Elliptic Curves over CPepijn Roos Hoefgeest

July 8, 2016

Bachelor thesis

Supervisor: Dr. Mingmin Shen

Korteweg-de Vries Instituut voor Wiskunde

Faculteit der Natuurwetenschappen, Wiskunde en Informatica

Universiteit van Amsterdam

Abstract

This thesis is focused on elliptic curves over the complex plane. We first discuss whatelliptic curves are and use the Riemann-Roch theorem to show there is a group structureon elliptic curves. We show that elliptic curves over C are complex Lie groups and thatevery elliptic curve over C is isomorphic as Lie group to some complex torus, and viceversa. We use this identification to discuss elliptic curves with complex multiplication,and show that for an imaginary quadratic field extension K of Q, there exists a bijectionbetween the set {E = C/Λ |End(E) = OK} /∼= and the class group. In the final chapterwe discuss an application of this bijection to class field theory.

Title: Elliptic Curves over CAuthors: Pepijn Roos Hoefgeest, P.E.R.RoosHoefgeest@uva.nl, 10574433Supervisor: Dr. Mingmin ShenSecond grader: Prof. Dr. Gerard van der GeerDate: July 8, 2016

Korteweg-de Vries Instituut voor WiskundeUniversiteit van AmsterdamScience Park 904, 1098 XH Amsterdamhttp://www.science.uva.nl/math

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Contents

1 Introduction 4

2 The group law 52.1 The geometric approach . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 The algebraic approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 The Lie group structure . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Elliptic curves are tori 153.1 Elliptic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.3 How to get back . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 Elliptic curves with complex multiplication 274.1 Endomorphism rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.2 Complex multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.3 The class group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

5 Applications to class field theory 335.1 The j-invariant is algebraic . . . . . . . . . . . . . . . . . . . . . . . . . . 335.2 Ramification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345.3 Hilbert class field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6 Conclusion and discussion 38

7 Popular summary 39

Bibliography 41

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1 Introduction

As the title suggests, this thesis will discuss elliptic curves over the complex plane. Anelliptic curve, roughly speaking, is the set of solutions (x, y) in the plane to an equationof the form

y2 = x3 + ax+ b.

Figure 1.1 shows what this looks like over the real numbers:

Figure 1.1: Examples over R (Source: mathworld.wolfram.com).

Elliptic curves arise in various areas in mathematics, such as number theory, algebraicgeometry and complex analyses, and are even applied in cryptography. In this thesis wewill discuss the geometric and arithmetic sides of elliptic curves over C. We will firsttake a closer look at the definition of an elliptic curve, and discuss the group structurethat can be defined on an elliptic curve. We then have two primary goals in this thesis.The first one is to show that elliptic curves over C can be identified with complex tori;every elliptic curve over C is isomorphic as complex Lie group to a complex torus, andevery complex torus is isomorphic to some elliptic curve. The second goal will be todiscuss what elliptic curves with complex multiplication are, and to show that thereexists a bijection between a special set of such curves and the class group of a numberfield. The final chapter will give an application of these objects to class field theory,concluding with a remarkable result.

This thesis is written at a level that should be understandable for third year studentof mathematics. Section 2.2 deals with some material that is typically not covered in anundergraduate curriculum. We have included references to some introductions to thismaterial.

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2 The group law

The aim of this chapter is to introduce elliptic curves over the complex plane and toshow there exists a group structure on these elliptic curves. We will approach this groupstructure from both the geometric and the algebraic point of view and we will showthat these approaches coincide. The idea of the geometric approach is to add a pointat infinity which will act as the identity, and define addition of two points by drawing aline between them, and to reflect the point at which this line intersects the elliptic curve.The algebraic approach will use the Riemann-Roch theorem, an important theorem usedin algebraic geometry and complex analysis. This chapter is largely inspired by chapter2 and 3 from [1].

2.1 The geometric approach

The proofs in this section are mostly my own, unless stated otherwise.Formally, an elliptic curve E over an algebraically closed field k is a smooth projectivecurve of genus 1 with a chosen point O ∈ E. We will take a closer look at this definitionin section 2.2. For now we will use the following definition:

Definition 2.1. An elliptic curve E(C) over C is the set of all points (x, y) ∈ C2 thatsatisfy y2 = p(x) for some polynomial p ∈ C[X] of degree 3, such that the curve isnon-singular, i.e. if we set f(x, y) = y2 − p(x), then for all P ∈ E(C) we have ∂f

∂x

∣∣P6= 0

or ∂f∂y

∣∣P6= 0.

It is important to note that the characteristic of C is 0, we will use this later. Intu-itively, that the curve is non-singular means that the graph of the curve looks smoothand behaves nicely, having no cusps, self intersections or other pathological behavior.The following lemma specifies the polynomial p.

Lemma 2.2. An algebraic plane curve curve over C defined by y2 = p(x) for somepolynomial p of degree 3 is an elliptic curve if and only if p has three distinct roots.

Proof.‘⇒’: Suppose E(C) = {(x, y) ∈ C2 : f(x, y) = y2 − p(x) = 0} is an elliptic curve, andsuppose p has a repeated root x0. Then (x0, 0) ∈ E(C) and p′(x0) = 0. This now gives

∂f

∂y

∣∣∣(x0,0)

= 0 = −p′(x0) =∂f

∂x

∣∣∣(x0,0)

,

contradicting the fact that the curve is non-singular.

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‘⇐’: Consider {(x, y) ∈ C2 : f(x, y) = y2 − p(x) = 0} for a polynomial p of degree3 with 3 distinct roots. Then ∂f

∂y= 2y only vanishes for points (x0, 0). Now suppose

∂f∂x

∣∣(x0,0)

= 0. Then p(x0) = p′(x0) = 0, and so x0 is a double root of p, contradicting the

fact that p has 3 distinct roots.

Note that through a change in variables and dividing by the leading coefficient, everyelliptic curve over C can now be defined by an equation of the form y2 = x3 + ax+ b.

As mentioned before, we want to add a point at infinity to the curve to define agroup structure on the group. For this we will have to consider the curve not in the flattwo-dimensional complex space C2, but in the projective two-dimensional complex spaceCP2.

Definition 2.3. The n-dimensional complex space CPn is given by Cn+1 \ {0} /∼ , wherethe equivalence is given by (x0, . . . , xn+1) ∼ (y0, . . . , yn+1) iff there exists a non-zeroλ ∈ C such that (x0, . . . , xn+1) = (λy0, . . . , λyn+1).

So CPn consists of all complex lines through the origin in Cn+1. Equivalence classesin CPn can be expressed by their projective coordinates [x0 : . . . : xn+1] and this classrepresents all lines through the origin and (x0, . . . , xn+1) in Cn+1.We will use CP2, together with an embedding C2 ↪→ CP2, (x, y) 7→ [x : y : 1]. Thismeans that all elements of the form [a : b : 0] could then be considered as points atinfinity.

If we want to consider a set of zeroes of some polynomial f ∈ C[X, Y ] in a projectivespace, we have to find a polynomial F ∈ C[X, Y, Z] such that F (x, y, 1) = f(x, y) andwe need that if F (x0, y0, z0) = 0, then also F (λx0, λy0, λz0) = 0 for all λ ∈ C, because[x0 : y0 : z0] and [λx0 : λy0 : λz0] are representatives of the same equivalence class in CP2.This last condition means that F can be written as sum of homogeneous polynomialsthat have a zero in (x0, y0, z0). This gives rise to the following definition:

Definition 2.4. A projective plane curve over C is a set of points in CP2 whose projectivecoordinates are the zeroes of a homogeneous polynomial F ∈ C[X, Y, Z].

To create such an F for a polynomial f , we will use the homogenization of f :

Definition 2.5. For a given polynomial f ∈ C[X, Y ], the homogenization of f is thepolynomial F ∈ C[X, Y, Z] given by F (X, Y, Z) = Zdeg ff(X/Z, Y/Z).

For the sake of clarity, we will use letters in lower case for polynomials of two variables,which define curves in affine space, and variables in upper case for their homogenizations,which define curves in projective space.

Example 2.6. If f(x, y) = x4+3xy+y3, the homogenization of f is given by F (X, Y, Z) =X4 + 3XY Z2 + Y 3Z.

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Elliptic curves are given by the zeroes of polynomials of the form f(x, y) = y2 − x3 −ax− b, so the homogenization becomes F (X, Y, Z) = Y 2Z −X3 − aXZ2 − bZ3. We seethat F (X, Y, 1) = Y 2−X3− aX − b as required. Now considering the points at infinityon an elliptic curve, we have F (X, Y, 0) = −X3 = 0 ⇒ X = 0. Hence the point O atinfinity on the elliptic curve is represented by [0 : 1 : 0].

We will now prove three lemmas that will be used to define the group structure on anelliptic curve.

Lemma 2.7. Let E(C) be an elliptic curve. The lines in C2 intersecting E(C) at thepoint O = [0 : 1 : 0] at infinity are precisely the lines x = λ for some λ ∈ C. In CP2

these are the lines X = λZ and the line at infinity.

Proof. Any line ` in C2 is given by ay = bx+ c, where a, b, c ∈ C and a, b not both zero.The homogenization of this equation is given by aY = bX + cZ. Now O = [0 : 1 : 0] lieson ` if and only if a = 0. So the lines through [0 : 1 : 0] are given by a homogeneousequation: bX + cZ = 0. The line at infinity is of this form. In C2 this becomes` : x = −c/b ∈ C.

Lemma 2.8. Let E(C) be an elliptic curve and let P = (xP , yP ) and Q = (xQ, yQ) betwo distinct points on E(C). Then the line ` through P and Q intersects the ellipticcurve at one other point R, or ` is tangent to E(C) in P or Q.

Proof. First suppose that (xP , yP ) 6= (xQ,−yQ). Let the line ` through P and Q begiven by y = αx+ β. Substituting this equation into y2 = x3 + ax+ b gives an equationp(x) = 0 for some polynomial p ∈ C[X] of degree 3. Two of its zeroes are given byxP and xQ. Since C is algebraically closed, there must be a third zero xR ∈ C. NowR = (xR, αxR + β) lies on E(C) and if R = P (w.l.o.g.) then ` is tangent to E(C) at P .Now suppose (xP , yP ) = (xQ,−yQ). Then the line ` through P and Q is given by x = xPand by lemma 2.7 this line meets E(C) at P , Q and O.

Lemma 2.9. Let E(C) be an elliptic curve and let p = (xp, yp) be a point on E(C).Then the line ` tangent to E(C) at p meets E(C) at another point q.

Proof. Suppose that the tangent ` is given by x = c for some c ∈ C. Then by lemma2.7 ` meets E(C) at p and O.Now suppose ` is given by y = αx+ β. Substituting this into y2 = x3 + ax+ b gives anequation P (x) = 0 for some polynomial P of degree 3. Since C is algebraically closed,there must be three solutions to this equation. (This equation may have a triple root,this is the case for the 3-torsion points.)

As mentioned in the introduction, the group structure arises through drawing linesthrough points and reflecting the third point at which this line meets the curve. In lightof the previous three lemmas, this approach makes sense. More formally:

Definition 2.10. If P and Q are points on an elliptic curve E(C), P ⊕Q is defined asfollows: Let ` be the line through P and Q (if P = Q, let ` be the tangent at P ), and let

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R be the third point at which ` meets E(C). Let `′ be the line through R and O. NowP ⊕ Q is given by the third point at which `′ meets E(C). If R = O, `′ is the tangentto E(C) at O, so then `′ only meets E(C) at O and P ⊕Q = O.

Figure 2.1 illustrates this operation when E is an elliptic curve over R:

x

y

y2 = x3 − 2x+ 2

P

QR

P ⊕Q

Figure 2.1: Addition of points on an elliptic curves over R.

Proposition 2.11. The operation ⊕ has the following properties:

1. If ` is a line intersecting E(C) at three points P,Q,R, then

(P ⊕Q)⊕R = O.

2. For all P ∈ E(C), P ⊕O = P .

3. For all P,Q ∈ E(C), P ⊕Q = Q⊕ P .

4. For all P ∈ E(C), there exists a point P ∈ E(C) such that

P ⊕ (P ) = O.

5. For all P,Q,R ∈ E(C), (P ⊕Q)⊕R) = P ⊕ (Q⊕R).

Proof. (Due to [1], see proposition 3.2.2.)1. The line through P ⊕ Q) and R intersects E(C) at O by the definition of ⊕. Thetangent at O meets E(C) only at O so (P ⊕Q)⊕R = O.2. If P,O,R are on one line `, then the line `′ through R and O coincides with ` and soP = P ⊕O. 3. This immediately follows from the symmetry in the definition of ⊕.

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4. Consider the line ` through P and O, suggestively denote the third point at which `meets E(C) by P . Then by 1. and 2. we see:

P ⊕ (P ) = (P ⊕O)⊕ (P ) = O.

5. This can be shown by writing out explicit formulas for the addition of points. We willinclude these formulas below, but will omit the verification of associativity as this comesdown to a tedious calculation that is not particularly insightful. In the next section wewill give a proof that will use the algebraic approach.

This can be summarized as follows:

Corollary 2.12. (E(C),⊕) is an abelian group with unit element O.

Remark: For an elliptic curve E(C) that is defined by y2 = x3 + ax + b and a pointO 6= P = (xP , yP ) on E(C), we see that P is given by (xP ,−yP ) as the line throughthese points is given by x = xP and this line meets E(C) at infinity.

Now we try to give explicit formulas for the addition of points. The slope of thetangent line at P can be found using implicit differentiation on the polynomial f(x, y) =y2 − x3 − ax − b. If yP = 0 the tangent is given by x = xP . Otherwise, the slope α of

the tangent is given by α =3x2P+b

2yP. The tangent line is then given by y = αx + β with

β =−x3

P+axP+2c

2yP.

If Q = (xQ, yQ) is another point on E(C) that is not P , then the line through P andQ is given by y = αx+ β with α =

yQ−yPxQ−xP

and β =xQyP−xP yQxQ−xP

.

We could now compute the coordinates of R = P ⊕ Q. If ` : y = αx + β is the linethrough P and Q (this is the tangent at P if P = Q), then ` meets E(C) at a thirdpoint R′. By our earlier remark, this point is then the inverse of R, so xR = xR′ andyR = −yR′ . So we could find the x-coordinate xR of R by solving f(x, αx + β) = 0.This is a polynomial of degree 3 in x, and we already know two solutions: xP and xQ.Therefore, f(x, αx+ β) = (x− xP )(x− xQ)(x− xR). Now by comparing the coefficientof the second order term, we obtain xR = α2−xP −xQ. Plugging this back into our line` and flipping the sign gives us yR = −αxR − β.

2.2 The algebraic approach

The aim of this section will be to deduce the general Weierstrass equation

a1y2 + a2xy + a3y = b1x

3 + b2x2 + b3x+ b4

for elliptic curves from the formal definition for elliptic curves, and to give an algebraicproof of the group law. Both approaches will make use of the Riemann-Roch theorem.

This section will use some notions from algebraic geometry that are typically not partof an undergraduate curriculum. It is beyond the scope of this thesis to work out all of

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this material. An understandable introduction to algebraic geometry can be found in[2]. The important parts are also summarized in the first two chapters of [1], which wewill follow. It will not be necessary to work over C, so we will work over an arbitraryalgebraically closed field k.As mentioned before, the formal definition of an elliptic curve is as follows:

Definition 2.13. An elliptic curve E over an algebraically closed field k is a smoothprojective curve of genus 1 with a chosen point O ∈ E.

To deduce the Weierstrass equation for elliptic curves from this definition, we willneed the following definition:

Definition 2.14. Let k be an algebraically closed field and let C be a curve over C. Adivisor on C is a function D : C → Z that is zero for all but finitely many points on C.We write

D =∑P∈C

D(P )P.

The degree of D is given by

deg(D) =∑P∈C

D(P ).

The set of divisors on a curve C is a free abelian group, denoted Div(C). The divisorsof degree zero form a subgroup of this group, denoted Div0(C).If C is a smooth curve, and f : C → k is a rational function, we have a divisor of fgiven by

div(f) =∑P∈C

ordP (f)P.

Here ordP (f) denotes the order of f at P . For the exact definition, see [2] (denotedvP (f)). Intuitively, div(f) = f−1({0})− f 1({∞}) counted with multiplicity.

Lemma 2.15. The divisors of the form div(f) are called principal divisors, and form asubgroup PDiv(C) of Div0(C).

Proof. For a proof, see [1] proposition 2.3.1.

In light of the previous lemma, the following definition makes sense:

Definition 2.16. Let C be a smooth curve. Two divisors D1 and D2 are linearlyequivalent, denoted D1 ∼ D2 if D1 −D2 ∈ PDiv(C). The Picard group is given by

Pic(C) :=Div(C)

PDiv(C).

We also set

Pic0(C) :=Div0(C)

PDiv(C).

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For divisors on a curve, we have a partial ordering ’≥’ given by

(D ≥ D′) ⇐⇒ (∀P ∈ C, D(P ) ≥ D′(P )).

For a divisor D on C, consider the following set:

H0(C,OC(D)) = {f : f is a rational function on C such that div(f) +D ≥ 0} ∪ {0}.

This is a vector space over k. We are now in a position to state the Riemann Rochtheorem. A proof for this theorem is given in chapters 10 and 11 of [2].

Theorem 2.17 (Riemann-Roch). Let C be a smooth curve, let K be a canonical divisoron C and let g be the genus of C. Then

dim H0(C,OC(D))− dim H0(C,OC(K −D)) = deg(D) + 1− g.

For an elliptic curve K = 0 is a canonical divisor. It is a fact that if D1 ∼ D2, thenH0(C,OC(D1)) ∼= H0(C,OC(D2)), so we have the following corollary:

Corollary 2.18. For an elliptic curve E over an algebraically closed field k, and D adivisor on E, we have:

dim H0(E,OE(D))− dim H0(E,OE(−D)) = deg(D).

Note that for rational functions f : E → k, deg div(f) = 0, so for a divisor D on Ewith D > 0, we have H0(E,OE(−D)) = {0}.

We can now deduce the Weierstrass equation for elliptic curves. Let E be an ellipticcurve over k and let D = n[O] for n ∈ Z≥1. By the previous remark, we see thatdim H0(E,OE(n[O])) = deg(n[O]) = n. Let x be a rational function on E with a doublepole at O, and let y be a rational function on E with a triple pole at O. Now {1} isa basis for H0(E,OE([O])), {1, x} is a basis for H0(E,OE(2[O])) and {1, x, y} is a basisfor H0(E,OE(3[O])). We see that

dim H0(E,OE(6[O])) = 6,

but there are 7 rational functions in H0(E,OE(6[O])), namely 1, x, x2, x3, y, y2 and xy.These must be linearly dependent, so we get a relation

a1y2 + a2xy + a3y = b1x

3 + b2x2 + b3x+ b4

for certain a1, a2, a3, b1, b2, b3, b4 ∈ k, for which a1b1 6= 0. Now the map

φ : E → P2

P 7→ [x(P ), y(P ), 1]

gives an isomorphism of E onto a curve C given by C : a1y2 +a2xy+a3y = b1x

3 +b2x2 +

b3x+ b4. For a proof of this, see [1] proposition 3.3.1.

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Since our field of interest is C, and char(C) = 0, we can divide this equation by 2 and3. We can now perform a change of variables in order to get an equation as in definition2.1. (The non-singular property is the same as saying the curve is smooth).First we can replace y by a1b

21y and x by a1b1x. Dividing by a3

1b41 then gives:

y2 + a′2xy + a′3y = x3 + b′2x2 + b′3x+ b′4

(y +1

2(a′2x+ a′3))2 = x3 + b′2x+ b′3x+ b′′4

(y +1

2(a′2x+ a′3))2 = (x+

1

3b′2)3 + b′′3(x+

1

3b′2) + b′′′4

So by replacing y + 12(a′2x + a′3) by y and x + 1

3b′2 by x, we get an equation of the form

y2 = x3 + Ax2 + B. Without loss of generality, we may also assume that elliptic curvesover C are given by equations of the form

y2 = 4x3 − ax− b.

We can now also prove the associativity of the group law using the Rieman-Rochtheorem. We will follow [4]. For a more detailed exposition, see lemma 3.5.3. andproposition 3.5.4. in [1].

Let E be an elliptic curve and let D be a divisor on E of degree 0. Then we seedeg(D + O) = 1, so Riemann-Roch tells us that dim H0(E,OE(D + O)) = 1. So up tomultiplication by a constant, there is a unique rational function f : E → k such that

div(f) +D +O ≥ 0.

Since divisors of rational functions are of degree zero, deg(div(f) + D + O) = 1. Itfollows that there exists a unique P ∈ E such that

div(f) +D +O = P.

This implies that D is linearly equivalent to P −O. Therefore, the map

ψ : E → Pic0(E)

P 7→ [P −O]

mapping a point P to the equivalence class of P −O in Pic0(E) is a bijection.Through this map, E inherits a group structure from Pic0(E). If we can show that thismap is a homomorphism, with on E the group structure given in definition 2.10, weautomatically see that this operation is associative.

Theorem 2.19. The map ψ : E → Pic0(E) is a homomorphism, therefore the operation⊕ defines a group structure on E with unit element O.

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Proof. (Due to [4], see the proof on page 13) It suffices to show that if P and Q arepoints on E, then

(P −O) + (Q−O) ∼ (P ⊕Q)−O.

Let `1 : aX + bY + cZ = 0 be the (homogeneous) line through P and Q, meeting theelliptic curve at a third point R, and let `2 : a′X + b′Y + c′Z be the (homogeneous) linethrough R and O meeting the curve at P ⊕ Q. Then h := `1/`2 is a rational functionthat has zeroes at the zeros of `1 and poles at the zeroes of `2. Hence,

div(h) = P +Q+R−R−O − (P ⊕Q)

=⇒P +Q ∼ (P ⊕Q) +O

⇐⇒ (P −O) + (Q−O) ∼ (P ⊕Q)−O.

2.3 The Lie group structure

We will now return to elliptic curves over C. So let E(C) be an elliptic curve over C. Wewill see that E(C) is not only a group, but even a complex Lie group. To show this, wewill have to show that E(C) is a complex manifold, and that the addition and inversionof points are holomorphic maps on this manifold.It is true in general that every smooth projective complex variety over C of dimension1 is a Riemann surface, i.e. a complex manifold of dimension 1. This follows from thefact that the analytification functor, from the category of smooth varieties over C to thecategoriy of complex manifolds, when restricted to the category of smooth projectivecomplex varieties over C of dimension 1, gives an equivalence of categories with thecategory of compact Riemann surfaces (see [2], chapter 9 and 10).It is however possible to give a more elementary proof:

Theorem 2.20. Every elliptic curve E(C) over C is a complex manifold.

Proof. (Proof by myself, based on an explanation from Dr. Shen) Let E(C) ⊂ CP2 be anelliptic curve over C, given by y2 = x3 + ax+ b. In projective coordinates, this becomes

Y 2Z = X3 + aXZ2 + bZ3.

For each point P on E(C) we will give local diffeomorphisms.CP2 is a complex manifold with coordinate charts

φi : C2 → Ui = {[x0, x1, x2] ∈ CP2 : xi 6= 0}.

These are diffeomorphisms. We will think of C2 being embedded into CP2 through(x, y) 7→ [1, x, y]. Let P = (xP , yP ) ∈ C2 be a point on E(C). If yP 6= 0, we use the factthat the curve is non-singular to see there is an open set UP ⊂ E(C) containing P for

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which the projection on the x-coordinate, πx|U : U → πx(U), is a diffeomorphism, withinverse x 7→ ±

√x3 + ax+ b. If yP 6= 0, we do the same, projecting on the y-coordinate.

This also works if xP = 0, for if (0, 0) ∈ E(C), then the line x = 0 is tangent to E(C) at(0, 0).It remains to give a coordinate chart for a open set containing O = [0 : 1 : 0]. For this,consider φ−1

2 : U2 → C2, [x0, x1, x2] 7→ (x0/x2, x1/x2). Consider the elliptic curve on thissubset of CP2. Now we have Y = 1, so the equation becomes

z = x3 + axz2 + bz3.

In this equation, the point at infinity is (x, z) = (0, 0). Because the curve is non-singular,we can now again use the projections to create coordinate charts. Hence we have foundcoordinate charts for open neighbourhoods of all points on E(C), showing E(C) is acomplex manifold.

We now conclude this chapter:

Theorem 2.21. Every elliptic curve over C is a complex Lie group.

Proof. We use the proof of theorem 3.3.6. from [1].Let E(C) : y2 = x3 + ax + b be an elliptic curve over C. The inverse map ρ : E(C) →E(C), P 7→ P , is clearly holomorphic, as ρ(x, y) = (x,−y).For the addition map µ : E(C) × E(C) → E(C), (P,Q) 7→ P ⊕ Q, we have deducedspecific formulas at the end of section 2.1. It is clear from these formulas that µ isholomorphic, except at pairs of the form

(P, P ), (P,−P ), (P,O), (O,P ).

So let (P1, P2) be of one of these forms. We will create a map φ : E(C)×E(C)→ E(C)that agrees with µ on E(C) and is holomorphic at (P1, P2), using the group law.Choose points Q1, Q2 such that the translation maps

τi : E(C)→ E(C), P 7→ P ⊕Qi, i∈{1,2}

are holomorphic at P1, P2 respectively, and such that (P1 ⊕ Q1, P2 ⊕ Q2) is not of theform (P, P ), (P,−P ), (P,O), (O,P ). These maps are clearly invertible. Now let φ bedefined as follows:

φ : E(C)× E(C) E(C)× E(C) E(C) E(C) E(C)τ1×τ2 µ τ−1

1 τ−12

Then we see φ(P,Q) = P ⊕ Q = µ(P,Q), for all (P,Q) ∈ E(C) × E(C). The first twosteps in φ are holomorphic at (P1, P2). It could be that τ−1

1 or τ−12 is not holomorphic

at (P1 ⊕ Q1 ⊕ P2 ⊕ Q2) or (P1 ⊕ P2 ⊕ Q2) respectively. But then these points must beof the form O,±Q1 or O,±Q2 respectively, and since Q1 and Q2 are arbitrary points,we could choose new points Q1, Q2 such that all maps are holomorphic. Thereforeφ is holomorphic at (P1, P2), and since φ agrees with µ on all of E(C) × E(C), µ isholomorphic. We conclude that E(C) is a complex Lie group.

14

3 Elliptic curves are tori

A lattice Λ = [ω1, ω2] ⊂ C is a free Z module generated by two complex numbersω1, ω2 ∈ C that are linearly independent over R. By taking the quotient of the complexplane C modulo a lattice Λ, we obtain a torus C/Λ, which forms a complex Lie group.

The main purpose of this chapter will be to prove the following theorem:

Theorem (Uniformization theorem).For every lattice Λ ⊂ C, the torus C/Λ is isomorphic as complex Lie group to someelliptic curve E(C).Conversely, every elliptic curve E(C) is isomorphic as complex Lie group to a torus C/Λfor some lattice Λ ⊂ C.

This will allow us to identify an arbitrary elliptic curve with a torus and vice-versa.To prove this theorem, we will first introduce elliptic functions to create a map betweenan arbitrary torus and an elliptic curve. This map will be holomorphic, and we will showthis map is an isomorphism of groups. To show this map is actually an isomorphismof Lie-groups, we will use the inverse function theorem. We will also give an outline ofan isomorphism that maps an elliptic curve to a torus. This will explain the name ofelliptic curves.

But first, let’s try to make sense of the Uniformization theorem topologically. Let’sassume, without loss of generality, that E(C) is an elliptic curve over C, given by

y2 = x(x− 1)(x− λ)

for some λ ∈ C. If we make branch cuts along the line segment joining 0 and 1 and theline segment joining λ and ∞, we get two single valued functions:

y = ±√x(x− 1)(x− λ).

Hence we get two copies of the Riemann sphere with branch cuts as in figure 3.1. Nowgluing these two copies of CP1 together along the corresponding branch cuts gives us aRiemann surface that is homeomorphic to a torus.

Sections 3.1. and 3.2. will entirely follow chapter 15 and 16 from [3].

3.1 Elliptic functions

To define functions on a torus C/Λ, we will need the notion of elliptic functions:

15

0

1

∞

λ

0

1

∞

λ

Figure 3.1: Branch cuts on the Riemann sphere

Definition 3.1. For a lattice Λ = ω1Z ⊕ ω2Z, an elliptic function with respect to Λ isa meromorphic function f : C→ C such that

f(z + ω) = f(z) ∀z ∈ C, ω ∈ Λ.

It follows that every elliptic function induces a well defined function f∗ : C/Λ → Cby taking f∗(z̄) = f(z) for some representative z ∈ C of z̄. We will usually denote thisinduced map by f as well.A very important elliptic function is the Weierstrass ℘-function, which will be used inthe mapping from a torus to an elliptic curve:

Definition 3.2. The Weierstrass ℘-function of a lattice Λ is defined by:

℘(z) := ℘(z; Λ) :=1

z2+∑ω∈Λ∗

(1

(z − ω)2− 1

ω2

).

Here Λ∗ = Λ \ {0}.

Theorem 3.6 states some properties of the Weierstrass ℘-function. To prove thistheorem, we will need the following propositions:

Proposition 3.3. For any lattice Λ ⊂ C and k > 2, the sum Gk(Λ) =∑

ω∈Λ∗1ωk

converges absolutely. This sum is called the weight-k Eisenstein series for Λ.

Proof. We could write C as union of annuli:

C =⋃n≥0

{z ∈ C : n ≤ |z| < n+ 1}.

16

Therefore, writing Cn = {z ∈ C : n ≤ |z| < n+ 1}, we have∑ω∈Λ∗

1

|ω|k=∑n≥0

∑ω∈Cn

1

|ω|k=∑|ω|<1

1

|ω|k+∑n≥1

∑ω∈Cn

1

|ω|k .

If δ is the minimum distance between any two lattice points, then for any annuluswith inner radius r and width δ/2, the amount of lattice points within the annulus isbounded by 4πr/δ. Therefore, the amount of lattice points in the annulus Cn is boundedby 8πn/δ2. It follows that for each n ≥ 1,∑

ω∈Cn

1

|ω|k≤∑ω∈Cn

1

nk≤ 8π

δ2nk−1

and so ∑n≥1

∑ω∈Cn

1

|ω|k≤ 8π

δ2

∑n≥1

1

nk−1<∞.

Since there are only finitely many points within C0, the sum∑|ω|<1

1|ω|k is also finite,

and so Gk(Λ) converges absolutely.

Proposition 3.4. For any lattice Λ ⊂ C, the sum in the definition of ℘(z; Λ) convergesabsolutely on all compact subsets of C \ Λ

Proof. Let S ⊂ C \ Λ be compact. Then S is bounded, so there exists an r > 0 suchthat for all z ∈ S, |z| ≤ r. Since Λ is infinite, we have that for all but finitely manyω ∈ Λ, |ω| ≥ 2r. The triangle inequality gives us |ω| = |ω − z + z| ≤ |ω − z| + |z|. If|ω| ≥ 2r, then |ω| ≥ 2|z| for all z ∈ S, so for all but finitely many ω ∈ Λ we have:

|ω − z| ≥ |ω| − |z| ≥ 1

2|ω|,

|2ω − z| ≤ |2ω|+ | − z| ≤ 5

2|ω|.

So, for all but finitely many ω ∈ Λ, we see that∣∣∣∣ 1

(z − ω)2− 1

ω2

∣∣∣∣ =

∣∣∣∣ z(2ω − z)

ω2(z − ω)2

∣∣∣∣ ≤ |z|52|ω|

|ω|2(12|ω|)2

≤r 5

2|ω|

|ω|2(12|ω|)2

=10r

|ω|3

By proposition 3.3,∑

ω∈Λ∗10r|ω|3 converges. As are only finitely many ω ∈ Λ∗ such that

|ω| ≤ r, the sum ∑ω∈Λ∗

(1

(z − ω)2− 1

ω2

)converges absolutely for all z ∈ S.

17

Proposition 3.5. Let Λ ⊂ C a lattice. For all n ∈ N, define fn : C→ C as follows:

fn(z) =1

z2+

∑ω∈Λ

0<|ω|<n

(1

(z − ω)2− 1

ω2

).

Then the sequence {fn} converges uniformly to ℘ on all compact subsets of C \ Λ.

Proof. Let S be a compact subset of C \ Λ, and choose r > 0 such that |z| < r for allz ∈ S. For all n ∈ N and z ∈ S, we have

|℘(z)− fn(z)| =

∣∣∣∣∣∣∑|ω|≥n

(1

(z − ω)2− 1

ω2

)∣∣∣∣∣∣ .For n ≥ r, the rate of convergence of this sum can be bounded in terms of r and Λ byproposition 3.4. Therefore, for all ε > 0, there exists an N ∈ N such that for all n ≥ N ,|℘(z)− fn(z)| < ε for all z ∈ S, as desired.

Theorem 3.6. For any lattice Λ = Zω1 ⊕ Zω2 ⊂ C, the function ℘(z) = ℘(z; Λ) hasthe following properties:

1. ℘(z) is an even meromorphic function with solely double poles at each z ∈ Λ.

2. ℘′(z) is an odd meromorphic function with solely triple poles at each z ∈ Λ.

3. ℘(z) and ℘′(z) are elliptic functions.

Proof.1. All functions fn : C → C from proposition 3.5 are clearly holomorphic on C/Λ. As{fn} converges uniformly to ℘ by proposition 3.5, ℘ is also holomorphic on C \ Λ. Itis clear that ℘(z) has poles of order 2 at all z ∈ Λ, so these are all poles. From thedefinition it is clear that ℘(z) = ℘(−z).2. As the partial sums defining ℘(z) converge uniformly to ℘(z) may differentiate ℘(z)term by term, so

℘′(z) = −2∑ω∈Λ

1

(z − ω)3.

As ℘(z) is holomorphic on C \ Λ, so is ℘′(z). It is clear that ℘′(z) has poles of order 3at all z ∈ Λ and that ℘′(z) = −℘′(−z) for all z ∈ C \ Λ.3. The only thing that remains to be shown is that ℘(z) and ℘′(z) are periodic in Λ. It isclear from the formula for ℘′(z) that ℘′(z+ω) = ℘′(z) for all z ∈ C, ω ∈ Λ. It will sufficeto show ℘(z+ω1) = ℘(z) = ℘(z+ω2). We already established that ℘′(z+ωi)−℘′(z) = 0for i ∈ {1, 2}. Integrating then yields

℘(z + ωi)− ℘(z) = ci

for some constant ci ∈ C. This holds for all z ∈ C, so in particular it holds for z =−1/2ωi. Since ℘(z) is an even function, we then see

ci = ℘(1/2ωi)− ℘(−1/2ωi) = 0.

Hence ci = 0 for i = 1, 2 and ℘(z) is an elliptic function.

18

The next theorem gives an interesting result about elliptic functions. We will not needthis result for our purposes, so we omit the proof.

Theorem 3.7. For any lattice Λ ⊂ C, all elliptic functions with respect to Λ can bewritten as a rational combination of ℘(z) and ℘′(z).

3.2 Isomorphism

Now that we have established that the Weierstrass ℘-function is in fact an elliptic func-tion, we are in a position to use it to construct a map from a torus to an elliptic curve.For this, we will first need the Laurent series for ℘(z) around z = 0.

Lemma 3.8. For any lattice Λ ⊂ C, the Laurent series for ℘(z) = ℘(z; Λ) around z = 0is given by

℘(z) =1

z2+∞∑k=1

(2k + 1)G2k+2(Λ)z2k.

Remember that that the weight-k Eisenstein seriesGk(Λ) is defined byGk(Λ) =∑

ω∈Λ∗1ωk

,which converges for k > 2 by proposition 3.3.

Proof. Around z = 0, | zω| < 1 for all ω ∈ Λ∗. Therefore we have

1

(z − ω)2− 1

ω2=

1

ω2

(1

(1− zω

)2− 1

)=

1

ω2

(∞∑k=0

(k + 1)( zω

)k− 1

)

=∞∑k=1

(k + 1)

(zk

ωk+2

).

Plugging this into the definition of ℘(z) gives us:

℘(z) =1

z2+∑ω∈Λ∗

∞∑k=1

(k + 1)

(zk

ωk+2

)

=1

z2+∞∑k=1

(k + 1)zk∑ω∈Λ∗

1

ωk+2

=1

z2+∞∑k=1

(2k + 1)z2k∑ω∈Λ∗

1

ω2k+2

=1

z2+∞∑k=1

(2k + 1)G2k+2(Λ)z2k.

In the first step we used uniform convergence to swap the summations, and in thesecond step we used that for odd k,

∑ω∈Λ∗

1ωk+2 = 0 as all terms with opposite sign

cancel out.

19

We will use this Laurent series to prove the following theorem, which will unveil ourinterest in Weierstrass ℘-functions.

Theorem 3.9. For any lattice Λ = ω1Z⊕ω2Z ⊂ C, the function ℘(z) = ℘(z; Λ) satisfies:

℘′(z)2 = 4℘(z)3 − 60G4(Λ)℘(z)− 140G6(Λ)

It is convention to write g2(Λ) = 60G4(Λ) and g3(Λ) = 140G6(Λ).

Note that this equation has the form of an equation defining an elliptic curve.

Proof. Using the Laurent series of ℘(z), we will determine the first couple of terms ofthe Laurent series for each function in the expression:

℘(z) = z−2 + 3G4(Λ)z2 + 5G6(Λ)z4 + ...

℘′(z) = −2z−3 + 6G4(Λ)z + 20G6(Λ)z3 + ...

℘(z)3 = z−6 + 9G4(Λ)z−2 + 15G6(Λ) + ...

℘′(z)2 = 4z−6 − 24G4(Λ)z−2 − 80G6(Λ) + ...

We now define f(z) = ℘′(z)2−4℘(z)3 +60G4(Λ)℘(z)+140G6(Λ). In the Laurent expan-sion of f(z) around z = 0, all terms of order 0 and lower vanish and f is holomorphicon C. It also follows that f(z) vanishes at z = 0.Let P = {t1ω1 + t2ω2 | 0 ≤ t1, t2 < 1}. P contains exactly one representative of eachequivalence class in C/Λ, so since f(z) is an elliptic function, f(C) = f(P) = f(P).Since P is closed and bounded it is compact, so since f is continuous, f is bounded.By Liouville’s theorem, all bounded functions that are holomorphic on C must be con-stant, and since f(0) = 0, it follows that f(z) = 0 for all z ∈ C and so we have℘′(z)2 = 4℘(z)3 − 60G4(Λ)℘(z)− 140G6(Λ), as desired.

If we show that the polynomial p(x) = 4x3 − g2(Λ)x− g3(Λ) has three distinct rootsfor all lattices Λ ⊂ C, then it follows from lemma 2.2 that each lattice Λ can be used todefine an elliptic curve. Recall the following theorem from complex analysis:

Theorem 3.10 (Argument principle). Let f be a meromorphic function on a neigh-bourhood of the closure of some open set Γ with a simple closed boundary γ, such that fhas no poles or zeroes on γ. Let Z and P be the number of zeroes and poles of f on Γrespectively. Then

1

2πi

∫γ

f ′(z)

f(z)dz = Z − P.

The argument principle can be used to prove the following proposition:

Proposition 3.11. Let Λ = ω1Z ⊕ ω2Z ⊂ C be a lattice and let f(z) be an ellipticfunction with respect to Λ. Then for all α ∈ C, the number of zeroes of f within thefundamental parallelogram Pα = {α+ t1ω1 + t2ω2 | 0 ≤ t1, t2 < 1} is equal to the numberof poles of f within Pα.

20

Proof. The poles and zeroes of f each form a discrete set, so we can choose a fundamentalparallelogram Pα such that f has no poles or zeroes on ∂Pα. Then by periodicity of fin Λ, f ′/f is periodic in Λ, so we have∫

∂Pα

f ′(z)

f(z)dz = 0

and so by proposition 3.11 the number of zeroes of f within Pα is equal to the numberof poles. Since f is periodic, this holds for any fundamental parallelogram.

Lemma 3.12. Let Λ ⊂ C a lattice and z ∈ C \ Λ. Then ℘′(z; Λ) = 0 if and only if2z ∈ Λ.

Proof. Suppose 2z ∈ Λ. Then because ℘′(z) is an odd function, we see

℘′(z) = ℘′(z − 2z) = ℘′(−z) = −℘′(z),

so ℘′(z) = 0. We can find three zeroes of ℘′(z) within P = {t1ω1 + t2ω2 | 0 ≤ t1, t2 < 1},namely ω1/2, ω2/2 and ω1/2 + ω2/2. Because ℘′(z) has one pole of order three withinP , these are all zeroes of ℘′(z) within P . Since ℘′ is periodic in Λ, this holds for anyfundamental parallelogram.

Lemma 3.13. The zeroes of the polynomial p(x) = 4x3 − g2(Λ)x− g3(Λ) are distinct.

Proof. Write Λ = ω1Z⊕ ω2Z and set

z1 =ω1

2, z2 =

ω2

2, z3 =

ω1 + ω2

2.

Then ℘(z1), ℘(z2) and ℘(z3) are the zeroes of p(x), as ℘′(zi) = 0 for i ∈ {1, 2, 3}. Toshow these zeroes are distinct, let gi(z) = ℘(z)− ℘(zi). This is an elliptic function witha pole of order 2 at z = 0 in P = {t1ω1 + t2ω2 | 0 ≤ t1, t2 < 1}. Therefore, it has only 2zeroes in P . We see that zi is a double zero as g′i(zi) = 0 by lemma 3.12. So gi(z) hasno other zeroes, and so ℘(zi) 6= ℘(zj) if i 6= j.

With lemma 3.13 we have established that each complex torus C/Λ defines an ellipticcurve given by y2 = 4x3 − g2(Λ)x − g3(Λ). This puts us in the position to give a welldefined map from any complex torus to an elliptic curve:

Let Λ ⊂ C be a lattice and consider the complex torus C/Λ. Let E(C) ⊂ CP2 be theelliptic curve defined by y2 = 4x3 − g2(Λ)x− g3(Λ). Define Φ : C/Λ→ E(C) by:

Φ(z) = [℘(z) : ℘′(z) : 1] (3.1)

Throughout the rest of the text, Φ will be used to refer to this map. If we can provethat Φ is an isomorphism of additive groups, we have established the first part of theUniformization theorem.

21

Theorem 3.14. For any lattice Λ = ω1Z ⊕ ω2Z ⊂ C, the map Φ : C/Λ → E(C) is anisomorphism of additive groups.

Proof. We will have to show Φ is a homomorphism that is both surjective and injective.First we note that as z tends to 0, both ℘(z) and ℘′(z) tend to infinity, and sincelimz→0 ℘(z)/℘′(z) =∞, Φ(0) = O and Φ preserves the identity.We will first consider how Φ acts on the points of order 2. There are precisely 3 of suchpoints in C/Λ, namely ω1/2, ω2/2 and (ω1 + ω2)/2. We already established in lemma3.12 that these points are zeroes of ℘′(z). Hence, the y-coordinates of the images of thesepoints are zero and as desired these are points of order 2 in E(C). Φ is also bijective onthe points of order 2, since the x-coordinate of the image of each point is mapped to adistinct root of 4x3 − g2(Λ)x− g3(Λ) as established in lemma 3.13. So Φ restricts to anisomorphism on points of order 2. So now we consider the points that are not of order2.

To show Φ is surjective, consider [x : y : 1] ∈ E(C). The function f(z) = ℘(z) − xis elliptic, and it has one pole of order 2 at z = 0 in P = {t1ω1 + t2ω2 | 0 ≤ t1, t2 < 1}.Therefore, it has 2 zeroes in P . Let z0 be such a zero. Then Φ(z0) = [1 : x : ±y]. Since℘ is an even function and ℘′ is odd, we have

Φ(−z) = [1 : ℘(−z) : ℘′(−z)] = [1 : ℘(z) : −℘′(z)] = Φ(z).

So if Φ(z0) = [1 : x : −y], then Φ(−z0) = [1 : x : y].To show Φ is injective, suppose Φ(z1) = Φ(z2). As we have already covered the 2-

torsion elements, we may assume 2z1 /∈ Λ, so ℘′(z1) 6= 0. The roots of f(z) = ℘(z)−℘(z1)in P are ±z1, so z2 ≡ ±z1 mod Λ. As we also have ℘′(z1) = ℘′(z2), and because ℘′ isan odd function, it follows that z1 ≡ z2 mod Λ.

What remains is a lengthy verification that Φ is a homomorphism, i.e. Φ(z1 + z2) =Φ(z1) ⊕ Φ(z2). So let z1, z2 ∈ P . We may assume z1 6= 0 and z2 6= 0 as this wouldbe trivial, and we may assume z1 + z2 /∈ Λ as in this case z1 = −z2 mod Λ, and soΦ(z1) = Φ(z2). Let ` : y = αx + β be the line through Q1 = Φ(z1) and Q2 = Φ(z2),and let Q3 be the third point at which ` intersects E(C). Then Q1 ⊕Q2 ⊕Q3 = O.Consider the function f(z) = −℘′(z) + α℘(z) + β. This is an elliptic function with onepole of order 3 in P at z = 0 so it also has 3 zeroes in P , two of which are z1 and z2.Denote the third by z3. Φ(z3) lies both on E(C) and on `, so it must be one of the pointsQ1, Q2, Q3. Since Φ is bijective, Φ(z3) = Q3.If we can show that z1 + z2 + z3 ∈ Λ, we have

Φ(z1 + z2) = Φ(−z3) = −Φ(z3) = Q3 = Q1 ⊕Q2 = Φ(z1)⊕ Φ(z2).

Pick a fundamental region Pζ = {ζ + t1ω1 + t2ω2 | 0 ≤ t1, t2 < 1} such that there are nopoles or zeroes of f(z) on the boundary of Pζ and replace z1, z2, z3 with points in Pζ ifthey are not already contained within Pζ . Using the generalized argument principle weobtain

1

2πi

∫∂Pζ

zf ′(z)

f(z)dz = z1 + z2 + z3.

22

Since g(z) = f ′(z)f(z)

is periodic with respect to Λ, we could also evaluate this integral asfollows:∫

∂Pζzg(z) dz =

∫ ζ+ω1

ζ

zg(z) dz +

∫ ζ+ω1+ω2

ζ+ω1

zg(z) dz +

∫ ζ+ω2

ζ+ω1+ω2

zg(z) dz +

∫ ζ

ζ+ω2

zg(z) dz

=

∫ ζ+ω1

ζ

zg(z) dz +

∫ ζ+ω2

ζ

(z + ω1)g(z) dz +

∫ ζ

ζ+ω1

(z + ω2)g(z) dz +

∫ ζ

ζ+ω2

zg(z) dz

= ω1

∫ ζ+ω2

ζ

g(z) dz + ω2

∫ ζ

ζ+ω1

g(z) dz (3.2)

Here we cancelled integrals along the same sides in opposite direction.If we can show that the remaining integrals in (3.2) are integers multiplied by 2πi, wehave z1 + z2 + z3 ∈ Λ as desired.To see this, recall that for any closed curve γ in C and some point z0 ∈ C, the integral

1

2πi

∫γ

1

z − z0

dz

gives the number of times γ winds around z0. In particular, it is an integer.Let γ1 = {f(ζ + tω2) | 0 ≤ t ≤ 1}. This is a closed curve as f(ζ) = f(ζ + ω2). We getthat the following expression is an integer:

1

2πi

∫γ1

1

z − 0dz =

1

2πi

∫ 1

0

f ′(ζ + tω2)

f(ζ + tω2)dt =

1

2πi

∫ ζ+ω2

ζ

f ′(z)

f(z)dz =

1

2πi

∫ ζ+ω2

ζ

g(z) dz

Analogously, using γ2 = {f(ζ + tω1) | 0 ≤ t ≤ 1}, we get that

1

2πi

∫ ζ

ζ+ω2

g(z) dz

is an integer. Hence, using our previous observation, z1 + z2 + z3 ∈ Λ and so Φ is ahomomorphism.

We have shown that every complex torus C/Λ is isomorphic as additive group tosome elliptic curve. This means that for every Φ : C/Λ→ E(C) there exists an inverseisomorphism Φ−1 : E(C)→ C/Λ. It is clear that Φ is holomorphic, as it is composed ofholomorphic maps. If we show that Φ−1 is also holomorphic, we have that every complextorus is isomorphic as Lie group to some elliptic curve.

Theorem 3.15. For every complex torus C/Λ and corresponding elliptic curve E(C),the isomorphism Φ−1 : E(C)→ C/Λ is holomorphic.

Proof. If we have that the derivative DΦ of Φ : C/Λ→ E(C) is non-zero on all of C/Λ,the inverse function theorem guaranties Φ−1 is holomorphic on E(C). By translation, it issufficient to show that DΦ is non-zero on some point Z0 ∈ C/Λ, as Φ is a homomorphism.That such a point z0 exists, follows from the fact that Φ is holomorphic and bijective,and that E(C) is non-singular.

23

It still remains to be shown that for each elliptic curve E(C), there exists some latticeΛ ⊂ C such that Φ gives an isomorphism between C/Λ and E(C).The question arises if we could find a lattice Λ ⊂ C such that for a given elliptic curveE(C) = {y2 = 4x3 − ax− b}, g2(Λ) = a and g3(Λ) = b.

To answer this question, we will need the so called j-invariant.

Definition 3.16. For an elliptic curve E = E(C) given by y2 = 4x3− ax− b, define thej-invariant by

j(E) =1728a3

a3 − 27b2.

Note that the denominator of the j-invariant is exactly the discriminant of the cubicpolynomial p(x) = 4x3 − ax− b. This polynomial has 3 distinct roots by lemma 2.2, sothe discriminant is non-zero and j is well defined.

We can also define the j-invariant for latices:

Definition 3.17. For a lattice Λ ⊂ C, define the j-invariant by

j(Λ) =1728g2(Λ)3

g2(Λ)3 − 27g3(Λ)2

Note that if we denote the elliptic curve corresponding to C/Λ by EΛ, that j(Λ) =j(EΛ). Hence, j(Λ) is well defined.

Proposition 3.18. For two lattices Λ,Λ′ ⊂ C, the following holds:

j(Λ) = j(Λ′) if and only if ∃α ∈ C such that Λ = αΛ′.

In that case, Λ and Λ′ are said to be homothetic.

Theorem 4.5. will tell us that such elliptic curves are isomorphic.

Proof. See the proof of theorem 16.5. in [3].

The j-invariant can be turned into a map j : H→ C, where H denotes the upper-halfplane: H = {z ∈ C | Im(z) > 0}.

Definition 3.19. For τ ∈ H, let Λτ = Z⊕ τZ. Then the j-function is defined by

j(τ) = j(Λτ ).

We know that the group SL2(Z) acts on H by(a bc d

)z =

az + b

cz + d.

The j-function is invariant under this action. To see this, let τ ∈ H and let τ ′ = aτ+bcτ+d

for a, b, c, d ∈ Z such that ad− bc = 1. Consider Λτ = Z⊕ τZ and Λτ ′ = Z⊕ τ ′Z. Thenwe see

(cτ + d)Λτ ′ = (cτ + d)Z⊕ (aτ + b)Z = Z⊕ τZ = Λτ .

24

So Λτ and Λτ ′ are homothetic, and hence j(τ) = j(Λτ ) = j(Λτ ′) = j(τ ′).

It follows that j gives a well-defined map j : H/SL2(Z) → C. We will state animportant proposition about this map that we will use later.

Proposition 3.20. The map j : H/SL2(Z)→ C is an analytic bijection.

Proof. See theorem 16.11. in [3].

We are now in the position to prove the following theorem, which is the last step toproving the Uniformization Theorem:

Theorem 3.21. For every elliptic curve E(C), there exists a lattice Λ ⊂ C such that Φgives an isomorphism between E(C) and C/Λ.

Proof. Let E(C) be defined by y2 = 4x3 − ax − b. As previously mentioned, it willsuffice to find a lattice Λ such that g2(Λ) = a and g3(Λ) = b, because the elliptic curvecorresponding with C/Λ through Φ is defined by y2 = 4x3 − g2(Λ)x− g3(Λ).Since the j-function is bijective, there exists τ ∈ H such that

j(τ) =1728a3

a3 − 27b2= j(E(C)).

Let Λτ = Z⊕ τZ. Then

1728a3

a3 − 27b2= j(τ) = j(Λτ ) =

1728g2(Λτ )3

g2(Λτ )3 − 27g3(Λτ )2.

Choose c ∈ C such that a = c−4g2(Λτ ). Multiplying both the numerator and thedenominator of j(Λτ ) by c−12 now yields b2 = c−12g3(Λτ ).If b = c−6g3(Λτ ) retain c. If b = −c−6g3(Λτ ), replace c by ci. So without loss of generalitywe have a = c−4g2(Λτ ) and b = c−6g3(Λτ ).For any α ∈ C, k > 2 and for any lattice Λ ⊂ C, we have that

Gk(αΛ) =∑

ω∈(αΛ)∗

1

ωk=∑ω∈Λ∗

1

(αω)k= α−k

∑ω∈Λ∗

1

ωk= α−kGk(Λ).

Because g2(Λτ ) = 60G4(Λτ ) and g3(Λτ ) = 140G6(Λτ ), it follows that a = g2(cΛτ )and b = g3(cΛτ ). We conclude that cΛτ = cZ ⊕ cτZ is our desired lattice and thatΦ : C/Λτ

∼−→ E(C).

Corollary 3.22 (Uniformization Theorem).For every lattice Λ ⊂ C, the torus C/Λ is isomorphic as complex Lie group to someelliptic curve E(C).Conversely, every elliptic curve E(C) is isomorphic as complex Lie group to a torus C/Λfor some lattice Λ ⊂ C.

Proof. This directly follows from theorem 3.14, theorem 3.15 and theorem 3.21.

25

3.3 How to get back

It is also possible to give a isomorphism of Lie groups from any elliptic curve E(C) toa torus. We will give an outline for this isomorphism, which is due to [4]. For thismap we will use integration along paths on E(C). Remember that all elliptic curves arehomeomorphic to a torus, and so H1(E(C),Z) ∼= Z⊕ Z. Let α, β be loops representingthe generators for H1(E(C),Z) and let ω be the differential 1-form given by

ω =dx

y.

This is a nowhere vanishing differential form on E(C) ⊂ CP2. Define the complexnumbers ω1 and ω2 by

ω1 =

∫α

ω, ω2 =

∫β

ω.

These complex numbers define a lattice Λ = ω1Z ⊕ ω2Z and a complex torus C/Λ.For each point P ∈ E(C), let γP : [0, 1] → E(C) be a path such that γP (0) = O andγP (1) = P . Now we define a map

Ψ : E(C)→ C/Λ

P 7→∫γP

ω + Λ.

This map is well defined: Suppose δ is another path joining O and P , then γP ∗ δ−1 isa loop and so [γP δ

−1] = m[α] + n[β] ∈ H(E(C),Z) for certain m,n ∈ Z. It follows that∫γP

ω =

∫γP δ−1δ

ω =

∫γP δ−1

ω +

∫δ

ω =

∫δ

ω +mω1 + nω2

which is the same equivalence class in C/Λ.It is also not difficult to see that Ψ(O) = 0 + Λ. It turns out that this map is actuallyan isomorphism of Lie-groups, but we will not show this here.This map does explain why elliptic curves are actually called elliptic curves: We areworking with integrals of the form∫

1√x3 + ax+ b

dx.

These integrals typically arise when calculating an arc length of an ellipse.

26

4 Elliptic curves with complexmultiplication

We will now turn our attention to elliptic curves with complex multiplication. These areelliptic curves that have a non-trivial endomorphism ring. The correspondence betweenelliptic curves and complex tori will allow us to consider these endomorphism rings asendomorphism rings of complex tori, which will make matters less complicated.We will determine which elliptic curves have a non-trivial endomorphism ring, we willcarefully calculate these endomorphism rings, and we will establish a bijection betweenelliptic curves that have a particular endomorphism ring and the class group. Thisbijection will be useful in the final chapter of this thesis.All material in this chapter was presented to me by Dr. Shen. Aside from the proofsof theorem 4.2. and proposition 4.7.,which were also presented to me by Dr. Shen, allproofs in this chapter are by myself.

4.1 Endomorphism rings

It turns out that the morphisms of elliptic curves are just holomorphic maps that preservethe identity. More formally:

Theorem 4.1. Let E and E ′ be elliptic curves over C. If φ : E → E ′ is a holomorphicmap such that φ(OE) = OE′, then φ is a homomorphism of Lie groups.

It can also be shown that the correspondence between elliptic curves and complex toridoes not only take objects to objects, but also morphisms to morphisms (see [3] theorem17.2.).Therefore, we only need to consider the endomorphism rings of complex tori if wewant to study the endomorphism rings of elliptic curves. Let’s try to characterize theseendomorphisms.

If E = C/Λ for some lattice Λ ⊂ C, then all elements of End(E) are in particularholomorphic maps.

Theorem 4.2. Let Λ ⊂ C be a lattice and let E = C/Λ. Let φ : E → E be a holomorphicmap such that φ(0) = 0. Then φ ∈ End(E).

Proof. Let p : C→ E be the quotient map. Because C is simply connected, there existsa unique continuous map φ̃ : C→ C determined by φ̃(0) = 0 such that φp = pφ̃, i.e. thefollowing diagram commutes:

27

C C

E E

φ̃

p p

φ

The continuous map φ̃ is actually holomorphic, as p is locally biholomorphic.Let z ∈ C and λ ∈ Λ. We then see that

p(φ̃(z + λ)− φ̃(z)) = p(φ̃(z + λ))− p(φ̃(z))

= φ(p(z + λ))− φ(p(z))

= φ(p(z))− φ(p(z)) = 0

We conclude that φ̃(z + λ)− φ̃(z) ∈ Λ.

For fixed λ ∈ Λ, the map ψ : z 7→ φ̃(z + λ)− φ̃(z) has to be constant. Evaluating in

z = 0 gives ψ(0) = φ̃(λ) and hence ψ(z) = φ̃(z) for all z ∈ C. It follows that

φ̃(z + λ) = φ̃(z) + φ̃(λ).

Now, taking the derivative of φ̃ shows

φ̃′(z + λ) = φ̃′(z + λ)

and so φ̃′(z) is holomorphic function that is periodic in Λ. If P is a fundamental paral-lelogram for Λ, then

φ̃′(C) = φ̃′(P) = φ̃′(P)

and since P is compact, we see φ̃′(z) is bounded. Since φ̃′(z) is holomorphic on C, φ̃′(z)must be a constant function by Liouville’s theorem. Therefore, integrating and usingthat φ̃(0) = 0, we see φ̃(z) = αz for some α ∈ C such that αΛ ⊂ Λ (this last statement

holds as we need φ̃(λ) ∈ Λ for all λ ∈ Λ.)It follows that φ : E → E is linear and hence φ ∈ End(E).

From the proof of theorem 4.2 we can conclude the following:

Corollary 4.3. Let E = C/Λ. Every holomorphic map f : E → E is given by

f(z) = φ(z) + z0,

where φ ∈ End(E) and z0 ∈ E.

Theorem 4.2 allows us to give an easier definition for the endomorphism ring of agiven elliptic curve:

Corollary 4.4. Let E = C/Λ. Then End(E) ∼= {α ∈ C | αΛ ⊂ Λ}.

Recall that two lattices Λ and Λ′ are said to be homothetic if there exists an α ∈ Csuch that Λ = αΛ′. It may come as no surprise that homothetic lattices give rise toisomorphic complex tori:

28

Theorem 4.5. For every lattice Λ ⊂ C and α ∈ C∗, we have

C/Λ ∼= C/αΛ.

Conversely, if C/Λ ∼= C/Λ′ for certain lattices Λ,Λ′ ⊂ C, then there exists an α ∈ C∗such that αΛ = Λ′.

Proof. The map C/Λ → C/αΛ, z 7→ αz is well defined, for if z and z′ are two repre-sentatives of the same equivalence class in C/Λ, then αz − αz′ = αλ ∈ αΛ for someλ ∈ Λ. Following the proof of theorem 4.2 we see that this map is holomorphic and ahomomorphism, and as it has an inverse (multiplication by α−1) it is an isomorphism.For the converse statement, let φ : C/Λ → C/Λ′ denote the isomorphism between thetwo complex Lie groups. The holomorphic group homomorphisms are linear maps, fol-lowing the proof of theorem 4.2. Since φ is an isomorphism, φ must be multiplicationby α for some α ∈ C∗. Because φ is bijective, we must have αΛ = Λ′.

This allows us to express every complex torus in a simpler manner:

Lemma 4.6. Up to isomorphism, every complex torus can be written as E = C/Λτ

where Λτ = Z⊕ τZ for some τ ∈ H.

Proof. Let Λ = ω1Z⊕ ω2Z ⊂ C be a lattice, and assume without loss of generality thatIm(ω2) > Im(ω1). Then ω2/ω1 ∈ H and

1

ω1

Λ = Z⊕ ω2

ω1

Z.

So setting τ = ω2/ω1, we see C/Λ ∼= C/Λτ . (Also compare this to definition 3.19.)

So without loss of generality, we will from now on assume that an elliptic curve C/Λis given by a lattice of the form Λ = Z⊕ τZ for some τ ∈ C.It might however still be that C/Λτ1

∼= C/Λτ2 for τ1, τ2 ∈ H. This happens preciselywhen:

C/Λτ1∼= C/Λτ2 ⇐⇒ ∃α ∈ C∗ such that αΛτ1 = Λτ2

⇐⇒ ατ1 = aτ2 + b and α = cτ2 + d

for a, b, c, d ∈ Z such that ad− bc = 1

⇐⇒ τ1 =aτ2 + b

cτ2 + d=

(a bc d

)τ2,

(a bc d

)∈ SL2(Z)

This gives the following result:

Proposition 4.7. There exists a bijection between H/SL2(Z) and {E = C/Λ} /∼= .

Compare this result to proposition 3.20.

29

4.2 Complex multiplication

For a given E = C/Λτ , it is clear that

Z ↪→ End(E) = {α ∈ C |αΛ ⊂ Λ}.

The question arises for which τ this is an isomorphism. Because 1 ∈ Λτ , we see that forall α ∈ End(E), α ·1 = α ∈ Λτ . So End(E) ↪→ Λτ and every α ∈ End(E) can be writtenas a+ bτ for certain a, b ∈ Z. We now see that

ατ = (a+ bτ)τ = aτ + bτ 2 ∈ Λτ .

This can be stated as aτ +bτ 2 = c+dτ for certain c, d ∈ Z, and so bτ 2 +(a−d)τ−c = 0,and we conclude the following:

Proposition 4.8. For a given E = C/Λτ , End(E) � Z if and only if Q(τ) is animaginary quadratic field extension of Q.

Definition 4.9. Elliptic curves that have a non-trivial endomorphism rings are said tohave complex multiplication.

Let’s try to find the endomorphism ring for a given τ explicitly:

Lemma 4.10. Let Q(τ) be a quadratic field extension and let E = C/Λτ . Then

End(E) = Z⊕ aτZ,

where a ∈ Z is the leading coefficient of the minimal polynomial of τ in Z[X].

Proof. We already established that End(E) ↪→ Λτ . Because Λτ is a free Z-moduleof rank 2, and End(E) is a submodule, End(E) must be free of rank ≤ 2. BecauseEnd(E) � Z, End(E) must be free of rank 2. Let ξ be a generator of End(E) suchthat End(E) = Z ⊕ ξZ. Then ξ = aτ for some a ∈ Z. It follows that aττ ∈ Λτ , soaτ 2 + bτ + c = 0 for some b, c ∈ Z. We conclude that a is the leading coefficient of theminimal polynomial of τ in Z[X], for if the minimal polynomial had a leading coefficienta′ that is smaller than a, then a′τ would be in Λτ , but there would be no n ∈ Z suchthat naτ = a′τ , giving a contradiction.

Finite field extensions K of Q have an important subring called the ring of integers.:

Definition 4.11. For a finite field extension K of Q, the ring of integers OK is definedas

OK = {α ∈ K | ∃f ∈ Z[X] with leading coefficient 1 and f(α) = 0}.

Any imaginary quadratic field extension K of Q can be written as K = Q(√−D) for

some D ∈ N. The following proposition will classify the ring of integers for such fieldextensions, but we will omit the proof:

30

Proposition 4.12. Let K = Q(√−D) for some D ∈ N. Then

OK =

{Z[1

2+ 1

2

√−D] if D ≡ 3 (mod 4)

Z[√−D] else

.

In particular, we see that OK is a lattice in C. In the rest of this chapter, K willdenote an imaginary quadratic field extension of Q and σ : K ↪→ C will denote anembedding of K into C.

Lemma 4.13. Let Λ = σ(OK) and let E = C/Λ. Then End(E) = OK.

Proof. This follows directly from the definition of OK and lemma 4.10.

4.3 The class group

We will now establish a bijection between the set of elliptic curves (up to isomorphism)whose endomorphism ring is equal to the ring of integers and the class group of K. Webegin with a definition:

Definition 4.14. A subset a of K is a fractional ideal if:

• a is a free Z-module of rank 2

• a is an OK-module.

Two fractional ideals a1 and a2 are said to be equivalent if there exists τ ∈ K such thata1 = τa2. This is denoted by a1 ∼ a2.

Proposition 4.15. Let a ⊂ K a fractional ideal and let Ea = C/σ(a). Then

End(Ea) = OK .

Proof. Clearly OK ⊂ End(Ea), as a is an OK-module.If a1 = τa2 for some τ ∈ C∗, then

End(Ea1) = {α ∈ C |αa1 ⊂ a1} = {α ∈ C |ατa1 ⊂ τa1} = End(Ea2).

Therefore, we may assume without loss of generality that a = Z⊕ωZ for some ω ∈ C, asthe endomorphism ring of Ea is the same as the endomorphism ring of Eτa. Now since1 ∈ a, we know End(Ea) ⊂ a and as established in lemma 4.10, End(Ea) = Z⊕ aωZ forsome a ∈ Z. Because (aω)2 ∈ End(Ea), we see (aω)2 + b(aω) + c = 0 for some b, c ∈ Z.Hence aω is a zero of some monic polynomial in Z[X], and hence aω ∈ OK . We concludethat End(Ea) ⊂ OK .

31

Definition 4.16. For an imaginary quadratic field extension K of Q, the class groupCl(K) is given by

Cl(K) := {a ⊂ K | a is a fractional ideal} /∼ .

Here ∼ is equivalence of fractional ideals.The group action is given by

Cl(K)× Cl(K)→ Cl(K)

([a1], [a2]) 7→ [a1a2].

We are now in a position to state and prove the main theorem of this chapter:

Theorem 4.17. Let K an imaginary quadratic field extension of Q. Then there existsa bijection

Ψ : Cl(K) −→ {E = C/Λ |End(E) = OK} /∼=

Proof. Define

Ψ : Cl(K)→ {E = C/Λ |End(E) = OK} /∼=[a] 7→ [C/σ(a)]

First we need to show this map is well defined:By proposition 4.15, End(C/σ(a)) = OK , so indeed

Ψ([a]) ∈ {E = C/Λ |End(E) = OK} /∼= .

Suppose a1 ∼ a2. Then by theorem 4.5 C/σ(a1) ∼= C/σ(a2) and as shown in the proof ofproposition 4.15 End(Ea1) = End(Ea2). It follows that Ψ does not depend on the choiceof representative and hence Ψ is well defined.Now we show Ψ is injective:Suppose Ψ([a1]) = Ψ([a2]). Then C/σ(a1) ∼= C/σ(a2). By theorem 4.5 there must existα ∈ C∗ such that a1 = αa2. Since a1 and a2 are subsets of K, it follows that α ∈ K.Hence a1 ∼ a2 and [a1] = [a2].Finally, we show Ψ is surjective:Let [E] ∈ {E = C/Λ |End(E) = OK} /∼= and let Λ ⊂ C be a lattice such that E = C/Λis a representative of [E]. Then Λ is a free Z-module of rank 2 by definition of a lattice,and as End(E) = OK by assumption, Λ is an OK-module. Hence, Λ is a fractional ideal,and Ψ([Λ] = [E].

Remark. The set J (OK) = {E = C/Λ |End(E) = OK} /∼= obtains a group structurethrough the bijection Ψ.

32

5 Applications to class field theory

In this chapter we will use the finiteness of the class group to show that for any E ∈J (OK) = {E = C/Λ |End(E) = OK} /∼=, the j-invariant j(E) is algebraic over Q. Thiswill be done by showing G = Aut(C/Q) acts on J(OK) = {j(E) |E ∈ J (OL)} and thatthe orbit of each j(E) is finite.We will discuss what it means for a field extension L/K to be ramified, and give someexamples. Then, we will briefly discuss the Hilbert class field and present a way toconstruct the Hilbert class field using elliptic curves with complex multiplication. Wewill conclude the chapter by showing that e

√163π is almost an integer.

The material and the proofs in this chapter were presented to me by Dr. Shen. I haveadded references to omitted proofs.

5.1 The j-invariant is algebraic

The aim of this section will be to show that the j-invariant of an elliptic curve that hasendomorphism ring OK is algebraic over Q.

In the previous chapter we have shown that for an imaginary quadratic field extensionK of Q, the set J (OK) is in bijection with the class group Cl(K). It has been shownthat the class group Cl(K) is finite (Minkowski), hence J (OK) is finite. We will usethis later in this section. First we will consider two propositions.

Proposition 5.1. Let G = Aut(C/Q) and let α ∈ C. Then

α is algebraic over Q ⇐⇒ |G · α| <∞.

Proof. (sketch)’⇒’: If α is algebraic over Q, then there exists a polynomial p ∈ Q[X] such that p(α) = 0.Then, for σ ∈ G, we have

p(σ(α)) =∑i

ciσ(α)i =∑i

σ(ci)σ(α)i = σ(p(α)) = σ(0) = 0,

So σ(α) is a root of p. It follows that |G · α| <∞ as p has only finitely many roots.’⇐’: Suppose α is transcendental over Q. Suppose t ∈ C is another transcendentalnumber over Q. Then the axiom of choice implies that there exists σ ∈ G such thatσ(α) = t. Since t was arbitrary, we see |G · α| =∞, contradicting our assumption.

Proposition 5.2. Let G = Aut(C/Q). Then G acts on J (OK).

33

Proof. Let E ∈ J (OK) be defined by y2 = 4x3 + ax+ b, for certain a, b ∈ C. Let σ ∈ Gand let Eσ be defined by y2 = 4x3 + σ(a)x+ σ(b). We claim that the map

G× J (OK)→ J (OK)

(σ,E) 7→ Eσ

defines an action of G on J (OK). For this we need to show that OK = End(E) =End(Eσ). Let f ∈ End(E) be given by f(x, y) = (p(x, y), q(x, y)). Now define

fσ : Eσ → Eσ

(u, v) 7→ σ(f(σ−1u, σ−1v))

= σ(p(σ−1u, σ−1v), q(σ−1u, σ−1v))

= (pσ(u, v), qσ(u, v))

It is easy to check that the map f 7→ fσ is a ring homomorphism. It also has ainverse, namely End(Eσ)→ End(E), f 7→ fσ

−1. Therefore, the endomorphism rings are

isomorphic. Hence G defines an action on J (OK).

We can now prove the following theorem:

Theorem 5.3. Let K be an imaginary quadratic field extension of Q. Then j(E) isalgebraic over Q for all E ∈ J (OK).

Proof. We have seen that G = Aut(C/Q) acts on J (OK). If E ∈ J (OK) and σ ∈Aut(C/Q) we see that

j(Eσ) =1728σ(a)3

σ(a)3 − 27σ(b)2= σ

(1728a3

a3 − 27b2

)= σ(j(E)).

Now we use the fact that #J (OK) = #Cl(K) <∞. It follows that

{σ(j(E)) |σ ∈ Aut(C/Q)} <∞.

Now it follows from proposition 5.1 that j(E) is algebraic over Q.

5.2 Ramification

To introduce the Hilbert class field and see an application of elliptic curves with complexmultiplication, we will first need to discuss what ramification is.We already used the ring of integers OK for a finite field extension K of Q. The ring ofintegers OK is a Dedekind domain, that is, it has the following properties:

• It is a Noetherian domain.

• It is integrally closed.

• All non-zero prime ideals are maximal.

34

OK is not necessarily a unique factorization domain, but unique factorization holds forideals [6]. This means that if a is an ideal in OK , then there exists a unique factorizationa = pe11 · . . . · penn for certain non-zero prime ideals p1, . . . , pn and integers e1, . . . , en.Now consider finite field extensions Q ⊂ K ⊂ L and their rings of integers:

B OL L

p = B ∩ OK OK K

Z Q

For any prime ideal p ⊂ OK there exists a unique factorization pOL = Be11 · . . . ·Ben

n

of prime ideals Bi, 1 ≤ i ≤ n in OL. If L/K is a Galois extension, then Gal(L/K) actstransitively on {B1, . . . ,Bn} and so e1 = e2 = . . . = en.A prime ideal p ⊂ OK is unramified if ei = 1 for all 1 ≤ i ≤ n in the factorizationpOL = Be1

1 · . . . ·Benn . A field extension L/K is unramified if all prime ideals p ⊂ OK

are unramified.

Example 5.4. Let t = s2 and consider the prime ideal (t− a) ⊂ C[t], a ∈ C:

C[s] C(s)

(t− a) C[t] C(t)⊂

We see (t− a)C[s] = (s2 − a)C[s] = (s−√a)(s+

√a)C[s]. So the prime ideal (t− a)

ramifies if a = 0, and is unramified if a 6= 0.

If K ⊂ L is a field extension, and OL = OK [α] for some α ∈ L, there is an easyway to detect ramification. Let f ∈ OK [X] be the minimal polynomial of α. Then

OL = OK [X]/f(X) . If p ⊂ OK is a prime ideal, then we see:

OL/pOL = OL

/∏Beii

= OL/Be1

1× . . .×OL

/Benn.

On the other hand, if we set f = f (mod p) = f e11 · . . . · f enn , we see that

OL/pOL = OL ⊗OK OK

/pOK = OK [X]

/f(X) ⊗OK OK

/pOK

= OK/p[X]/f(X) = OK/p[X]

/f e11 (X) × . . .×OK/p[X]

/f enn (X) .

Hence, all we need to detect ramification is the factorization of f (mod p).

35

Example 5.5. Let’s try to detect ramification in imaginary quadratic field extensionsof Q. So let K = Q(

√−D) for D ∈ Z≥1, and let p ∈ Z prime. We want to know for

which p the ideal (p) is ramified, and for which p we have that pOK = B1B2.First suppose D 6≡ 3 (mod 4). Then OK = Z[

√−D] = Z[X]/(X2 + D). Now we see

that:

(p) ramifies ⇐⇒ X2 +D = (X + τ)2 = X2 + 2τX + τ 2 ∈ Fp[X] for some τ ∈ Fp⇐⇒ τ 2 = D ∧ 2τ = 0 in Fp⇐⇒ p = 2 ∨ p | D

And, if p is unramified, we see that:

pOK = B1B2 ⇐⇒ X2 +D = 0 has a solution in Fp

⇐⇒(−Dp

)= 1

Now suppose that D ≡ 3 (mod 4). Then OK = Z[12

+ 12

√−D] = Z[X]/(X2−X + D+1

4).

Following the same line of reasoning, we see that:

(p) ramifies ⇐⇒ X2 −X +D + 1

4= (X + τ)2 = X2 + 2τX + τ 2 ∈ Fp[X] for some τ ∈ Fp

⇐⇒ 2τ = −1 ∧ τ 2 =D + 1

4∈ Fp

⇐⇒ p 6= 2 ∧ 1

4=D + 1

4⇐⇒ p 6= 2 ∧ p | D

And, if p is unramified, we see that for p = 2:

pOK = B1B2 ⇐⇒ X2 −X +D + 1

4= 0 has a solution in Fp

⇐⇒ D + 1

4≡ 0 (mod 2)

⇐⇒ D ≡ 7 (mod 8)

and for p 6= 2 ∧ p - D:

pOK = B1B2 ⇐⇒ X2 −X +D + 1

4= 0 has a solution in Fp

⇐⇒ (X − 1/2) +D/4 = 0 has a solution in Fp

⇐⇒

(D̃

p

)= 1, where D̃ ∈ Z is a representative of D/4

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5.3 Hilbert class field

We are now in a position to formulate an application of the theory discussed in this the-sis to class field theory. Let K = Q(

√−D) for some D ∈ Z≥1. A maximal unramified

abelian extension of K is a field extension L ⊃ K such that L is unramified, Gal(L/K) isabelian, and every other field extension of K containing L is not abelian or unramified.Such an extension exists and this extension is called the Hilbert class field. It can beshown that the Galois group Gal(L/K) of such an extension is canonically isomorphicto the class group Cl(K).

Elliptic curves can be used to construct the Hilbert class field of K. This can be doneby fixing an embedding σ : K ↪→ C, and considering the elliptic curve E = C/σ(OK).We have seen that End(E) = OK , so E ∈ J (OK). It turns out that the Hilbert classfield L is now given by:

L = K(j(E)).

A proof of this construction can be found in [7]. We also have the following theorem:

Theorem 5.6. With the above notation, j(E) ∈ OL.

Proof. Multiple proofs can be found in [7].

These two statements have a rather remarkable corollary:

Example 5.7. Let K = Q(√−163) and let τ = 1

2+ 1

2

√−163. As 163 ≡ 3 (mod 4), it

follows thatOK = Z[τ ].

Let L be the Hilbert class field of K. It is known that Cl(K) = {1}, hence the Galoisgroup Gal(L/K) is trivial and so L = K. Let σ : K ↪→ C be an embedding and letE = C/σ(OK). Then, by theorem 5.6, we have

j(E) ∈ OL = OK = Z⊕ τZ.

By definition of the j-function, we now see that j(τ) = a+ bτ = (a+ b/2) + (b√

163/2)ifor certain a, b ∈ Z. The Laurent series of the j-function is given by:

j(τ) =1

q+ 744 + 196884q + 21493760q2 + ...

where q = e2πiτ = −e−√

163π. It follows that j(τ) is a real number, so b = 0. The higher

order terms are all very close to zero, so −e√

163π + 744 is almost equal to a ∈ Z. Itfollows that the transcendental number e

√163π is almost equal to an integer. Indeed:

e√

163π = 262537412640768743.99999999999925007...

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6 Conclusion and discussion

We have introduced elliptic curves over C in the second chapter by giving both a geo-metric and an algebraic approach to the definition and the group law, and by showingelliptic curves over C are complex Lie groups. In the third chapter we have carefullyproven the uniformization theorem, which we used in the forth chapter to discuss ellipticcurves with complex multiplication. The bijection between J (OK) and Cl(K) estab-lished in this chapter has been used to carry out the construction of the Hilbert classfield in chapter five. We concluded the thesis with a remarkable result: That e

√163π is

almost an integer.There is plenty of room for further study, both within and outside of the realm of ellipticcurves. One could for example focus on torsion points of elliptic curves over C, on therational points of elliptic curves over Q, or on elliptic curves over finite fields. We haveonly scratched on the surface of class field theory, and there are many other applications,perhaps most noteworthy in the proof of Fermat’s last theorem.

The contents of the final version of this thesis are mostly in line with the original planDr. Shen set for me. I have predominantly diverged from this plan in the proof of theUniformization Theorem, as the original plan was to use the inverse map discussed insection 3.3. I personally intended to discuss the Riemann-Roch theorem as well, butthis proved to be too ambitious. I have studied the algebraic geometry behind it, but itwould have been too much to discuss in this thesis.The process of writing this thesis consisted of two main parts. In the first months, mysupervisor and I held regular meetings. In these meetings, Dr. Shen would present newmaterial to me, and would usually leave me an exercise. I would then try to make senseof what we had discussed, and prepare for the next meeting. I only started writing afterone of our last regular meetings. During this period, I would contact my supervisor whenI got stuck. Overall, I was very satisfied with this process, although it might have beenbetter to start writing a little sooner, as I often discovered that I did not understandthe material quite as well as I thought I did.

Finally, I would like to express my gratitude to my supervisor Dr. Mingmin Shen, forgiving me so much of his time, and for introducing me to this topic, which is related tomost of the areas of mathematics that have fascinated me so far. It has given me insightinto a very interesting branch of mathematics, and I am determined to learn more aboutelliptic curves and related fields. I would also like to thank Prof. Dr. Jan Wiegerinck,for refreshing my memory on some complex analysis, Dr. Arno Kret, for taking the timeto discuss some material with me, and Dr. Chris Zaal, for advising me to request Dr.Shen to be my supervisor.

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7 Popular summary

An elliptic curve is the set of solutions to an equation of the form

y2 = x3 + ax+ b.

The following figure illustrates what these curves look like:

What’s interesting about these curves, is that there is a way to ’add’ points on thecurve, much like we can add real numbers. We will illustrate how this works:Consider the elliptic curve E : y2 = x3− 2x+ 2, and choose two points P and Q on thiscurve. Draw a line through these points. This line will intersect the curve at a thirdpoint, we will call this point R. Now, the reflection of this point through the x-axis isthe sum of P and Q, denoted P ⊕Q. This process is illustrated below:

x

y

y2 = x3 − 2x+ 2

P

QR

P ⊕Q

39

In this thesis we discuss elliptic curves, for which x and y are not real, but complexnumbers. We are not able to draw these elliptic curves, but the same method of ad-dition still works. However, it is not hard to imagine that addition on these curves iscomplicated, so in this thesis we present a solution to this problem: We show that everycomplex elliptic curve is actually a torus (more commonly known as a doughnut). Theseare a lot easier to imagine, and addition on a torus is easier to understand.

This identification with tori allows us to also define a form of multiplication on certainelliptic curves, which are called elliptic curves with complex multiplication. We discusssome properties of such objects and in the final chapter we show how these objects canbe used in a related field of mathematics.

40

Bibliography

[1] J H Silverman, The Arithmetic of Elliptic Curves, Second Edition, Graduate Textsin Mathematics, Springer 2009.

[2] B Edixhoven, L Taelman, Algebraic geometry, Syllabus to the MasterMath course,Univeristy of Leiden, 2016, http://pub.math.leidenuniv.nl/ edixhovensj/ag.pdf

[3] A V Sutherland, Lecture Notes on Elliptic Curves, Course 18.783, MassachusettsInstitute of Technology, 2015, http://math.mit.edu

[4] T Ward, D Geraghty, Lecture notes on Elliptic Curves, Lecture 2 and 3, Notes forthe Part III Course, Warwick University, 2005, http://homepages.warwick.ac.uk

[5] N Boston, D Hast Notes on Elliptic Curves, Arithmetic Geometry, andModular Forms, Course 844, 2013, University of Wisconsin-Madison,https://www.math.wisc.edu/

[6] K Conrad, Ideal Factorization, University of Connecticuthttp://www.math.uconn.edu/ kconrad/blurbs/gradnumthy/idealfactor.pdf

[7] J H Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Graduate Textsin Mathematics, Springer, 1999

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