− bq)v12]y′+(v11v22 − v12v21)(xy′ − yx′)− (v11x+ v12y)

q(v21x′ + v22y

′)

+(v21x+ v22y)q(v11x

′ + v12y′),

it follows thatvP(S) = ordP x′ +

ordP[(a− aq)v21 − (b− bq)v11]+ [(a− aq)v22 − (b− bq)v12]y′/x′+ (v11v22 − v12v21)(xy′/x′ − y)− (v11x+ v12y)

q(v21 + v22y′/x′)

+ (v21x+ v22y)q(v11 + v12y/x

′).

(8.60)


THEOREM 8.86 Letn ≤ minpm − 4, ph−m.

(a) If j1 6= 3 for every branch ofF , then

Sq ≤(q − 1)n− 2(2g − 2)

2. (8.61)

(b) If p > 3, then

Sq <(q − 1)n− 2(2g − 2) + 12n

2. (8.62)

Proof. The idea is to computevP(S)− vP(R). By Proposition 8.85,

vP = j1 + j2 − 3

with just one exception:p = j1 = 3. Apart from this exception, in each of thethree cases in Theorem 7.72,j2 6= 0 (mod p). From (8.60),

(i) if a, b ∈ Fq, thenvP(S) ≥ j1 + j2 − 1;

(ii) if (a − aq)v21 − (b − bq)v11 = 0, but eithera 6∈ Fq or b 6∈ Fq, thenvP(S) = j2 − 1;

(iii) if (a− aq)v21 − (b− bq)v11 6= 0, thenvP(S) = j1 − 1.

The hypothesisn ≤ pm − 4 rules out case(iii). Hence, ifp = j1 = 3 does notoccur, then

vP(S)− vP(R) ≥

2 if (i) holds;−2 if (ii) holds andj1 = 3;0 if (ii) holds andj1 ≤ 2;

(8.63)

Now, sum this inequality asγ ranges over all branches ofF . Since

ordS=(2g − 2) + (q + 2)n,

ordR=3(2g − 2) + 3n,

this gives

2Sq ≤ (q − 1)n− 2(2g − 2) + 2N,

whereN denotes the number branchesγ satisfying (7.34) withj1 = 3.Assume thatN 6= 0. SincevP(R) = j2 = 1

2 (pm + 3) for anyγ counted inN ,an upper bound onN is

N ≤ 6(2g − 2) + 6n

pm + 3< 6n.

This completes the proof. 2

THEOREM 8.87 LetF be non-singular. Ifn < pm − 1, then

Sq =n(q − 2n+ 5)− I

2, (8.64)

whereI is the number of all non-Fq-rational inflexion points ofF .

296

Proof. The arguments in the proof of Theorem 8.86 can be used, but twomorepieces are needed.

First,vP(S) = j2 when(i) holds. To prove this, show that

vP(S) = ordP(xy′/x′ − y) = j2.

Sincej1 = 1, sox is a local parameter atP. Hence, there is a primitive represen-tationσ of P such that

σ(x) = t, σ(y) = ctj2 + . . .

with c 6= 0. Then

σ(xy′/x′ − y) = c(j2 − 1)tj2 + . . . .

Sincej2 = 12 (pm + 1) when (i) holds, the coefficientc(j2 − 1) 6= 0. Hence

ordt (σ(xy′/x′ − y)) = j2, and the assertion is proved.Secondly, (iii) does not occur by the hypothesisn < pm − 1. Therefore,

vP(S)− vP(R) =

2 if (i) holds;1 if (ii) holds;0 otherwise.

2

EXAMPLE 8.88 The Fermat curveF in Example 7.77 is Frobenius non-classicalwith respect to conics whose inflexion points are inPG(2, q). Hence (8.64) applieswith n = 1

2 (√q + 1), I = 0 giving

Sq = q + 1 + 14 (√q − 1)(

√q − 3)

√q.

Theorem 8.86 together with (8.58) gives the following result.

THEOREM 8.89 Letn ≤ minpm − 4, ph−m.

(i) If j1 6= 3 for every branch ofF , then

Sq ≤(6 + pm)(q − 1)n+ (q + 5)4n

22 + 2pm. (8.65)

(ii) If p > 3, then

Sq ≤(6 + pm)(q − 1)n+ (q + 3pm + 23)4n

22 + 2pm. (8.66)

REMARK 8.90 Let q ≥ 49 be a square power ofp with p > 2. The irreducibleplane curveG = v(g(X0,X1,X2)), with

g(X0,X1,X2) =X2√

q0 X2

1 +X21X

2√

q2 +X2

0X2√

q1

−2(X0X√

q+11 X

√q

2 +X√

q+10 X

√q

1 X2 +X√

q0 X1X

√q+1

2 ),

is classical with respect to lines, but non-classical with respect to conics, since theL2-orders are0, 1, 2, 3, 4,

√q. In fact, (7.22) holds with

h = 1, z0 = X2, z1 = −2XY, z2 = −2X, z3 = Y 2, z4 = −2Y, z5 = 1.


Since (8.55) does not hold,G is Frobenius classical with respect to conics. Also,G has three singular points, namely the vertices of the fundamental triangle. Eachof them is a double point and is the centre of only one branch. The(L1,P)-ordersof the corresponding placeP of such a branch is(0, 2, 2

√q). The genus ofG is

12 (q −√q). Now, the irreducible plane curveF = v(f(X0,X1,X2)) is defined tobe the image ofF under the projectivity ofPG(3,K) associated to the matrix,

a 1 aq+1

aq+1 a 11 aq+1 a

, (8.67)

wherea is an element of orderq2 + q+ 1 in the cubic extensionFq3 of Fq. Equiv-alently,

f(aX0 +X1 + aq+1X2, aq+1X0 + aX1 +X2,X0 + aq+1X1 +X2)

= g(X0,X1,X2).

Now, G andF have the sameL1- andL2-orders. In particular,F is classical withrespect to lines, but non-classical with respect to conics.Although the projectivity(8.67) is defined overFq3 , the curveF is defined overFq; this is not trivial. Itcannot be deduced thatF is also Frobenius classical with respect to conics.

Actually, this is false. In fact,F is the image of the Hermitian curve

Hq = v(X√

q+10 +X

√q+1

1 +X√

q+12 )

under a birational transformation defined overFq. In particular,Sq = q3 + 1. Thisis consistent with (8.58) only forpm =

√q, yielding that the last Frobenius order

is√q. This example also shows that the hypothesisn ≤ √q − 4 in Theorem 8.86

cannot be weakened ton ≤ 2√q + 2.

To obtain an upper bound onSq, begin with (7.43), evaluate ordPB(x, y) usingvP(S), and calculate ordS.

PROPOSITION 8.91 Letn < pm.

(i) If a, b ∈ Fq, then

ordPB ≤vP(S)− (3j1 − 1) if (7.33) holds,vP(S)− (j1 − 1) if (7.34) holds.

(8.68)

(ii) If eithera 6∈ Fq or b 6∈ Fq, then

ordPB ≤vP(S)− (j1 − 1) if (7.33) holds,vP(S) + 1 if (7.34) or (7.35) holds.

(8.69)

Proof. Let a, b ∈ Fq. From (8.60),vP(S) ≥ j1 + j2 − 1. By Proposition 8.82,either (7.33) or (7.34) occurs. In the former case, ordPB = 0 andj2 = 2j1; inthe latter case, ordPB = j2 and2j2 − j1 = pm. From this, (i) is obtained. Theargument is similar for (ii). 2

Summing up the inequalities in Proposition 8.91 gives∑

ordPB ≤∑vP(S)− 2Mq −M ′

q −∑ ′(r − 1) +N, (8.70)

298

whereMq andM ′q are as in Definition 8.75, while the summation

∑′ is over allbranches ofF of type (7.33), andN is the number branches of type (7.34) or (7.35).

An upper bound onN is

N ≤ 6(2g − 2 + d)/(pm − 3). (8.71)

To show (8.71), from the inequalityvP ≥ j2 + j1 − 3, it follows that∑

(j2 + j1 − 3) ≤∑vP(R) = 3(2g − 2) + 3n.

On the other hand,∑

(j2 + j1 − 3) = N0(pm − 3) +

∑′′

(pm + 3j1 − 6)/2,

whereN0 is the number of branches of type (7.33) while the summation∑′′ is

over theN −N0 branches of type (7.34). Therefore,∑

(j2 + j1 − 3) ≥ N0(pm − 3) + (N −N0)(p

m − 3)/2.

Hence,∑

(j2 + j1 − 3) ≥ 1212N(pm − 3)

and (8.71) follows.It should be noted that, since2g−2 ≤ n(n−3) andn ≤ pm−1, the expression

on the right of (8.71) is at most6n. Hence (8.70) can be put in a more manageablebut somewhat weaker form as follows:

∑ordPB ≤

∑vP(S)− 2Mq −M ′

q + 6n. (8.72)

This together with (7.44) gives

pm degZ < c+ (2g − 2) + [(q + 1)n− 2Mq −M ′q] + 6n. (8.73)

Also, from Theorem 8.83,

j1 ≤c+ (2g − 2) + [(q + 1)n− 2Mq −M ′

q] + 6n

q. (8.74)

Sincec ≤ n(n − 1) and2g − 2 ≤ n(n − 3), from (8.73) and (8.74), the follow-ing result is obtained. The second assertion together with Theorem 8.89 gives thefollowing upper bound onSq.

THEOREM 8.92 Letn < pm − 3 andpm ≤ √q. If (8.44) holds, then

Sq ≤(6 + pm)(q − 1)n+ (q + 5)4n

22 + 2pm.

PROPOSITION 8.93 Letn < pm. If (8.44) holds, then

pm degZ ≤ n(2n+√q + 2), (8.75)

and

j1 ≤ n(2n+√q + 2)/q. (8.76)


Now, the case thatq is a square andpm =√q, that is,

the order sequence is(0, 1, 2, 3, 4,

√q),

the Frobenius order sequence is(1, 2, 3,√q).

(8.77)

is closely investigated. Fromn ≤ √q − 1,

n(2n+√q + 2)/q ≤ 3− 3/

√q < 3.

By (8.76), this gives the following result

PROPOSITION 8.94 If n <√q and both (8.44) and (8.77) hold, then

(i) j1 = 1 for branches of type (7.34) whilej1 ≤ 2 for those of type (7.33) and(7.35);

(ii) for branches of type (7.35), there are only two cases:

(a) n =√q − 2 andj1 = 2;

(b) n =√q − 1 andj1 = 1.

It should be noted that the linearity of the branches of type (7.34) depends on thecondition thatp > 2. A first consequence of Proposition 8.94 is the followingresult.

PROPOSITION 8.95 Let τ1 andτ2 be the respective numbers of branches of type(7.34) and (7.35). Then

3 degZ = 2τ1 + τ2. (8.78)

Also, if

n <√q − 2, (8.79)

then3 degZ = 2τ1.

Proof. From Proposition 8.94,j1 ≤ 2 < p. So, by Corollary 7.57 the classicalformulavP(R) = j1 + j2 − 3 holds. Also, ordPdx = j1 − 1. From the discussionafter (7.43), the exact value of ordPB is known, namely

ordPB =

0 if the branch is of type (7.33),j2 if the branch is of type (7.34),j1 if the branch is of type (7.35).

This together with (7.42), (7.43) and Lemma 7.75 give (8.78). The second assertionis a consequence of (8.78) and (ii) of Proposition 8.94. 2

PROPOSITION 8.96 If each of the conditions (8.79), (8.44) and (8.77) is satisfied,then

degZ ≤ 2n < 2(√q − 2).

300

Proof. Sincen <√q − 2, it follows from Proposition 8.94 that no branch of type

(7.35) exists. Hence, for a branchγ,

3vP(S)− vP(R) =

6j1 if γ is of type (7.33) centred at a point inPG(2, q),0 if γ is of type (7.33) centred at a point not inPG(2, q),√

q + 3j1 if γ is of type (7.34) centred at a point inPG(2, q),√q if γ is of type (7.34) centred at a point not inPG(2, q).

Since∑3vP(S)− vP(R) = 3(2g − 2) + 3(q + 2)n− 3(2g − 2)− 3n = 3nq + 3n,

it follows that

3nq + 3n = 6Mq + 3M ′q + τ1

√q,

whence

2(nq − 2Mq −M ′q + n) =

√q degZ,

by Proposition 8.95 Now, from (8.44) the assertion follows. 2

PROPOSITION 8.97 If each of the conditions (8.79), (8.44) and (8.77) is satisfied,thenF is non-singular.

Proof. Suppose thatF has a singular point. Sinceph−m =√q, Proposition 8.84

implies thatdegZ ≥ 2√q; this contradicts Proposition 8.96. 2

THEOREM 8.98 If each of the conditions (8.79), (8.44) and (8.77) is satisfied, then

n = 12 (√q + 1),

and the number ofFq-rational points ofF is equal to

Sq = q + 1 + 14 (√q − 1)(

√q − 3)

√q.

Proof. As F is non-singular and classical with respect to lines, (7.20)applies.Since the inflexions ofF are exactly the points of type (7.34), it follows thatτ1 = 6n(n−2)/(

√q−3). Hence, by Proposition 8.95,degZ = n(n−2)/(

√q−3).

Therefore, the first assertion follows from Proposition 7.76. To prove the second itsuffices to show that every inflexion point ofF lies inPG(2, q) and apply Theorem8.87. Assume on the contrary thatF has an inflexion pointP outsidePG(2, q). If ℓis the tangent toF atP , then Theorem 7.74 shows thatℓ contains the Frobenius im-ageP q of P . SinceP 6= P q, Bezout’s theorem givesn = |F ∩ ℓ| ≥ 1

2 (√q+1)+1

contradictingn = 12 (√q + 1). 2

For the Fermat curveF = v(Xn + Y n + 1) with n = 12 (√q + 1), described in

Example 7.77,

Mq = 14 (√q + 1)(q −√q − 2) and M ′

q = 32 (√q + 1).

This shows that the conditions in Proposition 8.98 can be fulfilled. In fact, this Fer-mat curve is, up to projectivities inPG(2, q), the only non-singular plane curve ofdegreen defined overFq that satisfies both conditions in Proposition 8.98. There-fore, the following result holds.


THEOREM 8.99 If (8.79), (8.44) and (8.77) hold, thenn = 12 (√q + 1) andF is

the Fermat curve.

Theorem 8.99 shows that an irreducible plane curve defined over Fq is fullydetermined by three conditions, namely (8.79) on the degree, (8.44) on the numberof Fq-rational points, and (8.77) on the Frobenius order sequence. Actually, thefollowing proposition shows that the latter condition can be independent from theother two only for small values ofq.

PROPOSITION 8.100 Let q ≥ 25 be a power of an odd primep such that ifp = 3thenq is a square. LetG be an irreducible plane curve of degreen ≥ 3 and definedoverFq. If both conditions (8.44) and

n <

√q + 1 for q > 232, q 6= 36, 55,

23 for q = 36,

49 for q = 55,

min(q − 5√q + 45)/20, (q − 5

√q + 57)/24 for q ≤ 232.

(8.80)are satisfied, thenq is square and (8.77) holds.

Proof. By Theorem 8.76G is classical with respect to lines. Consequently, theL2-order sequence ofG is (0, 1, 2, 3, 4, ǫ), and FrobeniusL2-order sequence is(0, 1, 2, 3, ν) with eitherν = 4, or ν = ǫ.

To show thatν = ǫ, suppose on the contrary thatν = 4. Since

5Sq = 5(Mq +M ′q) ≥ 5

2 (2Mq +M ′q),

from (8.58),

10(n− 3) + (q + 5)n ≥ 52 (q −√q + 1),

whencen ≥ (q − 5√q + 45)/20, a contradiction.

Henceǫ = ν ≥ 5. If equality holds, then eitherp = 3, or p = 5. From Remark8.50,ǫ is a power ofp for p ≥ 5, apart from just one possibility forp = 3, namelywhenǫ = ν = 6. The latter case cannot actually occur under the hypothesis(8.80).In fact, forν = 6, (8.58) implies

12(n− 3) + (q + 5)2 ≥ 52 (q −√q + 1),

so thatn ≥ (q − 5√q + 57)/24 and henceq > 232 by (8.80). From

(q − 5√q + 57)/24 ≤ n ≤ √q,

it follows thatq − 29√q + 57 ≤ 0 whence529 < q < 722. This is a contradiction

as no power of3 lies in the interval[529, 722].To complete the proof it must be shown that neitherν >

√q, norn <

√q hold.

Suppose thatν >√q, that is,ν2 = peq with e ≥ 1. Sincep ≥ 3 andq is a

square forp = 3, this impliesν > 2√q. On the other hand,ǫ ≤ 2n < 2

√q, a

contradiction.Suppose thatν <

√q; that is,q = peν2, with e ≥ 1. Sincen(n− 3) ≥ 2g − 2,

n >√q, the hypothesis (8.44) together with (8.58) give

(6+ν)(√q−3)+(q+5)2 ≥ (6+ν)(n−3)+(q+5)2 ≥ 5

2 (q−√q+1). (8.81)

302

In particular,

n− 3 ≥ q − 5√q − 15

2(6 + ν). (8.82)

From (8.81) it follows that2ν√q − 6ν ≥ q − 17

√q + 21, whence

2ν − 6/√pe ≥ √q − 17 + 21/

√q.

Therefore,

17− 21√q− 6√

pe≥ ν(√pe − 2). (8.83)

Sinceν ≥ 5, this givespe < 36. As q is a square forp = 3 andν ≥ 5 is a powerof p, only five possibilities are left; namely,

(pe, ν, q) ∈ (9, 9, 729), (5, 5, 125), (5, 25, 3125), (7, 7, 343), 11, 11, 1331).But, for these, (8.82) implies that

n ≥ 23, 6, 49, 13, 37,

contradicting the hypothesis (8.80). 2

REMARK 8.101 If G is a non-singular plane curve defined overFq with q oddsquare such thatG has degreen = (

√q + 1)/2 andSq = q + 1 + n(n − 3)

√q

Fq-rational points, thenG is the Fermat curve, up to a projectivity overFq.

This result, together with Theorems 8.99 and 8.100 has the following corollary.

THEOREM 8.102 Letq ≥ 25 be a power of an odd primep such that ifp = 3 thenq is a square. LetG be an irreducible plane curve of degreen <

√q − 2 defined

overFq. If both (8.44) and (8.80) holds, thenn = 12 (√q + 1) andG is the Fermat

curve, up to a projectivity overFq.

REMARK 8.103 For the Fermat curve

Fn = v(Xn + Y n + 1))

of degreen with n 6≡ 0 (mod p), viewed as an irreducible curve overFq, thevalues ofn for whichFn is Frobenius non-classical with respect toL2 have beendetermined.

THEOREM 8.104 Let q = ph with p > 5. ThenFn is Frobenius non-classicalwith respect to conics if and only if one of the following holds:

(i) n− 1 ≡ 0 (mod p), andFn is Frobenius classical with respect to lines;

(ii) n − 2 ≡ 0 (mod p), andn = 2(ph − 1)/(pr − 1) with a divisorr of m,r < h;

(iii) 2n−1 ≡ 0 (mod p), andn = (ph−1)/2(pr−1) with a divisorr ofh suchthath/r is even.


The Stohr–Voloch theorem in its stronger form (8.18) can be applied toFn toshow that

Sq ≤ (M − 1)n(n− 3)

2+

1

Msn(q +M)− 3n(A−B),

where

A = 16 [(n− s− 1)s(s− 1)(s+ 4) + 1

4s(s− 1)(s− 2)(s+ 5)],M =(s+2

s

)− 1,

with 1 ≤ s ≤ n− 3, sn ≤ p, andB = sn−M .

REMARK 8.105 The numberNqk of Fqk -rational points ofFn when

N = (qk−1 + . . .+ q + 1)/n

is an integer andp is sufficiently large with respect toN is as follows:

Nqk = n2[(q − 2) + (d− 1)(d− 2)] + 3n,

whered = (N/n, t+ 1), q ≡ t (mod N/n) with 0 < t < N/n, and

p >

(2

t+1√

sin(nπ/2N))+ 1

)(t−1)(N/n−d)

.

8.8 SINGULAR PLANE CURVES DEFINED OVER Fq

Let f ∈ Fq[X,Y ] be an irreducible polynomial of degreen, and letF = v(f)be the corresponding irreducible, possibly singular, plane curve. The problem ofcounting the numberRq of points inPG(2, q) which lie onF is of interest notonly in the present but also in other contexts. Coding theory, cyclotomy, finitegeometry, graph colourings and Waring’s problem are just some of these. NotethatRq counts the solutions of the equationf(X,Y ) = 0 in Fq × Fq togetherwith the homogeneous non-zero solutions ofΦ(X,Y ) = 0, whereΦ(X,Y ) is thehomogeneous polynomial of all terms of degreen in f(X,Y ).

It is natural to compareRq with Sq, the number ofFq-rational branch points. IfBq is the number of branches ofF centred at points ofPG(2, q), andRq is similartoRq but counts eachr-fold point counted inRq with multiplicity r, then

Sq ≤ Bq ≤ Rq.

This shows that the problems of determiningSq, Rq, Bq, Rq are equivalent onlywhenF is non-singular. Nevertheless, some results onRq, Bq, Rq are similar tothose onSq. In fact, the proof of Theorem 8.40 remains valid whenRq replacesRq. This also finds confirmation in Theorem 8.106, which extendsTheorems 8.40and 8.65 to singular plane curves.

To see, in a simple case, the differences betweenRq, Sq, Bq, Rq, consider thethree singular plane cubics,N2,N1,N0, with respectively two, one, and zero tan-gents overFq at the singular pointP ; these are cubics with a node, a cusp, and anisolated double point atP , respectively. Then Table 8.1 is straightforward to verify.It may be noted thatBq = Rq whenP is a node. This holds true for curves with

only ordinary singularities centred at points inPG(2, q).

304

Table 8.1 Numbers of points on singular cubics

Rq Sq Bq Rq

N2 q q + 1 q + 1 q + 1

N1 q + 1 q + 1 q + 1 q + 2

N0 q + 2 q + 1 q + 3 q + 3

THEOREM 8.106 Let F be an irreducible plane curve of degreen and genusgdefined overFq.

(i) If F is either classical, or non-classical but Frobenius classical, then

Bq ≤1

2(2g − 2) + (q + 2)n. (8.84)

(ii) If F is Frobenius non-classical with order sequence(0, q′), then

Bq ≤1

2q′(2g − 2) + (q + 2)n. (8.85)

Proof. First, (i) is proved. The hypothesis says that the Frobeniusorder sequenceof F is (0, 1). From(i) of (8.53), it follows thatvP(S) ≥ 2 for everyP ∈ F(Fq),that is, for everyFq-rational branchP of F . By (8.13), it suffices to extend thisresult to every branchP of F centred at a point inPG(2, q). This is done afterstating a useful formula forvP(S).

For a generic pointQ = (x, y) of F = f(X,Y ), letK(F) = K(x, y) be thefunction field ofF . Let γ be any branch ofF . Without loss of generality, supposethat its centreA is affine. Assume thatA = (1, a, b) lies in PG(2, q), and denoteby ℓ the tangent toγ. As vP(S) is invariant underFq-linear transformations,Amay be taken to be the origin, and it may be assumed that ordPx ≤ ordPy. Then

ℓ = v(m21X −m11Y ),

with m11(6= 0),m21 ∈ K.Now introduce a new coordinate system(X, Y ) so that the tangentℓ becomes

theX-axis. This change of coordinates from(X,Y ) to (X, Y ) is given by thelinear substitutions

X = m11X −m12Y , Y = m21X −m22Y .

Definex, y by

x = m11x−m12y, y = m21x−m22y.

ThenF(K) = K(x, y).


Let ζ be a local parameter atP. Now,

D(ν)ζ x=m11Dζt

(ν)x+m12D(ν)ζ y,

D(ν)ζ y=m21D

(ν)ζ x+m22D

(ν)ζ y,

whence

(x− xq)D(ν)ζ y=(x− xq)(m21D

(ν)ζ x+m22D

(ν)ζ y),

(y − yq)D(ν)ζ x=(y − yq)(m11D

(ν)ζ x+m12D

(ν)ζ y).

Therefore

(x− xq)D(ν)ζ y − (y − yq)D

(ν)ζ x

=(x− xq) (m21D(ν)ζ x+m22D

(ν)ζ y)− (y − yq)(m11D

(ν)ζ x+m12D

(ν)ζ y)

= [m11x+m12y − (m11x+m12y)q] (m21D

(ν)ζ x+m22D

(ν)ζ y)

− [m21x−m22y − (m21x−m22y)q] (m11D

(ν)ζ x+m12D

(ν)ζ y)

= x[(m11m21 −m21m11)D(ν)ζ x+ (m11m22 −m21m12)D

(ν)ζ y]

+y[(m12m21 −m22m11)D(ν)ζ x+ (m12m22 −m22m12)D

(ν)ζ y] + . . .

=(m11m22 −m21m12)[xD(ν)ζ y − yD(ν)

ζ x] + . . . ,

where each omitted expression has order at leastrq atP. Hence, ifvP(S) < j1q,where the(L1,P)-orders are0, j1, j2, then

vP(S) ≥ ordt (xD(ν)ζ y − yD(ν)

ζ x). (8.86)

Now, it is shown that if the branchγ of F is centred at a point inPG(2, q), then

vP(S) ≥ 2j1. (8.87)

Assume first thatν1 = 1. SoD(ν)ζ y = nDζy, and ordPD

(ν)ζ y ≥ j2 − 1. From

(8.86),

vP(S) ≥ min2j1, j2 + j1 − 1.Sincej2 > j1, the assertion follows. Ifν = 2, then Proposition 8.53 gives theresult.

Suppose now thatF is Frobenius non-classical withν1 = ǫ = pm > 2. By(8.86), to prove the result, it suffices to show that ordP y ≥ pm + j1. To do this,write (7.17) and (8.16) in the form,

mpm

0 +mpm

1 x+mpm

2 y= 0, (8.88)

m0 +m1xq/pm

+m2yq/pm

= 0, (8.89)

wheremi = zi/zj with ordPzj = min ordPzi. Chooseµi ∈ K such thatordP(mi − µi) > 0. Then, one of theµi is not zero. From (8.89),µ0 = 0;more precisely, (8.89) implies that ordPm0 ≥ j1. Then eitherµ1 or µ2 is differentfrom zero, and it may be assumed without loss of generality thatµ1 6= 0. Also, ifµ2 were zero, then (8.88) would implyj1 = pmordPm0, while (8.89) would yieldj1q = pmordPm0. This contradiction shows that alsoµ2 6= 0.

306

Now, write (8.88) in the form

mpm

0 + (m1 − µ1)pm

x = −mpm

2 y − µpm

1 x.

As ordPmpm

0 ≥ j1pm and ordP((m1 − µ1)pm

x) ≥ pm + j1, this yields

ordP(−mpm

2 y − µpm

1 x) ≥ pm + j1.

Sinceµ2 6= 0, it follows that

ordP(−µpm

2 y − µpm

1 x) ≥ pm + j1.

This implies thatγ intersects the lineℓ = v(−µpm

2 Y − µpm

1 X) with multiplicityat leastpm + j1. Thusℓ is the tangent ofγ, which proves the assertion and hencecompletes the proof. 2

THEOREM 8.107 Let F be a non-classical irreducible plane curve of degreenand genusg defined overFq. If F is Frobenius non-classical, and has only tamebranches, then

Bq ≥ (q − 1)n− (2g − 2). (8.90)

Also, equality holds if and only if every non-linear branch ofF is centred at a pointin PG(2, q).

Proof. As in the proof of Theorem 8.65, the idea is to computevP(S)−vP(R) andapply (7.13) together with (8.13). Putν = ǫ = pm > 2. From (7.17) and (8.16),

zpm

1 (xq − x) + zpm

2 (yq − y) = 0,

zpm

1 D(1)ζ x+ zpm

2 D(1)ζ y=0.

These relations enable us to simplify both the Wronskian determinants,

det(D(ǫi)ζ xj) =

∣∣∣∣∣∣∣

1 x y

0 D(1)ζ x D

(1)ζ y

0 D(pm)ζ x D

(pm)ζ y

∣∣∣∣∣∣∣=

∣∣∣∣∣D

(1)ζ x D

(1)ζ y

D(pm)ζ x D

(pm)ζ y

∣∣∣∣∣ ,

det(D(νi)ζ xj) =

∣∣∣∣∣∣

1 x y1 xq yq

0 D(pm)ζ x D

(pm)ζ y

∣∣∣∣∣∣=

∣∣∣∣∣xq − x yq − yD

(pm)ζ x D

(pm)ζ y

∣∣∣∣∣ .

Multiplying the second column byzpm

2 and adding to itzpm

1 times the first column,the second column becomes

[0, zpm

1 D(pm)ζ x+ zpm

2 D(pm)ζ y]T

in both cases. Puttingu = zpm

1 D(pm)ζ x+ zpm

2 D(pm)ζ y, it follows that

vP(S) = ordP(xq − x) + ordPu− pmordPz2, (8.91)

vP(R)= ordP(D(1)ζ x) + ordPu− pmordPz2. (8.92)

Hence

vP(S)− vP(R) = ordP(xq − x)− ordP(D(1)ζ x). (8.93)


Up to a non-degenerate linear transformation

(X,Y ) 7→ (uX + vY,wX + zY ),

with u, v, w, z ∈ Fq, it may be supposed that the tangent to the branchγ is not thevertical line through the centre. Then

x(t) = a+ tj1 + . . . , y(t) = b+ bts + . . .

with s ≥ j1. Sinceγ is tame,p does not dividej1. Hence

j1 − 1 = ordP(D(1)ζ x) ≤ ordP(D

(1)ζ y).

Now ordP(xq − x) > 0 if and only if a ∈ Fq. If this is the case, then

ordP(xq − x) = j1,

and(1, a, b) ∈ PG(2, q).To prove this last assertion, it must be checked thata ∈ Fq implies b ∈ Fq.

SinceF is Frobenius non-classical,

(xq − x)D(1)ζ y − (yq − y)D(1)

ζ x = 0.

If b 6∈ Fq, then ordP(yq − y) = 0 and hence ordP(D(1)ζ x) > ordP(D

(1)ζ y); but

this contradicts ordP(D(1)ζ x) ≤ ordP(D

(1)ζ y).

Since ordP(xq − x) = 0 only if a 6∈ Fq, the following formula is obtained thatcan be viewed as an extension of (8.26) to singular branches:

ordP(xq − x) =

j1, if the centre ofγ is in PG(2, q);0, otherwise.

(8.94)

From (8.93) and ((8.94),

vP(S)− vP(R) =

1, if the centre ofγ is in PG(2, q);

−(j1 − 1), otherwise.

SincedegS − degR = (q − 1)n− (2g − 2), this implies that

Bq = (q − 1)n− (2g − 2) +∑

(r − 1),

where the sum is over all branchesγ of F whose centre(1, a, b) is not inPG(2, q).This completes the proof of Theorem 8.107. 2

EXAMPLE 8.108 (1) The Hermitian curveH√q = v(X

√q+1+Y

√q+1+1) attains

the upper bound (8.85) forq′ =√q, and the lower bound (8.90).

(2) Another example which illustrates Theorems 8.106 and 8.107 for q = p3

andp odd, is the dual curveC of the plane curveF of equation (8.54). The mainproperties ofC are as follows:

(i) C is a projective singular plane curve defined overFq birationally equivalentto C;

(ii) C has degree(p2 + p+ 1)(p+ 1), and genusg = (p2 + p)(p2 + p− 1)/2;

(iii) C is a Frobenius non-classical plane curve withǫ2 = ν1 = p2;

308

(iv) C has only one non-linear branch; it is centred at a point inPG(2, q) and hasorderp+ 1.

Apply Theorem 8.106:

Bp3 = (p2 + p+ 1)(p+ 1)(p3 − 1)− (p2 + p+ 1)(p2 + p− 2)

= (p2 + p+ 1)(p− 1)(p3 + 2p2 + p− 1).

Whenp = 3, this givesB27 = 1222, N27 = 208.(3) Now, letq = p = 2, andF = v((X2 + X)(Y 2 + Y ) + 1). As noted in

Remark 8.50,F is classical but Frobenius non-classical withǫ2 = ν1 = 2. Here,F has two points inPG(2, q), namelyX∞ andY∞, both singular. Also,X∞ isa double point and both branches ofF centred atX∞ areFq-rational; the similarproperty holds forY∞. Therefore,F has only linear and hence tame branches.SinceF has genusg = 1, soB2 = 4, as in (8.84).

(4) To illustrate Theorem 8.107, putq = 22e+1, q0 = 2e, with e ≥ 1, andconsider the Deligne–Lusztig–Serre irreducible plane curve,

F = v(Y q + Y +Xq0(Xq +X)),

of genusq0(q − 1).It has several interesting properties; see also Section 13.2. First,F is Frobenius

non-classical withǫ = ν = q0; in fact, both (7.62) and (8.52) hold for

z0(x, y) = xq/q0+1 + yq/q0 , z1(x, y) = x, z2(x, y) = 1.

Also,F has only one singular point, namelyY∞. More precisely,Y∞ is the centreof exactly one branch ofF , and henceBq = q2 + 1. This branchP has orderr = q0 and classs = q; in particular,P is a non-tame branch. Theorem 8.107 failsin the sense that here equality does not hold in (8.90); the unique singular branchof F is centred at a point inPG(2, q), but

q2 + 1 > q2 − qq0 − q − q0 + q0 + 2 = (q − 1)(q + q0)− (2g − 2).

8.9 THE DUAL OF A FROBENIUS NON-CLASSICAL CURVE

The notation is the same as Section 7.10. Also,Γ is defined overFq; then∆ is alsodefined overFq.

THEOREM 8.109 If ν1 > 1, thenK(Γ)/K(∆) is a purely inseparable extensionwith inseparability degreeq′ and withq′ ≤ q.Proof. From the definition of Frobenius orders,ν1 > 1 occurs if and only if thematrix

1 xq xq

2 . . . xqr

1 x x2 . . . xr

0 1 D(1)x x2 . . . D

(1)x xr

has rank2. Hence,ν1 > 1 yields

(x− xq)D(1)x xi = xi − xq

i , i = 2, . . . , r. (8.95)


Sincexi = xαi + βi, it follows that

(x− xq)αi = xαi + βi − (xαi + βi)q,

whencexq(αqi − αi) = βi − βq

i . Thus

xq =βq

i − βi

αqi − αi

.

AsK(Γ) = K(∆)(x) = K(α2, . . . , αr, β2, . . . , βr, x), the assertion follows. 2

This together with Theorem 7.84 has the following corollary.

THEOREM 8.110 If ν1 > 1, thenǫ2 = q′ with a power ofp not exceedingq.

Frobenius non-classicality has implications for the geometry of dual curves ofplane curves.

THEOREM 8.111 LetF = f(X,Y ) be an irreducible plane curve defined overFq

which is Frobenius non-classical. Ifǫ2 < q, the dual curveD ofF is non-reflexive.

Proof. By Theorem 5.81, it is assumed thatp > 2. From Remark 8.50,F is notonly Frobenius non-classical but also non-classical with order sequence(0, 1, q′)with ǫ2 = ν1 = q′. Sinceq > q′, the argument at the beginning of the proof ofTheorem 8.74 can be used. IfQ = (x, y) is a generic point ofF , from (7.62) and(8.16),

z0 + z1xq/q′

+ z2yq/q′

= 0 (8.96)

where at least one of the elementszi is a separable variable. SinceF is neither a linenor a strange curve by Exercise 3 of this chapter, soK(F) = K(D). Without lossof generality, suppose thatz2 6= 0, and letu = z0/z2, v = z1/z2 with d = q/q′.Then (8.96) readsyd + uxd + v = 0. Therefore,D is non-classical, and hence isnon-reflexive by Theorem 5.80. 2

8.10 ELLIPTIC CURVES

THEOREM 8.112 The rational points of an elliptic cubic curveF = v(F ) definedoverFq form an abelian groupGF .

Proof. Let S = F(Fq). For any pointT of S, let ℓT denote the tangent atT toF .LetO be any point ofS. ForP,Q ∈ S, letA = φ(P,Q) be such that

PQ.F = P +Q+A;

here, ifP = Q thenPQ = ℓP . Now, define

P ⊕Q = φ(O,A);

see Figure 8.1. Then, under this operation,S is an abelian groupGF with identityO.

310

Figure 8.1 Abelian group law on an elliptic curve

QQ

QQ

QQ

QQ

QQ

QQ

QQ

Q

P Q Aq q q

Oq

P ⊕Qq

Figure 8.2 Inverse of an element

@@

@@

@@

@@

@@

−P P

O O′

P ′

q q

q q q

q

First, the operation is well defined and abelian. Next,O is the identity, since iffor a pointP the pointP ′ is defined byφ(O,P ) = P ′, thenφ(O,P ′) = P ; that is,P ⊕O = P .

To show thatP has an inverse, letO′ be point whereℓO meetsF again; that is,φ(O,O) = O′. Thenφ(O′, P ) = −P ; see Figure 8.2. This incidentally meansthat, whenO′ = O, then−P = P ′.

It remains to show that the associative law holds. ForP,Q,R ∈ S, defineA, B, C, D, P ⊕Q, Q⊕R as follows:

φ(P,Q) = A, φ(A,O) = P ⊕Q, φ(P ⊕Q,R) = B, φ(Q,R) = C,

φ(Q⊕R,C) = O, φ(P,Q⊕R) = D.

The vertical and horizontal lines of Figure 8.3 make up two cubic curves with eightcommon points onF . By the Theorem of the Nine Associated Points, they have aninth point in common, namelyB = D, whence

(P ⊕Q)⊕R = φ(O,B) = φ(O,D) = P ⊕ (Q⊕R).

2


Figure 8.3 The associative law

q qq

q

q q q

q q q

C QR

B

Q⊕R D P

O P ⊕Q A

COROLLARY 8.113 (i) P,Q,R are collinear if and only ifP ⊕Q⊕ R = O′,whereℓO.F = 2O +O′.

(ii) If O is an inflexion, thenP,Q,R are collinear if and only ifP ⊕Q⊕R = O.

COROLLARY 8.114 (i) Any inflexion inF other thanO has order3 in thegroup.

(ii) If O is an inflexion, then the points of order2 are the points of contact of thetangents throughO other than the inflexional tangent.

ForP ⊕ P ⊕ . . .⊕ P (m times), write[m]P .Some further properties of the group structure are now summarised.

DEFINITION 8.115 (i) Given the group structure of a cubicF , theendomor-phism ringR(K) = EndK(F) is the set of all group homomorphisms ofF ,considered over an extensionK of Fq, where the sum and product inR aredefined as follows:

(S + T)P = S(P )⊕ T(P ), (ST)(P ) = S(T(P ));

the zeroO and the identityI are given by

O(P ) = O, I(P ) = P

for all P .

(ii) The non-singular cubicF defined overFq is supersingularif R(Fq) is non-commutative.

(iii) The Frobenius endomorphisminR(Fq) is the endomorphismΦ induced by

(x0, x1, x2) 7−→ (xq0, x

q1, x

q2).

THEOREM 8.116 LetF be non-singular cubic overFq, q = ph, withN1 rationalpoints.

312

(i) The endomorphismΦ satisfies a unique equationΦ2− tΦ+q = 0 inR(Fq);heret ∈ Z ⊂ R(Fq).

(ii) | t | ≤ 2√q.

(iii) N1 = q + 1− t.

(iv) p | t if and only ifF is supersingular.

With Sq = N1 the number of rational points on a curveF , consider the case thatF is an elliptic curve; equivalently,F is a non-singular plane cubic.

¿From Theorem 10.18,

(√q − 1)2 ≤ N1 ≤ (

√q + 1)2.

In fact, the precise values thatN1 can take are given by the next result.

THEOREM 8.117 (Waterhouse)There exists an elliptic cubic overFq, q = ph,with preciselyN1 = q + 1− t rational points, where| t | ≤ 2

√q, for precisely the

values oft in Table 8.2.

Table 8.2 Values oft

t p h(1) t 6≡ 0 (mod p)

(2) t = 0 odd

(3) t = 0 p 6≡ 1 (mod 4) even

(4) t = ±√q p 6≡ 1 (mod 3) even

(5) t = ±2√q even

(6) t = ±√2q p = 2 odd

(7) t = ±√3q p = 3 odd

Let Nq(1) denote the maximum number of rational points on any non-singularcubic overFq andLq(1) the minimum number. The prime powerq = ph is excep-tional if h is odd, h ≥ 3, andp divides⌊2√q⌋.REMARK 8.118 The only exceptionalq < 1000 is q = 128.

COROLLARY 8.119 The boundsNq(1) andLq(1) are as follows:

(i) Nq(1) =

q + ⌊2√q⌋, if q is exceptionalq + 1 + ⌊2√q⌋, if q is non-exceptional;

(ii) Lq(1) =

q + 2− ⌊2√q⌋, if q is exceptionalq + 1− ⌊2√q⌋, if q is non-exceptional.


Proof. This is an immediate consequence of Theorem 8.117. 2

COROLLARY 8.120 The numberN1 takes every value betweenq + 1 − ⌊2√q⌋andq + 1 + ⌊2√q⌋ if and only if (a) q = p or (b) q = p2 with p = 2 or p = 3 orp ≡ 11 (mod 12).

COROLLARY 8.121 For q ≤ 128, the values thatN1 cannot take betweenLq(1)andNq(1) are given in Table 8.3. The values ofLq(1) andNq(1) for q ≤ 128 aregiven in Table 8.5.

Table 8.3 The values thatN1 cannot take betweenLq(1) andNq(1) for q ≤ 128

q Forbidden values8 7, 11

16 11, 15, 19, 2325 2627 22, 25, 31, 3432 23, 27, 29, 31, 35, 37, 39, 4349 43, 5764 51, 53, 55, 59, 61, 63, 67, 69, 71, 75, 77, 7981 67, 70, 76, 79, 85, 88, 94, 97

106, 111, 116, 121, 131, 136, 141, 146128 109, 111, 115, 117, 119, 121, 123, 125, 127,

131, 133, 135, 137, 139, 141, 143, 147, 149

For any integern and any prime divisorℓ, let vℓ(n) be the highest power ofℓdividing n; that is,

∏ℓ ℓ

vℓ(n) is the prime decomposition ofn. Also write Z/(r)instead ofZr.

THEOREM 8.122 (Voloch, Ruck) The groupsGF of the cubicsF occurring inTheorem 8.117 in cases(1)− (7) are as follows:

(1) GF = Z/(pvp(N1))×∏ℓ 6=p (Z/(ℓrℓ)× Z/(ℓsℓ)) ,with rℓ + sℓ = vℓ(N1) andmin(rℓ, sℓ) ≤ vℓ(q − 1);

(2) (3) GF =

Z/(q + 1) whenq 6≡ −1 (mod 4),Z/(q + 1) orZ/(2)× Z/((q + 1)/2) whenq ≡ −1 (mod 4);

(4) GF = Z/(N1);

(5) GF = Z/(√N1)× Z/(

√N1), N1 = (

√q ± 1)2;

(6) GF = Z/(N1);

(7) GF = Z/(N1).

314

EXAMPLE 8.123 For F = X3 + Y 3 + Z3, Table 8.5 gives the numberN1 ofrational points ofv(F ) for 2 ≤ q ≤ 167 with also the values ofNq(1) andLq(1).

To calculateN1, note the following.

1. Whenq ≡ 0,−1 (mod 3), then

N1 = q + 1.

2. Whenq = p andp ≡ 1 (mod 3), then

N1 = q + 1 +A,

where4q = A2 + 27B2 with A,B ∈ N andA ≡ 1 (mod 3).

3. WithNi the number ofFqi-rational points ofv(F ), then

N2 = N12(q + 1)−N1.In particular, ifN1 = q + 1, then

N2 = (q + 1)2.

4. Whenq ≡ 1 (mod 3), then 9 dividesN1, as the nine inflexions form asubgroup.

Table 8.4 The numberA whenq = p andq ≡ 1 (mod 3)

q 4q = A2 + 27B2 A7 28 = 1 + 27 1

13 52 = 25 + 27 −519 76 = 49 + 27 731 124 = 16 + 108 437 148 = 121 + 27 −1143 172 = 64 + 108 −861 244 = 1 + 243 167 268 = 25 + 243 −573 292 = 49 + 243 779 316 = 289 + 27 −1797 388 = 361 + 27 19

103 412 = 169 + 243 13109 436 = 4 + 432 −2127 508 = 400 + 108 −20139 556 = 529 + 27 −23151 604 = 361 + 243 19157 628 = 196 + 432 −14163 652 = 625 + 27 25


Table 8.5 NumberN1 of points on the cubicv(X3 + Y 3 + Z3)

q 2 3 4 5 7 8 9 11 13 16 17N1 3 4 9 6 9 9 10 12 9 9 18Nq(1) 5 7 9 10 13 14 16 18 21 25 26Lq(1) 1 1 1 2 3 4 4 6 7 9 10q 19 23 25 27 29 31 32 37 41 43 47N1 27 24 36 28 30 36 33 27 42 36 48Nq(1) 28 33 36 38 40 43 44 50 54 57 61Lq(1) 12 15 16 18 20 21 22 26 30 31 35q 49 53 59 61 64 67 71 73 79 81 83N1 63 54 60 63 81 63 72 81 63 82 84Nq(1) 64 68 75 77 81 84 88 91 97 100 102Lq(1) 36 40 45 47 49 52 56 57 63 64 66q 89 97 101 103 107 109 113 121 125 127 128N1 90 117 102 1117 108 108 114 144 126 108 129Nq(1) 108 117 122 124 128 130 135 144 148 150 150Lq(1) 72 79 82 84 88 90 93 100 104 106 108q 131 137 139 149 151 157 163 167 169 173 179N1 132 138 117 150 171 144 189 168 171 174 180Nq(1) 154 161 163 176 176 183 189 193 196 200 206Lq(1) 110 115 117 126 128 133 139 143 144 148 154

8.11 CLASSIFICATION OF NON-SINGULAR CUBICS

In this section, two notions of equivalence are considered and the number of distinctcurves obtained in both cases.

The notion of projective equivalence of cubic curves must bedistinguished fromthe isomorphism of cubic curves. Under birational isomorphism, an elliptic curvehas an inflexion. Two non-singular cubicsF andF ′ are isomorphicif there is apolynomial mapT takingF to F ′ preserving the group law with respect to a aninflexion on each curve as zero. This in fact implies that an isomorphism class is aprojective class with respect to a fixed inflexion.

Let Aq be the total number of isomorphism classes andPq the total numberof projective equivalence classes. Here,ni, for i = 0, 1, 3, 9, is the number ofprojective equivalence classes with exactlyi rational inflexions. Hence,

Aq = n9 + n3 + n1, Pq = n9 + n3 + n1 + n0.

The values ofn0, n1, n3, n9 are given in Tables 11.28 to 11.30; see the end of thissection.

THEOREM 8.124 (i) Aq = 2q + 3 +

(−4

q

)+ 2

(−3

q

);

316

(ii) Pq = 3q + 2 +

(−4

q

)+

(−3

q

)2

+ 3

(−3

q

).

The number of inequivalent types of cubic with a fixed number of rational pointscan also be given. LetAq(t) andPq(t) be the number of inequivalent non-singularcubics with exactlyq + 1 − t rational points under isomorphism and projectiveequivalence. So

Aq =∑

tAq(t), Pq =∑

tPq(t).

In the formulas below, both certain Legendre and Legendre–Jacobi symbols areused:

(x3

)=

1 if x ≡ 1 (mod 3),0 if x ≡ 0 (mod 3),−1 if x ≡ −1 (mod 3);

(−4

c

)=

1 if c ≡ 1 (mod 4),0 if c ≡ 0 (mod 2),−1 if c ≡ −1 (mod 4);

(−3

c

)=

1 if c ≡ 1 (mod 3),0 if c ≡ 0 (mod 3),−1 if c ≡ −1 (mod 3).

It is necessary to define theclass numberof an integral quadratic form.Let S = f(X,Y ) = aX2 + bXY + cY 2 | a, b, c ∈ Z; a > 0. Consider

G = SL(2,Z) acting onS. ForT inG with T =

(A BC D

)andAD−BC = 1,

let

T(f) = a(AX + CY )2 + b(AX + CY )(BX +DY ) + c(BX +DY )2.

With ∆(f) = b2−4ac, also∆(T(f)) = ∆(f). So, all quadratic forms in the sameorbit have the the same discriminant.

The class numberH(∆) is the number of orbits ofG on quadratic forms inSwith discriminant∆.

THEOREM 8.125 Let (C1), (C2) be the following two conditions:(C1) : c > a and−a < b ≤ a;(C2) : c = a and0 ≤ b ≤ a.

The class number

H(∆) = |(a, b, c) ∈ Z3 | b2 − 4ac = ∆; a > 0; (C1) or (C2) is satisfied|.

For0 < −∆ ≤ 103, the class numberH(−∆) is given in Table 8.6.

L EMMA 8.126 Let q = ph.

(i) For p ≡ 1 (mod 3), there exists a unique solution fort such that

(t, p) = 1, t ≡ q + 1 (mod 9), t2 − 4q = −3x2 for somex ∈ Z.


Table 8.6 Values of the class number

−∆ H(∆) −∆ H(∆) −∆ H(∆)3 1 36 3 71 74 1 39 4 72 37 1 40 2 75 38 1 43 3 76 4

11 1 44 4 79 512 2 47 5 80 615 2 48 4 83 316 2 51 2 84 419 1 52 2 87 620 2 55 4 88 223 3 56 4 91 224 2 59 3 92 627 2 60 4 95 828 2 63 5 96 631 3 64 4 99 332 3 67 1 100 335 2 68 4 103 5

(ii) For p ≡ 1 (mod 4), there exists a unique solution fort such that

(t, p) = 1, t ≡ q + 1 (mod 9), t2 − 4q = −4x2 for somex ∈ Z.

Define the integerst0, t1 as follows. Whenq ≡ 1 (mod 3),

t0 =

the unique solution fort in Lemma 8.126 (i) ifp ≡ 1 (mod 3),

2(√

q

3

)√q if p 6≡ 1 (mod 3);

Whenq ≡ 1 or 4 (mod 12),

t1 =

the unique solution fort in Lemma 8.126 (ii) ifp ≡ 1 (mod 4),

2(√

q

3

)√q if p 6≡ 1 (mod 4);

THEOREM 8.127 The numberAq(t) of isomorphism classes of non-singular planecubics overFq, q = ph, with q + 1 − t points, where| t | ≤ 2

√q, is given by the

318

following values. In all other cases, Aq(t) = 0.

t Aq(t)(i) (t, p) = 1 H(t2 − 4q)(ii) h odd

(a) 0 H(−4p)(b) p = 2 ±√2q 1(c) p = 3 ±√3q 1

(iii) h even

(a) 0 1−(

−4p

)

(b) ±√q 1−(

−3p

)

(c) ±2√q Dp

whereDp = 112

[p+ 6− 4

(−3p

)− 3

(−4p

)]

The functionsB(t), C(t) are defined as follows. LetB(t) be the number ofisomorphism classes of elliptic curvesF with 3 dividingN1(F) = q + 1 − t; letC(t) be the number of isomorphism classes of elliptic curvesF with N1(F) =q+1− t andGF [3] ∼= Z3×Z3, whereGF [3] is the subgroup of elements of ordera power of 3 in the groupGF of the curve.

THEOREM 8.128 The numberPq(t) of projectively distinct non-singular planecubics overFq, q = ph, with q + 1− t points, where| t | ≤ 2

√q, is as follows:

Pq(t) = Aq(t) +B(t) + 3C(t)− ǫ(t),where

ǫ(t) = 2, if t = t0 or t1 with t0 6= t1;

= 3, if t = t0 = t1 andp = 2;

= 4, if t = t0 = t1 andp 6= 2;

= 0, otherwise.

When it is assumed under the isomorphism of elliptic curves that a curve alwayshas an inflexion, then the standard form for an elliptic cubicis the following:

v(Y 2Z + a1XY Z + a3Y Z2 −X3 − a2X

2Z − a4XZ2 − a6Z

3).

Putb2 = a2

1 + 4a2, b4 = a1a3 + 2a4, b6 = a23 + 4a6,

b8 = a21a6 − a1a3a4 + 4a2a6 + a2a

23 − a2

4,c4 = b22 − 24b4, c6 = −b32 + 36b2b4 − 216b6,∆ = −b22b8 − 8b34 − 27b26 + 9b2b4b6, j = c34/∆.

For p 6= 2, 3, 1728∆ = c34 − c26; for p 6= 2, 4b8 = b2b6 − b24. Here,∆ is thediscriminantandj is thej-invariant. The relationship betweenj andµ is given by

4µ =27j

j − 1278.

ThenF is harmonicfor µ =∞, j = 1728 andequianharmonicfor µ = 0, j = 0.


8.12 EXERCISES

1. For any two positive integersm,n, both prime top > 2, the irreducible planecurveF = v(Xn + Y m + 1) is viewed as a curve defined overFq. ShowthatF is Frobenius classical if and only ifm = n = (pk − 1)/(ps − 1),wheres dividesk.

2. Show by direct computation that, ifν1 > 1 andp > 2, thenǫ2 > 2.

3. Prove that, ifF is a strange curve defined overFq, thenν1 = 1.

4. Let p > 2. Assume thatf(X) ∈ Fq[X] has only simple roots, and thatits degreen is not divisible byp. Show that if the irreducible plane curveF = v(Y n − f(X)) defined overFq is Frobenius non-classical, then theroots off(X) belong toFq.

5. LetF be the curve considered in Example 1.40 and Exercise 3 in Chapter 7regarded as a curve defined overFq with q = ph andp ≥ 5.

(i) If k ≤ 12 (q−1) andk ≡ 1 (mod p), show thatF is Frobenius classical

with respect to the linear system of conics.

(ii) If q ≡ 1 (mod k) andk ≡ −1 (mod p), show thatF is Frobeniusclassical with respect to linear system of lines but it can beFrobeniusnon-classical with respect to the linear systems of conics.

The second case in (ii) occurs if and only if(q − 1)/k = pn + 1 with 2n | handu, v ∈ Fqn . For instance, this happens forh = 2, k = p + 1, n = 1,and also forh = 6, k = (p2 − p+ 1)(p3 − 1), n = 1.

6. LetF = f(X,Y ) be a Frobenius non-classical plane curve defined overFq,letQ = (x, y) be a generic point ofK(F) such thatx is a separable variableof K(F), and letǫ2 = q′.

(i) If q > q′, show that

D(q′)x yD(2q′+1)

x y −D(2q′)x y −D(q′+1)

x y = 0.

(ii) If q = q′, show that

(xq−x)(D(q)x yD(2q+1)

x y−D(2q)x yD(q+1)

x y) = (D(1)x y−D(1)

x

q)D2q

x y.

(iii) For q = q′ andp > 2, prove that the dual curveD of F is non-reflexiveif and only ifD(2q)

x y = 0.

7. Show that the number of pointsN1 on an elliptic cubic, for which the numbern of rational inflexions isn = 0, 1, 3, 9, satisfies the following:

(i) If n = 0, thenN1 ≡ 0 (mod 3);

(ii) If n = 1, thenN1 ≡ ±1 (mod 3);

(iii) If n = 3, thenN1 ≡ 0 (mod 3);

320

(iv) If n = 9, thenN1 ≡ 0 (mod 9).

8. Show that, for the curve in Exercise 6 of Chapter 4, the number ofF3-rationalpoints is13: three are non-singular points and ten arise from branches centredat singular points.

8.13 NOTES

A computer assisted search carried out by Luengo and Lopez [191] provided ir-reducible plane curves defined overF3 of genus4 ≤ g ≤ 8 that have manyF3-rational points. Their results imply that

N3(q) ≥

12 for g = 4,13 for g = 5,14 for g = 6,16 for g = 7,17 for g = 8.

On the other hand,N3(5) ≤ 14 andN3(7) ≤ 17. But the exact values ofN3(5)andN3(7) are not yet known.

The counter-example in Example 8.19 comes from [56]Theorem 8.62 is the central result in the book and comes from [276]A geometric interpretation ofSq in Example 8.67 is found in [164].Some particular families of Frobenius non-classical curves were investigated by

Garcia [85].The result that the curveF in Remark 8.90 is the image ofHq is shown in [54]

and [55].The characterisation of the Fermat curve in Theorem 8.98 is found in [53].The result in Remark 8.101 is shown in [53], that in Remark 8.103 is shown in

[97], and that in Remark 8.105 is shown in [170].For a similar but weaker result to Theorem 8.102 for curves ofdegree one less,

namely,n =√q − 2, see [9].

Theorem 8.117 giving the possible numbers of rational points on an elliptic cubicis due to Waterhouse [305]. See Ughi [293] for another proof of Waterhouse’stheorem. For the bounds onN1, see Zimmer [315, 316]. Theorem 8.122 giving thegroup structure of the set of rational points is due independently to Ruck [230] andto Voloch [302].

The formulas for the numbers of non-isomorphic and projectively distinct cubicswith a certain number of rational points are due to Schoof [242]; for the prime case,see also Deuring [60].

For a summary of isomorphic cubics, see also Menezes [198, Chapter 7].For singular plane cubics, see [127,§11.4].For more general accounts of elliptic curves, see Cassels [44], Koblitz [165],

Silverman [267], Tate [288]. For classical accounts of plane cubics, see Enriques[74], Hilton [125], Salmon [232], Seidenberg [252], Walker[303].

For Theorem 8.116, see Schoof [242], Tate [288].For the projective classification of plane cubics, see [127,Chapter 11]

Chapter Nine

Valuations and residues

9.1 VALUATION RINGS

Valuation of a fieldΣ over a base fieldF occurs frequently in algebraic geometry.Although, for the theory of algebraic curves as developed inthis book, this conceptis not really needed, it is useful to establish the relation between valuations andplaces. In the case thatF = K andΣ is a field of transcendency degree1 overK, any placeP of Σ gives rise to a valuation. in fact, all valuations ofΣ/K areobtained in this way.

DEFINITION 9.1 The groupG is orderedif it is an abelian group with a relation< satisfying the following properties:

(i) for all α, β, γ ∈ G,

α < β ⇒ α+ γ < β + γ; (9.1)

(ii) for every elementα ∈ G, precisely one of the relationsα = 0, α < 0, α > 0holds.

It may be observed that the additive groupZ with the natural order is the classicalexample. Another ordered abelian group consists of all ordered pairs(α, β) ofintegers with respect to the usual sum(α, β) + (γ, δ) = (α + γ, β + δ), where(α, β) < (γ, δ) if eitherα < γ, orα = γ, orβ < δ.

From (9.1) it follows that, ifα < β andγ < δ, thenα+ γ < β + δ.

DEFINITION 9.2 A valuation of a fieldΣ is a surjective mapping

v : Σ\(0)→ Γ,

whereΓ is an ordered abelian group, satisfying the following properties for alla, b ∈ Σ\0:

(i) v(ab) = v(a) + v(b);

(ii) v(a+ b) ≥ minv(a), v(b) whena+ b 6= 0.

If the symbol∞ is added toΣ with the natural rules,

α <∞, α+∞ =∞+ α =∞,then both (i) and (ii) hold for all elementsa, b ∈ Σ ∪ ∞. A valuation istrivial ifv(a) = 0 for anya 6= 0. If Γ = Z, the valuation isnormalised.

321

322

L EMMA 9.3 A valuationv satisfies the following properties:

(i) v(1) = v(−1) = 0;

(ii) v(a) = v(−a) for all a ∈ Σ;

(iii) if v(a) 6= v(b) thenv(a+ b) = minv(a), v(b).If F is a subfield ofΣ, the valuationv is overF provided thatv(a) = 0 for allF\0.

In the case whereF = K andΣ is a field of transcendency degree1 overK,every placeP of Σ defines a valuation by puttingv(f) = ordP f for everyf ∈ Σ.

EXAMPLE 9.4 Given a prime numberd, any non-zero rational numberc can bewritten in the form

c = dk · mn

with k,m, n integers andn positive such thatd ∤ mn. Here the integerv, indicatinghow oftend divides c, is uniquely determined byc. Thenv(c) = k defines anormalised valuation ofQ, thed-adic valuation ofQ. Any valuation onQ is eithertrivial or equivalent to ad-adic valuation for some prime numberd.

A major result result on valuations states that, ifΣ′ is a finite extension ofΣ, thenany normalised valuation ofΣ has an extension toΣ′, again a normalised valuation.

THEOREM 9.5 Letv be a valuation ofΣ. Then

(i) O = a | v(a) ≥ 0 is a ring, thevaluation ring ofv;

(ii) for every non-zero elementa of Σ, eithera ∈ O, or a−1 ∈ O;

(iii) conversely, if a ring O has this property, then there is a valuationv of Σ forwhichO is the ring.

Proof. First,O is a ring by (i) and (ii) of Definition 9.2. Since

0 = v(1) = v(a · a−1) = v(a) + v(a−1),

bothv(a) < 0 andv(a−1) < 0 cannot hold, because together these imply0 < 0.This proves (ii).

To show the converse, a relation∼ is defined fora, b ∈ Σ\0: a ∼ b if botha/b andb/a are inO, that is,a/b is a unit inO. This is an equivalence relation.For everya ∈ Σ\0, put

La = b | a ∼ b; b ∈ Σ\0;Γ′ = La | a ∈ Σ\0.

Then Γ′ is an ordered group with addition defined byLa + Lb = Lab and anordering given byLa < Lb whena/b is a non-unit inO. Both addition and orderdo not depend on the representatives chosen. Note thatL1 is the zero element ofΓ′. Also, the mappinga 7→ La is a valuation, and the seta | La ≥ 0 coincideswith O. 2

Valuations and residues 323

Let v be a valuation ofΣ overK and letΓ be its group. Ifτ : Γ → Γ′ is an orderpreserving isomorphism fromΓ to an ordered groupΓ′, thenv′ : a 7→ τ(v(a)) isalso a valuation overK, andv andv′ are regarded as the same valuation. Then,Theorem 9.5 shows that a valuation is uniquely determined byits ring.

To give a complete description of all valuations, the concept of morphism ofprojective fields is used.

DEFINITION 9.6 For a fieldF, the associatedprojective fieldisF ∪∞ togetherwith the following rules:

a+∞ =∞+ a =∞, for a ∈ F ;a · ∞ =∞ · a =∞, for a ∈ F ∪ ∞.

Here,∞+∞, ∞ · 0, 0 · ∞ are undefined.

DEFINITION 9.7 LetF∪∞ andF ′∪∞ be two projective fields. Amorphism,

τ : F ∪ ∞ → F ′ ∪ ∞,a 7→ a′,

has the following properties:

(i) 0 7→ 0, 1 7→ 1, ∞ 7→ ∞;

(ii) a+ b 7→ a′ + b′, when both sides are defined;

(iii) a · b 7→ a′ · b′, when both sides are defined.

Let F ′′ be the set of images underτ . If τ(a), τ(b) ∈ F ′′, then

τ(a+ b) = τ(a) + τ(b) ∈ F ′

and henceτ(a) + τ(b) ∈ F ′′. Similarly, if τ(a) ∈ F ′′ andτ(a) 6= 0, then

1 = τ(a · 1/a) = τ(a) · τ(1/a);soτ(1/a) = 1/τ(a) ∈ F ′ and hence1/τ(a) ∈ F ′′. From this it follows thatF ′′ isa field. This shows that every morphismF ∪ ∞ → F ′ ∪ ∞ can be assumedto be surjective. A morphism istrivial if it fixes ∞ and mapsΣ isomorphicallyontoΣ′. If K is a common subfield ofΣ andΣ′, the morphism is aK-morphismifτ(a) = a for everya ∈ K.

REMARK 9.8 In Number Theory, morphismsF ∪ ∞ → F ′ ∪ ∞ are usuallycalled places ofF onF ′.

EXAMPLE 9.9 Let t be an indeterminate overK, and letF be the rational functionfield K(t) overK. Any element inK(t) is represented by a fractionf(t)/g(t)with f(X), g(X) ∈ K[X] having no common root. Fora ∈ K with g(a) 6= 0, letf(t)/g(t) 7→ f(a)/g(a); if g(a) = 0, and hencef(a) 6= 0, let f(t)/g(t) 7→ ∞.This mapping is aK-morphismF ∪ ∞ → K ∪ ∞.

EXAMPLE 9.10 Let u = 1/t, so thatK(u) = K(t) = F . Puttingu 7→ 0 in theprevious example, another morphismF ∪ ∞ → K ∪ ∞ is obtained. It can beshown that these are all theK-morphisms ofF ∪ ∞ ontoK ∪ ∞ overK.

324

EXAMPLE 9.11 Let Σ be a field of transcendency degree1 overK. For everyplaceP of Σ, put

resPx =

a if ordP (x− a) ≥ 0,∞ if ordP x < 0.

Then theresidue map,x 7→ resP , associated toP is aK-morphism ofΣ ∪ ∞ontoK ∪ ∞. In other words, the following result is established.

THEOREM 9.12 Every residue map is aK-morphism.

A major result on valuations is that the converse of Theorem 9.12 also holds.The proof needs some preliminary results on aK-morphism,a 7→ a′.

L EMMA 9.13 If x 7→ ∞, thenx 6= 0 and1/x 7→ 0.

Proof. As 0 7→ 0 and hence0 7→ ∞ cannot occur,x 6= 0. Fromx · (1/x) = 1,it follows thatx′ · (1/x′) = 1. Therefore,∞ · (1/x)′ = 1, unless∞ · (1/x)′ isundefined. Now, whenevera 6= 0, then∞ · a is defined as∞. Since∞ · (1/x)′ isundefined, so(1/x)′ = 0. 2

L EMMA 9.14 If theK-morphism is non-trivial, thenΣ′ = K.

Proof. Let x ∈ Σ be transcendental overK. It suffices to show that if its imagex′

neither is∞ nor is inK, then theK-morphism is an isomorphism.Let y 6= 0 be another element inΣ. SinceΣ has transcendency degree1,

a0(x)ym + a1(x)y

m−1 + . . .+ am(x) = 0,

wherea0(X), a1(X), . . . , am(X) ∈ K[X] anda0(X) 6= 0. Hence

a0(x) + a1(x)y−1 + . . .+ am(x)(y−1)m = 0.

This shows thaty−1 7→ 0 is impossible, sincea0(x)′ = a0(x

′) 6= 0. Hencey 7→ ∞does not occur. Thus no element ofΣ is mapped into∞. Hence

(a+ b)′ = a′ + b′, (ab)′ = a′b′.

Thus theK-morphism is an isomorphism ofΣ onto Σ′ unlessa′ = 0 for somea 6= 0.

Now, if a 6= 0, thena ·(1/a) = 1; hencea′ ·(1/a)′ is defined anda′ ·(1/a)′ = 1,whencea′ 6= 0. This completes the proof. 2

THEOREM 9.15 Every non-trivialK-morphism ofΣ is the residue map of someplace ofΣ.

Proof. By Theorem 7.39, there is an irreducible non-singular curveΣ in PG(r,K)whose function fieldK(Γ) is Σ. Let Γ be given by the pointP = (x0, x1, . . . , xr)and, as usual, suppose thatx0 = 1. ThenΣ = K(x1, . . . , xr).

Now, letτ be a non-trivialK-morphism ofΣ. By Lemma 9.14,τ : Σ→ K. Notall τ(xj) = 0 or∞ sinceτ(x0) = τ(1) = 1, but it may happen thatτ(xi) = 0 fors ≥ 1 indices. Nevertheless, by changing the coordinate functions in the following


way, it may be assumed that this does not occur. Ifs ≥ 1 andτ(x1) = ∞, thenΓis also defined by the pointPi = (x−1

1 , 1, x2x−11 , . . . , xrx

−11 ). If τ(xj) 6=∞, then

τ(xjx−11 ) = τ(xj · (1/x1) = τ(xj)τ(1/x1) = 0.

Thus more finite quantities appear for the new form of the coordinate function thanfor the old;0 and1 are in the first two coordinates. There are now at mosts − 1indices at whichτ is infinite. In this ways can be reduced to0.

Suppose henceforth thats = 0; that is,τ(x1), . . . , τ(xr) are all finite. Then

(x1, . . . , xr) 7→ (τ(x1), . . . , τ(xr))

is a specialisation; see Section 7.17. HenceQ = (τ(x1), . . . , τ(xr)) is a point ofΓ, the centre ofτ . SinceQ is the centre of a branchγ of Γ, there is a placeP of Σarising fromγ. Then, theK-morphism resP is centred atQ.

It remains to prove that there is only oneK-morphism centred atQ, as this showsthatτ = resP .

Assume on the contrary thatτ ′ is anotherK-morphism centered atQ. Choosey ∈ Σ satisfyingτ(y) 6= τ ′(y). Actually, y may be chosen such thatτ(y), τ ′(y)are both finite. To show this, suppose thatτ ′(y) = ∞. Two cases occur accordingas eitherτ(y) 6= 0 or τ(y) = 0. In the former case,τ(y−1) 6= ∞ as well; also,τ(y·y−1) = τ(y)·τ(y−1) is defined and is equal to1. Thenτ(y−1) = τ ′(y−1) = 0,and replacingy by y−1 gives the result. In the latter case,τ(y) = 0 andτ ′(y) =∞.Replacingy by 1 + y, the former case occurs again. Therefore, in each case, thereexistsy with τ(y), τ ′(y) finite and unequal.

Now consider the irreducible curve∆ given by the pointP ′ = (x1, . . . , xr, y).ThenQ = (τ(x1), . . . , τ(xr), τ(y)) andQ′ = (τ(x1), . . . , τ(xr), τ(y)) are twodistinct points of∆. Let δ andδ′ be branches of∆ centred atQ andQ′. Then,δ 6= δ′ asQ 6= Q′. If P andP ′ are the corresponding places ofΣ, thenP 6= P ′.On the other hand, these places correspond to branches ofΓ having the same centreQ. But this is impossible sinceQ is a non-singular point ofΓ. 2

The relation between morphisms and valuations is describedin the followingtheorem. Recall that morphisms are tacitly assumed to be surjective.

THEOREM 9.16 Every morphismτ : Σ ∪ ∞ → Σ′ ∪ ∞ defines a valuationin a standard way; namely, the set of elements inΣ mapping toΣ′ form a valuationring, and this yields the valuation. Conversely, every valuation ring O yields amorphism in a standard way, and from the morphism the valuation ring can berecovered.

Proof. First, the elements inΣ mapping toΣ′ form a ringO. If a ∈ Σ buta 6∈ O,that is,τ(a) = ∞, thenτ(a−1) = 0; hencea−1 ∈ O. ThereforeO is a valuationring.

Before proving the converse, it is shown howΣ′ can be described up to an iso-morphism in terms ofO. For a, b ∈ O, write a ∼ b when τ(a) = τ(b); thisis an equivalence relation. Note thatτ(a) = τ(b) if and only if τ(a − b) = 0.In fact, τ(−b) = −τ(b) follows from the definition of a morphism, and thisimplies thatτ(a − b) = τ(a) − τ(b). Further,τ(x) = 0 if and only if x is

326

a non-unit inO. In fact, if x is a non-unit, thenτ(x−1) = ∞, as otherwise1 = τ(x · x−1) = τ(x) · τ(x−1) = 0, that is,x 6∈ O. In this way,τ is elimi-nated from the definition ofa ∼ b. Hencea ∼ b if and only if a − b is a non-unitin O.

LetLa = x | x ∼ a; x ∈ O. ThenLa 7→ τ(a) is a one-to-one correspondencebetween the setL consisting of allLa and the fieldΣ′. Defining

La + Lb = La+b, La · Lb = Lab,

the correspondence becomes an isomorphism.One more observation is needed to prove the converse. IfO is the valuation

ring arising from a morphismτ , then the non-units inO form an ideal. This canbe shown by observing that the non-units are those elementsx ∈ O for whichτ(x) = 0 holds. Hence, ifτ(x1) = τ(x2 = 0 anda ∈ O, thenτ(x1 +x2) = 0 andτ(ax1) = 0. Actually, a standard argument shows that this holds without assumingthatO arises fromτ . Also, in any valuation ringO, the non-units ofO form theunique maximal ideal ofO.

Now, starting from the valuation ringO, definea ∼ b if a − b is a non-unit inO; this is an equivalence relation. PutLa = x | x ∼ a;x ∈ O, and defineLa + Lb = La+b andLa · Lb = Lab. The setL consisting of allLa with thesetwo operations is a fieldΣ′. Also, puttingτ(a) = La for a ∈ O andτ(a) = ∞for a 6∈ O, the mapτ : Σ ∪ ∞ → Σ′ ∪ ∞ is a morphism such thatO is theassociated valuation ring. 2

REMARK 9.17 In the theory of function fields, the maximal ideal of a valuationring is usually called a place; see [275, I.1.8. Definition].

Theorem 9.5 shows that there is a standard one-to-one correspondence betweenthe valuations of a fieldΣ and its morphisms or, more precisely, the morphismsof Σ ∪ ∞. Also, if there is a base fieldF of Σ, then a valuation overF givesrise to anF -morphism, and vice versa. Thus, to find all valuations ofΣ overF , it is sufficient to find allF -morphisms ofΣ. WhenF = K andΣ is a fieldof transcendency degree1 overK, all K-morphisms are determined, and henceall valuations overK are known. TheK-morphisms all arise as residue maps ofplaces, by Theorem 9.15.

Let P be a place ofΣ. Then its valuation ringOP consists of all elements withfinite residue or, in other words, of the elements of non-negative order. Thus thering of the valuation ordP is the same as the ring of the residue map resP . Fromthis discussion the following result is obtained.

THEOREM 9.18 If Σ is a field of transcendency degree1 overK, then every valu-ation ofΣ is of the formordP for some place ofΣ.

REMARK 9.19 The concept of a place ofΣ is usually defined either via morphismsor via valuations. Thus the present section justifies the useof the termplacefor aformally different, although equivalent, concept.


9.2 RESIDUES OF DIFFERENTIAL FORMS

If y ∈ K((t)) with y =∑akt

k, thena−1; that is, the coefficient oft−1 is theresidue ofy with respect tot and denoted by Rest y. Note that the residue is afunctionK((t))→ K, which isK-linear. Also, Rest y = 0 wheny ∈ K[[t]].

L EMMA 9.20 If x, y ∈ K((t)), and ifu is a local parameter ofK((t)), then

Resu

(ydx

du

)= Rest

(ydx

dt

).

Proof. To simplify the computations, note that it suffices to prove the lemma in theparticular case thatx = t andy = tn with n any integer. Also, the lemma remainstrue whent 7→ at with a 6= 0. Hence it may be assumed that

t = u+ a2u2 + . . . , dt/du = 1 + 2a2u+ . . . .

It remains to prove the following:

Resu

(tndt

du

)=

1 for n = −1,0 otherwise.

Forn ≥ 0, the assertion follows, astndt/du has no term with negative powers of t.If n = −1, then

1

t

dt

du=

1 + 2a2u+ . . .

u+ 2a2u2 + . . .=

1

u+ . . . ,

and the residue is equal1.Whenn < −1, the casep = 0 is considered first. Here,

Resu

(tndt

du

)= Resu

(d

du

(1

n+ 1tn+1

)),

which is zero forn 6= −1. Whenp ≥ 0, putm = n−1 for a fixedn < −1. Thenthere is an integerN and a polynomialF (X1, . . . ,XN ) with integer coefficientssuch that

1

tmdt

du=

1 + 2a2u+ . . .

um(1 + a3u+ . . .)=F (a2, a3, . . . , aN+1)

u+ . . . .

Actually, F (X1, . . . ,XN ) is the same for any fieldK. Hence the truth of the as-sertion in positive characteristic is a formal consequenceof the result in zero char-acteristic, because in that case the assertion holds, showing thatF (X1, . . . ,XN )must be the zero polynomial. This completes the proof. 2

Note that, by Lemma 9.20, Rest(ydx), that is, theresidue of the differential formydx is meaningful.

Next, the case that ordt u = m > 1 is investigated. Up to a multiplication by aconstant,

u = tm + b1tm+1 + b2t

m+2 + . . . = tm(1 + b1t+ b2t2 + . . .).

If K((u)) is viewed as subfield ofK((t)), then [K((t)) : K((u))] = m; seeExercise 4 in Chapter 4. So, the trace functionT fromK((t)) toK((u)) is defined.

328

L EMMA 9.21 Let u ∈ K((t)) be a non-zero element of orderm ≥ 1 such thatm 6≡ 0 (mod p). If y ∈ K((t)), then

Rest

(ydu

dtdt

)= Resu(T(y)du). (9.2)

Proof. By Exercise 4 in Chapter 4, the powers1, t, . . . , tm−1 are linearly indepen-dent overK((u)). Multiplying u by a non-zero constant does not alter the validityof the lemma. Hence, there existb1, b2, . . . ∈ K such that

u = tm + b1tm+1 + . . . = tm(1 + b1t+ b2t

2 + . . .).

Hence, there existf0(u), f1(u), . . . , fm−1(u) ∈ K((u)) such that

tm = f0(u) + f1(u)t+ . . .+ fm−1(u)tm−1.

These elements can be found recursively and the coefficientsof fi(u) are polyno-mials inb1, b2, . . . with integer coefficients; that is,

fi(u) =∑

F (i)α (b)uα,

where eachF (i)v (b) is a polynomial with integer coefficients involving finitelymany

bj . Therefore, the matrix representing an arbitrary element

g0(u) + g1(u)t+ . . .+ gm−1(u)tm−1

in K((t)) is of type

G0,1(u) · · · G0,m−1(u)... · · ·

...Gm−1,1(u) · · · Gm−1,m−1(u)

,

with Gν,µ(u) ∈ K((u)), where the coefficients ofGν,µ(u) are polynomials withinteger coefficients in thebj and in the coefficients of thegi(u). This means that(9.2), if it is true, is a formal identity. The proof of (9.2) may therefore be carriedout in zero characteristic; the case of positive characteristic is then a corollary,since if two expression only involving integers are equal then they are also equal(mod p). So, takep = 0.

From Theorem 4.26,u = vm wherev = t + c2t2 + . . .. By Lemma 9.20, to

prove (9.2) it suffices to prove that

Resv

(ydu

dvdv

)= Resu(T(y)du). (9.3)

Now, (9.3) holds if ordt y is large enough, in which case both sides are equal tozero. Actually, it suffices to consider the casey = vj with j ∈ Z. This simplifi-cation is possible by linearity sincey can be written as a sum of a finite number oftermsajv

j and an element whose order is as high as required.Write j = ms + r with 0 ≤ r < m. Thenvj = usvr andT(vj) = usT(vr).

Also,

T(vr) =

0 for r 6= 0,m for r = 0,


whence

T(vj) =

mts for j = ms,0 otherwise.

Thus

Resu(T(vj)du) =

m for j = −m,0 otherwise.

(9.4)

Fory = vj , the left-hand side in (9.3) is Resv(vjmvm−1dv), which is equal to theright-hand side in (9.4). Note that, in characteristicp with p | m, both sides of theequation are zero. This completes the proof. 2

As before, letΣ denote a field of transcendency degree1 overK. In the sameway that the derivationd/dt of K((t)) gives rise to a derivation inΣ, it is possibleto define a residue on differential forms. To do this, choose adifferential formydxand a placeP of Σ. If σ : Σ → K((t)) is a primitive representation ofP, putz = σ(z) for z ∈ Σ. Then Rest(ydx) is the residue of the differential formydxatP. As already noted after the proof of Lemma 9.20, this is independent of thechoice ofσ.

DEFINITION 9.22 LetP be a place ofΣ. Then theresidue of the differential formydx atP is

ResP(ydx) = Rest(ydx).

The next lemma, which is an essential ingredient in the proofof the main result onresidue of differential forms, requires some preliminary observations.

Supposey ∈ Σ is a primitive element satisfying an irreducible monic polynomialg(Y ) ∈ K[x][Y ]. The idea is to viewg(Y ) as a polynomial overK((x)). This canbe done as follows. LetK((x)) be a field of power series. If

g(Y ) = Y n + a1(x)Yn−1 + . . .+ an(x),

with a1(X), . . . , an(X) ∈ K[X], put

g(Y ) = Y n + a1(x)Yn−1 + . . .+ an(x).

Since the subfieldK(x) of K((x)) is naturallyK-isomorphic toK((x)), the ratio-nal fieldK(x) may be assumed to be embedded inK((x)). This makes it possibleto replacex by x. With this convention,g(Y ) = g(Y ) andg(Y ) ∈ K[[x]][Y ].

By Theorem 4.15,g(Y ) splits into monic and irreducible factors overK[[x]]:

g(Y ) = g1(Y ) · · · gr(Y ). (9.5)

Putdj = deg gj(Y ) for j = 1, . . . , r. The factors are distinct,x being a separablevariable ofΣ.

Write

gj(Y ) =Y dj + α1(x)Ydj−1 + . . . ,

gj(X,Y ) =Y dj + α1(X)Y dj−1 + . . . ,

330

with α1(X), . . . ∈ K[[X]]. By Hensel’s lemma,gj(0, Y ) = 0 has only one solu-tion, saybj . Then

hj(X,Y ) = gj(X,Y + bj) ∈ K[[X]][Y ]

with hj(0, 0) = 0 andhj(X,Y ) is still monic and irreducible. By Theorem 4.87there is a finite extensionK((x))(vj)/K((x)) such thathj(x(vj), Y ) splits com-pletely into monic linear factors inK[[vj ]][Y ]. Each such factorY − f(vj) definesa branch representation(x = x(vj), y = f(vj)) of the irreducible analytic cyclehj(X,Y ) = 0.

From the proof of Theorem 4.87, one of these branch representations is primitive.By Theorem 4.85, it is of the form

(x = cvdj

j + . . . , y = f(vj))

with c 6= 0, andf(X) ∈ K[[X]]. In particular,K((x))(vj)/K((x)) is a finiteextension of degreedj . Then

(x = x(vj), yj = −bj + f(vj))

defines the unique branch of the analytic cyclegj(X,Y ) = 0. The same primitivebranch representation,

(x = x(vj), yj = −bj + f(vj)),

defines a branchγ of the irreducible plane curveF = v(g(X,Y )). Note thatγ isa centred at the pointPj = (0,−bj). Also, by Theorem 4.80, all thesedj factorsgive rise to the same branch ofgj(X,Y ) and hence the same branchγ of F . If Pj

is the associated place ofΣ, the subfieldK((vj)) of the algebraic closureK((x))ofK((x)) is denoted byΣPj

. Note that, ifσi is the primitive representation arisingfrom this primitive representation ofγ, that is,σi(x) = x, σi(y) = yj , thenσi is aK-isomorphism fromΣ to ΣPj

.Now, each of thedj elements−bj + f(vj) is a root ofg(Y ) Also, two such ele-

ments arising from distinct values ofj are different, by Theorem 4.73. Therefore,every root ofg(Y ) is one of these elements for somej with 1 ≤ j ≤ r.

Finally, every placeP of F overP ′ is obtained in this way, whereP ′ is theplace ofK(x) arising from the origin viewed as a point of the lineΓ = v(Y ), anon-singular model(Γ; (x, 0)) of K(x).

For eachj = 1, . . . , r, let Tj be the trace fromΣPjtoK((x)), andT the trace

from Σ toK(x).

L EMMA 9.23 With this notation,

T(y) =∑r

i=1Tj(σj(y)). (9.6)

Proof. If [Σ : K(x)] = n, thenT(y) is the coefficient ofY n−1 in the irreduciblepolynomialg(Y ) as above. The same holds for the local traces inΣPj

. Hence theequation follows from (9.5). 2

REMARK 9.24 Lemma 9.23 holds true for anyy ∈ Σ. In fact, ify is not a primitiveelement ofΣ overK(x), letz be such a primitive element. By Theorem 15.1, whichis the Theorem of the Primitive Element in its strong form, there existsc ∈ K suchthatw = y+ cz is a primitive element. Note that the formula (9.6) is valid for bothcz andw. Since (9.6) is linear on both sides, it follows that (9.6) also holds fory.


L EMMA 9.25 Let x be separable variable ofΣ. If P ′ is a place ofK(x) andP1, . . . ,Pr are the places ofΣ lying overP, then

ResP′(T(y)dx) =∑r

i=1 ResPi(ydx). (9.7)

Proof. Without loss of generality, suppose thatP ′ arises from the origin as in thepreceding discussion. By definition,

ResPi(ydx) = Resvi

(σi(y)

σi(x)

dvidvi

).

By Lemma (9.2), the right-hand side is equal to Resx(Ti(σi(y))dx). Taking (9.6)into account, this gives

∑ri=1 ResPi

(ydx) = Resx(∑r

i=1 Ti(σi(y))dx) = Resx(T(y)dx),

whence (9.7) follows. 2

The main result on residues of differential forms inΣ is the following.

THEOREM 9.26 Let Σ be a field of transcendency degree1 overK. If ydx is adifferential form, then

∑ResP(ydx) = 0. (9.8)

Note that the sum is taken over all placesP, but it is a finite sum since any differ-ential form has only a finite number of poles, by Lemma 5.52.

Proof. Consider first the rational case; that is, supposeΣ = K(x). There existh(X), k(X) ∈ K[X] with k(X) 6= 0 such thaty = h(x)/k(x). Places ofK(x)are in one-to-one correspondence with branches of the lineΓ = v(Y ). If P is aplace ofK(x) corresponding to the branchγ centred at the pointP = (c, 0) andhaving primitive representation(x = c+ t, η = 0), then

ResP(ydx) = Rest

(h(b+ t)

k(c+ t)

).

Expressing the rational functionh(X)/k(X) as partial fractions and replacingxby t+ b gives

h(b+ t)

k(b+ t)=∑ fµi(t)

(t− bµ)i+ f(t), (9.9)

wheref(X), fµi ∈ K[X] anddeg fµi < i. To calculate ResP(ydx), only thecoefficient oft−1 matters. The coefficientcν1 of t−1 is constant. With this notation,the sum of the residues taken over allP whose corresponding branches are centredat affine points is equal to

∑µ cµ1.

Suppose that the branchγ arising fromP is centred at the infinite point ofΓ.Then(x = t−1, η = 0) is a primitive representation ofγ. Sincedx/dt = −t−2,the coefficient oft−1 in (9.9) can be calculated. The residue of−f(1/t)/t2 at t iszero. The other terms in (9.9) can be written as

− 1

t2·∑ cµi

(t−1 − bµ)i

= −∑

µ,i

cµi1

ti−2(1 + bµt+ . . .)i.

332

This shows that the contribution to the residue only comes from the coefficient ofthe first term, which is precisely−∑µ cµi. Therefore,

∑ResP(ydx) = 0.

In the general case, the assertion follows from the rationalcase together withLemma 9.25. 2

REMARK 9.27 In the classical case,K = C, any irreducible curve has the naturalstructure of a Riemann surface, and ResP (ω) =

∮Pω/(2πi). Hence Theorem 9.26

for K = C follows from Stokes’ Theorem.

EXAMPLE 9.28 For an odd integern not divisible byp, with p odd, let

F (X,Y ) = Y n−2 −Xn + Y n.

ThenΣ = K(x, y), with F (x, y) = 0, hasx as a separable variable andydx as anon-trivial differential form. The placeP of Σ corresponds to a branchγ on themodel(F , (x, y)) of Σ, whereF = v(F ),

Assume first thatγ is centred at an affine point. Then, for any primitive repre-sentationσ, bothσ(x) andσ(y) are inK[[t]]. HenceResP(ydx) = 0. The pointsof F at infinity arePi = (ǫi, 1, 0), whereǫ is a primitiven-th root of unity andi = 0, 1, . . . , n− 1. If the centre ofγ is P = (0, ǫi, 1), thenσ can be chosen suchthat

σ(x) = x =ǫi + (1/n)ǫit2 + . . .

t, σ(y) = y =

1

t,

is a primitive representation ofγ. Then

ydx = −ǫi + (1/n)ǫit2 + . . .

t3.

This shows thatRest(ydx) = −(1/n)ǫi. Therefore, apart from the constant factor−(1/n), the right-hand side in (9.8) is

∑n−1i=0 ǫ

i, that is, the sum of alln-th root ofunity. Since this sum is equal to zero, (9.8) follows.

9.3 NOTES

For d-adic valuations in Remark 9.4, see [50, Proposition 4.3]. For normalisedvaluations, see [50, Proposition 2.4]. For Remark 9.8, see [50, Section 1.4].

Chapter Ten

Zeta functions and curves with many rational

points

In this chapter,K stands for the algebraic closure of the finite fieldFq of orderq.Let Γ be an irreducible algebraic curve ofPG(r,K) defined overFq and equippedwith the places of its function fieldΣ. By definition, as in Section 7.1,Γ arisesfrom an irreducible algebraic curveF defined overFq. If F = v(f(X,Y )) withf(X,Y ) ∈ Fq[X,Y ], thenΣ = K(x, y) with f(x, y) = 0.

The objective is to investigate birational properties ofΓ using itsFq-rationalpoints, in particular, the number ofFq-rational places ofΓ. For this purpose, anon-singular modelX of Σ is chosen which is birationally equivalent overFq toF . So,X is an irreducible non-singular curve in an appropriate projective spacePG(r,K), and places ofΣ are identified with points ofX . Consequently, divisorsare finite sums of points ofX .

SinceX can also be viewed as an irreducible algebraic curve overFqi for everyfinite extensionFqi of Fq, the numberNqi of theFqi-rational points ofX is de-fined, and every point ofX is anFqi-rational point for somei. It is conceivable, atleast intuitively, that from the infinite sequence

N1, N2, . . . , Ni, . . ., withNi = Nqi , (10.1)

it is possible to obtain information aboutX , including properties apparently inde-pendent of the behaviour of the setsF(Fqi) of theFqi-rational points ofX , suchas the class number and the fundamental equation. To move in this direction, thetechnical tool is thezeta function.

10.1 THE ZETA FUNCTION OF A CURVE OVER A FINITE FIELD

The Riemann zeta function

ζ(s) =

∞∑

n=1

n−s

is important in algebraic number theory. It is convergent for Re(s) > 1 and ex-tends to a meromorphic function ofs with a simple poles = 1. Equivalently, theRiemann zeta function can also be written as an infinite product, theEuler product,

ζ(s) =∏

p

1

1− p−s,

333

334

wherep ranges over all primes ofQ.A similar function for the curveX is the formal product,

∏

P

1

1−N(P )−s, (10.2)

whereP ranges over all closed points ofX andN(P ) = qdeg P . The aim is to givean expression for this product depending onFq-rational divisors ofX .

DEFINITION 10.1 Let s be a complex variable. Thezeta functionof X , viewed ascurve overFq, is

ζ(X , s) =∑

D

N(D)−s, (10.3)

whereD runs over all effectiveFq-rational divisors ofX , andN(D) = qdeg D isthenormof D.

Note that, ifD =∑nP P , then

N(D) =∏

P

N(P )nP .

Also,N(P ) = qi if P ∈ PG(r, qi) \ PG(r, qi−1).

THEOREM 10.2 For Re(s) > 1, the series (10.3) is convergent to the function

F (q−1) +hq1−gq(1−s)m

(q − 1)(1− qe(1−s))− h

(q − 1)(1− q−es),

where

(i) F (q−s) is a polynomial inq−s of degree at most2g − 2;

(ii) h is the class number ofX ;

(iii) g is the genus ofX ;

(iv) e is the smallest positive degree of divisors inDiv(Fq(X ));

(v)

m =

0, for g = 0,

2g − 2 + e, otherwise.

Proof. The minimality ofe implies that the degree of anyFq-rational divisor ofXis divisible bye, but it does not imply that every multiple ofe is the degree of someFq-rational divisor ofX . In fact,e = 1, but this is shown later, in Proposition 10.4;meanwhile no use is made of it.

TheFq-rational effective divisors ofX are partitioned into divisor classes, twosuch divisorsD1 andD2 being equivalent if and only if there exisysu ∈ Fq(F))such thatD1 = D2 + div u. Each class consists of allFq-rational divisors of acomplete linear series|C| for anFq-divisorC of X , that is, it coincides with the

Zeta functions and curves with many rational points 335

set|C| ∩Div(Fq(X )). By Theorem 8.38, the numbernq(|C|) of suchFq-rationaldivisors is equal to(qℓ(C)−1)/(q−1) with the usual notationℓ(C) = dim |C|+1.

From the Riemann-Roch Theorem 6.59,

ℓ(C) =

0, if degC < 0;1, if C is the zero divisor;0, if C is non-zero divisor anddegC = 0 ;

g − 1, if C is non-canonical divisor anddegC < 2g − 2;g, if C is canonical divisor;

degC − g + 1, if degC > 2g − 2.

By Remark 8.27,X has someFq-rational canonical divisor. Hencee divides2g−2.In the summations below,|C| runs over allFq-rational linear series ofX con-

taining some effectiveFq-rational divisors. Now,

ζ(X , s) =∑

D

N(D)−s =∑

|C|

∑

D∈|C|N(D)−s

=∑

|C|

∑

D∈|C|q−s deg D =

∑

|C|n(|C|)q−s deg C .

Hence

ζ(X , s) =∑

|C|q−s deg C · q

ℓ(C) − 1

q − 1

=1

q − 1

∑

deg C≥0

qℓ(C)−s deg C − 1

q − 1

∑

deg C≥0

q−s deg C .

If Re(s) > 1, then

∑

deg C≥0

q−s deg C =

∞∑

v=0

q−s deg C = h

∞∑

v=0

q−esv =h

1− q−es,

whence

ζ(X , s) =1

q − 1

∑

deg C≥0

qℓ(C)−s deg C − h

(q − 1)(1− q−es).

If g = 0, thenℓ(C) = degC + 1, andh = 1 by Example 8.36. So, forg = 0,

ζ(X , s) =1

q − 1

∞∑

v=0

∑

deg |C|=ve

qdeg C+1−s deg C − 1

(q − 1)(1− q−es)

=q

q − 1

∞∑

v=0

qe(1−s)v − 1

(q − 1)(1− q−es)

=q

(q − 1)(1− qe(1−s))− 1

(q − 1)(1− q−es).

336

Forg ≥ 1,

ζ(X , s) =1

q − 1

∑

0≤deg C≤2g−2

qℓ(C)−s deg C +1

q − 1

∑

deg C>2g−2

qℓ(C)−s deg C

− h

(q − 1)(1− q−es)=

1

q − 1

(2g−2/e∑

v=0

q−esvh∑

i=1

qℓ(Cvei )

=h

(q − 1)

∞∑

v=(2g−2)/e+1

qe(s−1)v−g+1 − h

(q − 1)(1− q−es),

where|Cvei | denotes anFq-rational divisor class of degreev · e. Hence

F (q−1) +hq1−gq(1−s)m

(q − 1)(1− q(2g−2)(1−s))− h

(q − 1)(1− q−es),

with

F (q−s) =1

q − 1

(2g−2/e∑

v=0

q−esv −h∑

i=1

qℓ(Cve),

which is a polynomial inq−s of degree at most2g − 2. 2

Theorem 10.2 shows thatζ(X , s) has an analytic continuation to the whole com-plex plane. Its poles of first order are of two types, namely

su =2πi

e log qu, sv = 1− 2πi

e log qv,

whereu, v range overZ.

THEOREM 10.3 For Re(s) > 1, the product (10.2) is absolutely convergent toζ(X , s). In particular, the product is independent of the order of its factors.

Proof. For any integerN ≥ 1,

∏

N(P )≤N

1

1−N(P )−s=

∏

N(P )≤N

( ∞∑

n=0

N(P )−ns

).

Since the product on the right-hand side has finitely many factors and each factoris an absolutely convergent series, it follows by multiplication that

∏

N(P )≤N

1

1−N(P )−s=

∑

N(D)≤N

N(D)−s +∑

N(D)>N

N(D)−s,

where the first summation on the right-hand side is over allFq-rational divisorswith N(D) ≤ N while the second is over all effectiveFq-rational divisorsD withN(D) > N which contain no closed point withN(P ) > N . Then

∣∣∣∣∣∣

∏

N(P )≤N

1

1−N(P )−s−

∑

N(D)≤N

N(D)−s

∣∣∣∣∣∣≤

∑

N(D)>N

N(D)−Re(s).


The sum on the right-hand side can be viewed as the error term of the convergentseriesζ(X , s), and hence it tends to zero. This proves the convergence of (10.2).The absolute convergence follows from

∣∣∣∣∣

∞∑

n=1

N(D)−ns

∣∣∣∣∣ ≤∞∑

n=1

N(D)−n(Re(s)).

2

Theorem 10.3 implies thatζ(X , s) has no zeros forRe(s) > 1. Also, it is thekey in the proof of the following proposition which has already been enunciated inSection 8.3 as well as in the proof of Theorem 10.2.

PROPOSITION 10.4 The curveX has anFq-rational divisor of degree1.

Proof. With the notation of Theorem 10.2, it suffices to prove thate = 1. By theminimality of e, the divisorD is Fq-rational of degreee if and only ifD is a closedpoint of degreee; that is,

D = P + Φ(P ) . . .+ Φe−1(P ),

for someFqe-rational pointP of X . Note that the Frobenius imagesPi = Φi(P )of P , with i = 1, . . . , e− 1, are alsoFqe-rational pointP of X . Let

ζ(X , s) =∏

P

1

1−N(P )−s

be the zeta function ofX regarded as a curve overFqe . Then

ζ(X , s) =∏

P

1

1− q−se deg P=

e∏

i=1

∏

Pi

1

1− q−se deg Pi

.

SincedegP = edeg Pi, it follows that

ζ(X , s) =e∏

i=1

(∏

P

1

1− q−s deg P

)= ζ(X , s)e.

As bothζ(X ; s) andζ(X , s) have a pole of the same order1 at s = 1, this impliesthate = 1. 2

Proposition 10.4 makes it possible to simplify Theorem 10.2.

THEOREM 10.5 The zeta function ofX can be written as

ζ(X , s) =L(q−s)

(1− q−s)(1− q1−s),

where

L(q−s) =

2g∑

j=0

ajq−js,

anda0 = 1, a2g = qg.

338

Proof. The proof is carried out in three parts according asg = 0, or g = 1, org ≥ 2. Forg = 0,

ζ(X , s) =q

(q − 1)(1− q1−s)− 1

(q − 1)(1− q−s)=

1

(1− q−s)(1− q1−s).

Forg = 1,

ζ(X , s) =1

q − 1

∑

deg C=0

qℓ(C)−s deg C +hq1−s

(q − 1)(1− q1−s)− h

(q − 1)(1− q−s)

=h− 1

q − 1+

q

q − 1+

hq1−s

(q − 1)(1− q1−s)− h

(q − 1)(1− q−s)

=1 + (h− q − 1)q−s + q · q−2s

(1− q1−s)(1− qs).

Forg ≥ 2,

ζ(X , s) =1

q − 1

∑

deg C=0

qℓ(C)−s deg C +1

q − 1

∑

1≤deg C<2g−2

qℓ(C)−s deg C

+1

q − 1+

∑

deg C≥2g−2

qℓ(C)−s deg C +hq1−sq(1−s)(2g−1)

(q − 1)(1− q1−s)

− h

(q − 1)(1− q−s)=h+ q − 1

q − 1+

1

q − 1

∑

1≤deg C<2g−2

qℓ(C)−s deg C

+(h− 1)qg − 1− s(2g − 2) + qg − s(2g − 2)

q − 1

+hq1−gq(1−s)(2g−1)

(q − 1)(1− q1−s)− h

(q − 1)(1− q−s)

=1

q − 1

2g−3∑

j=1

αjq−js +

(h+ q − 1)(qg−1q−2(g−1)s + 1)

q − 1

+hqgq−(2g−1)s

(q − 1)(1− q1−s)− h

(q − 1)(1− q−s)

=1 + α1q

−s + . . .+ α2g−1q−(2g−1)s + qgq−2gs

(1− q−s)(1− q1−s)).

2

The following result is the functional equation of the zeta function.

THEOREM 10.6 The zeta function ofX satisfies the equation

q(g−1)(2s−1)ζ(X , s) = ζ(X , 1− s). (10.4)

Proof. Again, three cases are treated separately according asg = 0, or g = 1, org ≥ 2. Forg = 0,

ζ(X , 1− s) =1

(1− qs)(1− qs−1)= g1−2sζ(X , s).


Forg = 1,

ζ(X , 1− s) =1 + (h− q − 1)qs−1 + q · q2s−2

(1− qs)(1− qs−1)= ζ(X , s).

Forg ≥ 2, let

ζ(X , s) = ζ1(X , s) + ζ2(X , s),with

ζ1(X , s) =1

q − 1

∑

1≤deg C<2g−2

qdeg C+1−s deg C

and

ζ2(X , s) = 1 + q−(g−1)(2s−1)

+h

q − 1

(1 + q(g−1)(2s−1) +

qgq−(2g−1)s

1− q1−s− 1

1− qs−1

).

Then

ζ2(X , s) = 1 + q(g−1)(2s−1)

+h

q − 1

(1 + q(g−1)(2s−1) +

qgq(2g−1)(s−1)

1− qs− 1

1− qs−1

)

= q(g−1)(2s−1)ζ2(X , s).It remains to show that

q(g−1)(2s−1)ζ1(X , s) = ζ1(X , 1− s).Setρ(C) = ℓ(C)− 1

2 degC. Then

ζ1(X , s) =1

q − 1

∑

1≤deg C<2g−2

qℓ(C)−s deg C

+1

q − 1

∑

1≤deg C<2g−2

qρ(C)−12 (s−1) deg C ,

and, from the Riemann–Roch Theorem 6.59,

ρ(C) = ρ(W − C)

for a Fq-rational canonical divisorW of X . Hence, ifC runs over the set of alldivisors classes satisfying the condition

1 ≤ degC < 2g − 2,

thenW − C does the same. Therefore

ζ1(X , s) =1

q − 1

∑

1≤deg(W−C)<2g−2

qρ(W−C)− 12 (s−1)(2g−2)−deg C)

=1

q − 1

∑

1≤deg C<2g−2

qρ(C)− 12 (s−1)(2g−2−deg C),

340

whence

ζ1(X , 1− s) =1

q − 1

∑

1≤deg C<2g−2

qρ(C)−12 (s−1) deg C

= q(g−1)(2s−1)

1

q − 1

∑

1≤deg C<2g−2

qρ(C)− 12 (s−1)(2g−2−deg C)

= q(g−1)(2s−1)ζ1(X , s).2

Now, the connection between the zeta functionζ(X , s) and the infinite sequence(10.1) is established. For this purpose, replaceq−s with t in ζ(X , s). The resultingformal power seriesZ(X , t) satisfies the equationζ(X , s) = Z(X , t).

THEOREM 10.7 If |t| < q−1 then

Z(X , t) = exp

( ∞∑

v=1

Ni

iti

).

Proof. The hypothesis|t| < q−1 holds if and only ifRe(s) > 1. ForRe(s) > 1,

ζ(X , s) =∏

P

1

1−N(P )−s

∏

P

1

1− q−s deg P=∏

P

1

1− tdeg P= Z(X , t).

Hence

logZ(X , t) = −∑

P

log(1− tdeg P ) =∑

P

∞∑

n=1

tm deg P

m

∞∑

n=1

∑

m deg P=n

1

m

tn =

∞∑

n=1

∑

deg P |ndegP

tn

n.

LetN∗n =

∑deg P |n degP . Then

Z(X , t) = exp

( ∞∑

n=1

N∗n

ntn

).

This shows that it suffices to prove the following equation:∑

deg P |ndegP = Nn. (10.5)

EveryFqn-rational pointQ of X gives rise to theFq-rational closed point

P = Q+ Φ(Q) + . . .Φn−1(Q)

of X . In this context, twoFqn-rational points ofX are equivalent if they definethe sameFq-rational closed point ofX . So, the set ofFqn-rational points ofXis partitioned into equivalence classes. If∆ is a representative system∆ of suchclasses, then

N∗q =

∑

P∈∆

degP.


On the other hand, if the closed pointP arises from a pointQ of X , that is,

P = Q+ Φ(Q) + . . .+ Φm−1(Q),

then the conditiondegP | n in the above summation means thatΦm(Q) = Q withm | n. Therefore,Q is anFqm-rational point ofX , and the divisibility conditionm | n implies thatFqm is a subfield ofFqn . From this argument, the desiredequation (10.5) follows. 2

As in Section 8.3, letAn denote the number of all effectiveFqn-rational divisorsof X of degreen. Then

Z(X , t) = 1 +

∞∑

n=1

Antn, (10.6)

This follows from Theorem 10.3, since∏

P

1

1−N(P )−s= ζ(X , s) = Z(X , t) =

∏

P

1

1− tdeg P,

whereP ranges over allFq-rational closed points ofX , and each factor can bewritten as a geometric series, giving

∏

P

1

1− tdeg P=∏

P

∞∑

n=0

tdeg(nP ) =∑

D

tdeg D = 1 +

∞∑

n=1

Antn,

whereD ranges over all effectiveFq-rational divisors ofX .

PROPOSITION 10.8 If g ≥ 1 thenZ(X , t) = G(t) +H(t) with

G(t) =1

q − 1

∑

0≤deg C≤2g−2

qℓ(C)tdeg C ,

H(t) =h

q − 1

(q1−g(qt)2g−1 1

1− qt −1

1− t

).

Proof. A straightforward computation depending on Theorem 8.38 and Proposition8.39 withe = 1 gives the following:

Z(X , t) =∞∑

n=0

Antn =

∑

deg C≥0

nq(|C|) tdeg C +∑

degC≥0

qℓ(C) − 1

q − 1tdeg C

=1

q − 1

∑

0≤deg C≤2g−2

qℓ(C)tdeg C +1

q − 1

∑

deg C>2g−2

qℓ(C)+1−g tdeg C

− 1

q − 1

∑

deg C≥0

tdeg C = G(t) +H(t).

2

Theorem 10.5 shows that the zeta functionZ(X , t) has the form

Z(X , t) =L(t)

(1− t)(1− qt) ,

342

where

L(t) = Lq(t) =

1 for g = 0,

1 +∑2g−1

i=1 aiti + qgt2g, for g ≥ 1,

(10.7)

is theL-polynomialof X viewed as a curve defined overFq. Note thatL(t) ∈ Z[t].Forg ≥ 1, from Proposition 10.8,

L(t) = (1− t)(1− qt)G(t) +h

q − 1(qgt2g−1(1− t)− (1− qt)). (10.8)

PROPOSITION 10.9 (i)

L(t) = qgt2gL

(1

qt

)= 1 + qg +

a2g−1

qg−1+ . . .+ qgt2g.

(ii) a2g−i = qg−iai for i = 0, . . . , g.

Proof. The first assertion is a consequence of the functional equation (10.4). Com-paring the coefficients ofti, the second assertion follows. 2

Let

L(t) =

2g∏

i=1

(1− ωit) (10.9)

be a factorisation ofL(t) into linear factors in some finite extension ofQ. Thefollowing result plays an essential role in the study ofFq-rational points ofX .

THEOREM 10.10 LetNn be the number ofFqn-rational points of an irreducible,non-singular algebraic curveX of genusg defined overFq. Then

Nn = qn + 1−2g∑

i=1

ωni (10.10)

whereω1, . . . , ω2g are the reciprocals of the roots of theL-polynomial ofX .

Proof. From the relation,

Z(X , t) = exp

( ∞∑

n=1

Nn

ntn

)=

∏2gi=1(1− ωit)

(1− t)(1− qt) ,

it follows that∞∑

n=1

Nn

ntn =

2g∑

i=1

log(1− ωit)− log(1− t)− log(1− qt)

=∞∑

n=1

1

n

(qn + 1−

2g∑

i=1

ωni

)tn.

Now, the assertion is shown by comparing coefficients. 2

Note thatω1, . . . , ω2g are algebraic integers asL(t) ∈ Z[t].


REMARK 10.11 If g = 1, that is,X is a non-singular cubic curve defined overFq,then

N1 = q + 1− 2√q cosϕ,

whereω1 =√q(cosϕ+ i sinϕ). From this, theHasse Bound,

q + 1− 2√q ≤ N1 ≤ q + 1 + 2

√q,

follows. There exists a non-singular elliptic cubic overFq, q = ph, with preciselyN1 = q + 1− t Fq-rational points, where| t | ≤ 2

√q, for precisely the values oft

see Table 8.2.

If the numbersNn are known forn = 1, . . . , r for sufficiently larger, then thecoefficients of theL-polynomial can be calculated as follows.

PROPOSITION 10.12 LetL(t) =∑2g

i=0 aiti be theL-polynomial ofX viewed as

a curve overFq, and letSn = Nn − (qn + 1).

(i) The logarithmetic derivative is the following:

dL(t)/dt

L(t)=

∞∑

n=1

Sntn−1.

(ii) For n = 1, . . . , g,

nan = Sn + Sn−1a1 + . . .+ S1an−1. (10.11)

In particular, fromN1, . . . , Ng, the coefficients ofL(t) can be calculated using(10.11) and the equationsa2g−n = qg−nan, for n = 1, . . . , g.

Proof. (i) From Theorem 10.10,

dL(t)/dt

L(t)=

2g∑

n=1

−ωn

1− ωnt=

2g∑

n=1

(−ωn) ·∞∑

k=0

(ωnt)k

=

∞∑

k=1

(2g∑

n=1

−ωkn

)tk−1 =

∞∑

k=1

Sktk−1.

(ii) From (i),

a1 + 2a2t+ . . .+ (2g − 1)a2g−1t2g−2 + 2gqt2g−1

=(1 + a1t+ . . .+ a2g−1t2g−2 + 2gqt2g) ·

∞∑

k=1

Sktk−1.

So (10.11) follows by comparing the coefficients ofti for i = 0, . . . , g − 1. 2

EXAMPLE 10.13 Let q = 2 and letX = v(X3 + Y 3 + Z3). The set

X (F2) = (0, 1, 1), (1, 0, 1), (1, 1, 0)).Then

a1 = S1 = N1 − 3 = 0.

344

So

Z(X , t) =1 + 2t2

(1− t)(1− 2t)(10.12)

and∑

Niti = logZ(X , t) =

∑ti/i+

∑(2t)i/i+

∑(−1)j−1(2t2)j/j.

Hence,

Nh =

1 + 2h for h odd,1 + 2h + 2.2h/2 for h ≡ 2 (mod 4),1 + 2h − 2.2h/2 for h ≡ 0 (mod 4).

Also, from (10.12),

Z(X , t) = 1 + 3∑

i>0

(2i − 1)ti,

whenceAi = 3(2i − 1) for i > 0.

EXAMPLE 10.14 LetX = v(X0X31 +X1X

32 +X2X

30 ) be the Klein quartic, see

Example 7.17, viewed as a curve defined overF2. The verticesUi with i = 1, 2, 3of the fundamental triangle are points ofPG(2, 2) lying onX . Since no other pointsof PG(2, 2) has this property,N1 = 3 holds. To count theF2-rational closed pointsof degree2, consider the quadratic extensionF4 = F2(ω) with ω2+ω+1 = 0. Thepoints ofX overF4 distinct from the pointsUi are two, namelyV1 = (ω, ω2, 1)andV2 = (ω2, ω, 1). There are 21 points ofX over the cubic extensionF8 = F2(ǫ)with ǫ3 + ǫ+ 1 = 0 distinct from the pointsUi, namely

W1 = (ǫ2, ǫ4, 1), W2 = (ǫ2, 1, 1), W3 = (ǫ2, ǫ5, 1),W4 = (ǫ, ǫ2, 1), W5 = (ǫ, 1, 1), W6 = (ǫ, ǫ6, 1),W7 = (ǫ4, ǫ, 1), W8 = (ǫ4, 1, 1), W9 = (ǫ4, ǫ3, 1),W10 = (ǫ5, ǫ5, 1), W11 = (1, ǫ2, 1), W12 = (1, ǫ, 1),W13 = (1, ǫ4, 1), W14 = (ǫ6, ǫ4, 1), W15 = (ǫ6, ǫ6, 1),W16 = (ǫ6, ǫ3, 1), W17 = (ǫ3, ǫ2, 1), W18 = (ǫ3, ǫ3, 1),W19 = (ǫ3, ǫ5, 1), W20 = (ǫ5, ǫ, 1), W21 = (ǫ5, ǫ6, 1).

Thus,

S1 = S2 = 0, S3 = 24− (8 + 1) = 15;

a1 = a2 = 0, a3 = 5, a4 = a5 = 0;

L8(t) = 1 + 5t3 + 8t6.

Any curveX defined overFq and can also be regarded as a curve overFqn for anextension ofFq of degreen. So, for everyn, the curveX viewed as a curve overFqn has its zeta functionZqn(X , t) andL-polynomialLqn(t). The relationshipbetweenZ(X , t) andZqn(X , t), as well as that betweenL(t) = Lq(t) andLqn(t),is described in the following results.


PROPOSITION 10.15

Zqn(X , tn) =∏

γ

Z(X , γt)

whereγ ranges over then-th roots of unity.

Proof. It suffices to prove the assertion for|t| < q−1. In this region,

Zqn(X , tn) =∏

P ′

1

1− tn deg P ′(10.13)

whereP ′ ranges over all closed points ofX viewed as a curve overFqn . WriteP ′ = Q+ Φn(Q) + . . .+ Φ(k−1)n(Q) with Q ∈ X wherek = degP ′. If

P = Q+ Φ(Q) + . . .+ Φn(Q) + . . .+ +Φm−1(Q)

with Φm(Q) = Q, then writeP ′ | P . Note thatP is a closed point ofX of degreem, and that

∏

P ′

1

1− tn deg P ′=∏

P

∏

P ′|P

1

1− tn deg P ′.

With d = gcd(n,m),∏

P ′|P(1− tn deg P ′

) = (1− tnm/d)d

=∏

γ

(1− (γt)m) =∏

γ

(1− (γt)deg P ),

whereγ ranges over then-th roots of unity. From (10.13) the assertion follows.2

PROPOSITION 10.16

LFqn (t) =

2g∏

i=1

(1− ωni t).

Proof. By Proposition 10.15, ifγ ranges over then-th roots of unity, then

LFqn (tn) = (1− tn)(1− qntn)Zqn(X , tn)

= (1− tn)(1− qntn)∏

γ

Z(X , γt)

= (1− tn)(1− qntn)∏

γ

L(γt)

(1− γt)(1− qγt)

=∏

γ

L(γt) =

2g∏

i=1

∏

γ

(1− ωγt =

2g∏

i=1

(1− ωni t

n).

2

The next result relatesL-polynomials of two curves.

THEOREM 10.17 (Serre)If X → X ′ is an Fq-rational covering, then theL-polynomial ofX ′ divides theL-polynomial ofX in Z[t].

Example 10 illustrates Theorem 10.17.

346

10.2 THE HASSE–WEIL THEOREM

The fundamental result on curves over finite fields is the Hasse–Weil Bound (10.14)which was originally proved by Hasse forg = 1 and then by Weil forg > 1.

THEOREM 10.18 (Hasse–Weil Bound)WithN1 the number ofFq-rational pointsof an irreducible non-singular algebraic curve defined overFq,

q + 1− 2g√q ≤ N1 ≤ q + 1 + 2g

√q. (10.14)

Here, without going into the details of its derivation, (10.14) is proved by using theStohr–Voloch Theorem 8.62 for the upper bound, and Galois Theory for the lowerbound. From the Hasse–Weil Bound, via zeta functions, the Hasse–Weil Theoremis deduced, which is actually the Riemann hypothesis for function fields over finitefields.

THEOREM 10.19 (Hasse–Weil Theorem)The reciprocals of the rootsω1, . . . , ω2g

of theL-polynomial of an irreducible non-singular curve defined over Fq satisfy

|ωi| =√q, i = 1, . . . , 2g.

The Hasse–Weil Bound (10.14) is stated for irreducible non-singular curves de-fined overFq, but it also implies bounds for irreducible singular plane curves. Re-finements and relevant improvements of (10.14) are also possible; all these aretreated in Section 10.3.

The first step is the following technical result.

L EMMA 10.20 The Hasse–Weil Theorem holds overFq if and only if it holds overeveryFqi with i ≥ 1.

Proof. Let ω1, . . . , ω2g be the reciprocals of the roots of theL-polynomialLq(t)of X . If Lqn(t) is theL-polynomial ofX , then its roots areωn

1 , . . . , ωn2g, by Propo-

sition 10.16. The result follows since|ωi| = √q if and only if |ωni | =

√qn. 2

The next step is to show that a somewhat weaker version of the bound (10.14)implies the Hasse–Weil Theorem 10.19.

L EMMA 10.21 If there is a constantc ∈ R such that, for everyn ≥ 1,

|Nn − (qn + 1)| ≤ c√qn, (10.15)

then the Hasse–Weil Theorem 10.19 holds.

Proof. Theorem 10.10 and (10.15) imply, for everyn ≥ 1, that∣∣∣∣∣

2g∑

i=1

ωni

∣∣∣∣∣ ≤ c√qn. (10.16)

Denote byρ the minimum of|ω−1i | for i ∈ 1, . . . , 2g. Then the radius of conver-

gence of the power series expansion of the meromorphic function,

H(t) =

2g∑

i=1

ωit

1− ωit,


is ρ, the only singularities ofH(t) beingω−11 , . . . , ω−1

2g . For |t| < ρ,

H(t) =∑2g

i=1

∑∞n=1(ωit)

n =∑∞

n=1

(∑2gi=1ω

ni

)tn.

By (10.16), this is a convergent series for|t| < 1/√q, whence it follows that√

q ≤ ρ. Therefore,√q ≥ |ωi| for i = 1, . . . , 2g. Finally, |ωi| =

√q since∏2g

i=1 ωi = qs, which is a consequence of (10.7) and (10.9). 2

In view of Lemma 10.21, the Hasse–Weil Theorem 10.19 can be proven in twofurther steps by showing the existence of constantsc1 > 0 andc2 > 0 satisfyingthe following inequalities:

Nn≤ qn + 1 + c1√qn, (10.17)

Nn≥ qn + 1− c2√qn. (10.18)

for anyn ≥ 1. Actually, the latter condition onn can be weakened to require onlyn ≥ n0 for somen0 ≥ 1; see Lemma 10.20. This is useful in proving (10.18).

The upper bound (10.17) is achieved in the following result which is essentiallya corollary to the Stohr–Voloch Theorem 8.62.

PROPOSITION 10.22 If qn is a square greater than4g4(g − 1)2, then

Nn ≤ qn + 1 + 2g√qn.

Proof. It may be supposed thatX has anFq-rational pointP . For any integerd ≥ 2g, let L = |dP|. ThenL = gd

r is a non-special base-point-freeFq-rationallinear series. In particular,r = d− g ≥ g. By Corollaries 8.60 and 8.61,νi = i fori < r− g andνi ≤ i+ g for r− g ≤ i < r− 1. Hence the Stohr–Voloch Theorem8.62 shows that

N1 ≤ q + 1 + (r + q/r)g + 2g2(g − 1)/r

for anyr ≥ g. If q is a square andq > 4g4(g − 1)2, then taker =√q to give the

result. 2

To obtain the lower bound (10.18), Galois Theory is also needed.

PROPOSITION 10.23 If q is a square then

Nn = qn +O(√qn),

where

limn=∞

O(√qn)√qn

= c

for a constantc.

Proof. Choose a separable variableu ∈ K(X ). From (VII) in Section 15.2,K(u)has a normal extensionΣ′ that is also a normal extension ofK(X ). LetG andHbe the corresponding Galois groups withH being a subgroup ofG. In geometricterms, ifX ′ is an irreducible non-singular curve ofPG(r′,K ′) which is a model ofΣ′, then bothX and the projective lineℓ = v(X) with generic pointP = (u) arequotient curves ofX ′; more precisely,X = X ′/H andℓ = X ′/G. Here,G and

348

H, are linear collineation groups ofPG(r′,K ′) which preserveX ′. Although thecurveX ′ and the groupG need not be defined overFqn , but only over some finiteextension ofFqn , hereFqn may be replaced by any of its finite extensions. So,X ′

is defined overFqn , andG is a subgroup of PGL(r′ + 1, qn).Let Φn be the Frobenius collineation ofPG(r′,K). For anyγ ∈ G, the product

γ′ = Φnγ−1

is also a collineation ofPG(r′,K ′) which preservesX ′. LetNn(X ′, γ) be the num-ber of all fixed points ofγ′ onX ′. Such points ofX ′ lie in ans-dimensional pro-jective subspace∆γ of PG(r′,K) consisting of all fixed points ofγ′ in PG(r′,K).Note that∆γ need not be consisting of points with coordinates inFqn since∆γ isthe image ofPG(s, qn) by a linear collineation belonging to PGL(r, qin) for somei ≥ 1. Again, for the present purpose, the finite extensionFqin may be replaced byFqn . Then the fixed points ofγ′ areFqn-rational points of theFqn-rational curveX ′.

To apply the Stohr–Voloch Theorem 8.62 in the same way as in the proof ofProposition 10.22, a further replacement ofFqn by a larger finite field may beneeded in order to ensure thatqn > 4g′4(g′−1)2. Doing so,Nn ≤ qn+1+2g

√qn

holds, whence

Nn(X ′, γ) ≤ qn + 1 + 2g′√qn. (10.19)

If Q is a point ofℓ, andP1, . . . , PN are the distinct points ofX ′ lying overQ inthe coveringX ′ → ℓ, thenQ is anFq-rational point ofℓ if and only if Φn(P1)coincides withγ(P ) for someγ ∈ G. Since the pointsP1, . . . , PN form an orbitof G, eitherN = |G| and the orbit is long, or|N | < |G| is a divisor of|G| and theorbit is short. The latter case only occurs for finitely many,more precisely, at mostg − 1 + |G| pointsQ, by Theorem 12.54 and Remark 12.57. Sinceℓ hasqn + 1Fq-rational points, it follows that

∑γ′∈G Nn(X ′, γ′) = |G|(qn + 1)− c, (10.20)

wherec > 0 andc only depends ong andG but is independent ofqn.From (10.19) and (10.20),

Nn(X ′, γ) = qn +O(√qn).

The above argument still works ifℓ is replaced byX , and it shows that∑

γ′∈H Nn(X ′, γ′) = |H|Nn +O(√qn).

Therefore,

Nn = qn +O(√qn).

2

The Hasse–Weil Bound also provides an estimate for the number Bn of closedFq-rational points of a given degreen. The idea is to use the equation,

Nn =∑

d|n d ·Bd, (10.21)


where the sum is extended over all positive divisorsd of n. The Mobius InversionFormula, see [188, Theorem 3.24], transforms (10.21) into

n ·Bn =∑

d|nµ(nd

)·Nd, (10.22)

whereµ : N → 0, 1,−1 is the Mobius function. Sincen > 1, from [188,Lemma 3.21],

∑

d|nµ(nd

)= 0,

substituting (10.10) into (10.22) gives the following result.

PROPOSITION 10.24 For anyn ≥ 2,

Bn =1

n

∑

d|nµ(nd

)(qd −

2g∑

i=1

ωni

),

whereωi, . . . , ω2g are the reciprocals of the roots of theL-polynomialLq(t) ofX .

From Proposition 10.24, an upper bound onBn is obtained which only depends onqn andg.

THEOREM 10.25 (i) For everyn ≥ 2,∣∣∣∣Bn −

qn

n

∣∣∣∣ ≤(

g

q − 1+ 2g

√q

√q − 1

)·√qn − 1

n< (2 + 7g) ·

√qn

n.

(ii) For anyn satisfying the condition,

2g + 1 ≤√qn−1(

√q − 1), (10.23)

the curveX has at least one closed point of degreen.

(iii) If n ≥ 4g + 3 thenBn ≥ 1.

Proof. (i) The assertion forn = 1 follows readily from the Hasse–Weil Bound(10.14). Forn ≥ 2, Proposition 10.25) gives

Bn −qn

n=

1

n

∑

d|n,d<n

µ(nd

)qn − 1

n

∑

d|n

2g∑

i=1

ωdi .

Let m = ⌊ 12n⌋ be the integer part of12n. From the Hasse–Weil Theorem 10.19,

|∑2gi=1 ω

di | ≤ 2g

√qd. Therefore,

∣∣∣∣Bn −qn

n

∣∣∣∣ ≤1

n

m∑

d=1

qn +2g

n

n∑

d=1

√qd =

q

n· q

m − 1

q − 1+

2g√q

n·√qn − 1√q − 1

≤(

q

q − 1+ 2g

√q

√q − 1

)·√qn − 1

n≤ (2 + 7g) ·

√qn

n.

350

(ii) By (i), the inequalityBn > 0 holds, provided that

qn

n>

(q

q − 1+ 2g

√q

√q − 1

)·√qn − 1

n. (10.24)

If g = 0, then (10.24) holds for alln ≥ 1. So, it may be assumed thatg ≥ 1. Adirect calculation shows that (10.24) can be written as

2g +

√q

√q + 1

<qn(√q − 1)

√q(√qn − 1)

. (10.25)

Taking into account the inequalities,

2g +

√q

√q + 1

< 2g + 1,√qn−1(

√q − 1) <

qn(√q − 1)

√q(√qn − 1)

,

the asumption (10.23) implies (10.25). ThereforeBn > 0.(iii) If n ≥ 4g + 3, then

2g + 1 < 22g+1(√q − 1) ≤

√2

n−1(√

2− 1) ≤√qn−1(

√q − 1),

whence it follows thatBn > 0. 2

It is of interest to compare the bounds (8.17) and (10.14), atleast for the seriesg2

n cut out by lines in the case of a plane non-singular curve of degreen.

L EMMA 10.26 The bound given by Theorem 8.62 is better than that given by The-orem 10.18 whenn ≥ 1

2

√q + 3.

Proof. Forg = 12 (n− 1)(n− 2), the bounds given by (8.17) and (10.14) are

12n(n+ q − 1) (SV),q + 1 + (n− 1)(n− 2)

√q (HW).

Then12n(n+ q − 1) ≤ q + 1 + (n− 1)(n− 2)

√q ⇐⇒

n2 + n(q − 1) ≤ 2q + 2 + 2(n2 − 3n+ 2)√q ⇐⇒

n2(2√q − 1)− n(q + 6

√q − 1) + 2(q + 2

√q + 1) ≥ 0. (10.26)

The discriminant of (10.26) is

δ=(q + 6√q − 1)2 − 8(2

√q − 1)(q + 2

√q + 1)

= (q − 2√q + 3)2.

Hence (10.26) holds when

n ≤ (q + 6√q − 1)− (q − 2

√q + 3)

2(2√q − 1)

= 2

or

n≥ (q + 6√q − 1) + (q − 2

√q + 3)

2(2√q − 1)

=(√q + 1)2

2√q − 1

=1

2

√q +

5

4+

9

4(2√q − 1)

,

which holds forn ≥ 12

√q + 3. 2


10.3 REFINEMENTS OF THE HASSE–WEIL THEOREM

The Hasse–Weil Bound 10.18 is a powerful result even though it only depends ontwo parameters, the genusg of the curveX and the orderq of the finite field overwhichX is defined. Even though the bound is sharp for an infinite number of valuesof q andg, see Chapter 11, some useful refinements of the bound are possible, andthis is the object of the present section.

An immediate refinement derives from the fact that the term2g√q in the Hasse–

Weil Bound (10.18) may be replaced by its integer part. Therefore,

|N1 − (q + 1)| ≤ ⌊2g√q⌋. (10.27)

In this direction, a non-trivial improvement of (10.27) is due to Serre.

THEOREM 10.27 (Serre Bound)

|N1 − (q + 1)| ≤ g⌊2√q⌋ (10.28)

Proof. For g = 0, equality holds in (10.28). Sog ≥ 1 is assumed and theL-polynomial (10.7) is considered. Its rootsω1, . . . , γ2g are algebraic integers, thatis, they belong to the subringA of C consisting of all complex numbers which areroots of polynomials with integer coefficients. All integers are inA, and

A ∩Q = Z. (10.29)

By the Hasse–Weil Theorem 10.19,|ωi| = √q for i = 1, . . . , 2g. Also, by the laststatement in Proposition 10.9, these roots may be ordered such thatωiωg+i = qholds fori = 1, . . . , g. So, ifωi denotes the complex conjugate ofωi, then

ωi = αg+i = q/ωi, for 1 ≤ i ≤ g.Let

αi =ωi + ωi + ⌊2√q⌋+ 1;

βi =−(ωi + ωi) + ⌊2√q⌋+ 1.

Thenαi, βi are real algebraic integers. Also, since|ωi| = √q,αi > 0, βi > 0, for i = 1, . . . , g. (10.30)

The monic minimal polynomial of the field extensionQ(ω1, . . . , ωg) of Q is

L⊥(t) =∏2g

i=1 (t− ωi),

whose roots are the reciprocals of the roots ofL(t). SinceL(t) ∈ Z[t], the sameholds true forL⊥(t).

Let σ be any embedding ofQ(ω1, . . . , ωg) in C. If σ(ωi) = ωj , then

σ(ωi) = σ(q/ωi) = q/σ(ωi) = σ(ωi) = ωj .

Therefore,σ induces permutations on both the setsα1, . . . , αg andβ1, . . . , βg.Hence, both algebraic integers

α =∏g

i=1 αi, β =∏g

i=1 βi

352

are are left invariant by all embeddings ofQ(ω1, . . . , ωg) into C. This implies thatα, β ∈ Q ∩A. Hence, by (10.29),α, β ∈ Z. From (10.30),α ≥ 1 andβ ≥ 1.

Since the arithmetic mean of positive real numbers is greater than or equal to thegeometric mean of the same numbers,

g−1∑gi=1 αi ≥ (

∏gi=1 αi)

1/g ≥ 1,

whence

g ≤∑gi=1 (ωi + ωi) + g⌊2√q⌋+ g =

∑2gi=1 ωi + g⌊2√q⌋+ g.

From (10.10),∑2g

i=1 ωi = (q + 1)−N1. Hence

N1 ≤ q + 1 + g⌊2√q⌋.Similarly, the inequality,

g−1∑gi=1 βi ≥ (

∏gi=1 βi)

1/g ≥ 1,

implies that

N1 ≥ q + 1− g⌊2√q⌋.2

THEOREM 10.28 In the Serre Bound, equality is attained if and only if theL-polynomial ofX is

Lq(t) = (1± ⌊2√q⌋t+ qt2)g, (10.31)

with + or − according as the bound is upper or lower.

Proof. The arithmetic mean of positive real numbersxi equals to their geometricmean if and only if allxi are equal. From the proof of Theorem 10.27, if equalityholds in the upper bound (10.28) thenα1 = . . . = αg and henceω1 = . . . = ωg.Let ω = ω1. Then

g(ω + q/ω) = q + 1−N1 = −⌊2√q⌋,whence (10.31) follows with+.

For the converse, (10.31) with+ implies that

−⌊2√q⌋ = ω1 + q/ω1 = . . . = ωg + q/ωg.

From (10.10),N1 − (q + 1) = g⌊2√q⌋.A similar argument usingβ shows the assertion for the lower bound in (10.28)

with −. 2

REMARK 10.29 The Klein quartic overF8, as in Example 10.14, shows that theupper Serre Bound is sharp. For another curves attaining theSerre Bound, seeExercise 4 and Section 13.5.

Now, another general formula forN1 is given. It may appear complicated butcan provide good estimates for appropriate choices of the polynomials.


THEOREM 10.30 (Serre’s Explicit Formula)Take any sequencec1, c2, . . . of realnumbers and let, for anym ≤ 1,

λm(t) =∑m

i=1 cmtm, fm(t) = 1 + λm(t) + λm(t−1). (10.32)

Then

λm

(1√q

)N1 = λm(

√q)+λm

(1√q

)+g−

g∑

i=1

fm(αi)−m∑

i=1

(Nn−N1)cn1√qn.

(10.33)

Proof. Let

αi = ωi/√q, i = 1, . . . , 2g, (10.34)

and observe that|αi| = 1 by the Hasse–Weil Theorem 10.19. As in the proof ofTheorem 10.27, these elements may be ordered in such a way that

αg+i = αi = α−1i , for i = 1, . . . , g.

From Theorem 10.10,

Nn√qn

=√qn +

1√qn−

g∑

i=1

(αn

i +1

αni

). (10.35)

Note thatfm(t) ∈ R for all t ∈ C with |t| = 1. Multiply (10.35) bycn and writethe result as follows:

cnN1√qn

= cn√qn +cn

1√qn−

g∑

i=1

cn

(αn

i +1

αni

)−(Nn−N1)cn

1√qn. (10.36)

Summing up the equations (10.36) forn = 1, . . . ,m gives the result. 2

A useful corollary is the following result.

PROPOSITION 10.31 (i) If c1, . . . , cm is a sequence of non-negative real num-bers not all zero, and the polynomialfm(t) defined in (10.32) is non-negativefor all t ∈ C with |t| = 1, then

N1 ≤g

λm(√q−1)

+λm(√q)

λm(√q−1)

+ 1, (10.37)

with λm(t) as defined in (10.32).

(ii) Equality holds in (10.37) if and only ifNn = N1, for n = 1, . . . ,m, and∑gi=1 fm(αi) = 0.

Proof. Note thatλm(√q−1) 6= 0, as the sequencec1, . . . , cm is supposed to con-

tain a non-zero element. SinceN1 ≤ Nn, Serre’s Explicit Formula implies that

λm(√q−1)N1 ≤ λm(

√q) + λm(

√q−1) + g.

Division byλm(√q−1) gives the result. 2

354

Before showing two important consequences of Proposition 10.31, an equationsimilar to (10.33) is stated.

With the notation of (10.9) and (10.34), let

αj = cosϕj + i sinϕj

for j = 1, . . . 2g. Then (10.10) can also be written as

Nn = qn + 1−√qn∑g

j=1 2 cosnαj . (10.38)

If Bd denotes the number ofFq-rational closed points ofX of degreed, then

Nn =∑

d|n dBd.

For a sequencec1, c2, . . . , of real numbers, define the trigonometric series,

f(α) = 1 + 2∑∞

j=1 cj cos jα,

and, for every positive integern, the power series,

Ψn(t) =∑∞

j=1 cjntjn.

From thisg∑

j=1

f(αj) +

∞∑

d=1

dBd Ψd(t) = g + Ψ1

(1√q

)Ψ1(√q), (10.39)

which is theWeil’s Explicit Formula.Now, two cases are examined, related to the maximal curves having the Suzuki

and Ree simple groups as their automorphisms.

PROPOSITION 10.32 Let q = 2q20 with q0 = 2s ands ≥ 1.

(i) If g = q0(q − 1), thenN1 ≤ q2 + 1.

(ii) If N1 = q2 + 1, then

Z(X , t) =(1 + 2q0t+ qt2)g

(1− t)(1− qt) ;

Nn =1 + qn − g√qnκ,

where

κ =

−2 if n ≡ 4 (mod 8),

−√

2 if n ≡ 1, 7 (mod 8) ,0 if n ≡ 2 (mod 4),√2 if n ≡ 3, 5 (mod 8),

2 if n ≡ 0 (mod 8).

Proof. Letm = 2, c1 = 12

√2, c2 = 1

4 . Then

λ2(t) = 1√2t+ 1

4qt2.

If |t| = 1, write t = cosϕ+ i sinϕ, and calculatef2(t). The result is that

f2(t) = ( 1√2

+ cosϕ)2,


and hencef2(t) ≥ 0 for |t| = 1. Applying (10.37) givesN1 ≤ q2 + 1.Suppose that equality holds and writeαj = cosφj + i sinφj . Then

∑gj=1 f2(αj) =

∑gj=1 ( 1√

2+ cosφj)

2 = 0

which implies thatcosφ1 = . . . = cosφg = − 1√2. Therefore,

α1 = . . . = αg = − 1√2

+ i 1√2, αg+1 = . . . = α2g = − 1√

2+ i 1√

2.

Direct computation shows the remaining parts. 2

REMARK 10.33 The DLS curve satisfies the hypothesis of Proposition 10.32.

PROPOSITION 10.34 Let q = 3q20 with q0 = 3s ands ≥ 1.

(i) If

g = 32 q0(q − 1)(q + q0 + 1),

thenN1 ≤ q3 + 1.

(ii) If N1 = q3 + 1, then

Z(X , t) =(1 + 3q0t+ qt2)q0(q

2−1)(1 + qt2)q0(q−1)(q+3q0+1)/2

(1− t)(1− qt) ,

Nn = 1 + qn

−q0√q(q − 1)(q + 3q0 + 1) cos 1

2πn+ 2(1 + q) cos 56πn.

Proof. The same argument in the proof of Proposition 10.32 is used, but with adifferent choice of the sequencec1, c2, . . ., namely,c1 = 1√

3, c2 = 1

3 . The calcu-lations give the following result: fort = cosφ+ i sinφ,

f2(t) = 13 (√

3 + 2 cosϕ)2 cos2 ϕ.

This and∑g

j=1 f2(αj) = 0 imply that eitherαj = ± 12 i, or αj = −

√3

2 ± 12 i.

Therefore,ωj + ωj can only assume two values, namely,0 and−√3q.Now, letk denote the number ofj with 1 ≤ j ≤ g such thatωj + ωj = −√3q.

Then (10.10) applied toN1 givesq3 − q = m2

√3q, whencek = q0(q

2 − 1). Thisalso shows thatωj + ωj = 0 occursg − k = 1

2q0(q − 1)(q + 3q0 + 1) times.Therefore, up to re-labelling the indices,

ω1 = . . . = ωk = 12

√q i, ωg+1 = . . . = ωg+k = − 1

2

√q i,

ωk+1 = . . . ωg = (−√

32 + 1

2 i)√q, ωg+k+1 = . . . = ω2g = (−

√3

2 − 12 i)√q.

From this and equation (10.10),Z(X , t) andNn can be obtained. 2

REMARK 10.35 The DLR curve satisfies the hypothesis of Proposition 10.32.

356

10.4 ASYMPTOTIC BOUNDS

Proposition 10.31 is also useful for investigating the asymptotic behaviour ofNn

as the genus goes to infinity.

DEFINITION 10.36 (i) Let

Ng(q) = maxN1

taken over all irreducible non-singular algebraic curves of genusg definedoverFq.

(ii) Let

A+(q) = lim supg→∞

Ng(q)/g.

(iii) Let

A−(q) = lim infg→∞

Ng(q)/g.

The Hasse-Weil Bound implies thatA+(q) ≤ 2√q. This bound was improved by

Ihara using the Hasse-Weil Bound forN1 andN2 and the inequalityN1 ≤ N2.

THEOREM 10.37 (Ihara Bound)

A+(q) ≤ 12 (√

8q + 1− 1).

Refining the above idea, a better bound can be obtained.

THEOREM 10.38 (Drinfeld–Vladut Bound)

A+(q) ≤ √q − 1.

Proof. For a fixed positive integerm, consider the sequencec1, . . . , cm, where

cn = 1− n

m, for n = 1, . . . ,m.

Then the functions in (10.32) are as follows:

λm(t) =m∑

n=1

(1− n

m

)tn =

t

(1− t)2(tm − 1

m+ 1− t

); (10.40)

fm(t) = 1 + λm(t) + λm(t−1) =2− (tm + t−m)

m(t− 1)(t−1 − 1). (10.41)

Sincet−1 = t for |t| = 1, equation (10.41) implies thatfm(t) ≥ 0 for all t ∈ C.From Proposition 10.31 applied to the sequencec1, . . . , cm,

N1

g≤ 1

λm(√q−1)

+1

g

(1 +

λm(√q)

λm(√q−1)

). (10.42)

Since (10.40) gives

limm=∞

λm(√q−1) =

√q−1

(1−√q−1)2

(1−√q−1) =

1√q − 1

,


for anyǫ > 0 there existsm0 such that

λm0(q−1/2) <

√q − 1 + 1

2 ǫ.

Now, choosingg0 such that

1

g0

(1 +

λm(√q)

λm(√q−1)

)< 1

2 ǫ,

(10.42) implies that

N1

g<√q − 1 + ǫ

for everyg ≥ g0. ThereforeA+(q) ≤ √q − 1. 2

On the other side,

A+(q) ≥ √q − 1 for q square. (10.43)

Therefore, the Drinfeld–Vladut Bound is sharp forq square.

DEFINITION 10.39 (i) A tower T over Fq, or anFq-tower, is an infinite se-quence of irreducible non-singular curves together with surjective and sepa-rable maps, both defined overFq,

. . .→ Xk+1 → Xk → X2 → X1

such that

limk→∞

g(Xk) =∞,

whereg(Xk) is the genus ofXk.

(ii) The limit of the towerT is

λ(T ) = limk→∞

|Xk(Fq)|g(Xk)

.

(iii) A tower is asymptotically goodif λ(T ) > 0, and isrecursiveif it uses thesame equation to construct all curves in the tower.

Note thatA+(q) ≥ λ(T ) for anyFq-towerT . Therefore, to show that, in theDrinfeld–Vladut Bound, equality holds over fields of square order, it suffices toconstruct anFq2-towerT satisfyingλ(T ) = q − 1.

As illustrative examples, two suchFq2-towers are presented by explicit algebraicequations.

EXAMPLE 10.40 λ(T ) = q − 1 for theFq2-towerT defined as follows.

(i) X1 is the linev(X1), that is, an irreducible non-singular model of the rationalfunction fieldK(x1).

(ii) X2 is the non-singular plane curvev(Y 2+Y −Xq+1), that is, a non-singularmodel overFq of the function fieldK(x1, z2) with zq

2 + z2 = xq1.

358

(iii) X3 is is a non-singular model overFq of the function fieldK(x1, x2, z2, z3)with

zq2 + z2 = xq

1, zq3 + z3 = xq+1

2 , x2 =z2x1.

(iv) X4 is a non-singular model overFq of the fieldK(x1, x2, x3, z2, z3, z4) with

zq2 + z2 = xq+1

1 , zq3 + z3 = xq+1

2 , zq4 + z4 = xq+1

3 , x3 =z3x2.

(v) X5,X6, . . . continue in this way.

EXAMPLE 10.41 The limit λ(T ) = q− 1 for theFq2-towerT defined as follows.


(ii) X2 is a non-singular model overFq2 of the fieldK(x1, x2) with

xq2 + x2 = xq

1/(xq−11 + 1).

(iii) X3 is is a non-singular model overFq2 of the fieldK(x1, x2, x3) with

xq2 + x2 =

xq1

xq−11 + 1

, xq3 + x3 =

xq2

xq−12 + 1

, xq4 + x4 =

xq3

xq−13 + 1

.

(iv) the curvesX4,X5, . . . continue in this way.

The mapXk+1 → Xk, that is, the rational transformation

K(Xk+1) = K(x1, . . . , xk, xk+1)→ K(Xk) = K(x1, . . . , xk),

is given by

(x1, . . . , xk, xk+1) 7→ (x1, . . . , xk).

The tower is recursively defined by the following equation:

yq + y = xq/(xq−1 + 1).

THEOREM 10.42 For arbitrary q, there exists a positive constantc such that

A+(q) > c log q. (10.44)

As a consequence,A+(q) > 0 for all q.

EXAMPLE 10.43 Let q > p, and letm = (q − 1)/(p− 1). Then

A+(q) ≥ λ(T ) ≥ 2

q − 2

for theFq-towerT defined as follows.



(ii) X2 is the non-singular plane curvev(Y m + (X + 1)m − 1), that is, a non-singular model overFq of the fieldK(x1, x2) with xm

2 + (x1 + 1)m = 1.

(iii) X3 is a non-singular model overFq of the fieldK(x1, x2, x3) with

xm3 + (x2 + 1)m = 1, xm

2 + (x1 + 1)m = 1.

(iv) the curvesX4,X5, . . . continue in this way.

REMARK 10.44 (i) For smallq, the following lower bounds are known:

A+(2) ≥ 29 ; A+(3) ≥ 1

3 ; A+(5) ≥ 12 .

(ii)

A+(q3) ≥ 2(q2 − 1)

q + 2.

(iii) If l is a prime number such thatq > 4l + 1 andq ≡ 1 (mod l), then

A+(ql) ≥ (√l(q − 1)− 2l)/(l − 1).

The significance ofA−(q) is that for it to be greater than some constantc it isnecessary that, for every sufficiently largeg, there is a curve of genusg definedoverFq with at leastcg points.

THEOREM 10.45 For everyq, the constantA−(q) is positive.

THEOREM 10.46 There exists a constantd such that, for allq,

A−(q) ≥ d log q.

For q square, this can be improved.

THEOREM 10.47

A−(q) ≥

√q − 1

2 + (log 2)/(log q)if q is an even square;

√q − 1

2 + (log 4)/(log q)if q is an odd square.

10.5 OTHER ESTIMATES

Ad hoc methods can produce estimates onN1 that are better than both the Hasse–Weil and the Stohr–Voloch Bounds in some rather special cases. Two such methodsfor producing upper bounds onN1 derive from the gonality of the curveX and thefirst non-gap at anFq-rational point.

For anyξ ∈ Fq(X ), theFq-rational transformationx′ = x, y′ = 0 defines anFq-rational covering of degreed = [K(X ) : K(x)].

360

PROPOSITION 10.48N1 ≤ d(q + 1) (10.45)

Proof. The projective lineℓ which is a non-singular model ofK(x) hasq + 1rational points. IfP is anFq-rational point ofX then the point ofℓ lying underPis anFq-rational point ofℓ. Although the points ofX lying overQ′ ∈ ℓ need notbeFq-rational, the bound is obtained by counting the points ofX lying over theFq-rational points ofℓ. 2

DEFINITION 10.49 Thegonalityof X istq(X ) = min d

taken over all coverings.

COROLLARY 10.50tq(X ) ≥ N1/(q + 1). (10.46)

REMARK 10.51 (i) The gonality ofX is one if and only ifX is rational.

(ii) The gonality ofX is two if and only ifX is either elliptic or hyperelliptic.

(iii) If N1 ≥ 1 andX is birationally equivalent overFq to an irreducible planecurve of degreem, then the gonality ism− 1.

PROPOSITION 10.52 (i) tq(X ) ≤ g + 1.

(ii) If N1 > 0 anddegE ≥ tq(X )− 1, thenℓ(E) ≤ degE + 2− t.The following result shows that, as long asN1 > 0, the parameterxmay be chosenso thatd ≤ g for g ≥ 2, and that this choice can be made in such a way that thenumberd(P ′

∞) of Fq-rational points lying over the infinite pointP ′∞ of ℓ is very

small.

L EMMA 10.53 LetX have genusg ≥ 2. If X has anFq-rational pointP, thenthere existsx ∈ Fq(X ) for which[K(X ) : K(x)] ≤ g andd(P ′

∞) ≤ 2.

Proof. From the Riemann–Roch Theorem 6.59, there is a canonical divisorE ofX such thatE ≻ (g − 1)P . By Remark 8.27, takeE to beFq-rational.

Write E = (g − 1)P + A and choose a pointQ ∈ SuppA. SinceA is anFq-rational divisor andD is the closed pointD = Q + Φ(Q) + . . . + Φk−1(Q) ofdegreek, thenA ≻ D. LetB = (g− d)P +D. SinceE ≻ B, the speciality indexi(B) is positive. From the Riemann–Roch Theorem 6.59,dimB ≥ 1. SinceB isanFq-rational divisor, there existsx ∈ Fq(X ) such that div(x)∞ = degB = g;this proves the first assertion.

Now, P ′∞ is the point ofℓ lying underP . Therefore,d(P ′

∞) ≤ 2, and equalityholds if and only ifk = 1, that is,Q ∈ X (Fq). 2

As a corollary of Lemma 10.53, the following upper bound onN1 is obtained.

PROPOSITION 10.54 If g ≥ 2, the following hold.

(i) N1 ≤ qd+ 2.

(ii) If n is a non-gap at anFq-rational point ofX , thenN1 ≤ qn+ 1.


10.6 COUNTING POINTS ON A PLANE CURVE

If F is a non-singular plane curve of degreen defined overFq, thenX can beidentified withF , and the Hasse–Weil Bound (10.14) reads as follows:

q + 1− (n− 1)(n− 2)√q ≤ Sq ≤ q + 1 + (n− 1)(n− 2)

√q, (10.47)

wheren is the degree ofF andSq = N1 is the number of allFq-rational points ofF . Now, (10.47) is extended to irreducible singular plane curves defined overFq.

If a curve is not identified with its non-singular model, thenthere is some am-biguity in the definition of anFq-rational point. For a non-singular plane curveF defined overFq, there is a one-to-one correspondence between theFq-rationalpoints ofF and theFq-rational places of the associated function fieldK(C); seeTheorem 8.18. IfF is singular andX is a non-singular model ofF , defined overFq as well, thenF andX have the same function field and hence they have thesame numberN1 of Fq-rational points, butN1 is, in general, different from thenumberRq of points ofF which lie inPG(2, q).

This already appeared in Chapter 8, in Section 8.8. For instance, if C3 is anirreducible cubic curve defined overFq with an isolated double point, thenSq =q + 1 but, from Table 8.1,Rq = q + 2.

From now on,F is any irreducible plane curve of degreen and genusg definedoverFq. To prove the desired result

q + 1− (n− 1)(n− 2) ≤ Rq ≤ q + 1 + (n− 1)(n− 2), (10.48)

it must be shown that (10.47) holds true whenN1 is replaced byRq.To compareRq with Sq = N1, certain other parameters forF must be defined:

Rq = number of pointsP ∈ C that lie inPG(2, q);

Sq =N1 = number ofFq-rational points ofF ;

R∗q = number of pointsP ∈ PG(2, q) which are centres ofFq-rational branches;

Bq = number of branches ofF centred at points ofPG(2, q);

Eq = number of singular points ofF ;

bP = number ofFq-rational branches ofF with centre atP ∈ PG(2, q);

cP = number of all branches centred ofF with centre atP ∈ PG(2, q);

mP = multiplicity of a pointP ∈ F .As in Chapter 8,F(Fq)

∗ stands for the set of all pointsP ∈ PG(2, q) lying onF .With this notation,

Sq =∑

P∈F(Fq)∗

bP , Rq = |F(Fq)|.

Also, R∗q ≤ Sq and equality holds if and only if no two distinctFq-rational

branches ofF have the same centre.The starting point is an upper bound and a lower bound forSq −Rq.

L EMMA 10.55

Sq −Rq ≤ 12 (n− 1)(n− 2)− g. (10.49)

362

Proof.

Sq −Rq =∑

P∈F(Fq)∗

(bP − 1) ≤∑

P∈Fq(F)

(cP − 1)

≤∑

P∈F(Fq)∗

(mP − 1) ≤∑

P∈F(Fq)∗

12 mP (mP − 1)

≤∑

P∈F

12 mP (mP − 1) ≤ 1

2 (n− 1)(n− 2)− g.

The first equality follows from the definitions. The first two inequalities followfrom the fact thatbP ≤ cP ≤ mP ; see (iii) of Theorem 4.34. The last inequalityfollows from Lemma 3.24 and Definition 5.54. 2

L EMMA 10.56

Rq − Sq ≤ 12 (n− 1)(n− 2)− g. (10.50)

Proof. SinceR∗q ≤ Sq it follows thatRq − Sq ≤ Rq − R∗

q . On the other hand,Rq−R∗

q ≤ S, as any pointP ∈ F(Fq)∗ for whichbP = 0 andcP > 0 is a singular

point ofF . Therefore

Rq − Sq ≤ Rq −R∗q ≤ S ≤ 1

2 (n− 1)(n− 2)− g. (10.51)

2

THEOREM 10.57 LetF be an irreducible plane curve defined overFq of degreen and genusg. If Rq is the number of points ofPG(2, q) lying onF , then

(i) |Rq − (q + 1)| ≤ g⌊2√q⌋+ 12 (n− 1)(n− 2)− g.

(ii) |Rq − (q + 1)| ≤ 12 (n− 1)(n− 2)⌊2√q⌋. (Serre Bound)

(iii) |Rq − (q + 1)| ≤ (n− 1)(n− 2)√q. (Hasse–Weil Bound)

Proof. (i) The Serre Bound (10.28) together with Lemmas 10.55 and 10.56 implythe following:

|Rq − (q + 1)| ≤ |Rq − Sq)|+ |Sq − (q + 1)|≤ 1

2 (n− 1)(n− 2)− g + g⌊2√q⌋.(ii) From (i) and the inequality,g ≤ 1

2 (n− 1)(n− 2), of Lemma 3.24,

|Rq − (q + 1)| ≤ g(⌊2√q⌋ − 1) 12 (n− 1)(n− 2)

≤ 12 (n− 1)(n− 2)(⌊2√q⌋ − 1) + 1

2 (n− 1)(n− 2)

= 12 (n− 1)(n− 2)⌊2√q⌋.

(iii) This is a consequence of (ii). 2

Under some additional hypotheses, equality can be attainedin the above bounds.

THEOREM 10.58 (i) Equality occurs in Lemma 10.56 if and only if every sin-gular point ofF is an isolated double point lying inPG(2, q).


(ii) Equality occurs in Lemma 10.55 if and only if every singular point ofF is anode.

Proof. (i) Assume that equality holds in (10.50). From (10.51),Sq = R∗q . Hence

every pointP ∈ F(Fq)∗ is the centre of at most oneFq-rational branch ofF .

Therefore,Rq−Sq gives the number of singular pointsP ∈ F(Fq)∗ which are not

the centre of anFq-rational branch ofF .Equality in (10.51) implies thatRq − Sq = S. Hence, every singular pointP

of F lies in PG(2, q), andP is not the centre of anyFq-rational branch ofF . Inparticular, at least two branches ofF are centred atP .

Now,∑

P∈F

12mP (mP − 1) ≥ g.

Since equality in (10.51) also implies thatS = 12 (n−1)(n−2)−g, every singular

pointP of F is an ordinary double point. Therefore, every singular point of F isan isolated double point lying inPG(2, q).

To prove the converse, suppose that every singular point ofF is an isolated dou-ble point. In particular,F has only ordinary singularities and hence, by Theorem5.56,S = 1

2 (n−1)(n−2)−g holds. AsRq−R∗q counts the points inF(Fq)

∗ thatare not the centre of anyFq-rational branch, it follows thatS = Rq − R∗

q . Thusonly the non-singular points inF(Fq)

∗ are the centres ofFq-rational branches ofF . HenceSq = R∗

q . Therefore,

Rq − Sq = Rq −R∗q = Eq = 1

2 (n− 1)(n− 2)− g.(ii) Assume that equality holds in (10.49). Then, from the proof of Lemma 10.55,

∑

P∈F(Fq)∗

(bP − 1) =∑

P∈F(Fq)∗

(cP − 1).

SincebP ≤ cP for all P ∈ F(Fq)∗, this implies thatbP = cP ; that is, every branch

of F centred at a point inF(Fq)∗ is Fq-rational. Also, from the proof of Lemma

10.55,∑

P∈F(Fq)

12 mP (mP − 1) =

∑

P∈F

12 mP (mP − 1),

which shows that every singular point ofF is inF(Fq)∗. From the proof of Lemma

10.55, every singular point ofF is a double point. Again from the proof of Lemma10.55,bP = cP = mP and hencevP = 2 for every singular pointP of F . Thisshows that each such pointP is a node.

Conversely, if the singularities ofF are nodes, thenS = 12 (n − 1)(n − 2) − g,

every singular point ofF belongs toF(Fq)∗, andbP = cP = mP for every point

P ∈ F(Fq)∗. Therefore,Sq −Rq = S and hence

Sq −Rq = 12 (n− 1)(n− 2)− g.

2

The following corollaries show that when equality occurs either in Lemma 10.55or in Lemma 10.56,q cannot be too small compared ton.

364

COROLLARY 10.59 If equality holds in Lemma 10.55, then

(i) (n− 1)(n− 2)− 2g ≤ Sq;

(ii) (n− 1)(n− 2)− 2g ≤ q + 1 + g⌊2√q⌋.Proof. From Theorem 10.58(ii), every singular point ofF is a node, and henceS = 1

2 (n − 1)(n − 2) − g. Therefore(n − 1)(n − 2) − 2g = 2S ≤ Sq. Thistogether with Theorem 10.57(ii) gives (ii). 2

COROLLARY 10.60 LetF be an irreducible singular plane curve of degreen ≥ 3defined overFq for which equality holds in Lemma 10.56.

(i) q ≥ n− 2 + (Sq − 2g − 2)/(n− 2).

(ii) If g = 0, thenq ≥ n− 1.

(iii) If n is odd andn ≥ 5, thenq ≥ n− 1− 2g/(n− 3).

Proof. (i) If P is any point inPG(2, q), the pencilΠP of lines ofPG(2, q) throughP cover every other point inPG(2, q) just once. Now, supposeP to be a singularpoint of F . From Theorem 10.58(ii),P is an isolated double point ofF . Also,from the proof of Lemma 10.55,Sq = R∗

q .For any lineℓ throughP , Bezout’s Theorem 3.14 becomes the following:

∑

Q∈ℓ∩F(Fq)∗

I(P, ℓ ∩ F) ≤ n.

Summing over all lines in the pencilΠP , this gives the inequality,∑

ℓ∈ΠP

∑

Q∈ℓ∩F(Fq)∗

I(P, ℓ ∩ F) ≤ n(q + 1).

Counting this double sum in a different way,∑

ℓ∈ΠP

∑

Q∈ℓ∩F(Fq)∗

I(P, ℓ ∩ F) =∑

ℓ∈ΠP

I(P, ℓ ∩ F)

+∑

Q∈Sing(F)\P

I((Q, ℓ ∩ F) +∑

Q∈F(Fq)∗\Sing(F)

I(Q, ℓ ∩ F),

whereSing(F) is the set of all singular points ofF . Also, since noFq-rationalbranch ofF is centred at a singular point ofF and the singularities ofF are doublepoints,

∑

ℓ∈ΠP

I(P, ℓ ∩ F)= 2(q + 1),

∑

Q∈Sing(F),Q6=P

I((Q, ℓ ∩ F)= 2(Eq − 1).

Also, sinceSq = R∗q ,

∑

Q∈F(Fq)∗\Sing(F)

I(Q, ℓ ∩ F) = Sq.


Therefore

2(q + 1) + 2(S − 1) + Sq ≤ n(q + 1).

As 2S = (n− 1)(b− 2)− 2g when equality holds in Lemma 10.56, so (i) follows.(ii) If g = 0, thenSq = q + 1 and (i) implies that

q ≥ n− 2 + (q − 1)/(n− 2).

Sinceq ≥ 2 andn > 2, part (ii) follows.(iii) SinceF is an irreducible plane curve, Bezout’s Theorem 3.14 implies that

no lineℓ contains more than⌊ 12n⌋ singular points ofF . As before, fix a singularpoint P and consider the pencilΠP of lines of PG(2, q) throughP . Each linecontains at most⌊ 12n⌋ − 1 singular points ofF other thanP . As already noted,Eq = 1

2 (n− 1)(n− 2)− g. Therefore, the inequality,(⌊ 12n⌋ − 1

)(q + 1) + 1 ≥ Eq = 1

2 (n− 1(n− 2)− g,follows.

If n is odd andn ≥ 5, then⌊ 12n⌋ = 12 (n− 1), which implies that

q ≥ n− 1− 2g/(n− 3).

Note that, ifn is even andn ≥ 4, then⌊ 12n⌋ = 12n, whence

q ≥ n− 2− (2g − 2)/(n− 2),

which is weaker than (i). 2

REMARK 10.61 The above results are sharp for curves of low degree and genus.Forg = 2 this is illustrated by two examples. Let

f(X0,X1,X2)

= X20 (X2

1 + 2X22 ) +X0(X

31 + 2X1X

22 +X3

2 ) + 3X31X2 + 3X1X

32 ,

g(X0,X1,X2)

= X20 (X2

1 −X22 ) +X0(X

21X2 −X3

2 ) + (6X43 + 6X2

1X2 − 4X42 .

Then the parameters are given in Table 10.1.

Table 10.1 Two curves

Curve q n g Rq Sq Bound onSq Bound onRq

F = v(f) 5 4 2 13 12 0 ≤ Sq ≤ 12 0 ≤ Rq ≤ 13

F = v(g) 13 4 2 1 2 2 ≤ Sq ≤ 26 1 ≤ Rq ≤ 27

An improvement on the bound in Proposition 10.45 is the following result.

366

PROPOSITION 10.62 If C is an irreducibleFq-rational curve of degreen > 1,thenRq ≤ (n− 1)q + ⌊n/2⌋.Proof. For each pointP ∈ C lying in PG(2, q) choose anFq-rational linetP suchthatI(P, C ∩ tP ) > 1. If P is non-singular, thentP is the tangent line toC atP . IfP is singular, then anyFq-rational line throughP has the required property. ThenumberuP of common points ofC with tP is at mostn − 1. Definek to be thesmallest value ofuP with P ranging over all points ofC lying in PG(2, q). ThenRq ≤ (n− 1)q + k as throughP there are exactlyq + 1 Fq-rational lines and oneof these lines istP .

A point P is said to “see a tangency” atQ ∈ C if tQ containsP ; hereP = Q isallowed. The points ofC lying in PG(2, q) can see at leastRqk tangencies, so theremust be some pointP ∈ C lying in PG(2, q) which sees at leastk tangencies. Onecan count the points ofC lying in PG(2, q) looking around fromP and finding thatRq ≤ q(n − 1) + n − k. Finally, sincemin(n − 1)q + k, (n − 1)q + n − k ≤(n− 1)q + ⌊n/2⌋, the assertion follows. 2

Some applications require information on the number of points ofPG(2, q) lyingon a plane curve not defined overFq. An upper bound on this number is as follows.

L EMMA 10.63 If an irreducible plane curve of degreen is defined overFqk butnot overFq, then the numberN of its points lying inPG(2, q) does not exceedn2.

Proof. By Bezout’s Theorem 3.14, two distinct irreducible curves of degreenmayintersect in up ton2 points. So, there can be no two such curves throughn2 + 1points. Thus an irreducible plane curve defined overFqk and containingn2 + 1points inPG(2, q) is unique and is defined overFq. HenceN ≤ n2. 2

In conclusion, two more bounds are proved that are applied inChapter 14.

THEOREM 10.64 LetC be any algebraic plane curve of degreen defined overFq

with no components defined overFq. Let S be the number of simple points ofCwhich lie inPG(2, q). If

√q > n− 1, (10.52)

then

S < n(q + 2− n). (10.53)

Proof. (i) Cn irreducible.Suppose (10.53) does not hold. Then, from (10.47), which is the Hasse–Weil

bound for plane curves,

n(q + 2− n) ≤ S ≤ Sq ≤ q + 1 + (n− 1)(n− 2)√q.

Hence

(n− 1)(√q + 1)(

√q + 1− n) ≤ 0,

and so√q ≤ n− 1, contradicting (10.52).

(ii) Cn reducible.


Suppose that the components ofCn areF1, . . . ,Fm, and letFi have degreeni

andSi simple points inPG(2, q) for i ∈ Nm. SinceS ≤∑Si, it suffices to showthat, for eachi,

Si < ni(q + 2− n). (10.54)

ThenS ≤∑Si <∑ni(q + 2− n) = n(q + 2− n).

Two cases must be distinguished, according as a component isFq-rational ornot.

(i) Fi anFq-rational component.By hypothesis,ni ≥ 2 and, sinceCn could not have just one notFq-rational

linear component,n ≥ ni + 2 ≥ 4.Suppose that (10.54) is false. Then, again by (10.47),

ni(q + 2− n) ≤ Si ≤ q + 1 + (ni − 1)(ni − 2)√q

⇐⇒ q(ni − 1)−√q(ni − 1)(ni − 2) ≤ ni(n− 2) + 1

⇐⇒√q(ni − 1)(√q − ni + 2) ≤ ni(n− 2) + 1

=⇒ (n− 1)(ni − 1)(n− ni + 1) ≤ ni(n− 2) + 1

=⇒ (n− 1)(ni − 1)(n− ni + 1) < ni(n− 1)

=⇒ (ni − 1)(n− ni + 1) < ni

=⇒ 3(ni − 1) < ni

⇐⇒ 2ni < 3,

a contradiction. SoSi < ni(q + 2− n).(ii) Fi a non-Fq-rational component.By Lemma 10.63,Si ≤ n2

i . Now,

(√q − 1)2 ≥ 0⇐⇒ 2

√q + 1 ≤ q + 2

=⇒ 2n− 1 < q + 2=⇒ n+ ni < q + 2⇐⇒ ni < q + 2− n⇐⇒ n2

i < ni(q + 2− n)=⇒ Si < ni(q + 2− n).

2

REMARK 10.65 It is not true thatn2 ≤ q + 1 + (n − 1)(n − 2)√q for all values

of n andq.

THEOREM 10.66 LetD2t be a plane algebraic curve of degree2t defined overFq

and letCn be a component ofD2t, with its degreen ≥ 3 if Cn is Fq-rational. LetS be the number of simple points ofCn lying in PG(2, q) ands the number of itssingular points lying inPG(2, q) of whichd are double points. If

√q > 2t+ n− 1, (10.55)

then

S + d < 12n(q + 2− t). (10.56)

368

Proof. (i) Cn anFq-rational component withn ≥ 3.Suppose that (10.56) is false. Then, from (10.47),

12n(q − t+ 2) ≤ S + d ≤ S + s ≤ q + 1 + (n− 1)(n− 2)

√q

=⇒ n(q − t+ 2) ≤ 2(q + 1) + 2(n− 1)(n− 2)√q

⇐⇒ (n− 2)q − 2(n− 1)(n− 2)√q ≤ n(t− 2) + 2

⇐⇒ (n− 2)√q[√q − 2(n− 1)] ≤ n(t− 2) + 2

=⇒ (n− 2)(2t+ n− 1)(2t− n+ 1) < n(t− 2) + 2 (using (10.6)

⇐⇒ (n− 2)[4t2 − (n− 1)2] < n(t− 2) + 2

=⇒ (n− 2)[4t2 − (2t− 1)2] < n(t− 2) + 2 (since2t ≥ n)

⇐⇒ (n− 2)(4t− 1) < n(t− 2) + 2

⇐⇒ (3n− 8)t+ n < 0,

which is impossible forn ≥ 3.(ii) Cn notFq-rational.Suppose the result is false. Then, by Lemma 2.24(i),

12n(q − t+ 2) ≤ S + d ≤ S + s ≤ n2

=⇒ q − t+ 2 ≤ 2n

=⇒ (2t+ n− 1)2 − (t− 2) < 2n (using (10.6)

=⇒ (2t+ n− 1)2 − (2t− 1)2 < 2n

⇐⇒n(4t+ n− 2) < 2n

⇐⇒ 4t+ n− 2 < 2

⇐⇒ 4(t− 1) + n < 0,

which is impossible ast ≥ 1, n ≥ 1. 2

10.7 FURTHER APPLICATIONS OF THE ZETA FUNCTION

The zeta function is a useful tool also in the study of the group Pic0(Fq(X )) ofall zero degreeFq-rational divisors classes ofX . This becomes apparent belowin the fundamental role of theL-polynomial in the proof of two major results onPic0(Fq(X )), previously stated without proof in Section 6.7.

Here, the notation of Sections 8.3 and 8.4 is changed as follows. If ℓ = Fqn withn ≥ 1, then

Div(l)(X ) = thel-divisor group Div(l(F));

Div(l)0 (X ) = the zero-degreel-divisor group Div0(l(F));

Pic(l)(X ) = thel-divisor class group Pic(l(F));

Pic(l)0 (X ) = the zero-degreel-divisor class group Pic0(l(F)).

By Theorem 8.34, Pic(Fq)0 (X ) is a finite abelian group. Its order, that is, the class

numberh of X , is closely related to the zeta function ofX via theL-polynomialLq(t).


THEOREM 10.67 h = Lq(1).

Proof. If g = 0, thenLq(1) = 1 andh = 1 by Example 8.36. Ifg ≥ 1, from (10.8)L(1) = h. 2

Theorem 6.91 was stated without proof. Now, using the zeta function, a proofis given in the case thatK is the algebraic closure ofFq. An essential tool in theproof is the algebraic number fieldQ(ω, . . . , ω2g) generated by the inverse rootsof theL-polynomial ofX . As suggested by Chapter 9, see especially Example9.4, function fields of transcendency degree1 are closely analogous to algebraicnumber fields, although number-theoretic methods have no direct counterpart infunction field theory. A detailed discussion of these methods is beyond the scopeof this book. The concepts and results used in the proofs below mostly concern thetheory of divisibility.

Two more pieces of notation are needed:

Pic(l)0 (X , d) = the subgroup of Pic(l)0 (X ) of all elements of orderd;

Pic(l)0 (X , d∞) = the subgroup of Pic(l)0 (X ) of all elements of order a power ofd.

The following lemma is the main ingredient in the proof.

L EMMA 10.68 Letd be a prime different fromp. Then, for the curveX of genusg,there exists a finite extensionk of Fq such that the following condition is satisfiedby all finite extensionsl of k, wherel1 is an extension ofl of degreedr :

[Pic(l1)0 (X , d∞) : Pic(l)0 (X , d∞)] = d2gr. (10.57)

Proof. The group Pic(l)0 (X , d) is finite and abelian, in which every element has

order a power ofd. Thus|Pic(l)0 (X , d)| = ds(l). By Theorem 10.67, the order of

the group Pic(l)0 (X ) containing Pic(l)0 (X , d) is

Lqm(1) =∏2g

i=1 (1− ωmi ),

wherel = qm; that is,m is the degree of the extensionFl/Fq. Therefore,qs(l) isthe largest power ofq which divides

∏2gi=1 (1− ωm

i ).

Replacingl by l1 in this argument shows thats(l1) is the largest power ofq dividing∏2g

i=1 (1− ωmdr

i ) =∏2g

i=1 (1− ωmi )(1 + ωm

i + . . .+ (ωmi )dr−1).

Choosen such thatωni ≡ 1 (mod d) for everyi = 1, . . . 2g, and letk = qn. Then

n dividesm, and the condition is satisfied. 2

Lemma 10.68 can be interpreted in terms of the norm map (8.6).Let φ be therestriction ofnorm0 l1|l on Pic(l1)0 (X , d∞). Sincenorm0 l1|l is surjective,φ is asurjective homomorphism. Then (10.57) reads:

| kerφ | = d2gr. (10.58)

Now, (10.58) implies thatkerφ ≤ Pic(l1)0 (X , d2gr). On the other side,kerφ con-

tainsPic(l)0 (X , dr). To show this, letl = Fqn , l1 = F(qn)dr , and choose any

370

[D] ∈ Pic(l)0 (X , dr). ThenΦn([D]) = ([D]) for then-th Frobenius map on divisor

classes, and hence

[0] = dr([D]) = [D] + Φn([D]) + . . .+ (Φn)dr−1([D]) = norm0 l1|l([D]),

which shows that[D] ∈ kerφ. Therefore,

Pic(l)0 (X , dr) ≤ kerφ ≤ Pic

(l1)0 (X , d2gr) (10.59)

To prove Theorem 6.91 it suffices to show the following result.

THEOREM 10.69 For all positive integersr and primesd distinct fromp,

Pic0(K(X ), dr) ∼= (Zdr )2g.

Proof. Choosek, l, l1 as in Lemma 10.68. Then

|Pic(ℓ)0 (X , dr)| ≤ | kerφ | = d2rg.

This shows that, for sufficiently largek, the following condition is satisfied by allfinite extensionsFl of Fk. If l′1 is a finite extension ofl′, then

Pic(l′)

0 (X , dr) = Pic(l′

1)0 (X , dr). (10.60)

Replacingdr by d2gr, this becomes

Pic(l′)

0 (X , d2gr) = Pic(l′

1)0 (X , d2gr). (10.61)

Now, choose a sufficiently largek such that both (10.60) and (10.61) hold wheneverFl′ is a finite extension ofFk andFl′1

is a finite extension ofFl′ . Now, from(10.59),

Pic(l′)

0 (X , dr) ≤ kerφ ≤ Pic(l′)

0 (X , d2gr).

Let [D] ∈ kerφ, andℓ′ = Fqn′ . Then

rd[D] = [D] + Φn′

([D]) + . . .+ (Φn′

)dr−1([D]) = norm0 l′1|l′([D]) = 0.

This implies that[D] ∈ Pic(l′)

0 (X , dr) showing thatkerφ ≤ Pic(l′)

0 (X , dr). There-

forekerφ = Pic(l′)

0 (X , dr). From (10.58),

Pic(l′)

0 (X , dr) ∼= (Zdr )2g.

This together with (10.60) imply the assertion. 2

A similar result enunciated in Chapter 6 is Theorem 6.93. Here, a proof is givenin the case thatK is the algebraic closure ofFq.

WriteLq(t) = 1 + a1t+ . . .+ aiti + . . .+ t2g, and denote byγ(X ) the highest

index i with 1 ≤ i ≤ g for which ai 6≡ 0 (mod p). To prove Theorem 6.93, itsuffices to show thatγ(X ) is equal to thep-rankγ of X . For this, two lemmas arerequired.

As before,ω1, . . . , ω2g are the inverse roots of theL-polynomialLq(t) of X .The idea is to use properties of the algebraic number fieldΩ = Q(ω1, . . . , ω2g).Let τ be a prime divisor ofΩ lying overp.


L EMMA 10.70 The indexγ(X ) is the number ofi for whichωi 6≡ 0 (mod τ).

Proof. Orderω1, . . . , ω2g so that

ωj 6≡ 0 (mod τ), j = 1, . . . , s;

ωj ≡ 0 (mod τ), j = s+ 1, . . . 2g.

From the equation,

1 + a1t+ . . .+ aiti + . . . qt2g =

∏2gi=1 (1− ωit),

it follows that

aj = (−1)j∏i1<...<ij

ωi1 · · ·ωij.

For j ≥ s+ 1, each term in the sum contains a factor fromτ while, for j = s, theproductω1 · · ·ωs is the only term not lying inτ . 2

The second preliminary result is analogous to Lemma 10.68.

L EMMA 10.71 There exists a finite extensionk of Fq such that the following con-dition is satisfied by all finite extensionsl of k. If l1 is an extension ofl of degreepthen

[Pic(l1)0 (X , p∞) : Pic(l)0 (X , p∞)] = pγ(X ). (10.62)

Proof. To τ there is associated an additive normalised valuationvτ of Ω. Choose apositive integern so that

ωni ≡ 1 (mod τvτ (p)+1)

for all i with ωi 6≡ 0 (mod τ). By Proposition 10.16,Lk(t) =∏2g

i=1(1− ωni t) for

an extensionk of Fq of degreen. Therefore, for any extensionl of k,

Ll(t) =∏2g

i=1 (1− ηit)

with

ηi ≡ 0 or 1 (mod τvτ (p)+1).

By Lemma 10.71, the latter case occurs as many times asγ(X ). So, (10.62) isequivalent to the equation,

vτLl1(1)

Ll(1)= γ(X ) · vτ (p), (10.63)

wherel1 is an extension ofl of degreep. Now, (10.63) is shown as follows:

vτLl1(1)

Ll(1)= γ(X ) · vτ (p) = vτ

2g∏

i=1

1− ηpi

1− ηi

=∑

ηi≡0 (mod τ)

(1 + ηi + . . .+ ηp−1i ).

Note that, ifηi 6≡ 0 (mod τ), then

1 + ηi + . . .+ ηp−1i ≡ p (mod τvτ (p)+1).

Therefore,

vτ (1 + ηi + . . .+ ηp−1i ) = vτ (p),


372

THEOREM 10.72 The indexγ(X ) equals thep-rankγ ofX .

Proof. The arguments in the proof of Theorem 10.69 remain valid whenLemma10.68 is replaced by Lemma 10.71. 2

A further characterisation of thep-rankγ of X is given by the following result.

THEOREM 10.73 (i) γ = 0 if and only ifAi ≡ 1 (mod p).

(ii) If γ 6= 0, thenγ = maxi | 1 ≤ i ≤ g; Ai−1 6≡ Ai (mod p).Proof. Comparing the coefficients inLFq

(t) = (1 − t(1 − qt)Z(X , t) modulopshows thatai ≡ Ai−Ai−1 (mod p) for i ≥ 1. Thus, Theorem 10.73 is a corollaryto Theorem 10.72. 2

10.8 THE FUNDAMENTAL EQUATION

The study of Pic0(K(X )) and its endomorphisms benefits from both group theoryand algebraic number theory, but it is best achieved by usingthe powerful tools ofalgebraic geometry.

To explain how to associate an algebraic variety withX , choose a pointP0 ∈ X ,and embedX in Pic0(K(X )) by the map sendingP ∈ X to the degree zero divisorclass[P − P0]. If X is a plane cubic curve, then such a map is bijective, thusallowing in this case the identification of the groupPic0(K(X )) with the points ofa plane cubic curve, which is a particular case of an algebraic variety. As a matterof fact, the groupPic0(K(X )) can always be identified in a functorial way with theset of points of an abelian varietyA which is an irreducible non-singular projectivealgebraic variety endowed with a commutative group compatible with the structureof the variety. Compatibility means that the group law,

µ : A×A → A,and the inverse map,

ι : A → A,are morphisms of varieties. Such an abelian variety is theJacobian varietyJX ofX . It has dimensiong and is defined overFq. If P0 is anFq-rational point ofX ,then the identification is carried out by a morphism of algebraic varieties,

X → JX ,defined overFq that sendsP0 to the identity element ofJX . This means that the

Fqn-rational points ofJX are the elements ofPic(qn)0 (X ), that is, theFqn-rational

degree zero divisor classes ofX .The action of the Frobenius mapΦ on the points ofX has a natural extension

on the divisors ofX : if D =∑nPP, thenΦ(D) =

∑nPΦ(P). This defines

an endomorphism of Div(K(X )), and hence gives rise to an endomorphism ofPic0(K(X )), theFrobenius endomorphism, that fixes each element of Pic(q)

0 (X ).


On the Jacobian varietyJX embedded in a projective space overFq, the FrobeniusmapΦ can be thought of as the map induced by the Frobenius collineation whosefixed points are exactly theFq-rational points ofJX . So,Φ is also theFrobeniusendomorphism ofJX .

The main results on the Frobenius endomorphism are summarised. The charac-teristic polynomialP (t) of the Frobenius endomorphismΦ is a monic polynomialof degree2g with coefficients inZ; it is related to theL-polynomial ofX .

THEOREM 10.74 The characteristic polynomialP (t) = t2gLq(1/t).

For any primed distinct fromp, Theorem 10.69 shows that Pic0(K(X ), d) canbe viewed as a2g-dimensional vector spaceV (2g, d) over the finite fieldFd. Thenthe Frobenius transformationΦ acts faithfully onV (2g, d) as an endomorphismwhose characteristic polynomial is described in the following result.

THEOREM 10.75 Letd be a prime distinct fromp. The Frobenius endomorphismacts linearly onPic0(K(X ), d) and the characteristic polynomial of the inducedendomorphism isP (t) with its coefficients reduced modulod.

THEOREM 10.76 If P (t) =∏P ri

i (t) is the factorisation of the characteristicpolynomialP (t) of the Frobenius endomorphism ofJX , then

∏Pi(Φ) = 0 on JX . (10.64)

Assume thatX (Fq) is non-empty, and choose anFq-rational pointP0 of X . Letm = |∏Pi(1)|, and write

∏Pi(t) =

∑ui=0 ait

u−i.

Then (10.64) is equivalent to theFundamental Equation,∑u

i=1 aiΦu−i(P ) + Φu(P ) ≡ mP0. (10.65)

EXAMPLE 10.77 Let q = 2q20 with q0 = 2s ands ≥ 1, as in Proposition 10.32. IfX has genusg = q0(q − 1) and the number ofFq-rational points isq2 + 1, thenthe Fundamental Equation reads as follows:

qP + 2q0Φ(P ) + Φ2(P ) ≡ (q + 2q0 + 1)P0.

The effectiveFq-rational divisormP0 ofX is theFrobenius divisorofX , and thecomplete linear series|mP0| is theFrobenius linear seriesof X . Some propertiesof the Frobenius linear series are established in the following propositions.

PROPOSITION 10.78 (i) If P,Q are Fq-rational points, thenmP0 ≡ mQ. Inparticular, m is a non-gap at eachP ∈ X (Fq) provided thatN1 ≥ 2.

(ii) If p ∤ m andN1 ≥ 2g + 3, then there existsP1 ∈ X (Fq) such that bothmandm− 1 are non-gaps atP1.

374

Proof. Part (i) follows from (10.65). To show (ii), note that Lemma 8.24 ensuresthe existence of an elementx ∈ Fq(X ) such that divx = mP0−mQ. Sincep ∤ m,sox is a separable variable. Then Hurwitz’s Theorem 7.26 applied to the rationaltransformationω : x′ = x, y′ = 0 gives2g − 2 = −2m+ degD, whereD is thedifferent. Note that

degD ≥ 2(m− 1) + (N1 − 2− k),wherek counts the unramifiedFq-rational points ofX . Therefore,

k ≥ N1 − (2g + 2).

So, the hypothesis implies thatk > 0. Let P1 ∈ X (Fq) be such an unramifiedpoint. ThenP1 is distinct fromP andQ and there existc ∈ Fq and an effectivedivisorD with P1, Q 6∈ supp(D) such that div(x − c) = P1 +D −mQ. Choosey ∈ Fq(X ) such that div(y) = mQ −mP1. Now, (ii) follows using the fact that(x− c)y ∈ Fq(X ). 2

The first assertion of Proposition 10.78 shows that the Frobenius linear series isindependent of the choice ofP0. Two more consequences are stated in the follow-ing proposition.

PROPOSITION 10.79 LetD be the Frobenius linear series ofX .

(i) If N1 ≥ 2, thenD is base-point-free.

(ii) If p ∤ m andN1 ≥ 2g + 3, thenD is simple.

There is a link between the(D, P )-orders and the Fundamental Equation. Nota-tion from Section 8.5 is used and three conditions are assumed:

|X (Fq)| ≥ 2;∏Pi(1) > 0;

a1 ≥ 1, au ≥ 1, ai ≥ ai−1 for i = 1, . . . , u− 1. (10.66)

Let r = dimD. Thenmr(P ) = m. From Section 6.6, the Weierstrass semi-groupH(P ) atP ∈ X contains the elements of the increasing sequence,

m0(P ) = 0, m1(P ), . . . , mi(P ), . . . ,

wheremi(P ) is thei-th non-gap atP .

PROPOSITION 10.80 The Frobenius linear seriesD ofX has the following prop-erties:

(i) the(D, P )-orders at anFq-rational pointP are ji(P ) = m−mr−i(P ) fori = 0, . . . , r;

(ii) if Φi(P ) 6= P for i = 1, . . . , u, thenj1(P ) = 1, and, in particular, ǫ1 = 1;

(iii) the integers1, a1, . . . , au areD-orders, so thatr ≥ u+ 1;

(iv) if Φi(P ) 6= P for i = 1, . . . , u+ 1, thenau is a non-gap atP ;


(v) if Φi(P ) 6= P for i = 1, . . . , u, but Φu+1(P ) = P, thenau−1 is a non-gapat P .

Proof. (i) Let m(P ) be a non-gap at anFq-rational pointP of X . There existsan effective divisorE for which E ≡ m(P )P andP 6∈ supp(E). From this,E + (m−m(P ))P ≡ mP . If m(P ) ≤ m, thenm−m(P ) is a(P,D)-order.

(ii) Since Φ and henceΦu is surjective onX , there is a unique pointP ′ ∈ Xwhose image underΦu is P. From (10.65) applied toP ′,

∑ui=1 aiΦ

u−i(P ′) + P ≡ mP0.

For 1 ≤ j ≤ u, the hypothesis thatΦj(P ) 6= P implies thatΦu−j(P ′) 6= P .Therefore,1 is a(P,D)-order.

(iii) Take a pointP ∈ X such thatΦj(P ) 6= P for j = 1, . . . , u. Then theassertion follow from (10.65).

(iv),(v) Again, the Fundamental Equation (10.65) applies,this time toP andΦ(P ), giving

mP0 ≡u∑

i=1

aiΦu−i(P ) + Φu(P ) ≡ a1Φ

u(P ) +

u−1∑

i=1

ai+1Φu−i(P ) + Φu+1(P ),

whence

auP ≡ Φu+1(P ) + (a1 − 1)Φu(P ) +u−1∑

i=1

(ai+1 − ai)Φu−i(P ).

Taking (10.66) into account, the results follow. 2

PROPOSITION 10.81 If N1 > q(m − au) + 1, then jr−1(P ) < au for everyFq-rational pointP ofX .

Proof. By (i) of Lemma 10.80,jr−1(P ) = m−m1(P ). On the other hand, Propo-sition 10.54 shows thatN1 ≤ 1 + m1(P )q. Taking the hypothesis into account,this gives

jr−1(P ) ≤ m− (N1 − 1)/q < au.

2

PROPOSITION 10.82 (i) The orderǫr = νr−1 = au, and everyFq-rationalpoint is in the support of the ramification divisorR ofD.

(ii) If N1 ≥ qau + 1, thenm1(P ) = q for everyFq-rational pointP ofX .

Proof. (i) First, ǫr−1 ≤ jr−1(P ) for anyP ∈ X . Sinceau is aD-order by Propo-sition 10.80, this implies thatǫr = au. From (10.65),Φ(P ) lies in the(r − 1)-thosculating hyperplane. Thusνr−1 = ǫr. By Lemma 10.80(i),jr(P ) = m for eachFq-rational point ofX . Sincem > au, the result follows from Theorem 7.52.

(ii) Let P ∈ X (Fq). Sincem1(P ) ≤ m1(Q) for all but finitely many pointsQof X , Lemma 10.80(iv) shows thatm1(P ) ≤ au. On the other hand,m1(P ) ≥ au

by Proposition 10.54 and the hypothesis on the size ofX (Fq). 2

376

Now, return to Pic0(K(X )). The machinery applied to any endomorphismα ofJX or, equivalently, of Pic0(K(X )) can give an equation of type (10.65) dependingon the characteristic polynomialPα(X) of α. To gain information onPα(X), it isappropriate to use a representation of the ring End(JX ) on a module of dimension2g over thel-adic integersZl with l a prime not equal top. Such a representationdepends on the Tate group.

LetA be an abelian group, written additively. Assume thatA is infinitely divis-ible by powers of a fixed primel, that is,lA = A. The Tate groupTl(A) of Aconsists of vectors

(a1, a2, . . . , an, . . .)

with denumerably many componentsai ∈ A satisfyinglai+1 = ai andla1 = 0.Addition in Tl(A) is carried out component-wise. It is possible to define onTl(A)the structure of a module over thel-adic integersZl as follows. Forz ∈ Zl, letmbe an integer such thatm ≡ z (mod ln). If lna = 0, defineza = ma, this beingindependent of the choice ofm. Then an operation ofz onTl(A) is defined:

z · (a1, a2, . . .) = (za1, za2, . . .).

From now onA is assumed to be Pic0(K(X ), l∞). This is possible since thegroup Pic0(K(X ), l∞) is a divisible group.

L EMMA 10.83 Tl(A) is a module overZl of dimension2g.

Proof. Let x1, . . . , x2g be vectors ofTl(A) whose first componentsa1,1, . . . , a2g,1

are linearly independent overFl. Then these vectors are linearly independent overZl. In fact, if a relation of linear dependence existed, it would be possible to assumethat not all the coefficients are divisible byl, and hence this relation considered onthe first component would contradict the hypothesis made on theai,j .

Now, it is shown by an inductive argument that these vectorsxi form a basis ofTl(A) overZl. Suppose that every elementw ∈ Tl(A) can be written as a linearcombination,

w = z1x1 + . . .+ z2gx2g (mod lnTl(A)), (10.67)

with integerszj ∈ Z. Letw = (b1, . . . , bn, bn+1, . . .). By definition, for the firstn+ 1 components

z1(a1,1,, . . . , a1,n+1) + . . .+ z2g(a2g,1, . . . , a2g,n+1)

= (b1, . . . , bn, bn+1) + (0, . . . , 0, cn+1)

for somecn+1 with ln+1cn+1 = 0. By the choice of the vectors, there exist integersd1, . . . , d2g such that

cn+1 = d1lna1,n+1 + . . .+ d2gl

na2g,n+1.

Replacingz1, . . . , z2g by z1 + d1ln, . . . , z2g + d2gl

n shows that the congruence(10.67) extends fromn to n+ 1. This gives the result. 2

If α ∈ End(JX ) thenα induces a linear map

αT : Tl(A)→ Tl(A),

(a1, a2, . . .) 7→ (αa1, αa2, . . .).


TheZ-moduleEndJX becomes a module over thel-adic integersZl by defining(zα)x = α(zx) for x ∈ Tl(A) andz ∈ Z. This gives the desired representation ofEnd(JX ). The main results in this area are summarised in the following theorems.

THEOREM 10.84 (i) If α1, . . . , αm are elements ofEnd(JX ) linearly indepen-dent overZ then they are independent overZl.

(ii) The mapα 7→ αT of End(JX ) into the ring ofZl-linear transformations ofTl(Pic0(K(X ), l) is a ring isomorphism.

(iii) End(JX ) is a module of finite type of dimension at most16g2.

THEOREM 10.85 The characteristic polynomial of an endomorphismα of JX isequal to the characteristic polynomial of the linear transformationαT induced byα onTl(Pic0(K(X ), l).

THEOREM 10.86 The coefficients of the characteristic polynomial of an endomor-phismα ofJX are integers.

10.9 EXERCISES

1. Show the following corollary of Weil’s Explicit Formula (10.39). Assumethatf(α) ≥ 0 for all α ∈ R and thatcn ≥ 0 for all n ≥ 1. Then

N1 ≤g

Ψ1(√q−1)

+Ψ1(√q)

Ψ1(√q−1)

+ 1;

equality holds if and only if∑g

j=1 f(αj) = 0,∑∞

d=2 dBdΨd(t)) = 0.

2. LetF be an plane algebraic curve of degreen which splits inton lines, allFq-rational. Prove the following upper and lower bounds on thenumberRq

of points ofPG(2, q) lying onF :

nq − 12n(n− 3) ≤ Rq ≤ nq + 1.

Show also that the lower limit is achieved if and only if no three of then linesare concurrent while the upper limit is achieved if and only if then lines areconcurrent.

3. Forn ≥ 1, let hFqn (X ) be the class number ofX . Also, for a primed, let

δ(d) =

lcm (d− 1, . . . , di − 1, . . . , d2g − 1) whend 6= p,

lcm (d− 1, . . . , di − 1, . . . , dg − 1) whend = p.,

Prove the following.

(i) Let d be a prime dividinghFq(X ). If d | n(qn − 1)/(q − 1) then also

d | hFqn (X )/hFq(X ).

378

(ii) If d ∤ hFq(X ) andn is relatively prime toδ(d) thend ∤ hFqn (X ).

(iii) If n is the smallest integer for whichd | hFqn (X ) thenn | δ(d).4. Letp ≡ 3 (mod 4) andq = p2n for n odd. Forλ 6= −3, 1, 0, letFλ be the

irreducibleFq-rational quartic of genus3 with

Fλ = v(X40 +X4

1 +X42 − (λ+ 1)(X2

0X21 +X2

1X22 +X2

2X20 )).

Show thatFλ attains the Serre Bound, that is,Nq = q + 1 + 6pn.

5. LetF be an irreducibleFq-rational curve of genus3 with more than2q + 6Fq-rational points. Show that then one of the following holds:

(i) q = 8, N1 = 24 andF is birationally equivalent overF8 to the curve,

v(X40 +X4

1 +X42 +X2

0X21 +X2

1X22

+X20X

22 +X2

0X1X2 +X21X0X2 +X2

2X0X1);

(ii) q = 9, N1 = 28 andF is is birationally equivalent overFq to theFermat curve,

v(X40 +X4

1 +X42 ).

6. Show that anF5-rational irreducible curve hasN1 ≤ 13.

7. Show that the irreducibleF5-rational curve,

v(X4 +X3Y 3 −X2 −XY 5 + Y 5 + 2Y ),

of genus3 hasN1 = 13.

8. Show that an irreducibleF8-rational curve hasN1 ≤ 25.

9. LetX = v(Y 3 + Y 2 + 1−X3Y ) be defined overF2. Show that

(i) X is non-singular and has genus3;

(ii) N1 = 3, N2 = 11, N3 = 9;

(iii) L2(t) = 1 + 3t2 + 6t4 + 8t6 = (1 + 2t2)(1 + t2 + 4t4);

(iv) the2-rank ofX is equal to2.

10. LetF = v(Y 6 + Y 5 + Y 4 + Y 3 + Y 2 + Y + 1 −X3(Y 2 + Y )), definedoverF2. Show that

(i) F has only one singular pointX∞;

(ii) X∞ is an ordinary triple point, and each of the branches ofF centredatX∞ is F2-rational;

(iii) F has genusg = 7;

(iv) N1 = 3, N2 = 9, N3 = 9, N4 = 45, N5 = 33,N6 = 69, N7 = 129;

(v) L2(t) = 1 + 2t2 + 9t4 + 16t6 + 32t8 + 72t10 + 64t12 + 128t14

= (1 + 2t2)(1− 2t2 + 4t4)(1 + t2 + 4t4)2;

(vi) the2-rank ofF is equal to4;

(vii) X , given as in Example 9, is covered overF2 byF .


10.10 NOTES

The general theory of zeta functions of an arbitrary fields offunctions was devel-oped by Schmidt in [236]; see also [227]. The presentation inSection 10.1, as inseveral other books, follows Hasses [115].

Theorem 10.17 is proved in [14, Proposition 5].The proof of Proposition 10.22 comes from [276]. Proposition 10.23 was proved

by Bombieri [33] via the approach initiated by Stepanov [268] and Schmidt [240].The original proof of Theorem 10.18 is due to Weil [308]. The problem of deter-mining the precise values ofN1 for a non-singular cubic as in Remark 10.11 wascompletely solved by Waterhouse [305]; for another proof, see Ughi [293].

Related to Remark 10.29, Fuhrmann and Torres [82] showed that, if q is square,then no curve of genusg with

14 (√q − 1)2 < g < 1

2 (q −√q)attains the Serre Bound. Also, Serre showed the non-existence of curves of genusg > 2 that miss his bound just by one. In other terms, if acurve of defectk is anFq-rational curve for whichN1 = q + 1 + ⌊2g⌋q − k with k ≥ 1, then no curveof genusg > 2 is a curve of defect1. Serre also provided a list of possible zetafunctions for curves of defectk ≤ 6 and genusg. Using this list together withresults on abelian varieties, Lauter [182] found that the Serre Bound can only beobtained if

g ≤ q2 − q⌊2√q⌋+ ⌊2√q⌋2 − 2q

.

She also proved various non-existence theorems for curves of defect2 and genusg > 2; for instance, no such curve exists whenq = 22s,s > 1.

On the positive side, Lauter [183] showed that for all finite fields Fq, eitherthere exists a curve of genus3 and defectk with k ≤ 3, or a curve is within threeof Serre’s lower bound. In [180] a method is also found for improving the Serrebound for a wide range of small genera, namely,

g ≤ (q2 − q)/(⌊2√q⌋+ ⌊2√q⌋2 − 2q),

provided thatq = 8, 32, 213, 27, 243, 125 and22s, s > 1. For more results, see[143] and, especially for a survey on curves of low defect andgenusg ≤ 5, see[148]. Curves of genus2 are investigated in [42, 210].

The lower bound (10.43) was proven by Ihara [150] and Tsfasman and Vladut[291, 292]. The constructive but not explicit proofs of these results are based onmodular curves and class field theory, see [150, 292, 203, 257, 317, 243, 311, 211,71, 173, 72]

In 1995 Garcia and Stichtenoth [88] give the first explicit example of a sequenceof curves overFq with squareq attaining the Drinfeld-Vladut Bound. The towerin Example 10.40 comes from that paper, see also Elkies [69].TheFq2-tower in10.41 is recursive; see Garcia and Stichtenoth [89].

Theorem 10.42 is due to Serre [257]. In the case thatq is a proper power ofp,that is,q > p, this also follows from Example 10.43, due to Garcia, Stichtenothand Thomas [92].

380

In Remark 10.44, the lower bounds in (i) forq = 2, 3, 5 are due to Schoof [243],Xing [311], Xing [311]; in (ii), the result is due to Zink [317]; in (iii), it is due toPerret [217].

Following the paper by Garcia and Stichtenoth [88], much work has been doneon towers given by explicit algebraic equations; see [299],[91], [31], [23], [24],[195], [310], [90]. The last paper also gives a survey on thissubject.

Surveys onA+(q) are found in [86], in [211, Chapter 5], and in [270, Chapters6, 9], as well as in [291, Chapter 4].

Theorems 10.45, 10.46 and 10.47 onA−(q) come from [72].For the gonality of a curve of degreem, see Homma [137]. For Proposition

10.52, see [215].Theorems 10.57, 10.58 and Corollaries 10.59, 10.60 are from[184], see also

[13]. Proposition 10.62 is found in [284]. For more results similar to Theorems10.64 and 10.66, see [127].

The proof of Theorem 10.69 comes from Frey’s paper [80]. Theorems 10.72 and10.73 and related proofs are due to Stichtenoth, see [273].

For proofs of the results on the theory of divisibility in Section 10.7, see [34].A proof of Theorem 10.76 via Castelnuovo’s inequality can befound in [67,

Chapter V, 5.4].The propositions on the Frobenius linear series in Section 10.8 come from [83].The presentation of the general theory of the abelian varieties is well beyond

the scope of this book. For background reading on this topic,see the books [204],[205], [178], and [199]. However, Theorems 6.91 and 6.93 proved in Section 10.7,describing the structure ofPic0(K(X )), extend to any abelian variety.

For the fact that Pic0(K(X ), l∞) is a divisible group, see [190, Theorem XI.3.2].Proofs, for abelian varieties, of the last results in Section 10.8 can be found in

[178, Chapter VII].For Exercises 4 and 5, see [16]. For Exercise 6, see [181]. ForExercise 7, see

[148, Chapter 2, Section 6]. For Exercise 8, see [233].

Chapter Eleven

Maximal and optimal curves

In the study of curves with manyFq-rational points the main problem is to deter-mineNq(g), the largest number ofFq-rational points that an irreducibleFq-rationalcurve of genusg can have. The asymptotic behaviour ofNq(g) with respect togtogether with the Drinfeld–Vladut Bound (10.38) are discussed in Section 10.3,where some examples withNq(g) ≈ (

√q− 1)g for a fixed squareq and forg large

enough are also constructed. In general,Nq(g) is known for only a few pairs(q, g).The aim of this chapter is threefold:

1. to describe the theory ofFq-maximal curves, that is, curves of genusg de-fined overFq whose numberSq of Fq-rational points isq + 1 + 2g

√q, the

Hasse–Weil Bound 10.18;

2. to give the known examples of maximal curves and some of their characteri-sations;

3. to describe some examples ofFq-optimal curves, that is, curves of genusgdefined overFq whose number ofFq-rational points equalsNq(g).

The Hasse–Weil Bound 10.18 implies thatNg(q) ≤ q + 1 + 2g√q. If equality

holds, then theFq-optimal curve isFq-maximal, andFq-maximal curves of genusg > 0 can only exist forq square. AnFq-maximal curve is, by definition,Fq-optimal.

Forg = 0, every curve isFq-optimal, as the number ofFq-rational points of theprojective line overFq is q + 1, whileNq(0) ≤ q + 1 by the Hasse–Weil Bound10.18.

Non-trivial examples of maximal and optimal curves appear in other chapters:

(a) the Hermitian curve, Example 1.38, is bothFq-optimal andFq-maximal forq = p2e;

(b) the Suzuki or DLS curve,

Example 5.24, isFq-optimal andFq4-maximal forq = 22e+1 with e ≥ 1;

(c) the Ree or DLR curve, Section 13.4, isFq-optimal andFq6-maximal forq = 32e+1 with e ≥ 1.

Notation and terminology are the same as in Chapter 10. To avoid trivial cases,it is assumed thatg > 0, unless otherwise stated.

First, maximal curves are investigated. So, the underlyingfield isFq2 .

381

382

11.1 BACKGROUND ON MAXIMAL CURVES

A characterisation of maximal curves can be deduced from theHasse–Weil Theo-rem 10.19 and (10.10) forn = 1.

THEOREM 11.1 WhenX be an irreducible non-singular curve of genusg definedoverFq2 , then the following conditions are equivalent:

(i) X is Fq2-maximal, that is,N1 = q2 + 1 + 2qg;

(ii) ωi = −q for i = 1, 2, . . . , 2g;

(iii) LFq2 (t) = (t+ q)2g.

Also, if X is Fq2-maximal andNm is the number of itsFq2m-rational points, then

Nj = q2m + 1 + (−1)m−12gqm, for m = 1, . . .. (11.1)

Theorem 11.1 shows that, if the curve is maximal then its zetafunction has par-ticularly simple form. Consequently, the zeta function viaits enumeratorLF

q2 (t),the corresponding fundamental equation, the Frobenius linear series, and the auto-morphisms combine to produce significant results.

Beginning withL-polynomials, Theorem 10.17 together with Theorem 11.1 hasthe following corollary.

THEOREM 11.2 If X ′ is anFq2-rational curve covered by anFq2-maximal curveX , with anFq2-rational covering, thenX ′ is alsoFq2-maximal.

EXAMPLE 11.3 (i) For q odd, the irreducible plane curve,

E1(q+1)/2 = v(Y q + Y −X(q+1)/2), (11.2)

is covered by the Hermitian curveHq = v(Y q + Y −Xq+1), sinceK(E1(q+1)/2)

is the subfieldK(x2, y) of K(Hq) = K(x, y). Note thatK(x2, y) is a propersubfield ofK(Hq) with yq + y − xq+1 = 0, as the genusg′ of E1

(q+1)/2 is smaller

than 12 (q2 − q), the genus ofHq. Therefore,[K(Hq) : K(E1

(q+1)/2)] = 2, and

the rational transformationω : (x, y) 7→ (x2, y) provides a two-fold covering ofE1(q+1)/2 byHq. From Lemma 13.1,g′ = 1

4 (q − 1)2. Sincex2, y ∈ Fq2(Hq), then

E1(q+1)/2, ω and the associated two-fold covering areFq2-rational, as well. From

Theorem 11.2,E1(q+1)/2 is anFq2-maximal curve. See also Example 11.33.

(ii) The corresponding example forq even is the irreducible plane curve,

T2 = v(Xq+1 + Y + Y 2 + Y 4 + . . .+ Y q/2), (11.3)

whose genus is14q(q−2) by Lemma 13.1. SinceK(T2) is the subfieldK(x, y2+y)of K(Hq), the same argument shows thatT2 is two-fold covered by the HermitiancurveHq = v(Y q + Y +Xq+1), and hence it is anFq2-maximal curve. See alsoExample 11.34.

Maximal and optimal curves 383

Table 11.1 Families of curvesF = v(F ) containingFq2-maximal curves

F F

Dr Xr + Y r + 1

E1m Y q + Y −Xm

Tp Xq+1 − (Y + Y p + Y p2

+ . . .+ Y q/p)

T ′3 (Y + Y 3 + . . .+ Y q/3)2 −Xq −XT ′′

3 Y + Y 3 + . . .+ Y q/3 + cXq+1, cq−1 = −1

K Y q+1 − f(X)

A Y q − Y − f(X)

F0 X(q+1)/3 +X2(q+1)/3 + Y q+1

F ′0 Y X(q−2)/3 + Y q +X(2q−1)/3

G Y q − Y X2(q−1)/3 + ωX(q−1)/3, ωq+1 = −1

Cn XnY + Y n +X

Cn,k XnY k + Y n +Xk

Cmi Xmi+m +Xmi + Y q+1

Xr Y 2r

+ a1Y2r−1

+ . . .+ ar−1Y2 + Y +Xq+1

(iii) See Example 11.31 for the curveF0 = v(F0) with

F0 = X(q+1)/3 +X2(q+1)/3 + Y q+1 ;

its genus is16 (q2 − q + 4); here,q ≡ 2 (mod 3).(iv) See Example 11.32 for the curveF ′

0 = v(F ′0) with

F ′0 = Y X(q−2)/3 + Y q +X(2q−1)/3 ;

its genus is16 (q2 − q − 2); here,q ≡ 2 (mod 3).(v) For curves of genus16 (q2 − q), examples areT ′

3 = v(T ′3) with

T ′3 = T(Y )2 −Xq −X, T(Y ) = Y + Y 3 + . . .+ Y q/3,

whenq ≡ 0 (mod 3) as in Example 11.35, andG = v(G) with

G = Y q − Y X2(q−1)/3 + ωX(q−1)/3, ωq+1 = −1,

whenq ≡ 1 (mod 3) as in Exercise 9.

As in Section 12.5, every non-trivial automorphism groupG of X gives rise to acovering ofX . In Example 11.3 (i),F = Hq/〈α〉 with α : (X,Y ) 7→ (−X,Y ).

384

Such a covering and the corresponding quotient curveX ′ = X/G areFq2-rationalif G is anFq2-automorphism group; that is,G is the restriction toX of a subgroupof PGL(r, q2). Each of the initial examples, (a), (b), (c), ofFq2-maximal curves hasa largeFq2-automorphism group with many non-conjugate subgroups; see Chapter13. From this, the existence of numerousFq2-rational maximal curves is deduced.As in Sections 13.2, 13.3 and 13.4, the genera of such quotient curves can often becomputed using the machinery developed in Chapter 12; see especially Theorems12.54 and 12.66, but the problem of finding an explicit equation has been solved sofar only in a few cases; see, for instance Theorem 13.27.

The existence of anFq2-maximal curve which is notFq2-covered by the Her-mitian curveHq is still unknown. The best candidates are the DLS and DLRcurves, or some of their quotient curves. In this vein, it would help to know ifany quotient curves of the DLS and the DLR curves are quotientcurves ofHq.There are, however,Fq2-maximal curves with simple equations such as those of

Kummer type,

K = v(Y q+1 − f(X)), (11.4)

and those of Artin–Schreier type,

A = v(Y q − Y − f(X)). (11.5)

The classification of maximal curves is currently out of reach. However, forlarger values ofg for which there exists anFq2-maximal curve, it seems that thereare few curves: see Table 11.2. This has been shown so far forg ≥ ⌊ 16 (q2−q+4)⌋.The proof is postponed to Section 11.5.

11.2 THE FROBENIUS LINEAR SERIES OF A MAXIMAL CURVE

The starting point is Theorem 11.1, showing that the fundamental equation (10.65)for an irreducible, non-singularFq2-maximal curveX is as follows:

qP + Φ(P ) ≡ (q + 1)P0, (11.6)

whereP0 ∈ X (Fq2) andΦ is the Frobenius map. LetD = |(q + 1)P0| be theFrobenius linear series ofX , and putr = dimD. From (11.6),dim |qP | = r − 1for everyP ∈ X . By the Weierstrass Gap Theorem 6.86 and Corollary 6.73, animmediate consequence of (11.6) for the non-gap sequence ofX at a pointP ∈ Xis the following:

0 < m1(P ) < . . . < mr−1(P ) ≤ q < mr(P ). (11.7)

Further, Propositions 10.78, 10.79, 10.80, 10.81, 10.82 applied to anFq2-maximalcurve, that is, foru = 1, m = q + 1, a0 = 1, a1 = q, provide a number of basicfacts that are collected in the next proposition.

PROPOSITION 11.4 With the notation above, the following hold:

(I) if P andQ areFq2-rational points, then(q+ 1)P ≡ (q+ 1)Q, andq+ 1 isa non-gap at eachP ∈ X (Fq2);


Table 11.2 KnownFq2-maximal curves of large genera

Genusg Condition onq Curves

1. 12q(q − 1) Hq = Dq+1

2. 14 (q − 1)2 q ≡ 1 (mod 2) E1

(q+1)/2

3. 14q(q − 2) q ≡ 0 (mod 2) T2

4. 16 (q2 − q + 4) q ≡ 2 (mod 3) F0

5. 16 (q2 − q) q ≡ 1 (mod 3) G

6. 16 (q2 − q) q ≡ 0 (mod 3) T ′

3

7. 16 (q2 − q − 2) q ≡ 2 (mod 3) F ′

0

8. 16 (q − 1)(q − 2) q ≡ 2 (mod 3) E1

(q+1)/3

9. 16q(q − 3) q ≡ 0 (mod 3) T ′′

3

10. 18 (q2 − 2q + 5) q ≡ 3 (mod 4)

11. 18 (q − 1)2 q ≡ 1 (mod 4)

12. 18q(q − 2) q ≡ 0 (mod 4)

13. 18 (q − 1)(q − 3) q ≡ 1 (mod 4) E1

(q+1)/4

14. 18 (q − 1)(q − 3) q ≡ 3 (mod 4) E1

(q+1)/4,D(q+1)/2

15. 18q(q − 4) q ≡ 0 (mod 2) X2

(II) there existsP1 ∈ X (Fq2) such that bothq + 1 andq are non-gaps atP1;

(III) the linear seriesD is complete, base-point-free, simple and defined overFq2 ;it gives rise to anFq2-rational curveΓ of PG(r, q) that isFq2-birationallyequivalent toX ;

(IV) the(D, P )-orders at anFq2-rational pointP are precisely

0 < q + 1−mr−1(P ) < . . . < q + 1−m1(P ) < q + 1;

that is, jr−i(P ) +mi(P ) = q + 1 for i = 0, . . . , r − 1;

(V) if P 6∈ X (Fq2), thenj1(P ) = 1 and soǫ1 = 1;

(VI) the integerq is aD-order, and sor ≥ 2;

(VII) if P ∈ X (Fq4)\X (Fq2) thenq− 1 is a non-gap atP ; if P 6∈ X (Fq4) thenqis a non-gap atP ;

386

(VIII) if P is anFq2-rational point ofX , thenjr−1(P ) < q;

(IX) ǫr = νr−1 = q, so Γ is Frobenius non-classical, and everyFq2-rationalpoint ofX is in the support of the ramification divisorR of Γ;

(X) If N1 ≥ q3 + 1, thenm1(P ) = q for everyFq2-rational pointP ofX .

Some useful refinements are now shown.

(XI) If P ∈ X is not anFq2-rational point, then

0 ≤ q −mr−1(P ) < . . . < q −m1(P ) < q

are (D, P )-orders atP . In particular, jr(P ) = q.

Arguing as in the proof of Proposition 10.80 (i) shows that, if m(P ) is a non-gap at a pointP 6∈ X (Fq2), then there exists an effective divisorE for whichE ≡ m(P )P andP 6∈ Supp(E). From this,

E + (q −m(P ))P + Φ(P ) ≡ qP + Φ(P ) ≡ (q + 1)P0.

If m(P ) ≤ q, thenq − m(P ) is a (D, P )-order. Assume on the contrary thatjr(P ) 6= q. Thenjr(P ) = q + 1, and hence(q + 1)P ≡ qP + Φ(P ) by (11.6).But thenP ≡ Φ(P ) which is not possible since the genus ofX is supposed to bepositive.

A similar argument shows that(II) is true for anyP1 ∈ X (Fq2).

(XII) If P is anFq2-rational point then bothq and q + 1 are non-gaps atP . Inparticular, j1(P ) = 1 for everyFq2-rational pointP .

As (IX) implies thatr ≤ q, this may be refined as follows.

(XIII) Either r = q − (g − 1) or r ≤ 12 (q + 1).

In fact, from the Riemann–Roch Theorem 6.59 it follows that either D is non-special andr = q + 1− g or it is special and Clifford’s Theorem 6.77 implies thatr ≤ 1

2 (q + 1).By (IV) and (XII), j1(P ) = 1 for everyP ∈ Γ; that is, the curveΓ has only linear

branches. This does not imply a priori thatΓ is non-singular. So the following resultis an improvement.

THEOREM 11.5 (i) The irreducibleFq2-rational curveΓ associated to the Fro-benius linear seriesD of anFq2-maximal curveX is non-singular.

(ii) The curvesX andΓ are isomorphic overFq2 .

Proof. Choosef0, . . . , fr ∈ Fq2(X ) such thatD consists of all divisors

Ac = div (c0f0 + . . . crfr) + (q + 1)P0

with c = (c0, . . . , cr) ∈ PG(r,K). By (III), let Γ be the irreducible curve arisingfrom the point(f0, . . . , fr); in other words, theFq2-rational transformation,

ω : x0 = f0, . . . , xr = fr,


defines anFq2-rational morphismπ : X −→ Γ. Since every branch ofΓ is linear,it must be shown that no point ofΓ is the centre of two or more branches.

Let P ∈ X andπ(P ) be a branch point ofΓ. After multiplying (f0, . . . , fr)by a suitable element ofFq2(X ), it may be assumed thatP is neither a pole ofany fk nor a zero of all thefk. Sinceǫr = q by (XI), Theorem 7.62 applied topm = q ensures the existence ofz0, . . . , zr in Fq2(X ) satisfying the followingthree properties:

(a) P neither a pole of anyzk nor the zero of all thezk;

(b) zq0f0 + . . .+ zq

rfr = 0; (11.8)

(c) the osculating hyperplaneLP at the branch pointπ(P ) of Γ is

v(z0(P )qX0 + . . .+ zr(P )qXr). (11.9)

SinceΓ is Frobenius non-classical, Theorem 8.52 gives forq2 = ph that

z0fq0 + . . .+ zrf

qr = 0. (11.10)

Suppose, on the contrary, thatπ(P ) andπ(Q) have the same centreA ∈ Γ. Thenthe osculating hyperplane toΓ at π(P ) contains the centre ofπ(Q). From (11.6),Φ(P ) = Q, andA ∈ PG(r, q2). After a suitable change of the coordinate systemof PG(r, q2), letA = (1, 0, . . . , 0). The coordinate functionsfk also change, but itis still true thatP is not a pole of anyfk, and thatP is not a zero of some coordinatefunction, sayf0; the same holds forzk. So takef0 = 1. If τ is a primitive placerepresentation of the branchπ(P ), thenπ(P ) has a primitive branch representation,

x0(t) = 1, x1(t) = τ(f1) = a1t+ . . . , . . . , xr(t) = τ(fr) = art+ . . . ,

with someak 6= 0 asπ(P ) is linear. Fork = 0, 1 . . . , r, write

zk(t) = τ(zk) = zk(P ) + uk(t)

with ord uk(t) ≥ 1. Apply τ to (11.8):

ord(z0(P )q + . . .+ zr(P )qxr(t)) = ord(u0(t)q + . . .+ ur(t)

qxr(t)).

Hence,z0(P ) = 0. From (11.10),u0(t) ≥ q > 1. This together with the precedingequation, implies that

ord(z0(P )q + . . .+ zr(P )qxr(t)) ≥ q + 1.

Therefore,jr(P ) = I(P,Γ ∩ LP ) = q + 1. But thenI(Q,Γ ∩ LP ) = 0, whichcontradicts the hypothesis that the centre ofπ(Q) lies inLP . 2

COROLLARY 11.6 LetX be anFq2-maximal curve of genusg and, for some pointP0 ∈ X (Fq2), suppose that there exist positive integersa, b ∈ H(P0) such that allnon-gaps atP which are at mostq + 1 can be written asλa+ µb for non-negativeintegersλ, µ. ThenH(P0) = 〈a, b〉 andg = 1

2 (a− 1)(b− 1).

388

Proof. Choosex, y ∈ Fq2(X ) such that div(x)∞ = aP0 and div(y)∞ = bP0.Sinceq, q + 1 ∈ H(P0) by (XII), the integersa, b are coprime, and so leta < b.ThereforeK(X ) = K(x, y) andFq2(X ) = Fq2(x, y). Let f(X,Y ) ∈ Fq2 [X,Y ]be an irreducible polynomial such thatf(x, y) = 0. Its Weierstrass normal form,see Theorem 6.89, is of type

f(X,Y ) = Xa + β Y b +∑

ai+bj<ab αijXiY j = 0

with β, αij ∈ Fq2 . Note thatF has a unique point at infinity, namelyX∞.AsK(X ) = K(x, y), theFq2-rational irreducible plane curveF = v(f(X,Y ))

is a birational model ofX over Fq2 . By Theorem 11.5 this holds true ifX isreplaced byΓ. It must be shown thatX∞ is the centre of a unique branch ofF andthat every other point ofF is non-singular. From this and Proposition 6.90(ii) itfollows thatF has genus12 (a− 1)(b− 1).

Choose the coordinate functionsf0, f1, . . . , fr of Γ such that

f0 = 1, div(fi)∞ = miP0, for i = 1, . . . , r.

Note that two of them arex andy. There exist polynomialsgi(X,Y ) ∈ Fq2 [X,Y ]for i = 1, . . . , r such that

f1 = g1(x, y), . . . , fr = gr(x, y).

Therefore, ifP ′ = (u, v) ∈ F andx = x(t), y = y(t) is a primitive representationof a branchγ′ of F centred atP , then the pointP = (1, g1(u, v), . . . , gr(u, v)) isthe centre of the branchγ of Γ with primitive representation,

x1(t) = g1(x(t), y(t)), . . . , xr(t) = gr(x(t), y(t)).

So, if P ′ were the centre of two distinct branches ofF , then the correspondingpoint P would be the centre of two distinct branches ofΓ, in contradiction withTheorem 11.5. Similarly, ifγ′ were non-linear, thenγ would also be non-linear,again contradicting Theorem 11.5.

Finally, if X∞ is the centre of a branchγ′ of F , then the corresponding place ofK(X ), that is, a pointP ∈ Γ is a pole of the coordinate functionx of Γ. But thenP = P0, and henceγ′ is uniquely determined. 2

From (XII), mr−1(P ) = q for everyFq2-rational pointP of X . This remainstrue for any pointP of X , but the analogous resultmr−2 = q− 1 can fail for somepoints.

PROPOSITION 11.7 LetX be anFq2-maximal curve.

(XIV) mr−1(P ) = q for everyP ∈ X .

(XV) If r ≥ 3 andmr−2(P ) < q − 1, thenP is a Weierstrass point ofX .Proof. (XIV) Assume thatP 6∈ X . AsD is base-point-free by (III), Theorems 11.5and 6.26 imply thatdim |(q − 1)P | = r − 2; so,mr−2(P ) ≤ q − 1 < mr−1(P ).This together with (11.7) shows (XIV).

(XV) From the definition of an ordinary place given in Section6.6, any twoordinary points ofX have the same non-gap numbers, that is,H(P ) = H(P ′) for


any two ordinary pointsP, P ′ ∈ X . Also, for every0 ≤ i ≤ r, the i-th non-gapmi(P ) at an ordinary pointP ∈ X is at least thei-th non-gapmi(Q) at any pointQ ∈ X .

Let r ≥ 3. If P is a non-Fq2 -rational point ofX , thenr−1 of the(D, P )-orderssmaller thanq are known from (XI). Ifmr−2(P ) < q − 1, the missing number in(XI) is 1. Actually, 1 is also a(D, P )-order by (V). Therefore, the(D, P )-ordersat such a pointP are completely determined. SinceX has only finitely manyWeierstrass points andFq2-rational points, so, ifmr−2(P ) < q− 1 for some pointP ∈ X which is neither a Weierstrass point nor anFq2-rational point, then theD-order sequence is the following:

(ǫ0 = 0, ǫ1 = 1, . . . , ǫ2 = q −mr−2(P ), . . . , ǫr−1 = q −m1(P ), q).

Now take anFq2-rational pointQ ∈ X . From (IV) and Theorem 7.52,

vQ(R) ≥∑ri=1 (ji(Q)− ǫi) =

∑ri=1 (q + 1−mr−i(Q)− ǫi).

Hence

vQ(R) ≥ 1 +∑r−2

i=1 (mi(P )−mi(Q) + 1) ≥ r − 1.

Since

|X (Fq2)| = q2 + 1 + 2gq = (q + 1)2 − (2g − 2)q,

it follows thatdegR ≥ (r − 1)(q + 1)2 − (2g − 2)q. From (7.13),

(∑

iǫi) (2g − 2) + (r + 1)(q + 1) ≥ (r − 1)(q + 1)2 + (r − 1)q(2g − 2) .

Sinceǫi ≤ ǫr = q, this implies that

(r + 1)ǫr(g − 1) + (r + 1)(q + 1) ≥ (r − 1)q(2g − 2) + (r − 1)(q + 1)2 .

Hence,

(q + 1)(q(r − 1)− 2) + q(g − 1)(r − 3) ≤ 0.

But this inequality does not hold forr ≥ 3. 2

From (XIV), a refinement of (XIII) can be deduced.

(XVI) If X has classical gap sequence at a general point, theng = q − (r − 1).

In fact, if X has classical gap sequence at a general pointP , then the positivenon-gaps atP are the numbersg + i for i = 1, . . ..

For a non-hyperelliptic curve of genusg > 2, X has classical gap sequence at ageneral point if and only if the canonical curve ofX is classical; see Remark 7.53.By Corollary 7.58, this is the case wheneverp > 2g − 2. The following exampleshows that (XVI) can occur.

EXAMPLE 11.8 The Klein curveF = v(X0X31 +X1X

32 +X2X

30 ), see Example

7.17, is a non-hyperelliptic curve of genus3, and is a canonical curve. Also, forp > 2g − 2 = 4, F has classical gap sequence at a general point. Now, letF beviewed as anFp2 -rational curve. By Theorem 11.75,F is Fq2-maximal if and onlyif q ≡ 6 (mod 7); see also Corollary 11.79 (ii). Note that the classical DirichletTheorem states that there exist infinitely many primes satisfying p ≡ 6 (mod 7).In particular,F is a classicalFp2 -maximal curve for infinite number of primesp.

390

It was noted in the proof of Theorem 11.7 that allD-orders but one can be de-duced from (XI) whenever the non-gaps at an ordinary point are known. LetJdenote the missingD-order; that is, at an ordinary pointP ∈ X ,

ǫJ−1 = q −mr−J(P ) < ǫJ < ǫJ+1 = q −mr−(J+1)(P ) . (11.11)

From Proposition 8.41, the FrobeniusD-orders are obtained from theD-orders byremoving one of them, sayǫI . In fact, the integersI andJ coincide.

PROPOSITION 11.9 The Frobenius orders ofD are theD-orders apart fromǫJwith J as in (11.11).

Proof. Let P be an ordinary point ofX and suppose thatP 6∈ X (Fq2). For t =1, . . . , r − 1, chooseξt ∈ K(X ) such that divξt = Dt −mt(P )P with Dt ≻ 0andP 6∈ SuppDt. Also, chooseξ, v ∈ K(X ) for which

div ξ = qP + Φ(P )− (q + 1)P0, div v = ǫJP +D − (q + 1)P0,

with D ≻ 0 andP 6∈ SuppD. From (XI),

div(ξ ξr−i) =

−(q + 1)P0 + ǫi−1P + Φ(P ) +Dr−i, i ≤ J,−(q + 1)P0 + ǫiP + Φ(P ) +Dr−i, i > J.

(11.12)

Let xi−1 = ξ ξr−i for i = 1, . . . , r − 1. From (11.12), divxi−1 + (q + 1)P0 ∈ D.More precisely,D consists of all divisors

Ac = div (

j−1∑

i=0

cixi + cjvj +

r−1∑

i=j

ci+1xi) + (q + 1)P0

with c = (c0, . . . , cr) ∈ PG(r,K). Since these divisors are cut out onΓ byhyperplanes, it also follows from (11.12) that every hyperplane∆ of PG(r,K) withI(P,∆ ∩ Γ) > J passes throughΦ(P ). On other hand, the hyperplane∆′ arisingfrom v does not have this property. Indeed, ifΦ(P ) ∈ ∆′, thenΦ(P ) ∈ SuppDand hence

div v = ǫJP + Φ(P )− (q + 1)P0 +D′

with D′ ≻ 0 andP 6∈ SuppD′. Thus there existsw ∈ K(X ) such that

divw = ǫJP + Φ(P )− (qP + Φ(P )) +D′ = (q − ǫj)P +D′.

But this implies thatq−ǫJ is a non-gap atP 6∈ X (Fq2), contradicting the definitionof J . So, the assertion is proved. Note that, in terms of the Frenet frame ofX atP ,see Section 7.6, this assertion can be reworded as saying that Φ(P ) ∈ (ΠJ\ΠJ−1)for any ordinary non-Fq-rational pointP of X . Using Proposition 8.41, the resultis established. 2

PROPOSITION 11.10 LetW be the set of all Weierstrass points of anFq2-maximalcurveX .

(i) If g > q − (r − 1), thenX (Fq2) ⊂ W.


(ii) If S is the Stohr–Voloch divisor ofX , then

SuppS ⊂ W ∪X (Fq2).

Proof. (i) By (XVI) and Exercise 7 in Chapter 7, the curveX is neither ellipticnor hyperelliptic. LetK be its canonical curve, and take an ordinary pointP of X .Then the ordersǫ0 = 0 < · · · < ǫj < . . . < ǫg−1 of K are also the orders ofK atP . From Section 7.6, the gaps atP are exactly the integersǫj +1, and henceH(P )consists of all non-negative integers distinct from

1, . . . , ǫj + 1, . . . , ǫg−1 + 1.

Also, for any integeri with 1 ≤ i ≤ q, Lemma 5.64 implies that(q + i− 1

q

)6≡ 0 (mod p).

Suppose thatq+ i 6∈ H(P ) for an integeri with 1 ≤ i ≤ q. Thenq+ i− 1 is anorder ofK atP . From Lemma 7.59,q is also an order ofK atP . Henceq+1 is a gapatP . By (I) P 6∈ X (Fq2). Therefore, ifP ∈ X (Fq2) thenq + 1, . . . , 2q ∈ H(P ).

Assume thatP ∈ X (Fq2). Then (XII) shows thatq, q + 1, . . . , ⊂ H(P ).SinceP is an ordinary point, from (11.7) it follows that the number of gaps atP isequal toq − (r− 1). By the Weierstrass Gap Theorem 6.86,g = q − (r− 1). Butby the hypothesis, this contradicts the assumptionP ∈ X (Fq2).

(ii) Let Di denote thei-th row vector in the generalised Wronskian determinant(7.12) of the coordinate functionsf0, . . . , fr of Γ, that is,

Di = (D(ǫi)f0,D(ǫi)f1, . . . ,D

(ǫi)fr) , i = 0, 1, 2, . . . , r ,

and letE be the first row vector of the Wronskian,detM(ν0, . . . , νr−1), fromwhich the Stohr–Voloch divisorS is derived, as in Section 8.5; that is,

E = (fq2

0 , fq2

1 , fq2

2 , . . . , fq2

r ).

If J is defined as in (11.11), Proposition 11.9 shows that

ν0 < ν1 < . . . < νJ−1 < ǫJ < νJ < . . . < νr−1

is the order sequence ofD. LetP 6∈ W ∪ X (Fq2). By the last remark in the proofof Proposition 11.9,Φ(P ) ∈ ΠJ\ΠJ−1. Therefore,E is a linear combination ofD0, . . . ,DJ but not ofD0, . . . ,DJ−1. HencedetM(ν0, . . . , νr−1) does not alterif its first rowE is replaced byDJ . In other words,

detM(ν0, . . . , νr−1) = det(D0, . . . ,Dr).

Now, on the contrary letP ∈ SuppS. This assumption means thatP is azero ofdetM(ν0, . . . , νr−1). ThenP is also a zero ofdet(D0, . . . ,Dr); that is,(j0(P ), . . . , jr(P )) 6= (ǫ0, . . . , ǫr). But thenP ∈ W ∪ X (Fq2), by (XI). 2

REMARK 11.11 LetX have classical gap sequence at a general point. Then, from(XVI), the first r − 1 positive non-gaps at any ordinary point areg + 1, . . . , g +r − 1 = q. Thus the orders ofD are0, . . . , r − 2, ǫr−1, q with r − 1 ≤ ǫr−1 <q. By Proposition 11.9, theFq2-Frobenius orders are0, . . . , r − 2, q. A directcomputation shows thatdegS = N1 + degR. It may be that the stronger equationS =

∑P∈X (F

q2 ) P +R also holds, but no proof has yet been found.

392

The following result regards hyperelliptic curves which areFq2-maximal.

PROPOSITION 11.12 LetX be anFq2-maximal curve of genusg ≥ 2.

(i) If X is hyperelliptic, thenq ≤ 2r − 2.

(ii) If there existsP ∈ X such thatjr−1(P ) = q − 1 thenX is hyperelliptic.

Proof. If X is hyperelliptic, thenm1(P ) = g + 1 for any ordinary pointP ; seeExercise 7 in Chapter 7. Then, by the Weierstrass Gap Theorem6.86, mr−1(P ) =g + r − 1. From (XI), it follows thatmr−1(P ) = q; henceg = q − r + 1. On theother hand,N1 ≤ 2(q2 +1) and the maximality ofX shows that2g ≤ q. Therefore(i) holds.

Let P ∈ X (Fq2) with jr−1(P ) = q − 1. Sincejr(P ) = q − 1, this and (IV)imply thatm1(P ) = 2. As g ≥ 2, this proves thatX is hyperelliptic.

Finally, letP 6∈ X (Fq2) with jr−1(P ) = q−1. Then (11.6) implies the existenceof a divisorE for which (q − 1)P + E ≡ qP + Φ(P ) with P 6∈ Supp(E).HenceE ≡ P + Φ(P ). This shows that the linear series|P + Φ(P )| has positivedimension. Sinceg ≥ 2, Theorem 6.79 (i) implies thatdim |P + Φ(P )| = 1.Therefore, by Theorem 7.41,X is hyperelliptic. 2

PROPOSITION 11.13 Suppose thatjr−1(P ) = r−1 for every pointP ∈ X . Then

(r − 1)r(g − 1) = (q + 1)(q − r) .Proof. Under the hypothesis, a pointP ∈ X is aD-Weierstrass point ofX if andonly if P is anFq2-rational point ofX . Also, ifRD is the ramification divisor ofX ,thenvP (RD) = 1 for P ∈ X (Fq2), see Theorem 7.52. SinceX is Fq2-maximal,the assertion follows from Definition 7.50. 2

11.3 EMBEDDING IN A HERMITIAN VARIETY

Although it should not matter which model of a curve is considered, some modelsare better than others in revealing birationally invariantproperties.

For Fq2-maximal curves, the model arising from the Frobenius linear seriesDappears to be a natural candidate, even though a slight change is necessary whenthe smallest linear subseriesL of D containing all divisorsqP + Φ(P ) is properlycontained inD, that is, whenm = dimL < dimD = r. As explained later,this replacement prevents the model from having higher-dimensional strangeness;that is, it ensures that there is no point lying in an infinite number of osculatinghyperplanes.

Therefore, throughout this section, letX be the curve given by Theorem 11.5.The idea is to use the geometric properties of the results of Section 11.2 to find geo-metric characterisations ofX as an irreducible non-singular curve lying on certainhypersurfaces.

Such a characterisation is the main aim in this section. Up toan isomorphismover Fq2 , Theorems 11.19 and 11.29 establish thatFq2-maximal curves are the


irreducibleFq2-rational algebraic curves of degreeq+1 lying on a non-degenerateHermitian variety ofPG(m,K) with 2 ≤ m ≤ r.

The Hermitian variety ofPG(n,K), withK = Fq, is a hypersurface which maybe thought of as the natural generalisation of the Hermitiancurve; canonically, it is

Hn,q = v(Xq+10 + . . .+Xq+1

n ).

By Theorem 11.5, the curveX satisfies the Linear General Position Principle.If m ≥ 3 andm is small, then Castelnuovo’s Bound provides an upper bound on

the genusg, which has applications for largerg in the case of maximal curves; seeSections 11.5, 11.6.

If m = 3 andg is large, such a curveX also lies on a low degree surface, byHalphen’s Theorem and its generalisations; see Section 7.16. In the known ex-amples, as at the end of this section,X is a component of degreeq + 1 of theintersection of the Hermitian surface with anFq2-rational quadric or cubic sur-face inPG(3,K). The characterisation, together with Castelnuovo’s BoundandHalphen’s Theorem, is the main ingredient in the classification of Fq2-maximalcurves withdimD = 3. For more details, see Section 11.6.

In the characterisation, the idea is to show that anFq2-maximal curveX comingfrom its Frobenius divisor lies on a non-degenerate Hermitian variety ofPG(r,K).This is demonstrated under the hypothesis thatX coincides with its dual curve upto the Frobenius collineation(X0, . . . ,Xr) 7→ (Xq

0 , . . . ,Xqr ) of PG(r,K). When

this hypothesis fails, a weaker result is obtained, namely that the dual curve lies ona degenerate Hermitian varietyH(m, q) of PG(r,K). However, it is still possibleto recover from the dual curve a non-singular modelX ′ of X lying on a non-degenerate Hermitian variety in somePG(m,K) with 2 ≤ m < r.

The starting point is the rational transformation,ω : x′0 = z0, . . . , x

′r = zr,

with z0, z1, . . . , zr satisfying (11.8), (11.9), (11.10), which defines a morphism π⋆

since the osculating hyperplanes ofX at distinct points are distinct.By (11.9), ρ π⋆ is the Gauss mapP 7→ L

(r−1)P , whereρ is the Frobenius

collineation(X0, . . . ,Xr) 7→ (Xq0 , . . . ,X

qr ). So, consider the irreducibleFq2-

rational curveX ′ = π⋆(X ) of PG(r,K). Note thatX ′ is the dual curve ofX upto ρ; see Section 7.10. Also,X ′ may be degenerate in the sense that it may becontained in a proper subspaceΠm = PG(m,K) of PG(r,K). If this is the caseandm is chosen to be minimal, then there is an(r − m)-dimensional subspaceΠ′

r−m of PG(r,K) which is the intersection of the osculating hyperplanes toX atgeneral pointsP ∈ X , except for finitely many pointsP .

Here, no point ofX lies in Π′r−m. To show this, letR ∈ Π′

r−m and assumeon the contrary thatR ∈ X . Choose a pointQ ∈ X such thatQ 6= R but theosculating hyperplaneLQ toX atQ containsΠ′

r−m. Since the common points ofLQ andX are only two, namelyQ andΦ(Q), it follows thatΦ(Q) = R. HenceQ is uniquely determined byR. But this is a contradiction, asQ can be chosen inmany different ways.

The Frobenius collineation(X0, . . . ,Xr) 7→ (Xq2

0 , . . . ,Xq2

r ) fixesΠm. Take aprojective frame inPG(r, q2) in such a way that that

Πm = v(Xm+1, . . . ,Xr).

394

Thenzm+1 = 0, . . . , zr = 0 and the coordinate functions ofπ⋆ : X → Πm are(z0, . . . , zm). Hence, by (11.9), the osculating hyperplane toX atQ is

LQ = v(γq0X0 + . . .+ γq

mXm),

whereC = (γ0, . . . , γm) is the centre of the branchπ⋆(Q). The linear seriesD′

cut out onX ′ by hyperplanes ofΠm consists of all divisors,

Ac = div(c0z0 + . . .+ cmzm) + E,

wherec = (c0, . . . , cm) ∈ Πm and, as usual,

E =∑

eP P, eP = −minordP z0, . . . ,ordP zm.

L EMMA 11.14 (i) The irreducibleFq2-rational curveX ′ = π⋆(X ) of Πm hasdegreeq + 1.

(ii) TheFq2-rational linear seriesD′ cut out onX ′ by hyperplanes ofΠm con-tains all divisorsqP + Φ(P ) with P ∈ X .

Proof. Choose a pointP0 = (α0, . . . , αr) ∈ X (Fq2). If αi = 0 for i = 0, . . . ,m,thenP0 lies in the osculating hyperplane at a general point ofX ; with the above no-tation,P0 ∈ Π′

r−m. But this has already been shown to be impossible. Therefore,αi 6= 0 for somei with 0 ≤ i ≤ m.

Now, consider the hyperplane,

H = v(αq0X0 + . . .+ αq

mXm),

of PG(r,K) which can also be regarded as a hyperplane ofΠm. For a pointP ∈ X ,let C = (γ0, . . . , γm) be the centre of the branchπ⋆(P ) of X ′. If C ∈ H thenαq

0γ0 + . . . + αqmγm = 0, and soγq

0α0 + . . . + γqmαm = 0. By (11.9), the

osculating hyperplane toX at P passes throughP0. SinceP0 ∈ X (Fq2), this isonly possible whenP = P0. ThusH ∩ X ′ contains no point other than the centreof π⋆(P0).

The next step is to show thatI(π⋆(P0),X ∩H) = q + 1; that is,

ordP0(αq

0w0 + . . .+ αqrwr) = q + 1 ,

where

wi = zi/zk, ordP0(zk) = minordP0

(z0), . . . ,ordP0(zr). (11.13)

After anFq2-linear transformation ofPG(r,K), let

P0 = (1, 0, . . . , 0).

Then it must be shown that ordP0(w0) = q+ 1. An Fq2-linear transformation may

be chosen so that

f0 = 1, ordP0fi < ordP0

fi+1, for i = 0, . . . , r − 1. (11.14)

Then the unique branchγ of X centred atP0 has a primitive representation,

x0(t) = 1, x1(t) = a1t+ . . . , xk(t) = aktjk + . . . , xr(t) = art

q+1 + . . . ,


where(0, 1, . . . , jk, . . . , q + 1) is the(D, P0)-order sequence ofX . If ρ is a primi-tive representation of the place arising fromγ, then (11.10) gives the following:

w0(t) = −w1(t)(a1t+ . . .)q + . . .+ wr(t)(artq+1 + . . .)q . (11.15)

Therefore it must be shown that botha1 6= 0 and ordtw1(t) = 1. From (11.8),

w0(t)q + w1(t)

q(a1t+ . . .) + . . .+ wk(t)q(aktjk + . . .)

+ . . .+ wr(t)q(art

q+1 + . . .) = 0.

Also, from the definition ofwi, it follows that ordtwi(t) = 0 for at most one indexi. Since1 = j1 < j2 < . . . < jr = q + 1 andjr−1 < q, the only possibility isi = r, whence the assertion follows.

Thus,(q + 1)π⋆(P0) is obtained as the intersection divisor ofX ′ with a hyper-plane ofΠm. ThereforedegX ′ = q + 1. Also, a1 6= 0 implies that the branchπ⋆(P ) of X ′ is linear.

This argument applied to a pointP ∈ X\X (Fq2) in place ofP0 shows that thehyperplaneH cuts out the divisorqπ⋆(P ) + Φ(π⋆(P )) onX ′. 2

By Lemma 11.14, the invariantsjm, ǫm of X ′ are equal to the correspondinginvariantsjr, ǫr of X . In the proof of Lemma 11.14, every branch ofX ′ is linear.From that proof,zm+1 = 0, . . . , zr = 0, sinceΠm = v(Xm+1 = 0, . . . ,Xr). Inparticular, (11.8) and (11.10) become the following:

zq0f0 + . . .+ zq

mfm = 0, z0fq0 + . . .+ zmf

qm = 0, (11.16)

and the osculating hyperplane at the branch pointπ⋆(P ) of X ′ is

v(f0(P )qX0 + . . .+ fr(P )qXr). (11.17)

Therefore, the proof of Theorem 11.5 may be used without alteration to prove thatX ′ has no singular points. This together with Theorem 11.5 gives the followingresult.

THEOREM 11.15 EveryFq2-maximal curve is isomorphic overFq2 to both thefollowing curves:

(i) the curveX arising from its Frobenius divisor;

(ii) the dual curveX ′ ofX .

REMARK 11.16 All proofs in this section remain valid forg = 0, that is, for theprojective lineℓ = v(Y ) overFq2 . The Frobenius divisorD of ℓ has dimensionr = q + 1 and consists of all effective divisors of degreeq + 1. The correspondingcurveX is the normal rational curve ofPG(q + 1,K) defined by the coordinatefunctionsfi = xi with i = 0, . . . , q + 1 andK(F) = K(x). Given any pointP (a) = (1, a, . . . , aq+1) ∈ X , the hyperplane

v(aq2+qX0 − aqX1 − aq2

Xq +Xq+1)

of PG(q + 1,K) cuts out the divisorqP + Φ(P ) onX . To show this, note thatΦ(P (a)) = P (aq2

), and that

x0(t) = 1, . . . , xi(t) = (a+ t)i, . . . , xq+1(t) = (a+ t)q+1

396

is a primitive representation of the unique branch centred at P (a). Then the resultfollows from the identities:

aq2+q − aq(a+ t)− aq2

(a+ t)q + (a+ t)q+1 = tq(a− aq2

+ t) ,

aq2+q − aqaq2 − aq2

(aq)q2

+ (aq+1)q2

= 0 .

Therefore, the smallest linear series containing all divisorsqP +Φ(P ) with P ∈ Xis cut out by the3-dimensional linear system of all hyperplanes,

v(u0X0 + u1X1 + uqXq + uq+1Xq+1);

that is,m = 3 in this case. Forq > 2, it follows thatm < r. No other suchexample is known to exist.

The next step is to show thatX ′ lies on a non-degenerate Hermitian variety ofΠm. SinceD′ is a linear subseries of the complete linear seriesD, Theorem 6.17implies the following result.

L EMMA 11.17 There exists a non-zero elementη ∈ K(X ) and a non-singularmatrixC = (cij) with entries inK(X ) such that

ηzi =∑r

j=0 cijfj , for i = 0, 1, 2, . . . , r. (11.18)

Since bothD andD′ areFq2-rational linear series, from the proof of Theorem6.17 it also follows thatη ∈ Fq2(X ) and that the entries inC are elements ofFq2 .Note thatcij = 0 for m + 1 ≤ i ≤ r and the matrixC has rankm + 1. Withoutloss of generality, let

f0 = 1, z0 = 1. (11.19)

Then, from (11.18) withi = 0,

η = 1, c00 = 1, c0j = 0 for j = 1, . . . r.

L EMMA 11.18 If (11.19) holds, then the matrixC in Lemma 11.17 is Hermitian;that is, cqij = cji for all i, j.

Proof. The method is to substitutezi in (11.10) by (11.18) and re-write the resultin the form,

1 +∑r

i=1(∑r

j=0 cijfj)fqi =1 +

∑rj=0(

∑ri=1 cijf

qi )fj

=1 +∑r

i=1(∑r

j=0 cjifqj )fi = 0.

This shows that the osculating hyperplaneHP of X at a general pointP is

v(X0 +∑r

j=0 (cj1fqj (P ))X1 + . . .+

∑rj=0 (cjrf

qj (P ))Xr).

But thenzqi (P ) =

∑rj=0(cjif

qj (P )) for i = 1, . . . , r. Hencezq

i −∑r

j=0 cjifqj has

an infinite number of zeros. By Theorem 5.33,

zqi =

∑rj=0 cjif

qj

for i = 1, . . . , r.On the other hand, (11.18) implies thatzq

i =∑r

j=0 cijfqj . Sincef1, . . . , fr are

linearly independent overK, the same holds forfq1 , . . . , f

qr . Therefore,cij = cqji

for all i, j with 0 ≤ i, j ≤ r. 2


L EMMA 11.19 Up to a projectivity overFq2 , the curveX ′ lies on the non-degener-ate Hermitian variety,

v(Xq+10 + . . .+Xq+1

m ),

of Πm.

Proof. From Lemma 11.18,cij = 0 for i = m + 1, . . . , r implies cij = 0 forj = m+ 1, . . . , r. Therefore, (11.18) becomes the following:

zi =∑r

j=0 cijfj , i = 0, 1, 2, . . . ,m, cij = cqji. (11.20)

Here, the(m+ 1)× (m+ 1) matrixC = (cij) is non-singular and Hermitian. Itsinverse matrixD is also Hermitian; hence

fi =∑r

j=0 dijzj , i = 0, 1, 2, . . . ,m, dij = dqji. (11.21)

Substitution forfi by (11.21) in (11.10) gives∑mi,j=0 dijz

qi zj = 0,

which shows thatX ′ lies on the hypersurface,

v(∑m

i,j=0 dijXqi Xj),

of Πm, with dij = dqji anddij ∈ Fq2 . The result follows since such a hypersur-

face is a non-degenerate Hermitian variety projectively equivalent overFq2 to theHermitian variety,v(Xq+1

0 + . . .+Xq+1m ). 2

Therefore, the following result is obtained.

THEOREM 11.20 (Natural Embedding Theorem)EveryFq2-maximal curveX ofgenusg ≥ 0 is isomorphic overFq2 to a curve ofPG(m,K) of degreeq + 1 lyingon a non-degenerate Hermitian varietyH = Hm,q defined overFq2 .

REMARK 11.21 The dimensionm is less then or equal to the dimensionr of theFrobenius linear seriesD of X . In particular, ifr = 2 then alsom = 2 andX isisomorphic overFq2 to the Hermitian curveHq.

REMARK 11.22 From the proof of Theorem 11.20, the osculating hyperplane ofX ′ at any pointP ∈ X ′ coincides with the tangent hyperplane atP to the non-degenerate Hermitian varietyH(m, q) in whichX ′ lies. As any point ofPG(m,K)is the common point of only finitely many tangent hyperplanesof H(m, q), thesame holds for the osculating hyperplanes atX ′. Therefore, theFq2-maximal curveX in the Natural Embedding Theorem 11.20 has no higher-dimensional strange-ness.

The Natural Embedding Theorem 11.20 and Castelnuovo’s Bound (7.66) givethe following result.

COROLLARY 11.23 LetX be anFq2-maximal curve of genusg. If the dimensionof the Frobenius linear series ofX is equal tor, then

g ≤ c0(q + 1, r) ≤ F (r) =

(2q − (r − 1))2 − 1

8(r − 1)if r is even,

(2q − (r − 1))2

8(r − 1)if r is odd.

(11.22)

398

It may be noted thatF (r) ≤ F (s) for s ≤ r. Thus,F (r) ≤ F (3) ≤ 14 (q − 1)2.

Next it is shown that the property given in Theorem 11.20 characterizesFq2-maximal curves. For this purpose, assume from now on thatY is anFq2-rationalirreducible, possibly singular, non-degenerate curve ofPG(m,K) having the fol-lowing property:

degY = q + 1 andY ⊂ Hm,q. (11.23)

Then the aim is to prove thatY is anFq2-maximal curve. It is also shown thatY isnon-singular and that the linear series cut out onY by hyperplanes is contained inthe Frobenius linear series ofY.

SinceY is not supposed to be non-singular, points ofY are thought of as branchpoints, that is, the term of a pointP of Y is used to indicate the branch ofYassociated to the placeP of K(Y). Such a convention was introduced in section7.6. Alternatively, choose a non-singular model ofY given by anFq2-rationalirreducible non-singular curveZ that is birationally equivalent toY overFq2 . Thenthere is a morphismπ : Z → Y defined overFq2 such that the branches ofY arisefrom the points ofZ, that is, the branches ofY areπ(A) with A ∈ Z.

The Hermitian varietyH(m, q) of PG(m,K) is taken in the canonical form

v(Xq+10 + . . .+Xq+1

m ).

If f0, . . . , fM are the coordinate functions ofY with f0, . . . , fM ∈ Fq2(Y), thehypothesis (11.23) means that

fq+10 + . . .+ fq+1

m = 0 . (11.24)

Let P ∈ Y. A primitive branch representation ofP is

x0 = f0(t), . . . , xm = fm(t).

LetQ = (a0 . . . , am) be the centre ofP , Then

fi(t) =∑∞

j=0 aijtj ∈ K[[t]], a00 = a0, . . . , am0 = am.

Hence (11.24) implies that∑m

i=0 (∑∞

j=0 aij)q+1 = 0.

The tangent hyperplaneHQ to the Hermitian varietyH(m, q) atQ = (a0, . . . , am)is v(aq

0X0 + . . .+ aqmXm).

The first step is to show that Lemma 11.5 (ii) holds forY.

L EMMA 11.24 The linear seriesR cut out onY by hyperplanes contains the divi-sor qP + Φ(P ) for everyP ∈ Y.

Proof. It is shown thatHQ cuts out the divisorqP + Φ(P ) onY. From (11.24),

(∑∞

j=0 a0jtj)qf0 + . . .+ (

∑∞j=0 amjt

j)qfm = 0 . (11.25)

Writing out the terms of lower order int,∑m

i=0 aqi0 fi + tq

∑mi=0 ai0 a

qi1 + tq+1∑m

i=0 aq+1i1 + tq+2g(t) = 0 .

Hence the intersection numberI(P,HQ ∩ Y) is at leastq and equality holds if andonly if

∑mi=0 a

qi1ai0 6= 0.


Now it is shown that, ifP ∈ Y(Fq2), then∑m

i=0 aqi1ai0 = 0. From (11.25),

∑mi=0 a

q+1i0 + t

∑mj=0 a

qi0 ai1 + tq h(t) = 0 .

Thus∑m

i=0 aqi0ai1 = 0. Since

(∑m

i=0 aqi0 ai1)

q =∑m

i=0 ai0 aqi1

whenP ∈ Y(Fq2), the assertion follows.Sincedeg(Y) = q + 1, this implies thatHQ cuts out the divisor(q + 1)P onY.

Therefore the lemma is established for everyP ∈ Y(Fq2).For the caseP 6∈ Y(Fq2), it must also be shown that the image point ofQ

under the Frobenius collineation(X0, . . . ,Xm) 7→ (Xq0 , . . . ,X

qm) lies inHQ. This

occurs when∑m

i=0 aq2+qi = 0.

Since this relation is a consequence of (11.25), the result follows. Therefore,HQ

cuts out the divisorqP + Φ(P ) onY for everyP 6∈ Y(Fq2). 2

Now, Lemma 11.24 makes it possible to use the proof of Theorem11.5 to showthatY is anFq2-rational irreducible non-singular curve.

If m = 2, thenY is the Hermitian curve and soY is Fq2-maximal. Therefore,letm ≥ 3.

The approach used for the proof is based on the relationship between the Wron-skian determinant ofY and the projectionY of Y to an(m− 1)-dimensional sub-space ofPG(m,K). More precisely, let

π : PG(m,K)→ PG(m− 1,K),(f0, . . . , fm) 7→ (f0, . . . , fm−1),

(11.26)

be the projectionY of Y from the pointUm = (0, . . . , 0, 1) to the hyperplanev(Xm). It may happen thatY and Y are not birationally equivalent overFq2 .However, it is always possible to avoid this situation by changing the coordinatesystem inPG(m,K).

L EMMA 11.25 (i) In PG(m,K) there is a pointP with coordinates inFq2 thatsatisfies each of the following conditions:

(a) P /∈ H;

(b) no tangent line toY at anFq2-rational point passes throughP ;

(c) no chord through twoFq2-rational points ofY passes throughP.

(ii) If ℓ is anFq2-rational line through anFq2-rational pointR of Y, then eachpoint ofℓ ∩ Y is anFq2-rational point ofℓ.

Proof. (i) Take anFq2-rational pointQ ∈ Y. Since the number ofFq2-rationalpoints ofY is at mostq2 +1+2gq ≤ q3 +1, there are at mostq3 chords joiningQto anotherFq2-rational point ofY. But, sincem ≥ 3, the number ofFq2-rationallines throughQ is at leastq4 + q2 + 1 and hence one of these lines is neither aline contained inH(m, q), nor a tangent line toY atQ, nor a chord throughQ and

400

anotherFq2-rational point ofY. Now, anyFq2-rational pointP outsideH(m, q) isa suitable choice forP .

(ii) Assume on the contrary thatℓ meetsY in a non-Fq2 -rational pointS ofℓ. Thenℓ is the line joiningS and the image pointS′ of S under the Frobenius

collineation(X0, . . . ,Xm) 7→ (Xq2

0 , . . . ,Xq2

m ). Therefore,ℓ is contained in theosculating hyperplane ofY at S. Hence the common points ofℓ andY are onlytwo, namelyS andS′. But this contradicts the hypothesis thatR ∈ ℓ ∩ Y. 2

Take a pointP as in (i) of this lemma. Since the linear collineation groupPGU(m + 1, q2) preservingH(m, q) acts transitively on the set of all points ofPG(m, q2) distinct from those ofH(m, q), it follows that a linear collineationof PG(m, q2) exists which preservesH(m, q) and takesP to the pointUm =(0, . . . , 0, 1). Lemma 11.25 now ensures thatY andY are birationally equivalentoverFq2 .

Choose a separable variableζ of K(Y), and consider the Wronskian matrices oftype investigated in Section 7.6:

W(f0, . . . , fm−1) = det

Dǫ0ζ f0 Dǫ0

ζ f1 . . . Dǫ0ζ fm−1

...... . . .

...D

ǫm−1

ζ f0 Dǫm−1

ζ f1 . . . Dǫm−1

ζ fm−1

,

W(f0, . . . , fm) = det

Dǫ0ζ f0 Dǫ0

ζ f1 . . . Dǫ0ζ fm

...... . . .

...Dǫm

ζ f0 Dǫm

ζ f1 . . . Dǫm

ζ fm

; (11.27)

hereD(i)ζ is thei-th Hasse derivative with respect to a separable variableζ inK(Y).

From Lemma 11.24,ǫ0 = 0, ǫ1 = 1, ǫm = q. So (11.27) becomes the following:

W(f0, . . . , fm) = det

f0 f1 . . . fm

Dζf0 Dζf1 . . . Dζfm

Dǫ2ζ f0 Dǫ2

ζ f1 . . . Dǫ2ζ fm

...... . . .

...D

ǫm−1

ζ f0 Dǫm−1

ζ f1 . . . Dǫm−1

ζ fm

Dqζf0 Dq

ζf1 . . . Dqζfm

. (11.28)

L EMMA 11.26

div(W(f0, . . . , fm))

= div(W(f0, . . . , fm−1))− q div(fm) + div(f0Dqζf

q0 + . . .+ fmD

qζf

qm) .

Proof. If the columns of the matrix in (11.28) arec0, c1, . . . , cm, for cm put

fq0 c0 + fq

1 c1 + . . .+ fqm−1cm−1 + fq

mcm.

Then

fqmW(f0, . . . fm) = det

f0 f1 . . . fq+10 + . . .+ fq+1

m

Dζf0 Dζf1 . . . fq0Dζf0 + . . .+ fq

mDζfm

...... . . .

...Dq

ζf0 Dqζf1 . . . fq

0Dqζf0 + . . .+ fq

mDqζfm

.


Each element in the last column is0 apart from the last: this follows from (11.24)by derivation. Also, theq-th Hasse derivative of the same relation gives the relation,

fq0Dζf0 + . . .+ fq

mDζfm + f0Dqζf

q0 + . . .+ fmD

qζf

qm = 0 ,

and this completes the proof. 2

Let Rm =∑vP (Rm)P be the ramification divisor of the linear series cut out

onY by hyperplanes ofPG(m,K). LetP ∈ Y and choose a local parameterζ atP as a separable variable. Then

vP (Rm) = ordP (W(f0, . . . , fm)) . (11.29)

Similarly, letRm−1 be the ramification divisor of the linear series cut out onY byhyperplanes ofPG(m,K).

L EMMA 11.27 If P ∈ Y andζ is a local parameter ofY at P, then

vP (Rm−1) = ordP (W(f0, . . . , fm−1)) .

Proof. By Definition 7.50,

vP (Rm−1) = ordP (W(f0, . . . , fm−1)) + (ǫ0 + ǫ1 + . . . ǫm−1)ordP (dζ) +meP ,

where

eP = −minordP (f0), . . . ,ordP (fm−1).Actually, eP = 0. In fact, if eP > 0 andeP = 0, thenUm = (0, . . . , 0, 1) lies onY, contradicting (11.24). Sinceζ is a local parameter atP , so ordP (dζ) = 0, andthe result follows. 2

L EMMA 11.28

ordP (f0Dqζf

q0 + . . .+ fMDq

ζfqM ) =

1 whenP ∈ Y(Fq2) ,

0 whenP 6∈ Y(Fq2) .

Proof. The proof of Lemma 11.24 also provides a proof of the following result.(i) For any pointP ∈ Y,

(a) P 6∈ Y(Fq2) if and only if∑m

i=0 aqi1am1 6= 0;

(b) P ∈ Y(Fq2) if and only if∑m

i=0 aqi1am1 = 0 and

∑mi=0 a

q+1i1 6= 0.

On the other hand,

f0Dqζf

q0 + . . .+ fmD

qζf

qm

=(∑∞

j=0 a0jtj)(aq

01 + tqg1(t)) + . . .+ (∑∞

j=0 amjtj)(aq

m1 + tqg2(t)) .

This implies the following.(ii)

(a) ordP (f0Dqζf

q0 + . . .+ fmD

qζf

qm) = 0 if and only if

∑ai0a

qi1 6= 0;

(b) ordP (f0Dqζf

q0 + . . . + fmD

qζf

qm) = 1 if and only if

∑ai0a

qi1 = 0 and∑

aq+1i,1 6= 0.

402

Now, comparison of (i) with (ii) proves Lemma 11.28. 2

THEOREM 11.29 Every irreducible algebraic curveY defined overFq2 satisfying(11.23) is a non-singularFq2-maximal curve.

Proof. By (7.13),∑

vP (Rm) = (ǫ0 + ǫ1 + . . . ǫm)(2g − 2) + (m+ 1)(q + 1) ,∑vP (Rm−1) = (ǫ0 + ǫ1 + . . . ǫm−1)(2g − 2) +m(q + 1) .

Hence∑

(vP (Rm)− vP (Rm−1)) = q(2g − 2) + q + 1.

Lemmas 11.26, 11.27, 11.28, (11.29) and the relation,∑

ordP (fM ) = q+1, finishthe proof. 2

REMARK 11.30 In (11.23) the condition ‘degreeq+ 1’ may be replaced by ‘min-imum degree with respect to the property of lying on a non-degenerate Hermitianvariety ofPG(m,K) defined overFq2 ’.

To show this, letf0, f1, . . . , fM be the coordinate functions of an irreduciblenon-degenerate algebraic curveΓ of PG(m,K) lying on the Hermitian varietyH(m, q) = v(Xq+1

0 + . . .+Xq+1m ) ⊂ PG(m,K); then (11.24) holds. If

(x0 = f0(t) = a0 + h0(t), . . . , xm = fm(t) = am + hm(t)), ordP hi(t) ≥ 1,

is a primitive representation of a branch ofΓ centred at the pointQ = (a0, . . . , ar),then (11.24) implies that

ordt(aq0f0(t) + . . .+ aq

mfm(t)) = ordt(h0(t)qf0(t) + . . .+ hm(t)qfm(t)),

whence

ordt(aq0f0(t) + . . .+ aq

mfm(t)) ≥ q.This shows that the tangent hyperplaneHQ = v(a0X0 + . . .+amXm) toH(m, q)atQ meetsY at the branchγ with multiplicity at leastq. If P 6∈ PG(r, q2), then itsimage,

P ′ = (aq2

0 , . . . , aq2

m ),

under the Frobenius collineation(X0, . . . ,Xm) 7→ (Xq2

0 , . . . ,Xq2

m ) is anothercommon point ofΓ andHP . In fact,aq+1

0 + . . .+ aq+1m = 0 implies that

aq0a

q2

0 + . . .+ aqma

q2

m = 0.

SinceY is a non-degenerate curve, it is not contained inHQ. ThusY must havedegree at leastq + 1.

As an illustration of this material, it is shown how the knownexamples ofFq2-maximal curves withr = m = 3 are embedded in a non-degenerate Hermitianvariety ofPG(3,K). In this way, an independent proof of the maximality of thesecurves is obtained.


EXAMPLE 11.31 Let q ≡ 2 (mod 3), and fix a primitive cube root of unityǫ inFq2 . For i = 0, 1, 2, let Fi = v(Fi) be theFq2-rational irreducible plane curvewith

Fi = ǫiX(q+1)/3 + ǫ2iX2(q+1)/3 + Y q+1 ,

and letK(x, y), with Fi(x, y) = 0, be its function field. LetΓi be theFq2-rationalirreducible algebraic curve ofPG(3,K), birationally equivalent overFq2 to Fi,defined by the coordinate functions

f0 = x, f1 = x2, f2 = y3, f3 = xy .

Note that the three curvesΓi are projectively equivalent inPG(3,K). In fact, theprojectivity overFq2 induced by the matrix,

T(i)4 =

ǫi 0 0 00 ǫ2i 0 00 0 1 00 0 0 ǫi

,

mapsΓ0 to Γi.Now, it is shown thatΓi is anFq2-rational irreducible non-singular algebraic

curve of degreeq + 1 lying on the Hermitian surface,

H = v(Xq+10 +Xq+1

1 +Xq+12 +Xq+1

3 ). (11.30)

To do this, note that the classical identity,

a3 + b3 + c3 − 3abc = (a+ b+ c)(a+ ǫb+ ǫ2c)(a+ ǫ2b+ ǫc) , (11.31)

with

a = Y q+1, b = X(q+1)/3, c = X2(q+1)/3,

becomes the identity,

(X(q+1)/3 +X2(q+1)/3 + Y q+1)

×(ǫX(q+1)/3 + ǫ2X2(q+1)/3 + Y q+1)

×(ǫ2X(q+1)/3 + ǫX2(q+1)/3 + Y q+1)

= Xq+1 +X2(q+1) + Y 3(q+1) − 3Xq+1Y q+1.

This yields:

xq+1 + x2(q+1) + y3(q+1) − 3xq+1yq+1 = 0,

whencefq+10 + fq+1

1 + fq+12 − 3fq+1

3 = 0. Therefore,Γi lies on the HermitianvarietyH(3, q), up to the projectivity,

(x0, x1, x2, x3) 7→ (x0, x1, x2, wx3), wq+1 = −3.

Also, Γi is contained in the cubic surface,

Σ3 = v(X33 + w3X0X1X2) ⊂ PG(3,K).

More precisely, the intersection ofH andΣ3 splits into the three curves,Γ0,Γ1,Γ2,each of degreeq + 1. By Theorem 11.29, eachΓi is a non-singular maximal curvedefined overFq2 . From Lemma 13.1, its genus is16 (q2 − q + 4).

404

EXAMPLE 11.32 For a similar but non-isomorphic example, again letq ≡ 2(mod 3), and fix a primitive cube root of unityǫ ∈ Fq2 . For i = 0, 1, 2, letF ′

i = v(F ′i ) be theFq2-rational irreducible plane curve with

F ′i = ǫiY X(q−2)/3 + Y q + ǫ2iX(2q−1)/3 ,

andK(x, y) with F ′i (x, y) = 0 be its function field. LetΓ′

i be theFq2-rational irre-ducible curve ofPG(3,K), birationally equivalent toF ′

i overFq2 , with coordinatefunctions

f0 = x, f1 = x2, f2 = y3, f3 = −3xy.

The three curvesΓ′i are projectively equivalent inPG(3,K). In fact, the projectiv-

ity induced by the matrixT (i)4 in Example 11.31 mapsΓ′

0 to Γ′i.

Arguing as in Example 11.31, it is shown now thatΓ′i is anFq2-rational irre-

ducible non-singular curve of degreeq + 1 lying on the Hermitian surfaceH as in(11.30). From the identity (11.31), with

a = Y q, b = Y X(q−2)/3, c = X(2q−1)/3,

it follows that

y3xq−2 + y3q + x2q−1 − 3xq−1yq+1 = 0,

whence

y3xq + y3q+2 + x2q+1 − 3xq+1yq+1 = 0.

Hencef2fq0 + fq

2 f1 + fq1 f0− 3fq+1

3 = 0 and this shows thatΓ′i lies on the surface,

Σq+1 = v(Xq0X1 +Xq

1X2 +Xq2X0 − 3Xq+1

3 ).

Also, Γ′i lies on the cubic surface,

Σ3 = v(X33 + 27X0X1X2).

Hence the intersection ofΣq+1 andΣ3 splits into the three curves,Γ′0, Γ′

1, Γ′2, each

of degreeq + 1.To prove thatΣq+1 is projectively equivalent toH, choose a rootα of the poly-

nomialXq+1 +X + 1. Thenαq2+q+1 = 1, and henceα ∈ Fq3 . From this,

αq+1 + αq2+q+1 + α = 0 αq2+q+2 + αq+1 + 1 = 0;

butαq+2 + αq2+1 + αq 6= 0 as(αq+2 + αq2+1 + αq)q−1 = α−1. Also, the matrix

M3 =

α 1 αq2+1

αq2+1 α 1

1 αq2+1 α

is non-singular. Choose an elementµ ∈ Fq2 satisfying

−3µq+1 = aq3+q+1 + αq2+1 + αq,

and defineT to be the projectivity ofPG(3,K) associated to the non-singular ma-trix,

M4 =

α 1 αq2+1 0

αq2+1 α 1 0

1 αq2+1 α 00 0 0 −µ

.


ThenT−1 mapsΣq+1 toH, andΣ3 to the cubic surfaceΣ3 = v(F3), with

F3 =(X30 +X3

1 +X32 ) + T(aq+1)(X2

0X1 +X21X2 +X2

2X0)

+T(α)(X20X2 +X2

1X0 +X22X1)

+(3 + T(αq−1))X0X1X2 − αq−1µ3X33 ,

whereT(u) = u+uq +uq2

is the trace ofu ∈ Fq3 overFq. Also,αq−1µ3 ∈ Fq2 ,and this shows thatΣ3 is defined overFq2 . Now, Γ′

i is mapped byT−1 to anFq2-rational irreducible curveC′i of degreeq + 1 lying onH. By Theorem 11.29,C′i isa non-singularFq2-maximal curve. From Lemma 13.1, its genus is1

6 (q2 − q − 2).

EXAMPLE 11.33 Let q be odd. Fori = 0, 1 andm = 12 (q + 1), let E i

m = v(Ei)be theFq2-rational irreducible plane curve, with

Ei = Y q + Y + (−1)iX(q+1)/2

andK(x, y) with Ei(x, y) = 0 its function field.LetΓ

′′

i be theFq2-rational irreducible curve ofPG(3,K), birationally equivalentto Ei overFq2 , with coordinate functions

f0 = 1, f1 = x, f2 = y, f3 = y2 .

The curvesΓ′′

1 andΓ′′

0 are projectively equivalent, since the projectivity induced bythe matrix,

T4 =

1 0 0 00 ǫ 0 00 0 1 00 0 0 1

,

with ǫ(q+1)/2 = −1, mapsΓ′′

0 to Γ′′

1 . The polynomial identity,

(Y q + Y −X(q+1)/2)(Y q + Y +X(q+1)/2) = Y 2q + 2Y q+1 + Y 2 −Xq+1,

implies thaty2q + 2yq+1 + y2 − xq+1 = 0 and sofq3 + f3 + 2fq+1

2 − fq+11 = 0.

This proves thatΓ′′

i lies on the surface

Σ = v(Xq3X0 +X3X

q0 + 2Xq+1

2 −Xq+11 ),

which is a non-degenerate Hermitian variety ofPG(3,K). Also, Γ′′

i lies on thequadric cone,

Q = v(X22 −X0X3),

and hence the intersection ofΣ andQ splits into the curvesΓ′′

0 andΓ′′

1 . By Theorem11.29,Γ

′′

i is a non-singularFq2-maximal curve. From Lemma 13.1, its genus is14 (q − 1)2.

EXAMPLE 11.34 Let q = 2t, and putT(Y ) = Y + Y 2 + . . .+ Y q/4 + Y q/2. Foreachi = 0, 1 ∈ F2 ⊂ Fq2 , let Ci = v(Ti) be theFq2-rational irreducible planecurve with

Ti = T(Y ) +Xq+1 + i,

406

andK(x, y) with Ti(x, y) = 0 its function field. It may be noted thatC0 = T2.LetΦi be theFq2-rational irreducible curve ofPG(3,K), birationally equivalent

to Ci overFq2 , with coordinate functions,

f0 = 1, f1 = x, f2 = y, f3 = x2 .

Since

(T(Y ) +Xq+1)(T(Y ) +Xq+1 + 1) = Y q + Y +Xq+1 +X2q+2,

soyq +y+xq+1 +x2q+2 = 0. Therefore,Φi lies on the non-degenerate Hermitianvariety,

H(3, q) = v(Xq2X0 +X2X

q0 +Xq+1

1 +Xq+13 ).

Also, the quadric coneQ = v(X0X3 −X21 ) containsΦi. HenceH∩Q splits into

Φ0 andΦ1. Note that the curvesΦ0 andΦ1 are projectively equivalent overFq2 inPG(3,K), and hence both have degreeq + 1. Again, by Theorem 11.29,Φi is anon-singularFq2-maximal curve. From Lemma 13.1, its genus is1

4q(q − 2).

EXAMPLE 11.35 Let q = 3t, and putT(Y ) = Y + Y 3 + . . . + Y q/3. For eachi = 0, 1, 2 ∈ F3 ⊂ Fq2 , let Ci = v(T ′

i ) be theFq2-rational irreducible plane curvewith

T ′i = T(Y )2 −Xq −X + i(T(Y ) + i) = 0 ,

andK(x, y) with T ′i (x, y) = 0 its function field. It may be noted thatC0 = T ′

3 .From the identity,

(T(Y )2 −Xq −X)

×(T(Y )2 + T(Y ) + 1−Xq −X)

×(T(Y )2 − T(Y ) + 1−Xq −X)

= (Y q − Y )2 − (Xq +X)(Xq +X − 1)2,

it follows that

(x3 + x2 − y2 + x)q + (x3 + x2 − y2 + x)− xq+1 − yq+1 = 0 .

Let Φ′i be theFq2-rational irreducible curve ofPG(3,K), birationally equivalent

to Ci overFq2 , with coordinate functions,

f0 = 1, f1 = x, f2 = y, f3 = x3 + x2 − y2 + x .

As in the preceding examples, the three curves,Φ′0, Φ′

1, Φ′2, areFq2-projectively

equivalent inPG(3,K). Also,Φ′i lies on the non-degenerate Hermitian variety,

v(X0Xq3 +Xq

0X3 −Xq+11 −Xq+1

2 ).

Finally, the cubic surface,

v(X3X20 −X3

1 +X21X0 +X2

2X0 −X1X20 ),

also containsΦ′i. ThereforeΦ′

i has degreeq + 1, and Theorem 11.29 ensures thateachΦ′

i is a non-singularFq2-maximal curve. From Lemma 13.1, its genus is16q(q − 1).


11.4 MAXIMAL CURVES LYING ON A QUADRIC SURFACE

The Natural Embedding Theorem 11.20 offers a geometric approach to the studyof Fq2-maximal curves. The idea is to use the specific properties ofthe model ofFq2-maximal curves described in that theorem and the successive three remarks.To do this, consider the following.

CONDITION 11.36 An Fq2-maximal curve is anFq2-rational, irreducible, non-singular curve of degreeq + 1 lying on a non-degenerate Hermitian variety ofPG(m,K), where2 ≤ m ≤ r = dimD withD the Frobenius linear series ofX .

As in many geometric problems, low dimensions can be tackledsuccessfully.Nevertheless, it is not easy to understand which of the low-dimensional resultsare of general type; this is the main difficulty in deciding what happens in higherdimensions.

Sincem = 2 only occurs whenX is a Hermitian curve, the smallest dimensionto consider ism = 3. Examples 11.33 and 11.34 show that there areFq2-maximalcurves ofPG(3,K) lying on a quadric. The aim of this section is to show that nootherFq2-maximal curve ofPG(3,K) has this property.

Therefore, letdimD = m = 3. Then the Frobenius linear seriesD of X is thelinear series cut out onX by hyperplanes. For a pointP ∈ X , the positiveD-ordersare

j1(P ) = 1 < j2(P ) < j3(P ),

where

j3(P ) =

q + 1, whenP is Fq2-rational ,

q, whenP is notFq2-rational ,(11.32)

L EMMA 11.37 LetdimD = 3 and suppose thatX lies on a quadric surfaceQ ofPG(3,K). If q ≥ 4, then the following properties hold for any pointP ∈ X :

(i) j2(P ) ∈ 2, 12j3(P ), 1

2 (j3(P ) + 1);

(ii) j2(P ) > 2 if and only if the tangent lineL1(P ) ofX at P lies onQ;

(iii) if j2 > 2 andP is a non-singular point ofQ, then

j2(P ) =

12q, whenq is even andP is notFq2-rational,

12 (q + 1), whenq is odd andP is Fq2-rational;

(iv) if there existsP ∈ X with j2(P ) > 2 thenQ is a cone.

Proof. Let the quadricQ = v(F ), with

F (X0,X1,X2,X3)

= a00X20 + a01X0X1 + a02X0X2 + a03X0X3 + a11X

21

+a12X1X2 + a13X1X3 + a22X22 + a23X2X3 + a33X

23 .

408

First it is shown thatQ is defined overFq2 . OtherwiseQ is distinct from its imageQ′ under the Frobenius collineation,

(X0,X1,X2,X3) 7→ (Xq2

0 ,Xq2

1 ,Xq2

2 ,Xq2

3 ). (11.33)

HenceX would be in the intersection of two distinct quadrics, anddegX = q + 1would be less than5 by Corollary 7.9, contradicting the hypothesis thatq ≥ 4.

The coordinate functionsf0, f1, f2, f3 ∈ Fq2(X ) of X may be chosen such thatordP (fi) = ji(P ) for i = 0, 1, 2, 3. The hypothesis onX to lie onQ implies thatF (f0, f1, f2, f3) = 0, by Theorem 7.5.

After a change of coordinates inPG(3,K), let the pointP be a vertex of theprojective frame, sayP = (1, 0, 0, 0), so that the osculating plane atP is v(X3)and the tangent line toX atP is v(X2,X3); see the proof of Theorem 7.45. Also,let f0 = 1.

To use the equation,

F (1, f1, f2, f3) = 0, (11.34)

properly, the orders atP of some other elements fromFq2(X ), namely,

f21 , f1f2, f1f3, f

22 , f2f3, f

23 ,

are required. By direct computation, these orders are

2j1, j1 + j2, j1 + j3, 2j2, j2 + j3, 2j3 (11.35)

Hence (11.34) together withP ∈ Q implies thata00 = a01 = 0.To show (i), note thatj2 + 1 < j3 by (VIII) and (XI). So, from (11.35) and the

inequalities,

2 ≤ j2 < j2 + 1 < j3 < j3 + 1 < j3 + j2 < 2j3,

(i) follows. Also, (11.35) shows thatj2 > 2 if and only if a11 = 0. Now, asF (X0,X1, 0, 0) = a11X

21 , the last condition is satisfied if and only if the tangent

lineL1(P ) toX atP is contained in the quadricQ. From this, (ii) follows.Also, if j2 > 2, from (i),

a11 = a02 = a12 = 0.

Now, P is a non-singular point ofQ if and only if a03 6= 0. Therefore2j2 = j3,and (iii) follows from the fact thatj1(P ) = 1 and (11.32).

To show (iv), note that ifP is a non-singular point ofQ thenQ is a cone whosevertex is the pointV = (−a13, a03, 0, 0). Otherwise,P is singular and henceQ isa cone with vertexP . 2

L EMMA 11.38 Let q ≥ 4 anddimD = 3. Then

(i) dim(2D) ≥ 8;

(ii) if dim(2D) = 8, the curveX lies on a quadric surface inPG(3,K);

(iii) the quadric surface in(ii) is uniquely determined and is defined overFq2 .


Proof. (i) Let P ∈ X (Fq2). Thenm2(P ) = q andm3(P ) = q + 1 by (IV) and(XIV). Now, as2m1(P ) ≥ m2(P ) = q andq ≥ 4, it follows that there are at leasteight non-gaps in the interval[m1(P ), 2m3(P )]. Hencedim 2D ≥ 8.

(ii) Geometrically, the linear system of all quadrics has dimension9 and thelinear seriesL cut out onX by quadrics is contained in2D. So if dim(2D) ≥ 8thenL = 2D and some quadrics must containX .

(iii) See the first part in the proof of Lemma 11.37. 2

THEOREM 11.39 Letq be odd withq ≥ 5, and letX be anFq2-maximal curve ofgenusg = 1

4 (q−1)2. ThenX is birationally equivalent overFq2 toF1 in Example11.33, that is, to theFq2-rational irreducible plane curveE1

(q+1)/2 = v(E1) with

E1(X,Y ) = Y q + Y −X(q+1)/2 . (11.36)

Proof. From Lemma 11.37,j2(P ) = 2 at each pointP ∈ X which is notFq2-rational while theFq2-rational pointsP of X are of two types:

(I) j2(P ) = 2;

(II) j2(P ) = 12 (q + 1).

In particular, the set of allD-Weierstrass points ofX coincides withX (Fq2).Since theD-orders are0, 1, 2, q, the weight ofP in the ramification divisorR

can be calculated from Theorem 7.52:

vP (R) =

12 (q − 1), whenP ∈ X (Fq2) andj2(P ) = 1

2 (q + 1),1, whenP ∈ X (Fq2) andj2(P ) = 2,0, whenP 6∈ X (Fq2).

Let T1 andT2 denote the numbers ofFq2-rational points of type (I) and (II). From(7.13),

12 (q + 1)T2 + T1 = (q + 3)(1

2 (q − 1)2 − 2) + 4(q + 1).

SinceX is Fq2-maximal,

T1 + T2 = q2 + 1 + 12 (q − 1)2q.

From these two equations,T1 = q + 1 and, in particular,T1 ≥ 4.Now the configuration of the points of type (I) is investigated. Such points are

not collinear, since no non-degenerate irreducible curve of PG(3,K) of degreekcontainsk collinear points. By Lemma 11.37 (iv), these points lie on a coneQ withvertexT . From the proof of that lemma, the tangent lineL1(P ) of X at a pointPof type (I) is contained inQ. Therefore, every line joiningT to a point of type (I)is a generator ofQ.

If P andP ′ are distinct points of type (I), thenL1(P ) does not containL1(P′).

HenceT is not a point of type (I). Also,T lies in every plane which osculatesX ata point of type (I).

Another property of a pointP of type (I), established implicitly in the proof ofLemma 11.37, is as follows.

L EMMA 11.40 The following three planes coincide:

410

(a) the osculating planeL2(P ) ofX at P ;

(b) the tangent planeα(P ) to the Hermitian surfaceH(3, q) at P ;

(c) the tangent planeβ(P ) to the coneQ at P .

A first consequence is thatT 6∈ H(3, q), as anFq2-rational lineℓ of α(P ) whichpasses throughP either lies inH(3, q) or meetsH(3, q) only in P .

The projective group PGU(4, q2) that fixesH(3, q) acts transitively on the pointsof PG(3, q2)\H(3, q). Therefore, a projective frame overFq2 may be chosen sothat

H(3, q) = v(Xq0X2 +X0X

q2 + 2Xq+1

1 +Xq+13 ), T = (0, 0, 0, 1). (11.37)

The polar planeΠ of T under the unitary polarity ofH(3, q) is v(X3). SoΠ cutsout onH(3, q) the non-degenerate Hermitian curve,

H(2, q) = v(Xq0X2 +X0X

q2 + 2Xq+1

1 ),

and onQ an Fq2-rational irreducible conicC. Another consequence of Lemma11.40 is thatC andH(2, q) have the same tangent at every pointP of type (I).From Bezout’s Theorem 3.14, the common points ofC withH(2, q) are exactly thepoints of type (I).

This shows thatC andH(2, q) are inpermutable positionwith respect to the twopolarities arising fromC andH(2, q). Given a non-degenerate Hermitian curve,the irreducible conics in permutable position with it form aunique orbit under theaction of the group PGU(3, q2) fixing the Hermitian curve.

TheFq2-rational irreducible conicC′ = v(X0X2 + X21 ) provides an example

of such a conic. There exists a projectivity(S) of Π preservingH(2, q) that takesC to C′. Also, δ can be lifted to an element(S)′ of PGU(4, q2); then(S)′ fixesT .Hence, the coneQmay be taken as

Q = v(X0X2 +X21 ). (11.38)

Now, letA = (a0, a1, a2, a3) be any point ofX . Then

aq0a2 + a0a

q2 + 2aq+1

1 + aq+13 = 0, a0a2 = a2

1.

If a0 = 0 thenA = (0, 0, 1, 0). Otherwise, leta0 = 1; then these equations implythat

(aq2 + a2)

2 + aq+13 = 0,

whence eitheraq2 +a2 +a

(q+1)/23 = 0 or aq

2 +a2−a(q+1)/23 = 0. Comparison with

Example 11.33 shows thatH(3, q) ∩ Q is Γ1 ∪ Γ2. From Exercise 8 in Chapter 7,eitherX = Γ1 or X = Γ2. SinceΓ1 andΓ2 are isomorphic overFq2 , the resultfollows. 2

Theorem 11.39 together with Corollary 11.6 gives the case thatm = 12 (q + 1)

of the following general result.

THEOREM 11.41 LetX be anFq2-maximal curve of genusg, and suppose thatthere is a pointP ∈ X (Fq2) with a non-gapm dividing q + 1. ThenX is bira-tionally equivalent overFq2 to the curveE1

m = v(E1) with

E1(X,Y ) = Y q + Y −Xm . (11.39)


PROPOSITION 11.42 Let q be even withq > 4, and letdimD = 3. If X lies on aquadricQ in PG(3,K), then

(i) Q is a cone;

(ii) the vertexV ofQ belongs toX ;

(iii) V ∈ X (Fq2) andj2(V ) = 12 (q + 2).

Proof. The following properties of quadrics ofPG(3,K) are used.For a non-singular pointP of Q let TP be the tangent plane toQ atP .

(a) If P ∈ X , thenTP ⊃ L1(P );

(b) If ℓ andℓ1 are distinct lines such thatP ∈ ℓ ⊂ Q andℓ1 ⊂ TP , thenTP isspanned byℓ andℓ1;

(c) There exist linesℓ andℓ1 such thatP ∈ ℓ ∩ ℓ1 andQ∩ TP = ℓ ∪ ℓ1;

(d) If Q is non-singular, then no two tangent hyperplanes ofQ at different pointscoincide.

(i) SinceX is non-degenerate,Q is irreducible. Also,Q is a cone if and onlyif Q is singular. Therefore letQ be non-singular, that is, a hyperbolic quadric ofPG(3,K).

From Lemma 11.37 (iii),j2(Q) = 2 for eachQ ∈ X (Fq2). Note that thereexistsP 6∈ X (Fq2) such thatj2(P ) > 2. Otherwise, by Lemma 11.13,

6(g − 1) = (q + 1)(q − 3);

but thenq would be odd, a contradiction. Hencej2(P ) = 12q again by Lemma

11.37 (iii).LetQ1 ∈ X (Fq2). ThenQ1 6∈ L1(P ) since

X ∩ L1(P ) ⊂ X ∩ L2(P ) = P,Φ(P ),and hence the planeH = HQ1

spanned byL1(P ) andQ1 is well defined. ThusH 6= L2(P ), and the intersection divisorX ·H is given by

X ·H = 12P +D , (11.40)

whereD = DQ1is a divisor onX of degree1

2 (q + 2) with Q1 ∈ Supp(D),andP 6∈ Supp(D). In addition, Lemma 11.37 (ii) ensures the existence of a lineℓ = ℓQ1

such that

Q∩H = L1(P ) ∪ ℓ . (11.41)

The lineℓ is defined overFq2 as isQ by Lemma 11.38 (iii); the fact thstQ1 is onX (Fq2) but not onL1(P ) implies thatQ1 ∈ ℓ.

Now it is shown that

X ∩ ℓ ⊂ X (Fq2). (11.42)

If there existsQ ∈ X ∩ ℓ\X (Fq2), thenΦ(Q) ∈ ℓ asℓ is defined overFq2 . Thusℓ ⊂ L2(Q), and henceℓ ∩ X ⊂ Q,Φ(Q). ThereforeQ1 6∈ ℓ; but this is acontradiction.

412

(A) If Q ∈ Supp(D)\Φ(P ), thenQ ∈ X (Fq2) andnQ(D) = 1.

To prove (A) note that from (11.42) and the relation,

Supp(D)\Φ(P ) ⊂ ℓ ∩ X ,it follows that Q ∈ X (Fq2). Now, if nQ(D) ≥ 2, thenH ⊃ L1(Q) sincej2(Q) = 2 and by (IV). Also,ℓ 6= L1(Q) becauseL1(Q) 6⊂ Q by Lemma 11.37(ii). Therefore the planeH is spanned by the linesℓ andL1(Q), and henceH is thetangent planeTQ1

toQ atQ1. If ℓ1 is the line defined by the property thatQ1 ∈ ℓ1,thenQ∩ TQ1

= ℓ∩ ℓ1. From (11.41),L1(P ) = ℓ1 and soQ1 ∈ L1(P ); but this isa contradiction.

(B) Φ(P ) 6∈ Supp(D).

To prove (B) suppose on the contrary thatΦ(P ) ∈ Supp(D). By (11.42), thisassumption is equivalent toΦ(P ) ∈ L1(P ). The argument in the proof of (A) canbe used to deduce thatH = TΦ(P ) sincenΦ(P )(D) 6= 1 andL1(P ) 6= L1(Φ(P )).But this leads to a contradiction with (11.6). Therefore,nΦ(P )(D) = 1. Hence, foreveryQ ∈ X (Fq2), the divisorD in (11.40) may also be written as

D = DQ = Φ(P ) +D′Q,

and this may be done so that not only (11.41) remains valid butalso that

Supp(D′Q) ⊂ X (Fq2), deg(D′

Q) = 12q.

This gives the following.

(C) HQ is spanned byL1(P ) andQ′ whereQ′ is any point ofSupp(D′Q).

Now, letQ1, Q2 ∈ X (Fq2) such thatQ2 6∈ Supp(D′Q1

). Then

Supp(D′Q1

) ∩ Supp(D′Q2

) = ∅;otherwiseHQ1

= HQ2by (C). So, 1

2q must divide the number ofFq2-rationalpoints ofX , which is a contradiction since|X (Fq2)| = q2 + 1 + 2gq is odd.

So far it has been shown that eachQ1 ∈ X (Fq2) gives rise to a planeHQ1

together with a lineℓ = ℓQ1and a divisorD = DQ1

such that (11.40) and (11.41)hold, with

D = Q1 +Q2 + . . .+Q(q+2)/2

being the sum of12 (q + 2) points allFq2-rational. Note thatSupp(D) = X ∩ ℓ.Let ℓ1 be chosen in such a way thatQ1 ∈ ℓ1 and that

Q∩ TQ1= ℓ ∪ ℓ1 . (11.43)

Here ℓ1 is Fq2-rational, and thusX ∩ ℓ1 ⊂ X (Fq2) as in the proof of (11.42).Therefore

X · TQ1= 2Q1 +Q2 + . . .+Q(q+2)/2 +D′ , (11.44)

whereD′ is a divisor ofX of degree12 (q−2) such thatQ1 6∈ Supp(D′) ⊂ X (Fq2).

Consider the following statement.


(D) Supp(D) ∩ Supp(D′) = ∅ andnS(D′) = 1 for eachS ∈ Supp(D′).

To show (D), letS ∈ Supp(D′). Suppose on the contrary thatS = Qi for somei.ThenTQ1

containsL1(Qi) which is different fromℓ asj2(Qi) = 2. HenceTQ1is

generated byL1(Qi) andℓ. These lines also spanTQiand soi = 1, contradicting

thatQ1 /∈ Supp(D′).Finally suppose on the contrary thatnS(D2) ≥ 2. Replacingℓ by ℓ1, the above

argument shows thatTS = TQ1, whenceS = Q1, again a contradiction. Thus, to

eachQ1 there is associated two linesℓ andℓ1 such that both (11.43) and (11.44)hold, whereD′ is a divisor of degree12 (q − 2), whereSupp(D′) ⊂ X (Fq2), andwhereSupp(D) ∩ Supp(D′) = Q1. A hyperbolic quadricQ contains exactlytwo families of lines and any two lines of the same family are disjoint. This impliesagain that|X (Fq2)|must be a multiple of12q, contradicting theFq2-maximality ofX .

(ii) As q is even, Lemma 11.13 ensures the existence of a pointP ∈ X withj2(P ) > 2. Suppose thatP 6∈ X (Fq2); then j2(Φ(P )) = j2(P ) > 2. Sincej2(P )P +D ≡ (q + 1)P0,

j2(P )Φ(P ) + Φ(D) ≡ (q + 1)P0

and soj2(Φ(P )) = j2(P ) > 2. Therefore, both the linesL1(P ) andL1(Φ(P ))are contained inQ by Lemma 11.37 (ii), and hence their common point isV . Now,sinceV is Fq2-rational by Lemma 11.38 (iii), soΦ(P ) 6= V , and henceL1(Φ(P ))is spanned byΦ(P ) andV . In particular,L1(Φ(P )) ⊂ L2(P ) and thus

1 = nΦ(P )(X · L2(P )) ≥ j2(Φ(P )),

a contradiction.ThereforeP must beFq2-rational and henceQ must have a singularity atP by

Lemma 11.37 (iii). ThenP = V andj2(P ) = 12 (q+2); these follow from Lemma

11.37 (i) and the assumption thatq is even. 2

THEOREM 11.43 Let q be even withq ≥ 16, and letX be anFq2-maximal curveof genusg = 1

4q(q − 2). ThenX is birationally equivalent overFq2 to T2 inExample 11.34, that is, to the curveT2 = v(T2), with

T2(X,Y ) = Y q/2 + Y q/4 + . . .+ Y 2 + Y +Xq+1 . (11.45)

Proof. From Castelnuovo’s Bound, Corollary 11.23,dimD ≤ 3. As dimD = 2only occurs whenX is a Hermitian curve, whose genus is1

2q(q−1), letdimD = 3.By Halphen’s Theorem 7.111,X lies on a quadric ofPG(3,K). From (IV) andProposition 11.42 (ii), there existsP0 ∈ X (Fq2) such that12q is a non-gap atP0.By (XII) the same holds forq + 1. Therefore the Weierstrass semigroupH(P0) isgenerated by12q andq+1. Choosex, y ∈ Fq2(X ) such that div(x)∞ = 1

2qP0 anddiv(y)∞ = (q+ 1)P0, and consider theFq2-rational linear series|12q(q+ 1)P0| oforder 1

2q(q + 1). Since12q(q + 1) > 2g − 2 = 1

2q(q − 2)− 2,

the Riemann–Roch Theorem 6.59 implies thatdim |12q(q + 1)P0| = 14q

2 + q. Onthe other hand, div(xq+1)+ 1

2q(q+1)P0, as well as each of the following14 (q+2)2

divisors, belongs to|12q(q + 1)P0|:div(xjyi) + 1

2q(q + 1)P0, 0 ≤ i ≤ 12q, 0 ≤ j ≤ q − 2i.

414

Therefore,x andy must satisfy a non-trivial polynomial relation; that is,

cyq/2 +∑q/2−1

i=0 Ai(x)yi = xq+1 , (11.46)

for c ∈ Fq2\0 and forAi[X] ∈ Fq2 [X] with degAi(x) ≤ q − 2i. A lengthycalculation reduces (11.46) to the following:

∑ti=1 aiy

q/2i

+ b = xq+1 , (11.47)

whereai ∈ Fq2\0 for all i, and the polynomial∑t

i=1 aiTq/2i

+ b has a rootα ∈ Fq2 . Replacingy by y + α, (11.47) becomes the following:

∑ti=1 aiy

q/2i

= xq+1 . (11.48)

Replacingx by a−11 x andy by at−1y give the desired equation:

∑ti=1 y

q/2i

= xq+1.

Therefore,X is birationally equivalent to theFq2-rational irreducible plane curveT2 of (11.45). 2

11.5 MAXIMAL CURVES WITH HIGH GENUS

In this section, the aim is to show that the upper part of the spectrum of genera ofFq2-maximal curves consists of the values ofg in Table 11.2. Although the wholespectrum depends on the nature ofq, the highest three genera in the spectrum havethe same positions independently ofq.

Since the Hermitian curveHq is Fq2-maximal, its genusg = 12q(q− 1) belongs

to the spectrum, and is the maximum.

PROPOSITION 11.44 LetX be anFq2-maximal curve of genusg. Then

g ≤ 12q(q − 1).

Proof. SinceN2 ≥ N1, from (11.1),

q4 + 1− 2gq2≥ q2 + 1 + 2gq,

q4 − q2≥ 2g(q2 + q),


To present the other results on the spectrum, let

gi = gi(q)

denote thei-th largest genus in the spectrum; then

gi − gi−1

is thei-th holeor gap in the spectrum.Proposition 11.44 and the existence ofHq show that

g1 = 12q(q − 1).

The following theorem implies the uniqueness of anFq2-maximal curve of genusg1. Also, it shows thatg2 = ⌊ 14 (q − 1)2⌋, as the curves in Example 11.3 achievethis value.


THEOREM 11.45 If X is anFq2-maximal curve of genusg, the following state-ments are equivalent:

(i) X is isomorphic overFq2 to the Hermitian curveHq;

(ii) g > ⌊ 14 (q − 1)2⌋;

(iii) r = 2;

(iv) ν1 > 1.

Proof. (i)⇒ (ii), (iii), (iv) This follows from properties ofHq.(iii)⇒ (iv) For r = 2, ν1 = νr−1 and the assertion follows from (IX).(iv)⇒ (ii) Theorem 8.65 together with the Natural Embedding Theorem 11.20

imply that if ν1 > 1 thenN1 = q3 + 1, whenceg = 12q(q − 1).

(ii)⇒ (iii) From Corollary 11.23,F (3) = (q − 1)2/4, and hencer = 2.(iii)⇒ (i) This follows from the Natural Embedding Theorem 11.20. 2

The second hole in the spectrum is determined in the next result.

THEOREM 11.46 For q ≥ 7, letX be anFq2-maximal curve of genusg satisfying(11.36). Then the following conditions are equivalent:

(i) ⌊ 16 (q2 − q + 4)⌋ < g ≤ ⌊ 14 (q − 1)2⌋;

(ii) dimD = 3, X lies on a quadric surface inPG(3,K), and

g 6= 16 (q2 − 2q + 3), for q ≡ 3, 5 (mod 6);

(iii) dimD = 3, dim(2D) = 8, and

g 6= 16 (q2 − 2q + 3), for q ≡ 3, 5 (mod 6);

(iv) dimD = 3 and there existsP ∈ X (Fq2) such that

j2(P ) =

12 (q + 1) for q odd,12 (q + 2) for q even;

(v) X is birationally equivalent overFq2 to the curve(i) or (ii) in Example 11.3according asq is odd or even;

(vi) g = 14 (q − 1)2 if q is odd andg = 1

4q(q − 2) if q is even; in particular, thegenusg = c0(q + 1, 3), Castelnuovo’s number.

Proof. (i)⇒ (ii) From the hypothesis ong, Halphen’s Theorem 7.111 and (11.6)imply thatdimD = 3 . Since, in Corollary 11.23,c0(q+1, 3) = ⌊ 16 (q2−q+4)⌋, theNatural Embedding Theorem 11.20 together with Halphen’s Theorem shows thatX lies on a quadric provided that eitherq /∈ 7, 8, 9, 11, 13, 17, 19, 23 or p > 2andX is reflexive. So, forq even, letq = 8. Theng > 1

6 (q2 − q + 4) = 10. ByLemma 11.38 (i),(ii), it is enough to show thatdim(2D) ≤ 8. Suppose, however,that dim(2D) ≥ 9. Then, by Lemma 7.107,g ≤ 1

4 (q − 1)(q − 2) = 21/2, acontradiction.

416

Now, letq be odd withq ≥ 7, and suppose thatX is not reflexive. By Theorem7.112,ǫ2 > 2. If P ∈ X (Fq2), then, from Propositions 11.45 and 8.53,

vP (S) ≥∑3i=1 (ji(P )− νi−1) ≥ ǫ2 + 1 ≥ 4.

So the Stohr–Voloch Theorem applied toX implies that

(3q − 1)(2g − 2) ≤ (q + 1)(q2 − 4q − 1).

On the other hand,2g − 2 > 13 (q + 1)(q − 2) by hypothesis, and thus5q + 5 < 0,

a contradiction.(iii)⇒ (ii) This follows from Lemma 11.38 (ii).(ii)⇒ (iv) Let q be odd. Then there existsP ∈ X such thatj2(P ) > 2; other-

wiseg = 16 (q2 − 2q + 3) by Proposition 11.13. If such a pointP ∈ X were not in

X (Fq2) then, by Lemma 11.37 (iii), bothP andΦ(P ) would be singular points ofthe quadric, a contradiction. ThereforeP ∈ X (Fq2) and hencej2(P ) = 1

2 (q + 1)by Lemma 11.37 (i).

If q is even, the result follows from Proposition 11.42 (ii).(iv)⇒ (v) From (IV) and the hypothesis,m1(P ) = 1

2 (q + 1) for q odd andm1(P ) = 1

2q for q even. In the odd case,q is another non-gap atP , and henceg = 1

4 (q−1)2 by Corollary 11.6. A similar argument works in the even case;sinceg = 1

4q(q − 2) andq + 1 is another non-gap atP , Corollary 11.6 applies.Finally, the implications(v)⇒ (vi), (vi)⇒ (i) and (v)⇒ (iii) follow read-

ily. 2

REMARK 11.47 Forq = 2, 3, 4, 5 the spectra of the genera ofFq2-maximal curvesare0, 1, 0, 1, 3, 0, 1, 2, 6, 0, 1, 2, 3, 4, 10.

11.6 CASTELNUOVO’S NUMBER

In this section, certainFq2-maximal curves whose genus is Castelnuovo’s numberc0(q + 1, r) for 4 ≤ r ≤ 5 are considered. First the case,

q ≡ 1, 2 (mod 3),

is investigated. In this casec0(q + 1, 4) = 16 (q − 1)(q − 2). The main result, see

Theorem 11.52 below, provides a complete classification ofFq2-maximal curvesof genus

g = 16 (q − 1)(q − 2), q ≡ 1, 2 (mod 3), q ≥ 11.

SuchFq2-maximal curves can only exist forq ≡ 2 (mod 3), and they are bira-tionally equivalent overFq2 to theFq2-rational curve,

E1(q+1)/3 = v(Y q + Y −X(q+1)/3) . (11.49)

The proof of Theorem 11.52 requires three preliminary lemmas.

L EMMA 11.48 For q ≡ 1, 2 (mod 3), let X be anFq2-maximal curve of genusg = 1

6 (q − 1)(q − 2). ThendimD = 4 andg = c0(q + 1, 4).


Proof. From (11.6) and Theorem 11.45,3 ≤ r ≤ 4. Suppose on the contrary thatr = 3.

If ǫ2 = 2, then

degR = (3 + q)(2g − 2) + 4(q + 1),

while theFq2-maximality ofX implies that

degR ≥ q2 + 1 + 2gq,

sincevP (R) ≥ 1 for everyP ∈ Fq2(X ). But theng ≥ 16 (q2−2q+3), contradicting

the hypothesis ong.If ǫ2 > 2, thenǫ2 ≥ 5 by Lemma 7.60 and sinceq 6≡ 0 (mod 3). Replacing the

ramification divisorR by the Stohr–Voloch divisorS, the previous argument againyields a contradiction. In fact,

degS = (1 + q)(2g − 2) + (q2 + 3)(q + 1),

while

degS ≥ (q2 + 1 + 2gq)(ǫ2 + 1),

by theFq2-maximality ofX and the lower boundvP (S) ≥ ǫ2+1 for P ∈ Fq2(X ),shown in the proof of Theorem 11.46. Sinceǫ2 ≥ 5, so

(5q − 1)(2g − 2) ≤ (q + 1)(q2 − 6q − 3),

whence2q2 − 3q + 13 ≤ 0 for g = 16 (q − 1)(q − 2), a contradiction. 2

The assumption that the genus ofX is Castelnuovo’s numberc0(q + 1, 4) isused via Lemma 7.107 (i). Indeed, from this lemma,dim(2D) = 11 and hencethe possibilities for(D, P )-orders can be determined. To show how to do this, letji = ji(P ) and denote by∆P the set of(2D, P )-orders. Then∆P contains bothsets

∆1 = 0, 1, 2, j3, j4, j4 + 1, j4 + j2, j4 + j3, 2j4 (11.50)

∆2 = j2, j2 + 1, j3 + 1, 2j2, j3 + j2, 2j3,wherej4 = q + 1 for P ∈ X (Fq2), andj4 = q otherwise.

L EMMA 11.49 LetP ∈ X with j2(P ) = 2. If r = dimD = 4, dim(2D) = 11,andq ≥ 9, thenj3(P ) = 3.

Proof. The hypothesis onq together with Proposition 11.13 implies that

j3 <

j4 − 2 for P ∈ X (Fq2),j4 − 1 for P ∈ X\X (Fq2).

(11.51)

Supposej3 > 3. If P ∈ X (Fq2), from (11.50) and (11.51),

∆P = Σ1 ∪ 3, j3 + 1, j3 + 2 ,and2j2, 2j3 ∈ ∆P . Hencej3 = 2j2 = 4, so that2j3 = 8 = j4 = q + 1; thusq = 7. If P /∈ X (Fq2) andj3 > 4, from (11.50) and (11.51) it follows that

∆P =∆1 ∪ 3, 4, j3 + 1,(j3 + 2, 2j3) ∈ (q, q + 1), (q, q + 2), (q + 1, q + 2).

418

Thenj3 ≤ 4, a contradiction. Finally, ifP 6∈ X (Fq2) andj3 = 4, then (11.50)together with (11.51) show that

∆P = ∆1 ∪ 3, 5, 6, 8 .Hencej4 = q = 8, and this completes the proof. 2

The previous lemma together with Lemma 11.13 gives the following result.

COROLLARY 11.50 If q ≥ 9, dim(D) = 4, dim(2D) = 11, j2(P ) = 2 for anyP ∈ X , thenq ≡ 1, 2 (mod 3) andg = 1

12 (q2 − 3q + 8).

Next, the case thatj2(P ) > 2 for some pointP ∈ X is considered.

L EMMA 11.51 LetP ∈ X with j2(P ) > 2, and suppose that

dimD = 4, dim(2D) = 11, q ≥ 7.

(i) If P ∈ X (Fq2) and g > 18q(q − 2) for q even, then one of the following

holds:

(a) q ≡ 2 (mod 3), j2(P ) = 13 (q + 1), j3(P ) = 1

3 (2q + 2);

(b) q ≡ 0 (mod 3), j2(P ) = 13 (q + 3), j3(P ) = 1

3 (2q + 3).

(ii) If P /∈ X (Fq2), then one of the following holds:

(a) q ≡ 1 (mod 3), j2(P ) = 13 (q + 2), j3(P ) = 1

3 (2q + 1);

(b) q ≡ 0 (mod 3), j2(P ) = 13q, j3(P ) = 2

3q;

(c) q ≡ 1 (mod 2), j2(P ) = 12 (q − 1), j3(P ) = 1

2 (q + 1);

(d) q ≡ 0 (mod 2), j2(P ) = 12q, j3(P ) = 1

2 (q + 2).

Proof. Suppose first thatj3 > j2 + 1. From (11.50) and (11.51), there are onlythree possibilities, namely

∆P =∆1 ∪ j2, j2 + 1, j3 + 1,(j3 + j2, 2j3) ∈ (j4, j4 + 1), (j4, j4 + j2), (j4 + 1, j4 + j2).

The first cannot occur sincej3 6= j2 + 1; in the second case,

j4 ≡ 0 (mod 3), j2 = 13j4, j3 = 2

3j4;

in the third case,

j4 ≡ 1 (mod 3), j2 = 13 (j4 + 2), j3 = 1

3 (2j4 + 1).

Suppose next thatj3 = j2 + 1. Then2j2 6∈ j3, j3 + 1 sincej2 > 2. Further,either2j2 6= j4 + 1 or j2 = 1

2 (j4 + 1), j3 = 12 (j4 + 3). In the latter case, from

(11.50) and (11.51),

∆P = ∆1 ∪ j2, j3 + 1, j4 + 2, j4 + 3.But this implies thatj4 + j2 = j4 + 3, whencej4 = 5 and soq ≤ 5. If 2j2 = j4,then eitherP 6∈ X (Fq2) or j3 = 1

2 (q + 3). Again, the latter case cannot actually


occur becausem1(P ) = 12 (q − 1) would then hold by (IV), and this would imply

thatdim(D) ≥ 5.Finally, assume that2j2 6∈ j3, j3 + 1, j4, j4 + 1. Then, from (11.50) and

(11.51),

∆P = j2, j3 + 1, 2j2,andj3 +j2 ∈ j4, j4 +1. If j3 +j2 = j4 +1, then2j2 = j4, whencej3 +j2 = j4.Thenj2 = 1

2 (j4 − 1) andj3 = 12 (j4 + 1). HereP 6∈ X (Fq2); otherwise,j2 = 1

2q,j3 = 1

2 (q + 2) and hencem1(P ) = 12q, m2(P ) = 1

2 (q + 2) by (IV). But theng ≤ 1

8q(q − 2), a contradiction. 2

THEOREM 11.52 Let q ≥ 11.

(i) If q ≡ 1 (mod 3), there is noFq2-maximal curve of genus16 (q − 1)(q − 2).

(ii) If q ≡ 2 (mod 3), the statements below are equivalent for anFq2-maximalcurveX of genusg :

(a)g = 16 (q − 1)(q − 2);

(b) there existP ∈ X (Fq2) andm ∈ H(P ) such that3m = q + 1;

(c)X is birationally equivalent overFq2 to the curve in (11.49).

Proof. (i) Suppose on the contrary thatX is anFq2-maximal curve of genus

g = 13 (q − 1)(q − 2), q ≡ 1 (mod 3).

Sinceq+1 = 13 (q−1) ·3+2, it follows from Lemma 11.48 thatg = c0(q+1, 3).

Hence, Lemma 7.107 implies thatdim(2D) = 11 and that|13 (q − 4)D| is thecanonical linear series onX . Then

1 + a1i1 + . . .+ a(q−4)/3i(q−4)/3 6∈ H(P ) , (11.52)

where the integersij are(D, P )-orders, and the coefficientsaj are non-negativeintegers such that

∑j aj ≤ (q − 4)/3.

Let P ∈ X with j2(P ) > 2 as in Corollary 11.50. By Lemma 11.51,P is not inX (Fq2). Recall thatm3(P ) = q by (XIV). So, three cases are considered.

Case (A): j2(P ) = 13 (q + 2), j3(P ) = 1

3 (2q + 1).

From (XI),

q −m2(P ), q −m1(P ) ⊂ 1, 13 (q + 2), 1

3 (2q + 1).This implies thatq −m1(P ) = 1

3 (q + 2), since otherwisem1(P ) = 13 (q − 1) and

henceq ≥ m4(P ), a contradiction. Thusm1(P ) = 13 (2q−2). But this again leads

to a contradiction as (11.52) implies that13 (q − 7) + 1

3 (q + 2) + 1 = 13 (2q − 2)

does not belong toH(P ).

Case (B): q ≡ 1 (mod 2), j2(P ) = 12 (q − 1), j3(P ) = 1

2 (q + 1).

From (11.52),2j2(P ) + 1 = q does not belong toH(P ), a contradiction.

Case (C): q ≡ 0 (mod 2), j2(P ) = 12q, j3(P ) = 1

2 (q + 2).

420

The argument used in Case (A) shows this time that eitherm1(P ) = 12q − 1

or m1(P ) = 12q. In the former case,q − 2 ∈ H(P ) and thus (XI) implies that

j2(P ) = 2. Since this contradicts the hypothesis, only the latter case can occur.Thenm1(P ) = 1

2q andm2(P ) = q− 1. Now, asdim(2D) = 11, som9(P ) = 2q.From this,2q − m4(P ) = 1

2q + 2. In fact, 2q − mi(P ) is a (2D, P )-order fori = 0, . . . , 9, and the set of(2D, P )-orders is

0, 1, 2, 12q,

12 (q + 2), 1

2 (q + 4), q, q + 1, q + 2, 32q, 2q .

Hencem4(P ) = 32q − 2.

Finally, letd = 13 (q − 4)(1

2q + 1) + 1. From (11.52),d /∈ H(P ). On the otherhand,d = m4(P ) + 1

6 (q − 10)m2(P ) ∈ H(P ), a contradiction.(ii) (a)⇒ (b) From Lemma 11.48,g = c0(q+1, 4). Sinceq+1 = 1

3 (q−2)·3+3,Lemma 7.107 shows thatdim(2D) = 11 and that13 (q− 5)D+D′ is the canonicallinear series, whereD′ is a base-point-free1-dimensional linear series of degree13 (q + 1). In particular, there existsx ∈ K(X ) with [K(X ) : K(x)] = 1

3 (q + 1).Let P ∈ X and assume by Corollary 11.50 thatj2(P ) > 2. If P ∈ X (Fq2), the

result follows from Lemma 11.51 (i). Otherwise,P /∈ X (Fq2), and there are twopossibilities according asq is odd or even; see Lemma 11.51 (ii).

Case (A): q ≡ 1 (mod 2), j2(P ) = 12 (q − 1), j3(P ) = 1

2 (q + 1).

A property similar to (11.52) holds; namely,

δ + 1 6∈ H(P ) for any( 13 (q − 5)D, P )-orderδ.

Hence2j2(P ) + 1 = q /∈ H(P ), a contradiction.

Case (B): q ≡ 0 (mod 2), j2(P ) = 12q, j3(P ) = 1

2 (q + 2).

From Case (C) in the proof of (i), it also follows thatm1(P ) = 12q. Let y

be an element ofK(X ) such that div(x)∞ = 12q. Sincegcd( 1

3 (q + 1), 12q), so

K(X ) = K(x, y). From Theorem 5.60,g ≤ 16 (q − 2)2, a contradiction.

Finally, (b) ⇒ (c) is Theorem 11.41 form = 13 (q + 1), and(c)⇒ (a) follows

immediately. 2

Now, Fq2-maximal curves of genus16q(q − 3) are considered in the case,

r = 4, q = 3h.

Note thatc0(q + 1, 4) = 16q(q − 3). The main result, see Theorem 11.55 below,

states that suchFq2-maximal curves are birationally equivalent overFq2 to theFq2-rational curve,

T3 = v(Y q/3 + Y q/9 + . . .+ Y 3 + Y −Xq+1). (11.53)

The proof of Theorem 11.55 depends on two preliminary lemmas.

L EMMA 11.53 For q = 3h, let X be anFq2-maximal curve of genus16q(q − 3).Thenr = dimD is either3 or 4. If r = 4, the following hold.

(i) dim(2D) = 11.

(ii) There exists a complete linear seriesD′ of order 23q and dimension2 such

that 13 (q − 6)D +D′ is the canonical linear series onX .


(iii) If j2 = 2, thenj3 = 3.

(iv) If P ∈ X (Fq2) and j2 > 2, thenj2 = 13 (q + 3), and j3 = 1

3 (2q + 3). Inparticular, the Weierstrass semigroup atP is generated by13q andq + 1.

(v) If q ≥ 27 andP 6∈ X (Fq2), thenj2 = 2.

Proof. This can be done using arguments and calculations similar tothose in thepreceding proofs for the caseq ≡ 1, 2 (mod 3). 2

L EMMA 11.54 For q ≥ 27, there existsP ∈ X (Fq2) such thatH(P ) is generatedby 1

3q andq + 1.

Proof. By Lemma 11.53 (ii), it is enough to show the existence ofP ∈ X (Fq2)with j2(P ) > 2. Suppose on the contrary thatj2(P ) = 2 for everyP ∈ X (Fq2).If P ∈ X (Fq2) then j4(P ) = q + 1 and this together with Lemma 11.53 (iii)implies that the(D, P )-orders are0, 1, 2, 3, q + 1. Likewise, ifP 6∈ X (Fq2), thenj4(P ) = q and, from this and Lemma 11.53 (v), the(D, P )-orders are0, 1, 2, 3, q.Therefore, by Corollary 8.55,

deg(R) = (6 + q)(2g − 2) + 5(q + 1) = |X (Fq2)| = (q + 1)2 + q(2g − 2) ,

whence2g − 2 = 16 (q − 4)(q + 1); that is,q2 − 3q − 8 = 0, a contradiction. 2

THEOREM 11.55 For q = 3h, letX be anFq2-maximal curve of genus16 (q−3)q.If dimD 6= 3 thendimD = 4 andX is birationally equivalent overFq2 to theFq2-rational curveT3, see (11.53).

Proof. From Lemma 11.53,r = dimD = 4. LetP ∈ X (Fq2) be as in Corollary11.54. Then the Weierstrass semigroupH(P ) is generated by13q andq + 1.

Choosex, y ∈ Fq2(X ) for which div(x)∞ = 13qP and div(y)∞ = (q + 1)P .

Then the Frobenius linear seriesD consists of all divisors

Ac = div(c0 + c1x+ c2x2 + x3x

3 + c4y) + (q + 1)P

with c = (c0, c1, c2, c3, c4) ∈ PG(4,K). Since ordP (xjyi) ≥ − 13q(q + 1) + 1

unless either(j, i) = (q + 1, 0) or (j, i) = (0, 13q), the complete linear series

|13q(q + 1)P )| comprises the divisors

Bb = div(b0xq+1 +

∑q/3i=0

∑q−3ij=0 bijx

jyi) + 13q(q + 1)P,

whereb = (b0, . . . , bij , . . .) ∈ PG(m,K) with m = 16 (q2 + 5q) + 1. By the

Riemann–Roch Theorem 6.59dim |13q(q + 1)P )| = 16 (q2 + 5q). Therefore,

xq+1 +∑q/3

i=0 Ai(x)y = 0, (11.54)

whereAi(X) ∈ Fq2 [X] with degAi(X) ≤ q− 3i and, in particular,Aq/3(X) is anon-zero constant belonging toFq2 . Finally, some ad hoc arguments depending onhigher Hasse derivatives and changes of coordinates reduce(11.54) to the simpler(11.53). 2

Now, consider the case,

g = 18 (q − 1)(q − 3), q odd.

422

Up to a birational transformation overFq2 , Theorem 11.61 below states that, forp ≥ 5 andq large enough, the two curves,

F = E1(q+1)/4 = v(Y q + Y −X(q+1)/4) , q ≡ 3 (mod 4) , (11.55)

F =D(q+1)/2 = v(X(q+1)/2 + Y (q+1)/2 + 1) , (11.56)

are the onlyFq2-maximal curves of genusg = 18 (q − 1)(q − 3) whendimD = 5.

It should be noted that the condition ondimD enters into play since, from theproof of Lemma 11.48, it can only be deduced that4 ≤ dimD ≤ 5, whereas18 (q − 1)(q − 3) is Castelnuovo’s number fordimD = 5 but not fordimD = 4.So the condition,dimD = 5, is necessary to take advantage of Lemma 7.107.Castelnuovo’s Bound It may also be noted that the two curves are not birationallyequivalent even overK since their Weierstrass semigroups are different; see Exer-cises 5 and 6 in Chapter 6.

THEOREM 11.56 Whenq ≥ 11, the Fermat curve in (11.56) is the uniqueFq2-maximal non-singular plane curve of degree1

2 (q + 1).

As dim(2D) = 14 by Lemma 7.107 (i), the possibilities for the(D, P )-ordersequences atP ∈ X can be determined. The two relevant results are stated belowbut their proofs are omitted as being similar to those of Lemmas 11.49 and 11.51,and Corollary 11.50.

L EMMA 11.57 For P ∈ X ,

(i) j1(P ) = 1;

(ii) j5(P ) = q + 1 if P ∈ X (Fq2) andj5(P ) = q if P ∈ X\X (Fq2).

L EMMA 11.58 For q ≥ 11 andP ∈ X , let dim(D) = 5, dim(2D) = 14.

(i) If j3(P ) = 3, thenj4(P ) = 4.

(ii) Let j2(P ) = 2, j3(P ) > 3;

(a) if P ∈ X (Fq2), thenq is odd, j3(P ) = 12 (q + 1), j4(P ) = 1

2 (q + 3);

(b) if P 6∈ X (Fq2), thenq is even, j3(P ) = 12q, j4(P ) = 1

2 (q + 2).

(iii) If P ∈ X (Fq2), j2(P ) > 2, and

g >

19 (q − 2)2, for q ≡ 2 (mod 3) ,19q(q − 3), for q ≡ 0 (mod 3),

then one of the following holds:

(a) q ≡ 3 (mod 4), j2(P ) = 14 (q + 1), j3(P ) = 1

2 (q + 1),

j4(P ) = 34 (q + 1);

(b) q ≡ 0 (mod 4), j2(P ) = 14 (q + 4), j3(P ) = 1

2 (q + 2),

j4(P ) = 14 (3q + 4).


(iv) If P 6∈ X (Fq2), j2(P ) > 2, then one of the following holds:

(a) q ≡ 1 (mod 4), j2(P ) = 14 (q + 3), j3(P ) = 1

2 (q + 1),

j4(P ) = 14 (3q + 1);

(b) q ≡ 0 (mod 4), j2(P ) = 14q, j3(P ) = 1

2q, j4(P ) = 34q;

(c) q ≡ 1 (mod 3), j2(P ) = 13 (q − 1), j3(P ) = 1

3 (q + 2),

j4(P ) = 13 (2q + 1);

(d) q ≡ 0 (mod 3), j2(P ) = 13q, j3(P ) = 1

3 (q + 3), j4(P ) = 23q.

COROLLARY 11.59 If q ≥ 11, dim(D) = 5, dim(2D) = 14, andj3(P ) = 3 foreveryP ∈ X , thenq ≡ 0, 4 (mod 5) andg = 1

20 (q2 − 4q + 15).

COROLLARY 11.60 If, for q ≥ 11 with q odd, X is an Fq2-maximal curve ofgenusg = 1

8 (q − 1)(q − 3) with dim(D) = 5, then the following hold:

(i) X is birationally equivalent overFq2 to the curveE1(q+1)4 in (11.55) if and

only if there existsP ∈ X (Fq2) with j2(P ) > 2;

(ii) X is birationally equivalent overFq2 to the curveD(q+1)/2 in (11.56) if andonly if there existsP ∈ X (Fq2) with j2(P ) = 2, j3(P ) > 3.

Proof. (i) From Lemma 13.1, the curveF in (11.55) has a unique branch centredatY∞ = (0, 0, 1). This branch isFq2-rational and hence it corresponds to a pointP ∈ X (Fq2). By a straightforward computation,m3(P ) = 3

4 (q + 1). Hencej2(P ) = 1

4 (q+1) by (IV). Conversely, from Lemma 11.58 (iii),j4(P ) = 34 (q+1)

and som1(P ) = 14 (q + 1) by (IV). Therefore the result follows from Theorem

11.41.(ii) The Fermat curveF in (11.55) is non-singular. By Theorem 6.49, the linear

system of conics cuts out onF a complete linear seriesg5q+1.

The pointP0 = (1, α, 0) of F , with α(q+1)/2 = −1, is an inflexion and thetangentF atP is ℓ = v(αX0 −X1). Therefore,I(P0,F ∩ ℓ) = 1

2 (q + 1). HenceI(P0,F ∩ C) = q + 1, whereC = v((αX0 −X1)

2) is the conic consisting of therepeated lineℓ. Thusg5

q+1 contains theFq2-rational divisor(q + 1)P , and henceD = g5

q+1 by Theorem 6.33.Choose a lineℓ1 throughP and a lineℓ2 disjoint fromP . Then

I(P0,F ∩ C′) =

2 when the conicC′ = ℓ21 ,

12 (q + 1) when the conicC′ = ℓℓ2 .

Therefore,j2(P ) = 2, j3(P ) > 3.Conversely, (IV) and Lemma 11.58 (ii) imply thatm1(P ) = 1

2 (q − 1), andm2(P ) = 1

2 (q + 1). Choosex, y ∈ K(X ) such that div(x)∞ = 12 (q − 1) and

div(y)∞ = 12 (q + 1). Since these numbers are coprime,K(X ) = K(x, y) and

there existsf ∈ Fq2 [X,Y ] such thatdeg f ≤ 12 (q + 1) andf(x, y) = 0. In other

words,X is birationally equivalent overFq2 to an irreducible plane curveF ofdegree at12 (q + 1). If F were not singular, from Lemma 3.24 its virtual genus

g∗ < 12 [(1

2 (q + 1)− 1)( 12 (q + 1)− 2] = 1

8 (q − 1)(q − 3).

424

But theng∗ < g, contradicting Lemma 3.28. HenceF is non-singular and theresult follows from Theorem 11.56. 2

THEOREM 11.61 If q is odd, p ≥ 5, andX is anFq2-maximal curve with genusg = 1

8 (q − 1)(q − 3) anddim(D) = 5, the the folowing hold.

(i) If q ≥ 17 andq ≡ 1 (mod 4), thenX is birationally equivalent overFq2 tothe curveD(q+1)/2 in (11.56).

(ii) If q ≥ 19 and q ≡ 3 (mod 4), thenX is birationally equivalent overFq2

either to the curveE1(q+1)/4 in (11.55) or to the curveD(q+1)/2 in (11.56).

Proof. As already observed,g = c0(q + 1, 5) and thusdim(2D) = 14. In particu-lar, by Corollary 11.59, there existsP ∈ X with j3(P ) > 3.

(i) Let q ≡ 1 (mod 4). If P ∈ X (Fq2), then Lemma 11.58 (ii), (iii) imply thatj2(P ) = 2. Hence the result follows from Corollary 11.60 (ii). To showthat this isthe only possible case, letP 6∈ X (Fq2). AsD′ = 1

4 (q−5)D is the canonical linearseries by Lemma 7.107 (ii), and henceδ + 1 /∈ H(P ) for any(K, P )-orderδ.

Sincem4(P ) = q by (XIV), Lemma 11.58 together with the hypothesisp ≥ 5leads to the following two cases (A) and (B).

Case (A): j2(P ) = 14 (q + 3), j3(P ) = 1

2 (q + 1), j4(P ) = 14 (3q + 1).

In this case,q −m3, q −m2, q −m1 ⊂ 1, 1

4 (q + 3), 12 (q + 1), 1

4 (3q + 1)by (XIV). Thus,

m1(P ) = 12 (q − 1), m2(P ) = 3(q − 1), m3(P ) = q − 1.

Now, δ = 14 (q−9)+ 1

4 (3q+1) = q−2 is a(D′, P )-order and henceq−1 /∈ H(P ),a contradiction.

Case (B):q ≡ 1 (mod 3), j2(P ) = 13 (q − 1), j3(P ) = 1

3 (q + 2),j4(P ) = 1

3 (2q + 1).

Since 14 (q − 5) ≥ 3 in this case,δ = 3j2(P ) must be a(D′, P )-order. Hence

q /∈ H(P ), a contradiction.(ii) q ≡ 3 (mod 4). Again, it suffices to show thatP ∈ X (Fq2), since the

result then follows from Corollary 11.60. IfP /∈ X (Fq2), Lemma 11.58 (ii), (iv)together with the hypothesisp ≥ 5 imply that j2(P ) = 1

3 (q − 1). From Lemma7.107 (ii) , δ + 1 /∈ H(P ) for every( 1

4 (q − 7)D, P )-orderδ. On the other hand,since1

4 (q − 7) ≥ 3 it follows that3j2(P ) + 1 = q ∈ H(P ), a contradiction. 2

REMARK 11.62 There existFq2-maximal curves whose genus equals Halphen’snumberc1(4, q + 1). The known examples are listed below, but in (ii) and (iii) noexplicit equation is known; see Section 11.12:

(i) for q ≡ 0 (mod 4), curves of genus18 (q2 − 2q);

(ii) for q ≡ 1 (mod 4), curves of genus18 (q − 1)2 constructed as a quotients ofthe Hermitian curveHq by a subgroup of the automorphism group;

(iii) for q ≡ 3 (mod 4), curves of genus18 (q2 − 2q+ 5) constructed similarly tothose in (ii).


11.7 PLANE MAXIMAL CURVES

In this sectionFq2-maximal non-singular plane curves are considered.

THEOREM 11.63 For every proper divisord of q + 1, the Fermat curve

F = D(q+1)/d = v(X(q+1)/d + Y (q+1)/d + 1),

is Fq2-maximal.

Proof. The rational transformationx′ = xd, y′ = yd of the function fieldK(Hq)of the Hermitian curveHq = v(Xq+1+Y q+1+1) is defined overFq2 and providesad-fold covering ofK(F). The result follows from Theorem 11.2. 2

Let F be anFq2-maximal non-singular plane curve of degreen with n ≥ 3.From Theorem 5.56, the genus ofF is g = 1

2 (n − 1)(n − 2). From Proposition11.44,n ≤ q+1, and equality holds if and only ifF is projectively equivalent overFq2 to the Hermitian curveHq, as in Theorems 11.45 and 11.20.

From now on, letn < q + 1. Upper bounds onn, and equivalently ong, areobtained from the complete linear seriesL1 = g2

n cut out onF by lines. Notationand terminology from Sections 7.8 and 8.6 are used. In particular, ǫ0 < ǫ1 < ǫ2with ǫ0 = 0, ǫ1 = 1 are theL1-orders andν0 < ν1 with ν0 = 0 are the Frobeniusorders ofL1.

L EMMA 11.64 Let F be anFq2-maximal non-singular plane curve of degreenwith n < q + 1. Then the following hold:

(i) ǫ2 ≤ q, ν1 = 1;

(ii) if

n2(q) =

⌊(q + 2)/2⌋ whenq ≥ 4 andq 6= 3, 5,

3 whenq = 3,4 whenq = 5,

thenn ≤ n2(q);

(iii) if n = 12 (q + 1) thenF is classical.

Proof. (i) For any lineℓ andP ∈ F ,

I(P, ℓ ∩ F) ≤ n < q + 1.

Henceǫ2 ≤ j2 ≤ n ≤ q. If ν1 > 1, then, from Theorem 8.65,

q2 + 1 + (n− 1)(n− 2)q=n(q2 − n+ 2)

n2 − (q + 2)n+ q + 1 =0,

whencen = 1 or q + 1, a contradiction. Henceν1 = 1; that is,F is Frobeniusclassical.

(ii) From the Stohr–Voloch Theorem 8.17,

2(q2 + 1 + (n− 1)(n− 2)q) ≤ n(n− 3) + (q2 + 2)n,

426

whencen ≤ n2(q).(iii) Suppose thatj2(P ) ≥ 3 for all P . From Corollary 8.55,vP (S) ≥ 3. Then,

from the Stohr–Voloch Theorem 8.17,

3(q2 + 1 + (n− 1)(n− 2)q) ≤ n(n− 3) + (q2 + 2)n.

But this is impossible forn = 12 (q + 1). 2

COROLLARY 11.65 (i) If n ≥ 3 is the degree of anFq2-maximal non-singularplane curve, then eithern = q + 1 or n ≤ n2.

(ii) For q 6= 3, 5, there is noFq2-maximal non-singular plane curve with genusg in the intervalI given as follows:

18q(q − 2) < g ≤ 1

4q(q − 2), for q even,18 (q − 1)(q − 3) < g ≤ 1

4 (q − 1)2, for q odd.

Proof. To show (ii), assume thatn ≤ n2(d). By Lemma 11.64 (i),F is Frobeniusclassical. From the Stohr–Voloch Theorem 8.17,

n ≤ q2 + 2q + 1

2q − 1= F (q).

It follows thatF (q) < 12 (q + 3) for q > 5 and thatF (3) = 16/5, F (4) = 25/7,

F (5) = 4. Finally, a straightforward computation shows thatn /∈ I for q 6= 3, 5. 2

REMARK 11.66 Let 3 ≤ d ≤ n2(q).

(i) If q is odd, then theFq2-maximal curve in (11.56) shows that the upper boundn2(q) = 1

2 (q + 1) in Corollary 11.65 is the best possible forq 6= 3, 5.

(ii) n2(q) = 3 is sharp forq = 3.

(iii) There exists anF25-maximal non-singular plane quintic; son2(q) = 4 issharp forq = 5.

(iv) From part (ii) of the lemma,n2(q) = 3 is sharp forq = 4. For q even andq ≥ 8, it is not known if the bound,n2(q) = 1

2 (q + 2), is good or bad.

Refinements to the above upper bound can be obtained by similar computationsdepending on the Stohr–Voloch Theorem 8.17 applied toL1.

PROPOSITION 11.67 If F is anFq2-maximal non-singular plane curve of degreen such that3 ≤ n ≤ n2(q), thenX is classical with respect toL1 when one of thefollowing conditions holds:

(i) p > n or n 6≡ 1 (mod p);

(ii) q = 4, 8, 16, 32;

(iii) p ≥ 3 and eitherq = p or q = p2;


(iv) p = 2, q ≥ 64, and eitherd ≤ 4 or

n ≤ n3(q) =

14q − 1, for q = 64, 128, 256,

14q, for q ≥ 512 andq odd;

(v) p ≥ 3, q = pv with v ≥ 3, andn ≥ n3(q) = q/p− p+ 2.

REMARK 11.68 For q = p3, p ≥ 3, the boundn3(q) in Proposition 11.67 issharp, as there exists a planeFp6 -maximal curve of degreep2 − p + 1 which isnon-classical with respect toL1; see Corollary 11.73 and Remark 11.83.

COROLLARY 11.69 If F is as in Proposition 11.67 and non-classical with respecttoL1, then

(i) q ≥ 64 for p = 2, andq ≥ p3 for p ≥ 3;

(ii) ǫ22 ≤ q/p.

The example in Remark 11.68 shows that Corollary 11.69 (i) issharp forp ≥ 3.

DEFINITION 11.70 (i) For q = 8 andq ≥ 11, let

n4(q) =2q2 + 15q − 20 +

√4q4 − 40q3 + 145q2 − 300q + 600

10(q − 2).

(ii) For q = pv, v ≥ 2, let

n4(p, q) =2q2 + 3(5− 1/p)q − 8

2(5− 1/p)q − 12

+

√4q4 − 8(5− 1/p)q3 + (113− 50/p+ 9/p2)q2 − 4(25− 17/p)q + 184

2(5− 1/p)q − 12.

THEOREM 11.71 If q = 8 or q ≥ 11 andn is the degree of a planeFq2-maximalcurveX , with 3 ≤ n < q + 1, then one of the following holds:

(i) n = ⌊(q + 2)/2⌋;

(ii)

n ≤ n5(q) =

n4(q) whenq = p,

n4(p, q) whenq = pv, v ≥ 2,

11.8 MAXIMAL CURVES OF HURWITZ TYPE

A Hurwitz curve of degreen+ 1 is a non-singular plane curve,

Cn = v(Xn0X1 +Xn

1X2 +Xn2X0) , (11.57)

wheren2 − n+ 1 6≡ 0 (mod p).

428

L EMMA 11.72 The Hurwitz curveCn is covered by the Fermat curve,

Dn2−n+1 = v(Xn2−n+10 +Xn2−n+1

1 +Xn2−n+12 ).

Proof. LetK(F) = K(x, y), with xn2−n+1 + yn2−n+1 + 1 = 0, be the functionfield ofF . Put

ξ = xn−1y−1, η = xyn−1;

then ξnη + ηn + ξ = 0. So the function fieldK(ξ, η) of Cn is a subfield ofK(F), and the rational transformationω, given byx′ = ξ, y′ = η, provides anFq2-rational covering ofCn byDn2−n+1. 2

Consider the following condition:

q + 1 ≡ 0 (mod (n2 − n+ 1)) . (11.58)

L EMMA 11.73 If (11.58) holds, then the curvesCn andDn2−n+1 are both coveredoverFq2 by the Hermitian curveHq, and hence areFq2-maximal.

Proof. The argument of Lemma 11.72 can be adapted to show thatDn2−n+1 isFq2-covered by the Hermitian curveHq = v(Xq+1

0 + Xq+11 + Xq+1

2 ). ThenTheorem 11.2 gives the result. 2

L EMMA 11.74 The Weierstrass semigroup ofCn at the pointP1 = (0, 1, 0) isgenerated by the setS = s(n− 1) + 1 | s = 1, . . . , n.Proof. LetP0 = (1, 0, 0) andP2 = (0, 0, 1). Then divx = nP2− (n− 1)P1−P0

and divy = (n− 1)P0 + P2 − nP1, so that

div(xs−1y) = ((n(s− 1) + 1)P2 + (n− s)P0 − (s(n− 1) + 1)P1 .

This shows thatS is contained in the Weierstrass semigroupH(P1) at P1, andhenceH(P1) ⊃ 〈S〉. Therefore, the result follows from a result on arithmeticprogressions:|N0\〈S〉| = 1

2n(n− 1); see [107]. 2

THEOREM 11.75 The Hurwitz curveCn is Fq2-maximal if and only if (11.58)holds.

Proof. If (11.58) holds, thenCn is Fq2-maximal by Lemma 11.73.Conversely, assume thatCn is Fq2-maximal. Then(q + 1)P1 ≡ (q + 1)P2 by

(11.6), and the cases = n in the proof of Lemma 11.74 gives that

(n2 − n+ 1)P1 ≡ (n2 − n+ 1)P2.

Thereforeu = gcd(n2 − n+ 1, q + 1) ∈ H(P1). From Lemma 11.74,

u = A(n− 1) +B with A ≥ B ≥ 1.

Now, there existsC ≥ 1 such that

(A(n− 1) +B)C = n2 − n+ 1,


and soBC = D(n− 1) + 1 for someD ≥ 0. Therefore,

AD(n− 1) +A+BD = Bn.

HereD = 0, as otherwise the left-hand side would exceedBn. So

B = C = 1, A = n;

that is,u = n2 − n+ 1. 2

COROLLARY 11.76 The Fermat curveDn2−n+1 is Fq2-maximal if and only if(11.58) holds.

Proof. If (11.58) is satisfied, the result follows from Corollary 11.73. Now ifDn2−n+1 is Fq2-maximal, thenCn is alsoFq2-maximal by Theorem 11.2. Thenthe corollary follows from Theorem 11.75. 2

REMARK 11.77 For a given positive integern, is there a powerq of a primep suchthatq + 1 ≡ 0 (mod m) with m = n2 − n+ 1?

Sincem 6≡ 0 (mod p), andp 6≡ 0 (mod m), a necessary and sufficient condi-tion for q to have property (11.58) is thatp ≡ x (mod m), wherex is a solution ofthe congruenceTw + 1 ≡ 0 (mod m), andw is defined byq = pφ(m)v+w, whereφ is the Euler function.

REMARK 11.78 The Hurwitz curveCn is Fq2-maximal in the following cases:

(1) n = 3, q = p6v+3 andp ≡ 3, 5 (mod 7);

(2) n = 4, q = p12v+6 andp ≡ 2, 6, 7, 11 (mod 13).

By using Theorem 11.75 and Remark 11.77, this is refined as follows.

COROLLARY 11.79 The Hurwitz curvesC2, C3, C4 are Fq2-maximal if and onlyif the one of the following conditions onp andq holds.

(i) C2 : q = p2v+1 andp ≡ 2 (mod 3).

(ii) C3 :

(a) q = p6v+1 andp ≡ 6 (mod 7);

(b) q = p6v+3 andp ≡ 3, 5, 6 (mod 7);

(c) q = p6v+5 andp ≡ 6 (mod 7).

(iii) C4 :

(a) q = p12v+1 andp ≡ 12 (mod 13);

(b) q = p12v+2 andp ≡ 5, 8;

(c) q = p12v+3 andp ≡ 4, 10, 12 (mod 13);

(d) q = p12v+5 andp ≡ 12 (mod 13);

(e) q = p12v+6 andp ≡ 2, 5, 6, 7, 8, 11 (mod 13);

430

(f) q = p12v+7 andp ≡ 12 (mod 13);

(g) q = p12v+9 andp ≡ 4, 10, 12 (mod 13);

(h) q = p12v+11 andp ≡ 12 (mod 13).

COROLLARY 11.80 For a positive integern, letm = n2 − n+ 1.

(i) If n = pe with e ≥ 1, then the curveCn is Fq2-maximal forq = pφ(m)v+3e.

(ii) If p ≡ 3 (mod 4), n ≡ 0, 1 (mod p), m is prime andm ≡ 3 (mod 4),then the curveCn is Fq2-maximal forq = p(m−1)v+(m−1)/2.

Proof. (i) This is a consequence of Theorem 11.75 and the identity,

p3e + 1 = (pe + 1)(p2e − pe + 1).

(ii) It is enough to show thatp(m−1)/2 + 1 ≡ 0 (mod m). Recall the Legendresymbol(a/p):

(a/p) =

1 if x2 ≡ a (mod p) has two solutions inZp,−1 if x2 ≡ a (mod p) has no solution inZp.

Sincem ≡ 1 (mod p), here(m/p) = 1. By the Quadratic Reciprocity Law,

(m/p)(p/m) = (−1)((m−1)/2)((p−1)/2) ,

and the condition thatm ≡ 3 (mod 4), it follows that (p/m) = (−1)(p−1)/2.Now, asp ≡ 3 (mod 4), this gives(p/m) = −1. In other words,p viewed asan element inFm is a non-square. Since−1 is also a non-square inFm, thisimplies thatp ≡ (−1)u2 (mod m) for an integeru with u 6≡ 0 (mod m). Hencep(m−1)/2 ≡ −1 (mod m) asm is odd andum−1 ≡ 1 (mod m). 2

REMARK 11.81 The condition thatm ≡ 3 (mod 4) in this corollary cannot berelaxed. Forn = 4,m = 13 but, from Corollary 11.79,C4 is notF36 -maximal.

REMARK 11.82 Under the hypotheses in Corollary 11.80 (ii) withm not necessar-ily prime, the congruence (11.58) may be investigated by means of the groupUm ofunits inZm. This group has orderφ(m), andp ∈ Um sincem ≡ 1 (mod p). Nowsuppose thatp, as an element ofUm, has even order2i. Thenp2i ≡ 1 (mod m)and hence(pi + 1)(pi − 1) ≡ 0 (mod m). Sincep has order greater thani, sopi− 1 6≡ 0 (mod m) unless bothpi + 1 andpi− 1 are zero divisors inZm. Whenthis does not happen, then (11.58) follows forq = pφ(m)v+i.

REMARK 11.83 Letp be a prime,n = peuwith e ≥ 1 andgcd(p, u) = 1. Assumethate ≥ 2 whenp = 2. Then the curvesCn andDn2−n+1 are both non-classicalwith respect toL1, and0, 1, pe are theirL1-orders.

DEFINITION 11.84 A generalised Hurwitz curve of degreen+k is the irreducibleplane curve

Cn,k = v(Xn0X

k1 +Xn

1Xk2 +Xn

2Xk0 ) ,

wheren ≥ k ≥ 2 andp does not divideQ(n, k) = n2 − nk + k2.


The singular points ofCn,k areP0 = (1, 0, 0), P1 = (0, 1, 0), andP2 = (0, 0, 1);each of them is the centre of a unique branch ofCn,k. The genus ofCn,k is

g =n2 − nk + k2 + 2− 3 gcd(n, k)

2,

TheFq2-maximality of these curves can be investigated by the same methods asfor Hurwitz curves, with analogous results.

L EMMA 11.85 The curveCn,k is Fq2-covered by the Fermat curveDn2−nk+k2 .

Consider the condition:

n2 − nk + k2 ≡ 0 (mod q + 1) . (11.59)

COROLLARY 11.86 The curveCn,k is Fq2-maximal provided that (11.59) holds.

COROLLARY 11.87 The curveDn2−nk+k2 is Fq2-maximal provided that (11.59)holds.

LetP1 be the place arising from the branch ofCn,k centred atP .

L EMMA 11.88 If gcd(n, k) = 1, then the Weierstrass semigroup atP1 is

H(P1) = (n−k)s+nt | s, t ∈ Z; t ≥ 0; −kt/n ≤ s ≤ (n−k)/(kt) . (11.60)

THEOREM 11.89 If gcd(n, k) = 1 andn2−nk+k2 is prime, then the curveCn,k

is Fq2-maximal if and only if(11.59) holds.

COROLLARY 11.90 If n, k are as in Theorem 11.89, then the curveDn2−nk+k2

Fq2-maximal if and only if (11.59) holds.

REMARK 11.91 There are infinitely many pairsn, k with n > k ≥ 1 such thatn2−nk+k2 is prime. In fact, for a primep′ such thatp′ ≡ 1 (mod 6), there existsuchn, k with p′ = n2 − nk + k2.

11.9 NON-ISOMORPHIC MAXIMAL CURVES

In this section, a 2-parameter family of curvesXmi is presented; for each fixedm,

there is a large number of non-isomorphic curves all with some identical properties.With K = Fq, letXm

i be a non-singular model overK of the plane curve,

Cmi = v(Xmi+m +Xmi + Y q+1) , (11.61)

wherem is a positive divisor ofq + 1 for which d = (q + 1)/m > 3 is prime.The curveXm

i is the quotient curve of the Hermitian curveHq arising from anautomorphism group ofH of the same orderd. Let D = |(q + 1)P | denote theassociated complete linear series at a pointP of Xm

i .

THEOREM 11.92 Assume1 ≤ i ≤ d− 2.

432

(i) (a) The curvesXmi and Xm

j are K-equivalent if and only if one of thefollowing equation holds modulod :

i ≡ j, ij ≡ 1, ij + i+ j ≡ 0 ,i+ j + 1 ≡ 0, ij + i+ 1 ≡ 0, ij + j + 1 ≡ 0 .

(b) The number ofK-isomorphism classes of curvesXmi is given by

n(d) =

16 (d+ 1) if d ≡ 2 (mod 3),16 (d− 1) + 1 if d ≡ 1 (mod 3).

(11.62)

(c) Each of these classes consists of six curves, apart from two exceptionsof sizes2 and3. The corresponding indicesi are as follows:

(1) i1, i2,wherei1 andi2 are the solutions oft2+t+1 = 0 (mod d),with d ≡ 1 (mod 3);

(2) 1, 12 (d− 1), d− 2.

(ii) The genus ofXmi is g = 1

2m(q − 2) + 1.

(iii) TheK-automorphism group ofXmi is

Aut(Xmi ) =

Z3 ⋊ (Zq+1 × Zm) in case(c)(1),

Z2 ⋊ (Zq+1 × Zm) in case(c)(2),

Zq+1 × Zm otherwise.

(iv) Whenm = 2, the seriesD has projective dimension12 (d + 3). There are atleast sixK-rational pointsP such that, if

(j0 = 0, j1 = 1, . . . , j(d+1)/2, j(d+3)/2)

is theD-order sequence atP, thenj(d+1)/2 = d, j(d+3)/2 = q + 1.

(v) Whenm = 2 andq is prime, then theD-order sequence at a generic point is(0, 1, . . . , 1

2 (d+ 1), q).

THEOREM 11.93 (i) The curvesXm0 andXm

d−1 areK-isomorphic.

(ii) Xm0 has genusg = 1

2 (m− 1)(q − 1), and is hyperelliptic whenm = 2.

(iii) The centreZ of theK-automorphism groupAutK(Xm0 ) is a cyclic group of

orderm, and the factor groupAutK(Xm0 )/Z is isomorphic toPGL(2, q).

(iv) The complete linear series onX 20 has projective dimensiond + 1, and the

D-order sequence at a Weierstrass point is one of

(0, 1, 2, . . . , d, q), (0, 1, 2, 4, 6, . . . , q − 1, q + 1).

(v) TheD-order sequence ofX 20 is (0, 1, . . . , d, q).


11.10 OPTIMAL CURVES

The previous sections contain examples of maximal curves such as the HermitiancurveHq that is bothFq2-maximal andFq2-optimal, as well as families of maximalcurves.

Optimal curves that are not maximal have only been classifiedfor g = 1. Thevalue ofNq(g) is known forg = 1, 2 and for various other pairs(g, q).

The aim of this section is a characterisation of the DLS curveas the uniqueoptimal curve overFq which has genusg = q0(q+ 1) whenq = 2q20 , q0 = 2s ands ≥ 1. The proof of Theorem 11.99 requires some preliminary results.

From Proposition 10.32, theL-polynomial ofX is Lq(t) = (t2 + 2q0t + q)g.Hence, as already observed in Example 10.77, the Fundamental Equation (10.65)of X becomes, forP, P0 ∈ X with P0 ∈ X (Fq):

qP + 2q0Φ(P ) + Φ2(P ) ≡ (q + 2q0 + 1)P0. (11.63)

Some results from Section 10.8 to be applied to the corresponding linear seriesD = |(q + 2q0 + 1)P0| are summarised below.

From Section 10.8,

1. r = dimD;

2. ǫ0 = 0 < ǫ1 = 1 < . . . < ǫr are theD-orders ofX ;

3. ν0 = 0 < . . . < νr−1 are the Frobenius orders ofX ;

4. m1(P ) < . . . < mr(P ) < . . . are the non-gaps ofX atP .

Then

(A) jr(P ) = mr(P ) = q + 2q0 + 1 for any P ∈ X (Fq), and there existsP1 ∈ X (Fq2) such thatmr−1(P1) = q + 2q0;

(B) ji(P ) = mr(P )−mr−i(P ) for anyP ∈ X (Fq) andi = 0, . . . r ;

(C) D is simple and base-point-free;

(D) 2q0 andq areD-orders, and sor ≥ 3;

(E) ǫr = νr−1 = q;

(F) m1(P ) = q for anyP ∈ X (Fq).

From (A), (C) and Proposition 10.80 (i), for anyP ∈ X (Fq),

jr−1(P ) = jr(P )−m1(P ) = 2q0 + 1,

whence

(G) 2q0 ≤ ǫr−1 ≤ 2q0 + 1 .

L EMMA 11.94 The orderǫr−1 = 2q0.

434

Proof. Suppose on the contrary thatǫr−1 > 2q0. Then

ǫr−2 = 2q0, ǫr−1 = 2q0 + 1.

By Corollary 8.55,

νr−2 ≤ jr−1(P )− j1(P ) ≤ 2q0 = ǫr−2,

and thus the Frobenius orders ofD are ǫ0, ǫ1, . . . , ǫr−2, andǫr. By Proposition8.53 (i), forP ∈ X (Fq),

vP (S) ≥∑ri=1 (ji(P )− νi−1) ≥ (r − 1)j1(P ) + 1 + 2q0 ≥ r + 2q0 . (11.64)

Hence

degS = (∑r

i=0 νi)(2g − 2) + (q + r)(q + 2q0 + 1) ≥ (r + 2q0)|X (Fq)|.Since2g− 2 = (2q0− 2)(q+2q0 +1) and|X (Fq)| = (q− 2q0 +1)(q+2q0 +1),

∑r−2i=1 νi =

∑r−2i=1 ǫi ≥ (r − 1)q0 .

Sinceǫi + ǫj ≤ ǫi+j for i+ j ≤ r by Exercise 8 in Chapter 7, it follows that

(r − 1)2q0 = (r − 1)ǫr−2 ≥ 2∑r−2

i=0 ǫi ≥ 2(r − 1)q0 .

Henceǫi + ǫr−2−i = ǫr−2 for i = 0, . . . , r − 2. In particular,ǫr−3 = 2q0 − 1and, by Lemma 7.59,ǫi = i for i = 0, 1, . . . , r − 3. Thenr = 2q0 + 2. NowCastelnuovo’s Bound, see Theorem 7.106, applied toD gives the following:

2g = 2q0(q − 1) ≤ q + 2q0 − (r − 1)/2)2

r − 1.

Sincer = 2q0 + 2, this implies that

2q0(q − 1) < 12 (q + q0)

2q0 = q0q + 12q + 1

2q0,

a contradiction. Castelnuovo’s Bound 2

COROLLARY 11.95 There existsP1 ∈ X (Fq) such that

j1(P1) =

1 for i = 1,

νi−1 + 1 for i = 2, . . . , r − 1.

Proof. Taking (11.64) into account, it suffices to show the existence of a pointP1 ∈ X (Fq2) for whichvP1

(S) = r+ 2q0. So, suppose thatvP (S) ≥ r+ 2q0 + 1for everyP ∈ X (Fq). Then (8.13) implies that

∑r−1i=0 νi ≥ q + rq0 + 1 .

Therefore,∑r−1

i=0 ǫi ≥ rq0 + 2 ,

sinceǫ1 = 1, νr−1 = q andνi ≤ ǫi+1. Then, from Exercise 8 in Chapter 7,rǫr−1 ≥ 2rq0 + 4. But this means thatǫr−1 > 2rq0, which contradicts Lemma11.94. 2

L EMMA 11.96 (i) The Frobenius orderν1 > ǫ1 = 1;


(ii) the orderǫ2 = 2e.

Proof. (i) Suppose on the contrary thatν1 = 1. Then, from Corollary 11.95, thereexists a pointP1 ∈ X (Fq) such thatj1(P1) = 1, j2(P1) = 2. Therefore

H(P1) ⊂ H = 〈q, q + 2q0 − 1, q + 2q0, q + 2q0 + 1〉 ,by (A), (B) and (F). In particular,g = q0(q − 1) ≤ g, whereg counts those thenon-negative integers which do not belong toH.

On the other hand,g > g. To show this, it suffices to show the following:

g = g − 14q

20 . (11.65)

The key idea is to observe thatΛ = ∪2q0−1i=1 Λi is a complete system of residues

moduloq, when

Λi = iq + i(2q0 − 1) + j | j = 0, . . . , 2i if 1 ≤ i ≤ q0 − 1,

Λq0= q0q + q − q0 + j | j = 0, . . . , q0 − 1;

Λq0+1 = (q0 + 1)q + 1 + j | j = 0, . . . , q0 − 1 for 2 ≤ i ≤ 12q0;

Λq0+i = (q0 + i)q + (2i− 3)q0 + i− 1 + j | j = 0, . . . , q0 − 2i+ 1∪(q0 + i)q + (2i− 2)q0 + i+ j | j = 0, . . . q0 − 1

for 1 ≤ i ≤ 12q0 − 1;

Λ3q0/2+i = (3q0/2 + i)q + (q0/2 + i− 1)(2q0 − 1) + q0 + 2i− 1 + j |j = 0, . . . , q0 − 2i− 1

Also, d ∈ Λ, d ∈ H andd− q 6∈ H for eachd ∈ Λ. Henceg can be calculated bysumming the coefficients ofq in this list. The result is as follows:

g=∑q0−1

i=1 i(2i+ 1) + q20 + (q0 + 1)q0 +∑q0/2

i=2 (q0 + i)(2q0 − 2i+ 2)

+∑q0/2−1

i=1 (3q0/2 + i)(q0 − 2i)

= q0(q − 1)− q20/4 .(ii) This follows from (i) and Lemma 7.59. 2

Now, let P0 = P1 be anFq-rational point satisfying Corollary 11.95, and letmi = mi(P1).

Lemma 11.96 (i) implies thatνi = ǫi+1 for i = 1, . . . , r − 1. Therefore, from(A), (B) and Corollary 11.95,

mi = 2q0 + q − ǫr−i for i = 1, . . . r − 2,

mr−1 = 2q0 + q, mr = 1 + 2q0 + q. (11.66)

Let x, y2, . . . , yr ∈ Fq(X ) be such that

div(x)∞ = m1P1, div(yi)∞ = miP1 for i = 2, . . . , r.

Note thatx is a separable variable ofK(X ). Sinceν1 > 1, Proposition 8.41 showsthat the matrix,

1 xq yq

2 . . . yqr

1 x y2 . . . yr

0 1 Dxy2 . . . Dxyr

,

has rank2. Therefore,

yqi − yi = Dxyi(x

q − x) for i = 2, . . . , r. (11.67)

436

L EMMA 11.97 (i) The divisor(2g − 2)P is canonical for anyP ∈ X (Fq2).

(ii) If m ∈ H(P1) satisfiesm < q + 2q0, thenm ≤ q + q0.

(iii) There existsgi ∈ Fq(X ) such thatDxyi = gǫ2i for i = 2, . . . , r, and

div(gi)∞ =qmi − q2

ǫ2P1.

Proof. (i) It may be assumed thatP = P1 from the identity (11.63) and the factthat2g− 2 = (2q0− 2)(q+ 2q0 + 1). Wheni = r in (11.67), ordP1

(dx) = 2g− 2and the result follows since ordQ(dx) ≥ 0 for Q 6= P1.

(ii) From (11.66),q, q + 2q0, q + 2q0 + 1 ∈ H(P1). Then

(2q0 − 2)q + q − 4q0 + j, j = 0, . . . , q0 − 2,

are also non-gaps atP1. Since (i) implies thatH(P ) is symmetric, it follows that

q + q0 + 1 + j, j = 0, . . . , q0 − 2,

are gaps atP1, and this completes the proof.(iii) Put fi = Dxyi. Then

D(j)x yi = (xq − x)D(j)

x fi +D(j−1)x fi,

for 1 ≤ j < q, by Lemmas 5.62 and 5.70. Then,D(j)x fi = 0 for 1 ≤ j < ǫ2, since

the matrix,

1 x y2 . . . yr

0 1 Dxy2 . . . Dxyr

0 0 D(j)x y2 . . . D

(j)x yr

,

has rank2 for eachj in 2 ≤ j < ǫ2. Consequently, asǫ2 is a power of2 by Lemma11.96 (ii), by Lemma 5.74, there existsgi ∈ K(X ) such thatfi = gǫ2

i . Sincefi ∈ Fq(X ), alsogi ∈ Fq(X ). Finally, from the proof of (i),x − x(P ) is a localparameter atP if P 6= P1. Then, by the choice of the elementsyi, the functiongi

has no pole other thanP1. From (11.67), ordP1(gi) = −(qmi − q2)/ǫ2. 2

L EMMA 11.98 The dimensionr = 4 and the orderǫ2 = q0.

Proof. From Theorem 11.45,r ≥ 3. Suppose thatr = 3. Then

ǫ2 = 2q0, n1 = q, n2 = q + 2q0, n3 = q + 2q0 + 1,

and hence ordP1g2 = −q for the sameg2 as in Lemma 11.97 (iii). Therefore, after

a change overFq of the coordinate system ofPG(3,K), the casei = 2 of (11.67)is as follows:

yq2 − y2 = x2q0(xq − x) .

Now, the elementz = yq0

2 −xq0+1 satisfieszq−z = xq0(xq−x). Thereforeq0 +qis a non-gap atP1. This contradiction implies thatr 6= 3.

Let r ≥ 4 and2 ≤ i ≤ r. By Lemma 11.97 (iii),

qni − q2ǫ2

∈ H(P1).


Since

qmi − q2ǫ2

≥ mni−1 ≥ q,

from (11.66),2q0 ≥ ǫ2 + ǫr−i for i = 2, . . . , r − 2. In particular,ǫ2 ≤ q0. On theother hand, by Lemma 11.97 (ii),mr−2 ≤ q + q0 and hence, by (11.66),ǫ2 ≥ q0;that is,ǫ2 = q0.

Note thatǫ2 = q0 implies ǫN−2 ≤ q0. Sincen2 ≤ q + q0 by Lemma 11.97)(ii), it follows from (11.66) thatǫr−2 ≥ q0. Thereforeǫr−2 = q0 = ǫ2 and hencer = 4. 2

THEOREM 11.99 If q = 2q20 , q0 = 2s ands ≥ 1, andX is an irreducible non-singularFq-rational curve of genusg such that

g = q0(q − 1), |X (Fq)| = q2 + 1 , (11.68)

thenX is birationally equivalent overFq to the DLS curve.

Proof. LetP1 ∈ X (Fq) be as above. From (11.67), Lemma 11.97 (iii) and Lemma11.98,

yq2 − y2 = gq0

2 (xq − x) ,whereg2 has no pole other thanP1. Also, from (11.66),m2 = q0 + q and soordP1

g2 = −q by Lemma 11.97 (iii). Thusg2 = ax + b with a, b ∈ Fq, a 6= 0.After a change overFq of the coordinate system ofPG(4,K), the result follows.

2

REMARK 11.100 The DLS curveX is also investigated in Section 13.2. Fromthe proof of Theorem 11.99, some further properties ofX related to its Frobeniuslinear seriesD = |(q + 2q0 + 1)P0|, P0 ∈ X (Fq), can be deduced.

(i) SinceD = g4q+2q0+1, the curveX has a complete, simple, base-point-free,

Fq-invariant linear series with orders0, 1, q0, 2q0, q and with Frobenius orders0, q0, 2q0, q. If K(X ) is given byK(x, y) with x2q0(xq + x) = yq + y as inSection 13.2, andu(x, y) = x2q0+1 + y2q0 , thenD consists of all divisors,

Ac = div(c0 + c1x+ c2y + c3u(x, y) + c4(xu(x, y) + y2)) + (q + 2q0 + 1)P0,

with c = (c0, c1, c2, c3, c4) ∈ PG(4,K). Therefore, the birational transformationπ : F 7→ X , whereF = v(X2q0(Xq +X)− (Y q +Y )) is given by the coordinatefunctions,

x0 = 1, x1 = x, x2 = y, x3 = u(x, y), x4 = xu(x, y) + y2.

From this representation ofX it can be verified thatX is a non-singular curve ofPG(4,K). To do this, note that the points ofX coming from the points ofF otherthanY∞ = (0, 0, 1) are non-singular sinceY∞ = (0, 0, 1) is the unique infinitepoint ofF and all other points are non-singular; see Theorem 13.12. AsY∞ is thecentre of a unique branch ofF , the corresponding pointP = (0, 0, 0, 0, 1) of X isthe only candidate to be a singular point ofX . But, if P were singular, then everyautomorphism ofX would fix P and the automorphism group Sz(q) of X could

438

not act transitively on the set of allFq-rational point ofX , contradicting Theorem13.12 (iv).

(ii) There existsP1 ∈ X (Fq) whose(D, P1)-orders are

j0(P1) = 0, j1(P1) = 1, j2(P1) = q0+1, j3(P1) = 2q0+1, j4(P1) = q+2q0+1.

They are also the(D, P )-orders for eachP ∈ X (Fq). To see this, note that

degS = (3q0 + q)(2g − 2) + (q + 4)(q + 2q0 + 1) = (4 + 2q0)|X (Fq)|.From (11.64),

vP (S) =∑4

i=1 (ji(P )− νi−1) = 4 + 2q0.

Now, using Corollary 8.55, it follows thatji(P ) = ji(P1) for all i.(iii) From (ii) and (B) it follows thatH(P ) contains the semigroup,

H = 〈q, q + q0, q + 2q0, q + 2q0 + 1〉,wheneverP ∈ X (Fq). Actually, H(P ) = H since the number of non-negativeintegers not inH is g = q0(q − 1). This can be shown as in the proof of Lemma11.96 (i).

(iv) From (i) and (ii),

degR =∑4

i=0 ǫi(2g − 2) + 5(q + 2q0 + 1) = (2q0 + 3)|X (Fq)| ,andvP (R) = 2q0 + 3 for P ∈ X (Fq). Therefore, theD-Weierstrass points ofX are precisely theFq-rational points ofX . In particular, the(D, P )-orders forP 6∈ X (Fq) are0, 1, q0, 2q0, q.

(v) Regarding the order sequence of the canonical curve ofX , from (iv) andLemma 11.97 (i), it follows that every integer,

m = a+ q0b+ 2q0c+ qd with a+ b+ c+ d ≤ 2q0 − 2,

is an order of the canonical curve at every non-Fq-rational point.(vi) From Proposition 10.80, the canonical curve ofX is non-classical asg ≥ q.

11.11 EXERCISES

1. If q = 3t ≥ 27 andX is anFq2-maximal curve over of genus16q(q − 3)with dimD = 4, show thatX is birationally equivalent overFq2 to the planecurveT ′′

3 = v(T′′

3 ), with

T′′

3 = Y q/3 + Y q/9 + . . .+ Y 3 + Y + cXq+1 ,

wherec ∈ Fq with cq−1 = −1.

2. Show that there is a maximal curve overFq2 whose genus is Halphen’s num-berc1(q + 1, 4) = ⌊ 18 (q2 − 2q + 5)⌋.

3. LetX be a non-singularFq2-rational plane curve and letL1 be the linearseries cut out onX by lines. Show that


(i) L1 is both non-classical and Frobenius non-classical if and only X isprojectively equivalent overFq2 to the Hermitian curve;

(ii) if q = p or q = p2, thenL1 is classical.

4. Letd be a positive integer such thatq ≡ −1 (mod d2). Show that the curveF = v(F ), with

F = Y (q+1)/d +X(d+1)(q+1)/d2

+X(q+1)/d2

,

is anFq2-maximal curve of genus1 + (q + 1)(q − d2 − 1)/(2d2).

5. Let d be a divisor ofq + 1. Use the Natural Embedding Theorem 11.20 toshow that the Fermat curveD(q+1)/d = v(F ), with

F = X(q+1)/d + Y (q+1)/d + 1,

is Fq2-maximal.

6. Prove Theorem 11.45 without using the Natural Embedding Theorem 11.20.

7. LetX be aFq2-maximal curve andP an ordinary point, that is, an ordinarypoint ofX . Show that, ifq + 1 ∈ H(P ), theng = q −N1 + 1.

8. Let q be a prime power of and letd > 2, i > 0 be integers such thatdi

dividesq + 1. Prove that there exists a sequence of curves,

X0 = H → X1 → . . .→ Xi,

such that each coveringXj−1 → Xj , j = 1, . . . , i, is unramified, whereH isthe Hermitian curve.

9. Forq ≡ 1 (mod 3), letG = v(G) with

G = Y q − Y X2(q−1)/3 + ωX(q−1)/3,

whereωq+1 = −1. Show that the genus ofG is g = 16 (q2 − q).

10. Find allF64-maximal curves of genus 3 up to isomorphism.

11. LetX be anFq2-maximal curve of genusg. Show that, ifX is hyperelliptic,theng ≤ 1

2q.

12. Show that noF64-maximal curve has genus 8.

13. Show thatF1 = v(X3 +X + Y 4) andF2 = v(X4 +X2 +X + Y 3) areF64-maximal curves which are not isomorphic overF64.

440

11.12 NOTES

Theorem 11.2 was originally stated by Serre; see Lachaud [175].In Table 11.2 the sources and characterisations are as follows:

1. g = 12q(q − 1), [231], [132];

2. g = 14 (q − 1)2, [81];

3. g = 14q(q − 2), [5];

4. g = 16 (q2 − q + 4), [57];

5, 6.g = 16 (q2 − q), [93];

7. g = 16 (q2 − q − 2), [56], [93];

8. g = 16 (q − 1)(q − 2), [172];

9. g = 16q(q − 3), [56], [5];

10. g = 18 (q2 − 2q + 5), [172];

11. g = 18 (q − 1)2, [172];

12. g = 18q(q − 2), [172];

13, 14.g = 18 (q − 1)(q − 3), [172];

15. g = 18q(q − 4), [3].

The minimality ofm in the preamble to Lemma 11.14 follows from a result dueto Kaji [156, Prop. 1]; see also [98, Prop. 2],

General properties of quadrics of a three-dimensional projective space over afinite field can be found in [126, Chapter 16]. For thorough studies of Hermitianvarieties, see [250], [126, Chapter 19] and [133, Chapter 23].

Section 11.2 is based on the survey paper [83]. The proof of Theorem 11.5 comesfrom [171].

For Remark 11.47, see [93, Remark 6.1].The results in Section 11.3, in particular, the Natural Embedding Theorem 11.20

and the other key Theorem 11.29, come from [171].Sections 11.4, 11.5 and 11.6 follow [172].The proof of Theorem 11.41 given in [81] is independent of theNatural Embed-

ding Theorem 11.20 but it uses Galois Theory. An alternativeproof may be carriedout by adapting the arguments in the proof of Theorem 11.41 and using results onHermitian varieties and quadrics in permutable position.

For the lengthy calculation omitted from Theorem 11.43, see[5].For Theorem 11.56, see [53].Sections 11.7 and 11.8 are based on [10].Theorem 11.99 and its proof come from [83].For the examples of maximal curves in Remark 11.62, see the following:

(i) [296, Prop. 5.2(ii)], [93, Theorem 3.3];


(ii) [57, Proposition 3.3(3)];

(iii) [57, Proposition 3.3(3)(1)], [93, Example 5.10].

For the quintic in Remark 11.66 (iii), see [258, Section 4].For Proposition 11.67, see [10].For Remark 11.78, see [40, Lemmes 3.3, 3.6].For generalised Hurwitz curves, see [28, Section 4], [27, Example 4.5], [10].The Weierstrass semigroupH(P1) in Lemma 11.88 is calculated fork = n− 1,

and(n, k) = (5, 2) in [28].For Remark 11.91, see [28, Remarque 4].Theorem 11.2 implies that the quotient curve of anFq2-maximal curve is also

Fq2-maximal. This makes it possible to find infinite families ofFq2-maximalcurves. Apart from those described here, others can be obtained not only fromthe Hermitian curve but also from the DLS and DLR curves; see Sections 13.3,13.2, 13.4, [93], [104], [38]. However, Theorem 11.2 fails for optimal curves ingeneral. But the curves attaining the Serre Bound 10.27 for anon-squareq are suchoptimal curves, such as the DLS and DLR curves overFq. Sporadic examples arethe Klein quartic overF8, as in Remark 10.29, and the curvev(Y 5 + Y 4 −X23)overF211 ; see Section 13.5. See also Remark 10.29.

The value ofNq(g) was found by Waterhouse forg = 1 and by Serre forg = 2and forg = 3 whenq ≤ 25. Also,Nq(g) is known for a number of sporadic pairs(g, q); see the survey in [148, Chapter 2]; van der Geer and van der Vlugt [298]maintain tables of known values ofNq(g).

For Remark 11.65, see [253].For an explicit description of the action of Sz(q) on the DLS curveX , see [104].

Also,X (Fq) is an ovoid of the non-singular quadricv(X0X4 −X1X3 −X22 ) of

PG(4, q); see [83].For other results on the orders of the canonical curve ofX , see [87, Section 4].

The non-classicality of the canonical curve is also shown in[87].

Chapter Twelve

Automorphisms of an algebraic curve

The concept of a symmetry, or automorphism in modern terminology, plays aprominent role in all branches of geometry. The saying thatthe larger its auto-morphism group, the richer its geometry,is especially appropriate for algebraiccurves. For instance, if there exists an automorphism of an irreducible algebraiccurveΓ, then another irreducible algebraic curve, the quotient curve, can be con-structed, and it inherits important properties ofΓ. Also, certain families of alge-braic curves have characterisations obtained from the structure and action of theirautomorphism groups. As in several concepts introduced previously, an automor-phism of an algebraic curve is a birational invariant. In thefunction fieldK(Γ), thecorresponding term is aK-automorphism, that is, a field automorphism ofK(Γ)which fixes every element ofK. So,K-automorphisms are mostly investigatedby using algebraic rather than geometric tools. However, sinceΓ is birationallyequivalent to an irreducible non-singular curveX embedded in a projective spacePG(r,K) such that everyK-automorphism ofK(Γ) can be represented as a linearcollineation ofPG(r,K) preservingX , so aK-automorphism can be viewed asa truly geometric concept. In this spirit, AutK(Σ) is theK-automorphism groupof X . More generally, the term,K-automorphism group ofF , is adopted for anycurveF birationally equivalent toΓ, that is, for every model ofK(Γ).

Since theK-automorphisms ofΓ form a group, their study greatly benefits fromgroup theory. A fundamental result is that theK-automorphism groups of an irre-ducible algebraic curve is finite, apart from rational and elliptic curves. So, finitegroup theory is relevant in this context, which is a further great advantage as it is inthe finite case that group theory has been most developed.

As, throughout the book,Σ denotes a function field of transcendence degree1,that is, the function field of an irreducible algebraic curveF . Further, AutK(Σ)stands for the group of allK-automorphisms ofΣ. For the special case thatF isdefined overFq, the concept ofFq-rationality forK-automorphisms is introduced.Let Fq(F) be theFq-rational function field ofF . A K-automorphismω of F isFq-rational if ω preservesFq(F), that is,ω viewed as a birational transformationof Σ is Fq-rational. Such elements form a subgroup of AutK(Σ), theFq-rationalK-automorphism group ofF .

12.1 THE ACTION OF K-AUTOMORPHISMS ON PLACES

The first result is a characterisation ofK-automorphisms as special birational trans-formations.

443

444

L EMMA 12.1 Letα be aK-automorphism ofΣ. If (F ; (x, y)) is any model ofΣ,then the birational transformationω

x′ = α(ξ), y′ = α(y)

preserves the curveF .

Proof. Sinceα ∈ AutK(Σ), from f(x, y) = 0 andα(f(x, y)) = f(x′, y′) itfollows thatf(x′, y′) = 0. Henceω preservesF . 2

L EMMA 12.2 Let (F ; (x, y)) be a model ofΣ. If a birational transformationω

(x, y) 7→ (x′, y′)

of Σ preserves the curveF then there is aK-automorphismα of Σ such thatx′ = α(x) andy′ = α(y).

Proof. Sinceω preservesF , P ′ = (x′, y′) is another generic point ofF . Now, letx′ = α(x), y′ = α(y), and extend this definition to any elementξ ∈ Σ in a naturalway; namely,

α(ξ) =u(x′, y′)

v(x′, y′)when ξ =

u(x, y)

v(x, y),

with u(X,Y ), v(X,Y ) ∈ K[X,Y ] andf(X,Y ) ∤ v(X,Y ). Note that the expres-sion ofξ as a rational function ofx, y is not unique; however,α(ξ) does not dependon the form of the expression. To verify this, let

u(x, y)

v(x, y)=u1(x, y)

v1(x, y).

Then

u(X,Y )v1(X,Y )− u1(X,Y )v(X,Y )

is a polynomial that vanishes at the generic pointP = (x, y) ofF . By the definitionof generic point, this polynomial is divisible byf(X,Y ). Sincef(x′, y′) = 0, thisyields

u(x′, y′)

v(x′, y′)=u1(x

′, y′)

v1(x′, y′).

So,α is correctly defined onΣ, and it is indeed aK-automorphism ofΣ. 2

It is intuitively apparent thatK-automorphisms ofΣ are an important tool ininvestigating places, divisors, and linear series by meansof the action of AutK(Σ)on such objects. In this context, the fixed fieldΣG consisting of all elements inΣwhich are fixed by every element inG plays an important role.

A geometric view of this is also possible, since there is a one-to-one correspon-dence between places ofΣ and points of any non-singular modelX of Σ, whichgives rise to a natural action of AutK(Σ) on the points ofX . In some circum-stances, such an action islinear in the sense that it arises from a linear collineationgroup of the projective space in whichX is embedded.

The first step is to define the action ofK-automorphisms on places.

Automorphisms of an algebraic curve 445

Letα ∈ AutK(Σ) and letσ be a primitive representation of a placeP of Σ. Thenthe imageσα of σ underα is defined by

(σα)(ξ) = σ(α(ξ)), ξ ∈ Σ.

As σα is a primitive place representation, it defines a place ofΣ: This place isindependent of the choice ofσ, since equivalent primitive place representationshave equivalent images. Such a place is called theimageof P by α, and denotedbyPα.

Once the action ofα on places has been defined, this extends toDiv(Σ) in anatural way. LetD ∈ Div(Σ) with D =

∑nPP; then

Dα =∑nPPα.

In particular,α preserves the set of all effective divisors ofΣ. For anyα-invariantsubfieldΣ′ = K(ξ, η) of Σ, it is appropriate to keep the notationα to denote theinducedK-automorphism ofΣ′. Thenα has an action, which may be trivial, onthe set of all placesPΣ′ of Σ′. It follows that, ifP ∈ PΣ andP ′ ∈ PΣ′ are placessuch thatP lies overP ′, thenPα lies overP ′α.

First, some elementary properties ofK-automorphisms ofΣ are collected.

L EMMA 12.3 Letα ∈ AutK(Σ) andP ∈ PΣ. Then, for anyz ∈ Σ,

ordPz = ordPα(α−1(z)). (12.1)

Proof. Let σ be a primitive representation ofP. Then ordP(z) = ordt σ(z) forz ∈ Σ. Also, ordt σ(α(α−1(z))) = ordt (σα)α−1(z) = ordPα(α−1(z)). Thisproves the assertion. 2

Note that (12.1) can also be written in the following way:

ordP(α(z)) = ordPα(z). (12.2)

Lemma 12.3 has some useful corollaries.

L EMMA 12.4 Letα ∈ AutK(Σ) andz ∈ Σ. Thendiv(z)α = div(α−1(z)). Moreprecisely,

(div(z)0)α = div(α−1(z))0, (div(z)∞)α = div(α−1(z))∞. (12.3)

Proof. First, div(z)0 =∑

P ordP(z)P, where the summation is over all placesP ∈ PΣ with ordP(z) > 0. Then(div(z)0)

α =∑

P ordP(z)Pα. By (12.1),∑

PordP(z)Pα =∑

PordPα(α−1(z))Pα.

By (12.1), ordP(z) > 0 if and only if ordPα(α−1(z)) > 0. Hence,∑

PordPα(α−1(z))Pα = div(α−1(z)),

which shows the second assertion. The equation for the divisor of poles is provedsimilarly. The first assertion is consequence of the second and third assertions.2

L EMMA 12.5 Let A ≡ B with A,B ∈ Div(Σ). ThenAα ≡ Bα for everyK-automorphismα of Σ.

446

Proof. A − B = div(z) for somez ∈ Σ. By (12.1),(A − B)α = div(α−1(z)).Since(A−B)α = Aα −Bα, the assertion follows. 2

A similar argument may be used to prove the following two results.

L EMMA 12.6 EveryK-automorphism ofΣ preserves the set of all Weierstrasspoints ofΣ.

L EMMA 12.7 Let F be a subfield ofΣ of transcendence degree1 such that thefield extensionΣ/F is separable. Ifα ∈ AutK(Σ) andF ′ = α(F ), then

D(Σ/F )α = D(Σ/F ′),

whereD(Σ/F ) andD(Σ/F ′) are the different divisors ofΣ/F andΣ/F ′.

L EMMA 12.8 Letα ∈ AutK(Σ). If div(z)α = div(z) for every non-zero elementz ∈ Σ, thenα is the identity automorphism ofΣ.

Proof. As α is aK-automorphism, it suffices to prove thatα(z) = z for anyz inΣ\K. Sinceα−1 also satisfies the hypothesis, so, from Lemma 12.4,

div(z) = div(α(z)),

whence div(z/α(z)) = 0. By Theorem 6.4,α(z) = cz for a non-zero constantc ∈ K. Replacingz by z+1, there isc′ ∈ K such thatα(z+1) = c′(z+1). Thusα(z) + α(1) = c′z + c′. These equations show thatcz + 1 = c′z + c′, whence(c − c′)z = c′ − 1. Sincez ∈ Σ \K, this is only possible whenc − c′ = 0 andc′ = 1. Hencec = 1, and the result follows. 2

L EMMA 12.9 Let α ∈ AutK(Σ). If Pα = P for every place ofΣ, thenα is theidentity automorphism ofΣ.

Proof. From the hypothesis,Dα = D for everyD ∈ Div(Σ). In particular, the hy-pothesis in the preceding lemma is satisfied. Hence the assertion is just a corollaryof that lemma. 2

Actually, Lemma 12.9 still holds if it is only required thatα fixes infinitely manyplaces ofPΣ. To show this the residue map resP associated to any placeP ∈ PΣ

as in Example 9.11, is required.

L EMMA 12.10 Letα ∈ AutK(Σ) andP ∈ PΣ. ThenresPz = resPαα−1(z) foranyz ∈ Σ.

Proof. Let σ be a primitive representation ofP. If z ∈ Σ, then

σ(z) = σ(αα−1(z)) = (σα)(α−1(z)),

and hence

resPz = res(σ(z)) = res(σα(α−1(z)) = resP α α−1(z).

2


THEOREM 12.11 Letα ∈ AutK(Σ). If Pα = P for infinitely many placesP ofΣ, thenα is the identity automorphism ofΣ.

Proof. Choose anα-invariant placeP in PΣ. From Lemma 12.10 applied toα−1,resPα(z) = resPz holds for allz ∈ Σ. If neitherz norα(z) has a pole atP, theadditive property of resP yields that

resPα(z)− resPz = resP(α(z)− z),giving resP(α(z) − z) = 0. Now, assume thatα has infinitely many fixed places,and remove those places that are poles for eitherz or α(z). The remaining fixedplaces ofα are still infinite, but none of them is a pole of any of the elements z,α(z) andα(z) − z. For each such placeP, the equation resP(α(z) − z) = 0implies thatP is a zero ofα(z) − z. By Theorem 5.33, there isc ∈ K such thatα(z)− z = c. So, sinceα is aK-automorphism,α(z)− z = 0 follows. 2

The number of fixed places that aK-automorphism can have is an importanttopic to which the whole Section 12.11 is devoted. Meanwhile, the following resultis required.

L EMMA 12.12 Every non-trivialK-automorphism ofΣ has at most2g + 2 fixedplaces.

Proof. Let α be a non-trivialK-automorphism ofΣ. By Theorem 12.11,α has afinite number of fixed places, says. Takeg + 1 distinct placesP1, . . . ,Pg+1 of Σsuch that the divisorD = P1 + . . .+Pg+1 and its imageDα share no place. By theRiemann–Roch theorem,Σ contains an elementz 6∈ K such that div(z)∞ = D.Now, considerw = z − α(z). Sincez andα(z) have different poles,w 6= 0. Moreprecisely,w has exactly2g + 2 poles, because each pole ofw is either a pole ofα(z) or a pole ofw. By Theorem 5.35,α has exactly2g + 2 zeros. On the otherhand, every fixed place ofα is a zero ofw. Hence, the number of such places is atmost2g + 2. 2

L EMMA 12.13 Letm be the smallest non-gap ofΣ at a placeP. Suppose thatξ ∈ Σ has a pole of orderm at P and is regular elsewhere. IfPα = P for anelementα ∈ AutK(Σ), thenα(ξ) = cξ + d with c, d ∈ K, c 6= 0.

Proof. Choose a primitive representationσ of P. Then

σ(ξ) = c1t−m + c2t

−m+1 + . . . ,

with c1 6= 0. By Lemma 12.3,

σ(α(ξ)) = d1t−m + d2t

−m+1 + . . . ,

with d1 6= 0. Let η = d1ξ − c1α(ξ). Then

σ(η) = e2tk + . . . , with k ≥ −m+ 1.

The minimality ofm implies thatk ≥ 0. This shows thatP is not a pole ofη. Sinceξ, and by Lemma 12.3 alsoα(ξ), is regular outsideP, the same holds forη. Henceη has no poles at all. By Theorem 5.33,η ∈ K, whence the assertion follows.2

448

THEOREM 12.14 LetΣ be rational.

(i) ThenAutK(Σ) ∼= PGL(2,K), andAutK(Σ) acts on the setPΣ of all placesof Σ asPGL(2,K) in its unique3-transitive permutation representation.

(ii) Writing Σ = K(ξ), then

(a) α ∈ AutK(Σ) if and only if

α(ξ) : ξ 7→ aξ + b

cξ + d; ad− bc 6= 0, a, b, c, d ∈ K;

(b) the identity ofAutK(Σ) is the onlyK-automorphism inAutK(Σ) fixingat least three places ofX ;

(c) everyK-automorphism ofAutK(Σ) fixes a placeP ∈ PΣ;

(d) if p ∤ ordα, thenα has exactly two fixed places.

Proof. The projective lineℓ is a non-singular model ofΣ. By Lemma 12.2, thelinear collineation groupL of ℓ is a subgroup of AutK(Σ). Note thatL is theprojective linear group PGL(2,K) consisting of all rational mappings

X 7→ aX + b

cX + d, ad− bc 6= 0, a, b, c, d ∈ K.

Also,L acts on the points or places ofℓ as PGL(2,K) acts onPG(1,K), that is, inits unique sharply3-transitive permutation representation. Assume thatℓ = v(Y )and letξ be a generic point ofℓ. Then the following rational transformationsαbelongs to AutK(Σ):

α : ξ 7→ aξ + b

cξ + d, ad− bc 6= 0, a, b, c, d ∈ K. (12.4)

Now, letβ be anyK-automorphism in AutK(Σ). LetP∞ denote the unique placecentred at the infinite point ofℓ. Since PGL(2,K) is transitive, there existsδ ∈ Lsuch that theK-automorphismβδ of Σ fixesP∞. SinceΣ = K(ξ) and sincediv(ξ)∞ = P∞, so1 is the smallest non-gap atP∞. Thus Lemma 12.13 applies,showing thatβδ ∈ L. But this shows thatβ itself belongs toL.

To show (ii)(c), letα be given as (12.4). If eitherc 6= 0, or c = 0 buta 6= d, thenthe equationaX + b = X(cX + d) has a solutionu in K, and the (unique) placePu centred atu is fixed byα. Otherwise,α(ξ) = ξ + b and henceα fixesP∞.

To show (d), a little more is needed. Since PGL(2,K) is transitive onℓ, soassume thatP∞ is fixed byα. Thenα(ξ) = aξ+ b with a, b ∈ K anda 6= 0. Sincep ∤ ordα, soa 6= 1. Therefore,α fixes not onlyP∞ but also the place centred atthe pointP = (u, 0) with u = −b/(a− 1). 2

REMARK 12.15 All finite subgroups of the group PGL(2,K) are known; see The-orem 12.82.

LetP be any place ofΣ. Choose a primitive place representationσ ofP. Assumethatα ∈ AutK(Σ) fixesP. Thenσα is a primitive place representationσ′ of P,andλσ = σ′ for aK-automorphismλ of K[[t]]. Here,λ depends onα, and this is


stressed by writingλα and saying thatλα is thecompanionK-automorphism ofαin Aut(K[[t]]). For anyK-automorphism group ofΣ which fixesP, this leads tothe introduction of the following map:

ϕ : G→ Aut(K[[t]]), α 7→ λα. (12.5)

L EMMA 12.16 The mappingϕ is an epimorphism.

Proof. Define the imageΛG = λα | α ∈ G of ϕ to be thecompanion subgroupofG. Letα, β ∈ G. Then

λαβσ = σ(αβ) = (σα)β = λα(σβ) = (λαλβ)σ,

whenceλαβ = λαλβ . Letα ∈ kerϕ. Thenσ = σα. Asσ is an isomorphism,α isthe the identity in AutK(Σ). This completes the proof. 2

L EMMA 12.17 Letα be a finiteK-automorphism ofΣ that fixes a placeP. Then,α has orderprs with r ≥ 0 andp ∤ s if and only ifλα = ut + · · · , wheres is thesmallest integer satisfyingus = 1.

Proof. LetG = 〈α〉. The elementsβ ∈ G with λβ = t + · · · form a subgroupHof G. To show thatH is ap-group, take a prime divisorr of |H|, and an elementβ ∈ H of orderr. If λβ = t + vti + . . ., thenλβr = t + rvti + . . ., which givesr = p.

On the other hand, every elementβ ∈ G whose order is a power ofp is in H.This follows from the fact that, ifλβ = ut + . . ., thenλβn = unt + . . ., whileupm

= 1 implies thatu = 1. SinceG is cyclic,G = H ×N with N a cyclic groupwhose ordern is prime top. If N = 〈δ〉, thenλδ = ut+ . . . andn is the smallestpositive integer such thatun = 1. From this, the assertion follows. 2

It may be noted thatΛG depends on the choice ofσ, even though only formally.In fact, if σ is replaced by another primitive place representationσ of P, then theresulting groupsΛG and ΛG are conjugate under in Aut(K[[t]]). To show this,write τσ = σ with τ ∈ Aut(K[[t]]). Thenτσα = λατσ yields τλασ = λατσ.Henceτλατ

−1 = λα for everyα ∈ G, whence the result.

12.2 LINEAR SERIES AND AUTOMORPHISMS

The following theorem shows howK-automorphisms ofΣ act on linear series.

THEOREM 12.18 Letα ∈ AutK(Σ) and let

L = div(∑cixi) +B | c = (c0, . . . , cr) ∈ PG(r,K), B ∈ Div(Σ), (12.6)

be a linear series ofΣ. The divisorsDα withD ranging overL are the divisors ofa linear series ofΣ. This linear series is

Lα = div(∑ciα(xi)) +Bα | c = (c0, . . . , cr) ∈ PG(r,K).

450

Note thatdegL = degLα anddimL = dimLα. Also, if bothLα = L andBα = B, thenα can be viewed as an element in EndK(E), whereE is the(r+1)-dimensional vector space overK with basisx0, . . . , xr.

By Theorem 6.70 (i), the canonical series ofΣ is the only effective linear seriesof Σ of order2g − 2 and dimensiong − 1. This gives the following result.

L EMMA 12.19 EveryK-automorphism ofΣ preserves the canonical series ofΣ.

Similarly, the following result holds.

L EMMA 12.20 If a K-automorphism ofΣ fixes a placeP, thenα preserves thecomplete linear series|kP| for every non-negative integerk.

If Σ is the function field of a hyperelliptic curve, then everyK-automorphismof Σ preservesg1

2 , the unique complete linear series of dimension1 and order2 ofΣ. By Theorems 7.88 and 7.94, the set consisting of all placesP of Σ such that2P ∈ g1

2 is finite. This has the following consequence.

L EMMA 12.21 LetF be a hyperelliptic curve. Then theK-automorphism groupofF preserves a finite set of places.

From a geometric point of view, it is important to have a linear representation ofα in the projective space in which a suitable irreducible model of Σ is embedded.This can be done by using the following theorem.

L EMMA 12.22 Let (12.6) be a simple, base-point-free linear series ofΣ, and letY be the irreducible curve ofPG(r,K) arising fromL. For any placeP ∈ PΣ, letC(P) ∈ Y denote the centre ofP. If α ∈ AutK(Σ) preservesL, then there is alinear collineationT of PG(r,K) such thatC(Pα) = T(C(P)). In particular, TpreservesY and acts on the points ofY asα onPΣ.

Proof. The seriesLmay be taken to be normalised. ThenB ∈ L by Theorem 6.22.Hence, there is a non-zero elementz =

∑cixi such thatBα = div(z) +B. From

this, for i = 0, . . . , r,

α(xi)z =∑aijxj ,

with aij ∈ K, anddet(aij) 6= 0. Let P be a place ofΣ. If σ is a primitiverepresentation ofP, it follows that

(σα)(xi)σ(z) =∑aijσ(xj). (12.7)

Let P = (u0, . . . , ur) andP ′ = (u′0, . . . , u′r) denote the centres ofP andPα.

It may be supposed that ordσ(xi) ≥ 0 for all indices i = 0, . . . , r and thatord σ(xi) = 0 for some indexi. Now, from (12.7),u′i =

∑aijuj . This shows

thatP ′ = T(P ), whereT is the linear collineation ofPG(r,K) associated to thematrixT = (tij). 2

A corollary of Lemmas 12.19 and 12.22 is the following.

THEOREM 12.23 LetF be an irreducible algebraic curve which is not rational,elliptic or hyperelliptic. Then theK-automorphism group ofF acts on the pointsof the canonical curve ofK(F) as a linear collineation ofPG(g − 1,K).


Theorem 12.23 makes possible the use of geometric methods ininvestigatingautomorphisms. Important results in this direction are thefollowing theorems thatwill be extended to all characteristicsp in Section 12.7.

THEOREM 12.24 LetF be an irreducible algebraic curve which is not rational,elliptic or hyperelliptic. If eitherp = 0 or p ≥ g, then everyK-automorphism ofF is of finite order.

Proof. By Theorem 7.61, the set of all Weierstrass points ofΣ = K(F) is finite.By Lemma 12.6 it suffices to prove the finiteness of anyK-automorphismα fixingevery Weierstrass point. By Corollary 7.58, the canonical curveΓ of Σ is classical.Let P ∈ Γ be a Weierstrass point. The hyperplanesH throughP with intersectionmultiplicity I(P,H ∩ Γ) at leastjg−2 constitute a pencil through a common pro-jective subspaceΠ of dimensiong− 3. Clearly,Γ is preserved byα. It may be thatΠ contains some more points fromΓ, and letk+1, with k ≥ 0, denote the numberof distinct points ofΓ lying in Π. Then every hyperplane in the pencil has at most2g − 2 − (g − 2) − k = g − k points outsideΠ. On the other hand, the lowerbound in (7.16) shows the existence of at least2g + 2 Weierstrass points. There-fore, at least three, but finitely many, distinct hyperplanes in the pencil contain aWeierstrass point not lying inΠ. Therefore, a powerβ of α preserves at least threehyperplanes in the pencil.

Since in the dual space ofPG(g − 1,K), the pencil corresponds to a line, itfollows from Theorem 12.14 (ii)(b) thatβ preserves every hyperplane throughΠ.If H is such a hyperplane, then a sufficiently large power ofβ fixes every commonpoint of H andΓ. Repeating this argument for another hyperplanes throughΠ,it turns out that, ifα is infinite, then a power ofα has more than2g + 2 fixedpoints inΓ. Therefore, to prove the assertion, it suffices to show that no non-trivialK-automorphism can have more than2g + 2 fixed points inΓ.

To do this, letδ be a non-trivialK-automorphism, and take a non-WeierstrasspointQ ∈ Γ such thatQδ 6= Q. Note that such a point exists by Theorem 12.11.Theng + 1 is a non-gap, and hence there is a functionx ∈ Σ such that

div(x)∞ = (g + 1)Q,whereQ is the place arising fromQ. Puty = x− δ(x). Then,

div(y)∞ = (q + 1)Q+ (g + 1)Qδ.

Hence,y has2g + 2 zeros. On the other hand, every fixed place ofδ is a zero ofy.This proves the assertion and completes the proof of the theorem. 2

Theorem 12.24 extends to the hyperelliptic case provided that eitherp = 0 or pis large enough with respect tog.

L EMMA 12.25 If p = 0 andg ≥ 2, then everyK-automorphism of an irreduciblealgebraic curveF of genusg is finite.

Proof. LetW be a canonical divisor ofΣ = K(F). Then the tri-canonical seriesL = |3W | of Σ is a non-special linear series of degreen = 3(2g − 2) > 2g anddimensionr = 5g − 6. By Theorem 7.39, the associated irreducible curveΓ in

452

PG(r,K) is non-singular. Asp = 0, Γ is a classical curve. By Corollary 7.58,vP(R) ≤ 1

2g(g + 1) for everyR ∈ PΣ, and the number ofL-Weierstrass points,each counted according to its weight, is

∑vP(R) = 25(g − 1)2g. ThusΓ has at

least25(g − 1)2g12g(g + 1)

L-Weierstrass points. Forg ≥ 2, this number is larger than2g + 2. Since everyK-automorphismα of Σ preserves the set∆ of all L-Weierstrass points, Lemma12.12 shows thatα cannot fix∆ element-wise. Henceα acts faithfully on∆, andα can be viewed as a permutation in the symmetric group on∆ which is a finite.Therefore ordα is finite. 2

In the elliptic case, Theorem 12.24 holds forK-automorphisms fixing a place. Thisis proven here forp 6= 2, 3. For the general case, see section 12.4.

L EMMA 12.26 Letp 6= 2, 3. If F is an elliptic curve, then everyK-automorphismF fixing a place is finite, its order being either4 or a divisor of6.

Proof. Sinceg = 1, the smallest non-gap atP is equal to1. Chooseξ ∈ K(F)with div(ξ)∞ = 2P. By Lemma 12.13,α(ξ) = aξ + b with a, b ∈ K anda 6= 0.Sincep 6= 2, 3, there isη ∈ Σ such that

η2 = 4ξ3 − g2ξ − g3. (12.8)

Thenη has a pole of order3 in P. Hence, there isc ∈ K such thatα(η) − cη hasa pole of order at most2. Therefore,α(η) − cη = dξ + e with d, e ∈ K. Hence,α(η) is aK-linear combination ofξ, η and1. Applying α to (12.8), this yields(cη + dξ + e)2 = 4(aξ + b)3 − g2(ξ + b)− g3 which can also be written as

c2ξ2 + d2η2 + e2 + 2ceξ + 2deη + 2cdξη

= 4(a3ξ3 + 3a2bξ2 + 3ab2ξ + b3)− g2(aξ + b)− g3.This differs from (12.8) only by a constant factor. Comparing coefficients ofξη, η,andξ2 shows thatc = e = b = 0. Henceα(ξ) = aξ, α(η) = dη andd2 = a3. Ifg3 = 0, g2 6= 0 then eithera = 1, d = ±1, or a = −1, d = ±j with j2 = −1. Ifg2 = 0, g3 6= 0, thena = ±1, d3 = 1. Otherwise,a = 1, d = ±1. 2

12.3 AUTOMORPHISM GROUPS OF PLANE CURVES

TheK-automorphism groups of rational, elliptic, hyperelliptic, and many otherirreducible algebraic curves are known; see Section 12.10,and Chapter 13. On theother hand, every finite group is isomorphic to theK-automorphism group of someirreducible algebraic curve; see [194] and [274].

For any irreducible algebraic curveF of genusg which is not rational, elliptic, orhyperelliptic, Theorem 12.19 shows that itsK-automorphism group is isomorphicto a subgroup of PGL(g,K) which preserves the canonical curveΓ of K(F) em-bedded inPG(g−1,K). This provides information aboutK-automorphisms, espe-cially when the dimensiong−1 is low. In fact,Σ = K(F) may have other, possible


singular, models inPG(r,K) such that AutK(Σ) is a subgroup ofPG(r + 1,K).If this is the case for an irreducible plane curveC, a plane model ofΣ, the struc-ture and the action of AutK(Σ) can be investigated by using the classification of allsubgroups of PGL(3,K). However, finding out whether a given function fieldΣhas such a plane model may be very involved and may depend on special featuresof Σ.

An important, general result in this direction is that this problem can be solvedfor those function fields that have a non-singular plane model. The elementaryproof depends on two lemmas of independent interest.

L EMMA 12.27 Let P1, . . . , Pn be a set ofn > 3 distinct points in the plane. Ifthere are at least12 (n− 2)(n− 3) linearly independent curvesCn−3 of ordern− 3passing all the pointsP1, . . . , Pn, then these points are collinear.

Proof. The assertion is true forn = 4; so proceed by induction onn.Let n > 4. On a lineℓ through one of these points, sayP1, disjoint from the

remaining pointsP2, . . . , Pn, choosen−3 distinct pointsQ1, . . . , Qn−3 which arealso distinct fromP1. The number of linearly independent curvesCn−3 of ordern− 3 passing all the pointsP1, . . . , Pn, Q1, . . . , Qn−3 equals

12 (n− 2)(n− 3)− n− 3 = 1

2 (n− 3)(n− 4).

Such curves are reducible as each of them containsn − 2 points fromℓ, namelythe pointsP1, Q1, . . . , Qn−3. It follows that they split intoℓ and 1

2 (n− 3)(n− 4)linearly independent curvesCn−4 of ordern − 4 passing through all the pointsP2, . . . , Pn. By induction,P2, . . . , Pn are collinear. Therefore, anyn − 1 of thepointsP1, . . . , Pn are collinear. This can only happens when all these points lie ona line. 2

L EMMA 12.28 Let Cn be a non-singular plane curve of ordern > 3. The linearseries cut out onCn by lines is the unique linear seriesg2

n ofCn.

Proof. Choose a divisorD = P1 + . . . + Pn in g2n such that the placesPi are

centred at distinct points, sayP1, . . . , Pn. SinceCn is non-singular, its genus isg = 1

2 (n − 1)(n − 2). From Theorem 6.59 applied tog2n, at leasti = g − n − 2

linearly independent curvesCn−2 of ordern − 2 pass throughP1, . . . , Pn. ByLemma 12.27, these points are collinear, and the assertion follows. 2

So the following theorem is a corollary of Lemmas 12.28 and 12.22.

THEOREM 12.29 TheK-automorphism group of any non-singular plane curveXof ordern > 3 is linear, that is, isomorphic to a subgroup ofPGL(3,K) preservingX .

It should be noted that Theorem 12.29 holds true forn = 2 but it fails for bothn = 1, 3; see Theorem 12.14 and the proof of Theorem 12.85. An application ofTheorem 12.29 is the following result.

PROPOSITION 12.30 LetK be the algebraic closure of the finite fieldFq2 of orderq2 with q > 2. TheK-automorphism group of the Hermitian curve

Hq = v(Xq +X + Y q+1)

454

is the projective unitary groupPGU(3, q).

Proof. SinceHq is a non-singular plane curve, Theorem 12.29 implies that itsK-automorphism groupG consists of all linear collineations preservingHq. Astraightforward computation shows that everyα ∈ G is defined overFq2 ; that is,a matrix associated toα has all entries fromFq2 . Therefore,G preserves the setHq(Fq2) of all Fq2-rational points ofHq. By Example 15.9,Hq(Fq2) coincideswith the setU of all self-conjugate points of a unitary polarity ofPG(2, q2), andPGU(3, q) is a subgroup ofG. It remains to show that everyα ∈ G belongs toPGU(3, q). As PGU(3, q) acts2-transitively onU , with the notation used in Exam-ple 15.9 it may be assumed thatα fixes both the pointsX∞ andO. After a changeof coordinates,U consists ofY∞ plus all pointsU = (1, u, cuq+1 +m) asu rangesoverFq2 andm ranges overM . By another simple computation, the hypothesisthatα preservesU implies thatα is associated to a matrix,

aq+1 0 0

0 a 00 0 1

,

for non-zero elementa in Fq2 . Therefore,α ∈ PGU(3, q). 2

12.4 A BOUND ON THE ORDER OF A K-AUTOMORPHISM

A major result on automorphisms of algebraic curves is that,apart from rationalcurves, everyK-automorphism fixing a place is finite. In characteristic zero, thisfollows from Lemmas 12.25 and 12.26. Forp > 0, the proof is much more involvedand requires some lemmas of independent interest on the invariant subfields ofK-automorphisms.

In this section it is assumed thatp > 0 although almost all results, but not allproofs, remain valid in zero characteristic.

L EMMA 12.31 If α ∈ AutK(Σ) fixes every element of a subfieldF of Σ of tran-scendence degree1, thenα is finite, and its order does not exceed[Σ : F ].

Proof. Since the field extensionΣ/F is algebraic, it is also simple. There existsz ∈ Σ such thatΣ = F (z). Let P (X) = Xn + . . . + an−1X + an be theminimal polynomial ofz overF . Sinceα fixes every coefficientai of P (X), itfollows thatα(z) is also a root ofP (X). Therefore, the setS = αi(z) | i ≥ 0has cardinality at mostρ. On the other hand,αi(z) = αj(z) holds if and only ifαi−j(z) = z. SinceΣ = F (z), the last condition can only occur whenαi−j is theidentity automorphism ofΣ. Hence, ordα ≤ |S| ≤ ρ. 2

L EMMA 12.32 Assume thatΣ is non-rational, and thatα ∈ AutK(Σ) preserves arational subfieldK(ξ) of Σ. If p ∤ [Σ : K(ξ)], thenα is finite and an upper boundonordα is 2p(g + 1)[Σ : K(ξ)].

Proof. By (iii) of Theorem 12.14 and by the first assertion in Lemma 12.13, it maybe assumed that, after aK-linear change ofξ, eitherα(ξ) = ξ + v or α(ξ) = cξ


for somev, c ∈ K, andc 6= 0. In the former case, asp 6= 0, soαp(ξ) = ξ, andhence ordα cannot exceedpρ by Lemma 12.31.

Henceforth, letα(ξ) = cξ, wherec ∈ K\0. By Theorem 15.5, there existsη ∈ Σ such thatΣ = K(ξ, η). LetF = v(f(X,Y )) be the irreducible plane curvewith generic pointP = (ξ, η). The projection ofF onto theX-axis defined by therational transformation,

τ : x′ = ξ, y′ = 0,

is not birational, asΣ is not rational. LetD = D(Σ/K(ξ)) be the associateddifferent divisor.

Note that, ifξ is a separating variable ofΣ, then Hurwitz’s Theorem 7.26 showsthat

degD = 2g − 2 + 2ρ, (12.9)

whereρ = [Σ : K(ξ)]. By Lemma 12.7,Dα = D sinceα preservesK(ξ). Also,α viewed as aK-automorphism ofK(ξ), preserves the set of placesQ ∈ PK(ξ)

which lie under the ramified places ofPK(X ).First, consider the case where div(ξ) = ρP1 − ρP2. As in the proof of Theorem

5.33, after replacingη by ac0(ξ)η with c0(ξ) ∈ K(ξ), it may be assumed that

f(X,Y ) = Y ρ + g(X,Y ),

whereg(X,Y ) ∈ K[X,Y ] has degree inY less thanρ. Since

fY (X,Y ) = ρY ρ−1 + gY (X,Y )

does not vanish, the hypothesisp ∤ ρ ensures thatξ is indeed a separating variableof Σ. Letσ1 andσ2 be primitive place representations ofP1 andP2. Then

σ1(ξ) = ctρ + . . . , c ∈ K\0,and the contribution ofP1 in D is ρ− 1 by the hypothesesp ∤ ρ. Similarly,

σ2(ξ) = ct−ρ + . . . , c ∈ K\0,and the contribution ofP2 in D is againρ − 1. ThusdegD = 2(ρ − 1) + m,with m ≥ 0. Comparison with (12.9) givesm = 2g. This yields that, apart fromP1, P2, at least one but at most2g places contribute toD. Therefore, a powerβ = αℓ, with ℓ ≤ 2g, of α fixes a placeP3 different from bothP1,P2. If Q1, Q2

andQ3 are the places ofK(ξ) underP1,P2 andP3, thenβ fixesQ1,Q2 andQ3.By Theorem 12.14, it follows thatβ fixes every placeQ ∈ PK(ξ). By Lemma12.9, the subfieldΣβ = z | β(z) = z; z ∈ Σ containsK(ξ). Hence ordβ ≤ ρ,yielding that ordα ≤ 2gρ.

Therefore, it may be assumed that either div(ξ)0 or div(ξ)∞, say the latter, con-tains two different placesP1 andP2. Put ordP1

(ξ) = −k1 and ordP2(ξ) = −k2,

wherek1, k2 are positive integers. Letn1 be the smallest non-gap ofΣ atP1. Thenthere existsζ ∈ Σ such that div(ζ)∞ = n1P1. For a powerβ = αℓ with ℓ ≤ ρ, itfollows thatPβ

1 = P1. After changingζ to uζ+ v, with suitableu, v ∈ K, Lemma12.13 allows thatβ(ζ) = dζ, d ∈ K \ 0.

456

Let f1(X,Y ) =∑aijX

iY j be an irreducible polynomial with coefficients inK such thatf1(ξ, ζ) = 0. Sinceβ(ξ) = cℓξ, β(ζ) = dζ,

f1(clξ, dζ) = ef1(ξ, ζ), e ∈ K\0.

Now, choose two distinct pairs(i, j) and(m,n) with n ≥ j such that bothaij andamn are distinct from zero. Thencℓidj = e andcℓmdn = e. Thusξ′ = ξi−mζj−n

is fixed byβ.It must be shown thatξ′ 6∈ K. Assume by contradiction thatζn−j = kξi−m with

k ∈ K. Then div(ξi−m)∞ = div(ζn−j)∞. On the other hand, sincen− j ≥ 0,

div(ξi−m) = (i−m)k1P1 + (i−m)k2P2 + . . . , (12.10)

div(ζn−j)∞ = (n− j)n1P1.

Hence,(i − m)k1 = −(n − j)n1 ≥ 0. Since(i, j) 6= (m,n), it follows thati−m < 0. Therefore,

div(ξi−m)∞ = (i−m)k1P1 + (i−m)k2P2 + . . . .

But this contradicts that div(ξi−m)∞ = div(ζn−j)∞.Finally, it is shown that[Σ : K(ξ′)] ≤ 2ρ(g + 1). Since the degree off1(X,Y )

in X is at most[Σ : K(ζ)] = g + 1, while in Y it is at most[Σ : K(ξ)] = ρ, so|i−m| ≤ g + 1 and|j − n| ≤ ρ. Then

[Σ : K(ξi−m)] = [Σ : K(ξ)] · [K(ξ) : K(ξi−m)] ≤ ρ(g + 1),

[Σ : K(ζj−m)] = [Σ : K(ζ)] · [K(ζ) : K(ζi−m)] ≤ (g + 1)ρ,

whence the assertion follows. From Lemma 12.31, ordβ ≤ 2ρ(g + 1). Thereforeordα ≤ 2pρ2(g + 1), and Lemma 12.32 is completely proved. 2

Now, the finiteness of everyα ∈ AutK(Σ) fixing a placeP of Σ is proven underthe hypothesis thatg ≥ 1. By Theorem 6.86, there are exactlyg gaps ofΣ atP,and each of them is at most2g. In particular,2g + 1 is a non-gap ofΣ atP. ByCorollary 6.73,|(2g+1)P| is a simple, based-point-free complete linear series. Itsdimensionr is at leastg + 1 ≥ 2, by the Riemann–Roch Theorem 6.59.

Let m0 = 0 < m1 < . . . < mr denote the firstr + 1 non-gaps ofΣ atP. Inparticular,mr−1 = 2g, mr = 2g + 1. Choosex0 = 1, x1, . . . , xr ∈ Σ such thatdiv(xi)∞ = mi, for i = 0, 1, . . . , r. Sincex0, x1, . . . , xr are linearly independentoverK, the linear series

L = div(∑cixi) + (2g + 1)P | c = (c0, . . . , cr) ∈ PG(r,K)

coincides with|(2g + 1)P|. Thenα preserves|(2g + 1)P|. By Lemma 12.3,αpreserves each subseries

Lj = div(∑r

i=jcixi) + (2g + 1)P | c = (c0, . . . , cr) ∈ PG(r,K).Now, considerα as a linear transformation of the(r+1)-dimensional vector spaceE overK with basisx0, . . . , xr. Note thatα ∈ End(Ej) whereEj is the(j+1)-dimensional subspace ofE with basisx0, . . . , xj. Write

α(xj) =∑j

i=0ajixi


for everyj = 0, . . . , r. In terms ofE, the matrixA = (aji) associated toα islower-triangular:

A =

a0 0 . . . 0

∗ a1

......

.. . 0∗ . . . ∗ an

.

THEOREM 12.33 EveryK-automorphism of a non-rational irreducible algebraiccurve that fixes a place is finite and its order is bounded aboveby2p(2g+1)2(g+1).

Proof. Let ξi andξj be eigenvectors ofα associated to the eigenvaluesai = aii

andaj = ajj . It is shown that, ifai 6= aj , then ordPξi 6= ordPξj . To do this,P isassumed not to be a pole ofξiξ

−1j . From Example 9.11,

resP ξiξ−1j = resPα(ξiξ

−1j ) = resP aia

−1j ξiξ

−1j = aia

−1j resP ξiξ

−1j .

As aia−1j 6= 1, this shows that resP ξiξ

−1j = 0. Therefore,P is a zero ofξiξ

−1j ,

and the assertion is proved.Now suppose that all eigenvalues are different from one other. Then, for every

non-gapmi of Σ atP, there is an eigenvectorξi of α such that ordPξi = mi. ThesubfieldsK(ξi) are left invariant byα. Here

[Σ : K(ξr−1)] = 2g, and [Σ : K(ξr)] = 2g + 1.

Sincep cannot divide two consecutive integers, Lemma 12.32 applies. This imme-diately implies that ordα ≤ 2p(2g + 1)2(g + 1).

Next, assume that two eigenvalues, sayai andaj , coincide. Then the Jordanform of α is either diagonal or contains a non-trivial block. So thereare elementsξ1, ξ2 ∈ Σ, such thatα(ξ1) = aiξ1 andα(ξ2) = aj(λξ1 + ξ2) with λ ∈ K. Putz = ξ2ξ

−11 . Thenα(z) = z + ǫ, and the subfieldK(z) is left invariant byα. Also,

ρ = [Σ : K(z)] is at most2(2g + 1) since both ordPξ1 and ordPξ2 are less than orequal to2g + 1. Sincep 6= 0, this yields thatαp(z) = z. Thus ordα ≤ 2p(g + 1).This completes the proof. 2

12.5 AUTOMORPHISM GROUPS AND THEIR FIXED FIELDS

For any subgroupG of AutK(Σ), the set,

ΣG = z ∈ Σ | σ(z) = z for all σ ∈ G,is a subfield ofΣ, thefixed subfield ofG. The most important result on such sub-fields states that the field extensionΣ/ΣG is finite if and only ifG is finite; also, ifthis is the case, then[Σ : ΣG] = |G|. Actually, as shown later, this occurs for anyg > 1, AutK(Σ) being infinite only for rational and elliptic function fields. For afinite subgroupG of AutK(Σ), this yields thatΣ/ΣG is a finite Galois extensionwith Galois groupGal(Σ/ΣG) isomorphic toG.

Geometrically speaking, ifΓ is a model ofΣ, the irreducible algebraic curvewhose function field isΣG is thequotient curve ofΓ with respect toG and denoted

458

by Γ/G. So, in the study of quotient curves using function field theory, the maintechnical tool is Galois theory. Here, the main goal is to prove that AutK(Σ) is finitefor g > 1; this is achieved after some preliminary results established in section 12.7.

In this section, some properties ofG that depend onΣG are discussed. First,a straightforward generalisation of Lemma 12.34 to any subgroupG of theK-automorphism group AutK(Σ) is stated.

L EMMA 12.34 LetG be a subgroup ofAutK(Σ) which fixes every element of asubfieldF of Σ of transcendence degree1. ThenG is finite and

|G| ≤ [Σ : F ].

For a finite subgroup of AutK(Σ), the fundamental result onΣG is the following.

THEOREM 12.35 LetG be a finite subgroup ofAutK(Σ). ThenΣ/ΣG is a finiteGalois extension, andG ∼= Gal(Σ/ΣG).

Proof. Let ξ ∈ Σ, and letα1, . . . , αr be a maximal set of elements ofG suchthatα1(ξ), . . . , αr(ξ) are pairwise distinct. Ifβ ∈ G, thenβα1(ξ), . . . , βαr(ξ)is just a permutation ofα1(ξ), . . . , αr(ξ); otherwise, the latter set is not maximal.Let f(X) =

∏ri=1(X − αi(ξ)). Since the coefficients of a monic polynomial are

symmetric polynomials of its roots, every coefficient off(X) is inΣG. Also,f(X)is a separable. Therefore, every elementξ ∈ Σ is a root of a separable polynomialof degree at mostn with coefficients inΣG. Also, f(X) splits into linear factorsin Σ. Thus, the field extensionΣ/ΣG is normal and separable, and hence a Galoisextension. By Lemma 12.34 it remains to show that[Σ : ΣG] ≤ n. To do this, thesimpler notationF to indicateΣG is used and an elementζ ∈ Σ is chosen such that[F (ζ) : F ] is as large as possible, saym.

Then it must be shown thatF (ζ) = Σ. If this is not true, then there existsan elementη ∈ Σ such thatη 6∈ F (ζ). By Theorem 15.5, there is an elementω ∈ F (ζ, η) such thatF (ω) = F (ζ, η). But, from the towerF ⊂ F (ζ) ⊂ F (ζ, η),it follows that [F (ζ, η) : F ] > m, whence[F (ω) : F ] > m; this contradicts thechoice ofm. 2

L EMMA 12.36 LetG be a finite subgroup ofAutK(Σ). Then two placesP1 andP2 of Σ lie over the same placeP ′ of ΣG if and only if there is aK-automorphismα ∈ G such thatP2 = Pα

1 .

Proof. Let τ1, τ2 ∈ Σ→ K((t)) be primitive representations of the placesP1,P2.Suppose first thatP2 = Pα

1 . Then τ1α = λτ2 with λ a K-automorphism ofK [[t]] . Sinceα fixes ΣG element-wise, it follows thatτ1(u′) = (λτ2)(u

′) foreveryu′ ∈ ΣG. This shows that the place ofΣG represented byτ1 coincides withthat represented byτ2. If P ′ is such a place, both placesP1 andP2 are overP ′.

Conversely, suppose thatP1 andP2 lie over the same placeP ′ of ΣG. There isaK-automorphismλ of K[[t]] such thatτ1(u′) = (λτ2)(u

′) for everyu′ ∈ ΣG.Hence,(τ−1

1 λτ2)(u′) = u′ for eachu′ ∈ ΣG. This shows thatβ = τ−1

1 λτ2 is aK-automorphism ofΣ which fixes the subfieldΣG element-wise. Henceβ ∈ G,andα = β−1 takesP1 toP2. 2

As a by-product, there is also the following result.


L EMMA 12.37 LetG be a finite subgroup ofAutK(Σ). If P1 andP2 be two placesofΣ lying over the same placeP ′ ofΣG, thenP1 andP2 have the same ramificationindexρ.

The next step is to investigate subgroupsG of AutK(Σ) by means of propertiesdepending on the action ofG onPΣ. Theorbit of a placeP underG is the set

PG = Pα | α ∈ G ,while thestabiliser ofP in G is the subgroup

GP = α ∈ G | Pα = P .An orbit PG is eitherlong or shortaccording asGP is trivial or not. For a finitegroupG, the orbitPG is short if and only if|PG| < |G|.

With this terminology, Lemma 12.36 can be reworded as follows.

L EMMA 12.38 LetG be a finite subgroup ofAutK(Σ). Then two places ofΣ lieover the same place ofΣG if and only if they are in the same orbit under the actionofG. In other words, there is a one-to-one correspondence between places ofΣG

andG-orbits of places ofΣ.

It may be noted that the number of the short orbits under a finite subgroupG ofAutK(Σ) is finite. If this were not true, then the stabiliserGP would contain somenon-trivialK-automorphisms. SinceG is finite, this could only occur when at leastone non-trivialK-automorphism inG fixes an infinite number of places ofΣ, acontradiction to Lemma 12.9.

A useful result on short orbits is given in the following theorem.

THEOREM 12.39 LetP be a place ofΣ lying over a placeP ′ of ΣG. If n = |G|andm = |GP |, then the number of distinct places lying overP ′ is n/m, and theramification index of each of them iseP = m.

Proof. If P = P1, . . . ,Ps are the places ofΣ overP ′, thens = n/m. In fact,

|G| = |GP | · |PG| = |GP | ·[Σ : ΣG

],

whences =[Σ : ΣG

]= |G|/|GP | = n/m. Finally, from Lemma 12.37 it follows

thatρ1 = . . . = ρs = m. 2

To end this section, a result applied in later sections is given.

L EMMA 12.40 Letα be a non-trivialK-automorphism ifΣ. If α fixes an elementξ ∈ Σ and a placeP of Σ, thenξ is not a uniformizing element ofΣ atP.Proof. By (7.1), ordPξ = eP ordP′ξ, whereP ′ is the place ofΣα underP, andePis the ramification index ofP with respect to the extensionΣ/Σα. Sinceα fixesP,soeP = ordα, whence ordPξ 6= 1. 2

12.6 THE STABILISER OF A PLACE

L EMMA 12.41 Letz ∈ Σ be a uniformizing element ofΣ atP.

460

(i) For anyα ∈ G, there is exactly onecα ∈ G such thatordP(α(z)−cαz) > 1.

(ii) (a) cα is independent of the choice of the uniformizing element;

(b) the mapping,

ϕ : G→ K,α 7→ cα,

(12.11)

is a group homomorphism with kernelN ;

(c) the factor groupG/N is a finite cyclic group whose order is prime top;

(d) if p = 0, thenN is trivial;

(e) if p > 0, thenN is a nilpotent group consisting of all elements inGPwhose order is a power ofp, andG = N ⋊H withH a subgroup ofGwhose order is prime top.

Proof. Choose a primitive place representationσ : Σ 7→ K((t)) of P such thatσ(z) = t. By (12.3),σ(α(z)) = kt + . . . . Thus, ordP(σ(α(z)) − cσ(z)) > 1 ifand only ifc = k. It is straightforward to check thatc is independent of the choiceof σ. For another uniformizing elementz′ atP, there existsu ∈ K\0 such thatσ(z′) = ut+ . . .. Hence,

ordP(α(z′)− uα(z)) = ordP(z′ − uz) > 1. (12.12)

Then

ordP(α(z′)− kz′) = ordP((α(z′)− uα(z)) + u(α(z)− kz) + k(uz − z′)),which is greater than1 by (12.12). Also, ordP(α(z)− kz) > 1. So, (ii)(a) holds.

Sincek depends on bothP andα, but is independent of the choice of the placerepresentationσ of P, soϕ is well defined. Letα, β ∈ G. Then

ordP((βα)(z)− cαcβ(z)) = ordP((βα)(z)− cαβ(z)) + cα(β(z)− cβz))≥minordP((βα)(z)− cαβ(z));

ordP(cα(β(z)− cβz)) = ordP(β(z)− cβz).Since both weights are greater than1, so

ordP((βα)(z)− cαcβ(z)) > 1.

By the uniqueness, proven previously,cβα = cβcα which shows thatϕ is indeed ahomomorphism. Hence,G/ kerϕ is isomorphic to a subgroup of the multiplicativegroup ofK. By Theorem 12.33, there is an integerM which is larger than theorder of each element inG. This holds true for any factor group ofG, and inparticular forG/ kerϕ. On the other hand, every finite multiplicative subgroup ofK is the group ofm-th roots of unity inK for a suitable integerm prime top.AsH ranges over the finite subgroups ofG/ kerϕ, only a finite number of distinctimagesϕ(H) are obtained, as|H|, and hence|ϕ(H)|, is bounded byM . So, allsuch subgroupsϕ(H) generate a multiplicative cyclic subgroupΓ of K whose


order is prime top and bounded above by the product of all integers smaller thanM. SinceΓ ∼= G/ kerϕ, so (ii)(c) follows.

To prove the last two assertions, it is first shown that every eigenvalue ofα is apower ofcα. With the notation in the proof of Theorem 12.33, every eigenvalueaj

can be extracted fromα(xj) =∑r

i=j ajixi by taking the first coefficientaj = ajj .

Since ordPxj = −mj , soxjzmj+1 is a uniformizing element ofΣ at P. Write

σ(xj) = bjt−mj + . . . , with bj ∈ K∗. Then

σ(α(xjzmj+1))− cαxjz

mj+1) =σ(α(xj)σ(α(zmj+1))− cασ(xj)σ(zmj+1)

= (ajbjcmj+1α t+ . . .)− (cαbjt+ . . .),

whenceaj = c−mjα ; this proves the assertion.

In particular, ifα ∈ kerϕ, then every eigenvalue ofα is equal to1. Hence,kerϕconsists of all thoseα ∈ G for which the associated upper-triangular matrixA hasthe form:

A =

1 ∗ . . . ∗0 1

......

.. . ∗0 . . . 0 1

.

However, if p = 0, such anα cannot have finite order unlessA is the identitymatrix, and henceα is the identity automorphism. This proves (ii)(d).

On the other hand, ifp is positive,kerϕ is a nilpotent group, and any non-trivialelement inkerϕ has order a power ofp. In fact,kerϕ is isomorphic to a group ofmatrices of the above form. This completes the proof of (ii)(e). 2

L EMMA 12.42 Let p > 0 and g > 0. Assume thatH is a non-trivial abeliansubgroup of the stabiliser ofP satisfying the following properties:

(a) the order of each non-trivial element inH is a power ofp;

(b) the fixed field of any non-trivial finite subgroup ofH is rational.

ThenH is a cyclic group of orderp or p2.

Proof. The result holds trivially whenH has orderp. Letm1 be the smallest non-gap ofΣ atP, and choose an elementz ∈ Σ such that div(z)∞ = m1P. Given asubgroupU of H of orderp generated byα ∈ H, an element inΣU having a poleof multiplicity p at P may be found in the following way. LetΣU = K(ξ). Bya suitable substitutionξ 7→ (aξ + b)/(cξ + d), ad − bc 6= 0, a, b, c, d ∈ K, it isensured thatP is a pole ofξ, and, in fact, the only pole ofξ. If Q ∈ PK(X ) were

another pole ofξ, then eachQαi

with 1 ≤ i ≤ p would be a pole ofξ. Sinceα, andhence its powers, does not fixQ, this would givep more poles ofξ, contradictingCorollary 5.35. ThusP is the only pole ofξ. So, in any case, there isξ ∈ Σ withΣU = K(ξ), div(ξ)∞ = pP, and[Σ : ΣU ] = p.

Now, to prove the assertion, assume by contradiction that there is an elementτof orderp such thatτ ∈ H but τ 6∈ U . SinceH is abelian,τ andα generate an

462

elementary abelian subgroupV of H of orderp2. By Lemma 12.31,ΣV 6= ΣU

showing thatτ does not fix all elements inΣU . On the other hand,ατ = τα yieldsthat τ preservesΣU . TakingPτ = P into account, it follows thatτ(ξ) = ξ + awith a ∈ K\0. Replacingξ by ξ/a gives the following:

α(ξ) = ξ, τ(ξ) = ξ + 1, div(ξ)∞ = pP. (12.13)

Interchanging the roles ofα andτ in the above argument shows the existence ofη ∈ Σ with the following properties:

α(η) = η + 1, τ(η) = η, div(η)∞ = pP. (12.14)

Since[Σ : K(ξ, η)] · [K(ξ, η) : K(ξ)] = [Σ : K(ξ)] = p, andK(ξ, η) 6= K(ξ),soΣ = K(ξ, η). Then the curveF with generic pointP = (ξ, η) has genusg, andbothα andτ act on the plane as translations.

In particular, the centre of the branch ofF associated toP lies on the line atinfinity. To find an equationf(X,Y ) = 0 of F , deduce from the first part of(12.14) thatηp−η ∈ K(ξ). Since bothηp−η andξ have no pole centred at a pointat finite distance, it follows thatηp − η = f(ξ) with a polynomialf(X) ∈ K[X].As div(ηp − η)∞ = p2P while div(ξ)∞ = p, so deg f(X) = p. Also, since(η − i)p − (η − i) = ηp − η, so

f(ξ − i) = f(ξ)

for everyi = 0, 1, . . . , p−1. Given a roota of f(X), it follows that the roots of thepolynomialf1(X) = f(X − a) are0, 1, . . . , p − 1. Hencef1(X) = c(Xp −X).Thus

ηp − η = c(ξp − ξ) + d c, d ∈ K. (12.15)

In other words,F has equationd − cX + Y + cXp − Y p = 0. This shows thatthe pointE = (0, 1, e), with ep = c, is a(p − 1)-fold point ofF . But thenF is arational curve, contradicting the hypothesisg > 0. 2

L EMMA 12.43 Let p > 0 and g > 0. Assume thatH is a non-trivial abeliansubgroup of the stabiliser ofP. If H has property(a) of Lemma 12.42, thenH isfinite and its order does not exceedp2(g − 1).

Proof. From (7.6),2(g − 1) = n(2g′ − 2) + deg(Σ/ΣU ) whereU is a subgroupof H of ordern, andg′ is the genus ofΣU . Further,deg(Σ/ΣU ) > n− 1, becauseP completely ramifies in the extensionΣ/ΣU . Forg′ > 0, it follows that

n ≤ 2(g − 1)/(2g′ − 1) < 2g.

Therefore, either the fixed field of every non-trivial finite subgroup ofH has genus0, or there is a subgroupU ofH of ordern < 2g such thatΣU has genusg′ ≥ 1. Inthe former case, Lemma 12.42 applies and the assertion follows. In the latter case,replaceU by a finite maximal subgroupV ofH containingU . Both properties holdtrue forV ; that is,|V | < 2g andΣV has genusg′ > 0.

Now, it is shown that Lemma 12.42 applies toΣV in so far as the factor groupH = H/V is regarded as aK-automorphism group ofΣV , and the place underP is taken as the fixed place ofH. Since both commutativity and property (a) are


inherited byH, it is only necessary to deal with property (a). Under the naturalhomomorphismH 7→ H, any subgroupW of H corresponds to a subgroupW ofH containingV . By the maximality ofV , the genus ofΣW is 0. ThenΣW viewedas a subfield ofΣV has genus0, showing thatH has property (a) of Lemma 12.42.Consequently, the order ofH is at mostp2 and hence the order ofH itself does notexceedp2(2g − 1). 2

At this point an elementary result from Group Theory is needed.

L EMMA 12.44 LetG be a finite or infinite group of order at leastn, containinga central subgroupZ of orderp, such that the factor groupG = G/Z is an ele-mentary abelianp-group. ThenG contains an abelian subgroup of order at least√pn.

Lemma 12.41 is improved by showing the finiteness ofN .

L EMMA 12.45 Let p > 0 and g > 0. ThenN is finite and its order does notexceedp2(g + 1)(2g − 1)2.

Proof. Choose an elementx1 ∈ Σ such that div(x1)∞ = m1P, wherem1 denotesthe smallest non-gap ofΣ at P. The subfieldK(x1) of Σ is invariant under anyK-automorphism fixingP. For everyα ∈ N , we have thatα(x1) = x1 + aα,with aα ∈ K. Then the mapα 7→ aα is a homomorphism ofN into the additivegroup ofK. Let N1 be its kernel. IfN1 is trivial, then, from Lemma 12.43,|N | ≤ p2(2g − 1).

So, for the rest of the proof,N1 may be assumed non-trivial. Then the factorgroupN/N1 is a, possibly trivial, elementary abelianp-group. Also, since anyα ∈ N1 fixesK(x1) element-wise, the order ofN1 is at mostm1 = [Σ : K(x1)]by Lemma 12.31. AsN is nilpotent,N1 has a normal subgroupN2 of index psuch thatN1/N2 is contained in the centre of the factor groupN/N2. By Theorem12.35,

p · [Σ : ΣN2 ] = p · |N2| = |N1| ≤ [Σ : K(x1)].

As [Σ : K(x1)] = [Σ : ΣN2 ] · [ΣN2 : K(x1)], it follows thatp ≤ [ΣN2 : K(x1)].By Lemma 12.13(ii),ΣN2 has positive genusg′. Now, as the factor groupN/N2

can be regarded as a group ofK-automorphisms ofΣN2 , from Lemma 12.43 itfollows that any abelian subgroup ofN/N2 has order at mostp2(2g′ − 1).

On the other hand, forZ = N1/N2, the quotient groupN/N2 has the structuredescribed in Lemma 12.44. Therefore, ifn′ is chosen such that the order ofN/N2

is not less thann′, thenN/N2 contains an abelian subgroup whose order is at least√pn′. It follows then that

√pn′ ≤ p2(2g′ − 1). Hence,N/N2 has order at most

p3(2g′ − 1)2, and the order ofN does not exceed

p3(2g′ − 1)2m1p−1 = p2m1(2g

′ − 1)2.

Asm1 ≤ g+1 andg′ ≤ g, the order ofN is bounded above byp2(g+1)(2g−1)2.2

To sum up, a complete description of the structure of the stabiliser of a placePof Σ is obtained.

464

THEOREM 12.46 Let Σ be a function field of positive genusg > 0. LetP be anyplace ofΣ. Then the stabiliser ofP is finite and its structure is described as follows.

(i) If p = 0, thenGP is a cyclic group of order at most8(2g + 1)3.

(ii) (a) If p > 0, then ap-Sylow subgroup ofN ofGP is a normal subgroup with

|N | ≤ p2(g + 1)(2g − 1)2, (12.16)

and the quotient groupGP/N is a cyclic group with

|GP/N | ≤ 2p(2g + 1)2(g + 1). (12.17)

(b) GP contains a cyclic subgroupH ∼= GP/N such thatGP = N ⋊ H,the semi-direct product ofN andH. All such subgroupsH are conjugate inGP .

In any case, the order ofGP has a bound depending only onp andg.

Both (12.16) and (12.17) are approximate bounds. As is shownbelow, (12.17)can be significantly improved to4g + 2, and (12.16) can be sharpened at least to|N | ≤ 4pg2/(p− 1)2.

12.7 FINITENESS OF THEK-AUTOMORPHISM GROUP

By Theorems 12.14 and 12.85, theK-automorphism group of any rational or el-liptic curve is infinite. These are actually the only irreducible algebraic curveswhoseK-automorphism group is infinite. This important result has already beenmentioned before. Now, a proof is given.

THEOREM 12.47 LetF an irreducible algebraic curve of genusg ≥ 2. Then theK-automorphism group ofF is finite,

Proof. Let Σ = K(F). By Theorem 12.46, it suffices to show that AutK(Σ)preserves a finite set of places ofΣ. In fact, if an AutK(Σ)-invariant set of placeshas size at mostm, and the stabiliser of a place in the set has ordern, thenmn is anupper bound on AutK(Σ). If Σ is hyperelliptic, Lemma 12.21 provides such a setof places of size at most2g+2. In the non-hyperelliptic case, the canonical series isuseful. By Lemma 12.23, AutK(Σ) is a linear collineation group ofPG(g − 1,K)which preserves the canonical curveΓ of Σ. In particular, AutK(Σ) preserves theset of all Weierstrass points ofΓ. This set is finite as its size does not exceed

(∑g−1i=0 ǫi

)(2g − 2) + g(2g − 2),

by (7.13). 2

REMARK 12.48 LetF be an irreducible, non-hyperelliptic algebraic curve whosegenusg bigger than2. PutΣ = K(F), and consider the canonical curve ofΣ.From the sequence,

ǫ0 = 0 < ǫ1 = 1 < . . . < ǫg−1 ≤ 2g − 2,


an upper bound on the number of its Weierstrass points is(g − 1)(2g − 1)(2g −2). Hence, the order of AutK(Σ) is less than4g3 multiplied by the order of thestabiliser of a Weierstrass pointP. Thus, an upper bound for AutK(Σ) followsfrom the upper bound for|AutK(Σ)P | given in Theorem 12.46.

The most important consequence of Theorem 12.47 is that, ifΣ has genusg ≥ 2,then Hurwitz’s Theorem 7.26 applies to every extensionΣ/ΣG such thatΣG is thefixed field of a non-trivial subgroupG of AutK(Σ). Using the geometric term ofquotient curve, the following result is obtained.

COROLLARY 12.49 LetF be an irreducible algebraic curve of genusg which isneither rational nor elliptic. For a non-trivialK-automorphism groupG ofF , letg′ be the genus of the quotient curveF/G. Then

2g − 2 = |G|(2g′ − 2) + d, (12.18)

whered = degD(Σ/ΣG) with Σ = K(F) andΣG = K(F/G).

REMARK 12.50 Corollary 12.49 holds true for rational and elliptic curvesas farasG is finite.

Although the aim of the proof of Theorem 12.47 was only to prove the finite-ness of AutK(Σ), an upper bound for the order of AutK(Σ) in terms ofp andgcan be extracted from it; see Remark 12.48. A further effort shows that such anupper bound can be improved significantly. In this direction, the first step is The-orem 12.53, which makes evident the importance of short orbits in investigatingalgebraic curves with largeK-automorphism groups. Two types of short orbits aredistinguished.

DEFINITION 12.51 A short orbit of aK-automorphism groupG is tameor non-tame(wild) according as the order ofGP for one and hence every placeP in theorbit is prime or not to the characteristicp of K.

REMARK 12.52 If p = 0, then every short orbit is tame.

THEOREM 12.53 Let F be an irreducible algebraic curve of genusg which isneither rational nor elliptic.

(i) If G is aK-automorphism group ofF , then Hurwitz’s upper bound,

|G| ≤ 84(g − 1),

holds in general with exceptions occurring in positive characteristic.

(ii) In terms of the function fieldΣ ofF , exceptions can only occur whenp > 0,the fixed fieldΣG is rational, andG has at most three short orbits, namely

(a) G has exactly three short orbits, two tame and one non-tame;

(b) G has exactly two short orbits, both non-tame;

(c) G has only one short orbit which is non-tame;

466

(d) G has exactly two short orbits, one tame and one non-tame.

Proof. The proof exploits Corollary 12.49 following the original idea of Hurwitz,see [149]. For everyG-orbitO of places ofΣ,

|G| =∑eP ,

with the summation over all places contained inO. Now, (12.18) may be rewrittenin the form,

2g − 2 = |G| · (2g′ − 2 + d′), (12.19)

whered′P = dPe−1P andd′ =

∑d′P with the summation only over a set of rep-

resentatives of places inPΣ, exactly one from each short orbit. So, it is necessaryto investigate the possibilities for|G| according to the numberr of short orbits of|G| onPΣ. Also, it is useful to take (7.7) into account; this givesdP ≥ eP − 1,with equality holding if and only if eitherp 6= 0 or eP is prime top. Therefore, ifdP > 0, thend′P ≥ 1

2 . Also, if d > 0, thend′ ≥ 12 .

If g′ ≥ 2, then|G| ≤ g − 1. Forg′ = 1, it follows thatd′ > 0 sinceg ≥ 2, andhence|G| ≤ 4(g − 1). So, assume thatg′ = 0; that is,ΣG is rational. Then

2g − 2 = |G| · (d′ − 2). (12.20)

In particular,d′ > 2. Therefore,G has some, sayr ≥ 1, short orbits of places ofΣ.

Take representatives,Q1, . . . ,Qr, from each short orbit, and letd′i = d′Qi. After

a change of indices, it may be assumed thatd′i ≤ d′j for i ≤ j.

(I) Whenr ≥ 5, thend′ ≥ 52 , and hence|G| ≤ 4(g − 1).

(II) When r = 4, thend′ > 2 andd′i >12 for at least one placeP. As d′i >

12

impliesd′i ≥ 23 , sod′ − 2 ≥ 1

6 , whence|G| ≤ 12(g − 1).

(III) When r = 3, then again used′ − 2 > 0. If d′1 = 23 thend′3 ≥ 3

4 and hence|G| ≤ 24(g−1). If d′1 = 1

2 , d′2 ≥ 3

4 , then|G| ≤ 40(g−1). Also,d′1 = 12 and

d′2 = 23 imply thatd′3 ≥ 6

7 , whence|G| ≤ 84(g−1). Finally, if d′1 = d′2 = 12 ,

thend′3 > 1 which can only occur whenp > 2 andeQ1= eQ2

= 2 whileeQ3

is divisible byp. This leads to (a).

(IV) When r = 2, thend′ = d′1 + d′2 > 2. This can only occur when eitherd′1, ord′2, or both are greater than1. Hence, one of (b) and (d) arises.

(V) Whenr = 1, thend′ = d′1 > 2, and case (c) occurs.

2

There are so many exceptions to Hurwitz’s bound that their classification appearsout of range. However, those with|G| greater thang3 seem not too numerous, atleast forg large enough, and their determination might be possible. Sofar, thishas been done for|G| ≥ 8g3, and the classification is stated in Theorem 12.113; itrequires some deeper preliminary results that are proved inthe next two sections.


12.8 TAME AUTOMORPHISM GROUPS

LetG be a non-trivialK-automorphism group of an irreducible algebraic curveF .In this and the next section, the important equation (12.18)linking the genusg ofthe curveF to the genusg′ of its quotient curveF ′ = F/G is given a more explicitform. The idea is to look at the action ofG on the set of places of the function fieldΣ of F . Here, the tame case is considered; that is, it is supposed that the order ofthe stabilisers are all prime top.

THEOREM 12.54 LetG be a finiteK-automorphism group of an irreducible al-gebraic curveF . If |GP | is prime top for every placeP of Σ = K(F), then

2g − 2 = n(2g′ − 2) +∑s

i=1 (n− li), (12.21)

wheren = |G|, andl1, . . . , ls denote the sizes of the short orbits ofG.

Proof. Choose a placeP of Σ, and letP ′ be a place ofΣG lying underP. ByTheorem 12.39, the ramification indexρ is equal tom, with m = |GP |. Since byhypothesisp ∤ n, by the remark made after the proof of Hurwitz’s Theorem 7.26,

ordP dξ = (ρ− 1) + ordP′ dξ.

Hence

D(Σ/ΣG) =∑

(m− 1)P, (12.22)

with P ranging over all places ofΣ. ThereforedegD(Σ/ΣG) =∑

(m− 1).To evaluate

∑(m−1), consider the finite sub-sum

∑(mi−1) restricted to those

placesP of Σ which lie over a given placeP ′i of ΣG. By Theorem 12.39, there are

preciselyn/mi distinct places overP ′i, each of them contributingmi − 1. Hence,

from every placeP ′i, a contributionn(mi − 1)/mi is obtained. Therefore,

∑(m− 1) =

∑ n

mi(mi − 1) =

∑(n− n

mi

).

As ℓi = n/mi, so

deg(Σ/ΣG) =∑

(m− 1) =∑s

i=1(n− li).From Corollary 12.49,

2g − 2 =[Σ : ΣG

]· (2g′ − 2) + degD(Σ/ΣG)

=n(2g′ − 2) +∑s

i=1(n− li).2

In the special case thatG has prime order, Theorem 12.54 reads as follows.

THEOREM 12.55 Letα be a finite non-trivialK-automorphism of an irreduciblealgebraic curve whose ordern is prime top. If ρ(α) is the number of fixed placesof α andn = ordα, then

2g − 2 = n(2g′ − 2) + ρ(α)(n− 1). (12.23)

468

For tameK-automorphism groups, a significant improvement on the bound (ii) inTheorem 12.46 is the following result.

THEOREM 12.56 LetF be an irreducible algebraic curve of genusg > 0. LetGPbe aK-automorphism group ofF fixing a placeP. If the ordern ofGP is primeto p, then

n ≤ 4g + 2. (12.24)

Proof. The placeP itself constitutes a short orbit of size1. Putℓs = 1 in (12.21):

2g − 1 = n(2g′ − 1) + n

s−1∑

i=1

(1− ℓi

n

), (12.25)

whereℓ1, . . . , ℓs−1 denote the sizes of the remaining short orbits ofGP . If g′ > 0,the assertion follows.

So, now assume thatg′ = 0. Then (12.25) becomes the following:

2g − 1 =∑s−1

i=1 (n− ℓi)− n. (12.26)

Sinceℓi dividesn, son − ℓi ≥ 12n. Hence, ifs ≥ 4, then2g − 1 ≥ 3

2n − n,whencen ≤ 2(2g − 1) ≤ 4g + 2. The same argument works fors = 3 as long asℓ1 + l2 ≤ 1

2n. Further, neithers = 2 nors = 1 can actually occur.So it remains to investigate the case that2g−1 = n−(ℓ1+l2) with ℓ1+ℓ2 >

12n.

Up to an index change,ℓ1 ≥ l2 may be assumed. Then the possibilities for(ℓ1, ℓ2)are

( 13n,

15n), ( 1

3n,14n), ( 1

3n,13n), ( 1

2n,≤ 13n). (12.27)

WriteG = GP for brevity, and denote byHi the subgroup ofG which fixes one,and hence every, place in the short orbit of sizeℓi wherei = 1, 2. LetH be thesubgroup ofG generated byH1 andH2. SinceG is cyclic by Theorem 12.46, so

|H| = n/gcd(ℓ1, ℓ2).

It will be shown thatG = H.By way of contradiction,ΣG is assumed to be a proper subfield ofΣH . Then

[ΣH : ΣG] = |G|/|H|. Let P ′ be the place ofΣH lying underP in the extensionΣ/ΣH . ThenP ′ completely ramifies inΣH/ΣG.

From the definition ofH, it can be deduced that no further place ofΣH ramifiesin ΣH/ΣG. To do this, note that, for any placeQ from the orbit of lengthℓ1,GQ = HQ and

|G||H| =

|GQ||HQ|

· ℓ1|QH | =

ℓ1|QH | =

|QG||QH | .

Then the same is true forℓ2. If Q is a place not in a short orbit, thenGQ is theidentity subgroup, and henceGQ = HQ holds. Therefore|G|/|H| = |QG|/|QH |,for every placeQ 6= P of Σ. This proves the assertion.

Corollary 12.49 gives the equation,

2g − 2 = −2m′ +m′ − 1,


whereg is the genus ofΣH , andm′ = |G|/|H|. But this can only occur form′ = 1, giving the desired result thatH = G. Henceℓ1 andℓ2 are prime to eachother. From (12.27), the following cases can occur:

(ℓ1, ℓ2) = (13n,

15n), |GP | = 15, g = 4;

(ℓ1, ℓ2) = (13n,

14n), |GP | = 12, g = 3;

(ℓ1, ℓ2) = (12n,

1kn), k odd, |GP | = 2k, g = 1

2 (k − 1).

In each of these cases,n ≤ 4g + 2. This completes the proof. 2

REMARK 12.57 If the hypothesis thatp ∤ n is dropped, then the proof of Theorem12.54 only provides a lower bound for2g − 2; that is,

2g − 2 ≥ n(2g′ − 2) +∑s

i=1(n− ℓi). (12.28)

However, for ap-groupG of ordern, if g is replaced by thep-rankγ, a relationshipsimilar to (12.21) holds.

THEOREM 12.58 (Deuring–Shafarevich)Let G be aK-automorphism group ofan irreducible algebraic curveF whose ordern is a power ofp. Then

γ − 1 = n(γ′ − 1) +∑s

i=1(n− ℓi), (12.29)

whereγ andγ′ are thep-ranks ofF and its quotient curveF/G, while ℓ1, . . . , ℓsare the sizes of the short orbits ofG on the places ofΣ.

If G has no short orbits, the possibilities for the structure ofG is described in thefollowing theorem.

THEOREM 12.59 LetG be aK-automorphism group of an irreducible algebraiccurveF whose ordern is a power ofp. Assume thatG has no short orbits. ThenG is isomorphic to a quotient group of the free group withγ generators.

12.9 NON-TAME AUTOMORPHISM GROUPS

As pointed out in Remark 12.57, the nice formula (12.21) depending only upon theaction ofG on places does not hold true for non-tame subgroups. However, a simi-lar, although more complicated, formula, relying on more concepts and argumentsexists for non-tame subgroups, is theHilbert Different Formulastated in Theorem12.65.

The goal in this section is to give a proof of the Hilbert Different Formula. Fur-ther, for non-tameK-automorphism groups, an significant improvement to Theo-rem 12.46(ii) is obtained by using Theorem 12.65.

Given a placeP of Σ, choose a primitive representationτ of P together with auniformizing elementx ∈ Σ atP, and assume thatτ(x) = t. By Lemma 12.40,α(x) 6= x for any non-trivialK-automorphismα ∈ GP . Also, n = |GP | dividesordt τ(ξ) for any ξ ∈ ΣGP . More precisely, there existss(t) ∈ K[[t]] such thatτ(ξ) = σ(s(t)) with σ ∈ K((s)). Such an elements(t) is independent of thechoice ofξ.

470

L EMMA 12.60 The set1, x, x2, . . . , xn−1 is a basis of[Σ : ΣGP ].

Proof. By Lemma 12.35, it suffices to prove that1, x, x2, . . . , xn−1 are linearlyindependent overΣGP . Suppose, on the contrary, that

ξ0 + ξ1x+ . . .+ ξn−1xn−1 = 0,

whereξ0, ξ1, . . . , ξn−1 ∈ ΣGP but not allξi = 0. Then

τ(ξ0) + τ(ξ1)τ(x) + . . .+ τ(ξn−1)(τ(x))n−1 = 0,

showing that two or more of then terms on the left-hand side have the same order.This cannot actually occur by the following argument. If0 ≤ i ≤ n− 1, then

ordt τ(ξi)τ(xi) = ordt τ(ξi) + iordt τ(x) = ki n+ i

for a non-negative integerki; the same holds for everyj with 0 ≤ j ≤ n−1.Henceordt τ(ξi)τ(x

i) = ordt τ(ξj)τ(xj), andn(kj − ki) = i− j. But this is impossible

since0 ≤ i, j < n, andi 6= j. 2

L EMMA 12.61 There existξ0, . . . , ξn−1 ∈ ΣGP satisfying

ξ0 + ξ1x+ . . .+ ξn−1xn−1 + xn = 0, (12.30)

with ordt τ(ξ0) = n, andordt τ(ξi) ≥ n, for i = 1, . . . n− 1.

Proof. By Lemma 12.60, there existη0, . . . , ηn ∈ ΣGP , ηn 6= 0, satisfying

η0 + η1x+ . . .+ ηn−1xn−1 + ηnx

n = 0. (12.31)

It follows that

τ(η0) + τ(η1)τ(x) + . . .+ τ(ηn)(τ(x))n = 0.

By the argument employed in the proof of Lemma 12.60, the equation,

ordt(τ(ηi)τ(xi)) = ordt(τ(ηj)τ(x

j)),

implies thatn(kj − ki) = i− j. Here, this is only possible fori = n, j = 0 and

ordtτ(η0) = n+ ordtτ(ηn).

It also follows that ordtηi ≥ n + ordtτ(ηn). Let ξi = ηi/ηn. Then ordtτ(ξ0) = nwhile ordtτ(ξi) ≥ n for 1 ≤ i ≤ n− 1. Dividing both sides in equation (12.31) byηn, the result follows. 2

From equation (12.30),

σ0 + σ1t+ . . .+ σn−1tn−1 + tn = 0. (12.32)

whereσi ∈ K[[s]] with ordsσ0 = 1, ands ∈ K[[t]] with ordts = n. Derivation ofequation (12.32) gives(

dσ0

ds+dσ1

dst+ . . .+

dσn−1

dstn−1

)ds

dt

+σ1 + σ2t+ . . .+ (n− 1)σn−1tn−2 + ntn−1 = 0.

Define the polynomialΦ(T ) ∈ K(s)[T ]:

Φ(T ) = σ0 + σ1T + . . .+ σn−1Tn−1 + Tn.

Then (dσ0

ds+dσ1

dst+ . . .+

dσn−1

dstn−1

)ds

dt= − dΦ(T )

dT

∣∣∣∣T=t

. (12.33)


L EMMA 12.62

Φ(T ) =∏

α∈GP(T − (τα)(x)). (12.34)

Proof. From (12.32),Φ(t) = 0. To find the other roots ofΦ(T ), note at first thatα(ξi) = ξi with i = 0, 1 . . . , n, for everyα ∈ GP . From (12.30), it follows that

ξ0 + ξ1α(x) + . . .+ ξn−1α(x)n−1 + α(x)n = 0.

Let δ(t) = (τα)(x) = ct+ . . . , c 6= 0. Then

σ0 + σ1δ(t) + . . .+ σn−1δ(t)n−1 + δ(t)n = 0.

As α ranges overGP , the number of distinctδ(t) obtained isn, as τ is aK-isomorphism fromΣ toK[[t]]. As Φ(T ) has degreen, it follows that the roots ofΦ(T ) in Σ are then elementsδ(t) = (τα)(x) with α ranging overGP . 2

¿From (12.34),

dΦ(T )

dT

∣∣∣∣T=t

=∏

α∈G∗

P

(t− (τα)(x)), (12.35)

with α ranging over the setG∗P of non-trivial K-automorphisms inGP . From

(12.33) and (12.35),(dσ0

ds+dσ1

dst+ . . .+

dσn−1

dstn−1

)ds

dt= −∏α∈G∗

P

(t− (τα)(x)).

Since ordsσ0 = 1, this gives

ordtds

dt=∑

α∈G∗

P

ordt((τα)(x)− t). (12.36)

As, by Theorem 12.39,d(Σ/ΣGP ) = d(Σ/ΣG), the following result is obtained.

THEOREM 12.63 For a subgroupG of AutK(Σ) and a placeP of Σ, letG∗P be

the set of all non-trivial elements ofGP . Then

D(Σ/ΣG) =∑

P∈PΣ

(∑α∈G∗

P

ordP(α(x)− x))P (12.37)

for any uniformizing elementx ∈ Σ atP.

For i = 0, 1, . . ., defineG(i)P to be the set of all elementsα ∈ GP such that

ordP(α(x)− x)) ≥ i+ 1.

Forα, β ∈ G(i)P ,

ordP(βα(x)− x)= ordP(βα(x)− α(x) + α(x)− x)≥minordP(β(α(x))− α(x)),ordP(α(x)− x).

Sinceα(x) is still a uniformizing element atP, so

ordP(β(α(x))− α(x)) ≥ i+ 1.

Hence ordP(βα(x)− x) ≥ i+ 1. This leads to the following definition.

472

DEFINITION 12.64 Thei-th ramification groupG(i)P atP to be

G(i)P = α | ordP(α(x)− x) ≥ i+ 1, α ∈ GP,

wherex ∈ Σ is a uniformizing element ofΣ atP.

From Theorem 12.63, the following important result is deduced.

THEOREM 12.65 (Hilbert’s Different Formula)

D(Σ/ΣG) =∑

P∈PΣdPP, with dP =

∑i≥0(|G

(i)P | − 1). (12.38)

This together with Theorem 7.26, see also Corollary 12.49, give following impor-tant result.

THEOREM 12.66 LetG be a finiteK-automorphism group of an irreducible alge-braic curveF , and letF/G be the quotient curve ofF arising fromG; equivalent-ly, Σ = K(F) andΣG = K(F/G). Then

2g − 2 = |G|(2g′ − 2) +∑

P∈PΣ

∑i≥0 (|G(i)

P | − 1), (12.39)

whereg andg′ are the genus ofF andF/G, respectively.

The subgroup chainG(0)P ≥ G(1)

P ≥ . . . is described in the following theorem.

THEOREM 12.67 (i) G(0)P is the stabiliser ofP.

(ii) G(1)P is the unique maximal normalp-subgroup ofGP .

(iii) For i ≥ 1, G(i)P is normal inGP and the quotient groupG(i)

P /G(i+1)P is an

elementary abelianp-group.

Proof. Statement (ii) follows from the definition ofG(0)P and Lemma 12.41.

To prove (iii), leti ≥ 1. From (12.35),

(τα)(x)− t = τ(α−1(x)) = cαti + . . .

for everyK-automorphismα ∈ G(i)P . The mapping

φ : G(i)P → (K,+),

α 7→ cα−1 , (12.40)

is a group homomorphism. In fact, if

(τα)(x) = t+ cαti + . . . , (τβ)(x) = t+ cβt

i + . . . ,

then

τ(αβ)(x) = τ((αβ)−1(x)) = τ(β−1(α−1(x)))

= τ(β−1τ−1(τ(α−1(x)))) = τ(β−1τ−1(t+ cαti + . . .))

= τ(β−1(x+ cαxi + . . .)) = (τβ)(x) + cατβ(x)i + . . .

= t+ cβti + cαt

i + . . . = t+ (cα + cβ)ti + . . . .


However, kerφ consists of allK-automorphisms belonging toG(i+1)P . Hence

G(i+1)P is a normal subgroup ofG(i)

P , and the factor groupG(i)P /G

(i+1)P is isomor-

phic to an additive subgroup ofK. Part(iii) now follows from the fact that finiteadditive subgroups ofK are abelian groups in which every non-trivial element hasorderp. 2

In the subgroup chainGP = G(0)P ≥ G

(1)P ≥ . . ., consecutive non-trivial sub-

groups may coincide. Sometimes, repetitions in the chain are deleted so that theHilbert Different Formula (12.38)only involves distinct ramification groups. To dothis, letρ−1 = 0 < ρ0 < ρ1 < . . . be the strictly increasing sequence of all non-negative integers which are of the form ordP(α(x) − x) for someα ∈ GP . Thenρi is thei-th ramification numberof GP . Note that

ρ0 = minordP(α(x)− x) | α ∈ GP, ρi = minordP(α(x)− x) | α ∈ U (i)P ,

where

U(i+1)P = α ∈ U (i)

P | ordP(α(x)− x) ≥ ρi + 1.Also, ρ0 = 1 if and only if GP > G

(1)P ; that is,GP = G

(1)P ⋊ H with H a

non-trivial tame subgroup.The subgroup chainGP = U

(0)P > U

(1)P > . . . is strictly decreasing, and

dP =∑

i≥0(ρi − ρi−1)(|U (i)P | − 1). (12.41)

Therefore, (12.39) can also be written as

2g − 2 = |G|(2g′ − 2) +∑

P∈PΣ

∑i≥0 (ρi − ρi−1)(|U (i)

P | − 1). (12.42)

Some further properties of the ramification subgroups may beproved by usingcompanionK-automorphisms, introduced in Section 12.1. This essentially de-pends on the fact that, in terms of the companionK-automorphismλα of α, theequation ordP(α(x)− x) = k means that

λα = t+ ctk + . . . , c 6= 0.

L EMMA 12.68 Letα ∈ GP andβ ∈ G(k)P , k ≥ 1.

(i) If α 6∈ G(1)P , then the commutator[α, β] = αβα−1β−1 belongs toG(k+1)

P if

and only if eitherαk ∈ G(1)P or β ∈ G(k+1)

P .

(ii) If α ∈ G(k)P , β ∈ G(l)

P , with k, l ≥ 1, then their commutator[α, β] belongs

toG(k+l+1)P .

(iii) Letα have orderprs with p ∤ s. If β 6∈ G(k+1)P andαβ = βα, thens divides

k.

(iv) Let k be an integer not divisible by|GP |/|G(1)P |. If GP is abelian, then

G(k)P = G

(k+1)P .

(v) The integersk ≥ 1 satisfyingG(k)P 6= G

(k+1)P are all congruent modulop;

that is,ρi ≡ ρj (mod p), for everyi, j ≥ 1.

474

Proof. SinceG(k)P is a normal subgroup ofGP , the commutatorγ = [α, β] belongs

toG(k)P . Write

λα =ut+∑

i≥i0uit

i, with ui0 6= 0,

λβ = t+∑

j≥j0wjt

j , with wj0 6= 0,

λγ = t+∑

m≥m0vmt

m.

By Lemma 12.17, ordα = prs, wheres is the smallest positive integer satisfyingus = 1; the hypothesis thatβ ∈ G

(k)P means thatk ≤ j0 − 1. Also, it may be

thatλγ = t, and this occurs if and only ifαβ = βα; otherwise,γ is a non-trivial

K-automorphism belonging to∈ G(m0−1)P . Then

λαλβ =ut+∑

i≥i0uit

i +∑

j≥j0wj(ut+

∑i≥i0

uiti)j ,

λγλβλα =ut+ u∑

j≥j0wjt

j +∑

i≥i0ui(t+

∑j≥j0

wjtj)i

+∑

m≥m0um(ut+ u

∑j≥j0

wjtj +

∑i≥i0

(t+∑

j≥j0wjt

j))m.

If u 6= 1, thenαβ = γβα implies

0 = w0(uj0 − u)tj0 + um0

um0tm0 + ctr + . . . ,

wherer ≥ j0 + 1. Hence, two cases can occur:

(a) uj0−1 = 1 and eitherγ = 1, orm0 ≥ j0 + 1;

(b) uj0−1 6= 1 andj0 = m0.

In the former case,αj0−1 has order a power ofp, while in the latter caseγ 6= 1.Now, a straightforward argument proves (i).

If u = 1, thenαβ = γβα impliesm0 ≥ i0 + j0 − 1. Hence

m0 − 1 ≥ (i0 − 1) + (j0 − 1) + 1,

whence(ii) follows.Let u 6= 1. If αβ = βα, thenλγ = t, and henceuj0−1 = 1. Thusαj0−1 ∈ G(1)

P .

This implies thats | (j0 − 1). If β 6∈ G(k+1)P , thenk = j0 − 1, and hences | k, as

stated in(iii).SinceGP is the semi-direct product of its normal subgroupG(1)

P by a subgroup

of orders = |GP |/|G(1)P |, someα ∈ GP has orders. If k does not divides, then

αk 6= 1. By (i), β ∈ G(k+1)P . This proves(iv).

Choose an elementβ from the last non-trivial ramification group, sayG(l)P . Then

G(l+1)P = 1. Letα ∈ G(k)

P . From (iv),αβ = βα. To show thati0 − j0 is divisibleby p, assume thatp does not divide bothi0 andj0. then,αβ = βα implies that

0 = (j0 − i0)uj0wj0tj0−1+i0 + . . . .

But thenp dividesj0 − i0, and (v) follows. 2

The following two results are corollaries of Lemma 12.68 (i).

L EMMA 12.69 LetA be an abelian subgroup ofAutK(Σ) that fixesm ≥ 1 places.


(i) If P is one of the fixed places andg′ is the genus ofΣA, then

2g − 2 ≥ |A|(2g′ − 2) +m(|A| − 1 +|A||AP |

(|AP | − 1)). (12.43)

(ii) If G is non-tame, then

2g − 2 ≥ |A|(2g′ + 32m− 2)−m. (12.44)

Proof. LetA = AP = A(1)P ⋊H with p ∤ |H|. SinceG is abelian, Lemma 12.68 (i)

implies thatA(1)P = . . . = A

(h)P with h = |H|. Now, (12.43) follows from Theorem

12.66 applied toA. If A(1)P is non-trivial, then|A(1)

P |−1 ≥ 12 |A

(1)P |. Hence (12.43)

implies (12.44). 2

L EMMA 12.70 LetG be an abelian non-tame subgroup ofAutK(Σ) fixing a place

P. If G = G(1)P ⋊ |H| with p ∤ |H|, then

2g ≥ 2g′|G(1)P |+ (|H| − 1)(|G(1)

P | − 1). (12.45)

Proof. As G is abelian, from (i) of Lemma 12.68,G(1)P = . . . = G

(h)P with h =

|H|. Then (12.45) follows from Theorem 12.66 applied toG(1)P . 2

From Theorem 12.46, ifg > 0, thenGP is finite and its order has an upperbound depending only onp andg. In the notation of this section, Theorem 12.46states that|G(1)

P | ≤ p2(g + 1)(2g − 1)2. The aim is to improve this bound.

THEOREM 12.71 LetF be an irreducible algebraic curve of positive genusg, andletGP be aK-automorphism group ofF fixing a placeP. Then

|G(1)P | ≤

4p

p− 1g2.

More precisely, if Fi is the fixed subfield ofG(i)P , then one of the following cases

occurs:

(i) F1 is not rational, and|G(1)P | ≤ g;

(ii) F1 is rational, GP has a short orbit onPΣ other thanP, and

|G(1)P ≤ p

p− 1g;

(iii) F1 andF2 are rational, P is the unique short orbit ofGP onPΣ, and

|G(1)P ≤ 4|G(2)

P

(|G(1)P − 1)2

g2 ≤ 4p

(p− 1)2g2.

Proof. From Theorem 12.66 applied toG(1)P ,

2g − 2 = |G(1)P |(2g1 − 2) + d, (12.46)

whereg1 is the genus ofF1 andd = degD(Σ/F1). SinceP completely ramifiesin Σ/F1, soeP = |G(1)

P |. Since|G(1)P | is a power ofp, hence|G(1)

P | is equal to the

476

order of the first ramification group. WriteD(Σ/F1) =∑dPP. By the Hilbert

Different Formula (12.38),

dP ≥ 2|G(1)P | − 2. (12.47)

Wheng1 > 0, (12.46) and (12.47) imply

2g − 2 ≥ d ≥ dP ≥ 2|G(1)P | − 2,

whence|G(1)P | ≤ g.

From now on, assume thatg1 = 0. Then (12.46) becomes

2g − 2 = −2|G(1)P |+ d. (12.48)

In case (ii), there is one more placeQ with the above properties. As usual,eQis its ramification index, anddQ is its weight in the different divisorD(Σ/F1). Inparticular,eQ ≤ |G(1)

P |. SinceeQ is a power ofp andF1 is rational, again from theHilbert Different Formula (12.38),2g− 2 ≥ 2eQ− 2. Taking (12.47) into account,it follows that

2g − 2≥−2|G(1)P |+ dP +

|G(1)P |eQ

dQ

≥−2|G(1)P |+ 2|G(1)

P | − 2 + 2|G(1)P |

eQ − 1

eQ;

hence

g ≥ eQ − 1

eQ|G(1)

P | ≥p− 1

p|G(1)

P |,

which is the desired result.Now, consider the case thatF1 is rational andP is the unique ramified place.

First, it is shown thatF2 is rational as well. Put

d1 = degD(Σ/F1), d2 = degD(Σ/F2), d∗ = degD(F2/F1).

From (7.8),

d1 = d2 + [Σ : F2]d∗ = d2 + |G(2)

P |d∗. (12.49)

Also,

d1 = 2(|G(1)P | − 1) +

∑i≥2(|G

(i)P | − 1), d2 = 2(|G(2)

P | − 1) +∑

i≥2(|Hi| − 1).(12.50)

whereH2,H3, . . . denotes the sequence of ramification groups ofΣ/F2 atP. AsF2 is the fixed field ofG(2)

P , the groupsHi andG(i)P coincide fori ≥ 2. Hence

d1 − 2(|G(1)P | − 1) = d2 − 2(|G(2)

P | − 1),

which, together with (12.49), gives

d∗ = 2|G(1)

P ||G(2)

P |− 2.

Then Theorem 12.66 shows thatF2 has genus0; that is,F2 is rational.


Let k denote the smallest integer such thatk ≥ 3 andG(k)P 6= G

(2)P . All ramifi-

cation groups, and thus bothG(2)P andG(k)

P are normal subgroups ofGP . Hence,

there is a normal subgroupG′ containingG(k)P which is a subgroup ofG(2)

P of indexp.

The next step is to compute the genusg′ of the fixed fieldF ′ of G′. Let

d′ = degD(Σ/F ′), d = degD(F ′/F2).

As before, the following equations hold:

d2 = d′ +|G(2)

P |p

d, d2 = k(|G(2)P | − 1) +

∑i≥k (|G(i)

P | − 1).

Also,

d′ = k

(1

p|G(p)

P | − 1

)+∑

i≥k

(|G(i)

P | − 1),

asG′ lies betweenG(2)P andG(k)

P . It follows thatd = k(p − 1). Now, Theorem12.66 shows that

g′ = 12 (k − 2)(p− 1). (12.51)

Sincek ≥ 3, soF ′ is not rational. LetP ′ the place ofF ′ lying underP in theextensionΣ/F ′, and letG(1)

P′ be the stabiliser ofP ′ in theK-automorphism group

of F ′. AsG′ is a normal subgroup ofG(1)P , the factor groupG(1)

P /G′ can be viewed

as a subgroup ofG(1)P′ . Since[G(2)

P : G′] = p, so[F ′ : F2] = p. Also,P ′ is the onlyplace which ramifies inF ′|F2. Choosex ∈ F2 havingP ′ as a pole. ThenF ′|F2

is an Artin–Schreier extension, and Theorem 13.5 ensures the existence ofy ∈ F ′

such thatyp − y = B(x) with B(X) ∈ K[X]. From Theorem 13.6,

|G(1)P′ | ≤ 4p

(p− 1)2g′

2.

Since the quotient groupG(1)P /G′ is a subgroup ofG(1)

P′ , from (12.51),

|G(1)P | ≤ |G′| 4p

(p− 1)2g′

2=

4|G(2)P |

(p− 1)2g′

2= |G(2)

P |(k − 2)2. (12.52)

From Theorem 12.66 applied toG(2)P ,

2g − 2 ≥ −2|G(2)P |+ k(|G(2)

P | − 1),

whence

k − 2 ≤ 2g

|G(2)P | − 1

.

By substituting this in (12.52) the final result is obtained:

|G(1)P | ≤

4|G(2)P |

(|G(2)P | − 1)2

g2.

2

The next objective is to obtain sharper estimates on the orders of abelianK-auto-morphism groups ofΣ.

478

THEOREM 12.72 LetG be an abelianK-automorphism groups of an irreduciblealgebraic curveF of genusg ≥ 2. Then

|G| ≤

4g + 4 for p 6= 2,4g + 2 for p = 2.

Proof. PutΣ = K(F). If Σ/ΣG does not ramify, then|G| ≤ g − 1 by Theorem12.66.

Let P be a place ofΣ at whichG ramifies. SinceG is abelian,GP fixes everyplace in the orbit ofP underG. LetA = GP anda = |A|. Then|G|/a is the sizeof the orbit ofP underG. Let g′ denote the genus ofΣA. From Theorem 12.66applied toA,

2g − 2 ≥ a(2g′ − 2) +|G|a

(a− 1) ≥ a(2g′ − 2) + 12 |G|.

If g′ ≥ 1, this gives|G| ≤ 4g − 4.ThereforeΣA is assumed to be rational. Then Theorem 12.66 applied toA shows

that

2g − 2 ≥ −2a+|G|a

(a− 1). (12.53)

Suppose thata = 2. Then (12.53) implies that|G| ≤ 4g+4. Assume thatp = 2and|G| = 4g + 3. ThenG is tame, and strict inequality occurs in (12.53). FromTheorem 12.54,G must also ramify at some placeQ outside theG-orbit of P. Inother words,GQ contains a non-trivial elementα. Sinceα has order at least3, thisimplies that (12.53) remains valid when2 is subtracted from the right-hand side.But then|G| ≤ 4g + 2, a contradiction.

Now, let p = 2 and |G| = 4g + 4. Then|G(1)P | ≤ 2, and hence (12.53) holds

true whena − 1 is replaced bya on the right-hand side. But this implies that|G| ≤ 2g + 4, contradiction. Therefore, (12.72) holds fora = 2.

Suppose thata = 3. From (12.53),|G| ≤ 3g+6 which implies|G| ≤ 4g+2 forg > 3. The preceding argument can be used to show that ifp = 2 then|G| = 14 isimpossible. Therefore, (12.72) holds fora = 3.

From now on, leta > 3. It is shown that (12.72) holds if the size of theG-orbitof P is at least4. To do this, suppose that

|G|a≥ 4 +

2

a− 2. (12.54)

A direct computation shows that (12.54) implies the right-hand side in (12.53) is atleast12 |G| − 3. Therefore, theG-orbit of P has size at most4. If the size is equalto 4, then|G| = 4a, and (12.53) shows that|G| ≤ 4g + 4. The possibility thatp = 2 and either|G| = 4g + 3 or |G| = 4g + 4, can be ruled out by the argumentpreviously used forg = 2.

Let ℓ1, . . . , ℓs be the shortG-orbits. From the above results,ℓi ≤ 3.Suppose thatA is tame. SinceΣG is rational, Theorem 12.54 applied toG gives

2g − 2 = −2|G|+ s|G| − (ℓ1 + . . . ℓs).

Therefore,s ≥ 3, and|G| ≤ 2g + 7. Thus, ifg ≥ 3, then|G| ≤ 4g + 2. If g = 2,then |G| ≤ 11 with equality occurring only forℓ1 = ℓ2 = ℓ3 = 3. But ℓ1 = 3


implies that|G| is divisible by3 showing that|G| 6= 11. Therefore, ifg = 2, then|G| ≤ 10 = 4g + 2.

Suppose thatA is non-tame. If a shortG-orbit has length3, thenA has at leastthree fixed places. Now, form = 3, (12.44) gives2g + 1 ≥ 5

2 |A| = 56 |G|, whence

|G| ≤ 15 (12g + 6) < 4g + 2. The same argument withm = 2 shows that, ifG has

a short orbit of length2, then2g ≥ |A| = 12 |G|, whence|G| ≤ 4g. Also, if G fixes

at least two places, then (12.44) applied to|G| gives|G| ≤ 2g − 2.It remains to consider the case thatG = A = GP , thatG(1)

P is non-trivial andthat the identity is the onlyK-automorphism inGP that fixes a place distinct fromP. WriteGP = G

(1)P ⋊H with p ∤ |H|.

Suppose thatH is non-trivial. From Lemma 12.70,

g ≥ 12 (|H| − 1)(|G(1)

P | − 1).

Since|H| = |G(1)P | = 2 is not possible,

(|H| − 1)(|GP | − 1) ≥ 12 |H||GP | − 1

also holds. Therefore,|G| = |GP | = |GP ||H| ≤ 4g + 2.Suppose thatG = GP = G

(1)P . A slightly sharper bound is proven, namely that

|G| ≥ 4g. By Theorem 12.88, this assertion holds forg = 2, but assume on thecontrary that it does not extend to everyg > 2. Let g be denote the minimum valueof such exceptional generag. Then there is an abelian subgroupG of AutK(Σ)such that|G| ≤ 4g + 1.

Let M be any non-trivial subgroup ofG. If ΣM has genusg′ ≥ 2, Theorem12.66 applied toM implies thatg ≥ g′|M |. Therefore,4g + 1 ≥ 4g′|M | +1, whence|G| ≥ 4g′|M | + 1. Since|G|/|M | is an integer, it follows that that|G|/|M | ≥ 4g′ + 1. But this is a contradiction, asG = G/M can be viewed as aK-automorphism group ofΣM whose genusg′ is smaller thang.

If ΣM is elliptic, Theorem 12.66 applied toM implies thatg ≥ |M |. SinceG = G/M can be viewed as aK-automorphism group ofΣM , Theorem 12.85shows that eitherp = 2 and|G|/|M | = 2, 4, or p = 3 and|G/M | = 3. Therefore,|G| ≤ 4g for p = 2, and|G| ≤ 3g for p = 3.

If ΣM is rational for every non-trivial subgroupM of G, Lemma 12.42 impliesthat either|G| = p, or |G| = p2. In particular,|G| ≤ 4g + 1 for p = 2. Therefore,p ≥ 3 is assumed. Sinceg > 0, Theorem 12.66 shows thatG(2)

P is not trivial. From

this, |G| ≤ 2g − 1 whenG = G(1)P = G

(2)P , in particular when|G| = p. It remains

to investigate the case that|G(1)P | = p2 and|G(2)

P | = p. By Lemma 12.68 (v),

G(2)P = G

(3)P = . . . = G

(p+1)P .

Therefore, Theorem 12.66 gives2g ≥ p(p− 1), whence4g > p2 = |G|. 2

The following lemmas are related to Hilbert’s Different Formula (12.38) andTheorem 12.66; they are used in later sections.

L EMMA 12.73 LetD(Σ/ΣG) =∑dPP. Then

dP ≥ |GP |+ |G(1)P | − 2.

480

Proof. By Theorem 12.67,|G(0)P | = m and |G(1)

P | = pv. The assertion followsfrom (12.38). 2

L EMMA 12.74 Assume that a subgroupM ofG(1)P contains no non-trivial element

from the second ramification groupG(2)P . If H ≤ GP is in the normaliser ofM

and|H| is prime top, then|H| ≤ |M | − 1.

Proof. Suppose thath ∈ H commutes with a non-trivial element ofM . By Theo-rem 12.68 (i), this can only occur whenh = 1. Thus,M contains more elementsthanH does. 2

L EMMA 12.75 LetG = GP be aK-automorphism group of an irreducible alge-braic curveF that fixes a placeP. Letv = dP − |GP | with dP = dP(Σ/ΣGP ),

andqi = [U(i)P : U

(i+1)P ]. Then the following hold:

(i) v = (µ+ 1)|H| − 1 with a non-negative integerµ;

(ii) |H| | (ρi − 1)(qi − 1);

(iii) qi | (ρi+1 − ρi)2, for every i ≥ 1.

Proof. SincedP(Σ/ΣH) = |H| − 1, anddP(Σ/ΣGP ) = |H||G(1)P |+ v, equation

(7.8) reads

|H| |G(1)P |+ v = |H| − 1 + |H| dP′(ΣH |ΣGP ),

whence (i) follows forµ = dP′(ΣH |ΣGP )− |G(1)P |. Note that

dP′(ΣH |ΣGP ) ≥ [ΣH : ΣGP ] = |G(1)P |,

by the remark after Theorem 7.26. Thus,µ ≥ 0.For i ≥ 1, putUi = U

(i)P ⋊H with vi = dP(Σ/ΣUi)− |Ui|, and apply (i) toUi.

This gives that

|H| | (dP(Σ/ΣUi)− |Ui|+ 1).

Hence,|H| also dividesvi − vi+1. This together with the equations,

vi − vi+1 = ρi(|U (i)P | − |U

(i+1)P |)− |U (i)

P |+ |U(i+1)P | = |U (i+1)

P |(ρi − 1)(qi − 1),

proves (ii).For the proof of (iii), it may be assumed thati = 1 after replacingGP byU (i)

P ⋊

H. Also, take|U (2)P | = p, after replacingΣ by ΣN , whereN is a normal subgroup

of U (2)P of indexp.

First, the case thatGP is not abelian is considered. Since the quotient groupG

(1)P /U

(2)P is an elementary abelianp-group,U (2)

P is the commutator subgroup of

G(1)P . Also, asZ(G

(1)P ) containsU (2)

P , the groupG(1)P /Z(G

(1)P ) is also an elemen-

tary abelianp-group. By a result from Group Theory,

[G(1)P : Z(G

(1)P )] = p2m.


If A is a maximal abelian normal subgroup ofG(1)P , then

Z(G(1)P ) ≤ A and |A| = pm|Z(G

(1)P |.

Since every non-trivial element inA has orderp or p2, the Main Theorem for FiniteAbelian Groups implies thatA is a direct product of cyclic groups:

A = A1 × 〈τ1〉 × . . .× 〈τm+n−2〉,where|A1| = p2, pn = Z(G

(1)P ), τp

1 = . . . = τpm+n−2 = 1, andA1 containsU (2)

P .

In particular,|G(1)P | = p2m+n andq1 = p2m+n−1.

LetA2 = 〈τ1〉 × . . . × 〈τm+n−2〉 andB = A2U(2)P . Then|A2| = pm+n−2 and

|B| = pm+n−1.Now, the contribution ofP to d = deg(D(Σ/ΣB)) is computed in two different

ways by applying (7.8) to the towersΣ ⊃ ΣU(2)P ⊃ ΣB andΣ ⊃ ΣA2 ⊃ ΣB . This

gives

ρ1(pm+n−2 − 1)p+ ρ2(p− 1) = ρ2(p− 1)pm+n−2 + ρ1(p

m+n+−2 − 1),

whereρ2 is the ramification number in the extensionΣA2/ΣB . It follows that

ρ2 = ρ2pm+n−2 − ρ1(p

m+n−2 − 1).

In particular,ρ2 > ρ1. Since the groupA/A2 is abelian andρ1 is also the ramifi-cation number in the extensionΣA2/ΣA, Lemma 12.68 (v) implies thatρ2 ≡ ρ1

(mod p); that is,ρ2 = ρ1 + kp for a positive integerk. Therefore,

ρ2 = ρ1pm+n−2 + kpm+n−1 − ρ1p

m+n−2 + ρ1 = ρ1 + kpm+n−1,

whenceρ2 − ρ1 = kpm+n−1. This implies thatq1 divides(ρ2 − ρ1)2.

If G(1)P is abelian, then the Main Theorem for Finite Abelian Groups applies

directly toG(1)P . Therefore,G(1)

P contains an elementary abelian subgroupA suchthat

A ∩ U (2)P = 1 and |A| = |U

(2)P |p2

.

Using the above argument, the sharper assertionq1 | (ρ2 − ρ1) follows. 2

L EMMA 12.76 Letz ∈ Σ be an element such thatp ∤ ordPz. If α ∈ U (i)P \U

(i+1)P ,

thenα(z) = z + z′ with z′ 6= 0, andρi = 1− ordPz + ordPz′.

Proof. Let a = ordPz. Thenα(z) = z + z′ with c ∈ K\0, and ordPz′ > a.Note thatz′ 6= 0, otherwisez ∈ Σα andp | a. Now, letx ∈ Σα with ordPx = bandgcd(a, b) = 1. Further, choose integersn1, n2 such thatn1a + n2b = 1. Asp | b implies thatp ∤ n1 and thatw = zn1xn2 is a uniformizing element atP, so

ρi = ordP(α(w)− w)

= ordP((z + z′)n1xn2 − zn1xn2)

= ordP(n1zn1−1z′wn1)

= (n1 − 1)a+ n2b+ ordPz′

=1− ordPz + ordPz′.

2

The following result is a refinement of Theorem 12.71.

482

THEOREM 12.77 Let γ be thep-rank ofΣ. Everyp-subgroupG of AutK(Σ) hasthe following properties.

(i) If γ ≥ 2, then|G| ≤ κp(γ − 1) with κp = p/(p− 2) for p ≥ 3, andκ2 = 4.

(ii) If γ = 1 andp ≥ 3, thenG is a cyclic group and|G| dividesγ − 1; also, Ghas no fixed place.

(iii) If γ = 1 andp = 2, then|G| ≤ 4(g − 1).

(iv) If γ = 0, then

|G| ≤ max

g,

4p

(p− 1)2g2

.

Proof. Let Σ′ = ΣG, n = |G| andλ = (γ′ − 1)/n whereγ′ denotes thep-rank ofΣ′. If G hass short orbits with lengthsℓ1, . . . , ℓs, then (12.29) reads

λ = γ′ − 1 +

s∑

i=1

(1− ℓi

n

). (12.55)

SinceG is ap-group,n/ℓi ≥ p holds fori = 1, . . . , s. Assume thatγ ≥ 2, thatis, λ > 0. Then (i) can be reworded asλ ≥ κ−1

p . If γ′ ≥ 2, (12.55) shows thatλ ≥ 1 > κ−1

p . Also, it shows that ifγ′ = 1 thens ≥ 1, and hence

λ ≥ 1− ℓi/n ≥ 1− p > κ−1p .

If γ′ = 0 andp ≥ 3, thens ≥ 2, and hence

λ ≥ −1 + 2(1− 1/p) = κ−1p .

If γ′ = 0 andp = 2, two cases occur: namely, either there are at least two shortorbits, one of length≥ 4, or there at least three short orbits. From (12.55),λ ≥ 1

4in the former case andλ ≥ 1

2 in the latter one. Thus (i) is established.Assume next thatγ = 1, that is,λ = 0. From (12.55), eitherγ′ = 0, or γ′ = 1.

If ζ = 1, no short orbit exists, andG must be cyclic by Theorem 12.59. Further,from Theorem 12.66,2g − 2 = |G|(2g′ − 2) which shows that|G| dividesg − 1.In particular,|G| ≤ g − 1. The caseg′ = 0 cannot occur whenp ≥ 3 since

23 ≤ 1− 1

/p ≤ 1− ℓ

/n < 1

for every short orbit of lengthℓ. Hence, ifp ≥ 3, then∑

(1 − ℓi/n) 6= 1. Thisproves (ii).

If γ′ = 0 andp = 2, there are exactly two short orbits both of length12 |G|, and

(12.19) reads

2g − 2 = |G|(2g′ − 2 + 12 (dP + dQ))

whereP andQ come from the two distinct short orbits. SincedP ≥ 2 anddQ ≥ 2are integers andg > 1, this gives (iii).

Finally, assume thatγ = 0, that is,λ < 0. Thenγ′ = 0 from (12.55). Ifthere were at least two short orbits, the right-hand side in (12.55) would be at least−1 + 2(1 + 1/p) ≥ 0 contradictingλ < 0. So, there is just one short orbit, andin this caseλ = −ℓ/n. But thenℓ = 1; that is,G fixes a placeP. From Theorem12.71, part (iv) follows. 2


REMARK 12.78 Theorem 12.77 shows that the order of aK-automorphism groupof an irreducible algebraic curve, whose order is a power ofp, is bounded above bya linear polynomial ing except when thep-rankγ is zero. A complete descriptionof the exceptions is given in Theorem 12.126.

THEOREM 12.79 Letg ≥ 2.

(i) If γ = 1 andp ≥ 3, thenAutK(Σ) is tame.

(ii) If p ≥ 5 and2 ≤ γ ≤ p− 2, then|AutK(Σ)| is not divisible byp.

Proof. Part (i) follows from Theorem 12.77 (ii). Also, Theorem 12.77 (i) showsthatγ ≤ p− 2, whence|G| ≤ p(p− 3)/(p− 2) < p. Therefore,|G| = 1 sinceGhas order a power ofp. This proves that AutK(Σ)| has nop-subgroups. 2

REMARK 12.80 Forp 6= 0, the fieldΣ is ordinary if its p-rankγ is equal tog. Forevery ordinaryΣ, the bound|AutK(Σ)| ≤ 84(g − 1)g holds.

THEOREM 12.81 Let p = 2. Assume thatG(1)P contains a cyclic characteristic

subgroupU . ThenGP has cyclic subgroupV such that[GP : V ] = [G(1)P : U ].

Proof. By Theorem 12.41,GP = G(1)P ⋊ H, whereH is a cyclic group of odd

order, andG(1)P is a normal2-subgroup ofGP . From the latter assertion, sinceU

is characteristic, it follows thatU is a normal subgroup ofGP . Therefore, the setV = UH is a subgroup ofGP . More precisely,V = U ⋊ H, andU is a cyclicSylow 2-subgroup ofV . Therefore,V = O(V ) ⋊ U . So, bothU andO(V ) arenormal subgroups, andV = O(V ) × U . Since bothO(V ) andU are cyclic,Vitself is cyclic. Also,

|GP | = |G(1)P |H| = [G

(1)P : U ]|U ||H| = [G

(1)P : U ]|V |,

whence (ii) follows. 2

12.10 K-AUTOMORPHISM GROUPS OF PARTICULAR CURVES

This section describes the groups of curves of low genus, namely rational and el-liptic curves, as well as those of hyperelliptic and Artin–Schreier curves.

Theorem 12.14 states thatAut(K(x)) ∼= PGL(2,K). The finite subgroups ofPGL(2,K) can be determined by the techniques used in Section 12.9, although adeeper result, namely Theorem 15.15, is also required. Herethe two main resultsare stated without proofs.

THEOREM 12.82 Let G be a non-trivial finite group ofK-automorphisms of arational function fieldK(x). Let s be the number of short orbits ofG on the set ofall places ofK(x), and letℓ1, . . . , ℓs be their lengths. ThenG is a group of thefollowing type:

(i) the cyclic groupZn of ordern with p ∤ n, s = 2, ℓ1 = ℓ2 = 1;

484

(ii) an elementary abelianp-group withs = 1, ℓ1 = 1;

(iii) the dihedral groupDn of order2n with p ∤ n, s = 2, ℓ1 = 2, ℓ2 = n, orp 6= 2, s = 3, ℓ1 = ℓ2 = n, e3 = 1;

(iv) the alternating groupA4 with p 6= 2, 3, s = 3, ℓ1 = 6, ℓ2 = ℓ3 = 4;

(v) the symmetric groupS4 with p 6= 2, 3, s = 3, ℓ1 = 12, ℓ2 = 4, ℓ3 = 6;

(vi) the alternating groupA5 withp = 3, s = 2, ℓ1 = 10, ℓ2 = 12, or p 6= 2, 3, 5,s = 3, ℓ1 = 30, ℓ2 = 2′, ℓ3 = 20;

(vii) the semi-direct product of an elementary abelianp-group of orderq with acyclic group of ordern with n | (q − 1), s = 2, ℓ1 = 1, ℓ2 = q;

(viii) PSL(2, q) with p 6= 2, q = pm, s = 2, ℓ1 = q(q − 1), ℓ2 = q + 1;

(ix) PGL(2, q) with q = pm, s = 2, ℓ1 = q(q − 1, ℓ2 = q + 1.

THEOREM 12.83 LetK = Fq and letG be the subgroup ofAut(K(x)) preserv-ingPΣ(Fq), the set of all places ofK(x) defined overFq. If G ∼= PGL(2, q), thenthe extensionK(x)/K(x)G has the following properties.

(i) Every place inPΣ(Fq) is wildly ramified withG0(P ) the semi-direct productof an elementary abelianp-group of orderq by a cyclic group of orderq −1, with G1(P ) an elementary abelianp-group of orderq, and withG2(P )trivial. All places inPΣ(Fq) form a single orbit of sizeq + 1 underG.

(ii) Every place defined overFq2 but not overFq is tamely ramified withG0(P )a cyclic group of orderq + 1; such places form a single orbit of sizeq2 − qunderG.

(iii) No other place ofK(x) ramifies.

The main result on Artin–Schreier curves in terms of the associated function fieldis the following theorem.

THEOREM 12.84 LetΣ/K(x) be an Artin–Schreier extension such that the genusof Σ is greater than1. Then theK-automorphism groupG of Σ/K(x) is an ex-tension of a cyclic group of orderp by a finite group ofK-automorphisms ofK(x)except in the following cases forp > 2 :

(i) Σ = K(x, y), (xp − x)(yp − y) = a with a ∈ K, andG is the semi-directproduct of an elementary abelianp-group of orderp2 with a dihedral groupof order2(p− 1);

(ii) Σ = K(x, y), y3 − y = i/(x(x − 1)) wherei2 = 2, andG is an extensionof a cyclic group of order2 byS4;

(iii) Σ = K(x, y), yp − y = xm with m | (p + 1), m < p + 1, andG is anextension of a cyclic group of orderm byPGL(2, p);


(iv) Σ = K(x, y), yp − y = xp+1; that is,Σ is the Hermitian function field overFp2 , andG ∼= PGU(3, p).

TheK-automorphism groups of elliptic curves are known. Here thefollowingresult is shown.

THEOREM 12.85 For an elliptic curveF , the following hold.

(I) TheK-automorphism group ofF is infinite, and it acts on the places as atransitive permutation group.

(II) For every placeP, the stabiliser ofP is finite. More precisely, if G is aK-automorphism group ofF , then

|GP | =

2, 4, 6 when p 6= 2, 3,2, 4, 6, 12 when p = 3,2, 4, 6, 8, 12, 24 when p = 2.

(12.56)

(III) For p = 2, the stabiliserG(1)P∞

is cyclic when|G(1)| ≤ 4, and it is the

quaternion group when|G(1)| = 8.

Proof. Let p 6= 2, 3. By Theorem 7.85,F is taken to be the plane cubic curve

v(Y 2 −X3 − uX − v)with 4u3 + 27v2 6= 0. As usual,Σ denotes the function field ofF . Let P∞be the place ofΣ arising from the branch ofF with centre atY∞. For anotherplaceP of Σ, let P = (a, b) be the centre of the corresponding branch ofF . It isstraightforward to show that the birational transformation ω(a, b) given by

x′ =

(y − bx− a

)2

− x− a, y′ =y − bx− a (x′ − a) + b

is aK-automorphism ofΣ. LetQ be the place ofΣ corresponding to the branchcentred at the the pointQ = (a,−b). Thenω(a, b) sendsQ to P∞. Henceω(a, b)−1 sendsP∞ to Q. Therefore, theK-automorphism group generated bythe aboveω(a, b) acts transitively on the set of all places ofΣ. The structure of thestabiliser of a placeP can be deduced from Lemma 12.26. In fact, Lemma 12.26shows that AutK(Σ)P contains nop-element, and hence it is tame. By Theorem12.46, AutK(Σ)P is cyclic. Therefore, Lemma 12.26 shows that the possibilitiesfor |AutK(Σ)P | are2, 3, 4, 6.

Now, letp = 3. By Theorem 7.85,Σ is assumed to be the function field of thenon-singular cubic curve

F = v(Y 2 −X3 − uX2 − vX + w)

with v3 6= u3w. To show (I), the previous argument still works, the only changebeing in the definition ofω(a, b) where−u is added inx′. Also, the stabiliserGof P∞ contains the above involutoryK-automorphismω. If G is tame, thenG isa cyclic group of order2 or 4 by Theorem 12.56. IfG is non-tame, thenG has anormal subgroupN of order3, and either|G| = 6 or |G| = 12.

486

Finally, let p = 2. Theorem 7.85 gives two possibilities. IfΣ is supersingular,then

F = v(Y 2 + uY +X3 + vX + w)

with u 6= 0 is a non-singular model ofΣ. It is possible to use the previous argument,provided thatω(a, b) is defined as follows:

x′ =

(y + b

x+ a

)2

+ x+ a, y′ =y − bx− a (x′ + a) + b+ u.

The stabiliserG of P∞ is non-tame, as it contains the involutoryK-automorphism

ω0 : x′ = x, y′ = y + u.

By Theorem 12.56, a Sylow2-subgroup ofG has order at most8. From (12.21),G has no element of order5. Hence, by Theorem 12.56,3 is the only possibilityfor the order of a non-trivialK-automorphism ofΣ of odd order. Hence|G| isan even divisor of24. From the proof of Theorem 13.6,GP∞

consists of lineartransformations of type,

ω : x′ = x+ d, y′ = y + ax+ b,

with a, b, d ∈ K. It is straightforward to show thatω is involutory if and only ifad = 0. As ω ∈ G, the conditionad = 0 implies thatd = b = 0, a = 1. It turnsout thatω0 is the unique involutory element ofGP∞

. In particular, if|G(1)P∞| ≤ 4,

thenG(1)P∞

is cyclic. If |G(1)P∞| = 8, Theorem 12.66 applied toG(1)

P∞implies that

|G(2)P∞| = 2. From this, the groupG(1)

P∞/G

(2)P∞

is elementary abelian of order4.

Therefore,G(1)P∞

is not cyclic, and hence a quaternion group of order8.Finally, if Σ is not supersingular, then

F = v(Y 2 +XY +X3 + uX2 + v),

with v 6= 0, is a non-singular model ofΣ. Again, the previous argument can beused with

ω(a, b) : x′ =

(y + b

x+ a

)2

+y + b

x+ a+ x+ a, y′ =

y + b

x+ a(x′ + a) + x′ + b,

ω0 : x′ = x, y′ = y + x,

to obtain the results. 2

REMARK 12.86 Theorem 12.85 implies that

|G| ≤

6 when p 6= 2, 3,12 when p = 324 when p = 2.

(12.57)

In each case, this upper bound is achieved by some elliptic elliptic curves. Forinstance, ifp 6= 2, 3, see the proof of Lemma 12.26. Here, a few examples aregiven forp = 2, 3.


Assume thatp = 3. LetF = v(Y 2 − X3 + X). If δ is a fourth root of unity,then both birational transformations,

ω : x′ = −x+ 1, y′ = δy,

ω′ : x′ = x+ 1, y′ = y,

areK-automorphisms ofΣ whose orders are4 and 3, respectively. The groupgenerated by them has order12.

Let p = 2, letF = v(Y 2 + Y +X3), and choose a cubic root of unityǫ in Σ.Then the birational transformation,

ω(i, k) : x′ = x+ ǫi, y′ = y + ǫ2ix+ ǫk,

with i = 0, 1, 2 andk = 1, 2 is aK-automorphism ofΣ. These, together with theK-automorphism,

ω : x′ = x, y′ = y + 1,

and the identity form a quaternion groupQ8 of order8. Also, the birational trans-formation,

ω′ : x′ = ǫx, y′ = y,

is aK-automorphism ofΣ of order3; this andQ8 generate a group of order 24.

Now, theK-automorphism groups of hyperelliptic curves are investigated. ByTheorem 12.47, such a group is finite. As in Section 7.11 two cases are distin-guished.

First, letK be of zero or odd characteristic. By Theorem 7.88,F is an irreducibleplane curvev(Y 2− f(X)), wheref(X) ∈ K[X] is monic, has degree2g+1, andits rootsα0, . . . , α2g are distinct. LetΣ = K(x, y) with y2 = f(x) be the functionfield Σ of F . The Weierstrass points are the places associated to the branches ofFcentred at the non-singular pointsUi = (αi, 0) with i = 0, . . . , 2g, together withthe place associated to the unique branch centred atY∞. As in Theorem 7.89,

w(X) =aX + b

cX + d, with ad− bc 6= 0,

is a rational function which maps the setα0, . . . , α2g,∞ onto itself. Such ra-tional functions constitute a subgroupW of PGL(2,K). Hence, ifw ∈ W fixesthree distinct elements from the setα0, . . . , α2g,∞, thenw(X) = X. Sinceg ≥ 2, this implies thatW is a finite group. From the proof of Theorem 7.89, theK-automorphism group ofF consists of all birational transformationsα of Σ suchthat one of the following hold:

(i) α(x) = ax+ b, with aX + b ∈W, α(y) = ǫy with ǫ2 = a2g+1;

(12.58)

(ii) α(x) =ax+ b

x− α0, with

aX + b

X − α0∈W,

α(y) = ǫy

(x− α0)g+1, with ǫ2 = (b+ aα0)

∏2gi=1 (α0 − αi). (12.59)

488

In particular,|AutK(Σ)| = 2|W |. More precisely, thehyperbolic involution, thatis, the birational transformationω : (x, y) 7→ (x,−y) is aK-automorphism ofΣsuch thatω ∈ Z(AutK(Σ)) andW ∼= AutK(Σ)/〈ω〉.

In even characteristic, let

F = v(Y 2 + h(X)Y + g(X)),

wheredeg h(X) ≤ g + 1 anddeg g(X) = 2g + 1. Then the function field ofF isΣ = K(x, y), with y2 + h(x)y + g(x) = 0.

If n = deg h(X) ≥ 1, let α1, . . . , αn denote the roots ofh(X), each countedwith multiplicity. As before,W is the subgroup of PGL(2,K) that preserves themultiset∞, α1, . . . , αn. The Weierstrass points are the places arising from thebranches centred at the non-singular pointsUi = (αi,

√αi) with i = 1, . . . n to-

gether with the place associated toY∞. From the proof of Theorem 7.95, AutK(Σ)consists of all birational transformationsα of Σ such thatα is one of the following:

(i) α(x) = w(x), α(y) = v(x)h(w(x)) +h(w(x))

h(x)y,

with w(X) ∈W andv(x)2 + v(x) =g(x)

h(x)2+

g(w(x))

h(w(x))2;

(ii) α(x) = w(x), α(y) = h(w(x)) +h(w(x))

h(x)y,

with w(X) ∈W andh(w(x))2

h(x)2g(x) + g(w(x)) = 0.

The centreZ(AutK(Σ)) of AutK(Σ) is non-trivial as thehyperbolic involution, thatis, the birational transformationω : (x, y) 7→ (x, y + 1) is aK-automorphism ofΣ lying in Z(AutK(Σ)). Therefore,W ∼= AutK(Σ)/〈ω〉, and|AutK(Σ)| = 2|W |.

If h(X) is constant, then AutK(Σ) fixes the unique Weierstrass point. Therefore,the following result is obtained.

THEOREM 12.87 LetF be a hyperelliptic curve. Then

(i) its hyperbolic involutionω is in the centre of theK-automorphism groupGofF ;

(ii) the quotient groupG/〈ω〉 is isomorphic to a subgroupW of PGL(2,K) thatpreserves the set, or for p = 2 the multiset, consisting of all Weierstrasspoints ofF ;

(iii) if F has at least three Weierstrass points, the identity is the onlyK-automor-phism ofF that fixes each Weierstrass point.

By Remark 12.15, all possibilities forW are known.Now, some particular classes of hyperelliptic curves are considered. The follow-

ing proposition is a special case of a more general result that is established later;see Theorem 12.124.

PROPOSITION 12.88 LetF be an irreducible hyperelliptic curve of genus2.


(i) Assume that eitherp 6= 2, or p = 2 andF has more than one Weierstrasspoint. LetG be aK-automorphism group ofF . Then

|G(1)P | ≤

1 for p = 0 andp ≥ 7,5 for p = 5,3 for p = 3,8 for p = 2.

(ii) If F has only one Weierstrass point andG is a abelianp-group, then|G| ≤ 8.

(iii) If G is a solvable group, then|G| ≤ 48.

Proof. By Theorem 12.71, (i) holds forp 6= 2, 3.To prove (i) for p = 3, Theorem 12.87 is relevant, asΣ has six Weierstrass

points. IfG had a subgroup of order at least9, then some non-trivial elements ofH would fix three Weierstrass points, in contradiction to Theorem 12.87 (iii).

To prove (i) forp = 2, let ω ∈ G, and Theorem 12.87 together with Remark12.15 are used. IfΣ has three Weierstrass points, then AutK(Σ)/〈ω〉 is a subgroupof the symmetric group of degree3. In particular, the order of a Sylow2-subgroupS2 of G is at most4. Similarly, |G/〈ω〉| ≤ 2, and hence|S2| ≤ 4 whenΣ has twoWeierstrass points.

To prove (ii), letp = 2. The essential tools are Theorem 12.85 and Lemma12.43. LetM be a non-trivial subgroup ofG. Suppose first thatΣM is elliptic.SinceM fixes the unique Weierstrass pointP, Theorem 12.66 applied toM impliesthat |M | = 2. Therefore,G/M is an abelianK-automorphism group of an ellipticcurve. Theorem 12.85 (II) implies that|G/M | ≤ 4, whence it follows that|G| ≤ 8.Finally, in the case that the fixed field of every non-trivial subgroup ofG is rational,Lemma 12.43 implies that|G| ≤ 4.

In proving (iii), Theorem 12.87 and Remark 12.15 play a role since they togetherwith the solvability condition imply that, ifΣ has more than two Weierstrass points,then|G|/〈ω〉| ≤ 24. Therefore, (iii) holds forp 6= 2,

To complete the proof forp = 2 it remains to examine the case thatΣ has exactlytwo Weierstrass points, sayP1 andP2. Then,G has a subgroupU of index2 fixingbothP1 andP2. SinceG containsω, the groupU = U/〈ω〉 is a subgroup ofPGL(2,K) fixing two objects, namely the placesP ′

1 andP ′2 of Σω lying underP1

andP2 in the extensionΣ/Σω. From Theorem 12.14,U is a cyclic group of oddorder. By Theorem 12.56,|U | ≤ 4g + 2 = 10. Therefore,|G| ≤ 18, and (iii)follows. 2

EXAMPLE 12.89 Let p > 2. For a powerq of p, the curve

F = v(Y 2 − g(X))

with g(X) = Xq−X is a special case of Example 5.58. So,F has genus12 (q−1).The fieldFq is viewed as a simple extension of the prime fieldFp of K, andFq

is assumed to be a subfield ofK. The distinct roots ofg(X) are elements inFq.Hence the groupW consists of all rational functions

w(X) =aX + b

cX + d, with a, b, c, d ∈ Fq, ad− bc 6= 0.

490

Therefore,W ∼= PGL(2, q) andW acts onFq ∪ ∞ as PGL(2, q) acts naturallyon PG(1, q). If ω : x′ = x, y′ = −y is the central involution in AutK(Σ),then AutK(Σ)/〈ω〉 ∼= PGL(2, q). It may be noted that ifq = 5 theng = 2 and|AutK(Σ)| = 240. This shows that the solvability condition in Proposition 12.88cannot be relaxed.

The following result is a characterisation of the curve in Example 12.89.

PROPOSITION 12.90 Let p > 2 and q = ph = 2g + 1. If W ∼= PSL(2, q) inTheorem 12.87 thenF is birationally equivalent to the curve in Example 12.89.

Proof. The notation is the same as in the proof of Theorem 12.87. Takeα0 = 0,α1 = 1 andf(X) monic, after a suitable substitution of typeX ′ = uX + v, withu, v ∈ K andu 6= 0.

SinceW ∼= PSL(2, q), the polynomialf(X) is invariant under the substitutionX 7→ cX for every 1

2 (q− 1)-st root of unityc ∈ K; it should be noted thatc ∈ Fq.Sinceα1 = 1, andcα1, . . . , cα2g = α1, . . . , α2g, this shows that everyc withc(q−1)/2 = 1 is a root off(X). Also, sinceW ∼= PSL(2, q) sof(X) is invariantunder the substitutionX 7→ X + 1. Choose a12 (q − 1)-st root of unityu ∈ K insuch a way thatv = u + 1 is not also such a root of unity. Then the elementscvwith c ranging over all12 (q − 1)-st roots of unity provide the remaining12 (q − 1)non-zero roots off(X). Hence, the roots off(X) are precisely the elements ofFq. Thus,f(X) = Xq −X. 2

EXAMPLE 12.91 Let p = 2, F = v(Y 2 + Y + X5) andΣ = K(F). The fieldF16 is viewed as a subfield ofK and as an extension of the prime fieldF2 of K.Sinceh(X) = 1, the placeP∞ of Σ arising from the unique branch ofF centred atY∞ is the only Weierstrass point. Therefore, AutK(Σ) fixesP∞. A straightforwardcomputation shows the following results.

For everyb, c ∈ F16 with c2 + c+ b5 = 0, the birational transformation,

αb,c(x) = x+ b, αb,c(y) = b8x2 + b4x+ c+ y,

of Σ is aK-automorphism fixingP∞. For b = c = 0, the identity is obtained;otherwise the order ofαb,c is either2 or 4, according asb10 + b5 is 0 or 1. The set,

αb,c | b, c ∈ F16; c2 + c+ b5 = 0,

is a groupU of order32 containing11 elements of order2 and20 elements oforder4. Such a group is uniquely determined up to isomorphism, and hence it isisomorphic toC4×D4. SinceU fixesP∞, Theorem 12.71 implies thatG(1)

P = U .Further, for everyd ∈ F16 with d5 = 1, the birational transformation,

αd(x) = dx, αd(y) = y,

is aK-automorphism ofΣ fixing P∞. These form a subgroupC5 of AutK(Σ) oforder5. By Theorem 12.56,C5 is a maximal subgroup of odd order in the stabiliserof P. Therefore, AutK(Σ) = U ⋊ C5; in particular, AutK(Σ) ∼= (C4 ⋊ D4) ⋊C5, and|AutK(Σ)| = 160. This shows that the Proposition 12.88 can fail whenAutK(Σ) has a fixed place.


EXAMPLE 12.92 Let q be a power ofp with p > 2. The irreducible plane curve

Fi = v(X(Y q − Y )− (aX2 + biX + 1)), a ∈ Fq\0, (12.60)

wherebi is a fixed element ofFq with traceTFq/Fp(bi) = i is a hyperelliptic curve

of genusp−1. Note thatFi = v(Y q−Y −R(X)) withR(X) = aX+X−1 + bi.ThenFi has the following properties:

(i) it is hyperelliptic;

(ii) the Fq-rationalK-automorphism group ofF is the direct product of the hy-perbolic involution by a dihedral groupDp of order2p;

(iii) the fixed places ofDp areFq-rational.

12.11 FIXED PLACES OF AUTOMORPHISMS

Some simplification in the statements of the results in this section is made by usingthe following notation. For a non-trivialK-automorphismα, the integerρ(α) de-notes the number of fixed places ofα in the action ofα on the setPΣ of all placesof Σ. More generally, for a non-trivial subgroupG of AutK(Σ), let ρ(G) be thenumber of places ofΣ fixed by everyK-automorphism inG. If G has prime orderandG = 〈α〉, thenρ(G) = ρ(α), but this is not always true for subgroupsG ofcomposite order.

L EMMA 12.93 Let α ∈ AutK(Σ) be a non-trivialK-automorphism such thatρ(α) > 2n, wheren = [Σ : K(z)] for a non-constant elementz of Σ. Thenα(z) = z, andn is divisible byordα. If, in addition, n is prime, thenordα is alsoa prime andΣα is a rational function field.

Proof. Considerh = α(z) − z, and assume thath 6∈ K. Then every pole ofhis either a pole ofz, or a pole of its imageα(z). Sincen = [Σ : K(α(z)], so[Σ : K(h)] ≤ 2n− r, wherer is the number of poles ofz fixed byα. In particular,r ≤ n. Each fixed place ofα that is not a pole ofh is a zero ofh. Thush hasρ(α) − r > 2n − r zeros. But this is a contradiction which shows thath ∈ K.More precisely,h = 0; that is,α(z) = z, becauseα fixes some places which arenot poles ofz.

To show the other assertions, choose a placeP fixed byα, and letP ′ be a placelying underP with respect to the extensionΣ/Σα. ThenP is completely ramified.Hence[Σ : γα] = ordα sinceeP = ordα by Theorem 12.39. From the relation,

[Σ : K(z)] = [Σ : Σα] · [Σα : K(z)],

n is the product of ordα and[Σα : K(z)]. If n is prime, this givesn = ordα andΣα = K(z). 2

L EMMA 12.94 Let α ∈ AutK(Σ) be a non-trivialK-automorphism of ordern,whereΣ has genusg. Then

ρ(α) ≤ 2 +2g

n− 1. (12.61)

492

Proof. Sinceρ(α) ≤ ρ(αj) for j ≥ 1, Remark 12.57 applied toG = 〈α〉 gives that

2g − 2 ≥ n(2g′ − 2) + ρ(α)(n− 1),

whereg′ is the genus of the function fieldΣα. Sinceg′ ≥ 0, the result follows. 2

L EMMA 12.95 If the ordern of aK-automorphismα is a prime different fromp,then

ρ(α) = 2 +2g − 2g′n

n− 1(12.62)

whereg andg′ are the genera ofΣ andΣα. In this case, equality holds in (12.61)if and only ifg′ = 0.

Proof. If n is prime, thenρ(α) = ρ(αj) for j ≥ 1. Then, the proof of Lemma12.94 gives the result. 2

THEOREM 12.96 If a prime numbern is the order of aK-automorphism ofΣ,then one of the following holds:

(i) n ≤ g + 1;

(ii) n = 2g − 1 andg > 2;

(iii) n = 2g + 1.

Proof. Using the formula (12.28) several times, an upper bound forn can be ob-tained.

If g′ ≥ 2, then2g − 2 ≥ 2n + ρ(α)(n − 1) ≥ 2n; hencen ≤ g − 1 yieldingn < g + 1. Again, by (12.23),g′ = 1 andρ(α) = 0 cannot occur simultaneously.If g′ = 1 and ρ(α) = 1, thenn = 2g − 1. If g′ = 1 and ρ(α) ≥ 2, then2g − 2 ≥ 2(n − 1), and hencen ≤ g, yieldingn < g + 1. Similarly, by (12.23),g′ = 0 andρ(α) ≤ 2 cannot occur simultaneously. Ifg′ = 0 andρ(α) = 3, thenn = 2g+1. If g′ = 0 andρ(α) = 4, then2g−2 ≥ −2n+4(n−1), which impliesthatn ≤ g + 1. This completes the proof. 2

REMARK 12.97 In the extremal case thatn = 2g + 1, all possibilities are known.Letn be as in Lemma 12.96. Thenn = 2g+1 holds if and only ifΣ is the functionfield of the curveF = v(F (X,Y )), whereF is one of the following:

(i) F (X,Y ) = Y m−r(Y − 1)r −Xn, with 1 ≤ r < m ≤ g + 1;

(ii) F (X,Y ) = Y 2 −Xn +X with n = p.

L EMMA 12.98 LetG be aK-automorphism group ofΣ with only one short orbito. If Σ has genusg ≥ 2, then|o| divides2g − 2.

Proof. Let g′ be the genus ofΣG. If the stabiliserGP of a placeP ∈ o is tame thenTheorem 12.54 reads2g − 2 = |G|(2g′ − 2) + |G| − |o|. As |G| = |GP | |o|, theassertion follows. In the non-tame case, such an equation may not exist. However,one still has in (12.38) thatdP = dQ for P,Q ∈ o and thatdQ = 0 for Q 6∈ o.Hence|o| dividesd, and the assertion can be proven as in the tame case. 2


The next two theorems are classical results linking Weierstrass points and fixedplaces ofK-automorphisms. For their validity in positive characteristic, some hy-potheses are needed.

THEOREM 12.99 Suppose the following hold:

(a) the canonical series has classical order sequence;

(b) p does not divide the order of theK-automorphismα.

If α has more than four fixed places, then every fixed place ofα is a Weierstrasspoint.

Proof. It suffices to consider the case thatn = ordα is prime, since a suitablepowerαj of α has prime order, and thenα may be replaced byαj , as every fixedplace ofα is fixed byαj .

Let P be a fixed place ofα, and take a placeP ′ of Σα underP. There existsz ∈ Σα such that div(z)P′ = m′P ′ with m′ ≤ g′ + 1. Theorem 12.35, togetherwith (7.1), shows that

ordPz = n · ordP′z = n ·m′ ≤ (g′ + 1)n,

and thatP is the unique pole ofz. On the other hand, Theorem 12.66 applied tothe subgroupG generated byα, together with the hypothesisρ(α) > 4, gives

2g − 2 = n(2g′ − 2) + (n− 1)ρ(α) > n(2g′ − 2) + 4(n− 1),

showing thatg+ 1 > n(g′ + 1). Hence ordP z < g+ 1. Since the canonical seriesis classical,P is a Weierstrass point. 2

THEOREM 12.100 Let α be a non-trivialK-automorphism contained in a tameK-automorphism group ofΣ. Suppose thatg > n2g′ +(n− 1)2, wheren = ordαandg′ is the genus ofΣα. If G also contains a non-trivialK-automorphismβ withρ(β) > 2n(g′ + 1), then the following hold:

(i) each fixed place ofβ is a fixed byα, andord β ≤ ordα = n;

(ii) if ord β = n, thenβ ∈ 〈α〉;

(iii) 〈β〉 is a normal subgroup ofG;

(iv) if n = 2, thenβ ∈ Z(G).

Proof. LetP1, . . . ,Pρ(α) denote the fixed places ofα. From the proof of Theorem12.99, for eachj there is a non-constantzj ∈ Σα such that div(zj)∞ ≤ n(g′ + 1).

From Proposition 12.93,zj = β(zj) for eachj, and thusPj = Pβj . So the first part

of (i) holds. Also,ρ(β) ≥ ρ(α).Rewriteg > n2g′ + (n− 1)2 in the form,

g − ng′n− 1

>g + n− 1

n. (12.63)

494

Now,

ρ(β) ≥ ρ(α) = 2 +2(g − ng′)n− 1

> 2 + 2(g + n− 1)

n

=2 +2g

n+ 2

(n− 1)

n> 2 +

2g

n.

This together with (12.61) gives ordβ ≤ n, completing the proof of (i).Bothα andβ are in the stabiliser ofP1. SinceG is tame,GP1

is a cyclic group,by Theorem 12.46. So, ordα = ord β yields (ii).

Next, supposeδ ∈ G is conjugate toα underG. Then ordδ = ordα and, by theinequality ong,

ρ(δ) = ρ(α) = 2 + 2(g − ng′)(n− 1) > 2n(g′ + 1).

Thus, by (ii),δ ∈ 〈α〉 showing that〈α〉 is a normal subgroup of AutK(Σ).Finally, (iv) is a consequence of (iii). 2

A finite groupG admits apartition if it contains a set of subgroupsG1, . . . , Gksatisfying the properties:

G =⋃k

i=1 Gi, Gi ∩Gj = 1.

The subgroupsGi are thecomponentsof the partition. A counting argument showsthatn =

∑ki=1 |Gi| − (k − 1).

THEOREM 12.101 LetG be a subgroup ofAutK(Σ) of ordern which has a parti-tion with componentsG1, . . . , Gk,where letni = |Gi|, for i = 1, . . . k. Letg, g′ g′ibe the genera ofΣ, ΣG, ΣGi for i = 1, . . . , k. Then

(k − 1)g + ng′ =∑k

i=1nig′i.

Proof. LetG0 = G andn0 = n. LetDi = D(Σ/ΣGi) denote the different divisorof the extensionΣ/ΣGi , for i = 0, 1, . . . , k. ThenD0 =

∑ki=1Di. From Corollary

12.49,

2g − 2 = ni(2g′i − 2) + degDi,

for i = 0, 1, . . . , k. Summing overi = 1, . . . k and subtracting fori = 0 sincedegD0 =

∑k1=1 degDi, gives that

(k − 1)(2g − 2) =∑k

i=1 ni(2g′i − 2)− n(2g′ − 2).

Hence

(k − 1)g=∑k

i=1 nig′i − ng′ − n+ (k − 1)−∑k

i=1 ni.

Sincen =∑k

i=1 ni − (k − 1), the result follows. 2

REMARK 12.102 Three important examples of finite groups with a partition arethe groups PSL(2, q), PGL(2, q), Sz(q).


12.12 LARGE AUTOMORPHISM GROUPS OF FUNCTION FIELDS

In this section Theorem 12.53 is refined.

THEOREM 12.103 In Theorem 12.53, the upper bounds can be sharpened for thecases(i), (ii), (iii) : by

(i) |G| < 24g2;

(ii) |G| < 16g2;

(iii) |G| < 8g3.

Proof. The notation used in the proof of Theorem 12.53 is maintained. Assumefirst that (i) occurs. Thend′1 = d′2 = 1

2 , d′3 ≥ 1, and hence

d′ − 2 =dQ3− eQ3

eQ3

. (12.64)

Write eQ3= e = e1e

′Q3

with e1 = |G(1)Q3| ≥ p andp ∤ e′Q3

. By Lemma 12.73,dQ3≥ e+ e1 − 2. Sincep ≥ 3, from (12.64),

d′ − 2 ≥ e− 1 + e1 − 1− ee

=1

e′Q3

(1− 2

e1

)≥ 1

e′Q3

(1− 2

p

)≥ 1

3e′Q3

.

From (12.20) and (12.24),

|G| ≤ 3e′(2g − 2) ≤ 3(4g + 2)(2g − 2) < 24g2.

Now, assume that (ii) occurs. There are two short orbits, both non-tame. Asabove, forp ≥ 3,

d′ − 2 =dQ1− eQ1

eQ1

+dQ2− eQ2

eQ2

≥ 1

3e′Q1

+1

3e′Q2

≥ 2

3(4g + 2),

whence|G| ≤ 3(2g + 1)(2g − 2) < 12g2.For p = 2, a further argument is required. Since|GQ1

| = eQ1, the Hilbert

Different Formula (12.38) implies that

dQ1= |GQ1

| − 1 + |G(1)Q1| − 1 +

∑i≥2(|G

(i)Q1| − 1)

= eQ1+G

(1)Q1− 2 +

∑i≥2(|G

(i)Q1| − 1).

Since the same holds forQ2, so

d′ − 2 =∑2

i=1

dQi− eQi

eQi

=∑2

j=1

1

eQj

(|G(1)

Qj| − 2 +

∑i≥2(|G

(i)Qj| − 1)

).

(12.65)Three cases (a), (b), (c) are investigated separately.

(a) For|G(1)Q1| ≥ 4 and, similarly foreQ2

≥ 4, (12.65) shows that

d′ − 2 ≥|G(1)

Q1| − 2

eQ1

=|G(1)

Q1|

eQ1

(1− 2

|G(1)Q1|

)≥ 1

2·|G(1)

Q1|

eQ1

.

496

Then (12.24) together with (12.20) gives the bound,

|G| ≤ 2(4g + 2)(2g − 2) < 16g2.

(b) The two equations|G(1)Q1| = |G(1)

Q2| = 2 and |G(2)

Q1| = |G(2)

Q2| = 1 cannot

occur simultaneously because of (12.65) andd′ − 2 > 0.(c) If |G(1)

Q1| = |G(1)

Q2| = 2, and either|G(1)

Q1| = 2 or |G(2)

Q2| = 2, from (12.65),

d′ − 2 ≥ 1

eQ1

=1

2·|G(1)

Q1|

eQ1

,

whence again|G| < 16g2, by (12.24) and (12.20).To investigate (iii), two cases are distinguished according as the unique short

orbit consists of just one place or more.

In the former case,G = GP . If F1 = ΣG(1)P is not rational, then|G| ≤ g(4g+2)

by Theorems 12.56 and 12.71. So, assume thatF1 is rational. It is shown thatG = G

(1)P . Assume on the contrary thatG contains a tame subgroupH whose

order is prime top. Choose a non-trivialK-automorphismα in H. Sinceα is inthe normaliser ofG(1)

P , it induces aK-automorphismα′ of F1. By Theorem 12.14(ii)(d), α′ has at least two fixed places, and hence one of them lies under aplaceQdistinct fromP. So the group generated byα together withG(1)

P has a non-trivialstabiliser atQ. Then the orbit of such a placeQ under the action ofG is short. ButthenG has at least two short orbits, a contradiction. Hence,G = GP . Now, fromTheorem 12.71,|G| = |G(1)

P | ≤ 8g2.In the latter case,G does not fixP. Let o denote theG-orbit of P, that is, the

unique short orbit ofG. Then|o| > 1, anddegD(Σ/ΣG) = |o|degD(Σ/ΣGP ).Since|G| = |GP ||o|, this yields the relation,

2g − 2 = |o|(degD(Σ/ΣGP )− 2|GP |). (12.66)

Therefore,|o| ≤ 2g−2. If F1 is not rational, then|G(1)P | ≤ g, by the first statement

of Theorem 12.71. This together with Theorem 12.56 give|GP | ≤ g(4g + 2),whence

|G| ≤ g(2g − 2)(4g + 2) < 8g3.

Assume thatF1 is rational. Then

degD(Σ/ΣGP ) = |GP | − |G(1)P |+ degD(Σ/ΣG

(1)P ) = |GP |+ |G(1)

P |+ 2g − 2.

Thus (12.66) reads

2g − 2 = |o|(2g − 2 + |G(1)P | − |GP |).

If |o| = 2g − 2, then2g − 2 + |G(1)P | − |GP | = 1 with GP = G

(1)P ⋊H, whence

2g − 2 = (|H| − 1)|G(1)P |+ 1.

This implies thatH is non-trivial. Also,2g − 2 > 12 |H||G

(1)P | = 1

2 |GP |, whence

|G| = |o||GP | < 2(2g − 2)2 < 8g3.


So,let|o| ≤ g − 1. If F1 is rational andG(1)P has a short orbit disjoint fromP, then

Theorem 12.71 (ii) again gives the desired upper bound,|G| < 8g3. Otherwise,ocontains a long orbit ofG(1)

P , and|o| ≥ 1 + |G(1)P |. As |o| ≤ g − 1, this shows that

|G(1)P | < g. As before, it follows that|G| < 8g3 and this completes the proof of

(iii). 2

The methods in this study ofG(1)P may also be used to investigate case (iv) of

Theorem 12.53. But the upper bound on|G| which may be shown in this way isonly aboutg5.

Therefore, a substantial improvement is needed. This requires a careful analysisusing deeper group-theoretical results. The final result isstated in Theorem 12.113.

In case (iv),G has two short orbits, one tame and another non-tame. Choose aplaceP in the non-tame orbit, and defineH again by meansGP = G

(1)P ⋊H with

|H| prime top. As before, writeF1 for ΣG(1)P , andF2 for ΣG

(2)P . The following

five cases are treated separately.

(iv.1) F1 is not rational.

(iv.2) F1 is rational, and there is a placeR distinct fromP such that the stabiliserofR in GP has orderpt with t ≥ 1.

(iv.3) F1 is rational, and there is a placeR distinct fromP such that the stabiliserofR in GP has orderhpt with t ≥ 1, h > 1 andh prime top.

(iv.4) F1 is rational, no non-trivial element ofG(1)P fixes a place distinct fromP,

and there is a placeR distinct fromP but lying in the orbit ofP in G suchthat the stabiliser ofR in GP is trivial.

(iv.5) F1 is rational, no non-trivial element ofG(1)P fixes a place distinct fromP,

and, for every placeR distinct fromP but lying in the orbit ofP in G, thestabiliser ofR in GP is non-trivial.

Before starting with (iv.1), another closed formula for theorder ofG is deduced.Choose a placeQ from the tame orbit ofG. ThendQ = eQ − 1. From Theorem12.66,

|G| = 2(g − 1)|G(1)

P | |H| |GQ|N

, (12.67)

where

N = 2g1 |GP | |GQ|+ dP |GQ| − |GP | |GQ| − |GP |. (12.68)

If ΣG is rational, which is the relevant case, thenN = |GQ|(dP − |GP |)− |GP |.Case(iv.1). If F1 is not rational, then its genusg1 ≥ 1. SinceP is fixed byG(1)

Pand|G(1)

P | is a power ofp,

dP ≥ 2(eP − 1) = 2(|G(1)P | − 1).

Hence, by Theorem 12.66 applied toG(1)P ,

2g − 2 ≥ |G(1)P |(2g1 − 2) + 2(|G(1)

P | − 1).

498

Thus, |G(1)P | ≤ g/g1. As the groupH ∼= GP/G

(1)P can be viewed as aK-

automorphism group ofF1, Theorem 12.56 shows that|H| ≤ 4g1 + 2. Also,N ≥ 6 since

N = |GP |2g1(|GQ| − 1) + |GQ|(dP − |GP |)with dP ≥ |GP |. By (12.67),

|G| ≤ 26 (g − 1)(4g + 2)(4g1 + 2)g/g1 <

86g(g − 1)(g + 1) (4 + 2/g1) < 8g3.

(12.69)Case(iv.2). Here,F1 is rational, and the orbit ofR underH has size|H|. Let

M denote the subgroup ofG(1)P fixingR. Then|M | = pt. Hence,

d(Σ/F1) ≥ 2(|G(1)P | − 1) +

|G(1)P |pt

(2pt − 2) |H|.

By Theorem 12.66,

g ≥ |G(1)P |(1− 1

pt ) |H| ≥ 12 |G

(1)P | |H|.

If |GQ| ≤ 2g + 2, this together with (12.67) implies that

|G| ≤ 2(g − 1)(2g + 2)2g/N ≤ 8(g3 − g) < 8g3.

If |GQ| > 2g+2, a much better upper bound is obtained from (12.67), as|GQ| < Nfollows from |GQ| > 2g + 2. That|GQ| > 2g + 2 implies |GQ| < N is shown inthe following way. Since|GP |/|GQ| ≤ 2g/(2g + 2) < 1,

N > |GQ|(dP − |GP | − 1).

On the other hand, Theorem 12.66 can be written as follows:

dP − |GP | − 1 = −2 +∑

i≥1 (|G(i)P | − 1).

Therefore,

N

|GQ|> |G(1)

P | − 3 + |G(2)P | − 1 + |G(3)

P | − 1.

If |G(1)P | ≥ 4, in particular forp ≥ 5, the assertion follows.

To show that this holds true for|G(1)P | ≤ 3, a little more is needed. For|GP | ≤ 3,

(12.67) implies that

|G| ≤ 3(2g − 2)(4g + 2)

which is smaller than8g3 for g ≥ 2. Therefore, let|GP | ≥ 4 and|G(1)P | ≤ 3.

If |G(1)P | = 3 and|GP | = 6, thenN = |GQ| − 6. Also g ≥ 4 and|GQ| > 10.

From (12.67),

|G| = 12(g − 1)|GQ|

|GQ| − 10< 36(g − 1) < 8g3.

If either |G(1)P | = 3 and|GP > 6, or |G(1)

P | = 2 and|GP | > 4, then

|H| > |G(1)P | − 1 with GP = G

(1)P ⋊H.


SinceG(1)P is a normal subgroup ofGP , this implies the existence of a non-trivial

elementα ∈ H in the centralizer ofG(1)P . From Theorem 12.68 (i),G(1)

P = G(2)P

whenp = 3, andG(1)P = G

(2)P = G

(3)P whenp = 2. Therefore,|GQ| < N . This

completes the proof for (iv.2).Case(iv.3). The stabiliser ofR in GP contains a non-trivial subgroupT of

GP,R of orderm. A subgroupH1 conjugate toH in GP containsT . Now, look

at the action ofT on F1, or, equivalently, on the set of all orbits ofG(1)P . By

Theorem 12.14 (ii)(d),T preserves exactly two of such orbits: one is the trivialorbit consisting ofP and the other is the orbito(R) containingR.

SinceH1 is cyclic and its order is prime top, a generator ofH1 has two invariantorbits ofG(1)

P , again by(iv) of Theorem 12.14. Therefore,P ando(R) are the

invariant orbits ofH1, as well. SinceGP = G(1)P ⋊H1, the same remains valid for

GP . So,|GP | = |GP,R| · |o(R)| with |o(R)| a power of ofp. Hence, a conjugateof H in GP , sayH2, fixesR.

Now, it is shown thatH fixes a place ofo(R). This is certainly true wheno(R)

coincides withR becauseGP = G(1)P ⋊ H yieldsGP = G

(1)P ⋊ H ′. Assume

that |o(R)| > 1. Then |o(R)| is a power ofp. SinceGP is transitive ono(R),the stabiliserGP ∩ GR of R has orderpt|H| with pt|o(R)| = |G(1)

P |. LetM be

the normalp-subgroup ofGP ∩ GR of orderpt; thenM = G(1)P ∩ G(1)

R . Since|M | is prime to[(GP ∩ GR) : M ], it has a complementH ′ in GP ∩ GR, that is,GP ∩GR = M⋊H ′ with |H ′| = [(GP ∩GQ) : M ]. In particular,H ′ ∼= H. SinceH ′ fixesP, soH andH ′ are conjugate underGP . Choose an elementα ∈ GPsuch thatαH ′α−1 = H. ThenRα ∈ o(R) is a fixed place ofH.

After replacingR withRα, it may assumed thatH fixesR. A first consequenceis the following result.

L EMMA 12.104 If o(R) is larger thanR, then|G(1)P | is divisible byp2.

Proof. If |G(1)P | = p, thenG(1)

P = M, and henceG(1)P = G

(1)R . In other words,G(1)

PfixesR. But then the wholeGP fixesR, sinceGP is generated byG(1)

P togetherwith H, andH fixesR. 2

Now, upper bounds on|H| and on|G(1)P | only depending upong are given.

L EMMA 12.105 Let (|M |, |H|) 6= (3, 2). Then

|G(1)P | ≤

43g, wheno(R) 6= R;

g + 1, wheno(R) = R;

|H| ≤g + 1, wheno(R) 6= R;g, wheno(R) = R.

Proof. First, assumeo(R) to be larger thanR. Then the non-tameG-orbit, that is,theG-orbit ofP, contains a placeS distinct fromP andR. By Theorem 12.66 andLemma 12.73,

2g − 2 ≥ −2|G|+ dP + dR + dS ≥ v − 1 + |H|.From Lemma 12.75,2g − 2 ≥ 2|H| − 2 whenceH ≤ g + 1.

500

Further,R contributes todegD(Σ|F1) with at least2(|M | − 1). SinceGP istransitive ono(R), the contribution todegD(Σ|F1) due to the places ino(R) isat leastc ≥ 2(|M | − 1)|o(R)| = 2(|M | − 1)|G(1)

P ||M |. On the other hand, by

Theorem 12.66 applied toG(1)P ,

2g − 2 ≥ −2G(1)P + 2G

(1)P − 2 + c,

whence2g ≥ c. Hence|G(1)P | ≤ 4

3g when |M | ≥ 4. Let |M | = 2. ThenHis contained in the centraliser ofM . By Lemma 12.68 (i),R contributes at least2(|M | − 1) + |M | − 1 ≥ 3 to degD(Σ/F1) at every place ino(R). By the aboveargument,c ≥ 3

2 |G(1)P | whence|G(1)

P | ≤ 43g. If |M | = 3 and|H| > 2, then some

non-trivial element ofH is in the centraliser ofM . In this case, the contribution ofR due to the places ino(R) is at least2(|M | − 1) + |M | − 1 ≥ 6, and the aboveargument givesc ≥ 2|G(1)

P |, whence|G(1)P | ≤ g ≤ 4

3g.Now, assumeo(R) = R. SinceGP fixes bothP andR, no non-trivial ele-

ment inGP can fix a further place. By Theorem 12.66 applied toGP ,

2g − 2 = −2|GP |+ 2[|GP | − 1 +∑

i≥1 (|G(i)P | − 1)).

Thus,g ≥ (ρ1 − 1)(|G(1)P | − 1), whence|G(1)

P | ≤ g + 1. By (b) of Lemma 12.75,

|H| ≤ (ρ1 − 1)(|G(1)P | − 1) which implies that|H| ≤ g. 2

It is possible to obtain the upper bound|G| < 8g3 from Lemma 12.105 underthe hypothesis that the size of the orbit ofP underG does not exceed6(g − 1).The next two lemmas show that this occurs in many cases; for instance, whenG(2)

Pis trivial, apart from a few exceptions described in the following lemma. However,such exceptional curves do not provide counterexamples forthe bound|G| < 8g3.

L EMMA 12.106 If the second ramification groupG(2)P is trivial and if the orbit of

P underG has sizer ≥ 6(g − 1), then|G| < 8g3.

Proof. From |G(2)P | = 1,

N = |GQ|(|G(1)P | − 2) + |G(1)

P ||H|.ThusN > 0 implies|GQ| = |H|+ ǫ with ǫ > 0. This together with the relation,

|G(1)P | − 2 = v = (µ+ 1)|H| − 1,

given in Lemma 12.75 (i), shows the following forµ ≥ 1 andǫ ≥ 2:

r

2g − 2=

|H|+ ǫ

|H|[(µ+ 1)ǫ− 2]− ǫ ≤|H|+ ǫ

|H|(2ǫ− 2)− ǫ ≤|H|+ 2

2|H| − 2≤ 2.

If ǫ = 1, then

r

2g − 2=

|H|+ 1

|H|(µ− 1)− 1≤ |H|+ 1

|H| − 1≤ 3.

If µ = 0, then|H| = |G(1)P | − 1 and, for|H| ≥ 6,

r

2g − 2=

|H|+ ǫ

|H|(ǫ− 2)− ǫ ≤|H|+ 3

2|H| − 3≤ 3,


while, for |H| = |G(1)P | − 1 < 6,

2 ≤ |H| ≤ 4.

In the last case,r ≤ 6(g−1) holds provided that either|GQ| < 7 and3 ≤ |H| ≤ 4,or |GQ| > 8 and |H| = 2. Otherwise, by Lemmas 12.104 and 12.105, one ofthe cases in Table 12.1 occurs. Since the first and the last case might provide a

Table 12.1 Order of automorphism group

|G(1)P | |H| |GQ|

r

2g − 2|G| o(R) 6= R o(R) = R

5 4 7 7 280(g − 1) impossible g = 4

4 3 7 7/2 84(g − 1) g ≥ 3 g = 3

3 2 7 7 84(g − 1) impossible g = 2

counterexample, some more is needed to complete the proof. These two cases aretreated separately, beginning with first. Here,|G| = 840 = 23 · 3 · 5 · 7, and theorbit o(Q) of Q, that is the tame orbit ofG, has size120. The key idea is to provethatGQ has some more fixed places ino(Q).

LetN be a minimal normal subgroup ofG. If N is an elementary abelian groupof prime power orderwk, thenwk ∈ 23, 3, 5, 7. If |N | < 7 = |GQ|, then anon-trivial elementα ∈ N is contained in the centraliser ofGQ. This implies thatQα is also fixed byGQ. Sinceα 6∈ GQ, it follows thatQ 6= Qα, and the assertionis proved.

If |N | = 7, thenGQ ∼= N , and henceGQ is a normal subgroup ofG. ThenG preserves the set of all fixed places ofGQ showing thatGQ fixes each place ino(Q). If |N | = 8, thenN is a Sylow2-subgroup ofN . Thus, up to conjugacy,NcontainsH. Therefore,N contains an element of order4, and henceN has at most5 involutions. So, at least one involution ofN is contained in the centraliser ofGQ.As before, this implies thatGQ has at least two fixed places.

If N is not an elementary abelian group, then it is either a simplegroup or thedirect product of isomorphic simple groups. The latter possibility does not occur inthis situation, since|G| is not a power of an integer with exponent greater than1.Also, there are only two simple groups whose order divides840, namelyA5 andPSL(2, 7). If N ∼= A5, thenGQ is contained in the centraliser ofN , as theK-automorphism group ofA5 has order120 and hence its order is prime to7. Again,this implies thatGQ has some more fixed places ino(Q).

A similar argument works for the caseN ∼= PSL(2, 7), as theK-automorphismgroup ofPSL(2, 7) has order336 which is prime to7. Hence,GQ coincides with

502

one of the eight Sylow7-subgroups ofG. Then, an element of order3 is containedin the normaliser ofGQ. As before, this implies thatGQ cannot have just one fixedplace. Once it has been shown thatGQ fixes not onlyQ but some more placesin o(Q), it is easy to show thatGQ must have at least seven fixed place. In fact,120 minus the number of fixed places ofGQ is divisible by7. By Theorem 12.66applied toGQ,

2g − 2 ≥ 7(2g′ − 2) + 7 · 6 ≥ 42− 14 = 28.

¿From this,g = 4, a contradiction. 2

L EMMA 12.107 Assume thatH is non-trivial but none of the cases in Table 12.1occurs. Then the orbit ofP underG is bounded above by3(2g − 2) provided thatat least one of the following conditions is satisfied:

(a) v < |G(1)P |, wherev = dP − |GP |;

(b) |GQ| > |H|;

(c) (µ+ 1)|GQ| 6= |G(1)P |+ 1, where(µ+ 1)|H| = v + 1, as in Lemma 12.75;

(d) v ≥ 2|G(1)P | − 3.

Proof. By (12.67), the bound to be established is the following:

r

2g − 2=|GQ|N

=|GQ|

|GQ|(dP − |GP |)− |GP |≤ 3.

By (12.38) and Lemma 12.68 (iv), (a) can only occur in one of two circumstances:

(i) v = |G(1)P | − 1 = 1, ρ1 = 3, |G(1)

P | = 2;

(ii) v = |G(1)P | − 2, ρ1 = 2, |G(2)

P | = 1.

In case (i), Lemma 12.75 (i) yields|H| = 1, whence|GP | = 2 andN = |GQ| − 2.Thereforer < 6(g−1). The latter case is described in the proof of Lemma 12.106.

Assume that (b) holds. By (a), takev ≥ |G(1)P |. Put|GQ| = |H|+ ǫ with ǫ > 0.

Then

|GQ|N

=|H|+ ǫ

|H|(v − |G(1)P |) + vǫ

≤ |H|+ ǫ

vǫ≤ |H|+ 1

v.

On the other hand, by Lemma 12.75 (i),|H| divides(v + 1). Thus|GQ| ≤ 2N ,and hencer ≤ 4(g − 1).

Now, assume that (c) holds. By (b), take|H| ≥ |GQ|. From Lemma 12.75 (i),

|GQ||GQ|v − |GP |

=|GQ|

|GQ|((µ+ 1)|H|)− |GP |

=|GQ|

|H||GQ|(µ+ 1)− |G(1)P |)− |GQ|

≤ |GQ|2|H| − |GQ|

≤ |GQ||H| .

Since|H| ≥ |GQ| by hypothesis, sor ≤ 2(g − 1).


Finally, assume that (d) holds. By (c), let|GQ|(µ + 1) = |G(1)P | + 1. Again,

from Lemma 12.75 (i),

|GP ||GQ|v − |GcP |

=|GQ|

|H| − |GQ|

=|GQ|(µ+ 1)

(|H| − |GQ|)(µ+ 1)=|G(1)

P |+ 1

v − |G(1)P |≤ |G

(1)P |+ 1

|G(1)P | − 3

.

Therefore,|G(1)P | 6= 2 andr ≤ 6(g − 1) for |G(1)

P | ≥ 5. Let |G(1)P | = 3. Then

v − |G(1)P | = dP = 2(ρ1 − 1)− 1− 3 = 2(ρ1 − 3).

If ρ1 = 3, thenv = |G(1)P |, and hencev − |GP | < 0 which is impossible. Thus,

ρ1 ≥ 4. Therefore,v−|G(1)P | ≥ 2, whencer ≤ 4(g−1). Finally, if |G(1)

P | = 4, then

r ≤ 5(g − 1). In fact,v − |G(1)P | = 1 is impossible, as then|H| = 3, contradicting

1 < |GQ| < |H. 2

Next, the possibility thato(R) 6= R occurs in case (iv.3) is investigated. Put

v′ = dP(Σ/F1)− |G(1)P |.

Then

v′ = −1 +∑s

i=1 (|M (i)| − 1),

whereM = M (1), whereM (i) is thei-th ramification group ofM atR, and where|M (s+1)| = 1. From Theorem 12.66 applied toG(1)

P ,

2g − 2 = −2|G(1)P |+ dP(Σ/F1) + |o(R)|(|M |+ v′).

Since|G(1)P | = |o(R)||M |, this gives the relation,

2(g − 1) = v + |o(R)|v′. (12.70)

Similarly, letv = dP′(ΣM/F1)−|G(1)P |/|M |, whereP ′ is the unique place ofΣM

lying underP. Then, from (7.8) applied to the towerΣ ⊃ ΣM ⊃ F1,

|GP |+ v = (|o(R)|+ v)|M |+ |M |+ v′. (12.71)

Also,

v′ ≥ |H| − 1, |H| ≤ |o(R)|. (12.72)

To show the first inequality, take|H| ≥ |M |. Also,H is contained in the normaliserof each ramification groupM (i). This and the inequality,|H| > |M (s)|−1, ensuresthe existence of a non-trivial element inM (s) which is contained in the centraliser anon-trivial subgroupH0 of H. LetH0 = 〈αℓ〉 with H = 〈α〉. Then|H0| = |H|/ℓ.From Theorem 12.68 (i),αℓ s = 1. By Lagrange’s Theorem, this implies that|H|dividesℓ s whence|H| ≤ ℓ s. LetH1 be the subgroup ofH of orderℓ. ReplaceHwith H1 in the above argument. Thenℓ = |H1| ≤ |M (s)| − 1. Thus,

|H| ≤ (|M (s)| − 1)s

= |M (s)| − 1 + |M (s)| − 1 + . . .+ |M (s)| − 1

≤ |M (1)| − 1 + |M (2)| − 1 + . . . |M (s)| − 1 = ρ′ + 1.

504

To prove the second inequality, assume on the contrary that|H| > |o(R)|. Thenthe stabiliserHS of S is non-trivial for any placeS ∈ o(R), but is trivial forS 6∈ o(R) becauseF1 is rational. This shows that the set of all places whereHramifies iso(R) ∪ P. This also holds true for any subgroupH ′ of GP whichis conjugate toH. If the roles ofP andR are interchanged,o(R) ∪ R alsorepresents the set of all places whereH ramifies. Therefore,o(P)∪R = o(R)∪P. Hence, the group generated byGP together withGR acts ono(P) ∪ R asa 2-transitive permutation group such that a non-trivial element fixesP,R andSfor anyS distinct from bothP andR. But such a2-transitive permutation groupcannot have a cyclic2-point stabiliser, by Theorem 15.16. Hence,|H| > |o(R)| isnot possible here.

These preliminary results provide enough information for aproof of the bound,|G| < 8g3, under the hypothesis thato(R) 6= R. Suppose2|o(R)| ≥ v + 1.Then (12.72) together with (12.70) imply that

2(g − 1) > |o(R)|v′ ≥ 12 |H|(|H| − 1) ≥ 1

2 |H||GQ|.Taking Lemma 12.105 into account,

|G| ≤ 2(g − 1)|H| |G(1)P | |GQ| ≤ 4(g − 1)2|G(1)

P | ≤ 4(g − 1)2g < 8g3.

Suppose2|o(R)| < v + 1. Then (12.71) gives the inequality,

|G(1)P |+ v ≥ (3|o(R)| − 1)|M |+ |M |+ v′ ≥ 3|G(1)

P |.Therefore, (d) in Lemma 12.107 occurs. By Lemma 12.105, thisshows that

|G| = r|H| |G(1)P | ≤ g(g + 1)6(g − 1) < 8g3.

If o(R) = R, the bound|G| < 8g3 can be proved by showing that either (b)or (d) of Lemma 12.107 holds. To do this, suppose (b) does not hold. By Lemma12.106, takeG(2)

P to be non-trivial; that is,ρ1 ≥ 2. If |H| ≤ 3, then also|GQ| ≤ 3,and hence (12.67) gives

|G| ≤ 9(2g − 2)|G(1)P | ≤ 18(g + 1)(g − 1) < 8g3.

Therefore,|H| ≥ 4. From Lemma 12.75 (i),[G(1)P : G

(2)P ] ≥ 5. On the other hand,

a careful study of the action ofG gives the missing result.

L EMMA 12.108[G

(1)P : G

(2)P

]≤

1 for p ≥ 3,4 for p = 2.

(12.73)

Proof. SinceF1 is rational, the identity automorphism is the only element in GPfixing a place ofΣ distinct from bothP andR. As these two places are in the sameorbit ofG, some elementα ∈ G takesP toR. Therefore,αGPα−1 = GP since|GP | = |GR| andαGPα−1 is a subgroup ofGQ. In other words,α belongs to thenormaliser ofGP . SinceP andR are the unique fixed places ofGP , it follows thatα takesR toP, and hence it interchanges them.

Two cases are treated separately, starting withp > 2. Here,α induces aK-automorphismα of F1. By Theorem 12.14 (iii),α fixes at least one place ofF1.


In other words, an orbit, sayo, of G(1)P is preserved byα. Sinceα does not fix

P orR, such a place is distinct from the place lying underP and, similarly, fromthat lying underR in the extensionΣ/F1. This implies thatα2 has at least threefixed places. By Theorem 12.14,α2 is the identity automorphism ofF1. Hence,α2 ∈ G(1)

P , showing that ordα2 is a power ofp. Sincep 6= 2, this implies that apowerδ of α has order2. Now, as the size ofo is equal to the order ofGP whichis a power ofp, soδ fixes a placeS ∈ o. In summary,

(i) δ is an involutoryK-automorphism of AutK(Σ);

(ii) δ ∈ NG(G(1)P );

(iii) Sδ = S, and soδ ∈ GS ;

(iv) Pδ = R, andRδ = P.

By (iii), a conjugate ofδ underG belongs toGP , and hence it fixes exactly twoplaces. This holds true forδ. Hence,δ has just one more fixed place, sayS. Let∆ denote the set of all places fixed byδ. If β ∈ G

(1)P commutes withδ, thenβ

preserves∆. Since|∆| = 2 < p, this implies thatβ is the identity automorphism.Thus, the groupG(1)

P is abelian, andδβδ−1 = β−1 for everyβ ∈ G(1)P . By (i) and

(ii), GS contains an involutory element, and hence its order is even.SinceP andSare in the same orbit ofG, this implies thatGP also has even order.

Now, the set of all Sylow2-subgroups ofGP is closed under conjugacy byδ.By Sylow’s theorem, such a set has odd size. Thus,δS2δ

−1 = S2 for at leastone Sylow2-subgroup ofG(1)

P . This ensures thatδ commutes with some non-

trivial involutory elementγ ∈ GP \ G(1)P . Hence,δ′ = γδ is an involutoryK-

automorphism interchangingP andR, andδ can be replaced byδ′ in the precedingargument. Therefore,δ′ has all the properties (i)–(iv) for some placeS ′ ∈ o. Also,the set∆′ of all fixed points ofδ′ has size2. If γβγ−1 = β−1 for some non-trivial K-automorphismβ ∈ G(1)

P , thenγδ commutes withβ. But this contradicts

|∆′| = 2. Therefore,γ is in the centralizer ofG(1)P . Then, by Lemma 12.68 (i), the

desired result,G(1)P = G

(2)P , follows.

Let p = 2. The basic idea of the proof is the same as forp 6= 2, although somechanges are required. With the same notation, the assertionthatα preserves anorbit o of G(1)

P holds true. But the existence of aK-automorphismδ satisfying thefour conditions requires a different proof.

Take a placeS ∈ o and aK-automorphismβ ∈ G(1)P such thatSα = Sβ . Then

δ = βα−1 fixesS. Henceδ2 has at least three fixed places, namelyP,R,S, show-ing thatδ2 is the trivial automorphism of AutK(Σ). Also, the idea of considering∆ still works even though some further elementary results on2-groups are needed.In fact, |∆| ≤ 2 implies that the centraliser ofδ inG(1)

P has order2. Then the group

generated byG(1)P together withδ is either dihedral or semi-dihedral. In both cases,

G(1)P has a cyclic subgroupC of index 2. Since the quotient groupG(1)

P /G(2)P is

elementary abelian, it follows thatC ∩G(2)P is a subgroup ofG(2)

P of index at most

2. Therefore,[G(1)P : G

(2)P ] ≤ 4. 2

506

Case(iv.4). Here,F1 is rational and no non-trivial element inGP fixesR. Thelatter hypothesis means that the orbito(R) of R underGP is long. Leto′(R)denote the orbit ofR underG. Then|o′(R)| · |GR| = |G|. SinceP andR liein the same orbit ofG, soGR ∼= GP . Also, o(R) is contained ino′(R). From(12.67),

|GP | ≤|G||GP |

= 2(g − 1) · |GQ|N≤ 2(g − 1)|GQ|.

Now, un upper bound onN is given. As

N = dP |GQ| − |GP ||GQ| − |GP | ≥ dP |GQ| − |GP ||GQ| − 2(g − 1)|GQ|,so

N ≥ |GQ|(dP − |GP | − 2(g − 1)) (12.74)

From Theorem 12.66 applied toGP ,

2(g − 1) = −2|G(1)P |+ (dP − |G(1)

P |(|H| − 1)) = dP − |GP | − |G(1)P |.

Hence

dP − |GP | − 2(g − 1) = |G(1)P |.

Taking (12.74) into account,N ≥ |G(1)P ||GQ|. Then (12.67) together with Theo-

rem 12.56 gives the desired result:

|G| ≤ 2(g − 1) · |G(1)P ||H||GQ||G(1)

P ||GQ| ≤ 8(g + 1)(g − 1) < 8g3.

Case(iv.5). The type of argument used for (iv.4) only gives an estimate|G| <cg5 in this case. So, a further investigation is needed which also requires a fewtechnical results, such as the inverse formula of (12.67):

2g − 2 =|G| (|GP | − |G(1)

P | |GQ|)|GQ|(|G| − |GP |)

. (12.75)

First, the possibility that the unique non-tame orbit consists of a single placePis considered. From Theorem 12.66 applied toG

(1)P ,

2g − 2 = −2 +∑

i≥2(|G(i)P | − 1).

Arguing as in the proof of (12.72), it follows that2g ≥ |H|. By Theorem 12.71,

|G| = |H||G(1)P | ≤ 2g

4|G(2)P |

(|G(2)P | − 1)2

g2,

which is greater than or equal to8g3 only if p = 2, |G(2)P | = 2, andF2 is rational.

Theorem 13.5 implies thatΣ = K(x, y) with y2 − y = B(x) whereB(X) ∈K[X] has odd degreem. Now, Theorem 13.6 provides all the necessary informa-tion:

pn = 2, m− 1 = 2g, |H| | m.


This shows that|G| ≥ 8g3 only occurs when|H| = m and|G(1)P | = (m− 1)2; in

particularm− 1 = 2k with k > 1. By Theorem 13.7,B(X) = cXm, with c 6= 0.It turns out that|G| < 8g3 apart from just one exception occurring for everyk > 1andp = 2, namely, the hyperelliptic curve

v(Y 2 − Y −X2k+1),

which has genus2k−1. In fact, if P is the unique place centred atY∞ = (0, 0, 1),thenGP has order22k(2k + 1), which is greater than8g3 = 23k.

Now, the possibility that the non-tame orbit is non-trivialis investigated. Permu-tation group theory will play an important role as in the nextresult.

L EMMA 12.109 The groupG acts on its non-tame short orbito as a2-transitivepermutation group. More precisely, o has sizeq + 1 with q = pt, and the possibil-ities for the permutation groupG induced byG ono are as follows:

(I) G is isomorphic to eitherPSL(2, q), or PGL(2, q);

(II) G is isomorphic to eitherPSU(3, n) or PGU(3, n), with q = n3;

(III) G ∼= Sz(n), with p = 2, n = 2n20, n0 = 2k, k odd, andq = n2;

(IV) G ∼= Ree(n) with p = 3, n = 3n20, n0 = 3k, andq = n3;

(V) a minimal normal subgroup ofG is solvable, and the size ofo is a primepower.

Proof. For a placeP ∈ o, let o0 = P, o1, . . . ok denote the orbits ofG(1)P con-

tained ino; then,o =⋃k

i=0 oi. To prove thatG acts2-transitively ono, it must beshown thatk = 1.

For anyi with 1 ≤ i ≤ k, take a placeR ∈ oi. In the present case,R is fixed byan elementα ∈ GP whose orderm is a prime different fromp, and hence divides|H|. By Sylow’s theorem, there is a subgroupH ′ conjugate toH in GP whichcontainsα; here,α preservesok. As noted previously,F1 being rational impliesthatα fixes at most two orbits ofG(1)

P . Therefore,o0 andoi are the orbits preservedby α. As H ′ is abelian and it fixeso0, so the orbitso0 andoi are also the onlyorbits ofG(1)

P which are fixed byH ′. SinceGP = G(1)P ⋊H ′, this implies that the

whole groupGP fixesoi. As i can be any integer between1 andk, it follows thatGP fixes each of the orbitso0, o1, . . . , ok. Therefore, eitherk = 1 orGP fixes atleast three orbits ofG(1)

P . The latter case cannot actually occur, by Theorem 12.14

applied toF1. Also, the size ofo is of the formq+1 with q = |G(1)P |; in particular,

q is a power ofp.Let G denote the2-transitive permutation group induced byG on o. The 2-

point stabiliser ofG is cyclic, as the subgroup ofG fixing two distinct places iscyclic when (iv.5) occurs; see Theorem 12.46. Two-transitivity together with thisspecial behaviour of the2-point stabiliser is enough to determine completely boththe abstract structure and the action ofG. Theorem 15.16 states indeed that, up toisomorphism,G is one of the groups in the list, withG acting in each of the firstfour cases in its natural2-transitive permutation representation. 2

508

A consequence of Lemma 12.109 is the following result.

L EMMA 12.110 The subgroupsGP andGQ have trivial trivial intersection.

Proof. Let α ∈ GP ∩ GQ be non trivial. Thenp ∤ ordα, and henceα ∈ H.This shows thatα fixes not onlyP but another place ino, sayR. SinceQ 6∈ o,this shows thatα has at least three fixed places. These places are in three differentorbits ofG(1)

P . SinceF1 is rational, Theorem 12.14 implies thatα fixes every orbit

of G(1)P , a contradiction. 2

Now, the cases in Lemma 12.109 are investigated further. LetM be the kernelof the above permutation representation ofG overo. ThenG = G/M . SinceMis a normal subgroup ofG, the groupG may also be viewed as an automorphismgroup ofΣM . WhenM is not trivial, then it has a great influence on the size ofG.

THEOREM 12.111 LetM be non-trivial.

(i) If ΣM is rational, then

(a) the genus

g = 12 (q − 1)(|M | − 1); (12.76)

(b) |G| ≥ 8g3 only occurs whenΣ = K(x, y) with yq − y = x2, and

q = pk odd, g = 12 (q − 1), |M | = 2, G = G/M ∼= PGL(2, q).

(ii) If ΣM is elliptic, theng = 12 (q + 1)(|M | − 1) + 1, and|G| ≤ 8g3.

(iii) If ΣM has genusg ≥ 2, theng = g|M |+ 12 (q − 1)(|M | − 1), and

|G|g3

<|G|g3

1

|M |2 <|G|g3.

Proof. SinceF1 is rational,M has no fixed place outsideo, by (iv) of Theorem12.14. From Theorem 12.66 applied toM ,

2g − 2 ≥ |M |(2g − 2) + (q + 1)(|M | − 1). (12.77)

For g ≤ 1, this shows the first statement in both (i) and (ii). Ifg > 1, write (12.77)as

g = g|M |+ 12 (q − 1)(|M | − 1).

Since|M | > 1 andq > 1, it follows thatg > g|M |. This together with the relation,|G| = |G||M |, proves (iii).

Let µ ∈ M be a non-trivialK-automorphism. From Theorem 12.46,µ is in thenormaliser ofG(1)

P . Also, µ fixes two orbits ofG(1)P , namelyP ando \ P.

SinceF1 is rational, they are all the orbits ofG(1)P fixed byµ. This implies that

µ fixes no place outsideo. Henceµ 6∈ GQ, showing thatM andGQ have trivialintersection; that is,|GQ| = |GQ|. So, (12.75) reads:

2g − 2 =|G|(|GP ||M | − |G(1)

P ||GQ)||GQ|(|G| − |GP |)

(12.78)


For the proof of the remaining assertions, assume first thatg = 0. Then, nonon-trivialK-automorphism inG fixes three distinct places ofΣM in o. From thelist in Theorem 12.109, the groups of type (II) and (IV) do notcomply with thisrequirement.

If G ∼= PSL(2, q) as in (I), then

|G| = 12 (q3 − q), |GP | =

1

2(q2 − q), |G(1)

P | = q.

From (12.78),

2g − 2 = 12 (q + 1)

(q − 1)|M | − 2|GQ||GQ|

.

This together with2g = (q − 1)(|M | − 1) gives|GQ| = 12 (q + 1).

Now, let |G| ≥ 8g3. Since|G| = 12 (q3 − q)|M | andg = 1

2 (q − 1)(|M | − 1), itfollows that

q(q + 1)

(q − 1)2> 2

(|M | − 1)3

|M | ,

whence|M | = 2. Thus,g = 12 (q − 1). In particular,p > 2. Also, ΣM = K(y)

with [Σ : K(y)] = 2, and hence div(y)∞ = 2P. By Theorem 13.4, part (i)(b)follows. If G ∼= PGL(2, q), this argument leads to the same conclusion.

If G ∼= Sz(n) as in (III), then|G| = (n2 + 1)n2(n− 1), |GP | = n2(n− 1), and|G(1)

P | = n2. From (12.78),

2g − 2 = (n2 + 1)(n− 1)|M | − |GQ|

|GQ|.

Comparison with (12.76) gives|GQ| = n− 1 + 2/(n+ 1) which is impossible forn > 1. This shows thatG ∼= Sz(n) cannot actually occur.

Similarly, G = Ree(n) is impossible in the present situation.If N is a minimal solvable normal subgroup ofG, thenN is an elementary

abelianw-group acting ono as a sharply transitive permutation group. In particular,|o| = wk given by the order ofN . This is possible in two cases only: eitherq = 8, wk = 9, or k = 1, q = 2r. Up to an isomorphism, there is only one2-transitive permutation group of prime degreew, namely, the group AGL(1, w) ofall permutationsx 7→ ax + b with botha 6= 0 andb ranging overFw. Hence, ifk = 1 thenGQ = d, whence2g − 2 = |M | − w by (12.78). Comparison with(12.76) givesw = 2 + 1/|M | which is impossible since|M | > 1.

It remains to rule out the casedk = 9. Let AΓL(1, 9) be the2-transitive permu-tation group overF9 which consist of all permutationsx 7→ axσ + b wherea 6= 0andb range overF9 andσ is either1 or 3. The groupAΓL(1, 9) has order144 andit contains two proper2-transitive subgroups, namely AGL(1, 9) andAγL(1, 9),both sharply2-transitive onF9. These three groups are the only2-transitive per-mutation groups of degree9. Now, assume that one of them occurs forG. Sincep = 2, either |GQ| = 3 or |GQ| = 9. Both possibilities are inconsistent with(12.76) and (12.78).

Finally, let g = 1. By Lemmas 12.56 and 12.56,|GP | divides24. This is con-sistent with the2-transitive action ofG ono only if |o| is one of3, 4, 5, 7, 9, 13, 25.

510

Since|o| divides |G| andG is aK-automorphism group ofΣM , Lemmas 12.56and 12.56 together with (12.55) also yield that the possibilities for the size ofo areactually only two, namely,3 and4. However, both are impossible.

If |o| = 3, thenq = p = 2 and|G| = 6, as the2-transitivity of G on o impliesthatG ∼= S3. Hence|G| = 6|M |. If |G| ≥ 8g3, then6|M | > 27(|M | − 1)3 by thefirst assertion in (ii), a contradiction.

If |o| = 4, thenq = p = 3. Also, from the2-transitive action ofG on o, either|G| = 12 or |G| = 24. If |G| ≥ 8g3, then24|M | > 64(|M | − 1)3 by (ii), again acontradiction. 2

THEOREM 12.112 If |G| ≥ 8g3, then(iv.5) only occurs in the following cases:

(I) p = 2, Σ = K(x, y) with y2 + y = x2k+1, Σ has genusg = 2k−1, G fixesa placeP and|G| = 22k(2k + 1);

(II) p > 2, Σ = K(x, y) with y2 = xq − x, Σ has genusg = 12 (q − 1), and

G/M ∼= PSL(2, q) or G/M ∼= PGL(2, q), whereq = pr and|M | = 2;

(III) p ≥ 2, Σ is the Hermitian function fieldK(x, y) with yq + y = xq+1, Σ hasgenusg = 1

2 (q2 − q), andG ∼= PSU(3, q) or G ∼= PGU(3, q);

(IV) p = 2, Σ = K(x, y) withxq0(xq +x) = yq +y, Σ has genusg = q0(q−1),andG ∼= Sz(q), whereq0 = 2r andq = 2q20 .

Proof. If the non-tame orbit consists of a single place, then (I) holds. This was thefirst result in case (iv.5).

Suppose that the non-tame orbito is non-trivial. The possible2-transitive actionsG of G on o are given in Lemma 12.109. First, the possible faithful actions ofGovero are investigated. So,G = G, and the possible cases are described in Lemma12.109.

Assume thatG ∼= PSL(2, q). Then

|G| = 12 (q3 − q), |GP | = 1

2 (q2 − q), |G(1)P | = q.

From (12.75),

2g = (q − 1)

(q − 1

2|GQ|− 1

).

Therefore,|GQ| divides 12 (q− 1)2. Take aK-automorphismα ∈ GQ whose order

is a power of a primew. Sincew divides|GQ|, Lemma 12.110 implies thatw = 2.Thus,|GQ| = 2. Hence,g = 1

8 (q − 1)(q − 5) with q ≥ 7. But then|G| < 8g3.ForG ∼= PGL(2, q), similarly g = 1

8 (q − 1)(q − 9), with q ≡ 1 (mod 4), andq ≥ 13. Again,|G| < 8g3. So, no example arises.

Next, letG ∼= PSU(3, q), and letµ = gcd(3, q + 1). Then

|G| = (q3 + 1)q3(q2 − 1)

µ, |GP | =

q3(q2 − 1)

µ, |G(1)

P | = q3.

¿From (12.75),

2g =(q2 − q + 1)(q + 1)(q2 − 1)

µ|GQ|− (q3 − 1).


Therefore,|GQ| divides(q2− q+1)(q+1)(q2− 1). Lemma 12.110 together withthe classification of subgroups of PSU(3, q) implies that no prime divisor of|GQ|divides(q2 − 1). Thus,

|GQ| =q2 − q + 1

mµ

for an odd integerm, and

2g = (q + 1)(q2 − 1)m− (q3 − 1) = (q − 1)(m(q + 1)2 − (q2 + q + 1).

Now, if |G| ≥ 8g3, thenm = 1, and2g = q2−q. To show that this is only possiblein case (III), observe first thatq+1 is a non-gap atP. This can be shown repeatingthe proof of Lemma 13.18 for

[Σ : F1] = q3, |H| = q2 − 1, 2g − 2 = (q + 1)(q − 2).

Then, Theorems 13.4 and 13.10 apply giving the result. The same argument worksfor G ∼= PGU(3, q).

In the case thatG ∼= Sz(2, q),

|G| = (q2 + 1)q2(q − 1), |GP | = q3 − q, |G(1)P | = q2.

From (12.75),

2g =(q2 + 1)(q − 1)

|GQ|− q2 + 1.

Again, Lemma 12.110 together with the classification of subgroups of Sz(q) im-plies that|GQ| is prime ton − 1. Hence,|GQ| divides eitherq − 2q0 + 1 orq + 2q0 + 1.

Let |GQ| = (q − 2q0 + 1)/m. Then

2g = (m− 1)(q2 − 1) + 2mq0(q − 1).

If |G| > 8g3, thenm = 1 andg = q0(q − 1).If |GQ| = (q + 2q0 + 1)/m, then

2g = (m− 1)(q2 − 1)− 2mq0(q − 1),

and hencem > 1. On the other hand,|G| < 8g3, form > 1. Hence,m = 1. FromTheorem 13.13, case (IV) follows.

Now, assume thatG ∼= Ree(q). Then

|G| = (q3 + 1)q3(q − 1), |GP | = q3(q − 1), |G(1)P | = q3.

¿From (12.75),

2g = (q − 1)

((q + 3q0 + 1)(q − 3q0 + 1)(q + 1)

|GQ| − (q2 + q + 1)

).

Lemma 12.110 together with the classification of subgroups of Ree(q) ensures thatno prime divisor of|GQ| dividesq − 1.

Suppose that|GQ| = (q + 1)/m. Then

2g = (q − 1)((m− 1)(q2 + 1)− (m+ 1)q);

512

hencem ≥ 2 and2g ≥ (q − 1)(q2 − 3q + 1). Thus|G| < 8g3.Suppose that|GQ| = (q + 3q0 + 1)/m. Then

2g = (q − 1)((m− 1)q2 −m(3q0q − 2q + 3q0 − 1)− q − 1);

hencem ≥ 2 and

2g ≥ (q − 1)(q2 − 3(2qq0 − q + 2q0) + 1).

Thus|G| < 8g3.Finally, suppose that|GQ| = (q − 3q0 + 1)/m. Then

2g = (q − 1)((m− 1)q2 +m(3q0q + 2q + 3q0 + 1)− q − 1);

hence2g ≥ 3(q − 1)(q + q0 + 1). Therefore|G| < 8g3.Assume now that a minimal normal subgroupN of G is solvable. Then the size

|o| = q + 1 is a power of a prime. It is shown in the proof of Theorem 12.111thatthis is only possible whenq+1 = 9 or q+1 is a prime. Also, eitherG ∼= AΓL(1, 9)orG is a sharply2-transitive group ono. SinceAΓL(1, 9) ∼= PSU(3, 2), the formercase has already been considered. In the latter case, (iv.4)occurs forG, and hence|G| < 8g3.

Finally, the possibility of a2-transitive permutation representationG of G on owith non-trivial kernelM is considered. Suppose that|G| ≥ 8g3. Theorem 12.111states that, ifΣM is rational, then (I) occurs and no example exists whenΣM iselliptic. Also, if the genusg of ΣM is at least2, then |G| > 8g3. Hence case(iv.5) occurs whereG is viewed as aK-automorphism group ofΣM . As G actsfaithfully on its 2-transitive non-tame orbit, from the first part of this proofthereare three possibilities forG. But all of them can be ruled out as follows.

If G ∼= PSU(3, q), then

2g = q2 − q, 2g = (q2 − q)|M |+ (q3 − 1)(|M | − 1)

by Theorem 12.111 (iii). Since|G| = |G||M |, it follows that

|G| = (q3 +1)q3(q2−1)|M | < (q3 +1)q3(q2−1)( 12 |M |)3(q−1)3(q2 +3q+1)3,

which is smaller than8g3. The same argument works forG ∼= PGU(3, q).If G = Sz(q), then

2g = 2q0(q − 1), 2g = 2q0(q − 1)|M |+ (q2 + 1)(|M | − 1).

Thus,

|G| = |G||M | = (q2 + 1)q2(q − 1)|M | < ( 12 |M)3(q2 + 4q0q − 4q0 + 1)3 < 8g3.

2

In geometric terms, the following result is obtained.

THEOREM 12.113 LetF be an irreducible curve of genusg ≥ 2. If |G| ≥ 8g3,thenF is birationally equivalent to one of the following:

(I) the hyperelliptic curvev(Y 2 + Y +X2k+1) with p = 2, g = 2k−1; also,Gfixes a placeP and|G| = 22k(2k + 1);


(II) the hyperelliptic curvev(Y 2 − (Xq −X)) with p > 2, g = 12 (q − 1); also,

G/M ∼= PSL(2, q) or G/M ∼= PGL(2, q), whereq = pr and|M | = 2;

(III) the Hermitian curvev(Y q + Y −Xq+1) with p ≥ 2, g = 12 (q2 − q); also,

G ∼= PSU(3, q) or G ∼= PGU(3, q);

(IV) theDLS curvev(Xq0(Xq +X)− (Y q +Y )) with p = 2, q0 = 2r, q = 2q20 ,g = q0(q − 1); also,G ∼= Sz(q).

12.13 K-AUTOMORPHISM GROUPS FIXING A PLACE

In this sectionK-automorphism groups ofΣ whose order is divisible byp and inwhich

every element of orderp has exactly one fixed place (12.79)

are investigated.

REMARK 12.114 A suitable powerβ of any non-trivial elementα of a p-grouphas orderp. Also, every fixed place ofα is fixed byβ as well. Therefore, condition(12.79) is satisfied by a subgroupG of AutK(Σ) if and only if everyp-element inG has exactly one fixed place. Also, since everyp-subgroup has non-trivial centre,the non-trivial elements of a Sylowp-subgroupSp of a K-automorphismG ofAutK(Σ) satisfying (12.79) have the same fixed place.

Condition (12.79) occurs in several circumstances, for instance whenG is oftype (iv.5) in Theorem 12.103, or whenG is theK-automorphism group of cer-tain curvesv(A(X) − B(Y ) = 0) investigated in Section 13.1. Another suchcircumstance is described in the following two lemmas.

L EMMA 12.115 If Σ hasp-rank0, then (12.79) holds inAutK(Σ).

Proof. Letα ∈ AutK(Σ) have orderp. Applying the Deuring–Shafarevich formula(12.29) to the group generated byα gives−1 = p(γ′ − 1) +m(p − 1), whereγ′

is thep-rank ofΣα andm is the number of fixed places ofα. This is only possiblewhenγ′ = 0 andm = 1. 2

REMARK 12.116 The converse of Lemma 12.115 is not true, a counter-examplebeing the curveF of 2-rank equal to4 given in Exercise 10 in Chapter 10. To showthis, let

Σ = K(x, y), y6 + y5 + y4 + y3 + y2 + y + 1 = x3(y2 + y)

be the function field ofF . Then the birational transformationx′ = x, y′ = y+1 isaK-automorphism ofΣ which has only one fixed place, namely that arising fromthe unique branch ofF tangent to the infinite line.

L EMMA 12.117 Assume thatAutK(Σ) contains ap-subgroupG of order pr. IfΣG hasp-rank zero, and (12.79) holds, thenΣ hasp-rank zero.

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Proof. Let P be the fixed place ofG. Sinceγ′ = 0, (12.29) applied toG givesγ − 1 = −|G|+ |G| − 1, whenceγ = 0. 2

REMARK 12.118 Exercises 9 and 10 in Chapter 10 show that the hypothesis onthep-rank ofΣG is essential.

Suppose that a subgroupG of AutK(Σ) has a Sylowp-subgroup with property(12.79). Then (12.79) is satisfied by all Sylowp-subgroups ofG. Suppose furtherthatG has no normal Sylowp-subgroup. LetSp andS′

p two distinct Sylowp-subgroups ofG. By Remark 12.114, bothSp andS′

p each have exactly one fixedplace, sayP andP ′. Also,Sp is the unique Sylowp-subgroup of the stabilizer of

P underG, andGP = G(1)P ⋊ H; see Theorem 12.46. Further,P 6= P ′. For, if

P = P ′ were true, thenSp andS′p would be two distinct Sylowp-subgroups in

GP ; but this is impossible asG(1)P is a normal subgroup ofG. Therefore, any two

distinct Sylowp-subgroups inG have trivial intersection.A subgroupH of a groupG is a trivial intersection setif for everyg ∈ G either

H = g−1Hg orH ∩ g−1Hg = 1. In the case whereH is a Sylow p-subgroup,this means thatH meets every other Sylowp-subgroup ofG trivially. So, in thiscase,Sp is a trivial intersection set inG. Hence, Lemma 12.115 has the followingcorollary.

THEOREM 12.119 If Σ hasp-rank0, then every Sylowp-subgroupSp in AutK(Σ)is a trivial intersection set.

THEOREM 12.120 (Burnside)If a Sylowp-subgroupSp of a finite solvable groupis a trivial intersection set, then eitherSp is normal or cyclic, orp = 2 andS2 is ageneralised quaternion group.

Finite groups whose Sylow2-subgroups are trivial intersection sets have beenclassified. This has been refined to groups containing a subgroup of even orderwhich intersects each of its distinct conjugates trivially.

THEOREM 12.121 (Hering)

(i) LetQ be a subgroup of a finite groupG with trivial normaliser intersection;that is, the normaliserNG(Q) ofQ in G has the following two properties:

(a) Q ∩ x−1Qx = 1 for all x ∈ G \NG(Q);

(b) NG(Q) 6= G.

If |Q| is even andS is the normal closure ofQ in G, thenS = O(S) ⋊ Q,the semi-direct product ofO(S) byQ, with Q is a Frobenius complement,unlessS isomorphic to one of the following groups:

PSL(2, n), Sz(n), PSU(3, n), SU(3, n), (12.80)

wheren is a power of2.

(ii) Let G be the permutation group induced byG on the setΩ of all conjugatesofQ underG. Then


(a) CG(S) is the kernel of this representation;

(b) S is transitive onΩ;

(c) |Ω| is odd;

(d) in the exceptional cases,

(1) |∆| = n+ 1, n2 + 1, n3 + 1, n3 + 1;

(2) G = G/CG(S) contains a normal subgroup isomorphic to one ofthe linear groups in 12.80;

(3) G is isomorphic to an automorphism group ofS containingS.

Hering’s result does not extend to subgroups of odd order with trivial normalintersection. A major result which can be viewed as a generalisation of Theorem12.120 is the following.

THEOREM 12.122 LetSd, with d > 11, be a Sylowd-subgroup of a finite groupG that is a trivial intersection set but not a normal subgroup.ThenSd is cyclicif and only ifG has no composition factors isomorphic to eitherPSL(2, dn) withn > 1 or PSU(3, dm) withm ≥ 1.

Now, some consequence of the above results are stated. Two cases are distin-guished according asp = 2 or p > 2.

Whenp = 2, if (12.79) is satisfied by a subgroupG of AutK(Σ), then Theorem12.121 applies toG with Q a 2-subgroup of a Sylow2-subgroupS2 of G. In fact,(i)(a) is true, while (i)(b) holds provided thatG is larger thanGP , whereP is thefixed place ofS2.

WhenGP = G, the structure ofG is fully described in Theorem 12.46. So, ifGhas no fixed place, Hering’s theorem determines the abstractstructure of the normalsubgroupS of G generated by all elements whose order is a power of2. Note that,sinceS2 has only one fixed point,NG(S2) = GP holds. IfS = O(S)⋊S2, thenS2

is a Frobenius complement, and hence it has a unique involutory element. Note thatit is not claimed thatS is a Frobenius group. In the exceptional cases, ifCG(S) isthe centralizer ofS in G, thenG/CG(S) is theK-automorphism group of a grouplisted in Hering’s theorem.

Therefore, the following result is obtained.

THEOREM 12.123 Let p = 2 and assume that (12.79) holds. LetS be the sub-groupG of AutK(Σ) of even order, generated by all elements whose order is apower of2.

(i) One of the following holds:

(a) S isomorphic to one of the following groups:

PSL(2, n), PSU(3, n), SU(3, n), Sz(n), with n = 2r; (12.81)

(b) S = O(S) ⋊ S2 with S2 a 2-Sylow subgroup ofG; hereS2 is either acyclic group or a generalised quaternion group.

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(ii) S is a Sylow 2-subgroup, andG = S ⋊H withH a subgroup of odd order.

Using Theorem 12.121, the action ofS on the setΩ of all places ofΣ fixed bysome involution or, equivalently, on the set of all involutions inG can be investi-gated. LetS be the permutation group induced byS onΩ. If M is the kernel of thepermutation representation ofS, thenS = S/M , andS2 = S2M/M is a2-Sylowsubgroup ofS. Here,S is transitive onΩ; also,S2 fixesP and acts onΩ \ P asa semi-regular permutation group. Further,M = CG(S). If S is isomorphic to oneof the groups in (12.81), thenS is 2-transitive onΩ, and|Ω| = n+1, n2+1, n3+1according asS ∼= PSL(2, n), Sz(n), PSU(3, n).

Theorem 12.120 is the main ingredient in the proof of the following result.

THEOREM 12.124 Let G be a solvableK-automorphism group ofΣ satisfyingcondition (12.79). Assume thatG fixes no place, and that

p2 divides|G| whenp > 2, while16 divides|G| whenp = 2. (12.82)

If Σ has genusg ≥ 2, then|G| ≤ 24g(g − 1).

Proof. Forg = 2, Theorem 12.124 follows from Theorem 12.88.Let N be a minimal normal subgroup ofG. SinceG is solvable,N is an ele-

mentary abelian group of orderdr with a primed. If d = p, then (12.79) impliesthatN fixes a unique place. But thenG itself must fixP, contradicting one of thehypotheses. Therefore,d 6= p. In particular, the Sylowp-subgroups ofG andG/Nare isomorphic, and hence Theorem 12.120 applies toG/N , while (12.82) remainsvalid forG/N .

This suggests to investigateG = G/N viewed as aK-automorphism group ofΣN . The goal is to show thatG also satisfies (12.79).

The placeP ′ lying below P in the extensionΣ/ΣN is fixed by a Sylowp-subgroup ofG. SinceG/N can be viewed as aK-automorphism group ofΣN ,this rules out the possibility thatΣN is elliptic, because (12.82) is inconsistent withTheorem 12.85. Similarly,ΣN cannot be rational. In fact, ifΣN were rational,then nop-subgroup ofG/N and hence ofG would be either cyclic or a quaterniongroup, by Theorem 12.14.

Therefore,ΣN has genusg ≥ 2. Assume that Theorem 12.124 does not hold,and choose a minimal counterexample with genusg as small as possible. Then|G| > 24g(g − 1), but Theorem 12.124 holds for all generag′ with 2 ≤ g′ < g.Theorem 12.66 applied toN gives2g − 2 ≥ |N |(2g − 2), whence

|G| > 12g(2g − 2) > |N | 12g(2g − 2).

Therefore|G/N | > 24g(g − 1). By the divisibility condition (12.82) on|G|, theorder ofG = G/N is not a prime. HenceG is a solvable group. Sinced 6= p, soG satisfies (12.82). On the other hand,G can be regarded as aK-automorphismgroup ofΣN .

To show thatG also satisfies condition (12.79), letα be any element ofG of orderp. Choose an elementα in G whose image isα under the natural homomorphismG 7→ N . By (12.79),Σ has a unique placeP fixed byα. Obviously,α fixes theplaceP lying underP in the extensionΣ/ΣN . Assume on the contrary thatα fixes


another place ofΣN , sayQ. Then the set of places ofΣ lying overQ is preservedbyα. The number of such places is prime top since their number divides|N |. Butthenα must fix one of these places, contradicting (12.79).

SinceG is a minimal counterexample,G fixes a place ofΣN . Equivalently, theplaces ofΣ fixed byp-elements ofG form a uniqueN -orbit. SinceSp has only onefixed place, so|Sp| ≤ |N | − 1. This together with (12.82) implies that|N | ≥ 10.As g − 1 ≥ |N |(g − 1), it follows thatg ≥ 11.

From (ii)(b) of Theorem 12.46,|G| = |GP | = |G(1)P | |H| with H a cyclic group

of order prime top. By Theorem 12.56,|H| ≤ 4g + 2. Let Sp be the Sylow

p-subgroup ofG; thenSp = G(1)

P . Sincep ∤ |N |, soSp is isomorphic toSp.

Assume thatSp is cyclic. From Theorem 12.72,|G(1)

P | ≤ 4g + 4. Therefore,

|G| ≤ (4g + 4)(4g + 2),

whence|G| ≤ (4g+4)(4g+2)|N |. Sinceg−1 ≥ |N |(g−1), g ≥ 2 and|N | ≥ 10,it follows that

|G| ≤ 16(g + 1)(g + 12 )|N | ≤ 16

10· g + 1

g − 1(g − 1)|N | · g + 1

2

g − 1(g − 1)|N | ≤

8

5· 2g

2 + 3g + 1

2(g − 1)2(g − 1)2 ≤ 12(g − 1)2 < 24g(g − 1).

But thenG is not a counterexample.If Sp is not cyclic, Theorem 12.120 implies thatp = 2 and thatS2 is a quaternion

group. Since a quaternion group has a cyclic subgroup of index 2, Theorem 12.72shows that|G(1)

P | ≤ 2(4g + 4). Now, the argument above can be used to show that12 |G| ≤ 12(g − 1)2 whence|G| ≤ 24(g − 1)2 < 24g(g − 1). Again,G is not acounterexample. 2

Finally, an alternative proof forp = 2 which does not require the hypothesis(12.82) is given.

THEOREM 12.125 Let p = 2, g ≥ 2, and assume that (12.79) holds. IfG is asolvableK-automorphism group ofΣ fixing no place ofΣ, then|G| ≤ 24g2.

Proof. Since84(g − 1) < 24g2, the groupG may be supposed to be one of theexceptions in Theorem 12.53. SinceG has no fixed place, from Theorem 12.123,S = O(S)⋊S2, andO(S) acts transitively on the seto consisting of all places fixedby involutory elements inG. In particularo is the unique non-tame orbit ofG. LetP ∈ o. By assumption (12.79),G(1)

P is a Sylow2-subgroupS2 of G. Therefore,|O(S)| = |o||O(S)P |, and

|G| = |GP ||O(S)||O(S)P |

. (12.83)

In particularo has odd size. AsO(S) is transitive ono andNG(S2) = GP , eachelement inG is the product of an element fixingP and an element fromO(S), thatis,G = GPO(S). Also, asO(S) is a characteristic subgroup ofS andS is a normalsubgroup ofG, O(S) is a normal subgroup ofG. HenceG = G/O(S) is a factor

518

group which can be viewed as aK-automorphism group ofΣO(S). Assume thatΣO(S) is rational. By Theorem 12.14, every2-subgroup ofG is elementary abelian.On the other hand, asS2

∼= S2O(S)/O(S), from Theorem 12.121 the factorgroupS2O(S)/O(S) has only one involutory element. Thus,|S2O(S)/O(S)| = 2whence|S2| = 2. Therefore,O(S) = O(G) andG = O(G) ⋊ S2. Hence

|G| = 2|O(G)| ≤ 168(g − 1) ≤ 24g2

for g ≥ 6. To show that this holds true for2 ≤ g ≤ 5, note that ifG has morethan three short orbits, then|G| ≤ 12(g − 1) < 24g2 by (I) in Theorem 12.53.SinceS2 has exactly one fixed place, it preserves only one orbit ofO(G). Hence,the number of short orbits ofO(G) is odd, and thus it may supposed to be equal toone or three. Sincep = 2, in the latter case|G| ≤ 84(g − 1) by (III) of Theorem12.53, and hence|G| < 24g2. If o is the unique short orbit ofG, then Lemma 12.98implies that|o| = 3, g = 4. Hence,O(G) acts ono as a group of order3. LetHbe the subgroup ofO(G) fixing every place ino. Then|O(G)| = 3|H| and fromTheorem 12.54,6 = −6|H|+ 3(|H| − 1) = −3|H| − 3, a contradiction.

Assume thatΣO(S) is elliptic. SincedP ≥ |O(S)P |−1 by (12.22) anddP = dQfor P,Q ∈ o, Corollary 12.49 implies that

2g − 2 ≥ |o|(|O(S)P | − 1) > 12 |o||O(S)P |,

whence|O(S)| < 4(g − 1). From Theorem 12.85,|G| < 96(g − 1) ≤ 24g2.It remains to consider the case where the genus ofΣO(S) is greater than1. From

Corollary 12.49,|O(S)| ≤ g − 1. Hence by (12.83)

|O(G)| ≤ (g − 1)|GP .

Assume thatF1 = ΣS2 has genusg′ > 1. Then |S2|(g′ − 1) < g − 1 by thesame Corollary 12.49. SinceGP/S2 is a tame automorphism group ofF1 fixingthe place ofF1 lying underP, from Theorem 12.56,|GP/S2| ≤ 4g′ + 2. Hence,

|GP | <4g′ + 2

g′ − 1(g′ − 1)|S2| ≤ 10(g − 1).

Therefore,|G| < 10(g − 1)2 < 24g2. If F1 is elliptic, then|GP/SP | ≤ 24 byTheorem 12.85 while|SP | ≤ g by Theorem 12.71. Therefore,|G| ≤ 24g(g−1) <24g2.

So, takeF1 to be rational. LetR ∈ o be a place distinct fromP. If the stabiliserGP,R ofR in GP is trivial, then|GP | < |o|, and hence

|G| = |GP ||o| < |o|2 ≤ |O(S)|2 ≤ (g − 1)2.

Therefore, let|GP,R| > 1 for everyR ∈ o. Then, from the proof of Theorem12.109,G acts ono as a2-transitive permutation groupG, ando comprisesPtogether with all places in the unique non-trivial orbit ofS2. The latter assertionimplies that|o| = |S2| + 1 = q + 1, q = 2r. On the other hand, asG is solvable,|o| = wk with an odd primew. ¿Fromq + 1 = wk, eitherk = 1, or k = 2, q = 8,w = 3. LetM be the kernel of the permutation representation ofG ono.

It may be thatM is trivial, that isG = G, and this possibility is investigatedfirst. Assume thatk = 1. ThenG = AGL(1, w) andG is sharply2-transitive ono.LetN be the normal subgroup ofG of orderw. If ΣN has genusg′ ≥ 2, then

2g − 2 ≥ w(2g′ − 2) ≥ 2w,


and hence

|G| = |o|(|o| − 1) = w(w − 1) ≤ (g − 1)(g − 2) < 24g2.

If ΣN is elliptic, then|S2| is either2, or 4, or 8 by Theorem 12.85. The last casecannot occur because9 is not a prime number. If|S2| = 2, thenw = 3, and hence|G| = 6, while |S2| = 4 occurs whenw = 5 and hence|G| = 20.

If ΣN is rational, thenS2, viewed as aK-automorphism group ofΣN , must bean elementary abelian group by Theorem 12.14. On the other hand, S2 is the1-point stabiliser of AGL(1, w) which is cyclic. It follows that|S2| = 2. Therefore,|o| = 3, whence|G| = 6.

If k = 2, w = 3, q = 8, thenASL(1, 9) < G ≤ AΓL(1, 9). In particular,G hasa normal subgroupN of order9, andS2 is the quaternion group of order8. Usingthe same argument as above, it follows thatΣN is neither rational, nor elliptic, andthat if its genus is at least2, then|G| ≤ 2(g − 1)(g − 2).

SupposeM be non-trivial. By Theorem 12.46,M is cyclic of odd order, andG = G/M can be viewed as aK-automorphism group ofΣM . The Sylow2-subgroupS2 = S2M/M of G is isomorphic toS2. Now, it is shown thatG hasproperty (12.79). If this did not hold, then some non-trivial K-automorphismα ∈S2 of ΣM would fix not only the placeP ′ underP but at least one more placeQ′.Equivalently, some non-trivial elementα ∈ S2 would fix P and the orbitθ of Mconsisting of all placesQ of Σ lying overQ′. Sinceθ has odd size, thenα wouldfix at least one place inθ, contradicting (12.79).

¿From above, if the genusg of ΣM is at least2, then|G| ≤ 24g2; in consequence|G| ≤ 24|M |g2. Since|o| ≥ 3, from Theorem 12.54

2g − 2 ≥ |M |(2g − 2) + 3(|M | − 1) ≥ 2|M |g − 2.

This shows that|G| ≤ 24gg < 24g2.If ΣM is elliptic, thenG ≤ 24 by Theorem 12.85 whence|G| ≤ 24|M |. On

the other hand,2g − 2 ≥ |o|(|M | − 1) ≥ 3(|M | − 1) > |M | by Theorem 12.54.Therefore,|G| < 24(2g− 2) < 24g2. Finally, if ΣM is rational, then the argumentabove depending on Theorem 12.14 shows that|S2| = |S2M/M | = 2 whence|G| ≤ 6 and|G| ≤ 6|M |. Therefore|G| < 6(2g − 2) < 24g2. 2

12.14 LARGEp-SUBGROUPS FIXING A PLACE

THEOREM 12.126 Let F be a curve of genusg ≥ 2 with function fieldΣ, andlet G be aK-automorphism group ofΣ such thatΣ has a placeP satisfying thefollowing condition:

|G(1)P | >

p

p− 1g. (12.84)

Then one of the following occurs:

• G = GP ;

• one of the cases(II), (III), (IV) in Theorem 12.109 holds;

520

• (V) p = 3, g = n0(n − 1), with n = 3n20, n0 = 3r, andF is theDLR

curve with

F = v(Y n2−(1+(Xn−X)n−1)Y n+(Xn−X)n−1Y −Xn(Xn−X)n+3n0);

here,G ∼= Ree(n).

Proof. Let Ω be the set of all placesR of Σ with non-trivial first ramification groupG

(1)R . SinceP ∈ Ω andG 6= GP , soΩ contains at least two places. By Section

15.5 (XXVIII), Ω is a fullG-orbit, and henceΩ is the only non-tameG-orbit.If there are at least two more shortG-orbits, sayΩ1 andΩ2, then, from Theorem

12.66,

2g − 2≥−2|G|+ (|GP |+ |G(1)P |+ |G

(2)P | − 3) |Ω|

+(|GQ1| − 1) |Ω1|+ (|GQ2

| − 1) |Ω2|,whereQi ∈ Ωi for i = 1, 2. Note that

(|GQ1| − 1)|Ω1|+ (|GQ2

| − 1)|Ω2| ≥ |G|since|GQi

| − 1 ≥ 12 |GQi

|, and |G| = |GQi| |Ωi|. Also, |G| = |GP | |Ω|, and

|G(2)P | > 1 by Theorem 12.71. Therefore,

2g − 2 ≥ (|G(1)P | − 1) |Ω|.

Since|Ω| ≥ 2, this implies thatg ≥ |G(1)P |.

If ΣG has genusg ≥ 2, then Theorem 12.66 gives|G| ≤ g− 1. If ΣG is elliptic,then

2g − 2 ≥ 2(|GP | − 1 + |G(1)P | − 1) ≥ 2|GP |,

whence|G| ≤ g − 1.From what has been shown so far, it may be assumed thatΣG is rational and that

one of the cases (iii), (iv.4), (iv.5) in Theorem 12.53 occurs.In case (iii),Ω is the unique shortG-orbit. By Theorem 12.66,

2g − 2 = |Ω| (degD(Σ/ΣGP )− 2|GP |). (12.85)

Therefore,|Ω| ≤ 2g − 2. SinceΣG(1)P is rational,

degD(Σ/ΣGP ) = |GP | − |G(1)P |+ degD(Σ/ΣG

(1)P )

= |GP |+ |G(1)P |+ 2g − 2.

Thus, (12.85) reads:

2(g − 1) = |Ω|(2g − 2) + |G(1)P | − |GP |).

First, the case|Ω| = 2g − 2 is considered. Here,2g − 2 + |G(1)P | − |GP | = 1

with GP = G(1)P ⋊H; hence2g − 2 = (|H| − 1)|G(1)

P | + 1. This implies thatH

is non-trivial. For|H| > 2, it follows that|G(1)P | ≤ g − 1.

Suppose that|H| = 2. Thenp 6= 2 and|G(1)P | = 2g − 3. Since|Ω| = 2g − 2

andGP only ramifies atP, this implies thatG(1)P acts onΩ \ P as a transitive


permutation group. Hence,|Ω| = q + 1 with q = |G(1)P |. SinceΩ is aG-orbit, it

follows thatG induces onΩ a 2-transitive permutation groupG whose one-pointstabiliser has order eitherq or 2q according as|G| = 2|G| orG = G.

If |G| = 2|G|, the subgroupH is the kernel of the permutation representation ofG onΩ; that is,H fixes every place inΩ. In particular,H is a normal subgroup ofG. Therefore,G can be viewed as aK-automorphism group ofΣH . LetP ′ be theplace ofΣH lying underP. Since2 ∤ |G(1)

P | it follows thatG(1)P∼= G

(1)P′ . Also, the

places ofΣH lying under the places inΩ form the unique shortG-orbit. Therefore,case (iii) can be considered withΣH andG. As has been shown before, this impliesthatGP′ = GP′ ⋊ H with H a non-trivial subgroup. But this is impossible as thestabiliser|GP′) has orderq.

If G = G, two cases are distinguished according asG is solvable or not.In the former case, Huppert’s classification of all finite solvable 2-transitive

groups applies. Sinceq + 1 is even, soq + 1 = 2k, andΩ can be identified withthe elements ofF2k in such a way thatG is a subgroup of the groupT (2k) of allsemi-linear permutationsX 7→ aϕ(X)+bwith a 6= 0, b ∈ F2k andϕ ∈ Aut(F2k).In particular,GP is a subgroup of the one-point stabiliser ofT (2k), and hence|GP |divideskq. On the other hand,2k = q + 1 can only occur whenk andq are bothprimes. Since|GP | = 2q, this implies thatk = 2. Henceq = g = 3; that is,p = 2

and|G(1)P | = g = 3.

Suppose thatG is not solvable. By Theorem 15.16,|Ω| = 6 andG ∼= PSL(2, 5).

Therefore,|G(1)P | = 5 andg = 4. Thus, (12.84) does not hold in case (iii).

In case (iv),G has not only a short non-tame orbitΩ but also a short tame orbit∆. LetP ∈ Ω andQ ∈ ∆.

For the case (iv.4), from the proof of Theorem 12.103,|GP | ≤ 2(g− 1)|GQ|/NandN ≥ |G(1)

P ||GQ|, whence

|GP ||G(1)P | ≤ 2(g − 1).

Since|GP | ≥ 2, this implies that|GP | ≤ g − 1 contradicting (12.84).Assume that case (iv.5) holds. From the proof of Theorem 12.103, the possible

actions ofG = G onΩ are those in Lemma 12.109.If G ∼= PSL(2, q), then|G| = 1

2 (q3−q), |GP | = 12 (q2−q), |G(1)

P | = q. From theproof of Theorem 12.103,g = 1

2 (q−1)(q−5) with q ≥ 7. Therefore, (12.84) doesnot hold. Since the Sylowp-subgroups of PGL(2, q) are contained in PSL(2, q),this also shows that (12.84) is impossible whenG ∼= PGL(2, q).

If G ∼= PSU(3, n) with q = n2, then |G| = 12 (n3 + 1)n3(n2 − 1)/µ, with

|GP | = n3(n2 − 1)/µ and|G(1)P | = n3, whereµ = gcd(3, q + 1). From the proof

of Theorem 12.103, there is a divisorm of (n2 − n+ 1)/µ, such that

2g = (n− 1)m((n+ 1)2 − (n2 + n+ 1), |GQ| = (n2 − n+ 1)/(mµ).

Sincem is odd, this implies that eitherm ≥ 3 andg ≥ |G(1)P | or m = 1 and

g = 12n(n − 1). From the proof of Theorem 12.103, this gives (III) of the same

theorem. 2

522

12.15 NOTES

The finiteness of theK-automorphism group of an irreducible algebraic curve ofgenusg ≥ 2 was proven forp = 0 by Schwarz in 1875. The result was extendedto p > 0 by Schmid [235]. The proof given in Sections 12.4 and 12.6 arebased on[151, 152, 153], see also [225].

Lemmas 12.27, 12.28 and Theorem 12.29 are due to Segre [244].For the proof of Lemma 12.44, see [151, Lemma 5].Related to Section 12.5, several good textbooks on the topicof quotient curves

using function field theory are available in the literature,such as [275] and [202].Theorems 12.53 12.56 12.71 are due to Stichtenoth, see [271].Theorem 12.58 was stated by Deuring fors ≥ 1 and then by Shafarevich for

s = 0. Later Rosen and, independently, Sullivan gave proofs for special cases.Subrao [278] proved the theorem in full generality. Madan [193] revised Deuring’soriginal proof; he fixed some errors and gave a proof for everys using Deuring’sapproach. A short proof valid forn = p andK = Fq is found in [229].

Theorem 12.59 is due to Witt [309] and independently to Shafarevich [261].The group-theoretic result required in Lemma 12.75 is [145,Theorem 13.7].Related to Section 12.9, a thorough investigation on ramification groups is found

in [256, Chapter IV], see also [275].Theorems 12.72, 12.77, 12.79 and Remark 12.80 are due to Nakajima, see [208],

[207] and the survey paper [209].Proofs of Theorems 12.82, 12.83 and 12.84, are found in [194], see also [169].

For further results on theK-automorphism group of an Artin–Schreier curve

v(Y ph − Y − f(X))

with f(X) ∈ K[X], see Section 13.1.There is a large literature on automorphism groups of complex hyperelliptic

curves whose study was initiated by Clebsch, Hurwitz and Bolza in the late Nine-teen century, see [8] and [77]. Many of the results hold true in zero characteristic,see for instance [37], but this does not happen in positive characteristic when non-tame automorphisms enter in play.

In recent years, important applications in cryptography have stimulated researchon hyperelliptic curves (especially of genus2) over finite fields with particular at-tention on their automorphism groups, see the seminal paper[166]. Recent re-sults improving Proposition 12.88 are found in a series of papers by Cardona, Du-ursma, Gonzalez, Gutierrez, Kiyavash, Lario, Quer, Rio, Shaska and Volklein, see[65, 41, 43, 265]. A relevant result is that ifp 6= 2, the possibleK-automorphismgroups of genus2 curves are

C2, C10, C2 × C2, D8, D12, C3 ⋊D8, GL(2, 3)

and an extension ofS5 by the hyperbolic involution. The classification of genus2 curves according to theirK-automorphism groups is being investigated. Forp 6= 2, the curves of genus2 whoseK-automorphism groupG is isomorphic toC3 ⋊D8, GL(2, 3), orC10 are determined. Such curves are birationally equivalent


to the irreducible plane curvesv(Y 2 − f(X)) with

f(X) =

X6 − 1 when G ∼= C3 ⋊D12,X5 −X when G ∼= GL(2, 3),X5 − 1 when G ∼= C10.

Results onK-automorphisms of hyperelliptic curves of genus3 are found in [109].In [314] and [263] hyperelliptic curves whose hyperbolic involution is the only

K-automorphism are considered. In [110] and [263] hyperelliptic curves with aK-automorphism group isomorphic toC2 × C2 are studied.

Duma’s paper [64] is a survey on results contained in the unpublished thesis byBrandt who gave a complete classification ofK-automorphism groups of Kummerextensions of the rational fieldK(x), by exhibiting explicit equations for the curvesto which they correspond. The main ingredients in Brandt’s analysis are a refine-ment of Dickson’s classification and the computation of all central extensions ofcyclic groups with these groups.

Section 12.11 comes from [8]. For the result in Remark 12.97,see [136].Section 12.12 is based on [121] and [271]. Theorem 12.113 relying on previous

results of Stichtenoth is due to Henn. As a consequence, Hurwitz’s upper bound|G| ≤ 84(g − 1) remains valid in positive characteristic provided that thecharac-teristic is big with respect to the genus, namelyp > g + 1. This was pointed outat first by Roquette in [226] who was also able to show that ifp = g + 1 and|G|exceeds Hurwitz’s upper bound then (II) of Theorem 12.113 holds.

For a proof of Theorem [106], due to Burnside, see [106].In finite group theory, trivial intersection sets plays an important role. Suzuki

[283] classified all finite groups whose Sylow2-subgroups are trivial intersectionsets. Hering [124] refined and extended Suzuki’s classification to finite groupscontaining a subgroup of even order which intersects each ofits distinct conjugatetrivially. Here, Hering’s result Theorem 12.121 is quoted.In the special case thatQ is a Sylow2-subgroup ofG, this is essentially Suzuki’s classification. Theorem12.122 is stated in [313].

For more on the curves given in Theorems 12.113 and 12.126, see Sections 13.2,13.3 and 13.4.

For Huppert’s classification referred to in the proof of Theorem 12.126, see [147,Theorem 7.3]

In connection with Example 12.92 and Kloosterman sums, see van der Geer andvan der Vlugt [295], where it is shown that the irreducible plane curves,

v(A(Y )−R(X)),

with A(Y ) ∈ Fq[Y ] andR(X) ∈ Fq(X), are precisely the curvesΓ overFq ofgenusp− 1 with the three properties listed.

Chapter Thirteen

Some Families of Algebraic Curves

The aim of this chapter is to provide families of algebraic curves in positive char-acteristic with properties that a complex algebraic curve cannot have.

13.1 PLANE CURVES GIVEN BY SEPARATED POLYNOMIALS

In this section, consider the curveC overK = Fq, q = ph, with

C = v(A(Y )−B(X)), (13.1)

where

(I) deg C ≥ 4 andgcd(p,degB(X)) = 1;

(II) A(Y ) = anYpn

+ an−1Ypn−1

+ . . .+ a0Y, aj ∈ K, a0, an 6= 0;

(III) B(X) = bmXm + bm−1X

m−1 + . . .+ b1X + b0, bj ∈ K, bm 6= 0;

(IV) m 6≡ 0 (mod p);

(V) n ≥ 1, m ≥ 2.

Note that (II) occurs if and only ifA(Y + a) = A(Y ) + A(a) for everya ∈ K,that is, the polynomialA(Y ) is additive. The basic properties ofC are collected inthe following lemma.

L EMMA 13.1 LetC be an irreducible plane curve with at most one singular point.

(i) If |m− pn| = 1, thenC is non-singular.

(ii) (a) If m > pn + 1, thenP∞ = (0, 0, 1) is an(m− pn)-fold point ofC.(b) If pn > m+ 1, thenP∞ = (0, 1, 0) is a (pn −m)-fold point of C.(c) In both cases,P∞ is the centre of only one branch ofC. Also,P∞ is

the unique infinite point ofC.

(iii) If Σ = K(x, y), withA(y) = B(x), is the function field ofC, andP∞ is theplace associated to the branch centred atP∞, then

(a) div(dx) = ((pn − 1)(m− 1)− 2)P∞;

(b) X has genusg = 12 (pn − 1)(m− 1);

(c) a translation(x, y) 7→ (x, y + a) preservesC if and only ifA(a) = 0;

525

526

(d) these translations form an elementary abelian group of order pn, andAutK(Σ) contains an elementary abelianp-groupG of orderpn thatfixes a unique placeP∞, and acts transitively on the zeros ofx;

(e) the sequence of ramification groups ofG atP∞ is

G = G(1)P∞

= G(2)P∞

= . . . = G(m)P∞

, G(m+1)P∞

= 1;

(f) P∞ is the unique short orbit ofG, and

div(Σ/ΣG) = (pn − 1)(m+ 1)P∞;

(g) ΣG is rational, andΣ hasp-rank zero.

Proof. PutF (X,Y ) = A(Y )−B(X). Since∂F/∂Y = a0 6= 0, no affine point isa singular point ofC.

The casem > pn is considered first. First,P∞ = (0, 0, 1) is the unique pointof C at infinity and is, in fact, an(m − pn)-fold point, with a unique tangent linev(X0).

It is now shown thatP∞ is the centre of only one branch ofC. To do this,P∞ istaken to the origin by interchangingX0 andX2. ThenC = v(A(Y ) − C(X,Y )),where

C(X,Y ) = bmXm + bm−1X

m−1Y + . . .+ b1XYm−1 + b0Y

m.

Let x(t) = uti + . . . , y(t) = vtj + . . . , with u, v 6= 0, i < j, be a primitiverepresentation of a branch ofC centred atO. SinceA(y(t)) = C(x(t), y(t)), itfollows that j(m − pn) = im. As gcd(m, pn) = 1, this leaves only one case,namelyi = m − pn, j = m. SinceO is an(m − pn)-fold point, the assertionfollows. As a consequence,C is irreducible.

Going back the original frame,

(x(t), y(t)) =

(utm−pn

+ . . .

vtm + . . .,

1

vtm + . . .

)=(t−pn

(u1 + . . .), t−m(v1 + . . .))

is a primitive representation of the unique branch ofC centredP∞. As ordtx(t) isdivisible byp, the computation of ordt(dx(t)/dt) is not simple. To overcome thedifficulty, take derivatives inA(y(t)) = B(x(t)). This gives

a0dy

dt=((m− 1)bmx(t)

m−1 + . . .+ b1) dxdt.

Since ordty(t) is not divisible byp andordtx(t) < 0, so

−(m+ 1) = −pn(m− 1) + ordt(dx/dt).

By assumptionC has degree at least4 and this implies that(pn − 1)(m − 1) > 2.Therefore, ordt(dx/dt) = (pn − 1)(m − 1) − 2. In particular,x is a separablevariable ofΣ. Also, as no vertical line is tangent toC, so ordγ(dx(t)/dt) = 0 forevery branchγ of C centred at an affine point. SinceC has degreem, so (iii)(b)follows from Definition 5.54. LetP∞ be the place associated with the branchγ.So (iii)(a) holds.

The roots of the polynomialA(Y ) are the elements of an additive subgroupA ofK of orderpn. Also, for everya ∈ A, the linear collineation(x, y) 7→ (x, y + a)

Some Families of Algebraic Curves 527

preservesC; that is,αa : (x, y) 7→ (x, y + a) is aK-automorphism of the functionfield Σ = K(x, y) of C which fixes the placeP∞. LetG be the group consistingof theseαa with a ranging overA. Fora 6= 0, the automorphismαa fixes no moreplace ofΣ showing (iii)(c) and the first part of (g).

To prove (e), takeP∞ again to the origin by interchanging the coordinatesX0

andX2. ThenΣ is assumed to beK(x, y) with A(x) = C(x, y) and, for everya ∈ A,

αa : (x, y) 7→(

x

1 + ay,

y

1 + ay

).

A uniformizing element atO can be found by using the following argument. Sincegcd(m, pn) = 1, there exist integersk, ℓ such thatk(m − pn) + ℓm = 1. Letξ = xkyℓ.As shown before,x(t) = uym−pn

+. . . , y(t) = vtm+. . . , is a primitiverepresentation of the unique branch centred atO. If O is the corresponding placeof Σ, then

ordOξ = ordt(tk(m−pn)+ℓm + . . .) = 1,

showing thatξ = xkyℓ is a uniformizing element ofΣ atO.As gcd(, pn,m) = 1), for everya ∈ A, a 6= 0,

ordt(α(ξ(t))− ξ(t)) = ordt

(x(t)ky(t)ℓ

(1 + ay(t))ℓ+m− x(t)ky(t)ℓ

)

= ordt(aty(t) + . . .)

is equal tom+ 1 whence (e) follows. Hilbert’s Different Formula (12.38) togetherwith (e) give the second part of (f). From this, by Theorem 12.66, the first part in(g) follows. Finally, Lemma 12.117 shows the last assertion.

The above proof can be adapted for the casepn > m. 2

L EMMA 13.2 (i) For everyk > 1,

|kP∞| = div(∑

cijxiyj) + kP∞,

summed over alli, j ≥ 0 satisfying bothj ≤ pn − 1 andipn + jm ≤ k.

(ii)

gg−12g−2 = div(

∑cijx

iyj) + (2g − 2)P∞,

summed over alli, j ≥ 0 satisfyingipn + jm ≤ (pn − 1)(m− 1)− 2.

Proof. With the notation in the previous proof, ordtx(t)iy(t)j = −(ipn + jm).

Since every place distinct fromP∞ arises from a branch ofC centred at an affinepoint, xiyj has no more poles, this implies that div(

∑cijx

iyj) + kP ∈ |kP|provided that

ipn + jm ≤ k whencij 6= 0. (13.2)

Thosexiyj satisfying not only (13.2) but alsoj < pn provide a linearly indepen-dent set overK.

528

Every element inΣ has a representationa(x, y)/b(x) wherea(X,Y ) ∈ K[X,Y ]andb(X) ∈ K[X]. As C = v(A(Y ) − B(X)), with degA(Y ) = pn, it may beassumed thatdegY a(X,Y ) < pn. Now, suppose that

div

(a(x, y)

b(x)

)≻ kP. (13.3)

Thena(x, y)/b(x) has no pole at an affine point. To show thatb(X) is a constantpolynomial, assume on the contrary thatu ∈ K be a root ofb(X). Then everypointP = (u, v) of C is a zero ofa(x, y) ∈ Σ; that is, every root of the polynomialA(Y ) − B(u) is a root of the polynomiala(u, Y ). The former polynomial haspn

distinct roots sincedA(y)/dY = 1, but the latter one has fewer roots, as its degreedoes not exceedpn − 1. This contradiction proves the assertion. So, it may beassumed thatb(X) = 1.

Let a(X,Y ) =∑cijX

iY j . If some(i, j) satisfies (13.2), replacea(X,Y ) bya(X,Y ) − cijXiY j . In doing so,ipn + jm > k with cij 6= 0. To show that thisleads to a contradiction, let a primitive representation ofthe unique branch centredatP∞ be

(x(t), y(t)) =(t−pn

(u1 + . . .), t−m(v1 + . . .)).

Therefore, ifcij 6= 0 then ordt x(t)iy(t)j = −(ipn + jm). As k < ipn + jm forcij 6= 0, it turns out that either ordta(x(t), y(t)) < −k, or there are two terms ina(X,Y ), saycijXiY j andcı,X ıY such that

ordt x(t)iy(t)j = ordt x(t)

ıy(t).

In the former case, the contradiction is with (13.3), while in the latter case,

(ı− i)pn = (− j)m,which givesj ≡ (mod pn). SincedegY a(X,Y ) < pn this implies thatj = and hencei = ı, a contradiction.

The second assertion is a corollary to the first when applied to k = 2g − 2, as(2g − 2)P∞ is the canonical series onX . 2

From Lemma 13.2 it also means that the Weierstrass semigroupof Σ at P isgenerated bypn andm. Hence, the following result holds.

L EMMA 13.3 There are exactly12 (pn−1)(m−1) gaps, every other positive integerbeing a linear combination ofm andpn with positive integer coefficients.

A characterisation of the curveC of (13.1) is given in the following theorem.

THEOREM 13.4 LetF be a projective irreducible plane curve such that a placeP of its function fieldΣ = K(F) satisfies the following conditions:

(i) Σ has aK-automorphism groupN of orderpn whose non-trivial elementsfixP but no other place ofΣ;

(ii) ΣN is rational;


(iii) Σ contains an elementy such thatdiv(y)∞ = mP with gcd(m, p) = 1;

(iv) Σ has genusg ≥ 12 (pn − 1)(m− 1).

ThenF is birationally equivalent to the projective irreducible plane curveC of(13.1) withA(Y ), B(X) satisfying(I)− (V ).

Proof. Choosex ∈ Σ such thatΣN = K(x). Let σ be a primitive representationof the placeP. If σ(x) = cit

i + . . . with ci 6= 0, i ≥ 0, replacex by x−1 or(x − c0)−1 depending on whetheri > 0 or i = 0. ThenP may be taken to be apole ofx and is the unique pole ofx. If Q were another pole ofx, then every placein the orbit ofQ underN would be a pole ofx as well. Sox would have more thanpn poles. But this contradicts Theorem 5.34 since

pn = |N | = [Σ : ΣN ] = [Σ : K(x)].

Thus, div(x)∞ = pnP. From this two facts follow. First, the Weierstrass semi-group atP contains the semigroup generated bypn andm, wherem is defined asin (iii). By Lemma 13.3 and (iv), it follows thatg = 1

2 (pn − 1)(m − 1). Sec-ondly, div(x)∞ = pnP, together with (iii), implies thatΣ = K(x, y). In fact,[Σ : K(x, y)] dividesgcd(m, pn) = 1, as

[Σ : K(x, y)][K(x, y) : K(x)] = [Σ : K(x)] = pn

while

[Σ : K(x, y)][K(x, y) : K(y)] = [Σ : K(y)] = m.

It remains to prove thatx andy satisfy an equation of typeA(y) = B(x), withA(Y ), B(X) as in(I)− (V ). To do this, two cases are distinguished.

Assume first thatm < pn, and chooseα ∈ N . Sincem is the smallest non-gapat P, there existsdα ∈ K such thatα(y) = y + dα, by Lemma 12.13. Then,dα + dβ = dαβ for any twoα, β ∈ N . As N is a group,dα | α ∈ N is anadditive subgroup ofK of orderpn. Then the polynomialA(Y ) =

∏(Y + dα)

with α ranging of all elements inN , is of type (II). Also,A(y) ∈ K(x), becauseA(y) is fixed byN . Therefore,A(y) = u(x)/w(x) whereu(X), w(X) ∈ K[X]are polynomials without common roots.

In fact,w(X) must be constant by the following argument. Ifc is a root ofw(X),replacex byx−c. LetQ be a zero ofx; thenQ is not a zero ofu(x). Also,Q 6= PbecauseP is a pole ofx. HenceQ is a pole ofA(y). As A(Y ) ∈ K[Y ], this isonly possible wheny itself has a pole atQ, contradicting that div(y)∞ = mP. So,there existsB(X) ∈ K[X] such thatA(y) = B(x). Then

div(A(y))∞ = div(B(x))∞.

As div(A(y))∞ = pnmP, and div(x)∞ = pnP, this implies thatdegB(X) = m.So, the proof is complete form < pn.

Now, assume thatm > pn, and take a non-trivialK-automorphismα ∈ N . If σis a primitive representation ofP, thenσ(y) = ct−m+. . .. If λα : t 7→ t+uti+. . .is the companion automorphism ofα, then

σ(α(y)) = c(t+ uti + . . .)−m + . . . .

530

Thus,σ(α(y)− y) = dt−j + . . . with j < m. Asm > pn, this implies that

div(α(y)− y)∞ = ℓ pnP. (13.4)

Since every non-gap atP smaller thanm is a multiple ofpn,

|ℓpnP| = div(∑ℓ

i=0 cixi) + ℓpnP.

Hence, there is a polynomialPα(X) such that

α(y)− y = Pα(x), s = deg Pα(X) ≤ ℓ. (13.5)

Changey to a uniformizing elementη by puttingη = yux−v with−um+vpn = 1.Sinceα(x) = x, so (13.4) becomes

div(α(η)− η)∞ = (m+ 1− spn)P. (13.6)

In fact,

α(η)− η = x−v(α(y)u − yu) = x−v(uyu−1Pα(x) + . . .+ Pα(x)u),

which implies that

ordt σ(α(η)− η) = pnv −m(u− 1)− pns = m+ 1− pns.

On the other hand, the Hilbert Different formula (12.38) applied toN gives

(m+ 1)(pn − 1) =∑

i≥0 (|N (i)P | − 1),

whereN (i)P is the i-th ramification group ofN at P. By (13.6) andℓpn < m it

follows thatN (i)P is trivial for any i > m. HenceN must coincide withN (i)

P for0 ≤ i ≤ m. Again by (13.6),Pα(X) is a constant, saydα ∈ K, for everyα ∈ N .Now the proof can be completed in the same way as in the casem < pn. 2

Forn = 1, Theorem 13.4 becomes the following.

THEOREM 13.5 LetF be a projective irreducible plane curve such that its func-tion fieldΣ = K(F) satisfies the following conditions:

(i) Σ has an automorphismα of orderp fixing only one place;

(ii) Σα is rational;

(iii) P is the fixed place ofα and andm is the smallest pole number atP whichis prime top.

ThenF is birationally equivalent to the curve,

C = v(Y p − Y −B(X)),

withB(X) ∈ K[X] of degreem.

Proof. The background and arguments from the preceding proof are used. LetΣ = K(x, y), Σα = K(x), div(x)∞ = pP, div(y)∞ = mP.

If m < p, thenα(y) = y + d with d ∈ K. Thereforeαi(y) = y + id, fori = 0, 1, . . . , p− 1. Since

∏p−1i=0 α

i(y) is fixed byα,

yp − dp−1y =∏p−1

i=0 (y + id) ∈ K(x).


Replacingy by yd−1 givesyd − y ∈ K(x). This implies thatyp − y = B(x) withB(X) ∈ K[X]. In particular,degB(X) = m. ThereforeC has equation of type(13.1) withA(Y ) = Y p − Y .

If m > p, thenα(y) = y+P (x) with P (X) ∈ K[X] andp ·degP (X) < m. Toprove thatP (X) is a constant polynomial, suppose on the contrary thatP (X) hasa root, saya. Replacingx by x − a, it may be assumed without loss of generalitythatP (X) = XQ(X) with Q(X) ∈ K[X]. This implies that

yp −Q(x)p−1xp−1y = B(x)

for a polynomialB(X) ∈ K[X] whose degree ism,since

ordP y = −m, ordP x = p, p · degQ(X) < m− p.Also, B(0) = 0 after replacingy by y − u whereup = B(0). Therefore thefollowing equations are valid:

yp −Q(x)p−1xp−1y = amx + . . .+ a1x;

α(x) = x;

α(y) = y + xQ(x), p · degQ(x) < m− p.The plane curve

G = v(Y p −Q(X)p−1Xp−1Y − (amXm + . . .+ a1X))

is a birational model ofΣ. If the placeQ is a zero ofx, thenQ is a zero ofy, aswell. Hence the branch ofG corresponding toQ is centred at the originO. Sincethe set of zeros ofx is preserved byα, but none of these is fixed byα, the originis the centre of at leastp distinct branches ofG. On the other hand,O is at most ap-fold point ofG. Hence the places corresponding to branches centred atO form asingle orbit underα, and hence

div(x)0 =∑p−1

i=0 Qαi

.

As div(y)0 ≻∑p−1

i=0 Qαi

, it follows thatP is the unique pole ofy/x. Further,ordP(y/x) = −(m − p). Sincegcd(p,m − p) = 1, this contradicts the choice ofy, and proves thatP (X) is a constant polynomial. Now, the proof finishes in thesame way as in the casem < p. 2

THEOREM 13.6 Let G be theK-automorphism group of a plane curveC as in(13.1). For a placeP of Σ = K(C), letGP∞

= G(1)P∞

⋊H with p ∤ |H|. Then

(i) |H| dividesm(pn − 1);

(ii) |G(1)P∞| ≤ pn(m− 1)2 =

4pn

(pn − 1)2g2;

(iii) |G(1)P∞| = pn whenm 6≡ 1 (mod pn) and sog 6≡ 0 (mod pn);

(iv) |G(2)P∞| = pn whenm ≡ 1 (mod pn) and sog ≡ 0 (mod pn).

532

Proof. Let α ∈ GP∞. By Lemma 12.20,α preserves both|mP∞| and|pmP∞|.

First, consider the casem < pn. By Lemma 13.2,

|mP|= div(c0 + c1y) +mP | c = (c0, c1) ∈ PG(1,K),|pnP|= div(cx+

∑ri=0 diy

i) + pnP | c = (d0, . . . , dr, c) ∈ PG(r + 1,K),with r the integer defined byrm < pn < (r + 1)m. Hence,

α(y) = ay + b, α(x) = cx+ P (y),

whereP (Y ) ∈ K[Y ] such thatm · degP (Y ) < pn, anda, b, c ∈ K with ac 6= 0.As α(A(y)) = α(B(x)), the polynomial

H(X,Y ) =A(aY + b)−B(cX + P (Y ))

= anapn

Y pn

+ an−1apn−1

Y pn−1

+ . . .+ a0aY +A(b)− bmcmXm

−cm−1(bmmP (Y ) + bm−1)Xm−1 − . . .−B(P (Y ))

is divisible byA(Y )−B(X). Also,

degH(X,Y ) = pn = deg(A(Y )−B(X)),

which implies thatH(X,Y ) coincides withA(Y )−B(X) up to a constant factor;that is,H(X,Y ) = k(A(Y )−B(X)) with k ∈ K\0. Since

k = apn

= a = cm, cm−1(bmmP (Y ) + bm−1) = kbm−1,

soP (Y ) is a constant; namely,P (Y ) = (c− 1)bm−1/(mbm). Hence,

α(x) = cx+(c− 1)bm−1

mbm, α(y) = ay + b with apn

= a = cm. (13.7)

The mapα 7→ c is a homomorphism ofGP∞into the multiplicative group ofK,

whose kernel isG(1)P∞

asc = 1 implies thata = 1.

To calculate the order ofG(1)P∞

, the number of elementsb whenα varies inG(1)P∞

needs to be found. These are given by the roots of the polynomial,

A(Y ) = kB

((c− 1)bm−1

mbm

),

whose number ispn. Further,H is isomorphic to a subgroup of the group of allm(pn − 1)-st roots of unity, ascm(pn−1) = apn−1 = 1. This completes the prooffor m < pn.

The proof for the casem > pn is similar. Again, by Lemma 13.2,

|pnP|= div(c0 + c1x) + pnP | c = (c0, c1) ∈ PG(1,K),|mP|= div(cy +

∑ri=0 dix

i) +mP | c = (d0, . . . , dr, c) ∈ PG(r + 1,K),with r defined byrpn < m < (r + 1)pn. Therefore,

α(x) = cx+ d, α(y) = ay +Q(x), (13.8)

whereQ(X) ∈ K[X] such thatpn · degQ(X) < m, anda, c, d ∈ K with ac 6= 0.The polynomial

H(X,Y ) =A(aY +Q(X))−B(cX + d)

= anapn

Y pn

+ an−1apn−1

Y pn−1

+ . . .+ a0aY +A(Q(X))

−bmcmXm − cm−1(bmmd+ bm−1)Xm−1 − . . .−B(d)


is A(Y ) − B(X) up to a constant factor; that is,H(X,Y ) = k(A(Y ) − B(X))with k ∈ K\0. As before, this givesapn

= a = cm. Again, the mapα 7→ c is ahomomorphism ofGP∞

into the multiplicative group ofK. Since

cm(pn−1) = apn−1 = 1,

it follows thatH is isomorphic to a subgroup of the group of allm(pn−1)-st rootsof unity. Asm < pn, the kernel of this homomorphism coincides withG(1)

P∞since

c = 1 implies thata = 1.To calculate the order ofG(1)

P∞, first consider the casem 6≡ 1 (mod pn). Here,

no monomial inA(Q(X)) has degreem− 1. Therefore, from the equation,

H(X,Y ) = k(A(Y )−B(X)), (13.9)

it follows that

cm−1(bmmd+ bm−1) = 0;

that is,d = −bm−1/(mbm). Then,d = 0 after substitutingx − bm−1/(mbm) forx. As before, (13.9) implies thatA(Q(X)) = 0 and this only occurs whenQ(X)

is constant. More precisely,Q(X) = u with A(u) = 0. Henceα ∈ G(1)P∞

if andonly if

α(x) = x, α(y) = y + u, u ∈ K, A(u) = 0.

AsA(Y ) haspn distinct roots, so|G(1)P∞| = pn. This completes the proof of (iii).

For the casem ≡ 1 (mod pn), more is needed. Letm = vpk + 1 with k ≥ n

andgcd(p, v) = 1. By (13.8), anyα ∈ G(1)P∞

has an equation of type,

α(x) = x+ d, α(y) = y +Q(x),

with

Q(x) = f1x+ . . .+ fvxv + . . .+ fvpk−nxvpk−n

+ b.

Since

ll0 = A(α(y))−B(α(x)) =A(y) +A(Q(x))−B(x+ d)

=A(Q(x)) +B(x)−B(x+ d),

it follows that

A(f1X + . . .+ fvXv + . . .+ fvpk−nXvpk−n

+ b) +B(X)−B(X + d)

is the zero polynomial, and hence

anfpn

vpk−n −mbmd = 0.

To calculate the2nd ramification groupG(2)P∞

, note that|G(2)P∞| ≥ pn by Lemma

13.1 (iii)(f). To see that equality holds, letα ∈ G(2)P∞

, that is,

ordt σ(α(z)− z) ≥ 3,

wherez = xvpk−n

/y is a local parameter atP∞. As

α(z)− z =y(xpk−n

+ dpk−n

)v − yxvpk−n −Q(x)xvpk−n

(y +Q(x))y,

534

straightforward computation shows thatfvpk−n = 0, whenced = 0. Therefore,

α ∈ G(2)P∞

if and only if

α(x) = x, α(y) = y + u, u ∈ K, A(u) = 0.

SinceA(Y ) haspn distinct roots, so|G(2)P∞| = pn. By Theorem 12.71 the proof is

complete. 2

THEOREM 13.7 If equality holds in both(i) and(ii) in Theorem 13.6, then

B(X) = bmXm.

Proof. Let α ∈ H. From the proof of Theorem 13.6,

α(x) = cx+ d, α(y) = ay +Q(x),

with apn

= a = cm andQ(X) ∈ K[X]. Also,Q(X) is linear form < pn whileits degree is at mostm/pn for m > pn. Assume thatα has orderm(pn − 1). Thenβ = αpn−1 has orderm. After replacingx by x+ u with a suitableu ∈ K,

β(x) = cx, β(y) = y +Q(x), with cm = 1, Q(X) ∈ K[X]. (13.10)

The key idea in the proof is to show thatQ(X) is the zero polynomial. To dothis it is shown first thatβ fixes every place arising from a branch centred on the

Y -axis. As, by Theorem 12.71,ΣG(2)P is rational, Lemma 12.14 (ii)(d) implies that

β fixes two places ofΣG(2)P , one of them is the place underP∞, the other lies under

pn distinct places ofΣ, sayR1, . . . ,Rpn . Since theK-automorphisms inG(2)P act

as

(x, y) 7→ (x, y + u), with u ∈ K andA(u) = 0,

they commute withβ. Thus,G(2)P ⋊ H is an abelian group. Asp does not divide

the order ofβ, this can only occur whenβ fixes each of the placesR1, . . . ,Rpn .In particular,Q(0) = 0. By β(x) = cx, the centres of the corresponding brancheslie on theY -axis, and they are indeed all the places ofΣ centred at an affine pointon theY -axis. Now, by Theorem 12.66 applied to theK-automorphism groupgenerated byβ,

2g − 2 = (2g1 − 2)m+ (pn + 1)(m− 1),

whereg1 denotes the genus ofΣβ . This implies thatg1 = 0. HenceΣβ is rational.Let Σβ = K(η). Then[Σ : K(η)] = m is prime top, and henceΣ = K(x, η).

After replacingη by y in the proof of Theorem 13.6, the argument used there showsthatA(x) = B(η) with A(X) andB(Y ) as in (13.1). The advantage is that theaction ofβ onΣ has become as simple as possible, that is,

β(x) = cx, β(η) = η.

As β is aK-automorphism ofΣ, soB(cx) = B(x). SinceB(X) has degreemandc is an element of orderm in the multiplicative group ofK, this only happenswhenB(X) = bmX

m. 2


THEOREM 13.8 Suppose that

(a) C is a projective irreducible plane curve of degree≥ 4 as in (13.1) with

A(Y ) = Y pn

+ Y, B(X) = Xm, gcd(p,m) = 1;

(b) G is theK-automorphism group ofC;

(c) P∞ is the place arising from the unique branch centred at the unique infinitepoint of C;

(d) GP∞= G

(1)P∞

⋊H with p ∤ |H|.Then

(i) |H| = m(pn − 1);

(ii)

|G(1)P∞| =

pn, if m− 1 is not a power ofpn,pn(2k+1), if m− 1 = pnk.

Proof. For a primitivem(pn − 1)-st root of unity, the birational transformationαof Σ = K(C) defined by

α(x) = cx, α(y) = cmy,

is aK-automorphism generating a cyclic group of orderm(pn− 1). Theorem 13.6(i) implies thatH = 〈α〉, showing (i).

By Theorem 13.6 (iii), part (ii) holds form 6≡ 1 (mod pn). Therefore, letm ≡ 1 (mod pn) and s = (m − 1)/pn. As in the proof of Theorem 13.6, ifα ∈ GP∞

, then

α(x) = x+ d, α(y) = y +Q(x), pn · degQ(X) < m. (13.11)

Hence,α ∈ GP∞if and only if Q(x)pn

+ Q(x) = (x + d)m − xm. So, thepolynomial equation

Q(X)pn

+Q(X) = (X + d)m −Xm. (13.12)

is to be solved. Ford = 0, only the trivial solution exists; that isQ(X) = c withcp

n

+ c = 0.Now, letd 6= 0. To find which coefficients of(X + d)m − Xm do not vanish,

consider thep-adic representation ofs: writing

s = sipi + si+1p

s+1 . . . , si 6= 0.

Suppose first thats is not a power ofp; thens − pn ≥ 12s. Putr = m − pi+n.

The coefficients in thep-adic representation ofr are smaller than or equal to thoseof m. Hence,

(mr

)does not vanish, by Lemma 5.64. Therefore,Xr has no zero

coefficient on the right-hand side in (13.12). So, the same holds for the left-handside. Asgcd(r, p) = 1,Q(X) itself containsXr which implies thatr ≤ degQ(X).On the other hand,degQ(X) ≤ s by pn · deg Q(X) < m. However, this leads toa contradiction, since

r = m− pi+n = (s− pi)pm + 1 > 12sp

n ≥ s.

536

This shows thatd = 0 provided thats is not a power ofp.It remains to investigate the case thats = pi; that is,m = pn+i + 1. This time,

(X + d)m −Xm = dXpn+i

+ dpn

X + dpn+i

.

With

Q(X) = q0 + q1X + . . .+ qrXr, qr 6= 0,

it follows from (13.12) that

qpn

0 + q0 = dpi+n+1, q1 = dpn+i+1, q2 = . . . = qpn−1 = 0.

In fact, the non-constant terms inQ(X) have degree of the formpjn, as a directargument shows. Therefore,m = pkn + 1.

From Theorem 13.6 (ii),

|G(1)Pinfty

| ≤ pnp2kn = p(2k+1)n.

To finish the proof, it suffices to exhibitp(2k+1)n K-automorphisms of type (13.11).To do this, choose elementsd, e ∈ K such that

dp2kn

+ (−1)kd = 0, epn

+ e = dpn+1, (13.13)

and let

α(x) = x+ d

α(y) = y + e+ dpkn

x− dp(k−1)n

+ . . .+ (−1)k−1dp(2k−1)n

xp(k−1)n

,(13.14)

It is straightforward to verify thatα is indeed aK-automorphism ofC:

α(y)pn

+ α(y) = ypn

+ y + epn

+ e+ dpkn

x+ (−1)k−1dp2kn

xpkn

=xpkn+1 + dpkn+1

+ dpkn

x+ dxkn

= (x+ d)pkn+1

= α(x)pkn+1.

For each of thepkn possible values ofd, there exist exactlypn values ofe that sat-isfy (13.13). This shows that (13.14) providesp(2k+1)n distinctK-automorphismsof C. 2

Let C again be the curve (13.1), and letΣ = K(C). Also, define the followingnotation for Weierstrass semigroups:

H(P) = the Weierstrass semigroup atP;

H0(P) = the Weierstrass semigroup atP generated bym andpn;

S0 = P | H(P) = H0(P).

THEOREM 13.9 Let Σ have a placeP 6= P∞ with P ∈ S0. Then, up to a linearsubstitution onX andY, one of the following holds:

(i) B(X) = Xm withm < pn, pn ≡ −1 (mod m), andS0 consists of thepn

distinct zeros ofx, together withP∞;


(ii) A(Y ) = Y pn

+ Y, B(X) = Xpn+1, that is, C is the Hermitian curve overFp2n , andS0 consists of the those places centred at thep3n + 1 rationalpoints ofC;

(iii) A(Y ) = Y pn

+ajYpj

+ · · ·+Y, B(X) = Xpn+1, with 1 ≤ j ≤ n− 1 andaj 6= 0, andS0 consists of thepn distinct zeros ofx, together withP∞.

Proof. Take the centre of the branch associated withP to the originP = (0, 0) bya translation. SinceC remains of type (13.1), all previous results hold true. Also,ordP x = 1 becauseP is a simple point ofC and the tangent toC is not the verticalline.

By Lemma 13.2, the canonical series ofΣ is |W |, where

W = ((pn − 1)(m− 1)− 2)P∞.

Hence,

div(xiyj) +W ≻ 0 whenever ipn + jm ≤ (pn − 1)(m− 1)− 2. (13.15)

First, take the casem > pn + 1. If pn > 2, then (13.15) implies that

div xpn−1 +W ≻ 0.

Actually, this is true forpn = 2 as well, sincegcd(m, p) = 1 together withdegC ≥ 4 implies thatm ≥ 5.

From the beginning of Section 7.6, the Weierstrass gaps are characterised bymeans of the canonical series as follows: a positive integeri is not in the WeierstrasssemigroupH(P) atP if and only if there existsz ∈ Σ such that div(z) +W ≻ 0and ordP z = i. This characterisation is used several times in this proof.Hence,pn does not belong toH(P), which gives a contradiction.

Assume thatm ≤ pn − 1 with pn > 4. From (13.15),

div xi +W ≻ 0 for 0 ≤ i ≤ m− 2,

div yj +W ≻ 0 for 0 ≤ j ≤ pn−1.

SinceA(y) = B(x),

z = anypn − bmxm − bm−1x

m−1

satisfies divz +W ≻ 0. If bm−1 6= 0, thenm < pn implies that ordP z = m− 1,which is a contradiction by the above characterisation of Weierstrass gaps. HenceeitherB(X) = bmX

m or

B(X) = bmXm + bm−2X

m−2 + · · ·+ bkXk with 1 ≤ k ≤ m− 2, bk 6= 0.

The latter case cannot in fact occur. Form = pn − 1, this is a consequence ofordP z = m = pn − 1 and the above characterisation of Weierstrass gaps. Simi-larly, for m ≤ pn − 2, there exists a rational functionη such that divη +W ≻ 0and ordPη = m− 1. Letting

η =

ym−1, if k = 1,yxm−k+1, if k ≥ 2,

this again gives a contradiction to the characterisation ofWeierstrass gaps. ThusB(X) = bmX

m. In particular,I(P, C∩ℓ) = mwhereP is the centre of the branchassociated toP andℓ = v(Y ).

538

Now, choose positive integerss < m andt such thatpn + s ≡ 0 (mod m); solm = pn + s. Note that

div(xy−l) = P2 + · · ·+ Pn + sP∞ − (pn + s− 1)P,whereP2, . . . ,Pn are the zeros ofx other thanP. Hencepn + s− 1 ∈ H(P) andthis gives a contradiction fors > 1. If s = 1, then

ordP(xy−t) = −pn, ordP(y−1) = −m.ThusH(P) is generated bym andpn. This result holds true for every zero ofx, thatis, for every place ofΣ arising from a branch centred at a common point ofC andthe lineX = 0. This depends on the fact that each translation(x, y) 7→ (x, y + c),for a rootc of A(Y ), preservesC.

The possibility of having another place with the same Weierstrass semigroup isruled out by using the invariance of the intersection multiplicity under translations.If such a place exists and it corresponds to a branch centred at the pointP = (u, v),the translation(x, y) 7→ (x− u, y − v) takesP to the originO and the image ofCunder any translation is a curveC′ of the same type. Hence what has already beenshown forC applies toC′. Therefore, ifℓ′ = v(Y ) andℓ = v(Y − v), then

I(O, C′ ∩ ℓ′) = I(P, C ∩ ℓ), I(O, C′ ∩ ℓ′) = m;

soI(P, C ∩ ℓ) = m. But a straightforward computation shows that this only occurswhenu = 0.

Letm ≤ pn−1 with pn ≤ 4. The casepn ≤ 3 cannot occur becausedeg C ≥ 4.Sincegcd(m, p) = 1, if pn = 4 thenm = 3, and

A(Y ) = a2Y4 + a1Y

2 + a0Y, B(X) = b3X3 + b2X

2 + b1X.

By (13.15), divz+W ≻ 0, wherez = a0y+ b1x. On the other hand, a calculationshows that

ordP z =

2, if a1a

−10 b21 + b2 6= 0,

3, if a1a−10 b21 + b2 = 0.

But again the characterisation of Weierstrass gaps leads toa contradiction sincebothpn = 4 andm = 3 are inH(P).

Letm = pn + 1. Without loss of generality, leta0 = 1. With η = y + b1x,

A(η)−B(x) =A(y) +A(b1x)−B(x) = A(b1x),

A(η) =B(x) +A(b1x) = bmxm + · · ·+ b′2x

2,

with b′2 = b2 for p > 2 andb′2 = b2+a1b21 for p = 2. After the linear transformation

(x, y) 7→ (x, a0y+b1x), it is possible to putb1 = 0, or equivalently, to assume that

pn + 1 ≥ ordPy ≥ 2.

In fact, ordP y = pn + 1. To show this, suppose on the contrary that

ordP y < pn + 1.

If ordP y = pn = m− 1, there is a contradiction to the characterisation of Weier-strass gaps since divy +W ≻ 0 by (13.15).


If ordPy < pn, let

z = yxpn−k−1, with k = ordP y.

As ordP z = pn − 1 and divz +W ≻ 0, once again there is a contradiction to thecharacterisation of Weierstrass gaps.

Therefore,B(X) = Xm. In other words, the pointP = (0, 0) of C has theproperty thatI(P, C ∩ ℓ) = m whereℓ is the tangent line toC atP .

By a preceding argument depending on the invariance of the intersection multi-plicity under translations, this holds true for any point ofC which is the centre of abranch corresponding to a place ofΣ with Weierstrass semigroupH0(P)

Now, if A(y) = anypn

+ a0y, thenC is the Hermitian curve overFp2n . Other-wise, there existsj with 0 < j < n such that

A(y) = anypn

+ ajypj · · ·+ a0y, aj 6= 0.

Since ordP(y−1) = −m, soH(P) = H0(P). As before, this holds true for everyzero ofx, again by the fact that every translation(X,Y ) 7→ (X,Y + c), for a rootc of A(Y ), preservesC.

It remains to show that no other place ofΣ has Weierstrass semigroupH0(P).Again, if P = (u, v) is the centre of a branch corresponding to such a place, thenI(P, C ∩ ℓ) = m, whereℓ is tangent line toC atP . But another calculation showsthat this is not the case as soon asu 6= 0. 2

THEOREM 13.10 If C is a curve of type (13.1), then theK-automorphism groupAutK(Σ) of C fixes the placeP∞ except in the following two cases.

(i) (a) C = v(Y pn

+ Y −Xm), withm < pn, pn ≡ −1 (mod m);

(b) the groupAutK(Σ) contains a cyclic normal subgroupCm of ordermsuch thatAutK(Σ)/Cm

∼= PGL(2, pn);

(c) Cm fixes each of thepn+1 places with the same Weierstrass semigroupasP∞;

(d) AutK(Σ)/Cm acts on the set of suchpn + 1 places asPGL(2, pn).

(ii) (a) C = v(Y pn

+ Y −Xpn+1), the Hermitian curve;

(b) AutK(Σ) ∼= PGU(3, q);

(c) AutK(Σ) acts on the set of all places with the same Weierstrass semi-group asP∞;

(d) AutK(Σ) acts on the set of such places asPGU(3, q) on a Hermitiancurve.

Proof. Assume thatG = AutK(Σ) does not fixP∞. The invariance of the Weier-strass semigroup under the action of AutK(Σ) implies that the orbit ofP∞ is S0.From this, Theorem 13.9 shows that there are only a few cases to consider.

First the following case is considered:

B(X) = Xm, m < pn, pn ≡ −1 (mod m), b0 = 0;

A(Y ) = Y pn

+ ajYpj

+ . . . Y, 1 ≤ j ≤ n− 1, aj 6= 0.

540

TheK-automorphisms,

(x, y) 7→ (x, y + d), A(d) = 0,

of Σ form a group acting transitively on the set of all places which are zeros ofx.If G 6= GP∞

, then the orbit ofP∞ consists of all zeros ofx together withP∞,and one of them, sayPO, corresponds to a branch centred at the originO. TheWeierstrass semigroupH(PO) is generated bym andpn. Also, dim |mPO| = 1anddim |pnPO| = s wheres = (pn + 1)/m. Since the tangent toC atO is notthe the linev(X), a primitive representation of the unique branch centred atO is(x = t, y(t) = vti + . . .). SinceA(y(t)) = B(x(t)),

y(t) = a−1O tm + . . . ;

so ordPO(y) = m. From the hypothesism < pn,

|mPO|= div(c0 + c1y−1) +mPO | c = (c0, c1) ∈ PG(1,K),

|pnPO|= div(c0 + c1y−1 + . . .+ cs−1y

−(s−1) + csxy−s) +mPO |

c = (c0, . . . , cs) ∈ PG(s,K),Assume thatα ∈ AutK(Σ) takesP∞ to PO. Thenα takes|mP∞| to |mP|, and|pnP∞| to |pnP|. Hence,

α(x) = y−1(cx+ d1y + . . .+ dsys), α(y) = y−1(ay + b). (13.16)

Here,b 6= 0, c 6= 0. Then,

A(α(y)) =A(a) + bpn

y−pn

+ ajbpj

y−pj

+ . . .+ by−1,

A(α(x)) = y−ms(cx+ d1y + . . .+ dsys)m.

On the other hand,

A(α(y)) = α(A(y)) = α(B(x)) = B(α(x))

implies thatymsA(α(y)) = ypn+1B(α(x)). Hence,

A(a)ypn+1 + bpn

y + ajbpj

ypn−pj+1 + . . .+ bypn

= cmxm +mcm−1xm−1(d1y + . . .+ dsys) + . . .

= cmA(y) +mcm−1xm−1(d1y + . . .+ dsys) + . . . .

SinceP = (x, y) is a generic point ofC, the polynomial,

H(X,Y ) =A(a)Y pn+1 + bpn

Y + ajbpj

Y pn−pj+1 + . . .+ bY pn

(13.17)

−(cmA(Y ) +mcm−1Xm−1(d1y + . . .+ dsYs) + . . .),

is divisible byA(Y )−B(X). Note that

degX H(X,Y ) < m, degX = A(Y )−B(X) = m.

Hence, no term inX can actually occur inH(X,Y ). Therefore,

A(a)Y pn+1 + bpn

Y + ajbpj

Y pn−pj+1 + . . .+ bY pn

= cmA(Y ).

SinceA(Y ) has only terms of typeY pi

, eitherb = 0 or aj = 0, a contradiction.


The next case is the following:

b(X) = Xm, m = pn + 1, b0 = 0,

A(Y ) = Y pn

+ ajYpj

+ . . . Y, 1 ≤ i ≤ n− 1, aj 6= 0.

This can be ruled out in a similar way, since

|pnPO|= div(c0 + c1xy−1) + pnPO | c = (c0, c1) ∈ PG(1,K),

|(pn + 1)PO|= div(c0 + c1xy−1 + c2y

−1) + (pn + 1)PO |c = (c0, c1, c2) ∈ PG(2,K).

In the case,

C = v(Y pn

+ Y −Xm), pn ≡ −1 (mod m),

the rational transformation ofΣ,

α(y) = y−1, α(x) = xy−s, (13.18)

with s = (pn + 1)/m is aK-automorphism ofΣ. In fact,

A(α(y)) =1

ypn +1

y=ypn

+ y

ypn+1=

xm

ysm= B(α(x)).

This shows thatG is larger thanGP∞. Also, the orbit ofP∞ consists of all zeros

of x together withP∞. To give an explicit permutation representation forG, thezeros ofx are identified with the elements ofFpn in the following way. Choose anelementu ∈ K satisfyingupn−1 = −1, and change(X,Y ) to (X,uY ). Then

C = v(Y pn − Y −Xm).

The place arising from the branch centred at the pointPv = (0, v) of C and theY -coordinatev of Pv are in one-to-one correspondence which can be extended toP∞ by associating the symbol∞ to it. Note thatvpn

= v; that is,v ∈ Fpn .To prove thatG acts onFpn ∪ ∞ as PGL(2, pn) on PG(1, pn), let G be the

permutation group induced byG on Fpn ∪ ∞. By Lemma 13.1 (iii)(c), everypermutation ofFpn ∪ ∞ of type,

v = v 7→v + b for v ∈ Fpn ,∞ for v =∞, (13.19)

with b ∈ Fpn is in G. Also, every permutation ofFpn ∪ ∞ of type,

v 7→cv for v ∈ Fpn ,∞ for v =∞, (13.20)

with c ∈ Fpn\0 is in G. Further,α defined as in (13.18) acts onFpn ∪ ∞ asthe permutation,

v 7→

v−1 for v ∈ Fpn ,∞ for v = 0,0 for v =∞.

(13.21)

The group generated by the permutations (13.19), (13.20), (13.21) is PGL(2, pn).In fact,G = PGL(2, pn). This is a consequence of Theorem 15.16.

542

An elementary proof is also possible, using Theorem 13.6 in the following way.TheK-automorphisms(x, y) 7→ (dx, y) with d ranging over them-th roots ofunity form a cyclic groupCm of orderm. Also, theK-automorphisms in (13.20)form a cyclic subgroupCpn−1. These two groups generate an automorphism groupof orderm(pn−1). On the other hand, from Theorem 13.6,|G∞| ≤ m(pn−1)pn,which implies that|GP∞,PO

| ≤ m(pn − 1). Therefore,|GP∞,PO| = m(pn − 1).

SinceG acts2-transitively on the orbit ofP∞, it follows that

|G| = m(pn + 1)pn(pn − 1).

As every place in this orbit is fixed byCm, this shows thatG/Cm∼= PGL(2, pn).

Finally, if

C = v(Y pn

+ Y −Xpn+1),

thenC is the Hermitian curve, and (ii) follows from Theorem 12.30. 2

REMARK 13.11 Let A(Y ) be a separable additive polynomial of degreepn, andB(X), C(X) polynomials with no common roots such that the rational functionR(X) = B(X)/C(X) is admissible, that is, no pole ofR(x) regarded as an ele-ment of the rational function fieldK(x) has multiplicity divisible byp. If

R(X) = f0(X) +∑r−1

i=1 fi(X)(X − ei)−1,

with fi ∈ K[X] for 0 ≤ i ≤ r − 1, thenR(X) is admissible if and only if thedegreedi of eachfi is relatively prime top. If, also,di ≥ 1 for 0 ≤ i ≤ r− 1, thenR(x) has exactlyr distinct poles inK(x). In this situation, the curve,

C = v(C(X)A(Y )−B(X)),

is irreducible and has genus,

g = 12 (pn − 1)(d0 − 1 +

∑r−1i=1 (di + 1)).

Also, if γ is thep-rank of the function fieldΣ = K(C), then

γ = (pn − 1)(r − 1).

In particular,g = γ if and only if di = 1 for 0 ≤ i ≤ r − 1; that is, every pole ofR(x) is simple.

13.2 CURVES WITH SUZUKI AUTOMORPHISM GROUP

Let p = 2, q0 = 2s, with s ≥ 1 andq = 2q20 = 22s+1. As before,K = Fq.LetF be the projective irreducible plane curve overFq with

F = v(X2q0(Xq +X)− (Y q + Y )). (13.22)

The Deligne-Lusztig curve associated with the Suzuki groupor DLS curve, forshort, is any algebraic curve birationally isomorphic toF .

So, the function field of a DLS curve isK(F) = K(x, y) with

x2q0(xq + x) = yq + y.


Note thatF is birationally isomorphic to the curve

v(Xq0(Xq +X)− (Y q + Y )).

In fact, if z = x2q0+1 + y2q0 , that is,yq = zq0 + xq+q0 , thenK(F) = K(x, z).Also,

zq0 = (xq∗q0 + yq) = xq0+1 + y,

and hencezq = xq+q0 + y2q0 . Now, since

zq + z = xq+q0 + y2q0 + x2q0+1 + y2q0 = x2q0(xq + x),

the assertion follows. So, the definition here is equivalentto the usual definition ofa DLS curve.

THEOREM 13.12 If F is the curve (13.22),P∞ = (0, 0, 1) andP∞ is the placeassociated to the branch centred atP∞, then the following properties hold:

(i) P∞ is a q0-fold point, it is the unique infinite point ofF , and is the centre ofjust one branch ofF ;

(ii) div(dx) = (2q0(q − 1)− 2)P∞;

(iii) F has genusg = q0(q − 1);

(iv) the setF(Fq) of all Fq-rational places ofF has sizeq2 + 1;

(v) theK-automorphism groupG of F is Fq-rational and is isomorphic to theSuzuki groupSz(q);

(vi) G acts onF(Fq) asSz(q) on the Tits ovoid inPG(3, q);

(vii) |G| = (q2 + 1)q2(q − 1), and|GP∞| = q2(q − 1);

(viii) |G(1)P∞| = q2, andG(2)

P∞is an elementary abelian group of orderq;

(ix) G(2)P∞

= . . . = G(2q0−1)P∞

= G(2q0+1)P∞

butG(2q0)P∞

= 1.Proof. It is immediate thatP∞ is the only singular point ofF and thatP∞ is a2q0-fold point ofF . First, it is shown that just one branch ofF is centred atP∞.As a consequence, the irreducibility ofF follows.

With h = xy + x2q0+2 + y2q0 , the functionh has a zero at the placeO ofthe fieldΣ = K(F) arising from the unique branchγ of F centred at the originO = (1, 0, 0). Actually,h has no more zeros corresponding to branches centred atan affine point. Equivalently, the system,

x2q0(xq + x)− (yq + y) = 0, (13.23)

xy + x2q0+2 + y2q0 =0, (13.24)

has only one solution overK, namely(x, y) = (0, 0), as follows From these equa-tions,

xq0yq0 + xq+2q0 + yq = xq0yq0 + x2q0+1 + y = 0,

544

whence

y2 = x2q0y2q0 + x4q0+2 = x2q0(xy + x2q0+2) + x4q0+2 = yx2q0+1.

This shows that eitherx = y = 0, or y = x2q0+1. If y = x2q0+1, from (13.22)xq = x. But then (13.24) givesx = 0, and againx = y = 0.

The utility of h depends on the fact that the rational transformationϕ, given by

(x, y) 7→ (y/h, x/h) (13.25)

is aK-automorphism ofΣ. From above,ϕ takes any branch centred at an affinepoint to a branch centred at an affine point, except forγ. Sinceϕ has order2 andhence coincides with its inverse, it takesγ to a branch centred atP∞.

Let (x′(t), y′(t)) be a primitive representation ofγ. Then

x′(t) = t, y′(t) = t2q0+1 + tq+2q0 . . . .

Therefore,

x(t) =t2q0+1 + . . .

tq+2q0+1 + t2q+2q0 + . . .= t−q 1

1 + tq−1 + . . .= t−q(1 + . . .),

y(t) =t

tq+2q0+1 + t2q+2q0 + . . .= t−q−2q0

1

1 + tq−1 + . . .= t−q−2q0(1 + ...).

The resulting branch has order2q0. SinceP∞ is a2q0-fold point, it is the uniquebranch ofF centred atP∞. So, in terms of places,ϕ interchangesP∞ andO. Inparticular,

ordP∞x = −q, ordP∞

y = −(q + 2q0), ordP∞h = −(q + 2q0 + 1).

To calculate div(dx), begin from

x(t)2q0(x(t)q + x(t)) = y(t)q + y(t).

Taking derivatives,

dy(t)

dt= x(t)2q0+1 dx(y)

dt.

Since ordt(dy(t)/dt) = (−2q0 − 2), this gives

ordtdx(t)

dt= 2q0(q − 1)− 2.

Therefore,P∞ contributes weight2q0(q − 1)− 2 to div(dx).To prove (ii), it must be shown that ordPdx = 0 for every placeP whose cor-

responding branch is centred at an affine point. By Lemma 5.55this only requiresthat no vertical line is tangent toF , or, equivalently, that the polynomialY q +Y +chasq distinct roots.

By Definition 5.54, (ii) implies (iii). Further,F(Fq) comprisesP∞ and everyplace which arises from a branch centred at a point inA(2, q). This proves (iv).

It can be shown that, for anya, c, d ∈ Fq with d 6= 0, the following transforma-tions areK-automorphisms ofK(F):

ψa,c : x 7→ x+ a, y 7→ a2q0x+ y + c; (13.26)

γd : x 7→ dx, y 7→ d2q0+1y. (13.27)


Let

Ψ = ψa,c | a, c ∈ Fq.It is straightforward to verify thatΨ is a group of orderq2 which preservesF(Fq).Also, Ψ fixesP∞ and acts on the other places inF(Fq) as a sharply transitivepermutation group. The involutory elements inΨ together with the identity forman elementary abelian subgroupΨ′ of orderq. Also,

H = γ | d ∈ Fq\0is a cyclic group of orderq − 1 which fixesP∞ and preservesF(Fq). The groupgenerated byΨ andH has orderq2(q − 1) and is the semi-direct productΨ ⋊H.This group together with theK-automorphismϕ of (13.25) generate a larger groupG which acts onF(Fq) as a2-transitive permutation group. From Theorem 15.16,G ∼= Sz(q), andG acts onC(Fq) as Sz(q) in its natural2-transitive representation.

Note thatG is Fq-rational as it is generated byFq-rationalK-automorphisms.

Also, |G| = (q2 + 1)q2(q − 1). HenceΨ = G(1)P∞

, andGP∞= G

(1)P∞

⋊H.Now, the ramification groups atP∞ can be determined by using Hilbert’s Dif-

ferent Formula 12.38. Asy/h is a uniformizing element, it suffices to calculate

ordP∞(u/h+ ψa,c(y/h)).

By a straightforward calculation,

u/h+ ψa,c(y/h)

=ay2 + a2q0xh+ bh+ (a2q0+1 + b)xy + (a2q0+2 + ab+ b2q0)y

h2 + ahy + (a2q0+1 + b)hx+ (a2q0+2 + ab+ b2q0)h,

which shows that ordP∞(u/h+ ψa,c(y/h)) equals either2 or q0 + 2 according as

a 6= 0 or a = 0. Therefore,Ψ′ = G(2)P∞

= . . . = G(2q0+1)P∞

. 2

A group-theoretic characterisation of the DLS curve is established in the follow-ing theorem.

THEOREM 13.13 With q0 = 2s, q = 2q20 , letF be an irreducible algebraic curveof genusq0(q − 1). If a K-automorphism group ofF contains a subgroupGisomorphic toSz(q), thenF is the DLS curve.

The proof is organized in a series of lemmas and uses some results obtained inSection 12.12. The same notation is employed here. Since|G| > 8g3, the relevantcase is (iv.5). Here,G has one non-tame orbito of places, and its faithful action ono is the natural2-transitive permutation representation of Sz(q). ForP ∈ o,

(i) |GP | = q2(q − 1);

(ii) G(1)P is a group of orderq2 acting ono\P as a sharply transitive permuta-

tion group;

(iii) GP = G(1)P ⋊H, whereH ∼= Cq−1;

(iv) F2 = ΣG(2)P is rational.

546

L EMMA 13.14 The groupG(2)P has orderq and consists of all involutory elements

in GP together with the identity. In particular,[G(1)P : G

(2)P ] = q.

Proof. Each involutory element in Sz(q) is the square of an element of order4 inSz(q). Since the factor groupG(1)

P /U(2)P is an elementary abelian2-group, every

involutory element inG(1)P belongs toU (2)

P . In particular,|U (2)P | ≥ q.

Choose any non-involutory elementα ∈ G(1)P , and look at the structure of the

1-point stabiliser of Sz(q). Every non-involutory element inG(1)P can be written as

βαβ−1δ with β ∈ H andδ2 = 1. Now, assume on the contrary thatα is in U (2)P .

SinceH is contained in the normaliserU (2)P , it follows thatβαβ−1δ is also inU (2)

P ,

contradicting thatG(1)P 6= U

(2)P . If ρ1 > 2, that is, ifG(1)

P = G(2)P , then Theorem

12.66 applied toG(1)P would imply that

2q0(q − 1)− 2 = 2g − 2 ≥ −2q2 + 3(q2 − 1) = q2 − 1,

which is impossible sinceq = 2q20 . Therefore,ρ1 = 2. 2

L EMMA 13.15 ρ2 = 2q0 + 2.

Proof. In the1-point stabiliser of Sz(q), any two involutory elements are conjugate.Thus, if i ≥ 3, then eitherG(i)

P = G(2)P , orG(i)

P = 1. By Theorem 12.66 applied

toG(2)P , it follows that

2q0(q − 1)− 2 = −2q + 2q − 2 + (ρ2 − 2)(q − 1),

whenceρ2 = 2q0 + 2. 2

L EMMA 13.16 The smallest non-gap atP is q.

Proof. Letm denote the smallest non-gap atP. Take an elementx ∈ Σ for whichdiv(x)∞ = mP. The groupG(1)

P induces an automorphism group onK(x). The

kernel of such a permutation representation containsG(2)P as every non-trivialK-

automorphism ofK(x) whose order is a power ofp has orderp. HenceK(x) ⊂ F2.Since

m = [Σ : K(x)] = [Σ : F2][F2 : K(x)],

and|G(2)P = [Σ : F2] by Theorem 12.35, it follows thatm ≥ q. On the other hand,

everyG(2)P -orbit distinct toP has lengthq. Therefore,m ≥ q whencem = q. 2

This result together with Theorem 12.14 also shows the following lemma.

L EMMA 13.17 There is an elementx ∈ Σ such thatdiv(x)∞ = qP, F2 = K(x),and, for everyα ∈ GP ,

α(x) =

x+ a, a ∈ K, a 6= 0, for α ∈ G(1)P \G

(2)P ,

bx, bq−1 = 1, b 6= 1, for α ∈ H,x, for α ∈ G(2)

P .

(13.28)


L EMMA 13.18 The smallest non-gapm at P which is not divisible byq does notexceedq + 2q0.

Proof. SinceF1 is rational,F1 = K(x1) with x1 ∈ Σ. After a change ofx1 inK(x1), letP be a pole ofx1. Apart from the trivial orbitP, all orbits of placesunderG(1)

P have maximum sizeq2. Since[Σ : F1] = q2, sox1 has at mostq2

distinct poles, each counted with multiplicity. Hence,P is the unique pole ofx1.SinceGP = G

(1)P ⋊H, an automorphismδ ∈ H of orderq − 1 preservesK(x1).

By Theorem 12.14,δ(x1) = ax1 with a ∈ K, a 6= 0. Henceδ preserves the setof zeros ofx1. Since the order ofδ is prime toq2, a power ofδ fixes at least onesuch zero, sayR. AsF1 is rational, ando is the onlyG(1)

P -orbit preserved byδ, itfollows thatR ∈ o. Hence,o consists of all zeros ofx1; that is,

div(x1) =∑

R∈o\P R− q2P.Now, take an elementβ ∈ Gwhich movesP, and putQ = Pβ . Note thatQ ∈ o.

ReplacingP byQ in the preceding argument gives

div(y1) =∑

R∈o\Q R− q2Qfor an elementy1 ∈ Σ. Let z1 = x1/y1; then div(z1) = q2Q− q2P.

Similarly, there existsz2 ∈ Σ such that

div(z2) = (2g − 2)Q− (2g − 2)P.To show this, note the following two properties of the differentialdx1:

(1) dx1 can have at most one pole, namelyP;

(2) the weight ofP in div(dx1) does not exceedq2, by div(x1)∞ = q2P.

As a consequence,dx1 has at most2g − 2 + q2 < 2q2 zeros. Sincex1 ∈ F1 and|G(1)

P | = n2 this implies that the zeros ofdx1 areq2. Also,

(div(dx1))δ = div(d(δ(x1))) = div(d(ax1)) = div(dx1);

soδ fixes div(dx1). Now, the above arguments can be used to prove thato\P isthe set of all zeros ofdx1 each with the same weight. Again, movingP toQ by anelementβ ∈ G, it can also be shown that there existsy1 such that

div(dx1)− div(dy1) = (2g − 2)Q− (2g − 2)P.Sincez2 = dy1 − dx1 ∈ Σ, it follows that

div(z2) = (2q0(q − 1)− 2)Q− (2q0(q − 1)− 2)P.Sinceq2 + 1 = (q + 2q0 + 1)(q − 2q0 + 1),

gcd(q2 + 1, 2(q(q0 − 1)− 1)) = q + 2q0 + 1.

Hence,

div(z3) = (q + 2q0 + 1)Q− (q + 2q0 + 1)Pfor some elementz3 ∈ Σ.

Finally, if α ∈ G(1)P \G

(2)P , then div(α(z3) − z3)∞ = (q + 2q0)P. This follows

from Lemma 12.76 and the fact thatρ1 = 2. 2

Choose an elementz ∈ Σ such that div(z)∞ = mP with m as in Lemma 13.18.

548

L EMMA 13.19 There exists a polynomialf(X) ∈ K[X] with deg f = m suchthatΣ = K(x, z), wherezq + z = f(x).

Proof. For any elementu ∈ F2, the equation[Σ : K(u)] = [Σ : F2][F2 : K(u)]

shows that if div(u)∞ = kP thenq | k. Hence,G(2)P must contain some elementδ

such thatδ(z) 6= z. Chooseα ∈ G(1)P \G

(2)P such thatα2 = δ. Thenα(z) 6= z, and

the minimality ofm implies that ordP(α(z)− z) > −m. As 2q > q + 2q0, this isonly possible when ordP(α(z)− z) = −q. Therefore,

α(z) = z + cx+ d with c, d ∈ K, c 6= 0.

Hence,

δ(z) = z + eδ with eδ ∈ K\0. (13.29)

Replacingz with e−1δ z ensures thatδ(z) = z + 1.

Now, take a generatorβ of H. From Lemma 12.17, eitherβ(z) = ǫ−mz, ordiv(β(z) − ǫ−mz)∞ < mP, whereǫ denotes a primitive(q − 1)-st root of unity.As before, this implies thatβ(z) = ǫ−mz + dx + e with d, e ∈ K. Replacingzwith z+ d′ whered′ = e/(c− 1) givesβ(z) = ǫ−mz+ dx. and replacingβ2 withβ, this becomes

β(z) = ǫ−mz. (13.30)

From this,eγ can be calculated for all non-trivial elementsγ ∈ G(2)P using two

facts:

(1) β−kδβk(z) = z + ǫ−mkeδ;

(2) any non-trivial element inG(2)P is conjugate toδ by an element inH.

So,eγ = ǫ−mkeδ showing thateγ 6= 0. If two of these values, sayeγ andeg′ were

equal, thenδ′ = γγ′ would fix x; that is,eδ′ would be zero. Sinceδ′ ∈ G(2)P , this

is impossible. As a consequence,ǫ−m is a(q − 1)-st of unity; that is, all non-zeroelements of a subfieldFq of K are obtained. Also, as eachα ∈ G

(1)P \G

(2)P is a

product of elements fromH andG(1)P , the coefficients of the equation ofα are all

in Fq. Therefore,GP is defined overFq. Now, for everyγ ∈ G(2)P ,

γ(zq + z) = (z + eγ)q + z + eγ = zq + z.

SinceK(x) = F2, this implies thatzq + z ∈ K(x). Hence,zq + z = f(x) withf(X) ∈ K[X] anddeg f = m, since bothz andx have only one pole, namelyP,and ordP(z)∞ = m, ordP(x)∞ = q. 2

Note thatm is even by Lemma 12.76.

L EMMA 13.20 f(X) = Xq+q0 +Xq0+1.

Proof. By (13.30) andm < 2q, applyingβ to zq + z + f(x) = 0 gives

f(X) = uXm + vXm+1−q, u 6= 0.


Note thatv 6= 0, because an equation of typezq + z = xm can be written asz = (zq/2 +

√uxm/2)2, showing that

div(zm/2 +√uym/2)∞ = 1

2mP,

in contradiction to the definition ofm. Now, takeα ∈ G(1)P \G

(2)P . Then

α(x) = x+ a, α(z) = z + bx+ c, a, b, c ∈ Fq, ab 6= 0.

Applying α to zq + z + uxm + vxm+1−q gives(mq

)6= 0 but

(m

q+i

)= 0 for

1 ≤ i ≤ m− q. Hencem− q is a power of2. Taking Lemma 13.18 into account,it follows that

f(X) = X2k

(rXq + sX), 2k ≤ q0, r, s ∈ K\0.Also, raq + sa = 0 for everya ∈ Fq. Hencer = s. Further,ca2k

= bq, whencer, s ∈ Fq. Replacingz by rz gives

f(X) = X2k

(Xq +X), with 2k ≤ q0.Applyingβ ∈ H to zq+z+f(X), whereβ(x) = ηx, β(z) = ǫz with ǫ, η primitive(q − 1)-st roots of unity, givesǫ = η2k+1.

For brevity, putq1 = 2k, q2 = q/q1, andq3 = q2/q1 = q/q21 . Note that each ofthese numbersq1, q2, q3 is an integer. Letv = zq2 + xq2+1; then

vq + v= zqq2 + x(q2+1)q + zq2 + xq2+1

=(zq + z)q2 + x(q2+1)

= [xq1(xq + x)]q2 + x(q2+1)q + xq2+1

=xq(xqq2 + xq2) + xqq2+q + xq2+1

=xq+q2 + xq2+1,

andvq1 = zq + xq+q1 = z + yq1+1. Also, in succession,

(zq + z)xq1(q3−1) = (xq + x)xq1xq1(q3−1),

zqxq1(q3−1) + zxq1(q3−1) =xq+q1q3 + xq1q3+1,

(zqx(q3−1) + xq2+q3)q1 = zxq1(q3−1) + xq1q3+1,

(zq2x(q3−1) + vq2)q1 = zxq1(q3−1) + v.

Letw = zq2x(q3−1) + vq2 . Since

ordP z = −m = −(b+ q1), ordP x = −q, ordP v ≤ −q(q2 − 1),

wq1 = zxq1(q3−1) + vq2 ,

it follows that

ordPw =1

q2(−q − q2 − q1q3q − q1q) = −(1 + q1 + q(q3 − 1)).

Also, if δ ∈ G(2)P , thenδ(w) = w + eq2

δ xq3−1 + e

q22

δ . From Lemmas 13.15 and12.76,

2 + 2q0 = 1 + 1 + +q2 + q(q3 − 1)− q(q3 − 1) = 2 + q2,

whenceq0 = 2q2. Also, q1 = q0, which gives the result. 2

550

THEOREM 13.21 If F ′ = F/G is a tame quotient curve of the DLS curve, then itis one of the following types, whereg′ is the genus ofF ′ andr = |G| :

(i) With r | (q − 1) andg′ = q0(q − 1)/r,

F ′ = v(Y (q−1)/rf(X)− (1 +Xq0)(Xq−1 + Y 2(q−1)/r));

(ii) With r | (q + 2q0 + 1) andg′ = (q + 2q0 + 1)(q0 − 1)/r + 1,

F ′ = v(Y (q+2q0+1)/rg(X)−Xq+2q0+1 + Y 2(q+2q0+1)/r);

(iii) With r | (q − 2q0 + 1) andg′ = (q − 2q0 + 1)(q0 + 1)/r − 1,

F ′ =v(bY (q−2q0+1)/rh(X)

−(Xq0−1 +X2q0−1)(Xq−2q0+1 + Y 2(q−2q0+1)/r)).

Here

f(X) = 1 +∑s−1

i=0 X2i(2q0+1)−(q0+1)(1 +X)2

i

,

g(X) = 1 +∑s−1

i=0 X2iq0(1 +X)2

i(q0+1)−q0 +Xq/2,

h(X) = 1 +∑s−1

i=0 X2i(2q0+1)−(q0+1)(1 +X)2

i

,

andb = λq0 + λq0−1 + λ−q0 + λ−(q0−1) with λ ∈ Fq4 of orderq − 2q0 + 1.

Since Sz(q) contains a large number of non-isomorphic subgroups of evenorder,especially2-subgroups, a complete classification for non-tame quotient curves isout of reach.

THEOREM 13.22 There exist non-tame quotient curvesF ′ = F/G of the DLScurveF given by the following examples, whereg′ is the genus ofF ′ andr = |G|.

(i) With r = q2 andg′ = q0(q/q2 − 1),

F ′ = v(∑q−1

i=0 Yqi2 −X2q0(Xq +X)),

whereFq2is a proper subfield ofFq. TheK-automorphismsψ0,c defined

in (13.26) form an elementary abelian groupG of order q2 = 2s′+1 with(2s′ + 1) | (2s+ 1).

(ii) With r = 4 andg′ = 14q0(q − 2),

F ′ = v(∑2s

i=0 Y2i − (f1(X) + f2(X) + f3(X))),

where

f1(X) =∑s

i=0

(∑sj=i X

2j)X2i

,

f2(X) =∑2s

i=0X2i

,

f3(X) =∑2s

i=s+2

(∑i−s−2j=0 X2j

)2q0

X2i

.

HereG ∼= C4.


13.3 CURVES WITH UNITARY AUTOMORPHISM GROUP

Let q = ph > 2. In this section,K = Fq andHq is the irreducible projective planecurve defined overFq2 with

Hq = v(Y q + Y −Xq+1). (13.31)

A Hermitian curvedefined overFq2 or theDeligne–Lusztig curve associated withPSU(3, q) is any algebraic curve birationally isomorphic toHq. In projective coor-dinates, equivalent forms of (13.31) are thev(Fi), i = 1, 2, 3, 4 with the followingFi:

(M1) F1 = Xq+10 +Xq+1

1 +Xq+12 ;

(M2) F2 = Xq2X0 −X2X

q0 + ωXq+1

1 , whereωq−1 = −1;

(M3) F3 = X1Xq −Xq

1X2 + ωXq+10 , whereωq−1 = −1;

(M4) F4 = Xq0X1 +Xq

1X2 +Xq2X0.

Each of the first three is obtained from (13.31) by a linear substitution defined overFq2 , but for (M4) this can be done only overFq3 ; see Remark 8.19. The functionfield ofHq isK(Hq) = K(x, y) with yq + y + xq+1 = 0.

The relevant properties ofHq are collected in the following theorem.

THEOREM 13.23 The curveHq is non-singular withP∞ the place associated tothe branch centred atP∞ = (0, 0, 1), and has the following properties:

(i) div(dx) = (q + 1)(q − 2)P∞;

(ii) Hq has genusg = 12q(q − 1);

(iii) the setF(Fq2) of all Fq2-rational places ofF has sizeq3 + 1;

(iv) theK-automorphism groupAutK(Σ) ofHq is Fq2-rational and isomorphicto PGU(3, q), acting projectively onHq(Fq2).

Proof. The curveHq is of type (13.1) withA(Y ) = Y q + Y and with the degreeq + 1 of B(X) not divisible byp. So the results obtained in Section 13.1 apply.Parts (i) and (ii) follow from Lemma 13.1, while (iii) is Theorem 13.9 (ii), and (iv)is Proposition 12.30. 2

THEOREM 13.24 For q = ph > 2, let F be an irreducible curve algebraic ofgenus1

2q(q− 1). If aK-automorphism group ofF contains a subgroupG isomor-phic toPSU(3, q), thenF is the Hermitian curve.

The proof is similar to that of Theorem 13.13. Since|G| > 8g3, case (iv.5) ofSection 12.12 occurs. From Theorems 12.109 and 12.111,G has just one non-tame orbito of places, and its action ono is faithful and2-transitive, the same asPSU(3, q) on its permutation representation of the classical unital.ForP ∈ o,

(i) |GP | = q3(q − 1);

552

(ii) G(1)P is a group of orderq3 acting ono\P as a sharply transitive permuta-

tion group;

(iii) GP = G(1)P ⋊H, whereH ∼= Cq−1;

(iv) F2 = ΣG(2)P is rational by Theorem 12.71.

The following lemmas can be proved using the same arguments as in the proofsof the analogous Lemmas 13.14 and 13.16. The necessary changes are as follows:

involutory element7−→ element of orderp;

n2 7−→ q3;

n 7−→ q2;

2g − 2 = 2n0(n− 1)− 2 7−→ 2g − 2 = q(q − 1)− 2.

L EMMA 13.25 G(2)P has orderq and consists of all elements of orderp in GP

together with the identity.

L EMMA 13.26 The smallest non-gap atP is q.

Theorem 13.24 now follows from these two lemmas and Theorems13.4 and13.10.

Since PGU(3, q) is particularly rich in subgroups, the corresponding quotientcurves ofHq provide a large family ofFq2-maximal curves. However, their ex-plicit determination appears to be difficult, especially for non-tame subgroups. Forthe simplest case, whereG has prime order, the following result is a complete de-scription of the quotient curves comes.

THEOREM 13.27 LetHq/G be the quotient curve ofHq with respect to a sub-groupG of PGU(3, q) of prime orderd.

(i) One of the following holds:

(a) d = 2 6= p;

(b) d = p;

(c) d ≥ 3 and(q2 − 1)(q2 − q + 1) ≡ 0 (mod d).

(ii) The curveF ′ = Hq/G is birationally Fq2-equivalent to one the followingcases of projective irreducible plane curves of genusg′ defined overFq2 .

(I) F ′ = v(Y q + Y −X(q+1)/2)

with d = 2, p > 2, g′ = 14 (q − 1)2 .

(II) (a) F ′ = v(∑h

i=1Yq/pi

+ ωXq+1)with d = p, q = ph, g′ = 1

2q(q/p− 1), whereωq−1 = −1;

(b) F ′ = v(Y q + Y − (∑h

i=1Xq/pi

)2)with d = p, q = ph, p ≥ 3, g′ = q(q − 1)/(2p).


(III) F ′ = v(Y q − Y X2(q−1)/d + ωX(q−1)/d)

with d ≥ 3, q ≡ 1 (mod d), g′ = q(q − 1)/(2d), whereωq−1 = −1 .

(IV) (a) F ′ = v(Y q + Y −X(q+1)/d)with d ≥ 3, q ≡ −1 (mod d), g′ = 1

2 (q − 1)((q + 1)/d− 1);

(b) F ′i = v(Xi(q+1)/d +X(i+1)(q+1)/d + Y q+1),

with d ≥ 3, q ≡ −1 (mod d), g′ = (q + 1)(q − 2)/(2d) + 1,1 ≤ i ≤ d− 1.

(V) F ′i = v(S(Xq/d, Y 1/d,X1/dyq/d)),

whereS(X,Y,Z) =

∏(βX + βqY + Z),

with β ranging over alld-th roots of unity, d ≥ 3, (q2 − q + 1) ≡ 0(mod d), g′ = 1

2 ((q2 − q + 1)/d− 1).

(iii) In (IV)(b) two such curves,F ′i andF ′

j , are birationallyFq2-equivalent if andonly if one of the following relations holds modulod :

i ≡ j, ij ≡ 1, ij + i+ j ≡ 0 ,i+ j + 1 ≡ 0, ij + i+ 1 ≡ 0, ij + j + 1 ≡ 0 .

(13.32)

The number of birationallyFq2-equivalent classes of curvesFi is

n(d) =

16 (d+ 1) for d ≡ 2 (mod 3) ,

16 (d− 1) + 1 for d ≡ 1 (mod 3) .

(13.33)

13.4 CURVES WITH REE AUTOMORPHISM GROUP

In this section,p = 3, q = 3q20 , with q0 = 3s, s ≥ 1, andK = Fq.LetF = v(F ) be the projective irreducible plane curve defined overFq with

F (X,Y ) = Y q2 − [1+(Xq−X)q−1]Y q +(Xq−X)q−1Y −Xq(Xq−X)q+3q0 .(13.34)

TheDeligne-Lusztig curve associated with the Ree groupor DLR curve, for short,is any algebraic curve birationally isomorphic toF . The function field ofF isΣ = K(x, y) with F (x, y) = 0.

To obtain a simpler description ofΣ, let

y1 =

(yq − yxq − x

)q0

− xq0(xq − x), (13.35)

y2 =xq0

(yq − yxq − x

)q0

− x2q0(xq − x)− yq0 . (13.36)

Hence

yq1 − y1 =xq0(xq − x), (13.37)

yq2 − y2 =xq0(xq − x). (13.38)

Also, y = xy3q0

1 − y3q0

2 . Therefore,Σ = K(x, y1, y2). The relevant properties ofF are collected in the following two theorems.

554

THEOREM 13.28 If F is the curve (13.34),P∞ = (0, 0, 1), andP∞ is the placeassociated to the branch centred atP∞, then the following properties hold:

(i) P∞ is a q0-fold point, it is the unique infinite point ofF , as well as the onlysingular point, and is the centre of just one branch ofF ;

(ii) div(dx) = (3q0(q − 1)(q + q0 + 1)− 2)P∞;

(iii) F has genusg = 32q0(q − 1)(q + q0 + 1);

(iv) the setF(Fq) of all Fq-rational places ofF has sizeq3 + 1;

(v) TheK-automorphism groupG ofF is Fq-rational and is isomorphic to theRee groupRee(q);

(vi) G acts onF(Fq) asRee(q) in its unique2-transitive permutation represen-tation on the Kantor ovoid;

(vii) |G| = (q3 + 1)q3(q − 1), and|GP∞| = q3(q − 1).

Proof. (ii) The rational subfieldK(x) of Σ = K(F) is regarded as the functionfield of the lineℓ = v(X). Then[Σ : K(x)] = q2. LetP ′

∞ be the unique place ofK(x) centred at the infinite point ofℓ. By Lemma 7.19, some places ofΣ lie overP ′. If P∞ is one of them, by (7.1),e∞ = −eP , wheree∞ denotes the ramificationindex ofP. From (7.3),e∞ ≤ q2. On the other hand, from Lemma 13.29 (I),

ordP∞(w8) = −(1 + q0

−1 + 2q−1 + (q0q)−1 + q−2)e∞.

Since ordP∞(w8) is an integer, soe∞ = q2. This together with (7.3) implies that

P∞ is the unique place ofΣ overP ′∞. Also,

ordP∞x = −q2, ordP∞

y1 = −q(q + q0), ordP∞y2 = −q(q + 2q0). (13.39)

Now, letP be any place ofΣ other thanP∞. Denote byP ′ the place ofK(x)lying underP. Let P ′ = (u, 0) be the centre ofP ′. Choose a primitive represen-tationσ of P. For everya, b ∈ Fq, let σ(a,b) be theK-isomorphism fromΣ intoK((t)) defined by

σ(a,b)(x) = σ(x), σ(a,b)(y1) = σ(y1) + a, σ(a,b)(y2) = σ(y2) + b.

Then,σ(a,b) is also a primitive place representation. By (13.37), (13.38), the corre-sponding placeP(a,b) of Σ lies overP ′. Note thatP = P(0,0). Since[Σ : Fq] = q2,this shows that there areq2 places ofΣ over every place ofK(x). Therefore, noplace ofΣ distinct fromP∞ ramifies. From (7.1) applied toξ = x − u, it followsthat ordP(x− u) = 1. Hence,x− u is a local parameter ofΣ atP. Foru 6= 0,

σ(x) = u+ t,σ(y1) = v + uq0t− (uq − u)tq0 + tq0+1 + . . . ,σ(y2) = w + u2q0t+ uq0(uq − u)tq0 − uq0tq0+1 + . . . ,

(13.40)

with

vq − v = uq0(uq − u), wq − v = u2q0(uq − u), (13.41)


while, foru = 0,

σ(x) = t,σ(y1) = v + tq0+1 + . . . ,σ(y2) = w + t2q0+1 + . . . ,

(13.42)

with v, w ∈ Fq.From parts (G) and (I) of Lemma 13.29,ζ = w6/w8 is a local parameter atP∞.

Hence,ζ is a separable variable, and it defines a differentialdξ for everyξ ∈ Σ.Again from (G) and (I) of Lemma 13.29,

dw6 = w3q0

4 dx, dw8 = w3q0

7 dx.

Since

1 = d(ζ) = d

(w6

w8

)=w6dw8 − w8dw6

w28

=w6w

3q0

7 − w8w3q0

4

w28

dx,

it follows that

dx =w2

8

w6w3q0

7 − w8w3q0

4

.

From parts (D) and (G) of Lemma 13.29,

ordP∞(w6w

3q0

7 − w8w3q0

4 ) = ordP∞(w6w

3q0

7 )

=−3(q0q2 + q2 + 2q0q + q + q0),

ordP∞w8 =−(q2 + 3q0q + 2q + 3q0 + 1).

Hence

ordP∞(dx) = 3q0q

2 + q2 − q − 3q0 − 2.

By (13.40) and (13.42), ordP(dx) = 0 for everyP 6= P∞. Therefore, (ii) holds.From this and Definition 5.54, (iii) follows.

As Σ = K(x, y1, y2), the pointQ = (1, x, y1, y2) defines a model ofΣ givenby an irreducible curveΓ of PG(4,K). Note thatΓ is defined overFq. From(13.37),(13.38), and (13.40) to (13.42),Γ consists of the pointP∞ = (0, 0, 0, 1) to-gether with all pointsP(u,v,w) = (1, u, v, w) whereu, v, w ∈ K satisfying (13.41).The pointP∞ is the unique singular point ofΓ. More precisely,P∞ is a (q0q)-fold point, and it is the centre of a unique branch ofΓ, namely that arising fromthe placeP∞ of Σ. Also, theFq-rational places ofΓ are those centred at pointswith coordinates inFq, that is, either atP∞ or atP(u,v,w) with u, v, w ∈ Fq. Thisproves (iv).

A computation shows that ifa, b, c, d ∈ Fq with a 6= 0, then the birationaltransformationφ(a,b,c,d),

x 7→ ax+ b, y1 7→ abq0x+ aq0+1y1 + c,y2 7→ ab3q0x− aq0+1bq0y1 + a3q0+1y2 + d,

(13.43)

of Σ is aK-automorphism ofΣ. Also,

φ1,b,c,d)φ(1,b′,c′,d′) = φ(1,b′′,c′′,d′′)

556

with

b′′ = b+ b′, c′′ = c+ c′, d′′ = b′3q0b− bq0c+ d′ + d.

TheseK-automorphisms form a subgroupΦ of AutK(Σ) of orderq3(q− 1). Also,

Φ1 = φ(1,b,c,d) | b, c, d ∈ Fq, Φ2 = φ(1,0,c,d) | c, d ∈ Fq,Φ3 = φ(1,0,0,d) | d ∈ Fq, Ψ = φ(a,0,0,0) | a ∈ Fq

are subgroups ofΦ; their orders are

|Φ1| = q3, |Φ2| = q2, |Φ3| = q, |Ψ| = q − 1,

andΦ2 is the commutator subgroup ofΦ1. In addition, the set of cubes inΦ1 is itscentreΦ3. Also,Φ = Φ1 ⋊ Ψ.

A furtherK-automorphism ofΣ is the birational transformationθ:

x 7→ w6/w8, y1 7→ w10/w8, y2 7→ w9/w8 (13.44)

Also,θ is an involution. As a consequence, a primitive representation τ of the placeP∞ is

τ(x) = σ(w6/w8), τ(y1) = σ(w10/w8), τ(y2) = σ(w9/w8), (13.45)

whereσ is given by (13.42).TheseK-automorphisms ofΣ all fix F(Fq). Also, everyφa,b,c,d fixesP∞, and

Φ1 acts on the remainingq3 places inF(Fq) as a sharply transitive permutationgroup. The mapθ interchangesP∞ and the placeP(0,0,0) corresponding to thebranch centred atP(0,0,0). Therefore, the groupG generated byΦ andθ acts onF(Fq) as a2-transitive permutation group.

To prove (v), the first step is to show that this permutation representation ofGon F(Fq) is faithful. Since|Φ1| = q3 andΦ is a subgroup ofG(0)

P∞, Theorem

12.71 together with (iii) imply that no non-trivial elementof G(1)P has more than

one fixed place. Hence the kernel of the representation is a tame subgroup ofGP∞.

By Theorem 12.56, this subgroup is cyclic, that is, generated by an elementα.From Lemma 12.94, ifα were non-trivial, then it would have at most2g + 2 fixedplaces while (iii) implies that2g+2 < q3. This contradiction shows that the kernelis trivial.

The possibilities for the structure ofG are given by Theorem 15.16, and it re-mains to show that only the caseG = Ree(q) can occur. IfG had a regular normalsubgroup, thenq3 + 1 = 2m for somem. But this equation has no solution, asqis a power of3. Also, PSL(2, q) has an abelian Sylow3-subgroup, and the sameholds for PGL(2, q). So, the final step in the proof of (v) consists in showing thatG cannot contain PSU(3, q). Since|PSU(3, q)| = q3(q3 + 1)(q2 − 1)/µ withµ = gcd(3, q + 1), it follows from (iii) that |G| > 8g3. From Theorem 12.113,Σis the Hermitian function field of genus12q(q − 1). But this is a contradiction asthis value is different from the genus ofΣ given by (iii). Finally,G is Fq-rationalas it is generated byFq-rationalK-automorphisms. 2

L EMMA 13.29 (i) Withx, y1, y2 ∈ Σ as in (13.35), (13.36),

ordP∞(y1) = −(1 + (3q0)

−1)e∞, ordP∞(y2) = −(1 + 2(3q0)

−1)e∞.


(ii) Define the further functions onΣ :

w1 = x3q0+1 − y3q0

1 , w2 = xy3q0+11 − y3q0

2 , w3 = xy3q0

2 − w3q0

1 ,

w4 = xw3q0

2 − y1w3q0

1 , v = xwq0

3 − y2wq0

1 , w5 = y1wq0

3 − y2wq0

2 ,

w6 = v3q0 − w3q0

2 + w3q0

4 ,

w7 = y1wq0

2 − xwq0

3 − w3q0

6 , w8 = w3q0

5 − xw3q0

7 ,

w9 = wq0

2 w4 − y1wq0

6 , w10 = y2wq0

6 − wq0

3 w4.

Then

(A) wq0

1 = xq0+1 − y1, wq1 − w1 = x3q0(xq − x),

ordP∞(w1) = −(1 + q0

−1)e∞;

(B) wq0

2 = xq0y1 − y2, wq2 − w2 = y3q0

1 (xq − x),ordP∞

(w2) = −(1 + q0−1 + q−1)e∞;

(C) wq0

3 = xq0y2 − w1, wq3 − w3 = y3q0


(w3) = −(1 + q0−1 + 2q−1)e∞;

(D) w3q0

4 = x3q0w2− y3q0

1 w1, wq4 −w4 = wq0

2 (xq − x)−wq0

1 (yq1 − y1),

ordP∞(w4) = −(1 + 2(3q0)

−1 + q−1)e∞;

(E) v3q0 = x3q0w3 − y3q0

2 w1, vq − v = wq0

3 (xq − x)− wq0

1 (yq1 − y1),

ordP∞(v) = −(1 + q0

−1 + q−1)e∞;

(F) w3q0

5 = y3q0

1 w3 − y3q0

2 w2, wq5 − w5 = wq0

3 (yq1 − y1)− wq0

2 (yq2 − y2),

ordP∞(w5) = −(1 + q0

−1 + q−1 + (3q0q)−1)e∞;

(G) wq0

6 = −w2 + xq0w4, wq6 − w6 = w3q0


(w6) = −(1 + q0−1 + 2q−1 + (q0q)

−1)e∞;

(H) w3q0

7 = y3q0

1 w2−x3q0w3−w6, wq7−w7 = wq0

2 (yq1−y1)−wq0

3 (xq−x),ordP∞

(w7) = −(1 + 2(3q0)−1 + q−1 + (3q0q)

−1)e∞;

(I) wq0

8 = w5 + xq0w7, wq8 − w8 = w3q0


(w8) = −(1 + q0−1 + 2q−1 + (q0q)

−1 + q−2)e∞;

(J) w3q0

9 = w2w3q0

4 − y3q0

1 w6, wq9−w9 = wq0

2 (wq4−w4)−wq0

6 (yq1 − y1),

ordP∞(w7) = −(1 + q0

−1 + 2q−1 + (3q0q)−1)e∞;

(K) w3q0

10 = y3q0

2 w6−w3w3q0

4 , wq10−w10 = wq0

6 (yq2−y2)−wq0

3 (wq4−w4),

ordP∞(w10) = −(1 + q0

−1 + 2q−1 + (3q0q)−1)e∞.

Proof. This is by straightforward calculation. 2

A major result on the DLR curve is the following characterisation, analogous toTheorem 13.13.

THEOREM 13.30 Letp = 3, q = 3q20 with q0 = 3s, s ≥ 1. LetF be an algebraiccurve of genusg = 3

2q0(q − 1)(q + q0 + 1), such that theK-automorphism groupofF contains a subgroupG isomorphic toRee(q). ThenF is the DLR curve.

As in the proof of Theorem 13.13, a series of preliminary results are first estab-lished.

558

L EMMA 13.31 There is a placeP of Σ such that the ramification groups ofGPhas the following properties:

(i) |G(1)P | = q3, andρ1 = 2, that is, G(1)

P 6= G(2)P ;

(ii) G(2)P is the commutator group ofG(1)

P , an elementary abelian group of orderq2, andρ2 = 3q0 + 2, that is,

U(2)P = G

(2)P = G

(3)P = . . . = G

(3q0+2)P ;

(iii) G(3q0+3)P is the centre ofG(1)

P , an elementary abelian group of orderq, andρ3 = q + 3q0 + 2, that is,

U(3)P = G

(3q0+3)P = G

(3q0+4)P = . . . = G

(q+3q0+2)P ;

(iv) U(4)P = G

(q+3q0+3)P = 1;

(v) ΣG(i)P is rational for0 ≤ i ≤ 3q0 + 2, butΣG

(3q0+3)

P has genus32q0(q − 1).

Proof. A Sylow 3-subgroupS3 of G has orderq3. Since3g < q3 and the3-rankof F is at mostg, Theorem 12.77 implies that the3-rank of F is zero. FromLemma 12.115,S3 fixes a placeP of Σ and no non-trivial element inS3 fixes aplace distinct fromP. In particular,S3 = G

(1)P for some placeP. SinceG has

q3 + 1 Sylow 3-subgroups, the setΩ of all places fixed by some3-element ofGhas sizeq3 + 1, andG acts onΩ as Ree(q) in its natural2-transitive permutationrepresentations.

From Theorem 12.71,ΣG(1)P is rational. This implies thatΣGP also rational.

The normaliser ofS3 in G coincides withGP , and henceGP = S3 ⋊ H withH ∼= Cq−1.

Theorem 12.66 applied toGP gives

2(g − 1) ≥ q3(2g1 − 2) + 2q3 − 2 + |G(2)P | − 1,

whereg1 denotes the genus ofΣG(1)P . Sinceg < q3, this implies thatg1 = 0. Hence

G(1)P 6= G

(2)P , that is,ρ1 = 2. From Lemma 12.75,

(q − 1) | (ρi − 1)(ni − 1). (13.46)

whereni = |U (i)P |/|U

(i+1)P |. For i = 1, this givesn1 = q; that is,|U (2)

P | = q2.

By Lemma 12.68 (ii),G(2)P contains the commutator subgroup ofG(1)

P . On theother hand, the commutator subgroup of a3-subgroup of Sylow of Ree(q) is ele-mentary abelian of orderq2. Therefore,G(2)

P is the commutator subgroup ofG(1)P .

The genusg2 ofG(2)P is calculated from Theorem 12.66 applied toG(1)

P regarded

as aK-automorphism group ofΣG(2)P :

2g2 − 2 = −2q2 + 2(q2 − 1) + ρ1(q − 1),

which givesg2 = 0.


Further,ρ2 ≡ 2 (mod 3). This follows from the fact thatρ1 = 2 and Lemma12.68 (iv). The congruenceρ2 ≡ 2 (mod 3) together with the equation,

2g = (ρ2 − 2)(q2 − 1) +∑

i≥3 (ρi − ρi−1)(|U (i)P | − 1) (13.47)

which is actually a consequence of (12.41), provide the bound ρ2 ≤ 3q0 + 2. Onthe other hand, Lemma 12.75 (iii) shows thatq | (ρ2 − 2)2, whence3q0 | (ρ2 − 2).Thus,ρ2 = 3q0 + 2. From Lemma 12.75 (ii),(q − 1) | (3q0 + 1)(n2 − 1), whichimplies thatn2 = q. Therefore,|U (3)

P | = q.

Since the centraliser ofH inGP intersectsG(1)P trivially, every normal subgroup

of GP contains at leastq− 1 non-trivial elements. In particular, the last non-trivialramification group ofGP has order at leastq. Therefore,|U (4)

P | = 1, andU (3)P

coincides withZ(G(1)P ).

Now (13.47) reads:

2g = (ρ2 − 2)(q2 − 1) + (ρ3 − ρ2)(q − 1). (13.48)

It follows thatρ3 = q + 3q0 + 2.

Finally, the genusg3 of ΣU(3)P can be calculated by applying Theorem 12.66 to

U(2)P regarded as aK-automorphism group ofΣU

(3)P :

2(g3 − 1) = −2q + ρ2(q − 1).

The result is thatg3 = 32q0(q − 1). 2

To find generators ofΣ with the desired polynomial relations, the first step con-sists in showing the existence of elementsx, y1 ∈ Σ satisfying (13.37) such that

Fq(x, y1) is K-isomorphic toΣU(3)P = ΣG

(3n0+3)

P . After that an elementy2 ∈ Σis found satisfying bothΣ = K(x, y1, y2) and (13.38). Finally, it is shown thatGpreservesFq(x, y1, y2).

LetP∞ denote a place ofΣ satisfying the properties listed in Lemma 13.31. Thearguments used in the proofs of Lemmas 13.16 13.17 and 13.18 can be adapted toprove analogous results onΣ. The necessary changes are as follows:

q2 7→ q3; q 7→ q2;

2g − 2 = 2q0(q − 1)− 2 7→ 2g − 2 = 3q0(q − 1)(q + q0 + 1)− 2

= (q + 1)(q + 3q0 + 1)(3q0 − 2);

q2 + 1 = (q + 2q0 + 1)(q − 2q0 + 1)

7→ q3 + 1 = (q + 1)(q + 3q0 + 1)(q − 3q0 + 1);

q + 2q0 + 1 7→ (q + 1)(q + 3q0 + 1).

L EMMA 13.32 (i) The smallest non-gap atP∞ is q2.

(ii) There existsx ∈ Σ such that

(a) div x = −q2P∞ +∑q2

i=1 Pi;

(b) K(x) = U(2)P∞

.

560

L EMMA 13.33 For α ∈ GP∞,

α(x) =

x+ a, a ∈ K\0, for α ∈ G(1)P∞\U (2)

P∞,

bx, bq−1 = 1, b 6= 1 for α ∈ H,x, for α ∈ U (2)

P∞.

(13.49)

L EMMA 13.34 There existsw ∈ Σ such that

div(w)∞ = (q + 1)(q + 3q0 + 1)P∞.

By Lemma 13.34, the smallest non-gapm atP which is not divisible byq doesnot exceed(q + 1)(q + 3q0 + 1).

From Lemma 13.31,ρ1 = 2, ρ2 = 3q0 + 2, ρ3 = q + 3q0 + 3. This togetherwith Lemma 12.75 gives the following result.

L EMMA 13.35 Letw be as in Lemma 13.34. Ifα is a non-trivial element inU (3)P∞

andw′ = α(w)− w, then

div(w′)∞ = q(q + 3q0 + 1)P∞.

L EMMA 13.36 Let y1 ∈ Σ\K(x1) with div(y1) = kP∞, such thatk is minimalandk ≥ 1.

(i) If β is a generator ofH, thenβ(y1) = bβ y1 for a primitive (q − 1)-st rootof unitybβ .

(ii) y1 ∈ ΣU(3)P∞ .

Proof. (i) If z ∈ Σ\K(x) with div(z) = kP∞, then, underβ,

x 7→ bx, for a primitive(q − 1)-st of unityb ∈ K;z 7→ uβx+ vβy1 + wβ , uβ , vβ , wβ ∈ K, uβ 6= 0.

(13.50)

A calculation shows that there existv′β , w′β ∈ K for which y1 = z + v′βx + w′

β

satisfies (i).(ii) Sincey1 6∈ K(x1), from Lemma 13.35,q2 < k ≤ (q+ 1)(q+ 3q0 + 1). For

α ∈ G(1)P∞\G(2)

P∞,

x 7→ x+ bα, bα ∈ K\0;y1 7→ y1 + dαx+ eα, dα, eα ∈ K, dα 6= 0,

(13.51)

whenceα3(y1) = y1. Sinceα3, with α ranging overG(1)P∞\G(2)

P∞, produces all

elements ofZ(G(1)P ) = U

(3)P∞

, so (ii) follows. 2

From now on, letx andy1 have the properties stated in Lemmas 13.32, 13.33and 13.36. Choose an elementα from U

(2)P∞

. SinceU (3)P∞

is a normal subgroup of

U(2)P∞

, soα induces aK-automorphism ofΣU(3)P∞ fixing every element inΣU

(2)P∞ .

By Lemma 13.31 (i), the latter function field is rational. From Lemma 12.14, itfollows thatα(y1) = y1 + aα with aα ∈ K\0. Replacingy1 by a−1

α y1, it maybe supposed thatα(y1) = y1 + 1.


PROPOSITION 13.37 The fieldΣU(3)P = K(x, y1), andyq

1 − y1 = xq0(xq − x).Proof. Let αj = βjαβ−j for j = 1, . . . , q − 1. From (i) of the previous lemma,αj(y1) = y1 + bj . Sincebq−1 = 1, this shows thatαj(y1) = y1 + bj with bj ∈ Fq.

Since|U (2)P∞

/U(3)P∞| = q, every non-zero element inFq is obtained exactly once in

this way. Puttingb0 = 0, this implies that∏q−1

j=0 (y1 − bj) = yq1 − y1 ∈ Fq(x).

From the definition ofx and Lemma 13.36 (ii),K(x, y1) is a subfield ofΣU(3)P∞ .

On the other hand,

[ΣU(2)P∞ : K(x, y1)] = q, |U (2)

P∞/U

(3)P∞| = q.

Therefore,K(x, y1) = ΣU(3)P∞ . Sincey1 is assumed to have no pole other thanP∞,

this implies that

yq1 − y1 = f(x), f(X) ∈ Fq[X].

Putr = deg f(X); thenr = m/q asq ordP∞(y1) = −r eP∞

. Now, since

q2 < m ≤ (q + 1)(q + 3q0 + 1),

it follows thatq < r ≤ q + 3q0 + 2. Therefore,

yq1 − y1 =

∑ri=0 aix

i, q < r ≤ q + 3q0 + 2. (13.52)

Without loss of generality, letar = 1. Let b and bβ be as in Lemma 13.36.Applying the mapβ, in whichx 7→ bx, y1 7→ bβy1, to (13.52) givesbβai = bibifor i = 0, 1, . . . r. Note thatbβ = br asar = 1. Thus, (br − bi)ai = 0 fori = 0, 1, . . . , r. Sinceai = 0 for i neitherr − q + 1 nor r, (13.52) can be writtenas follows:

yq1 − y1 = xr + cxr−q+1, r ∈ Fq. (13.53)

Choose an elementα from G(1)P∞\G(2)

P∞; then (13.51) holds. Applyingα to

(13.53) gives the following:

yq1 − y1 + dα(xq − x) = (x+ eα)r + c(x+ eα)r−q+1. (13.54)

From this,c = −1. Now, replaceyq1 − y1 in (13.54) byxr + cxr−q−1 and use the

binomial expansion. Then (13.54) becomes the following:

xr−q + dα = xr − q + er−1α +

∑r−q−1i=1

(r−q

i

)xier−1−i

α .

This implies that(r−q

i

)≡ 0 (mod 3), i = 1, . . . , r − q − 1;

that is,r − q is a power of3. From (13.52),1 ≤ r − q ≤ 3q0. From (13.53), theonly possibility isr − q = q0 asm is chosen to be minimal. Also, the genus of

K(x, y1) must be32q0(q − 1), the genus ofΣU

(3)P∞ . 2

Using the arguments in the proof of Lemma 13.36 the followingresult can beobtained.

562

L EMMA 13.38 Let y2 ∈ Σ\ΣU(3)P∞ for which div y2 = kP∞ with k ≥ 1 and

minimal. Theny2 can be chosen such that

β(y2) = cβy2, cβ ∈ Fq\0.

In this lemma, it follows from Lemma 13.35 that

q(q + q0) < k ≤ q(q + 3q0 + 1),

asw′ 6∈ ΣU(3)P∞ .

So far, the following results have been shown. The cyclic groupH is generatedby b and, for a primitive elementb of Fq, the mapβ acts as follows:

x 7→ bx, y1 7→ bq0+1y1, y2 7→ bsy2, for 1 ≤ s ≤ q − 1, (13.55)

whileG(1)P∞

consists ofq3 elementsα that act as follows:

x 7→ x+ bα, y1 7→ y1 + bqαx+ eα y2 7→ y2 + fαy1 + gαx+ hα, (13.56)

wherebα, eα, fα, gα, hα ∈ Fq. Note thatα ∈ GP∞if and only if bα = 0. As the

cube of any element inG(1)P∞\G(2)

P∞belongs toU (3)

P∞, the necessary and sufficient

condition forα to be inU (3)P∞

is that it acts as follows:

x 7→ x, y1 7→ y1, y2 7→ y2 + u, for u ∈ Fq. (13.57)

Now, the proof of Theorem 13.30 is given. By (13.57),∏

u∈Fq(y2 − u) = yq

2 − y2 ∈ ΣU(3)P∞ . (13.58)

SinceK(x, y1, y2) is a subfield ofΣ, this implies thatΣ = K(x, y1, y2) wherex, y1, y2 satisfy (13.37),(13.38) and (13.58).

The unique involutory element inH is β′ = β(q−1)/2. By (13.55), its action isthe following:

x 7→ −x, y1 7→ y1, y2 7→ vy2,

where eitherv = 1 or v = −1. Using (13.56), a calculation shows thatβ′ com-mutes withψ ∈ G(1)

P∞if and only ifψ has the following action:

x 7→ x, y1 7→ y1 + u for u ∈ Fq, y2 7→ y2.

In particular, the centraliserC of ψ in G(1)P∞

has orderq and is a subgroup ofG(2)P∞

.

SinceC ∩ U (3)P∞

= 1, this implies thatG(2)P∞

= C × U (3)P∞

. Hence,G(2)P∞

consists

of all τ ∈ G(1)P∞

acting as follows:

x 7→ x, y1 7→ y1 + u, y2 7→ y2 + w, for u,w ∈ Fq.

Therefore,∏

w∈Fq(y2 − w) = yq

2 − y2 ∈ Fq(x).

The assumption thaty2 has no pole other than fromP∞ implies that

yq2 − y2 = f(x), with f(X) ∈ Fq[X].


Arguing as before, putt = deg f(X). Thent = k/q asq ordP∞(y1) = −t eP∞

.Sinceq2 < k ≤ (q + 1)(q + 3q0 + 1), soq < t ≤ q + 3q0 + 2. Therefore,

yq2 − y2 =

∑ti=0 bix

i, bi ∈ Fq, q < t ≤ q + 3q0 + 2. (13.59)

Without loss of generality, letbt = 1. Also, letb andbβ be as in Lemma 13.36.Applying the mapβ, with x 7→ bx, y2 7→ bsβy2, to (13.52) givesbβbi = bibi fori = 0, 1, . . . t. Note thatbβ = bt asbr = 1. Thus,(br−bi)bi = 0 for i = 0, 1, . . . , t.As bi = 0 for i 6= t− q + 1, t, so (13.59) can be written as

yq2 − y2 = xt + cxt−q+1, c ∈ Fq. (13.60)

As above, choose an elementα fromG(1)P∞\G(2)

P∞; then (13.51) holds. Applyingα

to (13.60) gives

yq2 − y2 + dα(yq

1 − y1) + eα(xq − x) = (x+ fα)s−q(xq − x). (13.61)

From this,c = −1.In (13.61), replaceyq

2 − y2 by xs + cxs−q+1, as well asyq1 − y1 by xq0(xq − x),

and use the binomial expansion. Then (13.54) becomes the following:

xt−q + dαxq0 + eα = xt−q + f t−1

α +∑t−q−1

i=1

(t−q

i

)xiet−1−i

α . (13.62)

In particular,eα = f t−qα . Suppose thatdα = 0. Then (13.62) implies that

(s−q

i

)≡ 0 (mod 3), i = 1, . . . , r − q − 1;

that is, r − q is a power of3. Since,q0 ≤ s − q ≤ 3q0 + 1, this yields thatr− q = 3q0. However, in this case,y2 ∈ U (3)

P∞, a contradiction. Therefore,dα 6= 0.

Now, from (13.62), modulo3,(t−qq0

)6≡ 0,

(t−q

i

)≡ 0 for i 6= q0, 1 ≤ t− q − 1,

showing thatt−q = sq0 with s not divisible by3. Sinceq0 < t−q ≤ 3q0 +1, andt− q is distinct from bothq0 and3q0, the desired result, thatt− q = 2q0, follows.

13.5 A CURVE ATTAINING THE SERRE BOUND

In this section,K = F2. Consider the curve,

C = v(X180 X5

2 +X190 X2 +X23

1 ), (13.63)

together with its function fieldΣ = K(x, y), where

y5 + y4 = x23. (13.64)

PROPOSITION 13.39 The curveC has the following properties:

(i) it is an irreducible plane curve of genus11 with exactly two singular points,the18-fold pointY∞ = (0, 0, 1) and the4-fold pointO = (1, 0, 0), each thecentre of a unique branch ofC;

(ii) if P∞ andP0 are the associated places ofΣ, then

div(dx) = 26P0 − 6P∞. (13.65)

564

(iii) with

z = y−1, x0 = z, x1 = xz, x2 = x2z,x3 = x3z, x4 = x5z2, x5 = x6z2, x6 = x7z2,x7 = x8z2, x8 = x11z3, x9 = x12z3, x10 = x17z4,

the canonical series ofΣ is

|dx| = div(∑10

i=0 cixi)+26P0−6P∞ | c = (c0, . . . , c11) ∈ PG(11,K);(13.66)

(iv) theK-automorphism group ofC contains a cyclic group of order23.

Proof. The arguments and calculations are similar to those in the proof of Lemma13.1. The pointY∞ is the centre of a unique branch ofC with a primitive represen-tation,

(x(t) = t−5 + . . . , y(t) = t−23 + . . .).

From this,

ordP∞(dx) = ordt(dx/dt) = −6. (13.67)

The other singular point ofC is the originO; the unique branch ofC centred herehas a primitive representation,

(x(t) = t4 + . . . , y(t) = t23 + . . . .)

From (13.64),y4dy = x22dx. Sincedy/dt = t22 + . . ., this gives

ordP0(dx) = ordt(dx/dt) = 4ordty(t)− 22ordtx(t) + ordt(dy/dt) = 26.

(13.68)SinceC has no more singular points, a first consequence is that everypoint ofC

is the centre of just one branch ofC. Hence,C is an irreducible plane curve and, fora powerq of 2, a place ofΣ is Fq-rational if and only if the corresponding branchof C is centred at a point with coordinates inFq.

Also, ordP(dx) 6= 0 only at these two places,P∞ andP0. Therefore, (13.67)and (13.68) give (13.65). From Definition 5.54,2g − 2 = 26 − 6 = 20, whenceg = 11.

LetP1 be the place ofΣ associated to the unique branch ofC whose centre is thepointA = (1, 1, 0). Then,

div x = 4P0 + P1 − 5P∞, div y = 23P0 − 23P∞.

From this, (13.66) follows.Finally, for a primitive twenty-third root of unityλ ∈ K, the birational transfor-

mation,

ω : x′ = λx, y′ = λy,

of Σ is aK-automorphism of order23. 2

From (13.66), the gap sequences ofΣ at the placesP∞,P1 andP0 are the fol-lowing:

P∞ : (1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18);

P1 : (1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 18);

P0 : (1, 2, 3, 4, 5, 6, 8, 9, 12, 13, 16).


Next, the Weierstrass points ofΣ are determined. From (13.66), the canonicalcurveΓ of Σ is given by the pointP = (x0, x1, . . . , x11) of PG(10,K) with thexi

as defined above. The11×11 Wronskian matrixW = (D(i)x xj) with 0 ≤ i, j ≤ 10

in Figure 13.1 can be calculated using the method illustrated in Example 5.71. Init, let

fi = yi + yi−1 + . . .+ y + 1, g4 = y4 + y + 1,

h3 = y8 + y3 + 1, h4 = y8 + y4 + 1.

From the matrix,

detW =x17(y3 + y2 + 1)(y8 + y7 + y6 + y5 + y4 + y + 1)

×(y35 + y34 + y32 + y30 + y29 + y28 + y27 + y26 + y19

+y18 + y15 + y11 + y8 + y5 + y4 + y3 + 1)(y + 1)3y−76.

SinceW is not identically zero, the canonical series is classical;that is,ǫi = i fori = 0, 1, . . . 10. Further, the zeros ofW are exactly the Weierstrass points ofΣ;their total number is3 + (3 + 8 + 35) · 23 = 1061.

The weights of the placesP∞,P0,P1 in the ramification divisorR of Σ are asfollows:

ordP∞W = 536, ordP∞

(dx) = −6, eP∞= −6,

vP∞(R) = 536− 330− 66 = 140;

ordP0W = −1680, ordP0

(dx) = 26, eP0= 26,

vP0(R) = −1680 + 1430 + 286 = 36;

ordP1W = 86, ordP1

(dx) = 0, eP1= 0,

vP1(R) = 86 + 0 + 0 = 86.

From (7.13),∑vP(R) = 1320. Hence∑1058

i=1 vP(R) = 1320− (140 + 36 + 86) = 1058.

SinceΣ has1058 Weierstrass points other thanP∞,P0,P1, this is only possiblewhenvP(R) = 1 for each of these 1058 places. Also, since the canonical seriesof Σ is classical, Theorem 7.52 shows thatvP(R) = 1 if and only if the ordersequence ofΓ at the branch point associated toP is

(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11).

Therefore, the gap sequence ofΣ at any Weierstrass pointP other thanP∞,P0,P1

is

(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12).

This shows that each of the three places,P∞,P0,P1, differs from any other placeof Σ in its Weierstrass gap sequence.

As a consequence, AutK(Σ) has three fixed places. From Theorem 12.66 appliedto AutK(Σ),

20 ≥ −2|AutK(Σ)|+ 3(|AutK(Σ)| − 1) = |AutK(Σ)| − 3,

566

whence|AutK(Σ)| ≤ 23. By Proposition 13.39 (iv),|AutK(Σ)| = 23. It may benoted that|AutK(Σ)| = 23 = 2g + 1, which agrees with Remark 12.97 (ii) form = 5, r = 1, n = 23.

Calculating the number of solutions of the equationY 4(Y +1) = X23 overF211

it is found thatC has exactly3039 F211 -rational places. Since

211 + 1 + ⌊2√

211⌋ = 3039,

soΣ attains Serre’s bound, and Theorem 10.28 applies to calculate theL-polynom-ial of Σ overF211 .

These results are summarised in the following theorem.

THEOREM 13.40 (i) The number of Weierstrass points ofΣ is 1061.

(ii) |AutK(Σ)| = 23.

(iii) The number ofF211-rational points ofC attains the Serre Bound, 3039.

(iv) TheL-polynomial ofC overF211 is (1 + 90t+ 2048t2)11.

13.6 NOTES

The properties of the curveC in Remark 13.11 are due to Sullivan [279].For Theorems 13.21 and 13.22, see [104].The explicit determination of quotient curves ofHq appears to be difficult, espe-

cially for non-tame subgroups; Garcia, Stichtenoth and Xiang [93] calculated thegenera of many such quotient curves. For the simplest case whereG has prime or-der the complete description of the quotient curves in Theorem 13.27 comes from[57].

Using the method developed by Garcia, Stichtenoth and Xing [93] for the Her-mitian curve, Cakcak andOzbudak determined many quotient curves of the DLRcurve and their genera. Their paper [38] also contains numerous results on thearithmetic and automorphism groups of the these curves.

Section 13.4 is based on [214] and [111].Section 13.5 is based on [160]

Som

eFam

iliesofA

lgebraicC

urves567

Figure 13.1 The Wronskian for the curveC

W =

2666666666666666666666666666666666666664

1y

xy

x2

yx2

yx3

y2x6

y2x7

y2x8

y2x11

y3x12

y3x17

y4

x22

y61

y2(y+1)x

y2x2

y2x4

y2 0x8

y2 0x10

y4(y+1)x11

y4x16

y4

x21

y7(y+1)x22

y7y2+y+1

y3x

y3(y2+1)x3

y4x4

y4x5

y4(y2+1)x6

y4(y+1)x10

y4(y+1)x10

y4 0

x20

y8(y+1)x21

y8(y2+1)x27

y81

y4(y2+1)x2

y4 0x4

y4 0x8

y4(y+1)x9

y4 0

g4x19

y10(y4+f2)x20

y10(y4+y3+y+1)x21

y10f3x22

y10(y4+y2+1)x

y6x2

y6x3

y6(y2+1)x4

y6(y4+1)x7

y8x8

y8(y4+1)x13

y8

x18

y6(y+1)x19

y10(y2+1)x20

y8f3x21

y10(y4+y2+1)

y6 0x2

y6 0(y4+1)x6

y8 0(y4+1)x12

y8

g4x17

y12(y8+g4)x18

y12(y6+f4)x10

y12f3x20

y12(y5+g4)x22

y121

y8x

y8(y2+1)x2

y8(y4+1)x5

y8 0 0

x16

y8(y+1)x17

y12(y2+1)x18

y8f3x19

y12(y5+g4)x21

y12 01

y8 0(y4+1)x4

y8 0 0

(y4+f2)x15

y15(y8+y7+f2)x16

y15(y8+y7+f4)x17

y15(y9+y8+f4)x18

y15(y6+1)f3x20

y16(y8+1)x21

y16(y+1)

(y8+1)x22

y16(y+1)h4y12

h3x3

y11(y8+y7+1)x4

y11(y4+1)x9

y12

(y8+1)x14

y16(y9+y8+f3)x15

y16x16

y16(y9+y8+f5)x17

y16(y6+1)f3x19

y16 0(y8+1)x21

y16(y+1)0

h4x2

y12(y8+1)(y+1)x3

y12(y4+1)x8

y12

(y9+1)x13

y16(y+1)x14

y8f3x15

y16f2x16

y15(y5+g4)x18

y12(y8+1)x19

y16(y+1)

(y8+1)x20

y16(y+1)0

h3x

y11(y8+1)(y+1)x2

y11 0

3777777777777777777777777777777777777775

Chapter Fourteen

Applications: codes and arcs

Algebraic curves over finite fields play a prominent role in the recently emergedarea which combines algebraic geometry, finite geometry andthe theory of error–correcting codes.

Here, only a brief exposition of the main construction due toGoppa and a fewillustrative examples are presented in section 14.1.

Coding theory is also connected with algebraic curves via finite geometry, sincecomplete arcs inPG(r, q) are the geometric counterpart of MDS (Maximum Dis-tance Separable) codes which are linear codes correcting the greatest number oferrors with respect to their parameters. In the other sections, an account of thissubject, especially on the main conjecture on MDS, is given.

The key definition using terminology from linear algebra is the following.

DEFINITION 14.1 An [n, k, d]q codeC is a k-dimensional subspace ofV (n, q),that is,n-dimensional vector space overFq, such that the minimum number of non-zero co-ordinates of any non-zero element ofC is at leastd, with preciselyd forsome element ofC.

14.1 ALGEBRAIC GEOMETRY CODES

In this sectionK is the algebraic closure ofFq, andF is an algebraic curve definedoverFq.

Assume thatF has someFq-rational points, that is, the setF(Fq) is not empty.For an ordered set of distinctFq-rational placesPi of K(F), let

D = P1 + . . .+ Pn

be the associatedFq-rational divisor. Further, forQ1, . . . ,Qs not necessarily dis-tinct Fq-rational places ofK(F), let

E =∑s

j=1mjQj ,

with mj ≥ 0 and∑mj = m. Suppose that the associatedFq-rational divisorE is

disjoint fromD, that is,

Supp(D) ∩ Supp(E) = ∅.As in section 6.4,L(E) denotes the space of functions associated toE. Then theevaluation mapθ atP is

θ : L(E) −→ (Fq)n,

569

570

given by

f 7−→ (resP1f, . . . , resPn

f).

Now,

im θ = C = C(D,E) = (F ,D,E)L.

is analgebraic geometrycode.If f1, . . . , fk is a basis forL(E), then a generator matrix forC is

G =

f1(P1) . . . f1(Pn)

......

fk(P1) . . . fk(Pn)

.

THEOREM 14.2 If n > m > 2g − 2, then

(i) k = m− g + 1;

(ii) d ≥ n−m.Proof. (i) This is precisely part (iii) of Theorem 6.59.

(ii) If f ∈ L(E) andw(θ(f)) = d, thenn − d placesPi1 , . . . ,Pin−dare zeros

of f and their sum defines the divisorD′, wheredegD′ = n− d andD′ ≺ D. Sodiv(f) ≻ D′ − E, whence

deg div(f) ≥ degD′ − degE;

that is,0 ≥ n− d−m. 2

EXAMPLE 14.3 Consider the case that the curveF is the linev(X2), and writeFq = t1, . . . , tq. Then theFq-rational pointsP , the centres of theFq-rationalplacesP of F , are

Pti= (1, ti, 0), i = 1, . . . , q, andP∞ = (0, 1, 0).

In particular,P0 = (1, 0, 0). LetE = mP∞. For a generic pointQ = (1, x, 0) ofF , the divisor divxr = rP0 − rP∞ andL(E) contains the functions

1, x, . . . xm, (14.1)

but not the functionxm+1.Also,

D = Pt1 + . . .Ptq.

So,

n = q, k = m+ 1, d = n− k + 1 = n−m,and

G =

1 1 . . . 1t1 t2 . . . tq...

......

tm1 tm2 . . . tmq

1x...xm

.

Applications: codes and arcs 571

This is a Reed–Solomon code and is MDS.Under the equivalence of an[n, k, d]q code and a projective[n, k] system ofn

points inPG(k − 1, q) with at mostn− d in a hyperplane, this case givesq pointsin PG(m, q) on a normal rational curve. The system can be extended to includethe point(0, 0, . . . , 0, 1), the remaining rational point on the normal rational curve,and the code can be extended to the MDS codeC ′ by adding the transpose of thisvector as an extra column ofG. ForC ′,

n = q + 1, k = m+ 1, d = n−m = q + 2− k = n− k + 1.

EXAMPLE 14.4 Takeq = 4 andF = v(X30 +X3

1 +X32 ). LetQ = (x, y, 1) be a

generic point ofF . TheFq-rational points ofF are given in Table 14.1. LetSi bethe place associated withSi.

Table 14.1F4-rational points ofv(X30 +X3

1 +X32 )

0 0 0 1 1 1 1 1 11 1 1 0 0 0 1 ω ω2

1 ω ω2 1 ω ω2 0 0 0

S0 S1 S2 S3 S4 S5 S6 S7 S8

Considerf1 = 1/(x+ y) andf2 = x/(x+ y). Then

div f1 = (S0 + S1 + S2)− 3S0 = (S1 + S2)− 2S0.div (f2) = (S3 + S4 + S5)− 3S0.

(14.2)

As in Section 13.1, it is possible to verify (14.2) by direct computation using prim-itive representations of the branches centred at the pointsSi. Here, another methodis illustrated. To calculate ordS0

(f1), note that the tangent atS0 isX1 +X2. Since

x+ y =1

x2 + xy + y2,

sof1 = 1/(x2 + xy + y2), whence ordS0(f1) = −2. For ordS0

(f2), note that

x+ y

x=

1

x(x2 + xy + y2);

sof2 = 1/x(x2 + xy + y2), whence ordS0(f2) = −3.

With S1, . . . , S8 as in Table 14.1, letE = 3S0 andD = S1 + . . . + S8. Thegenusg = 1 andC(D,E) has parameters

n = 8, k = 3, 5 ≤ d ≤ 6.

Also,

G =

1 1 1 1 1 1 1 10 0 1 ω2 ω 1 ω2 ωω ω2 0 0 0 1 1 1

11/ (x+ y)x/ (x+ y)

As the last row ofG reveals a word of weight 5, sod = 5. HenceC is an[8, 3, 5]4code.

572

EXAMPLE 14.5 As in the previous example, letq = 4 andF = v(X30 +X3

1 +X32 ).

Here, letD = S0 + S1 + . . . ,+S8. Also, withG = v(X0X1 +X0X2 +X1X2),let E = F · G; that is,E is the intersection divisor of the curvesF andG. SoEis anF4-rational divisor, even though no point in its support isF4-rational. In fact,the six points of the divisorE are, withτ a primitive element ofF64,

(τ9i + 1, τ−9i + 1, 1),

for i = 1, . . . , 6. As degE = 6, sok = 6. Let z = x + y + xy. Then a basis forL(E) is the following:

1

z,x2

z,y2

z,x

z,y

z,xy

z.

Hence, evaluating the placesSi at these functions ofL(E) gives the generatormatrixG below:

S0 S1 S2 S3 S4 S5 S6 S7 S8

0 0 0 1 1 1 1 1 11 1 1 0 0 0 1 ω ω2

1 ω ω2 1 ω ω2 0 0 0

G =

0 0 0 1 ω2 ω 1 ω2 ω1 ω2 ω 0 0 0 1 ω ω2

1 ω ω2 1 ω ω2 0 0 00 0 0 0 0 0 1 1 10 0 0 1 1 1 0 0 01 1 1 0 0 0 0 0 0

1/zx2/zy2/zx/zy/zxy/z

Here,C is a[9, 6, 3]4 code.

COROLLARY 14.6 (i) n− k + 1− g ≤ d ≤ n− k + 1;

(ii) R+ δ ≥ 1− (g − 1)/n.

LetC⊥, the dual ofC(D,E), be an[n, k⊥, d⊥] code. It turns out thatC⊥ is alsoan algebraic geometry code.

COROLLARY 14.7 If n > m > 2g − 2, then

(i) k⊥ = n−m+ g − 1;

(ii) d⊥ ≥ m− 2g + 2;

(iii) n− k⊥ + 1− g ≤ d⊥ ≤ n− k⊥ + 1.

In the last example, withC a [9, 6, 3]4 code, the dualC⊥ is a[9, 3, 6]4 code. Thisis evident since a generator matrix forC⊥ is

H =

0 0 0 1 ω ω2 1 ω ω2

1 ω ω2 0 0 0 1 ω2 ω1 ω2 ω 1 ω2 ω 0 0 0

.


Now, the nine columns ofH are the nineF4-rational points of the cubic curveF .If the rows ofH areh1, h2, h3, then an element ofC⊥ is a1h1 +a2h2 +a3h3. So acoordinate of this codeword is zero ifa1x1+a2x2+a3x3 = 0 for the correspondingcolumn(x1, x2, x3)

T . Any line ofPG(2, 4) meetsF in at most 3 points and some,namely 12, in precisely 3 points. Hence any line misses at least 6 points ofF and12 miss precisely 6; sod⊥ = 6.

14.2 MAXIMUM DISTANCE SEPARABLE CODES

The following four concepts are equivalent forn ≥ k:

1. (CODING THEORY) amaximum distance separable(MDS) linear codeCof lengthn, dimensionk and hence minimum distanced = n − k + 1, thatis, an[n, k, n− k + 1] code overFq;

2. (MATRIX THEORY) a k × (n − k) matrixA with entries inFq such thatevery minor is non-zero;

3. (VECTOR SPACE) a setK′ of n vectors inV (k, q), the vector space ofkdimensions overFq, with anyk linearly independent;

4. (PROJECTIVE SPACE) ann-arc in PG(k−1, q), that is, a setK of n pointswith at mostk− 1 in any hyperplane of the projective space ofk− 1 dimen-sions overFq.

To show the equivalence, consider a generator matrixG for such a codeC in canon-ical form:

n

k

1 0 . . . 0 a11 . . . a1,n−k

0 1 . . . 0 a21 . . . a2,n−k

......

......

...0 0 . . . 1 ak1 . . . ak,n−k

= G.

SinceC has minimum distancen− k+ 1, any linear combination of the rows ofG has at mostk − 1 zeros; that is, considering the columns ofG as a setK′ of nvectors inV (k, q), anyk are linearly independent. Regarding the columns ofG asa setK of points ofPG(n, q) means that nok lie in a hyperplane; equivalently, anyk points ofK are linearly independent. This, in turn, implies that everyminor ofAis non-zero.

For givenk andq, letM(k, q) be the maximum value ofn for such a code. Then

M(k, q) = k + 1 for q ≤ k.A suitable set of vectors inV (k, q) is

(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1), (1, 1, . . . , 1);

that is, forq ≤ k, every element ofV (k, q) is a linear combination of at mostk− 1of thesek + 1 vectors.

574

The Main ConjectureMCk for MDS Codes, always takingq > k, is the follow-ing:

M(k, q) =

q + 2 for k = 3 andk = q − 1 both withq even,q + 1 in all other cases.

It is convenient to have the notationm(k − 1, q) = M(k, q).In the rest of this section, the main problems and related results on MDS codes,

in terms of projective geometry, are presented.In his seminal paper [246] Segre enunciated three problems:

I. For givenk andq, what is the maximum value ofn such that ann-arc existsin PG(k − 1, q)? What are then-arcs corresponding to this value ofn?

II. For what values ofk andq with q > k is every(q + 1)-arc ofPG(k − 1, q)is a normal rational curve?

III. For givenk andq with q > k, what are the values ofn(≤ q) such that eachn-arc is contained in a normal rational curve ofPG(k − 1, q)? In how manysuch curves is then-arc contained?

An n-arc iscompleteif it is maximal with respect to inclusion; that is, it is notcontained in an(n + 1)-arc. Implicit in Problem III is Problem IV, which may beenunciated as follows:

IV. What are the values ofn for which a completen-arc exists inPG(k− 1, q)?In particular, what is the size of the second largest complete arc in this space?

From above,m(r, q) is the maximum size of an arc inPG(r, q); also, letm′(r, q)denote the size of the second largest complete arc inPG(r, q). Then ann-arc inPG(r, q) with n > m′(r, q) is contained in anm(r, q)-arc. This is an importantinductive tool.

EXAMPLE 14.8 Let Γ be the rational normal curve ofPG(r,K) viewed as a curveoverFq. Then the setΓ(Fq) of all Fq-rational points is a(q+ 1)-arc ofPG(r,K),theclassicalarc ofPG(r,K).

If q ≥ r+2, then any(r+3)-arc ofPG(r, q) is contained in a unique classical arc.The next result gives simultaneous information on pairs of dimensions.

THEOREM 14.9 (i) The dual code of an MDS code is also MDS.

(ii) An n-arc exists inPG(k − 1, q) if and only if ann-arc exists in the spacePG(n− k − 1, q).

Proof. If G = [ Ik | A ] is a generator matrix for the codeC, then a generator matixfor the dual codeC⊥ isH = [−A∗ | In−k ], whereA∗ is the transpose ofA. 2

COROLLARY 14.10 (i) A (q + 1)-arc exists inPG(k − 1, q) if and only if a(q + 1)-arc exists inPG(q − k, q).


(ii) A (q + 2)-arc exists inPG(k − 1, q) if and only if a(q + 2)-arc exists inPG(q − k + 1, q).

(iii) A (q + 3)-arc exists inPG(k − 1, q) if and only if a(q + 3)-arc exists inPG(q − k + 2, q).

THEOREM 14.11 Let an,r be the number ofn-arcs inPG(r, q) and letνr be thenumber of normal rational curves inPG(r, q). Then

an,n−2−r/an,r = νn−2−r/νr.

The crucial result for approaching the Main Conjecture is the fact that, in a cer-tain dimension, a(q + 1)-arc is classical. In the case ofq odd, this occurs in theplane, see Theorem 14.31, but forq even, it is necessary to go to a higher dimen-sion.

THEOREM 14.12 In PG(4, q), q even, a(q + 1)-arc is a normal rational curve.

The following result is discussed in Section 14.3.

m(2, q) = M(3, q) =

q + 1 for q odd,q + 2 for q even.

Theorems 14.31 and 14.12 imply a dependency ofm(r, q) andm′(2, q) for q oddand ofm(r, q) andm′(4, q) for q even. Define the integer functionsF andG bythe purely arithmetic conditions:

q + 1 > m′(2, q) + r − 2⇐⇒ q > F (r);

q + 1 > m′(4, q) + r − 4⇐⇒ q > G(r).

There are two results that produce an inductive argument on dimension.

THEOREM 14.13 LetK be ann−arc in PG(r, q) with q + 1 ≥ n ≥ r + 3 ≥ 6and suppose there existP0, P1 ∈ K and a hyperplaneπ containing neitherP0 norP1 such that, fori = 0, 1, the projectionKi ofK ontoπ is Fq-rational in π. Thenthe arcK is contained in one and only one classical arc ofPG(r, q).

THEOREM 14.14 LetK be a(q + 2)-arc in PG(r, q) with q + 1 ≥ r + 3 ≥ 6.If a hyperplaneπ of PG(r, q) contains neither of the pointsP0, P1 of K, then itcannot happen that both projectionsKi of K from Pi, i = 0, 1, ontoπ are Fq-rational in π. In particular, if every(q + 1)-arc in PG(r − 1, q) is classical, thenm(r, q) = q + 1.

These two results have as corollaries the following way of approaching solutions toproblems I, II, III.

THEOREM 14.15 In PG(r, q), q odd, r ≥ 3,

(i) if K is ann-arc withn > m′(2, q) + r− 2, thenK lies on a unique classicalarc;

(ii) if q > F (r), then every(q + 1)-arc is classical;

576

(iii) if q > F (r − 1), thenm(r, q) = q + 1.

THEOREM 14.16 In PG(r, q), q even, q > 2, r ≥ 4,

(i) if K is ann-arc withn > m′(4, q) + r− 4, thenK lies on a unique classicalarc;

(ii) if q > G(r), then every(q + 1)-arc is classical;

(iii) if q > G(r − 1), thenm(r, q) = q + 1.

THEOREM 14.17 If q is even andq > 2, thenm′(4, q) ≤ q − 12

√q + 13

4 .

Theorems 14.13 and 14.14 can be applied to give the following.

THEOREM 14.18 For q odd, r ≥ 3,

(i) F (r) ≤ (4r − 234 )2;

(ii) for q prime, F (r) ≤ 5(9r − 19).

THEOREM 14.19 For q even, q > 2, r ≥ 4,

G(r) ≤ (2r − 72 )2.

However, it should be noted that the classification of(q+1)-arcs forr ≥ 4 will notproduce a simple result.

THEOREM 14.20 In PG(4, 9), the number of projectively distinct10−arcs is pre-cisely two, a normal rational curve and a non-classical arc.

A result on the Main Conjecture is the following.

THEOREM 14.21 For q odd andq ≤ 27, the largest size of square matrix overFq

with every minor non-zero is12 (q + 1).

As an illustration,

A =

1 1 1 1 1 1 1 1 1 11 10 13 5 4 16 11 12 17 2

10 13 5 4 16 11 12 17 2 713 5 4 16 11 12 17 2 7 85 4 16 11 12 17 2 7 8 34 16 11 12 17 2 7 8 3 15

16 11 12 17 2 7 8 3 15 1411 12 17 2 7 8 3 15 14 612 17 2 7 8 3 15 14 6 917 2 7 8 3 15 14 6 9 18

is a 10 × 10 matrix with the property that every minor is non-zero modulo19;further, there is no matrix of larger size overF19 with this property.

Near-MDS codes are linear codes whose parameters differ only slightly fromthose of MDS codes.


DEFINITION 14.22 A linear [n, k]q-code is anear-MDScodeif d(C) = n−k andd(C⊥) = k.

Like MDS codes, near-MDS codes can be investigated within finite projectivegeometry, as a near-MDS code[n, k]q can be viewed as a point setC of sizenin PG(k − 1, q) satisfying the following conditions:

(I) everyk − 1 points inC generate a hyperplane ofPG(k − 1, q);

(II) there existk points inC lying in a hyperplane ofPG(k − 1, q);

(III) any k + 1 points inC generatePG(k − 1, q).

In the special case of near-MDS codes of dimension4, this geometric represen-tation shows that a near-MDS[n, 4]q code is a point setC of sizen in PG(3, q)such that

(I) C is a cap, i.e. no three points inC are collinear;

(II) C is not an arc, i.e.C contains four coplanar points;

(III) no five points inC are coplanar.

Let K be the algebraic closure ofFq. The set ofFq-rational points of an ellipticquarticΓ4 of PG(3,K) defined overFq is a natural example of such a point setCwith n equal to the numberNq of all Fq-rational points ofΓ4. Since such an ellipticquartic is birationally equivalent overFq to a non-singular plane cubic defined overFq, an[n, 4]q near-MDS code exists forn provided that there exists a non-singularplane cubic overFq with exactlyn Fq-rational points.

14.3 ARCS AND OVALS

As explained in Section 14.2, MDS codes and arcs in finite projective spaces areequivalent objects. In this section, arcs in planes are investigated by using previousresults on algebraic curves with manyFq-rational points.

Some standard notation from finite geometry used in this chapter is as follows.

NOTATION 14.23 (1) In PG(2, q), the coordinates arex0, x1, x2 and corresp-ndingly the indeterminatesX0,X1,X2.

(2) In PG(2, q), let

U0 = (1, 0, 0), U1 = (0, 1, 0), U2 = (0, 0, 1), U = (1, 1, 1).

(3) WriteNs = 1, 2, . . . , s for s > 0.

(4) For binomial coefficients, write

c(n, r) =

(n

r

).

578

The concept of an oval in finite geometry arises from two combinatorial propertiesof a closed convex curve in the real plane.

DEFINITION 14.24 A setΩ of points inPG(2, q) is anoval if

(i) no three points inΩ are collinear;

(ii) at every pointP ∈ Ω there is exactly oneunisecant, that is, a line ofPG(2, q)that meetsΩ only in P .

In this setting, ak-arc inPG(2, q) is a setΩ of k-points that has property (i). In par-ticular, there are three types of lines respect to an arc, namely secants (or chords),unisecants and external lines (or lines disjoint from the arc). A combinatorial argu-ment shows that

k ≤q + 1 whenq odd,

q + 2 whenq even,

for everyk-arc inPG(2, q).A dual k-arc in PG(2, q) is a k-arc in the dual plane, that is, a set ofk lines

on PG(2, q) no three of them concurring at a point. In the study of arcs, the setof points ofPG(2, q) lying on exactly one of the lines of a dualk-arcK plays animportant role. Such a set∆(K) has sizekt with t = q − k + 2, each line ofKcontaining exactlyt points from∆(K).

Ovals are exactly the(q + 1)-arcs inPG(2, q). An example of an oval is the setof all Fq-rational points of an irreducible conic defined overFq. For q odd, everyoval in PG(2, q) is a complete arc. This does not hold true forq even because anirreducible conic in even characteristic is a strange curve, and hence, adding thenucleus to theFq-rational points of the conic, a (complete)(q+ 2)-arc inPG(2, q)is obtained. More generally, every ovalΩ in PG(2, q) with q even has a nucleus;that is, its unisecants have a common pointN ; the ovalΩ, together withN , formsa (complete)(q + 2)-arc ofPG(2, q), called ahyperoval. Therefore,

m(2, q) =

q + 1 whenq odd,

q + 2 whenq even,

wherem(2, q) is the maximum value ofk for which ak-arc inPG(2, q) exists. Itappears much more difficult to find the sizem′(2, q) of the second largest completearc inPG(2, q). Note that ifK is ak-arc withk > m′(2, q), thenK is contained inan oval ofPG(2, q).

By Segre’s famous theorem of1955, every oval inPG(2, q) with q odd consistsof theFq-rational points of an irreducible conic defined overFq. Segre noted thatthis theorem does not extend toq even withq ≥ 8. Although ovals forq even havebeen classified forq ≤ 32, the classification of ovals forq ≥ 64 appears difficultand is a major challenge in finite geometry.

Subsequently, Segre used hisLemma of Tangents, the key idea in his ingeniousproof, to obtain a deep generalisation of Menelaus’ classical theorem to show theexistence of a plane algebraic curve passing through each point in ∆(K), the pointset arising from a dual arcK of PG(2, q). Such a curveF , which may be singular


and even reducible, is defined overFq; it has degreet, as small as possible, forq even, and2t, a bit larger but still quite low, forq odd. If k is large, thenF isirreducible and an interesting case occurs; namely,F has low order, and hence lowgenus, compared to its large number ofFq-rational points.

Referring to the results in Chapter 10 on the maximum number of Fq-rationalpoints that an algebraic curve defined overFq can have, it is conceivable that this isan unusual occurrence. Large completearcs are rare, but they can be investigated bystudying particular algebraic curves defined overFq which have manyFq-rationalpoints. This is done in section 14.4. Here, two examples are presented.

EXAMPLE 14.25 In PG(2, 13) there exists exactly one complete12-arc, up to aprojectivity, as in Table 14.2. In this case,∆(K′) has size36 and its points are

Table 14.2 The 12-arcK and its dualK′

K K′

P1 (1, 1, 1) X + Y + Z t1P2 (1,−1, 1) X − Y + Z t2P3 (−1, 1, 1) −X + Y + Z t3P4 (−1,−1, 1) −X − Y + Z t4P5 (3, 4, 1) 3X + 4Y + Z t5P6 (3,−4, 1) 3X − 4Y + Z t6P7 (−3, 4, 1) −3X + 4Y + Z t7P8 (−3,−4, 1) −3X − 4Y + Z t8P9 (4, 3, 1) 4X + 3Y + Z t9P10 (4,−3, 1) 4X − 3Y + Z t10P11 (−4, 3, 1) −4X + 3Y + Z t11P12 (−4,−3, 1) −4X − 3Y + Z t12

listed in Table 14.3.

Table 14.3 The points of∆(K′)

(±1,±2, 1), (±2,±1, 1), (±2,±5, 1),(±5,±2, 1), (±3,±5, 1), (±5,±3, 1),(±4,±6, 1), (±6,±4, 1), (±6,±6, 1).

By Theorem 14.42, no plane cubic contains all points in∆(K′), but there is aplane sextic that does, namely the non–singular plane curveΓ6 defined overF13:

Γ6 = v(X60 +X6

1 +X62 + 3X2

0X21X

22 ).

In addition,Γ6 contains18 points on the coordinate axes, precisely, the points

Bt = (0, t, 1), Ct = (t, 0, 1), Dt = (t, 1, 0)

580

with t6 = −1, that is,t = ±2,±5,±6.These54 points inPG(2, 13) are allF13-rational points ofΓ6. It should be

noted thatΓ6 is optimal in that it attains the upper bound in Theorem 8.62 appliedto anF13-rational linear series of dimension2. Here,g = 10, q = 13, n = 6andN13 = 54. Therefore,54 is the largest number ofF13-rational points of anirreducible plane curve defined overF13 that has genus10.

EXAMPLE 14.26 For a primitive polynomiala0X2 + a1X + a2 of Fq2 [X], the

projectivityT of PG(2, q) associated to the matrix

0 1 00 0 1a2 a1 a0

,

is a Singer cycle; that is,T acts on the set of points ofPG(2, q2), as well as on theset of lines ofPG(2, q2) as a sharply transitive permutation group. In particular,T

has orderq4 + q2 +1, and henceS = Tq2+q+1 is a projectivity of orderq2− q+1.Also, every point orbit ofS is a complete(q2 − q + 1)-arcK in PG(2, q2), andevery line orbit ofβ is a complete dual(q2−q+1)-arcK′ in PG(2, q2). For evenq,the plane curveΓq+1 associated toK′ is the Hermitian curve; see Theorem 14.46.This does not hold true for oddq, the associated curveΓ2(q+1) being projectivelyequivalent overFq6 to the irreducible plane curve in affine form,

v(Y 2 +X2Y 2q +X2q − 2(Xq+1Y q +XqY +XY q+1)).

However,Γ2(q+1) is birationally equivalent overFq6 to the Hermitian curve; seeTheorem 14.59.

The starting point of the study of ovals in odd characteristic is a series of combi-natorial results.

L EMMA 14.27 In PG(2, q) for q odd, every point off an ovalΩ lies on exactly twoor no unisecants ofΩ.

Proof. SinceΩ consists ofq+1 points, there is exactly one unisecant through eachpoint of Ω. Let ℓ be the unisecant atP to Ω and letQ ∈ ℓ\P. If σi(Q), withi = 1, 2, denotes the number ofi-secants toΩ throughQ, then

σ1(Q) + 2σ2(Q) = q + 1

and, sinceq is odd,σ1(Q) is even. Asℓ itself is a unisecant throughQ, it followsthatσ1(Q) ≥ 2. Since this is true for each of theq pointsQ in ℓ\P and sinceΩ has exactlyq + 1 unisecants,σ1(Q) = 2 for all such pointsQ; that is, throughevery point ofℓ\P, there is exactly one other unisecant. 2

A point of PG(2, q) is externalor internal to the ovalΩ according as it lies ontwo or no unisecants ofΩ. Hence, with respect toΩ, the q2 + q + 1 points ofPG(2, q) are partitioned into three classes:

(a) q + 1 points onΩ;

(b) 12q(q + 1) external points;


(c) 12q(q − 1) internal points.

Since ak-arc in the dual plane is a dualk-arc, theq2 + q + 1 lines ofPG(2, q) arepartitioned similarly into three classes with respect toΩ:

(a′) q + 1 unisecants;

(b′) 12q(q + 1) bisecants;

(c′) 12q(q − 1) external lines.

For anyk-arcK with 3 ≤ k ≤ q + 1, choose three of its points as the triangleof referenceU0U1U2 of the coordinate system. A unisecant toK through one ofU0,U1,U2 has the respective form

v(X1 − dX2), v(X2 − dX0), v(X0 − dX1),

with d 6= 0. Here,d is thecoordinateof such a line.Suppose thet = q + 2− k unisecants toK at each ofU0,U1,U2 are

v(X1 − aiX2), v(X2 − biX0), v(X0 − ciX1),

i ∈ Nt.

L EMMA 14.28 (Lemma of Tangents)The coordinatesai, bi, ci of the unisecantsat U0,U1,U2 to ak-arcK through these points satisfy

∏ti=1aibici = −1.

Proof. If P = (d0, d1, d2) is any point ofK other thanU0,U1,U2, then the linesPU0, PU1, PU2 are respectively

v(X1 − eX2), v(X2 − fX0), v(X0 − gX1),

wheree = d1/d2, f = d2/d0, g = d0/d1 and so

efg = 1. (14.3)

ThroughU0 there areq− 1 lines other thanU0U1,U0U2; they consist oft unise-cants

v(X1 − aiX2), i ∈ Nt,

andk − 3 bisecants

v(X1 − ejX2), j ∈ Nk−3.

Since the product of the non-zero elements ofFq is−1, so∏ai

∏ej = −1.

Similarly, for theq−1 lines throughU1 and throughU2 other than the sides of thetriangle of reference,

∏bi∏fj = −1,

∏ci∏gj = −1.

Hence∏

(aibici)∏

(ejfjgj) = (−1)3 = −1.

However, asejfjgj = 1 for eachj by (14.3), so∏

(aibici) = −1. 2

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COROLLARY 14.29 If the unisecants to a(q+1)-arcK at the pointsU0,U1,U2

ofK arev(X1 − aX2),v(X2 − bX0),v(X0 − cX1), thenabc = −1.

L EMMA 14.30 For q odd, the triangles formed by three points of a(q + 1)-arcKand the unisecants at these points are in perspective.

Proof. Choose the three points asU0,U1,U2; then the unisecants are the linesv(X1−aX2),v(X2−bX0),v(X0−cX1),with abc = −1. The verticesV0, V1, V2

of the triangle with the unisecants as sides are(c, 1, bc), (ca, a, 1), (1, ab, b).

Hence the linesU0V0,U1V1,U2V2 arev(X2 − bcX1), v(X0 − caX2), v(X1 − abX0).

These three lines are concurrent if and only ifa2b2c2 = 1, which is satisfied sinceabc = −1. It may be noted that, since the three unisecants are not concurrent,Lemma 14.27, soabc 6= 1. 2

THEOREM 14.31 (Segre)In PG(2, q), with q odd, every oval consists of allFq-rational points of an irreducible conic defined overFq.

Proof. LetK be the oval and choose any three of its points asU0,U1,U2. Also,let the unit pointU be the point of perspectivity of the triangleU0U1U2 andits circumscribed triangle as in the previous lemma. AsU is the intersection ofv(X0−caX2) andv(X1−abX0), which is(ca,−a, 1), soa = b = c = −1. Thusthe unisecants toK atU0,U1,U2 are

v(X1 +X2), v(X2 +X0), v(X0 +X1).

Let P = (y0, y1, y2) be any other point ofK and let the unisecant toK at P beℓ = v(d0X0 +d1X1 +d2X2). Lemma 14.28 applied to the trianglePU1U2 givesthe condition:

(d0 − d1 − d2)[d1(y0 + y1)− d2(y0 + y2)] = 0.

However, by Lemma 14.27, the lineℓ cannot pass through the meet of the unise-cants atU1 andU2; this point is(1,−1,−1) and sod0 − d1 − d2 6= 0. Thus

d1(y0 + y1) = d2(y0 + y2);

similarly, from the trianglesPU2U0 andPU0U1,d2(y1 + y2) = d0(y1 + y0),

d0(y2 + y0) = d1(y2 + y1).

Henced0 : d1 : d2 = y1 + y2 : y2 + y0 : y0 + y1.

As d0y0 + d1y1 + d2y2 = 0 and2 6= 0, soy1y2 + y2y0 + y0y1 = 0.

Hence all points ofK lie on the irreducible conicC = v(X1X2 +X2X0 +X0X1)defined overFq. As the setC(Fq) of all Fq-rational points ofC has sizeq + 1, itcoincides withK. 2

The above proof is typical in this subject. A known property of conics is provedfor ovals and this property is used to show that an oval is a conic. In other words,a property which characterizes conics among algebraic curves also characterizesovals amongk-arcs.


14.4 SEGRE’S GENERALIZATION OF THE THEOREM OF MENELAUS

An n-point in PG(2,K) is a set ofn points, no three of which are collinear, togetherwith the 1

2n(n − 1) lines that are joins of pairs of the points. The points and linesare calledverticesandsidesof then-point. The vertices form ann-arc.

Dually, ann-line in PG(2,K) is a set ofn lines, no three of which are concur-rent, together with the12n(n− 1) points that are intersections of pairs of the lines.The points and lines are again calledverticesandsidesof then-line. The sidesform adualn-arc.

An n-gon is an ordered set(P1, P2, . . . , Pn) of n points, no three of which arecollinear: thePi are theverticesand then linesP1P2, P2P3, . . . , Pn−1Pn, PnP1

are thesidesof then-gon.For smalln, the following terms are used:

n-point n-line n-gonn = 3 triangle triangle trianglen = 4 (complete) (complete) quadrangle,

quadrangle quadrilateral quadrilateraln = 5 pentastigm pentagram pentagonn = 6 hexastigm hexagram hexagon

A triangle with verticesA,B,C and sidesa, b, c is denotedABC orabcwhen thereis no risk of confusing it with the entire plane.

Let Cn be an algebraic curve of degreen defined overFq. Certain problemson the intersection of two curvesCn andCk are considered, particularly whenCkis completely reducible and its linear components are sidesof ak-line. Menelaus’theorem gives a necessary and sufficient condition for threepoints, one on each sideof a triangle, to be collinear. In its Euclidean form, this isexpressed as a product ofratios of distances being−1. This is now generalized projectively.

Let K = l1, l2, . . . , lk with li = v(Li) andLi a linear polynomial overFq.Let Ck = v(L1 · · ·Lk. Consider thek-line K′ that hasK as its set of sides. LetAij = li ∩ lj , let Ai = Aij | j 6= i be the set of vertices onli, and also letA = Aij | i, j ∈ Nk be the set of all the vertices ofK′. Consider on eachli amultisetGi

n whose support is disjoint fromA, and let

G =

k⋃

i=1

Gin.

For anyCk such thatCn contains no point ofA, the multisetG consists of allcommon pointsP of Cn andCk, each countedI(P, Cn ∩ Ck) times.

Conditions are required under which a givenG is the complete intersection ofCkwith someCn. In fact, givenG onK, necessary and sufficient conditions are foundfor the setMG = Cn | Cn ∩ Ck = G to be non-empty.

Let L be the linear system of all curvesCn through the points ofG, that is,I(P, Cn, Ck) ≥ nP for every pointP with weightnP in G. Geometrically speak-ing,Cn ∈ L if Cn meetsK at least inG. Then,L contains as a linear subsystem thesetMG consisting of all curvesCn that meetsK exactly inG.

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The dimension ofL is found, as well as the number of conditions for a planecurveCn to belong toL.

L EMMA 14.32 (i) Cn meetsK in G and some further pointP if and only if allthe linesli ofK are components ofCn.

(ii) The setL\MG of reducible curves containingCk as a component is a linearsubsystemLK ofL whose dimension is equal toc(n− k + 2, 2)− 1.

Proof. If P ∈ l1, thenCn containsn+ 1 pointsP ∪G1n of l1; soCn containsl1.

However, any other lineli meetsl1 in A1i. So eachli meetsCn in at leastn + 1points. Hence,Cn contains allli. Thus any curve ofL not inMG containsK, andLK consists of all curvesv(L1L2 . . . LkFn−k), whereFn−k is a form of degreen − k; such a curve has as components thek lines li andCn−k = v(Fn−k). ThusLK is a linear subsystem of the required dimension. 2

THEOREM 14.33 SupposeL is not empty, and letC′n = v(F ′) be a curve inL.

(i) If k > n, thenLK = ∅ andL =MG = C′n.

(ii) If LK 6= ∅, thenMG is not a linear subsystem ofL. However, if Cn ∈ LandC′n ∈MG, there is a curveCn−k = v(Fn−k) such that

Cn = v(F ′ + L1L2 . . . LkFn−k).

(iii) If d is the dimension ofL, thend = c(n− k + 2, 2).

Proof. (i) If k > n, noCn can haveCk as a component. SoLK 6= ∅. From Lemma14.32,L =MG andL consists of the single curveC′n.

(ii) If LK 6= ∅, thenLK andMG form a partition ofL. SoMG cannot be alinear subsystem. Take a pointP ∈ l1 not inG. If Cn = v(F ) ∈ L andC′n ∈MG,then some member of the pencilv(sF + tF ′) contains and so is inLK. HenceλF = F ′ + L1L2 . . . LkFn−k for someλ and someFn−k.

(iii) From (ii) and Lemma 14.32,d = dimLK + 1 = c(n− k + 2, 2). 2

COROLLARY 14.34 (i) If some curve meetsK exactly inG, thenG imposes

c(n+ 2, 2)− c(n− k + 2, 2) = 12k(2n− k + 3)− 1

conditions on aCn to meetK in G.

(ii) If n ≥ k, thenK imposes

c(n+ 2, 2)− c(n− k + 2, 2) = 12k(2n− k + 3)

conditions on aCn to meetK in G.

THEOREM 14.35 If k = 1, thenMG is never empty and then conditions imposedbyG are independent.

Proof. If K = l1, G = P1, . . . , Pn, P 6∈ l1, andPPi = V(L′i), all i, then

v(L′1L

′2 . . . L

′n) is inMG. 2


THEOREM 14.36 If k = 2, thenMG is never empty and the2n conditions im-posed byG are independent.

Proof. If K = l1, l2, Gin = P i

1, . . . , Pin, andP 1

j P2j = V(L′

j), all j, thenv(L′

1L′2 . . . L

′n) is inMG. 2

Let the coordinate system inPG(2, q) be fixed. If the pointP lies on a side ofU0U1U2 and is not a vertex, it is the meet of that side and one of

v(X1 − cX2), v(X2 − cX0), v(X0 − cX1);

thenc is thecoordinateof P .

THEOREM 14.37 LetK be the triangleU0U1U2. ThenMG 6= ∅ if and only if∏c = (−1)n,

where the product ranges over all not-necessarily-distinct 3n pointsP of G andc is the coordinate ofP . If MG 6= ∅, thenG imposes3n − 1 conditions on thecurves of ordern to meetK in G; if n = 1 or 2, thenMG consists of a singleCn.

Proof. If Cn ∈ MG andCn = v(c0Xn0 + c1X

n1 + c2X

n2 + . . .), thenc0c1c2 6= 0

sinceCn contains no vertex ofU0U1U2. Let

C0n =v(c1Xn1 + . . .+ c2X

n2 ),

C1n =v(c0Xn0 + . . .+ c2X

n2 ),

C2n =v(c0Xn1 + . . .+ c1X

n1 ).

Then fori = 0, 1, 2,

Gin = nPP | P ∈ ui; nP = I(P,ui ∩ Ci

n).Hence

∏c = (−1)nc2/c1, (−1)nc0/c2, (−1)nc1/c0,

according as the product is taken overG0n, G

1n, G

2n. So

∏c = (−1)n, where the

product is taken over all points ofG.Now suppose that the condition is satisfied by the points ofG; then 3n − 1

points ofG determine the remaining one,P say. LetLP be the linear system ofCnthrough the points ofG\P. ThenL ⊂ LP and, ifd is the dimension ofLP ,

0 ≤ c(n+ 2, 2)− 1− (3n− 1) = c(n− 1, 2) ≤ d ≤ c(n+ 2, 2)− 1.

Since the dimension ofLK is c(n− 1, 2)− 1 < d, soLP \LK 6= ∅. Now,

MG = L\LK ⊂ LP \LK.

However, ifC′n ∈ LP \LK, it meetsK in G′, whereG′ = (G\P) ∪ Q andQ ∈ (

⋃li)\A. ButG′ also satisfies the given condition. SoQ = P andG′ = G.

HenceC′n ∈MG andMG = LP \LK 6= ∅. 2

Forn = 1, this is the theorem of Menelaus and, forn ≥ 2, the theorem of Carnot.This theorem is now generalized to the case thatk ≥ 4.

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THEOREM 14.38 LetK = l1, l2, . . . , lk be a set ofk lines ofPG(2,K) and letGj be a multiset of sizen whose support is onlj , such that no point of intersectionof any two of thek lines lies in the support of anyGj . Suppose that for eachJ ⊂ Nk of the following special types there is an algebraic plane curve CJ

n ofdegreen that meetslj in Gj exactly:

(a) lj | j ∈ J is a triangle;

(b) lj | j ∈ J is a pencil.

Then

(i) there is an algebraic plane curveCn that meetslj in Gj exactly, for allj;

(ii) if there is such a curveCn andk > n, thenCn is unique.

Proof. If all the lines are concurrent there is nothing to prove. Also the result isestablished for a triangle in Theorem 14.37. So assume thatk ≥ 4 and that not allthe lines are concurrent; the proof is by induction.

Let Pi = li ∩ lk for i = 1, 2, . . . , k − 1 and suppose thatP1 6= P2. Let lk = u0

andPi = (0, ci, di), i ∈ Nk−1. By induction, there is a curveCin = v(Fi) with

the correct intersection with all lineslj , j 6= i.Now,F1(0,X1,X2) andF2(0,X1,X2) have degreen, and both meetlk in Gk;

henceF1(0,X1,X2) andF2(0,X1,X2) only differ by a constant factor, and maybe assumed equal. Also,P2 does not lie on the curveC1n, and soF1(0, c2, d2) 6= 0;similarly,Fk(0, c2, d2) 6= 0. Let

β′ =−Fk(0, c2, d2)/F1(0, c2, d2),

F ′(X0,X1,X2) =Fk(X0,X1,X2) + β′F1(X0,X1,X2),

and letC′ = v(F ′); if F ′(X0,X1,X2) is the zero polynomial, the result is proved.Note thatβ′ 6= 0.

The curveC′ meetsl2 in G2 and containsP2; hence, by Bezout’s theorem,l2 isa linear component ofC′. Then for eachj = 3, 4, . . . , k − 1 the curveC′ meets theline lj in Gj and also contains the pointlj ∩ l2; hencelj is a linear component ofC′ for j = 3, 4, . . . , k − 1. Thus, ifPi = v(X0) ∩ v(Ri) with Ri = diX1 − ciX2,then

F ′(0,X1,X2) =Fk(0,X1,X2) + β′F1(0,X1,X2)

is divisible byR2R3 . . . Rk−1.Now, interchange the roles ofl1 andl2 in the above argument. Put

β′′ =−Fk(0, c1, d1)/F2(0, c1, d1),

F ′′(X0,X1,X2) =Fk(X0,X1,X2) + β′′F2(X0,X1,X2),

and letC′′ = v(F ′′). As before, ifF ′′ = 0 the result is proved. Now,

F ′(0,X1,X2)− F ′′(0,X1,X2) =β′F1(0,X1,X2)− β′′F2(0,X1,X2)

= (β′ − β′′)F1(0,X1,X2),


and soF ′(0,X1,X2)−F ′′(0,X1,X2) = 0 meetslk in Gk and the pointP3, sincek ≥ 4. Hence(β′ − β′′)F1(0,X1,X2) is the zero polynomial, and soβ′ = β′′. Itfollows that

F ′′(0,X1,X2) =Fk(0,X1,X2) + β′′F2(0,X1,X2)

=Fk(0,X1,X2) + β′F1(0,X1,X2)

is divisible byR1R3R4 . . . Rk−1. HenceF ′(0,X1,X2) = F ′′(0,X1,X2) is di-visible byR1R2 . . . Rk; here, the fact thatP1 6= P2 and soR1 6= cR2 is used.So

Fk(0,X1,X2) + β′F1(0,X1,X2) =R1R2 . . . Rk−1R.

Now, let li = v(Li) with Li = eiX0 + Ri for i ∈ Nk−1 and defineCn = v(F ),where

F (X0,X1,X2) =Fk(X0,X1,X2)− L1L2 . . . Lk−1R.

ThenCn meetslj in Gj for j ∈ Nk−1. The curveCn meetslk where

Fk(0,X1,X2)−R1R2 . . . Rk−1R = 0,

that is, whereβ′F1(0,X1,X2) = 0; henceCn meetslk in Gk. HenceCn is therequired curve.

Finally, suppose thatk > n, and thatCn = v(F ) andDn = v(G) are curvessatisfying the hypotheses. To obtain a contradiction, suppose that

H(X0,X1,X2) = F (X0,X1,X2) + λG(X0,X1,X2),

with λ = −F (0, c1, d1)/G(0, c1, d1), is not the zero polynomial. Then the curveEn = v(H) has degreen. By Bezout’s theorem,En contains the linelk as alinear component and so contains the pointsP1, P2, . . . , Pk−1. Again, by Bezout’stheorem,En contains the lineslj for j = 1, 2, . . . , k − 1. HenceEn of degreencontainsk > n distinct linear components, a contradiction. HenceH is the zeropolynomial, whenceCn = Dn. 2

14.5 THE CONNECTION BETWEEN ARCS AND CURVES

The objective of this section is to associate a plane algebraic curve defined overFq

to a dualk-arc inPG(2, q). This shows that the dual view is useful in the study ofk-arcs.

The essential result in dualk-arcs is stated and proved separately forq even andodd. As before, letK be ak-arc, ketK′ be a dualk-arc, and lett = q − k + 2.Then∆(K′) stands for the set consisting of thekt points inPG(2, q) which lie onone but not on two of the lines inK.

THEOREM 14.39 Let K′ be a dualk-arc in PG(2, q) with q even. Then thetkpoints on∆(K′) belong to an algebraic plane curveΓt of degreet with the follow-ing properties:

(i) Γt is unique ifk > t, that is, k > 12q + 1;

588

(ii) the points ofK′ distinct from those in∆(K′) do not belong toΓt, and so noline ofK′ is a component ofΓt;

(iii) if ℓ is any line ofK′ andP ∈ ℓ∩∆(K′), thenI(P, ℓ∩Γt) = 1; in particular,P is a non-singular, and hence anFq-rational, point ofΓt;

(iv) every linear component ofΓt is defined overFq; if k ≥ 5 this also holds forevery irreducible component ofΓtofdegreetwo.

Proof. By Lemma 14.28,∏c = 1 for each triangle ofK′, where the product ranges

over all points of∆(K′) on the sides of the triangle andc is the coordinate of sucha point. By Theorem 14.38, there exists an algebraic plane curveΓt define overFq,which is unique ifk > t.

Also, by one of the hypotheses in Theorem 14.38, the points ofthe lines ofKdistinct those from∆(K′) do not belong toΓt, and so no line ofK′ is component ofΓt. Hence, every line ofK′ contains at mostt points ofΓt. As each of thet pointsin ∆(K′) lying on a lineℓ of K is in Γt, none is counted twice in the intersectionof ℓ andΓt.

If a line ℓ′ is a linear component ofΓt, then it meets every line ofK′ at a pointin ∆(K′). HenceΓt contains a point inPG(2, q) from each line ofK′. SinceK′

consists of more than one line,Γt contains two distinct points ofPG(2, q), andhence is defined overFq. The same argument works for a quadratic componentCof Γt whenk ≥ 5, the assumption being necessary as there exist irreducibleconicsnot defined overFq which contain only four points ofPG(2, q). 2

COROLLARY 14.40 In PG(2, q) with q even, a (q + 1)-arc Ω is incomplete. Theq + 1 unisecants toΩ are concurrent andΩ lies in a unique(q + 2)-arc.

Proof. In the dual plane ofPG(2, q), the setΩ is a dual(q+ 1)-arc. Sincek+ t =q+1, sot = 1. Therefore,Γt is a line. InPG(2, q), the corresponding point addedto Ω gives a(q + 2)-arc. 2

COROLLARY 14.41 If K is an incompletek-arc with3 ≤ k ≤ q+1 in PG(2, q), qeven, and ifk > 1

2q + 1, then

(i) there is a pointR on no bisecant ofK;

(ii) if K′ is the dualk-arc arising fromK in the dual plane ofPG(2, q) andΓt is an algebraic curve associated toK′ via Segre’s generalisation of theMenelaus theorem, then the lineℓ corresponding toR in the dual plane is acomponent ofΓt;

(iii) Γt is unique.

Proof. (i) SinceK is incomplete, there is such anR.(ii) Let K′ be the dualk-arc arising fromK in the dual plane ofPG(2, q). In

that plane,R defines a lineℓ such that the points ofℓ cut out by the lines ofK′ aredistinct. There arek such points, each lying in∆(K′). By Theorem 14.39, thereexists an algebraic plane curveΓt defined overFq that contains thesek points ofℓ.


SinceΓt has degreet and the hypothesisk > 12q + 1 implies thatk > t, the lineℓ

must be a component ofΓt.(iii) Again by Theorem 14.39, sincek > t, soΓt is unique. 2

A result similar to Theorem 14.39 is established forq odd. Now,

−1 = 1 for q even; (−1)2 = 1 for q odd.

This is the simple reason that the results forq odd are weaker than forq even;Lemma 14.28 can no longer be directly applied to Theorems 14.37 and 14.38.

THEOREM 14.42 Let K′ be a dualk-arc in PG(2, q) with q odd. Then thetkpoints of∆(K′) wheret = q + k − 2, belong to an algebraic plane curveΓ2t

defined overFq of degree2t with the properties:

(i) Γ2t is unique ifk > 2t, that is, k > 13 (2q + 4);

(ii) the points of the lines ofK′ distinct from those in∆(K′) do not belong toΓ2t

and so no line ofK is a component ofΓ2t;

(iii) if ℓ is any line ofK′ andP ∈ ℓ ∩∆(K′), thenI(P, ℓ ∩ Γ2t) = 2;

(iv) Γ2t may contain components of multiplicity at most two but does not consistentirely of double components;

(v) every linear component ofΓ2t is defined overFq; if k ≥ 5 this also holds forevery irreducible component ofΓ2tofdegreetwo.

Proof. By Lemma 14.28,∏c = −1 for each triangle ofK′, where the product

ranges over all points of∆(K′) on the sides of the triangle andc is the coordinateof such a point. So

∏c2 = 1. By Theorem 14.38, there exists an algebraic plane

curveΓ2t define overFq, which is unique ifk > 2t. Also, by one of the hypothesesin Theorem 14.38, the points of the lines ofK′ distinct those from∆(K′) do notbelong toΓ2t, and so no line ofK′ is component ofΓ2t. Hence, every line ofK′

contains at most2t points ofΓ2t.As each of thet points in∆(K′) lying on a lineℓ of K′ is in Γ2t and counted

twice in the intersection ofℓ andΓ2t, there are no components of multiplicity three.If Γ2t consisted entirely of double components, the points in∆(K′) would belongto a curve of degreet and the condition

∏d = 1 would be satisfied for any triangle

of K′. Finally, the proof of (iv) of Theorem 14.39 also shows (v). 2

COROLLARY 14.43 LetK be an incompletek-arc with3 ≤ k < q + 1, q odd. Ifk > 2

3 (q + 2), that is, k > 2t, then

(i) there is a pointR on no bisecant ofK;

(ii) if K′ is the dualk-arc arising fromK in the dual plane ofPG(2, q) andΓ2t

is the curve associated toK′, then the lineℓ corresponding toR in the dualplane is a component ofΓ2t;

(iii) Γ2t is unique.

590

Proof. The arguments used to prove Corollary 14.41 remain valid foroddq whenk > 2t, this assumption being necessary to ensure the validity of (ii). By Theorem14.42, there exists a plane curveΓ2t defined overFq such thatI(P,Γ2t ∩ ℓ) ≥ 2for every pointP ∈ ℓ ∩ ∆(K′). SinceΓ2t has degree2t, the hypothesisk > 2timplies that the lineℓ must be a component ofΓt. 2

REMARK 14.44 Segre’s Theorem 14.31 can be deduced from Theorem 14.42. Ask = q + 1, sot = 1. Hence, by Theorem 14.42, the points of the dual(q + 1)-arcK′ belong to a conicΓ2 defined overFq which does not split into two, possiblycoincident, lines, as otherwise the(q + 1)-arcK would be incomplete. Also, sinceI(P,Γ2 ∩ l) = 2 for P ∈ l ∩ ∆(K′), so the lines ofK′ are tangents toΓ2 at thepoints of∆(K′). Since the number ofFq-rational points ofΓ2 is exactlyq + 1, soK′ can be viewed as the envelope of the conicΓ2t at itsFq-rational points. Afterdualizing, Theorem 14.31 follows.

14.6 ARCS IN OVALS IN PLANES OF EVEN ORDER

In this section, the sizem′(2, q) of the second largest complete arc inPG(2, q), qeven, is obtained.

THEOREM 14.45 In PG(2, q), q even,

m′(2, q) ≤ q −√q + 1; (14.4)

whenq is also square, then

m′(2, q) = q −√q + 1. (14.5)

Proof. Let K be a dualk-arc inPG(2, q) with k > q − √q + 1. Without loss ofgenerality,K is assumed to be complete. LetΓt be the associated curve of degreetas in Theorem 14.39. Note thatt = q − k + 2 <

√q + 1.

Suppose thatΓt has anFq-rational linear component, that is, there is a lineℓ inPG(2, q). Then the common point ofℓ and any line inK is in ∆(K). Therefore,ℓ added toK is a dual(k + 1)-arc, a contradiction toK being complete. ThusΓt

does not contain anFq-rational linear component.LetNq be the number ofFq-rational points ofΓt. From Theorem 14.39 (iii),

|Nq| ≥ |∆(K)| = kt.

On the other hand, Theorem 10.64 applied toΓt implies thatNq < kt. This con-tradiction proves thatm′(2, q) ≤ q − √q + 1. If q is square, then Example 14.26implies (14.5). 2

It is still an open problem to determine all complete(q−√q+1)-arcs inPG(2, q)with q even and square. The following result is a step in this direction.

THEOREM 14.46 LetK be a complete dualq2 − q + 1-arc in PG(2, q2) with qeven. Then the associated curveΓt is the Hermitian curveHq.


Proof. Here, t = q + 1. So Γt has genusg ≤ 12 (q2 − q). From the proof of

Theorem 14.45, the number ofFq-rational points ofΓt is at leastkt = q3 + 1. Bythe Hasse–Weil theorem, this is only possible wheng = 1

2q(q− 1) and the numberof Fq-rational points ofΓt is exactlyq3 + 1. In particular,Γt is a non-singularmodel of anFq2-maximal curve of genusg = 1

2q(q−1). Thus,Γt is the HermitiancurveHq. 2

14.7 ARCS IN OVALS IN PLANES OF ODD ORDER

In this section, a similar treatment to the previous sectionis given for planes of oddorder. An upper bound is obtained form′(2, q) in the case thatq is odd.

The next result is both weaker and stronger, in the sense thatthe arithmetic con-dition is stronger, as is the specification of the complete arc containingK.

THEOREM 14.47 LetK′ be a dualk-arc with3 ≤ k ≤ q + 1 in PG(2, q), q odd,and letΓ2t be the associated curve. If

(i) Γ2t has an irreducible conicC defined overFq as a component,

(ii) k > 14 (3q + 5),

then the lines ofK′ are tangents toC.

Proof. The lines ofK′ are of two types with respect toC according as the line is atangent toC or not. The lines of the first type form a dualk1-arcK′

1, and those ofthe second type form a dualk2-arcK′

2 such thatk1 +k2 = k = q− t+2. Also, theFq-rational tangents toC distinct from the lines inK′

1 form a dualk3-arcK′3 with

k3 = q + 1 − k1. SinceC meets a line at most two points, sok1 + 2k2 ≤ q + 1.Hence

k1 ≥ q + 3− 2t, k2 ≤ t− 1, k3 ≤ 2t− 2.

Let ℓ be a line inPG(2, q) which is not a tangent toC. At least 12 (q − 1) points of

ℓ in PG(2, q) are the intersections of two tangents toC atFq-rational points. Also,k > 1

4 (3q + 5) implies that12 (q − 1) > 2t− 2. As k3 ≤ 2t− 2, at least one pointP ∈ ℓ must be the meet of twoFq-rational tangents toC not contained inK′

3. Soboth are lines inK′

1. As ℓ is not anFq-rational tangent toC, soℓ is not inK′. Thus,K′ consists ofFq-rational tangents toC. 2

The next result is a key one.

THEOREM 14.48 In PG(2, q), q odd,

m′(2, q) ≤ q − 14

√q + 7

4 .

Equivalently, if K is ak-arc withk > q − 14

√q + 7

4 , that is,√q > 4t− 1, thenK

is contained in the set ofFq-rational points of a unique irreducible conic definedoverFq.

592

Proof. If k > q − 14

√q + 7

4 , thenk ≥ 5. So, if K is contained in the set ofFq-rational points of a conic, then the conic is unique.

Consider the dual plane ofPG(2, q). LetK′ be the dualk-arc arising fromK.Theorem 14.42 gives all the necessary information on the curve Γ2t defined overFq associated toK′. For t = 1, such a curve is an irreducible conicC defined overFq, andK′ consists of lines tangents toC. So it may be assumed thatt ≥ 2.

Choose an irreducible componentΓ′n of Γ2t, wheren ≤ 2t, and distinguish the

following three cases:

(i) Γ′n is anFq-rational linear component ofΓ2t;

(ii) Γ′n is anFq-rational component of degree two;

(iii) Γ′n is anFq-rational component of degree at least three orΓ′

n is not definedoverFq.

Case(i).Here,Γ′

n is anFq-rational lineℓ which is not inK′. SoK′1 = K′ ∪ ℓ is a dual

(k+1)-arc. If the envelopeΓ′n associated withK′

1 is again of type (i), the process iscontinued until the dual arc becomes the set of all tangents to anFq-rational conicor the respective plane curve becomes one of the other two types.

Case(ii).Now, Γ′

n is anFq-rational conicC. Also,

k > q − 14

√q + 7

4 ⇐⇒ q > 4t− 1⇒ q > 4t− 3⇐⇒ k > 14 (3q + 5).

So, by Theorem 14.47,K′ consists ofFq-rational tangents toC.Case(iii).

Here,Γ′n is Fq-rational with3 ≤ n ≤ 2t or Γ′

n is notFq-rational. Suppose thatΓ′

n hasS non-singular andd double points inPG(2, q). From Theorem 10.66,

S + d < 12n(q − t+ 2) = 1

2nk. (14.6)

By Theorem 14.42,I(P,Γ2t ∩ ℓ) = 2 holds for each of thet pointsP ∈ ∆(K′) ofa lineℓ in K′. Therefore,I(P,Γ′

n ∩ ℓ) ≤ 2, andI(Q,Γ′n ∩ ℓ) = 0 for Q 6∈ ∆(K′).

Since∑

Q∈ℓ I(Q,Γ′n ∩ ℓ) = n,

this implies thatΓ′n has at least12n, possibly double, points fromℓ in PG(2, q).

Hence,

S + d ≥ kn/2,contradicting the previous estimate (14.6).

Thus,K is ever contained in the set ofFq-rational points of an irreducible conicdefined overFq. 2

COROLLARY 14.49 If K is a k-arc in PG(2, q), q odd, with k > q − 14

√q + 7

4 ,thenK is contained in the setC(Fq) of all Fq-rational points of an irreducibleconic defined overFq, and the only points for whichK ∪ Q is a (k + 1)-arc aretheq + 1− k points ofC(Fq)\K.


Proof. If K ∪ Q is a (k + 1)-arc, then it lies in some complete arc, which isnecessarilyC(Fq) by the theorem. HenceQ ∈ C(Fq)\K. 2

COROLLARY 14.50 In PG(2, q), q odd andq > 49, a q-arc is contained in theset ofFq-rational points of a conic.

Proof. The inequalityq > q − 14

√q + 7

4 holds whenq > 49. 2

REMARK 14.51 This result in fact holds true forq ≤ 49, giving the similar resultin the case thatq is even.

Next, a small improvement to Theorem 14.48 is given. There the result is equiv-alent tot < 1

4 (√q+1) for ak-arc that is contained in a conic, wheret = q+2−k.

THEOREM 14.52 In PG(2, q), q odd,

(i) a k-arc K for which t < (√q + 1)2/(4

√q + 1) is contained in the set of

Fq-rational points of an irreducible conic defined overFq;

(ii) m′(2, q) < q − 14

√q + 25

16 .

Proof. (i) If t = 1, thenK is a (q + 1)-arc, that is, an oval inPG(2, q), and theassertion follows from Segre’s Theorem 14.31. So assume that t > 1. In particular,q > 3.

LetK′ be the dualk-arc ofK. Assume that the algebraic plane curveΓ2t definedoverFq, associated toK′ via Segre’s generalisation of the Menelaus Theorem, isirreducible. WithR the number of points ofΓ2t with coordinates inFq, (iii) ofTheorem 10.57 gives thatR ≤ q + 1 + (2t − 1)(2t − 2)

√q. So, asK haskt

unisecants, that is,|∆(K′)| = kt,

q + 1 + (2t− 1)(2t− 2)√q≥ kt,

(t− 1)q + 1 + 2√q − t(1 + 4

√q)≤ 0.

Hencet ≥ (√q + 1)2/(4

√q + 1), a contradiction. Therefore,Γ2t is reducible.

At least one irreducible componentΓ′n has degreen ≤ t. Consider the same

three cases as in the proof of Theorem 14.48.If Γ′

n is Fq-rational of degreen = 1, thenΓ′n is a lineℓ not inK′. SoK′ ∪ ℓ

is a dual(k + 1)-arc.If Γ′

n is Fq-rational of degreen = 2, thenΓ′n is an irreducible conicC. Since

k < 14 (3q + 5), that is,t < 1

4 (q + 3), Theorem 14.47 implies thatK′ is a set oftangents toC atFq-rational points.

Finally, suppose thatΓ′n is notFq-rational or isFq-rational withn ≥ 3. Then

3(√q + 1)2/(4

√q + 1) − 1 ≤ √q,

sinceq > 3, and so3t− 1 <√q. By Theorem 10.66,

R < 12n(q − t+ 2) = 1

2nk.

From the proof of Theorem 14.48,Γ′n has at least12n, possibly double, points from

ℓ in PG(2, q), whereℓ is a line ofK′. HenceR ≥ 12nk, a contradiction.

594

ThusK′ is contained in a dual(k + 1)-arc. Continuing the process shows thatK′ is contained in a dual(q + 1)-arc. Then Theorem 14.47 implies thatK′ is a setof tangents toC atFq-rational points.

(ii) The inequalityt < (√q + 1)2/(4

√q + 1) is equivalent to

k > q − 14

√q + 25

16 − 916/(4

√q + 1).

2

14.8 THE SECOND LARGEST COMPLETE ARC: FURTHER RESULTS

In this section similar techniques to the previous section are used, but the Stohr–Voloch theorem is applied instead of the Hasse–Weil theorem. This makes it pos-sible to obtain improvements on upper bounds form′(2, q).

THEOREM 14.53 In PG(2, p), with p an odd prime andp ≥ 5,

m′(2, p) ≤ 4445p+ 8

9 .

Proof. Suppose that there exists a completek-arcK with k > 4445p+ 8

9 .As in the proof of Theorem 14.48, the same three possibilities for an irreducible

componentΓ′n of Γ2t are considered, whereΓ2t is theFq-rational curve associated

to the dualk-arcK′.(i) If Γ′

n is Fq-rational andn = 1, thenK′ is incomplete.(ii) If Γ′

n is Fq-rational andn = 2, thenΓ′n is an irreducible conicC defined over

Fq. However,

k > 4445p+ 8

9 >14 (3p+ 5).

So, by Theorem 14.47,K is contained in the setC(Fq) of all Fq-rational points ofC.

(iii) Suppose thatΓ′n hasS non-singular andd singular points inPG(2, q). Then,

as in the proof of Theorem 14.48,

S + d ≥ kn/2.If Γ′

n is notFq-rational, thenS + d ≤ n2; sok ≤ 2n ≤ 4t = 4(p+ 2− k). Hencek ≤ 4

5 (p+ 2) < 4445p+ 8

9 for p ≥ 5.Now, letΓ′

n beFq-rational withn ≥ 3. The linear system of all conics cuts outon Γ′

n the fixed-point-freeFq-rational linear seriesL2 that defines the Veronesemap. The resulting curve is an irreducibleFq-rational curveΓ of PG(5,K) ofdegree2n whereK is the algebraic closure ofFq. Let ν0 = 0, ν1, . . . , ν4 be theFrobenius orders ofΓ. From the fact thatνi ≤ deg Γ = 2n = 4t andk > 1

4 (3p+8)for p ≥ 5, it follows that2n < p. Asp ≥ 5, Proposition 8.48 (ii) implies thatΓ is aFrobenius classical curve, that is,νi = i for 0 ≤ i ≤ 4. Therefore, the Stohr-VolochTheorem 8.62 gives that

12kn− d ≤ 2

5n5(n− 2) + p − 4d.

So

k≤ 455(n− 2) + p

≤ 455(2t− 2) + p

= 4510(p+ 1− k) + p

= 8(p+ 1− k) + 45p.


Hencek ≤ 4445p+ 8

9 .If t > 1

4p, thenk = p + 2 − t < 34p + 2 < 44

45p + 89 , unlessp = 3; but this is

ruled out by the hypothesis. 2

Now this theorem is extended to arbitrary odd powers of the characteristic.

THEOREM 14.54 In PG(2, q), q = p2e+1 with p an odd prime ande ≥ 1,

m′(2, q) ≤ q − 14

√pq + 29

16p+ 1.

Proof. LetK be a completek-arc withk greater than the bound in the theorem. LetK′ be the dualk-arc arising fromK in the dual plane ofPG(2, q). As in the proofof the previous theorem, it is only necessary to consider case (iii) and a componentΓ′

n of Γ2t of degreen ≥ 3 that is Frobenius non-classical with respect to the linearsystem of conics.

If the points in∆(K′) are are all double points ofΓ′n, then

12kn ≤ 1

2 (n− 1)(n− 2) ≤ 12n

2;

sok ≤ n ≤ 2t, and hence

k ≤ 2

3(q + 2) < q − 1

4

√pq + 29

16p+ 1,

as required. Otherwise, someP ∈ ∆(K′) is a non-singular point ofΓ′n. If ℓ is

the line inK′ throughP , thenI(P,Γ′n ∩ ℓ) = 2. Therefore, the(P,L1)-orders are

0, 1, 2, whereL1 is the linear series onΓ′n cut out by lines. By the argument in the

third paragraph of Section 7.9,Γ′n has order sequence(0, 1, 2, 3, 4, ǫ5 ≤ 2n) with

respect to the linear seriesL2 cut out by conics. IfΓ′n is Frobenius non-classical for

L2, then the Frobenius order sequence is(0, 1, 2, 3, ǫ5) such thatǫ5 = pm unlessp = 3 andǫ5 = 6. From the Stohr–Voloch Theorem 8.62 applied toL2, see also(8.58), ifd ≥ 0 points in∆(K′) are double points, then

12kn− d ≤ 1

5 [(6 + ǫ5)(n(n− 3)− 2d) + 2n(q + 5)].

Hence

k ≤ 25 [(6 + ǫ5)(n− 3) + 2(q + 5)] ≤ 2

5 [(6 + ǫ5)(2t− 3) + 2(q + 5)].

As k = q + 2− t, it follows that

t ≥ q

4ǫ5 + 29+

6ǫ5 + 26

4ǫ5 + 29≥ q

4ǫ5 + 29+ 1. (14.7)

If ǫ5 ≥ pe+1 then, asǫ5 ≤ 2n ≤ 4t, it follows that

k ≤ q + 2− 14p

e+1 = q + 2− 14

√pq,

a contradiction.Suppose, therefore, thatǫ5 < pe+1. If p 6= 3, thenǫ5 is a power ofp; soǫ5 ≤ pe

and, from (14.7),t ≥ q/(4pe + 29) + 1, whence

k > q − 14

√pq + 29

16p+ 1,

the required contradiction.If p = 3, the same argument gives the result unlessǫ5 = 6, which also satisfies

ǫ5 ≤ pe unlesse = 1, that is,q = 27. However, the result is trivially true forq = 27. 2

596

THEOREM 14.55 In PG(2, q), q = 22e+1 with e ≥ 1,

m′(2, q) ≤ q −√

2q + 2.

Proof. Assume that there exists a completek-arcK with k > q − √2q + 2. Asin Theorem 14.39, letΓt be the curve of degreet = q + 2 − k is associated tothe dualk-arc. Such a curveΓt is defined overFq and has a non-linear irreduciblecomponentΓ′

n of degreen with n ≤ t. Let ν0 = 0, ν1 = ν be the Frobenius ordersof Γ′

n with respect to the linear series cut out by lines.If Γ′

n is not defined overFq, thenkn ≤ n2. Sok ≤ n ≤ t = q + 2− n; that is,k ≤ 1

2 (q + 2) ≤ q −√2q + 2.If Γ′

n is defined overFq, then, using (8.19),

kn ≤ 12 [ν(2g − 2) + n(q + 2)] ≤ 1

2n[ν(n− 3) + q + 2].

Hence

2k ≤ ν(n− 3) + q + 2 ≤ ν(t− 3) + q + 2

and, sincek = q + 2− n,

t ≥ q + 2 + 3v1ν + 2

.

If ν ≤√q/2, then, forq > 8,

t ≥ 2(q + 2) + 3√

2q√2q + 4

>

⌊2(q + 2) + 3

√2q√

2q + 4

⌋=√

2q − 1.

Therefore,k = q + 2 − t < q + 3 − √2q. As√

2q is an integer, the theorem isproved in this case.

Sinceν ≤ n ≤ t andν is a power of 2, ifν >√q/2, then

√2q ≤ ν1 ≤ t; so

k ≤ q + 2−√2q, as required.For q = 8, the bound is sharp as the only complete arcs other than hyperovals

are6-arcs. 2

Table 14.4 gives the known values ofm′(2, q) apart from those withq ≥ 64implicit in Theorem 14.45. Here,m′ is written form′(2, q).

Table 14.4 Size of the second largest complete arc in small planes

q 7 8 9 11 13 16 17 19 23 25 27 29 31 32m′ 6 6 8 10 12 13 14 14 17 21 22 24 22 24

From Theorems 14.45 and 14.45 and Table 14.4, aq-arc inPG(2, q) is alwaysincomplete.

As an application of the theorems in this section, the statusof (q − 1)-arcs canbe clarified.

THEOREM 14.56 In PG(2, q), a (q − 1)-arc is incomplete forq > 13 exceptpossibly for the fourteen values ofq consisting of49, 81 and the twelve primes37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83.


Proof. From Table 14.4, onlyq > 29 need be considered. Forq even, Theorem14.45 shows thatm′(2, q) < q− 1 for q ≥ 8. Forq odd, Theorem 14.52 shows thatm′(2, q) < q − 1 for q > 103; Theorem 14.53 improves this toq > 83. 2

To resolve the question of the completeness of(q − 1)-arcs for the remainingfifteen values ofq, the plane sexticΓ6 associated to the dual(q − 1)-arc can beconsidered.

THEOREM 14.57 If q > 9, the plane sexticΓ6 is irreducible.

Proof. The possibilities forΓ6 are examined in three parts.(I) Γ6 contains noFq-rational linear components.

If Γ6 did contain such a component,K would be incomplete.(II) Γ6 contains no multiple components.

Suppose thatΓ6 = v(F ) andF = G1G2, whereC1 = v(G1) has no multi-ple components andC2 = v(G2) consists of double components. From Theorem14.42,Γ6 does not consist entirely of double components. By (I), it may be as-sumed thatΓ6 has just one double component, anFq-rational irreducible coniccounted twice. HenceC1 is an irreducible conic defined overFq. By Theorem14.47, the lines ofK′ areFq-rational tangents toC′2r, andK′ is incomplete.(III) Γ6 is irreducible.

If Γ6 were reducible, then one of the following three cases would apply:(a)Γ6 splits into three irreducible conics defined overFq;(b) Γ6 splits into an irreducible conic and an irreducible quartic, both defined

overFq

(c) C6 splits into two irreducible, distinct cubics.Again, by Theorem 14.47, (a) and (b) are impossible, since the fact that an irre-

ducible conic defined overFq is a component ofΓ6 implies thatK is incomplete.If, in case (c),Γ6 has componentsΓ3 andΓ′

3, thenΓ3 andΓ′3 have a setR of

at most nine common points inPG(2, q). Also, each line in the dual(q − 1)-arcK′ passes through three points ofΓ6 with intersection multiplicity two. If such aline ℓ passed through two points ofΓ3 each with multiplicity two, thenΓ3 wouldbe reducible. Similarly, ifℓ met Γ3 at one point not inR with multiplicity twoand passed through two points ofR, thenΓ3 would be reducible. Soℓ must passthrough three points ofR. Since no two lines inK′ intersect onΓ6, it follows thatq − 1 ≤ 3, a contradiction. 2

CONJECTURE 14.58 In PG(2, q), with q square andq > 9,

m′(2, q) = q −√q + 1.

Further, the corresponding arc is projectively unique.

The lowest value ofq for which this conjecture is unresolved isq = 49; that is, itmust be shown thatm′(2, 49) = 43. Related to this conjecture are Example 14.26and the following result analogous to Theorem 14.46.

THEOREM 14.59 LetK′ be the dual of the cyclic(q2 − q + 1)–arc in PG(2, q2)with q odd, defined in Example 14.26. Then the associated curveΓ2(q+1) is bira-tionally equivalent overFq6 to the Hermitian curveHq.

598

The proof requires some preliminary results.For a primitive(q2 + q + 1)-th t root of unity inFq6 ,

α :

ρX0′ = tX0,

ρX1′ = tq

2+1X1,ρX2

′ = X2,

is a projectivity of order(q4 + q2 + 1) that fixes each vertex of the fundamentaltriangleU0U1U2 of a projective coordinate system inPG(2,K), whereK is thealgebraic closure ofFq. LetA denote the collineation group generated byα.

The orbitΠ of the pointU = (1, 1, 1) underA has sizeq4 + q2 + 1 and consistsof all pointsPc = (c, cq

2+1, 1) with cq4+q2+1 = 1. Note thatΠ lies in PG(2, q6)

as the(q4 + q2 + 1)-th roots of unity are inFq6 .The orbitΠ may be viewed as a subgeometry ofPG(2,K) induced by the lines

meetingΠ in at least two points.

PROPOSITION 14.60 The orbitΠ ∼= PG(2, q2). More precisely,Π is a projectivesubplane ofPG(2, q6) lying in a non-classical position; that is, Π 6= PG(2, q2)but there is a projectivity ofPG(2, q6) that mapsΠ to PG(2, q). The lines ofΠ are

v(tX0 + tq2+1X1 +X2),

with t running over the(q4 + q2 + 1)−th roots of unity, form the line orbit ofv(X0 +X1 +X2) underA.

Proof. The linev(X0 +X1 +X2) meetsΠ in exactlyq2 + 1 distinct points, sinceuq2+1 + u + 1 = 0 implies thatu is a (q4 + q2 + 1)−th root of unity. In fact, ifuq4+q2

+ uq2

+ 1 = 0, thenuq4+q2+1 + uq2+1 + u = 0 whenceuq4+q2+1 = 1.Also, the polynomialF (X) = Xq2+1 +X + 1 has no multiple roots asF (X) andits derivativeF ′(X) = Xq2

+ 1 have no common root.For any(q4 + q2 + 1)−th roott ∈ Fq6 , the linelt = v(tX0 + tq

2+1X1 +X2)meetsΠ in the points

Pi = (ci, c(q2+1)i, 1) with (cit)q2+1 + cit+ 1 = 0.

As before, there are exactlyq2 + 1 such points. ThereforeΠ equipped with all thelineslt is a projective plane of orderq2. This is shown by a direct calculation.2

Next, intersections ofΠ with conics and Hermitian curves are considered.

PROPOSITION 14.61 The conicC = v(X20 + X2

1 + X22 ) and the plane curve

D = v(X0Xq2 −Xq+1

1 ) have2(q+ 1) distinct points in common, and half of thesepoints lie inΠ.

Proof. First,v(X2)∩C ∩D = ∅. Now, choose any pointP = (u, v, 1) off the linev(X2). ThenP is a point ofD if and only if u = vq+1. In this case,P is also apoint ofC if and only if v2(q+1) + v2 + 1 = 0. The polynomialf(X), with

f(X) = X2(q+1) +X2 + 1,

has no double roots since its derivative

f ′(X) = 2X2q+1 + 2X = 2X(X2 + 1)q


has no root in common withf(X).Now, q + 1 of the rootsv of f(X) give rise to pointsP = (vq+1, v, 1) lying in

Π. Since

f(X)qX2 = X2(q2+q+1) +X2(q+1) +X2,

each rootv of f(X) is also a root of the polynomialg(X) = X2(q2+q+1) − 1.Sincef(X) = f(−X) andg(X) = g(−X), half of the rootsv of f(X) are alsoroots of the polynomialh(X) = Xq2+q+1 − 1. This shows that the correspondingq + 1 pointsP = (vq+1, v, 1) lie in Π. 2

PROPOSITION 14.62 The conicC = v(X20 +X2

1 +X22 ) and the Hermitian curve

Hq = v(X0Xq1 +X1X

q2 +X2X

q0 ) haveq + 1 distinct common points, and all lie

in Π. Also, the curvesC andHq have the same tangent line at each of these points.

Proof. From the proof of Proposition 14.61, the setΠ containsq+1 distinct pointsP = (vq+1, v, 1) of C, wherev is a common root off(X) andh(X). Each of thesepoints also lies onHq, since

v2q+1 + vq2+q + v =1

v(vq2+q+1 + v2(q+1) + v2) =

1

v(1 + v2(q+1) + v2) = 0.

A calculation shows that the tangent line toHq at the pointP = (vq+1, v, 1) hasslope equal to−√v. The same value is obtained whenHq is replaced byC. HenceC andHq are tangent at each of theseq + 1 points. By Bezout’s Theorem 3.14,there is no other common point ofC andHq, and this completes the proof. 2

Now, consider the family of conicsCt = v(Ft), with

Ft = tX20 + tq+1X2

1 +X22 (14.8)

with t a (q2 + q + 1)-th root of unity. This family consists of all conics which areimages of the conicC = v(X2

0 + X21 + X2

2 ) under the action of the collineationgroupA. Sinceq4 +q2 +1 = (q2 +q+1)(q2−q+1), the groupA has a subgroupB of orderq2 − q + 1. The orbit ofC underB consists of the conicsCt with tranging over the(q2 − q + 1)−th roots of unity. The next result connects the setΦof points ofHq lying in Π with the sets∆t of points ofCt lying in Π.

PROPOSITION 14.63 The sets∆t, with t a (q2 − q+ 1)-th root of unity, partitionΦ into (q2 − q + 1) point sets each of sizeq + 1.

Proof. It is shown first thatB leavesHq invariant. The image ofH underαi is thecurve of equationX0X

q1 + uq(q2−q+1)X1X

q2 + u(q+1)(q2−q+1)X2X

q0 = 0 where

u = ti. If αi is a generator ofB, theni = q2 + q+ 1 anduq2−q+1 = 1 whence theassertion follows.

Proposition 14.60 remains valid whenC is replaced by any conicCt in the family.Therefore, no point ofΦ lies on more than one such conic. In fact, if botht andu are(q2−q+1)-th roots of unity, andP = (a, b, 1) is a point inΦ∩∆t∩∆u, Proposition14.60 implies that the conicsCt andCu have the same tangent atP = (a, b, 1). But,this can only occur when−tq = −uq, and hencet = u. From Proposition 14.62,

600

each conicCt in the family meetsΦ in q + 1 points. SinceΦ has sizeq2 + q + 1,the assertion follows. 2

A relationship between conicsCt and lineslt is given in the following result.

L EMMA 14.64 The quadratic transformation,

τ :

X ′0 = X2

0 ,X ′

1 = X21 ,

X ′2 = X2

2 ,

leavesΠ invariant. The restriction ofτ to Π is a bijection and transforms∆t intothe setΨt of the common points oflt andΠ.

Proof. The invariance ofΠ follows from a direct calculation. The fact that themappingx → x2 is a permutation of the set of all(q2 + q + 1)−th roots of unityshows that∆t is transformed intoΨt. 2

The proof of Theorem 14.59 consists in showing that the transformationτ turnsHq into Γ2t. Let Γ be the image ofHq underτ . From Propositions 14.62, 14.63and 14.64, each point of theq3 + 1 points ofΓ lying in Π is a non-singular point ofΓ, and the linelt with t a (q2 − q + 1)−th root of unity is a(q + 1)−tangent toΓ;that is,lt is the tangent line toΓ at each of their common points. Such lineslt forma cyclic dual(q2− q+ 1)-arcK′ in Π viewed as projective plane defined overFq2 .By Theorem 14.42,Γ is the curveΓ2(q+1) associated toK′.

To complete the proof of Theorem 14.59 it remains to show thatτ mapsH ontoΓ birationally. Consider the linear systemΦ of all conics

v(c0X20 + c1X

21 + c2X

22 ), with c = (c0, c1, c2) ∈ PG(2,K).

It must be shown that the linear seriesg22(q+1) cut out onHq by Φ is simple, that

is, not composed of an involution. For a pointP of Hq, the conics inΦ throughPform a linear subsystem ofΦ. To show thatg2

2(q+1) is simple, it suffices to provethe existence of a pointP ∈ Hq such that the conics inΦP have no further commonpoint onHq.

It is shown that such a condition is satisfied by every pointP = (a, b, c) ∈ Hq

with abc 6= 0. The conics ofΦP have exactly four common points inPG(2,K),namely,

P = (a, b, c), P1 = (a,−b, c), P2 = (−a, b, c), P3 = (a, b,−c).

None of the last three points belongs toHq. Thus, the pointP = (a, b, c) is the onlycommon point of the conics ofΦP which lies onHq. This completes the proof.

Theorem 8.102 can be applied to give better bounds form′(2, q).

THEOREM 14.65 Let q = ph with p ≥ 3, and letq = 32e whenp = 3.

(i) If q ≥ 232 andq 6= 36 or 55, then

m′(2, q) ≤ q − 12

√q + 3.


(ii)

m′(2, q) ≤

q − 22 whenq = 55,

q − 9 whenq = 36,

q − 9 whenq = 232,

q − 5 whenq = 192.

Proof. Given ak-arcK in PG(2, q) with k > q − 12

√q + 3, the aim is to prove

thatK is contained in an irreducible conic defined overFq. LetK′ be the completedualk-arc arising fromK in the dual plane ofPG(2, q).

From Theorems 14.42 and 14.47, the curveΓ2t associated toK′ does not havelinear components. Further, if it has a quadratic componentthenK′ is a dual(q+1)-arc consisting of all tangents to anFq-irreducible conicC at its points inPG(2, q).So, assume that each component ofΓ2t has degree at least three. For such a compo-nentΓ = v(f(X,Y )) of degreed, two cases are distinguished according as eitherΓ is defined overFq or over a non-trivial finite extension ofFq.

In the latter case,f(X,Y ) is distinct from its first conjugatef (1)(X,Y ) and alsofrom cf (1)f(X,Y ) for c ∈ K; see Section 8.3. HenceΓ is distinct from its firstconjugate which is the irreducible plane curveΓ′ = v(f (1)(X,Y )) but ans-foldpointP of Γ with coordinates inFq is also anr-fold point ofΓ′.

Let γ1, . . . , γN be the branches ofΓ centred at the points in∆(K′), and letPi bethe corresponding places ofK(D). If ri = j1(Pi) denotes the order ofγi, then thecentre ofγi is a point ofΓ whose multiplicity is at leastri. The same holds for thebranchesγ′1, . . . , γ

′N of Γ′ centred at the points in∆(K′). By Bezout’s Theorem

3.14,∑N

i=1 r2i ≤

∑P∈D(K′)I(P,Γ ∩ Γ′) ≤ d2 ≤ d(√q − 2) < 1

2d(q −√q + 1).

If Mq andM ′q, with respect toΓ, have the same meaning as in Definition 8.75, it

follows that

2Mq +M ′q < d(q −√q + 1). (14.9)

Now, let Γ be anFq-rational irreducible component ofΓ2t of degreed. If (14.9)does not hold, then Theorem 8.102 applies toΓ, showing thatΓ is projectivelyequivalent overFq to the Fermat curve of degreed = 1

2 (√q + 1); in particular,

2Mq +M ′q = d(q −√q + 1).

Thus, for every irreducible componentΓ of Γ2t, also

2Mq +M ′q ≤ d(q −

√q + 1).

Therefore,∑

(2Mq +M ′q) ≤ 2t(q −√q + 1). (14.10)

where the summation is over all irreducible components ofΓ2t. On the other hand,Theorem 14.42 provides a lower bound for the above sum. In fact, every pointP ∈ ∆(K′) is either a non-singular point or a double point ofΓ. In the formercase,P counts inMq asI(P,Γ2t ∩ ℓ) = 2 for the unique lineℓ in ∆(K′) whichpasses throughP . In the latter case,P is the centre of either one branch of order2

602

or of two branches of order1. ThusP counts at least twice in2Mq + M ′q. Since

∆(K′) consists ofkt points, so

2kt ≤∑ (2Mq +M ′q). (14.11)

Comparison of (14.10) and (14.11) givesk ≤ q − √q + 1, contradicting the hy-pothesis thatk > q − 1

2

√q + 3. 2

REMARK 14.66 Sinceq is odd, in the case in whichq is a square, the3 can belowered to5

2 . The best result for the caseq = 32e+1 is that

m′(2, q) ≤ q −√

34

√q + 103

16 .

By Theorem 14.45, the exact value ofm′(2, q) is known for every even squareq.In the study of the spectrum of large arcs ofPG(2, q) with even squareq, the nextquestion is to investigate the maximum sizem′′(2, q) of the third largest completearcs. The following result depending on Theorem 8.74 provides an upper bound.

THEOREM 14.67 Let q = 2h, q ≥ 16 andq square. Then

m′′(2, q) ≤q − 2

√q + 6 for q ≥ 64,

q − 2√q + 4 = 12 for q = 16.

(14.12)

Proof. LetK′ be the a complete dualk-arc inPG(2, q) with q = 2h, q ≥ 16 andqsquare. It must be shown that, ifk is larger than the right hand side in (14.12), thenone of the following holds:

(i) k = q −√q + 1;

(ii) k = q + 2.

Let Γt be the curve defined overFq associated toK′. It must be shown thatΓt isirreducible.

Let Γ be an irreducible component ofΓ of degreen. From Theorem 14.39 (iii),Γ containsn distinct points on every line inK′, and such points lie in∆(K′).

As in the proof of Theorem 14.65, letΓ′ be the first conjugate ofΓ. These curvesare distinct if and only ifΓ is not defined overFq. Since every point ofΓ withcoordinates inFq is also a point ofΓ′, each of thekn points in∆(K′) counts in theintersection ofΓ andΓ′. Sincekn = k(q − k + 2) > n2, Bezout’s Theorem 3.14shows that the two curves coincide, unlessn2 ≥ nk. This inequality implies thatn ≥ k. Sincet ≥ n, alsoq+ 2− k ≥ k; but this contradicts thatk > q− 2

√q+ 6.

Therefore,Γ is defined overFq.Next, it is shown thatΓt = Γ. For, if not, then

2n ≤ t = q + 2− k < q + 2− (q − 2√q + 6) = 2

√q − 4

andn <√q − 2. Let g2

n be the linear series cut out onΓ by lines. It may be thatΓ is Frobenius non-classical, and if this occurs then its Frobenius orderν1 = q′ isbounded above by12

√q asq′ does not exceed the largest power of2 smaller than

n <√q − 2. From Theorem 8.71 applied tog2

n,

nk ≤ N ≤ 12n(n− 3)q′ + (q + 2)n;


so

2k ≤ (n− 3)q′ + (q + 2).

Now, apply the conditionsk > q − 2√q + 6, n <

√q − 2, q′ ≤ 1

2

√q:

2(q − 2√q + 6) < 2k ≤ (

√q − 5) 1

2

√q + q + 2,

whence

4(q − 2√q + 6) < (

√q − 5)

√q + 2(q + 2),

and

q − 3√q + 20 < 0,

a contradiction. Hencen = t andΓt = Γ.Now, Theorem 8.74 is applied. Sincek > q − 2

√q + 6, so

t = q + 2− k < 2√q − 4.

Hence, by Theorem 8.74, the number ofFq-rational simple pointsN0 of Γt satisfiesN0 ≤ t(q+2− t). Since∆(K′) has sizet(q+2− t), soN0 ≥ t(q+2− t). HenceN0 = t(q+ 2− t) and again, by Theorem 8.74,Γt is projectively equivalent to theHermitian curve of degree

√q + 1, which is case (i). 2

14.9 NOTES

Several books, such as [105, 203, 291, 275, 269, 218], offer adetailed treatment ofthe algebraic–geometry codes.

Algebraic curves defined overFq with manyFq-rational points, in particularthe Hasse-Weil Theorem, have important applications to a wider variety of com-binatorial problems, especially in finite geometry, see thesurvey paper of Szonyi[285].§13.2For MDS algebraic geometry codes,MCk has been proved so far for some par-

ticular cases, namely for codes arising from elliptic curves, and curves of genustwo whenq > 83, by Munuera [206], see also [46]. For any fixed genusg there isa q0 such that for allq > q0, MC holds for MDS algebraic geometry codes arisingfrom hyperelliptic curves defined overFq. This was shown by de Boer [58] usingMunuera’s approach.

The formal definition of near-MDS codes is due to Dodunekov and Landjev [62].The exact values of the maximum possible length of a near-MDScode forq =2, 3, 4, 5 were computed by Dodunekov and Landjev in [63]. Marcugini, Milaniand Pambianco went on to investigate near-MDS codes fork = 3, 4 and obtained acomplete classification of those of maximal length for everyq ≤ 11, see [196]. Formore results on near-MDS codes, see Abatangelo and Larato [2], Giulietti [102]and Landjev [176].

Fisher, Hirschfeld and Thas [79], and independently Boros and Szonyi [35], Cos-sidente [52], Kestenband [162] and Ebert [66], showed that every point orbit ofS is

604

a complete(q2− q+1)-arcK in PG(2, q2), and every line orbit ofS is a completedual(q2 − q + 1)-arcK′ in PG(2, q2). In [79] it is also shown for evenq that theplane curveΓq+1 associated withK′, via Segre’s generalisation of the Menelaustheorem, is the Hermitian curve; see Theorem 14.46.

For more results on cyclic arcs, see [277] and [197].For a proof that aq-arc is contained in the rational points of a conic, see [247].Direct proofs that aq-arc in PG(2, q) is always incomplete depending on the

Lemma of Tangents are found in [246] for oddq and in [286] for evenq.

Chapter Fifteen

Appendix on Field Theory and Group Theory

15.1 FIELD THEORY

In this section, fundamental concepts and results from fieldtheory are listed. Apartfrom two generalisations of the Theorem of the Primitive Element, all the materialwith proofs can be found in standard textbooks.

(I) The characteristicof a fieldF is the smallest positive integern, if it exists,such that

1 + 1 + . . .+ 1︸︷︷︸n times

= 0.

In this casen is prime and is denoted byp. If there is no suchn, thenF is ofcharacteristic0; write p = 0.

(II) Every field has aprime field, that is, a subfield containing no subfield otherthan itself. Ifp > 0, the prime field is isomorphic toZp = Z/pZ = Fp, thefield of integers modulop, while, if p = 0, the prime field is isomorphic toQ, the field of rational numbers.

(III) Let L be a field that containsF as a subfield. ThenL is anextension ofF ,and the symbolL/F is used to express this. IfL is regarded as a vectorspace overF , its dimension is thedegree ofL/K and denoted by[L : K]. If[L : K] = n < ∞, thenL/K is afinite extension. If both L/F andM/Lare finite extensions thenM/F is also finite, and its degree is

[M : F ] = [M : L][L : F ].

(IV) An elementα ∈ L is eitheralgebraicor transcendentalover a subfieldF ofL, according as there is or is not a non-constant polynomialf(X) ∈ F [X]satisfied byα; that is,f(α) = 0.

Every (algebraic) elementα satisfying the polynomialXn− 1 is annth rootof unity. If, in addition, α does not satisfy any polynomialXk − 1 with1 ≤ k < n, thenα is aprimitiventh root of unity.

(V) The fieldL/F is analgebraic extension, andL is analgebraic field overF ,when all elements inL are algebraic overF ; otherwise,L is atranscendentalextension. In the latter case,L has at least one transcendental element overF .

605

606

(VI) If L1/F andL2/F are any two extensions, then a mappingσ : L1 → L2 isafield homomorphismif

σ(x+ y) = σ(x) + σ(y), σ(xy) = σ(x)σ(y),

for all x, y ∈ L1.

(VII) A field homomorphismσ : L1 → L2 is anembeddingof L1 intoL2 overFprovided thatσ(x) = x for all x ∈ F .

Such a homomorphism must be injective, giving an isomorphism ofL1 ontothe subfieldσ(L1) of L2. A surjective, and hence bijective, embedding ofL1

intoL2 is anF -isomorphism.

(VIII) A field M is algebraically closedif every polynomialf(X) ∈ M [X] ofpositive degree has a root inM . For every fieldF , there exists an algebraicfield extensionF /F such thatF is algebraically closed. Then,F is uniquelydetermined up to anF -isomorphism, and it isthe algebraic closure ofF .If L/F is an algebraic extension, then there exists an embeddingL → FoverF . There are several different embeddings but their number isat most[L : K].

(IX) If α1, . . . , αr ∈ L, then the smallest subfieldM of L containingF and allelementsα1, . . . , αr is denoted byF (α1, . . . , αr), andα1, . . . , αr are theadjoined elementsof F (α1, . . . , αr). The field extensionF (α1, . . . , αr)/Fis algebraic if and only if everyαi is algebraic overF .

(X) If α is an algebraic element ofL/F , there is a unique, monic polynomialf(X) overF of minimal degree satisfied byα. This minimal degree is thedegree ofα overF . The polynomialf(X) is uniquely determined, and it istheminimal polynomial ofα overF . Other polynomialsg(X) ∈ F [X] canbe satisfied byα, but this only occurs whenf(X) | g(X).

(XI) An elementα ∈ L is algebraic overF if and only if [F (α) : F ] = n < ∞.Let f(X) be the minimal polynomial ofα overF , and putr = deg f(X).Then[F (α) : F ] = r, and a basis ofF (α)/F is 1, α, . . . , αr−1.

(XII) Conversely, letf(X) ∈ F [X] be a non-constant polynomial; then, thereexists an algebraic extension fieldL = K(α) such thatf(α) = 0. If f(X)is irreducible overF , this extension field is uniquely determined up to anF -isomorphism. In other words, ifL′ = K(α′) is another algebraic extensionfield with f(α′) = 0, then there exists anF -isomorphismσ : L → L′ withσ(α) = α′.

(XIII) More generally, if f1(X), . . . , fr(X) ∈ F [X] are monic polynomials ofdegreed1, . . . , dr, then there is an algebraic extensionM , unique up to anF -isomorphism, containingF such that allfi(X) split into linear factorsoverM ; that is,

fi(X) =∏di

j=1 (X − aij), aij ∈M.

Appendix on Field Theory and Group Theory 607

The fieldF (a11, . . . , ardr) is uniquely determined up to anF -isomorphism

and is thesplitting field off1, . . . , fr overF . Over the prime field ofF , thedecomposition,

Xn − 1 =∏

d|nFd(X),

holds for every natural numbern not divisible byp for p > 0, where

Fn(X) =∏

d|n(Xd − 1)µ(n/d)

is a polynomial of degreeϕ(n) whose zeros in a splitting field ofXn− 1 areexactly the primitiven-th roots of unity. Hereµ(i) is the Mobius function:

µ(i) =

1 if i = 1,0 if n is divisible by the square of a prime,

(−1)j if n is the product ofj distinct primes.(15.1)

andϕ(n) is the number of all primitiventh roots of unity. The polynomialFn(X) is then-th cyclotomic polynomial.

(XIV) Let f(X) ∈ F [X] be a monic polynomial of degreed ≤ 1. There is analgebraic extensionL of K in whichf(X) splits into linear factors:

f(X) =∏d

i=1 (X − αi).

The polynomialf(X) is separableif the roots off(X) are distinct. If, how-ever,αi = αj for somei 6= j, thenf(X) is inseparable.

If F has zero characteristic, then all irreducible polynomialsover F areseparable. In the case of positive characteristic, an irreducible polynomial,f(X) =

∑aiX

i ∈ F [X], is separable if and only ifai 6≡ 0 (mod p) for atleast one indexi.

(XV) The derivativeof a polynomialf(X) =∑aiX

i ∈ F [X] is the polynomialf ′(X) =

∑iaiX

i−1.

A separability criterion is thatf ′(X) is not the zero polynomial.

(XVI) Given an algebraic extensionL/K, an elementα ∈ L is separableoverF ifits minimal polynomialf(X) ∈ F [X] is a separable polynomial.L/K is aseparable extensionif all α ∈ L are separable overF . In zero characteristic,every algebraic extension is separable.

(XVII) If L/K is a finite extension of degree[L : K] = n, thenL/K is separable ifand only if there aren distinct embeddingsσ1, . . . , σn of L into F overF . Inthis case, an elementα ∈ L is inF if and only ifσi(α) = α for i = 1, . . . , n.

For a towerM ⊃ L ⊃ F of algebraic extensions, the fieldM/F is separableif and only if bothM/L andL/F are separable.

(XVIII) In positive characteristic, inseparable extensions exist. An elementα ∈ Lis purely inseparableoverF if some powerαpr

of α with r ≥ 0 lies inF .This occurs if and only if the minimal polynomial ofα overF is of typeXpe − c with c ∈ F ande ≤ r. The extensionL/F is purely inseparableifall elementsα ∈ L are purely inseparable.

608

(XIX) Given any algebraic extensionL/K, there exists a unique intermediate fieldS, that is,F ⊂ S ⊂ M , such thatS/K is separable butL/S is purelyinseparable. This depends on the fact that the set of all separable elementsform a subfield.

(XX) If L is a field containingF as a subfield, the elementsα1, . . . , αr ∈ L areal-gebraically dependentoverF if they satisfy a non-trivial polynomial relationoverF . Otherwise,α1, . . . , αr arealgebraically independentoverF .

(XXI) An infinite subsetA is algebraically independentoverF if every finite subsetin A is algebraically independent; otherwise,A is algebraically dependent.If α1, α2, . . . are algebraically independent overF andL/F (α1, α2, . . .) isan algebraic extension, thenα1, α2, . . . is atranscendency basis ofL overF .

(XXII) If α1, . . . , αs is a finite transcendency basis ofL overF , then anys + 1or more elements ofL are algebraically dependent overF . Any two finitetranscendency bases ofL overF have the same number of elements, and thisnumber is thetranscendency degreeof L overF .

Note that the function fieldK(F) of an irreducible plane curveF is a finiteextension of an algebraically closed fieldK of transcendency degree1 overK. So, any two elements inK(F) are algebraically dependent overK.

(XXIII) A field F is perfectif all algebraic extensionsL/F are separable. In zerocharacteristic, every field is perfect. In positive characteristicp, a fieldF isperfect if and only if every element ofF is ap-th power of some element inF .

Important examples of perfect fields are the algebraically closed fields andfinite fields.

(XXIV) An algebraic field extensionL/F is simpleif L = F (α) for someα ∈ L. Aprimitive elementof L overF is any elementβ ∈ L such thatL = F (β).

It should be noted that the elements of a simple extension fieldF (α) may bewritten in the formg(α), whereg(X) is a polynomial overF of degree atmostn− 1, with n = [F (α) : F ].

Here,F (α) is an algebra of rankn with basis1, α, . . . , αn−1. Using theminimal polynomial ofα, sayf(X) = Xn+b1X

n−1+. . .+bn with bi ∈ F ,the product of two arbitrary elements of the algebra may be calculated.

(XXV) In positive characteristic, a finite algebraic extensionL/F may not be simple.A sufficient condition for a finite extensionL/F to be simple is that theadjoined elementsα1, . . . , αr be, with at most one exception, separable overF .

(XXVI) A complex number isalgebraicif it is algebraic overQ; that is, it satisfies anon-zero polynomial equation with coefficients inQ. The setA of algebraicnumbers is a subfield ofC. A subfield ofA which is an algebraic extension


of Q is an algebraic number field. An algebraic number is analgebraicintegerif it satisfies a monic polynomial equation with coefficientsin Z. ThesetB of algebraic integers is a subring ofA. An algebraic integer is a rationalnumber if and only if it is a rational integer; that is,B ∩Q = Z.

The following result is an essential ingredient of other chapters.

THEOREM 15.1 (Theorem of the Primitive Element: Strong Form)Let F be afield containing infinitely many elements, and letF (α, β) in an algebraic extensionof F . If α, β are separable overF, then there existsλ ∈ F such thatα + λβ is aprimitive element ofF (α, β)/F .

Proof. Let z be an indeterminate overF (α, β). Every irreducible polynomialG(X) ∈ F [X] is also irreducible when regarded as a polynomial inF (z)[X].This can be shown indirectly.

If G(X) = U(z,X)V (z,X)/d(z) with U, V ∈ F (z,X) andd(z) ∈ F (z),thenG(X) = U(λ,X)(V λ,X)/d(λ), for anyλ ∈ F such thatd(λ) 6= 0, is afactorisation inF [X]. Such aλ exists asF contains infinitely many elements.Therefore,α, β are separable overF (z) as well, because they satisfy the sameirreducible equations overF (z) as overF . Thenw = α + zβ is separable overF (z).

Consider the simple extensionF (z)(w)/F (z) and denote byg(W ) ∈ F (z)the minimal polynomial ofw overF (z). Theng(W ) comes from a polynomialG(X,W ) ∈ F (X,W ) in the usual way; that is,g(W ) = G(z,W ). Sincew isseparable,∂G/∂W 6= 0. From the fact that

degW

∂G

∂W< degW G,

it follows that(∂G

∂W

)

W=w

6= 0.

Now,G(z, α+ zβ) = 0, whence(∂G(z,W )

∂z

)

W=α+zβ

+

(∂G(z,W )

∂W

)

W=α+zβ

· β = 0.

¿From this,

β =H(z, α+ zβ)

d1(z),

with H(X,W ) ∈ F [X,W ] andd1(z) ∈ F [z]. So, forλ ∈ F\0,

β =H(λ, α+ λβ)

d1(λ).

Thusβ ∈ F (α+ λβ). Sinceα ∈ F (α+ λβ), the assertion follows. 2

A major result on primitive elements is the following.

610

THEOREM 15.2 (Theorem of the Primitive Element)Every finite separable exten-sionL/K is simple.

This theorem holds without the hypothesis on separability provided thatL has tran-scendency degree1. A proof is given, for which two technical lemmas are required.

L EMMA 15.3 For p > 0, let x be a transcendental element overK, let v be thep-th root ofx in the algebraic closureK(x) ofK(x), and putv = x1/p. Then

(i)

[K(x1/p) : K(x)] = p;

(ii) more generally, if e is a positive integer andx1/pe

is a root ofx of orderpe,

[K(x1/pe

) : K(x)] = pe.

Proof. The first assertion is a consequence of the irreducibility ofthe polynomialZp − x ∈ K(x)[Z] overK. Applying this tox1/p gives the following:

[K(x1/p2

) : K(x)] = [K(x1/p2

) : K(x1/p] · [K(x1/p) : K(x)] = p · p = p2.

Repeating the argument shows thatK(x1/pe

: K(x)] = pe. 2

L EMMA 15.4 Make the following assumptions:

(a) p > 0;

(b) the elementu is in an extension field ofK(x) which is algebraic overK(x),wherex is transcendental;

(c) the elementx is not separable inK(x, u), whereu is a root of an irreduciblepolynomialF ∈ K(x)[Upe

] for somee ≥ 1;

(d) the polynomialF does not belong toK(x)[Upe+1

].

Then

(i) K(x, u) contains a rootx1/pe

of orderpe of x;

(ii) in particular, u is a separable element overK(x1/pe

).

Proof. From the hypothesis, putG(x,Upe

) = F (x,U). If degV G(x, V ) = d,thendegU F (x,U) = dpe and[K(x, u) : K(x)] = dpe. Without loss of generality,suppose thatF ∈ K[x,U ] and writeG =

∑cijx

i(Upe

)j . In K(x), the elementusatisfies the polynomial

G(1/pe) =∑c1/pe

ij x1/pe

U j

of degreed which is the minimal polynomial ofu overK(x1/pe

). In fact, if usatisfied a polynomialH(x1/pe

, U) of degreed′ < d, thenuwould satisfyH(1/pe),which is a polynomial overK(x) of degreed′ep < dep, a contradiction. Hence

[K(x1/pe

, u) : K(x1/pe

)] = d


and, from Lemma 15.3, it follows that

[K(x1/pe

, u) : K(x)] = dpe.

ThusK(x, u) = K(x1/pe

, u), whencex1/pe ∈ K(x, u). Finally, (ii) is a conse-quence of the definition ofe. 2

THEOREM 15.5 (Theorem of Primitive Element for Function Fields of Transcen-dency Degree1) Let Σ be a field of transcendency degree1 over an algebraicallyclosed subfieldK. If x ∈ Σ is transcendental overK, then there isy ∈ Σ such thatΣ = K(x, y).

Proof. By Theorem 15.2, takex to be not separating. Letu be a separating functionof Σ. Then Theorem 15.2 ensures the existence ofv ∈ Σ such thatΣ = K(u, v).With the notation of Lemma 15.3, from (ii) in that lemma,upe

is separable overK(x) for somee ≥ 0. Similarly, there is an integern ≥ 0 such thatvpn

is separableoverK(x). Let k = minm,n. By the sufficient condition for an extension to besimple,K(x, upe

, vpn

, x1/pk

) = K(x, y) for somey ∈ Σ.To complete the proof, it suffices to show thatu, v ∈ K(x, upe

, vpn

, x1/pe

). Inproving this foru, the hypothesise > 0 may be assumed. Thenx1/pe

is a root ofthe polynomialZpe − x overK(x), and hence overK(x, upe

). This polynomial isirreducible overK(x, upe

) by the following argument.Suppose

Zpe − x = (Zm1 + . . .)(Zm2 + . . .), 0 < m1,m2 < pm.

In the algebraic closureK(x, upe),

Zpe − x = (Z − x1/pe

)pe

,

which shows thatZm1 + . . . andZm2 + . . . must each be a power ofZ − x1/pe

.Let ps be the largest power ofp dividing m1. Sincem1 < pe, sos < e. Thenps = tm1 + rps with t, r integers. Therefore,

x1/pe−s

= (xm1/pe

)sxr ∈ K(x, upe

).

The minimal polynomial ofx1/pe−s

overK(x) is Zpe−s − x contradicting theseparability ofK(x, upe

) overK(x).It follows that[K(x, upe

, x1/pe

) : K(x, upe

] = pe. Putting

d = [K(x, upe

) : K(x)],

this gives[K(x, upe

, x1/pe

) : K(x)] = dpe. SinceK(x, upe

, x1/pe

) is a sub-field of K(x1/pe

, u) and the latter coincides withK(x, u) by Lemma 15.4, from[K(x, u) : K(x)] = dpe the assertionK(x, u) = K(upe

, x1/pe

) follows. Henceu ∈ K(x, upe

, x1/pe

). Similarly,v ∈ K(x, vpe

, x1/pe

). 2

15.2 GALOIS THEORY

(I) An extensionL/F is normal if, wheneverf(X) ∈ F [X] is an irreduciblepolynomial which has one root inL, then all roots off(X) lie in L. A finiteextensionL/F is normal if and only ifL is the splitting field of a polynomialin F [X]. In particular, a finite normal extension is separable if andonly if Lis the splitting field of a separable polynomial overF .

612

(II) A finite, normal, separable extensionL/F is a Galois extension. Galoistheory relates field extensions to groups via the concept of an automorphism.

(III) An isomorphism of a fieldL onto itself is anautomorphism. The automor-phisms ofL form a group, theautomorphism group ofL, written Aut(L).If H ⊂ Aut(L), then the set of all fixed elements ofH, written LH , is asubfield ofL, thefixed fieldof H.

(IV) If L/F is finite, then Aut(L/F ) is also finite as its order is less than or equalto [L : F ]. If L/F is a Galois extension then equality holds, and Aut(L/F ) istheGalois groupofL/F . Usually, the Galois group is denoted by Gal(L/F ).

(V) The most important result on Galois extensions is theGalois correspondencewhich is a connection between subgroupsH of Gal(L/F ) and subfieldsMof L intermediate betweenF andL. More precisely, the maps

M 7→ Gal(L/F ),H 7→ HL,

are mutually inverse bijections between the subfields ofL containingF andthe subgroups of Gal(L/F ).

(VI) (a) For every subgroupH of Gal(L/F ),

[L : LH ] = |H|, [LH : F ] = |Gal(L/F )/H|, H = Gal(L/LH).

(b) For every two subgroupsH,G of Gal(L/F ),

H ⊂ G⇐⇒ LH ⊃ LG.

(c) For every intermediate fieldM of L/F , there is a subgroupH ofGal(L/F ) such thatH = Gal(L/M).

(d) For every two intermediate fieldsM1,M2 of L/F ,

Gal(L/〈M1,M2〉) = Gal(L/M1) ∩Gal(L/M2),

where〈M1,M2〉 is the subfield generated byM1 andM2, and

Gal(L/(M1 ∩M2) = 〈Gal(L/M1),Gal(L/M2)〉,where〈G1, G2〉 is the group generated byG1 andG2.

(e) A subgroupH of Gal(L/F ) is a normal subgroup if and only ifLH/Fis a Galois extension.

(f) If the equivalent conditions of(e) hold, then Gal(M/F ) is isomorphicto the factor group Gal(L/F )/Gal(L/M).

(VII) Let L/F be a finite separable extension, and letL be the algebraic closure ofL. Then there exists a unique fieldM with following properties:

(a) L ⊂M ⊂ L;

(b) M/F is a Galois extension;

(c) if L ⊂ N ⊂ L andN/F is Galois extension, thenM ⊂ N .


The fieldM is theGalois closure ofL/F .

(VIII) A Galois extensionL/F is cyclic if Gal(L/F ) is cyclic. Here, two relevantspecial cases are considered, namelyn = [L : F ] with n prime top, and[L : F ] = p.

(IX) If gcd(n, p) = 1, thenL/F is a Kummer extensionif F contains alln-throots of unity; that is, the polynomialXn − 1 splits into linear factors inF [X].

Every Kummer extension can be described as a simple algebraic extension inthe following way. Take an elementc ∈ F which is not anm-th power inFfor any divisorm of n. Then the polynomialXn − c is irreducible inF [X],andL/F is the simple extensionF (v) with vn− c = 0. The automorphismsσ in Gal(L/F ) are given byσ(u) = ξu asξ ranges over then-th roots ofunity in F . Conversely, IfF contains alln-th roots of unity,n being primeto p, andL = F (v) wherevn − c = 0 with c satisfying the above condition,thenL/F is a cyclic extension of degreen.

(X) When [L : F ] = p, take an elementc ∈ F that cannot be written asap − afor anya ∈ F . ThenXp −X − c is an irreducible polynomial inF [X], andthe simple extensionL = F (v) with vn − v − c = 0 is anArtin–Schreierextensionof degreep.

The automorphismsσ in Gal(L/F ) are given byσ(u) = u + η asη rangesover thep elements in the prime field ofF . Conversely, ifL = F (v) is asimple extension wherevp − v − c = 0 andc satisfies the above condition,thenL/K is cyclic of degreep.

15.3 NORMS AND TRACES

(I) Let L/F be a finite extension of degreen = [L : F ]. SinceL may beviewed as a vector space overF , every elementu ∈ L induces a lineartransformationνu of L given byνu(x) = ux for x ∈ L.

(II) The normof u with respect to the extensionL/F is

NL/F (u) = det(νu)

in F .

(III) The trace ofu is TL/F (νu) = Trace(νu) with respect toL/F . In otherwords, ifu1, . . . , un is a basis ofL/F and

uui =∑n

j=1aijui

with aij ∈ F , then

(a) NL/F (u) = det(aij),

(b) TL/F (u) =∑n

j=1aii.

614

(IV) For x, y ∈ L andα ∈ F , with n = [L : F ],

(i) NL/F (α) = αn;

(ii) TL/F (α) = n · α;

(iii) NL/F (x · y) = NL/F (x) ·NL/F (y);

(iv) NL/F (x) = 0 if and only if x = 0;

(v) TL/F (x+ y) = TL/F (x) + TL/F (y);

(vi) TL/F (α · x) = α · TL/F (x);

(vii) for any finite extensionM/L andu ∈M ,

(a) TM/F (u) = TL/F (TM/L(u))

(b) NM/F (u) = NL/F (NM/L(u)).

(V) A finite extensionL/F is inseparable if and only ifTL/F (x) = 0 for everyx ∈ L.

(VI) L/F is separable if and only ifTL/F : L→ F is surjective.

(VII) If f(X) = Xr + a1Xr−1 + . . .+ ar ∈ F [X] is the minimal polynomial of

a ∈ L overF , and[L : F ] = n = rs with s = [L : F (a)], then

(a) NL/F (a) = (−1)nasr;

(b) TL/F (a) = −sa1.

(VIII) Given a separable field extensionL/F of degreen, let σ1, . . . , σn be thedistinct embeddings ofL into the algebraic closureF of F . Then, for everya ∈ L,

(a) NL/F (x) =∏n

i=1σi(x);

(b) TL/F (x) =∑n

i=1σi(x).

In the case of a Galois extension with Galois groupG = Gal(L/F ), thesebecome the following:

(a) NL/F (x) =∏

σ∈Gσ(x);

(b) TL/F (x) =∑n

σ∈Gσ(x).

15.4 FINITE FIELDS

Finite fields, also called Galois fields, are all of positive characteristic and havebeen classified. Their structure is described below.

Forq = pn, the splitting fieldFq of the irreducible polynomialXq −X over thefield Z/pZ has orderq; that is, it consists ofq elements, and is of characteristicp.

(I) Up to isomorphism, the fieldsFq are the only finite fields.


(II) The multiplicative group ofFq is a cyclic group of orderq − 1.

(III) The additive group ofFq is an elementary abelian group of orderq.

(IV) Every subfield ofFq is isomorphic to someFpi for a divisori of n. Con-versely, for every divisori of n, the fieldFq has a unique subfield of orderpi.

(V) The automorphism group ofFq is the cyclic group of ordern, generated byσ : x 7→ xp.

(VI) Fq ⊂ Fqm for m ≥ 1.

(VII) The extensionFqm/Fq is a Galois extension of degreem.

(VIII) The Galois group Gal(Fqm/Fq) is cyclic and is generated by the FrobeniusautomorphismΦ of Fqm given byΦ(x) = xq.

(IX) The norm

NFqm/Fq(x) = x1+q+q2+...+qm−1

.

(X) The trace

TFqm/Fq(x) = x+ xq + xq2

+ . . .+ xqm−1

.

(XI) When x ∈ Fqm , thenTFqm (x) = 0 if and only if x = yq − y for somey ∈ Fqm .

L EMMA 15.6 (Lucas)

(i) Letp be a prime and let the positive integersn,m have thep-adic expansionsn =

∑nip

i, m =∑mip

i.

with 0 ≤ ni ≤ p− 1, 0 ≤ mi ≤ p− 1. Then(n

m

)6≡ 0 (mod p)

if and onlyni ≥ mi for all i.

(ii) With q = ph, (nq

q

)≡ n (mod p).

Proof. (i) For an indeterminateX, consider the following expansion overFp:

(1 +X)n = (1 +X)n0+n1p+...+nrpr

= (1 +X)n0(1 +Xp)n1 . . . (1 +Xpr

)nr .

Now, pick out the coefficient ofXm from both sides:(n

m

)=

(n0

m0

)(n1

m1

)· · ·(nr

mr

).

This gives the result.(ii) With n =

∑nip

i andq = ph thep-adic expansions, compare the coefficientof Xq on both sides of

(1 +X)nq = (1 +X)n0ph · · · (1 +X)nrpr+h

.

Also, note thatn ≡ nq (mod p) to obtain the result. 2

616

15.5 GROUP THEORY

The purpose here is to state definitions and results from group theory which areapplied in Chapter 12.

(I) The order of a groupG, written |G| or ordG, is the cardinality of the setG.If this cardinality if finite, then|G| is the number of elements inG.

(II) Two elementsx, y in a groupG areconjugateif y = g−1xg for someg inG. Similarly, two subgroupsG, H areconjugateif every element inH isconjugate to some element inG.

(III) A cyclic groupG is a group generated by a single elementg; that is,Gconsists of all the powers ofg, and is denotedG = 〈g〉. Theorder ofg is theorder of〈g〉, written ordg. Hence

G= 1, g, g2, . . . , gn−1 | gn = 1 if G is finite,

G= . . . , g−2, g−1, 1, g, g2, . . . if G is infinite.

(IV) An groupG is abelianor commutativeif xy = yx for all elementsx, y of G.

(V) The centreZ = Z(G) of a groupG is Z = x ∈ G | xg = gx all g ∈ G.So a group is abelian if it coincides with its centre. Cyclic groups are abelian.

(VI) The direct productof the groupsG andH is

G×H = (g, h) | g ∈ G, h ∈ Hsuch that(g1, h1)(g2, h2) = (g1g2, h1h2). The direct product of abeliangroups is abelian.

(VII) By Lagrange’s theorem, ifH is a subgroup of a finite groupG, then |H|divides|G|, and|G|/|H| is the number of all cosets ofH in G; in particular,if N is a normal subgroup of the finite groupG, then|G|/|N | = |G/N |.

(VIII) The converse of Lagrange’s theorem holds for subgroups of prime powerorder by Sylow’s theorem. Letd be a prime dividing|G|. If dn divides|G|butdn+1 does not, thenG has a subgroup of orderdn. Such a subgroup is ad-Sylow subgroupof G. Any two d-Sylow subgroups ofG are conjugate inG and their number is congruent to1 modulod.

(IX) Given a subgroup ofH of G, the left cosetof an elementg ∈ G is the setgH = gh | h ∈ H and theright cosetof g is the setHg = hg | h ∈ H.If G is finite, then the number of right or left cosets is theindexof H.

A subgroupN of G is normal, writtenN ⊳G, if gN = Ng for everyg ∈ G.For instance, subgroups of index2 are normal. A subgroup is normal if andonly it is conjugate only to itself. Thetrivial normal subgroups ofG areGand1 = I.

If G has only trivial normal subgroups, thenG is simple.


(X) If N is a normal subgroup ofG, then the set of all left or right cosets ofN inG form thequotientor factor groupG/N under the operation,

(g1N)(g2N) = (g1g2)N.

(XI) Given two groupsA andB, an extensionof A by B is a groupG whichcontains a normal subgroupN such thatN is isomorphic toA while G/Nis isomorphic toB. If G = NH andN ∩ H = 1, then the extensionis split, or, equivalently,G is thesemi-direct productof N by H, writtenG = N ⋊ H. In this context,H is a complement. A sufficient conditionfor an extension to be split is that|N | and[G : N ] have no common divisor.If a normal subgroupS of G contains a Sylowd-subgroupSd of G, thenG=NG(Sd)S.

(XII) A minimal normal subgroupof G is a normal subgroupN with N 6= Iwhich does not contain properly any other non-trivial normal subgroups ofG. Every minimal normal subgroupN of a finite groupG is a direct productof N = T1 × . . . × Tk, where theTi are simple normal subgroups ofN ,pairwise conjugate underG. Thesocleof G is the subgroup generated by allminimal normal non-trivial subgroups ofG, written soc(G).

(XIII) If x, y ∈ G, the elementx−1y−1xy ∈ G is thecommutatorof x andy. ThesubgroupG′ generated by all commutators ofG is thederivedgroup ofG.The factor groupG/G′ is abelian and, ifN is a normal subgroup ofG suchthatG/N is abelian, thenN ⊃ G′.

(XIV) A group is solvableif the sequenceG ⊃ G′ ⊃ . . . ⊃ G(i) . . ., in which eachgroupG(i) is the derived group of the preceding, terminates withI.

(XV) A group isnilpotentif it has a finite sequenceG = G0 ⊃ G1 ⊃ . . . ⊃ Gi . . .,in which each groupGi is normal in subgroup of the preceding, such thatGi−1/Gi ⊂ Z(G/Gi) for everyi ≥ 1. A finite group is nilpotent if and onlyif it is the direct product of its Sylow subgroups.

(XVI) For g ∈ G, thecentraliserCG(g) = x ∈ G | xg = gx. The intersectionof all centralisers inG is the centreZ = Z(G) and is a normal subgroupG.If |G| = pn with p prime andn > 1, thenZ(G) 6= I.

(XVII) For a subgroupH of G, thenormaliserNG(H) = g ∈ G | gH = Hg.

(XVIII) The Frattini subgroup ofG is the intersection of all maximal subgroups ofG.

(XIX) The group of all permutations of a finite setΩ = ω1, . . . , ωn is thesym-metric groupSn of Ω. Its order|Sn| is equal ton!.

The normal subgroup ofSn of index2 consisting of all even permutations ofΩ is thealternating groupAn of Ω.

A permutationgroupG is a group whose elements are permutations of afinite set, that is, a subgroup of a symmetric group.

618

(XX) If H is a finite group andG is a permutation group with a homomorphismρ : H → G, thenH/ ker ρ ∼= G andG is the permutation groupinduced byH on Ω, orG is apermutation representation ofH. If ker ρ is trivial, thenthis representation isfaithful.

(XXI) Let G be a permutation group on a finite setΩ whose elements are calledpoints. Thedegreeof G is |Ω|. Theorbit of ω ∈ Ω underG, writtenG(ω),consists of the images ofω underG; that is,

G(ω) = ω′ | ω′ = g(ω), g ∈ G.Theorbit structureof G is the list of lengths of its orbits, each counted withmultiplicity.

(XXII) The stabiliserof a pointω ∈ Ω, writtenGω, is the subgroup consisting of allelements inG that fixω; that is,

Gω = g | g(ω) = ω, g ∈ G.More generally, if∆ ⊂ Ω, then the stabiliser of∆ underG, writtenG∆, isthe subgroup consisting of all elements inG which fix each point in∆; thatis,

G∆ = g | g(ω) = ω, ω ∈ ∆.Similarly, theglobal stabiliserof ∆, writtenG∆, consists of all elementsin G which preserve∆; that is,G∆ = g | g(ω) ∈ ∆, ω ∈ ∆. If ∆ hassizen, thenG∆ is ann-point stabiliser ofG.

(XXIII) The length of an orbit and the stabiliser of a point inthe orbit are connected:

|G| = |G(ω)| |Gω|.In particular, ifGω = I, then |G(ω)| = |G| and the orbit is along orbit;otherwiseG(ω) is ashort orbit.

If every orbit is long, that is, no non-trivial element inG fixes a point inΩ,thenG is asemi-regularpermutation group onΩ. Theorbit structureof Gis the list of the lengths of its orbits.

(XXIV) The groupG is transitiveif it has only one orbit, and it isregular or sharplytransitiveif it is also semi-regular. So,G is transitive onΩ if and only if, forany ordered pair(ω, δ) with ω, δ ∈ Ω, there existsg ∈ G such thatδ = g(ω),andG is regular if there is exactly one such permutationg ∈ G.

(XXV) More generally,G is n-fold transitiveon Ω if, for every n ordered pairs(ωi, δi), there existsg ∈ G such thatδi = g(ωi) for i = 1, . . . , n, andsharplyn-fold transitiveif there is exactly one such permutation inG.

Note that, ifG is (n− 1)-fold transitive onΩ and then-point stabilisers aretransitive on the remaining points, thenG isn-fold transitive, and vice versa.Also, if G is sharplyn-fold transitive, then its order is equal to

|Ω|(|Ω| − 1) · · · (|Ω| − n+ 1).


(XXVI) A transitive permutation groupG onΩ is imprimitive if Ω has aG-invariantpartition; otherwiseG is primitive. Forn ≥ 2, everyn-fold transitive groupis primitive.

A transitive permutation groupG is primitive if and only ifGω is a maximalsubgroup ofG for everyω ∈ Ω. If G is a primitive group andN is a minimalnormal subgroup, then one of the following holds:

(i) there is a powerdn of a primed such thatN is regular and elementaryabelian of orderdn, and soc(G) = N = CG(N);

(ii) N is a regular non-abelian group,CG(N) is a minimal normal sub-group ofG andsoc(G) = N × CG(N);

(iii) N is not abelian,CG(N) = I and soc(G) = N .

Also, soc(G) is a direct product of pairwise isomorphic simple groups. IfH is the normaliser of soc(G) in the symmetric groupSΩ, then soc(G) is aminimal normal subgroup ofH. If soc(G) is not regular, then it is the onlyminimal normal subgroup ofH.

(XXVII) A Frobeniusgroup is a transitive group which is not regular, but in whichonly the identity has more than one fixed point. Those elements in a Frobe-nius groupF which have no fixed point together with the identity constitutea regular normal nilpotent subgroupN , thekernelof F . Also,G = N ⋊Gω

for anyω ∈ Ω. If Gω has even order thenN is an abelian group.

(XXVIII) Let G be a permutation group ofΩ andd a prime number. Assume that everypoint ofΩ is fixed by a permutation inG of orderd which has no other fixedpoint; thenG is transitive onΩ. If, in addition,d = 2, every point inΩ isfixed by exactly one involutory element inG, and no involutory element inG fixes more than one point ofΩ then the subgroup ofG consisting of allproducts of even number of involutory elements inG has odd order and actstransitively onΩ.

(XXIX) For n ≥ 2, all n-transitive permutation groups are known, their classificationbeing a corollary to that of all finite simple groups. In Chapter 12, doubly-transitive permutation groups whose2-point stabilisers are cyclic play animportant role. Below, the relevant classification theorems are stated af-ter describing the permutation groups involved given in their usual doubly-transitive permutation representations.

EXAMPLE 15.7 Projective linear groups

1. LetΩ = Fq ∪ ∞. The linear fractional permutations

ϕ(a,b,c,d) : x 7→ ax+ b

cx+ d, ad− bc 6= 0,

with a, b, c, d ∈ Fq form a sharply triply-transitive group, theprojective lin-ear groupPGL(2, q).

620

2. The2-point stabilisers are isomorphic to the multiplicative groupF∗q of Fq.

3. If q is odd, then PGL(2, q) contains a normal2-transitive subgroup consistingof all ϕ(a,b,c,d) with ad−bc square inFq, the special linear groupPSL(2, q);it has order12 (q3 − q).

4. The2-point stabiliser of PSL(2, q) is isomorphic to the subgroup of all squareelements inF∗

q .

5. Apart from the casesq = 2, 3, the group PSU(2, q) is simple.

THEOREM 15.8 For q = pm, the following is a complete list of subgroups of thegroupPGL(2, q) up to conjugacy:

(i) the cyclic group of ordern with n | (pm ± 1);

(ii) the elementary abelianp-group of orderpf with f ≤ m;

(iii) the dihedral groupDn of order2n with n | (q ± 1);

(iv) the alternating groupA4 for p > 2, or p = 2 andm even;

(v) the symmetric groupS4 for p > 2;

(vi) the alternating groupA5 for 5 | (q2 − 1);

(vii) the semi-direct product of an elementary abelianp-group of orderph by acyclic group of ordern with h ≤ m andn | (q − 1);

(viii) PSL(2, pf ) for f | m;

(ix) PGL(2, pf ) for f | m.

EXAMPLE 15.9 Projective unitary groups

1. LetU be the classical unital inPG(2, q2), that is, the set of all self-conjugatepoints of a unitary polarityΠ of PG(2, q2). The projective unitary groupPGU(3, q) comprises the projective transformations inPG(2, q2) commut-ing with Π. Equivalently, PGU(3, q) preservesU and can be viewed as a per-mutation group onU , since the only collineation in PGU(3, q) fixing everypoint inU is the identity.

2. With Ω = U , the group PGU(3, q) is a doubly-transitive permutation grouponΩ and its2-point stabiliser is isomorphic toF∗

q .

3. |PGU(3, q)| = (q3 + 1)q3(q2 − 1).

4. With d = gcd(3, q + 1), the group PGU(3, q) contains a normal subgroupPSU(3, q), the special unitary group, of indexd which is still2-transitive. Ifq > 2, the group PSU(3, q) is non-abelian and simple.


5. For another representation of PGU(3, q), let

M = m ∈ Fq2 | mq +m = 0.Take an elementc ∈ Fq2 such thatcq+c+1 = 0. A homogeneous coordinatesystem inPG(2, q2) can be chosen so that

U = X∞ ∪ U = (1, u, cuq+1 +m) | u ∈ Fq2 , m ∈M.Then,U consists of allFq2-rational points of the Hermitian curve

Hq = v(cXq0X2 + cqX0X

q2 +Xq+1

1 ).

6. A Sylowp-subgroupSp of PGU(3, q) consists of the projective transforma-tions associated to the matrices,

Ta,m =

1 aq(1 + cq−1) aq+1 + c−1m0 1 a0 0 1

,

for a ∈ Fq2 andm ∈M .

7. The setT0,m | m ∈M is an elementary abelianp-group, which isZ(Sp).Also, Sp is a normal subgroup of PGU(3, q)Y∞

and acts transitively on theotherq3 points ofU .

8. The pointO is in U , and the stabiliserT of O, Y∞ under PGU(3, q) consistsof the projective transformations associated to the matrices,

aq+1 0 0

0 a 00 0 1

,

for a ∈ Fq2\0.

9. The group PGU(3, q) is generated bySp andT , together with the projectivetransformationW associated to the matrix,

0 0 c0 1 0c−1 0 0

,

that interchanges the pointsY∞ andO.

10. The group PGU(3, q) also fixesHq, and hence is aK-automorphism groupof Σ = K(Hq).

THEOREM 15.10 The following is the list of maximal subgroups ofPSU(3, q) upto conjugacy:

(i) the one-point stabiliser of orderq3(q2 − 1)/d;

(ii) the non-tangent line stabiliser of orderq(q2 − 1)(q + 1)/d;

(iii) the self-conjugate triangle stabiliser of order6(q + 1)2/d;

622

(iv) the normaliser of a cyclic Singer group of order3(q2 − q + 1)/d.

Further, forq = pk with p > 2,

(v) PGL(2, q) preserving a conic;

(vi) PSU(3, pm), withm | k andk/m odd;

(vii) the subgroup containingPSU(3, pm) as a normal subgroup of index3 whenm | k, k/m is odd, and3 divides bothk/m andq + 1;

(viii) the Hessian groups of order216 when9 | (q + 1), and of order72 and36when3 | (q + 1);

(ix) PSL(2, 7) when eitherp = 7 or√−7 6∈ Fq;

(x) the alternating groupA6 when eitherp = 3 andk is even, or√

5 ∈ Fq butFq contains no cube root of unity;

(xi) the symmetric groupS6 for p = 5 andk odd;

(xii) the alternating groupA7 for p = 5 andk odd.

For q = 2k,

(xiii) PSU(3, 2m) with k/m an odd prime;

(xiv) the subgroups containingPSU(3, 2m) as a normal subgroup of index3 whenk = 3m with oddm;

(xv) a group of order36 whenk = 1.

EXAMPLE 15.11 Suzuki groups

1. AnovoidO in PG(3, q) is a point set with the same combinatorial propertiesas an elliptic quadric inPG(3, q); namely,O consists ofq2 + 1 points, nothree collinear, such that the lines through any pointP ∈ O meetingO onlyin P are coplanar.

2. Now, assume thatq = 2q20 with q0 = 2s ands ≥ 1. Thenxϕ = x2q0 is anautomorphism ofFq, andϕ2(x) = x2.

Let T be theTits ovoid in PG(3, q), which is the only known ovoid inPG(3, q) other than an elliptic quadric. TheSuzuki groupSz(q), also written2B2(q), is the projective group ofPG(3, q) fixing T . The group Sz(q) canbe viewed as a permutation group onT as the identity is the only projectivetransformation in Sz(q) fixing every point inT .

3. With Ω = T , the group Sz(q) is a2-transitive permutation group onΩ, andits 2-point stabiliser is isomorphic toF∗

q .

4. |Sz(q)| = (q2 + 1)q2(q − 1).


5. The group Sz(q) is simple. IfS2 is a 2-Sylow subgroup of Sz(q), the ele-ments ofS2 of order2 together with the identity form an elementary abeliannormal subgroupN of orderq, and the quotient groupS2/N is isomorphictoN .

6. In a suitable coordinate system ofPG(3, q) with Z∞ = (0, 0, 0, 1),

T = Z∞ ∪ (1, u, v, uv + u2ϕ+2vϕ) | u, v ∈ Fq.A 2-Sylow subgroupS2 consists of the projective transformations associatedto the matrices,

1 0 0 0a 1 0 0c aϕ 1 0

ac+ aϕ+2 + cϕ aϕ+1 + c a 1

,

for a, c ∈ Fq.

7. The subgroupS2 is normal in Sz(q)Z∞and regular on the remainingq2

points ofT . The stabiliser Sz(q)Z∞,O with O = (1, 0, 0, 0) consists of theprojective transformations associated to diagonal matrices ,

diag(d−q0−1, d−q0 , dq0 , dq0+1),

for d ∈ Fq2\0.

8. The group Sz(q) is generated byS2 and SzZ∞,O together with the projectivetransformationW associated to the matrix,

0 0 0 10 0 1 00 1 0 01 0 0 0

,

that interchangesZ∞ andO.

THEOREM 15.12 Up to conjugacy,Sz(q) has the following maximal subgroups:

(i) the one-point stabiliser of orderq2(q − 1);

(ii) the normaliser of a cyclic Singer group of order4(q + 2q0 + 1);

(iii) the normaliser of a cyclic Singer group of order4(q − 2q0 + 1);

(iv) Sz(q′) for everyq′ such thatq = qm withm prime.

Further, the subgroups listed below form a partition ofSz(q) :

(v) all subgroups of orderq2;

(vi) all cyclic subgroups of orderq − 1;

(vii) all cyclic Singer subgroups of orderq + 2q0 + 1;

624

(viii) all cyclic Singer subgroups of orderq − 2q0 + 1.

EXAMPLE 15.13 Ree groupsThe Ree group can be introduced in a similar way using the combinatorial con-

cept of an ovoid, this time in the context of polar geometries.

1. An ovoid in the polar space associated to the non-degenerate quadricQ inthe spacePG(6, q) is a point set of sizeq3 + 1, with no two of the pointsconjugate with respect to the orthogonal polarity arising fromQ.

2. Now, letq = 3q20 andq0 = 3s with s ≥ 1. Thenxϕ = x3q0 is an automor-phism ofFq, andϕ3(x) = x3.

LetK be theKantor ovoidofQ. TheRee groupRee(q), also written2G2(q),is the projective group ofPG(6, q) fixing K.

3. The group Ree(q) can be viewed as a permutation group onK as the identityis the only projective transformation in Ree(q) fixing every point inK. WithΩ = K, the group Ree(q) is a2-transitive permutation group onΩ, and its2-point stabiliser is isomorphic toF∗

q .

4. |Ree(q)| = (q3 + 1)q3(q − 1).

5. The group Ree(q) is simple. LetS3 be a3-Sylow subgroup of Ree(q). Theelements of order3 in Z(S3) together with the identity constitute an elemen-tary abelian normal subgroupN of orderq2, and the factor groupS3/N isan elementary abelian group of orderq.

6. In a suitable coordinate system ofPG(6, q) with Z∞ = (0, 0, 0, 0, 0, 0, 1),the quadricQ = v(X2

3 +X0X6 +X1X5 +X2X4) and

K = Z∞ ∪ (1, u1, u2, u3, v1, v2, v3),with

v1(u1, u2, u3) = u21u2 − u1u3 + uϕ

2 − uϕ+31 ,

v2(u1, u2, u3) = uϕ1 u

ϕ2 − uϕ

3 + u1u22 + u2u3 − u2ϕ+3

1 ,

v3(u1, u2, u3)

= u1uϕ3 − uϕ+1

1 u2 − uϕ+31 u2 + u2

1u22 − uϕ+1

2 − u23 + u2ϕ+4

1 ,

for u1, u2, u3 ∈ Fq.

7. Put

w1(u1, u2, u3) = −uϕ+21 + u1u2 − u3,

w2(u1, u2, u3) = uϕ+11 u2 + uϕ

1 u3 + u22 + u3,

w3(u1, u2, u3)

= uϕ3 + (u1u2)

ϕ − uϕ+21 u2 − u1u

22 + u2u3 − uϕ+1

1 u3 − u2ϕ+31 ,

w4(u1, u2, u3) = uϕ+31 − u2

1u2 − uϕ2 − u1u3.


Then a Sylow3-subgroupS3 of Ree(q) consists of the projective transfor-mations associated to the matrices

1 0 0 0 0 0 0a 1 0 0 0 0 0b aϕ 1 0 0 0 0c b− aϕ+1 −a 1 0 0 0

v1(a, b, c) w1(a, b, c) −a2 −a 1 0 0v2(a, b, c) w2(a, b, c) −ab b −aϕ 1 0v3(a, b, c) w3(a, b, c) w4(a, b, c) c −b+ aϕ+1 −a 1

for a, b, c ∈ Fq. Here,S3 is a normal subgroup in Ree(q)Z∞and regular on

the remainingq3 points ofO.

8. The stabiliser Ree(q)Z∞,O with O = (1, 0, 0, 0, 0, 0, 0) is the cyclic groupconsisting of the projective transformations associated to the diagonal matri-ces,

diag(1, d, dϕ+1, dϕ+2, dϕ+3, d2ϕ+3, d2ϕ+4)

for d ∈ Fq2 .

9. The group Ree(q) is generated byS3 and Ree(q)Z∞,O, together with theprojective transformationW of order2 associated to the matrix,

0 0 0 1 0 0 00 0 −1 0 0 0 00 −1 0 0 0 0 00 0 0 0 0 0 −10 0 0 0 0 −1 00 0 0 0 −1 0 01 0 0 0 0 0 0

,

that interchangesZ∞ andO.

THEOREM 15.14 Up to conjugacy, Ree(q) has the following maximal subgroups:

(i) the one-point stabiliser of orderq3(q − 1);

(ii) the centraliser of an involutionz ∈ Ree(q) isomorphic to〈z〉 ×PSL(2, q) oforder q(q − 1)(q + 1);

(iii) a subgroup of order6(q + 3q0 + 1), the normaliser of a cyclic Singer groupof orderq + 3q0 + 1;

(iv) a subgroup of order6(q − 3q0 + 1), the normaliser of a cyclic Singer orderof order6(q − 3q0 + 1);

(v) a subgroup of order6(q + 1), the normaliser of a cyclic subgroup of orderq + 1;

(vi) Ree(q′) with q = q′m andm prime.

626

THEOREM 15.15 (Zassenhaus)LetG be a2-transitive permutation group whose3-point stabiliser is trivial. Suppose thatG has no regular normal subgroup. IfGhas degreen + 1 and order(n + 1)ns with d even andd ≤ 1

2 (n − 1) thenG isPSL(2, n) in the usual2-transitive permutation representation.

THEOREM 15.16 (Kantor–O’Nan–Seitz)Let G be a permutation group whose2-point stabiliser is cyclic. ThenG has either a regular normal subgroup, orG is one of the following groups in their natural doubly-transitive permutationrepresentations:

PSL(2, q), PGL(2, q), PSU(3, q), PGU(3, q),Sz(q), Ree(q).

15.6 NOTES

For Lemma 15.6 see E. Lucas, Bull. Soc. Math. France 6 (1877-78), 52 and L.E.Dickson, History of the Theory of Numbers I, 271

For more detail on projective unitary groups, see [144, II.8], [145, II.10], and[135].

For more details on Suzuki groups, see [280] [282], [289], [290], [192].For more details on Ree groups, see [289], [222], [147], [73].

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Index

birational transformation, 444

different, 446divisor group, 445

Weierstrass point, 446

647

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