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Lectures On Algebraic Number

Theory

Dipendra Prasad

Notes by Anupam

1 Number Fields

We begin by recalling that a complex number is called an algebraic number if

it satisfies a polynomial with rational coefficients or equivalently with integer

coefficients. A complex number is called an algebraic integer if it satisfies

a polynomial with integral coefficients having leading coefficient as 1. Let

Q be the set of all algebraic numbers inside C. It is well known that Q is asubfield ofC. Any finite extension ofQ is called an Algebraic Number Field.

Some of the most studied examples of number fields are:

1. Q, the field of rational numbers.

2. Quadratic extensions Q(

d); where d Z is a non-square integer.

3. Cyclotomic fields Q(n), n = e2in .

4. Kummer extensions K(a

1n

) where a K and K is a number field.Definition. A Dedekind domain is an integral domain R such that

1. Every ideal is finitely generated;

2. Every nonzero prime ideal is a maximal ideal;

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3. The ring R is integrally closed (in its field of fractions).

Let K be a number field. Let us denote the set of algebraic integers inside

K by OK.

Theorem 1.1. 1. If K is a number field then OK, the set of algebraic

integers inside K, forms a subring of K. This subring OK is a finitely

generatedZ module of the same rank as degQK.

2. The ring OK is a Dedekind domain. In particular, every ideal in OK

can be written uniquely as a product of prime ideals.

Unlike Z, OK need not be a PID and typically it is not. Gauss conjecturedthat there are exactly nine imaginary quadratic fields which are PID, and

this was proved in 60s by Baker and Stark. These are Q(d) for d =

3, 4, 7, 8, 11, 19, 43, 67, 163. Using Minkowski bound on class number (see

later section) it is easy to prove that these are PID. Gauss also conjectured

that there are infinitely many real quadratic fields which are PID. This is

NOT YET SOLVED. The following theorem will be proved in one of the

later lectures. Let us denote OK for the units in OK i.e. elements in OKwhich has inverse in OK. Then O

K is a group.

Theorem 1.2 (Dirichlet Unit Theorem). The group OK is finitely generated

abelian group. The rank of OK is = r1 + r2 1 , where r1 is number of realembeddings of K inR and r2 is number of complex embeddings of K inC up

to complex conjugation.

As some simple consequences of the theorem we note the following.

1. The group OK is finite if and only ifK = Q or is a quadratic imaginary

field.

2. If K is real quadratic, then OK is Z/2 Z. If K = Q(d) thenOK = Z(

d) if d 1(mod4) and OK = Z( 1+

d

2) if d 1(mod4).

Thus existence of units inside OK is equivalent to solving the Pells

equation a2 db2 = 1. (Observe that x OK is unit if and only ifxx = 1).

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Here are few Questions about quadratic fields which are still unsolved !

1. For which real quadratic fields does there exist a unit with norm = 1.

2. Are there estimates on the fundamental unit ofQ(

d), d > 1 in terms

of d?

Exercise: If K = Q[m1n ] , then OK contains Z[m

1n ] as a subgroup of

finite index. Calculate the index ofZ[m1n ] in OK.

2 Class Group of A Number FieldDefinition. A nonzero additive subgroup A K is called a fractional idealif there exists K such that A OK and AOK A.

It is easy to see that nonzero fractional ideals form a group under multi-

plication with OK as the identity and

A1A2 =

ii / i A1, i A2

.

Definition.

Class Group :=The group of all nonzero fractional ideals

group of all nonzero principal ideals.

Theorem 2.1. The Class Group of a number field is finite.

Remark: The finiteness of class group is not true for arbitrary Dedekind

domains. It is a special feature of number fields.

There exists an analytic expression for the class number as the residue

of the zeta function associated to the number fields, which is given by the

following theorem. The theorem will be proved later.

Theorem 2.2 (Class Number Formula). LetK be a number field. Then,

K(s) =2r1(2)r2hR

w

dK.

1

s 1 + holomorphic function

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where r1 is number of real embeddings of K inC, r2 is number of complex

embeddings of K inC upto complex conjugate, h is class number of K, R isthe regulator of number field K, w is number of roots of unity in K and dK

is the modulus of discriminant.

3 More Algebraic Background

Let K be a number field. It is sometimes convenient to write it pictorially

as followsOK

K

Z Q

Let p Z be a prime. Denote the decomposition of the ideal (p) = pOKin OK as pOK =

ni=1 p

eii with pi prime ideal in OK. This ei is called

ramification index ofpi over p.

One of the basic questions in algebraic number theory is the following.

Given a polynomial f(x) Z[x], look at the reduced polynomial modulop as polynomial in Z/p[x], say f(x), and write it as product of powers of

irreducible polynomials f(x) =

fi(x)ei in Z/p[x]. The question is whether

one can describe a LAW which tells us how many of fis occur and what are

the possible degrees in terms of p.

Example : Let f(x) = x2 5, then quadratic reciprocity law gives theanswer. Let p be an odd prime. Then x2 5 = 0 has solution in Z/p if andonly if there exist x0 Z/p such that x20 = 5. And x20 = 5 has solution inZ/p if and only if 5

p12

1(mod p) ( which is by defenition denoted as ( 5

p)

= 1). This can be seen by noting that (Z/p) is a cyclic group. So if p/5then f(x) = x2. I f ( 5

p) = 1 then f(x) = (x x0)(x + x0), where x20 = 5 in

Z/p. And if ( 5p

) = 1 then f(x) is irreducible in Z/p.We will later prove the following theorem. There exists a law of decom-

posing polynomials mod p as primes in arithmetic progression if and only if

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the polynomial defines an abelian extension ofQ.

Let us look at following field extension.

OK K Z Q

Let p Z be a prime and let p be a prime ideal lying over p in OK. LetpOK =

gi=1 p

eii . Then we get field extensions of finite fields Z/p

OK/p

of degree, say fi. This fi is called residual degree of pi over p.

Lemma 3.1. Let K be a number field. Theng

i=1 eifi = [K : Q], where the

notations are defined as above. If K is Galois extension ofQ, then Galois

group acts naturally on the set of primes pi lying above p. In this case fis are

the same, eis are the same and ef g = [K : Q], where e is the ramification

index, f is the residual degree and g is the number of primes lying over p.

4 Artin Reciprocity TheoremLet K be a number field which is Galois over a number field k. We keep

notations of previous section. Let p be a prime ideal in OK lying over a prime

ideal p in Ok. Let Dp G denote the subgroup of Galois group Gal(K/k)preserving the prime p; the subgroup Dp is called the Decomposition group

of the prime p. We have following exact sequence associated to the prime

ideal p.

0 Ip Dp GalOK/p

Ok/p 0The kernel Ip is called the inertia group of p.We call K/k is unramified if e = 1 Ip = 1. In case of the finite

field extension Fq Fqe, the map : Fqe Fqe, x xq is called theFrobenius element. It generates the Galois group ofFqe over Fq. In the case

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of unramified extension the decomposition group gets a prefered element,

called the Artin symbol p =< p, K/k > Gal(K/k).The following is the most fundamental theorem in algebraic number the-

ory, called the Artin Reciprocity Theorem.

Theorem 4.1 (Artin Reciprocity Theorem ). Let K be an abelian extension

overQ. Then Artin symbol defines a map from I(C) Gal(K/Q), whereC is some ideal in K and contains in its kernel PC (group of principal ideals

containing a generator which is congruent to 1 mod C). Thus Gal(K/Q) is

a quotient of I(C)/PC, generalised class group.

5 Quadratic form associated to a number field

Let K be a number field. Let x K. Considering K as a Q vector space,we get a Q-linear map lx : K K defined by lx() = x. Define trKQ (x)as trace of lx and N m

KQ (x) as determinant of lx. If there is no possibility of

confusion, we will denote trKQ (x), N mKQ (x) simply as tr and Nm.

Example : If p Q then tr(p) = p.[K : Q] and N(p) = p[K:Q].

Define a bilinear form B : K K Q by B(x, y) = tr(xy). One canrestrict this bilinear form to the ring of integers OK to get a bilinear form

B : OK OK Z. Recall that OK is a free Z module of rank same as degof K/Q.

Definition. Let B be a bilinear form B : Zd Zd Z. One can associatean integer, called the discriminant of the quadratic form, and denoted be dB

to be det(B(ei, ej)), where {e1,...ed} is an integral basis ofZd.

Note that dB does not depend on the basis chosen as can be seen as

follows. Let B be another integral basis ofZd. Then there exists an integeralmatrix A such that B = ABAt. Hence,

dB = det B = det(ABAt) = (det A)2 det B = det B

Here we have used that for A GL(n,Z), det A = 1 or 1. This discrim-inant of the trace form on the ring of integers of a number field is called the

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discriminant of number field.

The following theorem will be proved later.

Theorem 5.1 (Minkowski). Let K be a number field, with dK as its dis-

criminant. Then |dK| > 1 if K = Q.Problem of determining the fields of given discriminant is yet to be AN-

SWERED.

Exercise: Quadratic form associated above to a number field does not

determine number field uniquely. As specific examples, construct cubic num-

ber fields with same quadratic form. This question will be answered later.

Note, however, a quadratic number field is determined by the quadratic form

uniquely.

6 Dirichlet Unit Theorem1

Dirichlet Unit Thoerem is a statement about the structure of the unit group

of the ring of integers of a number field. What it says is that such a group

is necessarily finitely generated and it gives the rank as well as a description

of the torsion part. The proof however, in its natural set up, belongs to therealms of Geometric theory of numbers, also reffered as Minkowski Theory.

So let us take a birds eye view of Minkowski Theory.

The central notion in Minkowski Theory is that of a Lattice. A Lattice

for us means the following,

Definition. LetV be an n-dimensionalR-vector space. A Lattice inV is a

subgroup of the form

= Zv1 + + Zvmwith linearly independent vectors v1, , vm of V. The set

= {x1v1 + + xmvm| xi R, 0 xi 1}

is called a fundamental mesh of the lattice.

1This section is written by Purusottam Rath.

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The lattice is said to have rank m and is called a complete lattice if its

rank is equal to n. Let us enumerate some properties of a lattice.

Lemma 6.1. A subgroup ofV is a lattice if and only if it is discrete.

Note that Z+ Z

3 is not a lattice in R.

Lemma 6.2. A lattice inV is complete if and only if there exists a bounded

subsetM V such that the collection of all translates M + , coversthe whole space V.

Now suppose V is a euclidean vector space, hence endowed with an innerproduct

( , ) : V V RThen we have on V a notion of volume (more precisely a Haar measure).

Then if is a fundamental mesh of a lattice with basis vectors v1, , vmthen we define volume of as

vol := vol ()

Note this is independent of choice of basis vectors of the lattice as any suchtwo have a change of basis matrix in SLm(Z).

We now come to the most important theorem about lattices. A subset

X of V is called centrally symmetric, if for any x X, the point x alsobelongs to X . It is called convex if for any two points x, y X the linesegment {ty + (1 t)x| 0 t 1} joining x and y is contained in X. Withthese definitions we have,

Theorem 6.1 (Minkowski Theorem). Let be a complete lattice in a eu-

clidean vector space V and letX be a centrally symmetric, convex subset ofV. Suppose that

vol(X) > 2n vol()

ThenX contains at least one nonzero lattice point .

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The Minkowski Theorem stated above would be crucial to our proof of

the unit theorem. In fact it plays a pivotal role in number theory and hassome really non-trivial applications. For instance we can prove the famous

four-square theorem using this as is done below.

Theorem 6.2 (Four Square Theorem). Every positive integer is a sum of

four squares.

Proof : Suffices to show that any odd prime is a sum of four squares.

Let p be any odd prime. Now there exists positive integers a, b such that

a2

+ b2

+ 1 0 ( mod p). Conside the lattice in R4

with basis vectors

v1 = (p, 0, 0, 0), v2 = (0, p, 0, 0),

v3 = (a,b, 1, 0), v3 = (b, a, 0, 1).

Any vector v in has the form (r1p + r3a + r4b, r2p + r3b r4a, r3, r4). Thuswe see that the integer |v|2 is a multiple of p.

Consider the open ball B in R4 of radius 2p centered at origin. It is aconvex centrally symmetric set with volume 2

2

p2

which is strictly greaterthan 24p2. Since has volume p2, by Minkowski Theorem, B contains anon-zero point u = (A,B,C,D), say.

Now, 0 < |u|2 < 2p and |u|2 is a multiple ofp. Hence |u|2 = p, that is,A2 + B2 + C2 + D2 = p.

The basic idea in the proof of Minkowski Unit Theorem is to interpret

the elements of a number field K over Q of degree n as points lying in an

n-dimensional space. We consider the canonical mapping

j : K KC = C

a ja = ( a)

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which results from the n embeddings ofK in C. Let these embeddings be

given by the ordered set

X = {1, , r, 1, 1, , s, s}

where the s give the r real embeddings and s and s give the 2s pairwise

conjugate complex embeddings. We see that the image of K under this

canonical mapping into Cn =

X C actually lies inside the following set

known as the Minkowski Space.

KR = (z) XC | z R, z = z

We note that this space is isomorphic to

X R = Rr+2s given by the map

F : (z) (x) where x = z, x = Re(z), x = Im(z). However thismap is not volume preserving and

V olcanonical(Y) = 2s V olLebesgue(F(Y))

where the canonical volume is the volume on KR induced from the stan-

dard inner product onCn

. A little reflection will show that the mappingj : K KR identifies the vector space KR with the tensor product KQR,

KQ R KRa x (ja)x

Lemma 6.3. If I = 0 is an ideal in OK, the ring of integers ofK , then := j(I) is a complete lattice inKR with volume

|dK|.[OK : I], where dKis the discriminant ofK.

Proof : Just note that I has an integral basis i.e. a Z basis 1, , n,so that = Zj1 + + Zjn. Consider the matrix A =(l(i)) with srunning over the embeddings ofK in C. Then we have

ji, jk =

1lnl(i)l(k) = AA

t

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Then we get

V ol() = |det(ji, jk)| 12 = |det(A)| = |dK|.[OK : I] So via the canonical embedding j : K KC, we can identify the ideals

of OK with lattices in Rn. But since we are interested in the units of the

ring of integers, OK, we pass on to the multiplcative group K by using the

standard logarithm map

l : C R, z log|z|

It induces the surjective homomorphism

l : KC =

C

R , (z) (log|z|

Note that image ofKR under the above map lies in the following set

R

+:=

(x)

R | x R, x = x

Further the multiplicative group KC admits the homomorphism N : KC

C given by the product of the coordinates and we have the homomorphism

T r :

R

+ R

given by the sum of the coordinates. Also note that the R-vectorspace

[

R]+ is isomorphic to Rr+s. Now conside the following subgroups

OK = { OK | NK|Q = 1}S = {y KR | N(y) = 1}

H = {x [R]+ | T r(x) = 0}

It is clear that j(OK) Sand l(S) H. Thus we obtain the homomorphisms

OKj S l H

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and the composite := l o j : OK

H. The image will be denoted by

= (OK) Hand we obtain the

Theorem 6.3. LetK be a number field. Then the sequence

1 (K) OK 0is exact, where (K) is the group of roots of unity that lie inK.

Proof : The only non obvious part is to show that ker() lies in OK.

Now () = l(j) = 0 implies|

|= 1 for each embedding ofK. Thus

(j) = ( ) lies in a bounded domain of the R-vector space KR. But (j)

is a point in the lattice jOK ofKR. Therefore the kernel of can contain

only finitely many elements, and thus, being a finite group, contains only the

roots of unity in K.

Given this theorem, it remains to determine the group . For this we need

the following lemma which depends on Minkowski Lattice Point Theorem.

Lemma 6.4. Let I = 0 be an integral ideal of K, and let c > 0, for

Hom(K , C), be real numbers such that c = c and

c > A[OK : I]

where A = ( 2

)s|dK|. Then there exists a I, a = 0, such that

| a| < c for all Hom(K,C)Proof : We just note that the the set

Y = {(z) KR | |z| < c}is centrally symmetric, convex and has volume 2r+ss c. Now, = j(I)is a lattice with volume |dK|.[OK : I]. So we have

Vol(Y) > 2r+ss(2

)

s|dK|.[OK : I] = 2nVol()

So the assertion now follows from Minkowski Lattice Point Theorem.

Finally we determine the structure of .

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Theorem 6.4. The group is a complete lattice in the (r + s

1)- dimen-

sionalR-vector space H .

Proof : Since [

R]+ has dimension (r+s), H is of dimension (r+s1) ,

being the kernel of the linear functional Tr. The mapping := l o j : OK Harises by restricting the mapping

Kj

C l

R

Thus it suffices to show, that for any c > 0, the bounded domain

{(x) R | |x| c} contains only finitely many points of . Sincel((z)) = (log|z|), the preimage of this domain with respect to l is thebounded domain

(z)

C | ec |z| ec

But it contains only finitely many points of the set j(OK) as this is a subset

of the lattice j(OK). Therefore is a lattice in H. Finally we show that

is a complete lattice in H. We chose real numbers c > 0, for Hom(K ,C), satisfying c = c and

C =

c > A[OK : I]

where A = ( 2

)s|dK|, and we consider the set

W = {(z) KR | |z| < c}

For an arbitrary point y = (y) S, it follows that

W y = {(z) KR | |z| < c}where c = c|y| and one has c = c,

c =

c = C because

|y| =

|N(y)| = 1. Then, there exists a point

ja = ( a) Wy, a OK, a = 0

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Now, upto multiplication by units, there are only finitely many elements

OK of a given norm NK|Q = a. Thus we may pick up a system of1, , M OK, i = 0 such that every a OK with 0 < |NK|Q(a)| C isassociated to one of these numbers. The set

T = SM

i=1

W(ji)1

is a bounded set as W is bounded and we have

S =

KT j

Because for any y S, we find, by the above, an a OK, a = 0, such thatja W y1, so ja = xy1 for some x W. Since

|NK|Q(a)| = |N(xy1)| = |N(x)| 0}{0}. This is an ideal in A.Any element in A m is invertible and hence A is a local ring with m as itsmaximal ideal, in fact, A is PID. Without loss of generality one can assume

that image of v is Z. It can be seen that m is generated by any element

A such that v() = 1. Elements A such that v() = 1 are calleduniformizing elements. A is thus a local integral domain of dim 1, which can

be checked to be integrally closed in K.

Definition. An absolute value on K is a map to positive reals |.| : K Rsuch that

1. |xy| = |x||y| and |x| = 0 if and only if x = 0.

2. |x + y| |x| + |y|.If we have the stronger property,

3. |x + y| max{|x|, |y|}then the absolute value is called a non-archmedian absolute value.

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Examples :

1. The usual absolute value on Q.

2. For p-adic valuations on Q we can define absolute values as |x|p =pvp(x). This is a non-archmedian absolute value on Q.

The absolute values on fields define metric and hence we can talk of

completion of the field with respect to the metric. Two absolute values

|.| and |.| on field K are equivalent if there is a positive constant t suchthat

|a

| =

|a

|t

a

K. Equivalent absolute values give rise to equivalent

topology on the field. A place on the field K is an equivalence class of non-

trivial absolute values. Completion ofQ with respect to absolute value vp is

defined to be Qp, called p-adic field. These correspond to places which are

called finite places. Note that with respect to usual absolute value on Q the

completion is R, which corresponds to the infinite place.

An element in Qp is a sequence {an} of rational numbers such that forevery i > 0 ; pi|am an for all m,n >> 0 or equivalentily vp(am an) 0.

Another way to look at Qp is to define Zp by inverse limit:

Zp = lim Z/pn

Then Qp = Zp

ZQ is a field containing Zp such that Qp = Zp[1p

].

Theorem 7.1 (Ostrowskis Theorem). Let K be a prime global field i.e.

K = Q or K = Fq(t). Then

1. Suppose thatK = Q. Then every non-trivial place of K is represented

by either the usual absolute value, denoted as |.|, or ap-adic one |.|p, for some prime p.

2. Suppose that K = Fq(t) and let R = Fq[t]. Then every non-trivial

place of K is given by either the infinite place |.| defined by |f /g| =qdeg(f)deg(g) or by the finite place |.|p corresponding to an irreduciblepolynomialp(t) R.

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Let K be a number field with OK as its ring of integers. Let p be a

prime ideal is OK. Let x K be a nonzero element. Look at the fractionalideal xOK and write it as product of prime ideals xOK =

pnp(x). Define

a discrete valuation on K as vp : K Z as vp(x) = np(x). Using this

discrete valuation we can define absolute value on K as |.|p : K R0 by|x|p = (#OK/p)np(x). The completion of K with respect to this absolutevalue is written as Kp.

Any element of a discrete valuation ring A looks like

a0r + a1

r+1 + ....

where ais are representations of A/m , a0 = 0 in A/m and r Z.Example : Let us take K = C[|t|][t1], which is a field. Any element of

this field looks like

f(t) =

nmZant

n

where an C.We have thus defined completion of a number field at various prime ide-

als. These prime ideals are also called finite places of the number field. There

are infinite places (or archmedian absolute values) which correspond to em-beddings of the number field K in C. Two embeddings of K into C are said

to define the same infinite place if and only if they are complex conjugate of

each other. The set of finite places together with infinite places constitute

the set of places of the number field K.

There is another way of looking at completion of a number field. If K is

a finite field extension ofQ then K

QQp is a separable algebra which is a

product of fields.

KQ

Qp = v|p

Kv

Similarly,

KQ

R = Rr1

Cr2

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8 A sample calculation of the ring of integers

of a number field

We will calculate the ring of integers of the field K = Q(p), p = e2ip . The

Galois group in this case is Gal(K/Q) = (Z/p). In general Q(n) are called

cyclotomic fields. These have Galois group which are abelian and isomorphic

to (Z/n).

The following theorem will be proved later as a consequence of the class

field theory.

Theorem 8.1 (Kronecker). Let K be a number field which is an abelian

extension ofQ, i.e. K is Galois overQ with abelian Galois group. Then K

is contained inQ(n), for some n.

Proposition 8.1. LetK = Q(p), p = e2ip . Then the ring of integers of K

is OK = Z[p].

Remark: The theorem is true for p any positive integer. But we will

prove here for p a prime. Also ring of integers need not be generated by a

single element but it is always generated by at most two elements.Proof : Clearly Z[p] = Z[1 p] OK. Let x OK we write

x = a0 + a1(1 p) + ...... + ap2(1 p)p2 with ai Q. We need toshow ai Z. By using valuation theory we claim that there exists a val-uation with property that v(1 p) = 1 , v(p) = p 1. To see this wenote,

f(x) = 1 + x + x2 + ..... + xp1

=

p1

i=1(x ip).Putting x = 1 in this equation,

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p =

p1i=1

(1 ip)

= (1 p)p1. 1 ip

1 p

Note that 1p1ip Z[p] is unit. For this observe that (i, p) = 1, there existsj

such that ij = 1 mod p.Therefore

1 p1 ip

= 1 (ip)j1 ip

= 1 + ip + ..... Z[p]

proving that 1p1ip Z[p]. Clearly

1ip1p Z[p]. Thus

1p1ip Z[p] is a unit

in Z[p]. Thus the equation p = (1 p)i implies that v(p) = p 1. As thedegree of field is p 1 which is equal to ramification index of 1 p whichshows that < 1 p > is a prime ideal. There exists a valuation on Q(p)s.t. v(x) 0 x OK with v(1 p) = 1 , v(p) = p 1. Also

v(x) = min v(a0), .......v(ap2(1 p)p2)= min (v(a0), .......v(ap2)) 0

This proves that ais do not have p in the denominator. The rest of the

argument is simpler.

Let x = b0 + b1p + ...... + bp2(p)p2 with bi Q and bi have no p indenominator. Note that if x OK then tr(xip ) Z i. Using trjp =

1

j

= 0 and tr1 = p

1 we get

(p 1)bi (b0 + ....bi1 + bi+1 + .... + bp2) Z

pbi p2i=0

bi Z.

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This implies that p(b0

bi)

Z. Since bis do not have p in denominator we

get b0 bi Z. If we can prove that b0 Z then we will be done.

x = b0 + b1p + ...... + bp2p2

p

= b0 + (b1 b0)p + ...... + (bp2 b0)p2p + b0(p + ...... + p2p )= (b1 b0)p + ...... + (bp2 b0)p2p + b0(1 + p + ...... + p2p )= b0p1p +

(bi b0)ip

On taking trace we get b0Z. This completes the proof.

Remark : Not always the ring of integers OK of a number field is gener-

ated by one element but it is always generated by atmost two elements.

9 The different and the discriminant of a num-

ber field

Let a be a fractionl ideal in a number field K. We look at the map tr :

K K Q defined by tr(x, y) = trK

Q (xy).

Definition. The seta = {x K|tr(xy) Z, y a} is called complemen-try set.

It is not difficult to see that a is also a fractional ideal.

Definition. OK = {x K|tr(xy) Z, y OK}. Then K OK OKand OK is a fractional ideal. The inverse of O

K, denoted as , is called the

different, which is an ideal in OK.

Definition (Norm of an ideal). LetI OK is an ideal then we define Nm(I)to be the cardinality of OK/I.

Theorem 9.1. Let K be a number field. Then N m() = disc(K).

Exercise: disc(Q(p)/Q) = pp2

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Theorem 9.2. Let p be a prime ideal in OK lying over p. Let e be the

ramification index ofp over p. Thenpe1 divides . Hence, in a number fieldthe primes which are ramified (ramification index 2) are exactly primesdividing the discriminant of the number field.

Exercises : N m(IJ) = N m(I)Nm(J)

N m() = |NmK/Q()|.An example calculation of different :

Let p Z be a prime. Take K = Qp(p) then OK = Zp[p], for anyprime p. Let us take case p = 2. We claim K =< p >. Recall that1K = {x K|tr(xy) Zp, y OK}. Let a + bp,a,b Zp is an elementofZp[

p]. Then 1

p(a + b

p) = a

p+ b = tr( a

p+ b) = 2b Zpa, b. That

means 1p

1K .Let us take case when p = 2 the we claim that 2 K. 12 (a + b

2) =

12

a + b2

= tr( 12

a + b2

) = a Z2 = 2 K .

Lemma 9.1. IfL is a finite extension of a local field K, withK andL uni-

formizing parameter such thateL = K.u, u a unit in OL. Then e1L /L/K.

In fact L/K = e1L

p/e.

Proof: Let us prove that e1L /L/K. For this we show tr(e1L OL)

OK. But then tr(e1L OL) = tr

LK

OL

= 1

Ktr(LOL) 1K KOK OK.

10 The Riemann Zeta Function

The Riemann zeta function is defined as follows:

Q(s) = n1

1ns = p2

11 1ps

where the product is taken over all primes. The expression of Q(s) in the

product form is called Euler product of Q. The Euler product for Q(s) is

equivalent to the fundamental theorem of arithmetic: every positive integer

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is uniquely a product of primes. Both the expressions are valid for Re(s) > 1.

The function Q(s) has an analytic continuation to a meromorphic functionin s-plane with a simple pole at s = 1 with residue 1. That is

Q(s) =1

s 1 + holomorphic function around s = 1The Riemann zeta function satisfies a functional equation relating its value

at s to 1 s. It is best expressed by introducing another function (s).

(s) = s2 (

s

2)Q(s)

The function (s) satisfies (s) = (1 s).Riemann conjectured in 1850s that all the zeros of Q(s) are on the line

Re(s) = 12

. This is one of the most famous unsolved problems in mathemat-

ics, called the Riemann Hypothesis.

11 Zeta Function of a Number Field (Dedekind

Zeta Function)

Let K be a number field. We define Dedekind zeta function as follows.

K(s) =

I=0,IOK

1

(N I)s=P

1

1 1(NP)s

where I ranges over all nonzero ideals of OK, N I is the norm of the ideal I

and P is a prime ideal in OK. Both the expressions are valid for re(s) > 1.The function K(s) is called the Dedekind zeta function of the number field

K. It has an analytic continuation to a meromorphic function in s-plane with

a simple pole at s = 1.

Theorem 11.1 (Class Number Formula). LetK be a number field. Then,

K(s) =2r1(2)r2hR

w

dK.

1

s 1 + holomorphic function around s = 1

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where r1 is the number of real embeddings of K in C, r2 is the number of

complex embeddings of K inC upto complex conjugate, h is class number ofK, R is the regulator of the number field K, w is number of roots of unity in

K and dK is the modulus of discriminant.

This theorem will be proved later. Let us check convergence of

K(s) =

I=0,IOK

1

(NI)s

= P1

1

1

(NP)s

for Re(s) > 1. Since the equalityI=0,IOK

1

(NI)s=P

1

1 1(NP)s

is formal, so it suffices to check the convergence of the product.

log(K(s)) = P

log(1 1(NP)s )

= P,m1

1m(NP)ms

We have NP= pf where Pis prime lying over p and number of primes lyingover p of resudial degree f is N = [K : Q]. Then,

log(K(s))

p, m1, Nf1

N

mpfms

where p is prime.

Therefore upto a function which is bounded in a neighborhood of 1,

log(K(s)) = ap

ps

ap denotes number of primes above p of degree 1.

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Thus log(K(s)) for s > 1 is bounded by N

ps, thus it is convergent for

re(s) > 1. Actually we will be proving later that

log(K(s)) = log1

s 1 = 1

ps(up to a bounded function).

12 Exercises

1. (Problem of Erdos) If m and n are integers, such that the order of m

is same as order of n in (Z/p) for almost all primes, then m = n.

2. Define the zeta function A1(s) of the affine line A1 over the finite field

Fp as follows.

A1(s) =

p(x)

1

(1 1Np(x)

)s

where p(x) runs over irreducible polynomials in Fp[x] with leading term

1 and f(x) is monic polynomial over Fp. And N f(x) = pn where n =

deg f(x). But then

A1(s) = p(x)

1(1 1

Np(x))s

=f(x)

1

(N f(x))s

=

d

pd

pds

=1

1 pps

3. Let K = Q(i), then K(s) =

1(m2+n2)s

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13 Class Number Formula 2

13.1 Introduction

Let G be a finite cyclic group of even order. Then the number of squares

in G is equal to the number of non-squares in G. In particular, for a prime

p, taking G = (Z/pZ), number of quadratic residue modulo p is equal to

the number of quadratic non-residue mod p. One can ask how the quadratic

residues are distributed in Ip = {1, 2, , (p 1)/2}. Let Rp = Number ofquadratic residues in Ip and Np = Number of quadratic non-residues in Ip.

Then the question isIs Rp = Np ?

If p 1 mod 4, then1p

= 1. Hence the map

:

1, , p 1

2

p + 1

2, , p 1

defined by k k is an one-one correspondence, preserving squares. So,we conclude that exactly half of the quadratic residues are in Ip implying

Rp = Np. If p 3 mod 4 then the following result answers our question.Theorem 13.1. Let p > 3 be a prime such that p 3 mod 4. Also lethpdenote the class number of the quadratic fieldQ(

p). Then

hp =

Rp Np if p 7 mod 8,13

(Rp Np) if p 3 mod 8.

Remark 13.1. As hp 1, we have Rp > Np for prime p > 3 and p 3mod 4. Moreover 3 divides Rp

Np if p

3 mod 8.

Proof of Theorem 13.1 depends on the Class number formula for quadratic

extensions ofQ. Let K = Q(

d) where d is the discreminant of K. Let w

be the number of roots of unity in K. For an ideal class C of K and a real2This section is written by Anirban Mukhopadhyay.

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number X > 0, we define N(X,

C) = number of integral ideals I in

Csuch

that N(I) < X where N(I) denotes the norm ofI. We have

Theorem 13.2.

limX

N(X, C)X

=

where

=

2log d

if d > 0, is the unique fundamental unit > 1,

2

w|d| if d < 0.

Now we state the class number formula for the quadratic field K =Q(

d).

Theorem 13.3. For a quadratic number field K,

lims1+

(s 1)K(s) = h

where h =Class number of K and K is the Dedekind zeta function.

In the rest of the article we give proofs of these results.

13.2 Proof of Theorem 13.1

Here we write a sequence of lemmas leading to the proof of Theorem 1.

Throughout this section we assume K = Q(

d). Following lemma gives a

simple expression for K.

Lemma 13.1. For s > 1, we have

K(s) = (s)Ld(s)

where (s) is the Riemann zeta function and

Ld(s) =

m=1

d

m

ms

with

dm

denoting the quadratic residue symbol.

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The series Ld(s) converges for s > 0. Hence using the fact that

lims1+(s 1)(s) = 1 we derive from the above Lemma

lims1+

(s 1)K(s) = lims1+

(s 1)(s) = Ld(1).

Thus we have the following simplier expression for class number formula.

h =

2log

dLd(1) if d > 0, is the unique fundamental unit > 1,

2

w|d|Ld(1) if d < 0.

Now to prove Theorem 13.1 we consider the particular case d =

p where

p > 3 is a prime such that p 3 mod 4. It is easy to see that the numberof roots of unity in K, w = 2. For a Dirichlet character , the Dirichlet

series is defined to be L(s, ) =

n=1(n)

ns. In this case we define to be the

quadratic character ( .p

). Using above expression for class number one can

easily derive that hp =

p

L(1, ). Next Lemma expresses L(1, ) as a finite

sum.

Lemma 13.2. We have

L(1, ) = i1()

p2

p

1k=1

(k)k.

where

k() =

1a 1 we have

L(s, ) = n=1

(n)ns

= 1a

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where cn = 1 if n

a mod p and cn = 0 otherwise. If denote the p-th

primitive root of unity, we can write

cn =1

p

p1k=0

(an)k.

Thus

L(s, ) =1

p

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We split this sum as

hpp =(p1)/2

k=1

k

p

k +

p1k=(p+1)/2

k

p

k

=

(p1)/2k=1

k

p

(k (p k))

= 2

(p1)/2k=1

k

p

k p(Rp Np).

On the other hand we can write,

hpp =(p1)/2

k=1

2k

p

2k +

(p1)/2k=1

p 2k

p

(p 2k)

= 4

2

p

(p1)/2k=1

k

p

k p

2

p

(Rp Np).

from above two we get2

2

p

1

hpp = p

2

p

(Rp Np).

Now Theorem 13.1 follows from the fact that ( 2p

) = 1 or 1 according asp 7 mod 8 or p 3 mod 8.The next section is dedicated to proofs of Theorems 13.2 and 13.3.

13.3 Proofs of Theorems 13.2 and 13.3

We start with a lemma

Lemma 13.3. Let be a bounded open set in R2. For any real number

r > 0, let

r =

(1, 2) R2|

1r

,2r

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and N(r) = Number of lattice points in r. Then

limx

1

r2N(r) = Area of .

First we present a proof of Theorem 13.2. Let C be any ideal class ofK and J be an integral ideal in C1. Then for any integral ideal I C,IJ = OK with J. Conversely if J then J|OK, hence I = J1OKis an integral ideal in C. Further |NK/Q()| = N(I)N(J). Thus N(I) < X ifand only if|NK/Q()| < XN(J) = Y(say). We can conclude that N(X, C) =Number of pairwise non-associative elements J such that |NK/Q()| < Y.We divide the proof in two cases.

Case 1: d > 0 Here is the fundamental unit > 1. We observe that for

any J, = 0, there exist an integer m such that

0 log |N()|1/2

< log (13.3.1)where = m and we use N to denote norm without mentioning the fields

involved. It is easy to see that if two associates 1 and 2 satisfies (13.3.1)

then 1 = 2 with = 1. Hence

2N(X, C) = J | 0 < |N()| < y, 0 log |N()|1/2 < log .

Let 1, 2 be an integral basis for J and let

=

(1, 2) R2|0 < |1||1| < 1, 0 < log

|1|1/2|1|1/2 < log

where = 11 + 22 and = 11 + 2

2,

1,

2 denote the conjugates of

1, 2. We observe that is bounded.

2N(X, C) = Number of lattice points in y+Number of lattice points inAy

where

Ay =

(1, 2) Z | ||2 y, || = || = 0

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and is as above. One can easily see that

|Ay

|= O(

y) = O(X). Now

using Lemma 13.3 we derive

limX

2N(X, C)

X= lim

yN(

y)

yN(J) = N(J)Area of .

Area of can be computed to be 4log N(J)

d

. This completes the proof in this

case.

Case 2: d < 0

In this case we have

wN(X,C

) =|{

J|

0 < N() < y}|

.

Let

=

(1, 2) R2|0 < |11 + 22|2 < 1

.

Then wN(X, C) = N(y), hence

limX

wN(X, C)

X= lim

yN(

y)

yN(J) = N(J)Area of .

Here area of is calculated to be 2N(J)

d

, completing the proof of Theorem

13.2.

To derive class number formula we need the following lemma

Lemma 13.4. Let{am} be a sequence of real numbers. For a positive realnumber X, we set

A(X) =

m 1 and we have

lims1+

(s 1)f(s) = c.

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Let am = Number of integral ideals I such that N(I) = m. Then for a

real number X > 0, A(X) = number of integral ideals I such that N(I) < X.From Theorem 13.2 we get

limX

A(X)

X= h.

So, by the above Lemma, K(s) is convergent for s > 1 and

lims1+

(s 1)k(s) = h.

Which gives the required result.

14 Density of Primes (Dirichlet density)

Let K be a number field and M is set of primes, then the density of primes

in M is defined to be,

lims1+

pM

1(Np)s

log 1s1

In one of the earlier lecture we had observed that

log1

s 1 log K(s)

p

1

(Np)s

degp=1

1

(Np)s

where f1 f2 if (f1 f2)(s 1) 0 as s 1 and Np = pdegp ( in fact degp= residual degree ).

Theorem 14.1. Let K be a number field which is a galois extension of Q.

If K has degree d overQ then the set of primes inQ which split completely

in K has density = 1d

.

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Proof: We have M =

{p

Z

|pOK = p1...pd, p

s are completely split

}.

So using remark preceding theorem we get

dpM

1

(Np)s=

degp=1

1

(Np)s= log

1

s 1

Hence substituting in density formula we get density of completely split

primes ( primes in M ) is 1d

.

Theorem 14.2. Two number fields K1 and K2 which are Galois overQ are

the same if and only if they have the same set of split primes, up to density

zero primes.

Proof: We claim that primes which split in K1K2 are precisely those

which split in both K1 and K2. This follows by noting that a prime p is split

in a number field K if and only if all embeddings ofK in Qp land inside Qp.

Thus if p splits completely in K1 and K2, it does so in K1K2 also.

Let K1 be degree d1 extension ofQ, K2 be degree d2 extension ofQ and

K1K2 degree d extension ofQ. Then, split primes in K1 are of density1

d1,

split primes in K2 are of density1

d2and split primes in K1K2 are of density

1d . Since the split primes in K1, K2 and K1K2 are the same,

1d1 =

1d2 =

1d .

This implies d1 = d2 = d, hence K1 = K2 = K1K2.

Lemma 14.1. Let K be a number field. Let L denotes the Galois closure of

K. Then the primes inQ which split completely in K are exactly the primes

which split completely in L.

Proof: We have proved earlier that for field extensions K1, K2 the primes

in Q which split completely in K1K2 are exactly those which split completely

in K1 and K2. Now noting that L is compositium of conjugates of K, the

lemma follows.

Example: Calculation of density of primes.

Let K = Q(213 ), and L = Q(2

13 , ) be Galois clousre of K, where is

cube root of unity. We have Q K L. Let us denote primes in Q by p,prime ideals in K by p and prime ideals in L by q.

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1. Let us calculate the density of split primes in K, M =

{p

Z

|pOK =

p1p2p3, in K}. But from previous lemma density of completely splitprimes in K is same as density of completely split primes in its closure

L. Then by the theorem, density of M = 16

.

2. Let us calculate the density of primes which can be written as product

of two primes. Then M = {p Z|pOK = p1p2 in K} = {p Z|pOL =q1q2q3 in L} {p OK| pOL = q1q2}. Hence density of M is 12 .

3. Let us calculate the density of primes which remains prime in K. In this

case M = {p Z| pOK = p in K}. But we can see that deg 1 primesin Z falls in one of the above three categories. Hence the density inthis case will be 1 1

6 1

2= 1

3.

15 Decomposition of primes in the cyclotomic

fields

Let K = Q(n), where n = e2in and OK = Z[n].

Proposition 15.1. Let K = Q(n) be a cyclotomic number field. Then the

primes inZ which ramify in K are exactly the divisors of n.

Proof: It suffices to prove the proposition for n which is prime power.

So let us assume n = pm i.e. K = Q(pm). We need to show that p is the

only prime which ramifies in K. In fact, in this case, p is a totally ramified

prime i.e. pOK = pd where d = (pm).

Observe that the minimal polynomial satisfied by pm is

1 + xpm1

+ (x2

)pm1

+ ..... + (xp

1

)pm1

=

(x )

where product is over , all pmth roots of unity. Putting x = 1 we get

p =

(i,p)=1

(1 ipm). i Z/pm

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Let us take = (1

pm) then () = 1

(pm) for any embedding of K

in C. Since embeddings of K in C correspond to (i, p) = 1 i Z/pm whichsends pm to

ipm we get

p =

Gal(K/Q)().

Observe that1i

pm

1pm Z[pm] is a unit for (i, p) = 1. Hence we get p =(1 pm)(pm).u where u is a unit in Z[pm]. By taking p =< 1pm > we getpOK = p

(pm). The discriminant of field is power of p hence no other prime

ramifies.

Lemma 15.1. LetK = Q(n) be a cyclotomic field. Then the Galois groupGal(Q(n)/Q) is naturally isomorphic to (Z/n)

given by a (Z/n) actingon Q(n) by n an. Further, for p prime in Z with (p,n) = 1, theFrobenius element p Gal(K/Q) is the element p (Z/n).

Proof: For this note that if (m, n) = 1 then Q(n)Q(m) = Q. This

reduces the problem to the case when n is prime power.

Corollary 15.1. Let p be a prime inZ. Then p splits completely inQ(n)

if and only if p

1(mod n). In general, forp prime lying over p the residual

degree is exactly the order of the element p (Z/n).Proof: Let p be a prime such that p does not divide n and let p be a

prime lying over p in Z[n]. The frobenius automorphism p : Z[n] Z[n]is defined by px xp mod p for all x Z[n]. But p(n) = pn whereasorder ofp is the degree of residue field extension ofp which is Z[n]/p. Now

fp = 1 fp (n) = n (n)pf

= n pf 1 ( mod n)Since p is unramified we get p splits into (n)/f distinct prime ideals. Also

p splits completely (e=1 and f=1) if and only if p 1(modn). Another Proof of Quadratic Reciprocity Law:

Let p 1(mod4) be a prime. Then Q(p) Q(p). For this consider

G =

p1a=1

a

p

e2iap

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It is easy to see that G2 = (1p

)p = p hence G =

p belongs to Q(p).

Let q be a prime. Then q gal(Q(p)/Q) = (Z/p) and from previouslemma we get, q is a quadratic residue in Z/p if and only if the prime q splits

in Q(

p).

Now look at the field extension Q Q(p) and corresponding polyno-mial x2 p = 0. From this point of view, q splits in Q(p) if and only if pis a quadratic residue mod q. Combining with earlier statement we get p is

a quadratic residue mod q if and only ifq is quadratic residue mod p.

Remark: One can prove that G =

p if p 1(mod4) and G = ip ifp

3(mod4).

16 Subfields of Cyclotomic Fields

Let K be a number field which is contained in Q(n). Let n =

ipnii .

Then Gal(Q(n)/Q) = (Z/n) =

i(Z/p

nii )

. Using Galois theory there is

a one-one correspondence between the subfields ofQ(n) and subgroups of

Gal(Q(n)/Q) = (Z/n).

Definition. Let G be an abelian group. A group homomorphism : G C is called a character of a group G.Then G, the set of all characters of G, forms a group. This group is called

the character group of G.

Lemma 16.1. If G is a finite abelian group then there exists a bijective

correspondance between subgroups of G and subgroups of the character groupG. This correspondence is given by associating to a subgroup A of G, the setof all characters of G which are trivial on A.

Combining this lemma with Galois theory we get that subfields K ofQ(n)

correspond to the subgroups of group of characters of (Z/n) i.e. (Z/n).

A character of (Z/n) which is a homomorphism : (Z/n) C iscalled a Dirichlet Character. Let n|m are positive integers. Then we get ahomomorphism (Z/m) (Z/n).

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Definition. Let : (Z/n)

C be a Dirichlet character. Let n0 be the

smallest divisor of n such that factors through (Z/n0), i.e. there exist : (Z/n0) C such that following diagram commutes.

(Z/n) //

##HH

HH

HH

HH

HH

HH

HH

HH

HH

H

(Z/n0)

C

Then n0 is called the conductor of the Dirichlet character.

Theorem 16.1 (Conductor-Discriminant Formula). Let K be a number field

such that K Q(n), for some n. Let this abelian extension K be defined bya group of characters X = { : (Z/n) C}. Then

|dK| =X

cond()

Examples :

1. Let p be a prime and consider K = Q(p). Then the Galois group

G = (Z/p) and let X = { : (Z/p) C} be character group.Then conductor of = p, X if and only if is not trivial character.Hence using conductor-discriminant formula we get |dK| = pp2.

2. Let p be a prime and consider K = Q(p2). Then the Galois group

G = (Z/p2) = Z/(p2 p) and let X = { : (Z/p2) C} becharacter group. Then one can see that trivial character has conductor

1, p

2 of them have conductor p and p2

2p+1 of them have character

p2 . To see this observe that Z/(p2 p) = Z/p Z/(p 1). Hence weget |dK| = p2p23p.

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17 L-functions associated to a Dirichlet char-

acter

Let : (Z/n) C be a Dirichlet character. We can extend to define afunction on Z by

(a) = (a) if(a, n) = 10 otherwise

Definition. Let be a Dirichlet character. Then

L(s, ) = n1(n)

ns

is called the Dirichlet L-series associated to the Dirichlet character .

The Dirichlet L-functions have properties very similar to the Riemann

zeta function. For instance, these also have meromorphic continuation to the

complex plane and are holomorphic except if = 1. These series also have

functional equations like Riemann zeta functions.

We recall that if (s) = s2 ( s

2)Q(s), then the functional equation for

the Riemann zeta function is expressed as (s) = (1 s). Let K be numberfield and K(s) be the Dedekind zeta function of the number field K. Define

K(s) = (

|dK|)s(s2 ( s2

))r1(s(s))r2K(s)

Then the functional equation for the Dedekind zeta function is expressed by

the equality K(s) = K(1 s).There are two basic points here.

1. The L-function L(s, ) has a functional equations in which the conduc-

tor of appears.

2. For an abelian extension K ofQ, the zeta function can be written as a

product of L-functions as

K(s) =

L(s, )

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where product runs over all characters of the Galois group of K over

Q.

Using these two remarks one can get conductor discriminant formula.

18 Introduction to Tates thesis

In the following sections we will be discussing the Tates thesis which is

essentially about analytic contituation and functional equation of Dedekind

zeta functions, and also of L-functions associated to Grossencharacters. This

is the theory of abelian L-functions on the multiplicative group of number

fields as well as of local fields. We begin by recalling that local fields are

either archmedian (R or C) local fields or non-archmedian (finite extensions

ofQp and Fp((t))) local fields. Further the global fields are either number

fields (finite extensions ofQ) or function fields (finite extensions ofFp(t)).

19 Harmonic analysis on p-adic fields

Let G be a locally compact abelian topological group. Let us denote unitarydual of G by G = { : G S1}. Then G is a locally compact topologicalgroup on which the topology is the compact open topology defined by taking

basic open sets of G as W(C, U) = { G|(C) U},where C is a compactset in G and U an open set in S1.

Theorem 19.1 (Pontryagin duality). The map G G is an anti equiva-lence of categories.

Let us recall Fourier analysis onR

. Let us take the continuous grouphomomorphism : R C defined by (x) = exp(2ix). Then anyunitary character (continuous group homomorphisms from R to S1) on R

is of the form y : x (xy) for some y R. Let us define unitarydual ofR by R = { : R S1| unitary character ofR}. Then the map

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y

y defines isomorphism ofR with its unitary dual R. One defines foran integrable function f its Fourier transform by

f(y) = R

f(x)(xy)dx

where dx is a Haar measure on R which, in this case, is the usual Lebesgue

measure on R. For an appropriate choice of a Haar measure dx we have,f(y) = f(y).There are similar theorems for any local field. Consider the character

: Qp C given as follows. For x Qp write x as

x =

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Let k = (pk); then

pk+1 =

(pk1)p

= (p.pk1) = (pk) = k

Now 1 = 1 and 0 = 1 so 1 = exp(2ic0p1) for some c0 {1,...,p 1}.Proceeding by induction, suppose k = exp

2ik

j=1 ckjpj

. Since k=1

is pth root of k, there exists ck {0, 1,...,p 1} such that

k+1 = exp

2i

kj=1

ckjpj1

exp(2ickp1) = exp

2i

k+1j=1

ck+1jpj

.

(3) If Qp, (1) = 1, and (p1) = 1, there exists y Qp with |y| = 1such that = y.

Let us take {cj}0 as above and set y =

0 cjpj. Then |y| = 1 since

c0 = 0, and for k 1,

(pk) = exp

2i

kj=1

ckjpj

= exp

2i

1j=k

cj+kpj

= 1(

k

cj+kpj)

= 1(pky) = y(p

k)

From 1,2 and 3 surjectivity of lemma follows easily. Rest of the proof is

easy exrcise.

Now if we have any finite extension K ofQp then we define character

using trace map combining with character as follows

Ktr Qp C

This is a continuous unitary character on K.

Lemma 19.2 (Pontryagin Duality). LetK be a field which is finite extensionofQp. Let be a non trivial character of K (for example above defined ).

Then defines a topological isomorphism of K to its dual K. Which isK K defined as y y where y(x) = (xy).

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Proof: If we use the previous lemma and the fact that dual of direct sum

is direct product of duals, the lemma follows. But we give different proof.It is easy to see that the map is continuous additive homomorphism from

K to K. Also y = 1 if and only if y = 0. Since is a non trivial characterthere exists y0 such that (y0) = 1 and if y = 0 = (y0) = (y0y1y) =y(y0y

1) = 1 = y = 1. And if y = 0 = y = 1. This insuresinjectivity of the map. Moreover it gives that character distinguishes points

of K. It remains to prove that the map is surjective.

We claim to show that image of K in

K is dense as well as closed. We

have Image ={

y|

y

K} K. Let be a character in the closure ofthe image of K. Thus assume that there exists a sequence xn K such

that xn . We can assume that xns do not have a convergent sub-sequence (else if xn x then xn x = ) . Thus we assume thatxn . And xn means that for any compact set X K, xn(x) (x)x K uniformly on K. That is |xn 1| < x K for n sufficientlylarge. Any neighbourhood of 0 in K contains a subgroup (as it is locally

compact, look at its valuation ring or power of its unique maximal ideal) but

in C the neighbourhood of 1 does not contain any subgroup. (This is called

non-existence of non-trivial subgroups in a neighbourhood of 1.) This givesxn(K) = 1, a contradiction to xn .

From previous proposition K = K so fourier analysis of standard kindbecomes available on K. For any function f L1 we define,

f(x) = K

f(x)(xy)dy

where dy is a Haar measure on K and : K C is a fixed non-trivialcharacter.

Proposition 19.1. There exists a unique choice of Haar measure dx (de-

pending on ) on K such thatf(x) = f(x), for all f s(K), schwartz

space (space of locally constant compactly supported functions on K).

Now let us fix a Haar measure on K, a locally compact field, so that

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fourier inversion formula holds. Recall that the different ideal K/Qp of K is

defined by 1K|Qp = {x K|tr(xOK) Zp} .Lemma 19.3. If A is a compact abelian group and : A C is acharacter, then

A

(a)da = 0 if = 1 and A

(a)da = vol(A) if = 1.

Proof: If = 1 then

A(a)da =

A

1.da = vol(A). If = 1 thereexists x0 A such that (x0) = 1. Put I =

A

(a)da then I =

A(a)da =

A(a + x0)da = I.(x0) = I = 0. Let us denote the characteristic function of OK, the valuation ring of K,

by the OK . The fourier transform will be

OK(y) = K

OK (x)(xy)dx =

OK

(xy)dx

Then OK(y) = 0 i f y / 1K and OK(y) = vol(OK) if y 1K . Where(x) = exp 2i tr(x) is a character of K. Then,

OK(y) =

K

OK(x)(xy)dx = vol(OK)

1K

(xy)dx

OK(0) = vol(OK)1K

1.dx = vol(OK)N(K)vol(OK)

But OK(0) = (0) = 1. This implies that vol(OK) = 1N(K) , whereN(K) = [

1K : OK] = [OK : K].

20 Local Theory

Let K be a local field and let

S(K) denote the Schwartz space of locally

constant compactly supported functions on K. Let be a character on K.Then we define local zeta function by

Z(f , , s) =

Kf(x)(x)|x|sdx

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for s

C, where dx is normalised Haar measure on K. Initially

Z(f , , s)

is defined only for Re(s) > 0 but actually it has meromorphic continuationto whole of the s-plane. In fact, Z(f,,s) C(qs) where q = ||, isuniformizing parameter. Let us prove this in case is a ramified character

of K (i.e. (OK) 1), where OK is the group of invertible elements of thering OK.

Z(f,,s) =

Kf(x)(x)|x|sdx

= OK

(x)|x|sdx + a term belonging to C[qs, qs]

Now let us look at the termOK

(x)|x|sdx =n=0[nOKn+1OK ]

(x)|x|sdx

=

n=0

nO

K

(x)|x|sdx

=

n=0

OK

(nx)qns|x|sdx

=

n=0

(n)

OK

(x)qns|x|sdx

= 0

Therfore for ramified character Z(f , , s) C[qs, qs].If

|OK

1 i.e. is unramified then by a similar calculation

Z(OK , , s) =

n=0

(n)qnsvol(OK) =1

1 ()qs

vol(OK)

The local zeta functions satisfy a certain functional equation which is

basic to the whole theory.

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Theorem 20.1. Let K be a local field, a non-trivial character and f

S(K) then for all g S(K),Z(f,,s)Z(g, 1, 1 s) = Z(g,,s)Z(f , 1, 1 s).

Proof: This follows by definitions as we will see.

Z(f,,s)Z(g, 1, 1 s) = KK

f(x)(x)|x|sg(y)1(y)|y|1sdxdyWe apply change of variable (x, y) (x, xy).

= KK f(x)(x)|x|s

g(xy)1(xy)

|xy

|1sdxdy

=

K

K

f(x)g(xy)|x|dx(y1)|y|1sdy=

KKK

f(x)g(z)(xyz)|y|s(y1)dxdydz

This expression is symmetric in f and g; proving the theorem.

Corollary 20.1.

Z(f , , s) = (, s)Z(f , 1, 1 s)where (, s) is independent of f.

Definition. We define L(, s) = 1 if is ramified, and L(, s) = 11()

qs

if

is unramified.

Note that the value of an unramified character on a uniformizing ele-

ment does not depend on the choice of uniformizing element . Rearranging

the functional equation, we get

Z(f , 1, 1 s)L(1, 1 s) = (, s)

Z(f , , s)L(, s)

These epsilons are called local constants. These are monomial function

in qs.

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Proposition 20.1. The local constant (, s) is a monomial in qs.

Proof : Let assume that is ramified. In this case functional equation

is

Z(f , 1, 1 s) = (, s)Z(f , , s)We calculated in the beginning of the section that when is ramified Z(f,,s) C[qs, qs]. This implies that (, s) C[qs, qs]. Similarly (1, 1 s) C[qs, qs]. Using previous proposition we see that (, s)(1, 1 s) =(1), a constant. Hence (, s) is a unit in C[qs, qs] but the only units inC[qs, qs] are monomials. Hence the proposition.

Proposition 20.2.

(, s)(1, 1 s) = (1)

Proof follows from the functional equation for local zeta functions and

the definition of (, s).

21 Language of Adeles and Ideles

Harmonic analysis on the topological groups is most conveniently done when

the group is locally compact. By Tychonoffs theorem, an arbitrary product

of compact spaces is compact. However such a theorem is not true for lo-

cally compact spaces. There is a way out, and there is a general notion in

topology of restricted direct product of locally compact topological groups

which constructs a locally compact topological group. Suppose we are given

system of pairs (Gp, Hp) for p p (some index set), wher Gp are locallycompact groups and Hp are compact open subgroups of Gp. The restricted

direct product of the system (Gp, Hp) will be denoted as (Gp, Hp) or Gp.This product is defined as follows.

(Gp, Hp) = {(xp)|xp Gp, p and xp Hp for all but finitely many ps}

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A topology on this group is given by declaring the system of neighborhoods

of 1 to be:u1 u2 ..... ur

p={1,2,..r}

Hp

where uis are compact open subgroups of Gpi at finitely many places pi.

With this topology G =

(Gp, Hp) is a locally compact topological group.

The corresponding Schwartz space (space of locally constant and com-

pactly supported functions) will be given by

S(G) = p S(Gp), Hp = lim {1,..,r}pr

i=1 S(Gpi) r+1

i=1 S(Gpi)where the maps are defined as

ri=1

S(Gpi) r+1i=1

S(Gpi)

f1

....

fr f1

...

fr

Hr+1

Proposition 21.1. Let G = p(Gp, Hp) be the restricted direct product ofGp, Hp defined as above. Let dgp denote the corresponding Haar measure on

Gp normalized so that the volume of Hp is 1 for almost all p p. Then thereis a unique Haar measure dg on G such that for each finite subset of indices

S p the restriction of dg on GS =

pS Gp

p /S Hp is precisely the

product measure.

Let us take the field Q of rational numbers and all its completions at

finite places Qp and infinite place R = Q. The corresponding restricted

direct product of {(Qp,Zp)}{R, } is called the adeles ofQ, denoted asAQ = RQ where Q = QZZ. This is a locally compact topological ring.

Similar notion is associated to the multiplicative group in which we take

{(Qp,Zp)}{R, }. Corresponding restricted direct product is called the

idele group ofQ denoted by JQ, or AQ, or Gm.

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For a general number field K one can define adeles and ideles as follows:

AK = AQQ

K =

(Kv, Ov)

JK =

AQQ

K

=

(Kv , Ov)

where Kv denotes completion of K at place v and Ov is corresponding valu-

ation ring. On JK the topology is defined by the system of neighbourhoods

coming from Op = (ZpZ OK) in usual way. Note that JQ AQ is a

continuous map, but JQ is not closed in AQ. Thus the topology on JQ is notthe one it inherits as a subset ofAQ.

Lemma 21.1. A number field K embeds inside its adeles AK as a discrete,

cocompact subgroup, i.e. AK/K is a compact group.

Proof : The map K AK given by x (x, x, x....) (given any x Kit belongs to Ov for almost all v) will have discrete image if there exists a

neighbourhood of {0} AK which does not contain any nonzero element

of K. In fact let U = U v Ov where U is a small neighborhood of0 in KR. We have OK Rr1 Cr2 is discrete, where r1 number ofreal embedding of K and r2 is number of complex embeddings of K up

to conjugation. Using which we can pull out small neighbourhood for our

purpose.

The map K KAQ can be compared to the map Qr ArQ to showthe cocompactness. As we can see

AQQ

=ZpRZ

that is AQ = Q+Zp R.

Lemma 21.2. The multiplicative group K embedds insideJ

K as a discretesubgroup. Inside JK there exists a closed subgroup J

1K = ker{|.| : JK

C; |x| = v |x|v} which contains K.Theorem 21.1. The group J1K/K

is a compact topological group.

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Proof : The discreteness of K inside JK is analogous. We prove that

K J1K is cocompact. For this observe that K\JK/v Ov K isnothing but the class group of K. Look at the map

JK

v

Zv = divisors

xv

v

ordv(xv)v

Then JK/vOv.

K divisors, going modulo K we getK\JK/v Ov. K K\divisors, which is the class group of K. SinceJK is K

. (

v Ov

K) up to finite index, for the purpose of the proof

of compactness ofJ1K/K we can as well look K K.v Ov ( K)1.

Thus J1K/K is compact if and only if (

K

)

1 /units in K is compact.

Which is nothing but Dirichlet unit theorem.

22 Classical Language

Definition. Cycles: C = Cf.C where Cf is an ideal in OK and C is a setof embeddings of K R.

Corresponding to a cycle C one can define I(C) as ideals coprime to C anddefine PC to be the principal ideals which have generators x such that x > 0

in all embeddings ofK corresponding to primes in C and x 1 Cf i.e.Cf =

p

nii then x 1 Cf vpi(x 1) ni.

Notation: We write above one as x 1 mod Cf.Clearly I(C) is a group and PC is a subgroup and thus I(C)/PC is a group.

This is a finite group.Example : Let us take K = Q and C = n. then I(C)/PC = (Z/n).Proof : Let x = d1

d2be an element ofI(C) with d1, d2 coprime to n. Then

we define a map I(C) (Z/n) as x d1( mod n).(d2( mod n))1.This is a surjective map with kernel PC.

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Lemma 22.1. Let

C=

Cf.

C be a cycle. Then

I(C)/PC JK/K

(v,C)=1Uv

Kv/C

Uv(C)

where (v, C) = 1 means v coprime to C.

The main theorem of class field theory i.e. Artin Reciprocity will imply

that these are precisely the Galois group of abelian extensions of K. Proof

of the lemma is an exercise for readers.

23 Character ofAK

Definition. If a group G operates on a space X then X is called a principal

homogeneous space for G if G operates simply transitively on X.

Lemma 23.1. The characters of AK can be considered to be a principal

homogeneous space ofAK.

Let be any non-trivial character on AK then AK operates on AK and

give rise to a principal homogeneous structure.

AK AK AK(a, ) a

a(x) = (ax)

Lemma 23.2. If G =

(Gp, Hp) then

G =

(

Gp, H

p ) where H

p

Gp

consists of all characters on Gp which are trivial onHp which is isomporphic

to Gp/Hp, a compact group.

Proof : Since G =

(Gp, Hp), then : G C gives rise to a charac-ter p : Gp C. We claim that p is trivial on Hp for almost all p. Thisfollows because C has no small subgroups i.e. there exists a neighbourhood

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of 1 which contains no subgroup ofC exept 1 itself. This implies by conti-

nuity of and the definition of almost direct product that p = 1 for almostall p.Then we get Hp = Gp/Hp Gp is an open subgroup. To prove thiswe look at the continuous map Gp Gp C defined by (a, ) (a).Define U = { Gp|(Hp) a ball of radius 13 around 1} This set ofcharacers are open set in Gp. But since there are no small subgroups we getU = Hp .

Next objective is to construct a non-trivial character on AK/K. We will

do this for K = Q and then take the trace from AK to AQ to construct one

for all number fields K.

Construction of character on AQ/Q

The local characters we constructed has the required property. We had

Qp Qp/Zp = (Q/Z)p exp(2ix)

These mappings can be combined to give a mapping

AQ/(R

Z) =

p

Qp/Zp = Q/Z S1

But AQ = (R+Z).Q = AQQ = RZZThis gives a character (a) = e2ia on R

ZZ

and hence onAQQ

.

Lemma 23.3. For K, an algebraic number field we have AK/K = K as aprincipal homogeneous space.

Proof Let be a character of AK/K. For any element a K, onecan define another character a : AK/K S1 using a(x) = (ax). Weclaim to show that if is non-trivial character ofAK/K then all characters

ofAK/K are of this form. Note that since AK/K is compact its character

group is a discrete subgroup ofAK containing K. Since AK/K is compact

it follows that its discrete subgroup must contain K as a subgroup of finite

index. However a discrete subgroup is a vector space over K. hence the

index if not 1 must be infinite.

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Corollary 23.1 (Strong Approximation Theorem). LetK be a number field.

Let v0 be some place (finite or infinite). Then the image of

K

v=v0Kv

is dense.

Proof : If the image is not dense then there exists a non-trivial character

ofAK trivial on Kv0 K. However all characters ofAK, trivial on K are ofthe form (ax), a K and these are non-trivial on Kv0.

Corollary 23.2 (Weak Approximation Theorem). LetS be any finite set of

places of a global field K. Then K is dense in

vS Kv.

24 Grossencharacter

Characters of idele group JK C which are non-trivial on K play a veryspecific role.

Definition (Grossencharacters). A Grossencharacter is a character ofJK/K

which is a continuous group homomorphism JK/K C. A continuous

group homomorphismJK/K S1 is called unitary Grossencharacter.

Grossencharacters are trivial on Uv, for almost all v.

Definition (L-function). Let : JK/K C be a character. We define

the L-function associated to as

L(, s) =

(p,c)=11

1 (p)(N p)s where c denotes conductor of consists of those primes in K for which is

non-trivial on Uv and (p,c) = 1 denotes the prime ideals p of K which are

coprime to the conductor c of and the product is taken over all such p.

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We can also write L(, s) = p L(p, s) where L(p, s) =1

1

p(p)

(Np)s

ifp is

an unramified character of Kp i.e. it is trivial on O

p and define L(p, s) = 1

ifp is ramified. By comparing to K(s) one finds that L(, s) is holomorphic

in the domain Re(s) > 1 if is unitary.

The infinity-type of a Grossencharacter : If : JK C isa Grossencharacter, the restriction of to K = (K

R) is called the

infinity type of character.

Definition (Algebricity of Grossencharacter). The character is called al-

gebraic if and only if is algebraic i.e. (x) = xn if x

R+ and

(z) = znzm, m , n Z if z C.Example : For K = Q we have JQ = Q R+

Zp. This means that

JQ/Q = R+ Zp. Thus characters are easy to describe in this case. The

characters ofJQ/Q of finite order are identified to characters of (Z/m) for

some integer m. The trivial character on JK/K is a Grossencharacter and

the corresponding L-function is the Dedekind Zeta function of K.

Exercise :

1. Prove that any Grossencharacter can be written as = 0||

s where 0

is a unitary Grossencharacter.

2. Prove that by using formula L(, s) = L(.||s0 , s) = L(0, s + s0)

25 Fourier Analysis on Adeles and Ideles and

Global Zeta Function

We define Global zeta function for f S(AK), Schwartz space, and :JK/K C a character,

(f , , s) =

JK

f(x)(x)|x|sdx

where dx is a Haar measure on JK. Note that to define Haar measure on

an almost direct product one must have measure(Hp) = 1 for almost all p.

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In our case Hp = OP, we have volume

Np1Np

. thus to define Haar measure on

JK one multiplies local Haar measure by N pN p1 dx|x| .If is unitary character then (f,,s) is an analytic function of s in

the region Re(s) > 1. It suffices to check this for factorizable function f S(AK) =

v S(Kv). So it suffices to assume f =

fp, =

p, in which

case if we can prove that

p Z(fp, p, s) converges and defines an analyticfunction in the region Re(s) > 1, then the integral defining (f , , s) will

also converge to

p Z(fp, p, s) and will be analytic function in the regionRe(s) > 1. But we have calculated earlier that Z(fp, p, s) = 1

1p(p)(Np)sis

convergent for Re(s) > 1, hence the zeta integral make sense and is analyticin Re(s) > 1.

Theorem 25.1. If is a non-trivial Grossencharacter then (f , , s) has

analytic continuation to the entire complex-plane and has functional equation

(f,,s) = (f , 1, 1 s)Proof : It has two basic ingradients.

1. Poisson Summation formula : We recall that Poisson summation

formula ofR states that nZ

f(n) =nZ

f(n)in case ofR. where f S(R). We will be using the analogue of thisfor AK.

Kf() =

K

f()Hence for f S(AK) and a JK we get |a|K f(a) = Kf(a).

2. Fubini Theorem : If G is a discrete subgroup and dx is a Haarmeasure on G then it descends to give a Haar measure on G/ such

that iff L1(G) then F(x) = f(x + ) is a L1-function on G/and

G/

F(x)dx =

Gf(x)dx.

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We continue with the proof of the theorem. Write JK = J1K

R+ (we use

(xt) variable).0 J1K JK N m R+

(f , , s) =

JK

f(x)(x)|x|sdx

=

JK

f(xt)(xt)|xt|sdxdtt

=

R+

J1K

f(xt)(xt)|t|sdx dtt

= 10

J1K

f(xt)(xt)|t|sdxdtt

+ 1

J1K

f(xt)(xt)|t|sdxdtt

The second integral on the right hand side is automatically analytic in

the entire plane. We will use Poisson summation formula to transform the

first integral on the right hand side to a similar integral but one from 1 to

giving analytic continuation. Define,

t(f , , s) = J1K

f(xt)(xt)ts

dx.

Then

t(f , , s) =

J1K

/K

K

f(xt)(xt)

tsdx

=

J1K

/K

K

f(xt)

(xt)tsdx

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since is non-trivial,

t(f,,s) =

J1K

/K

Kf

xt

(xt)ts1dx

=

J1K

f 1xt

(xt)ts1dx

=

J1K

fxt

(x1t)ts1dx

= 1t

(

f , 1, 1 s)

Thus, 10

t(f , , s) =

10

1t

(f , 1, 1 s) = 1

t(f , 1, 1 s)Hence,

(f,,s) =

10

t(f , , s)dt

t+

1

t(f , , s)dt

t

= 1

1t

(f , 1, 1 s)dtt

+ 1

t(f,,s)dtt

This is valid initially for Re(s) > 1 but both integrals are entire giving

analytic continuation to (f,,s) and also functional equation.

Corollary 25.1. From this theorem we get analytic continuation and func-

tional equation for L-functions associated to Grossencharacter.

Proof of Corollary : Let : JK/K C be a Grossencharacterand f : AK C be factorizable f = fp function. We can assume that(fp, p, s) = L(p, s) for almost all place, say outside a finite set S of places

of K. Then from the theorem we have (f , , s) = (f , 1, 1 s).

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pS

p /S

(fp, p, s) =

pS

p /S

(fp, 1p , 1 s)pS

(fp, p, s)

p /SL(p, s) =

pS

(fp, 1p , 1 s)p /S

L(1p , 1 s)

=

pS

(fp, 1p , 1 s)L(1p , 1 s)

pS

L(1p , 1 s)

= pS(

fp,

1p , 1 s)

L(1p

, 1

s)L(1, 1 s)

But using local functional equation

(p, s) =(fp, p, s)

L(p, s)=

(fp, 1p , 1 s)L(1p , 1 s)

We get

L(, s)(, s) = L(1, 1 s)

with (, s) = p (p, s) where (p, s) = 1 at almost all places and is ofthe form abs.

26 Riemann-Roch Theorem

Let K be a function field over a finite field (i.e. finite extension of Fq(t), q

is power of some prime). The field K has valuations, denoted as v, which

are considered as points of the corresponding Riemann surface X or XK. By

a divisor D on X we mean an element of the free abelian group on X, i.e.

D = v nvv where nv Z and v belongs to the set of places of K suchthat nv is nonzero only for finitely many v. We define the degree of a divisor

D =

v nvv to be deg(D) =

nv. We call the finite set of places v with

nv = 0 the support of the divisor D.

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For an element f

K we associate a divisor, called the divisor of f and

denoted as (f), to be described as follows. For v X, let us denote nv = v(f)for the valuation off at the place v. It is easy to see that this is nonzero for

only finitely many v. Then we define (f) =

v v(f)v. One can check some

simple properties of it.

1. IfD1 and D2 are divisors, then deg(D1 + D2) = deg(D1) + deg(D2).

2. Iff1, f2 K, then (f1f2) = (f1) + (f2).

3. Let f

K. Then the degree deg((f)) = 0.

For a divisor D one defines certain vector space of functions,

L(D) = {f K|(f) + D 0}{0}.Lemma 26.1. For a divisor D, L(D) is a finite dimensional vector space

overFq.

There is a divisor on X, called the canonical divisor and denoted by

. Let be a non-trivial character on AK/K. This defines character v :

Kv

C. Define the order of at v tobe the minimal integer nv such that

|Ov .nvv 1. Then the canonical divisor = nv.v is a divisor on X, andis divisor class group.

Theorem 26.1 (Riemann-Roch Theorem). Let K be a function field and

L(D) defined as above. Then there exists g an integer 0 such thatdimL(D)dimL(k D) = deg(D) + (1 g).

Proof : The proof follows from Poisson summation formula applied to

the characteristic function ofOD AK where OD is defined to be the productof nvv Ov for all the places v of K.

27 Artin Reciprocity

Now we move towards algebraic number fields and Artin reciprocity law.

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Let K/k be an abelian extension of number fields. We note few facts from

Galois theory. Let G = Gal(K/k) be the Galois group of K over k, and letq be a prime in K lying over a prime ideal p of k. We define decomposition

group of q in G to be Dq = { G|(q) = q}. As residue field OK/q is afinite field and is a finite extension of Ok/p of degree f. We have canonical

homomorphism,

q : Dq Gal

OK/q

Ok/p

(x mod q (x) mod q)

This map is clearly well defined homomorphism of groups.

Proposition 27.1. The canonical map q : Dq Gal

OK/qOk/p

has the

following properties.

1. The map q is surjective.

2. The map q is injective if and only if the prime p in k is unramified in

K; i.e. if and only if the local extension Kq/kp is unramified.

3. Each Dq extends to an automorphism of the completion Kq that istrivial on the subfield kp. The induced map jq : Dq Gal(Kq/kp) isin fact an isomorphism.

One knows from elementry field theory that Gal

OK/qOk/p

is cyclic, gener-

ated by the Frobenius map x xq where #(k/p) = q. In case ofp beingunramified the Frobenius element F rp Dq G is called the Artin symbolattached to p.

In other language let u be an unramified place of the number field k and v

be a place of the number field K lying over u. Let u be uniformizing element

ofOu and v be that ofOv. Then u.Ov = v.Ov and the decomposition group

Du = Gal

Ov/vOu/u

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This defines the map

r :v /S

kvOv

Gal(K/k)

v F rv

where S denotes set of infinite places as well as set of finite places of k.

Theorem 27.1 (Artin Reciprocity). The map r :

v /S k

v Gal(K/k)

extends uniquely to a group homomorphism r : Jk

Gal(K/k) which is

trivial on k.

Proof : The uniqueness follows from weak approximation theorem i.e.

k vS kv is dense. This implies that k.v /S kv is dense in Jk, hence acharacter which is trivial on k.

v /S k

v is identically 1. In rest of the section

we try to prove Artin Reciprocity.

We have defined a map Jk Gal(K/k) for abelian extensions of num-ber fields. These maps as K varies over all abelian extensions of k form a

compatible system of maps i.e. if K1

K2

k then the following diagram

commutes.

Jkr1

//

r2

##FF

FF

FF

FF

FF

FF

FF

FF

FF

FGal(K1/k)

restriction

Gal(K2/k)

Thus we have a map

Jk lim Gal(K/k) = Gal(kab/k)Theorem 27.2. Letk be a number field. Then,

Jk/(k.k+) = Gal(kab/k)

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i.e. finite abelian extensions of k are in bijective correspondence with open

subgroups of finite index of Jk/k. Where k+ denotes identity component ofk.

Hilbert Class Field : These are maximal abelian everywhere unramified

extension of a number field k. These are finite extensions of k with the

property that Galois group is isomorphic to the class group of k. In terms of

previous theorem these fields correspond to the subgroup

kk.

vkfOv Jk

More generally if U Jk is an open subgroup containing k, then the corre-sponding abelian extension kU of k is called the class field associated to U.

Also kU is unramified at a place v if and only if Ov U.

Ray Class Field : Let C be a cycle in number field k. We defineU(C) = vC Uv(nv) where C = {v1, v2....vl, v} and Uv(nv) = {x Ov|x 1mod (nvv )}. Then we take U = k.U(C) and the corresponding field exten-sion is called Ray class field. Just as kv has filtration by congruence sub-

groups the field k has a distinguished family of abelian extensions generating

Q(n)/Q, which corresponds to Ray class field associated to the ideal ( n).Theorem 27.3 (Cebotaraev Density Theorem). If K/k is a finite Galois

extension of global fields with Galois group G and C is a conjugacy class in

G then the density (Dirichlet density) of primes p in k such that F rp Cis of density |C||G| . In particular, the set X = {p|F rp C} is nonempty andinfinite.

Corollary 27.1 (Dirichlet Theorem). There are infinitely many primes in

any arithmetic progression a + nd,n Z if (a, d) = 1.Proof : Let us look at field extension Q(n) ofQ. Its Galois group is

Z/n and F rp = p. The Cebotaraev density theorem implies that there are

infinitely many primes in arithmetic progression.

For the purpose of the proof of Artins reciprocity law its useful to go

back and forth between ideal theoretic and adelic language.

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Proposition 27.2. LetK be a Galois extension of a number field k. Then

Jk/(k.NKk JK) =

Ik(C)PC.NKk (IK(C))

where C is a cycle in k divisible by ramified primes such thatOv(Cv) NKwkv (Kw) (called admissible cycles) and Ik(C) is ideals in k whichare coprime to C.

Proof : We have

JC =v/C

Ov(C)

(v,C)=1kv

Let kC = JC k, then JC/kC Jk/k. This is injective (by definition) andsurjective by weak approximation theorem (

v/C O

v(C).k =

v/C k

v). This

gives rise to following diagram.

IC/kC Jk/k

Ik(C)PC.NKk (IK(C))

Jk/(k.NKk JK)

Check that the isomorphism in the top horizontol arrow descends to give anisomorphism of the bottom horizontol arrow.

In the light of this proposition we need to define map (called Artin map)

from IK(C) to Gal(K/k). Which we define by v F rv. Observe thatNKk (IK(C)) lies in the kernel of the Artin map for K/k (Galois extension ofdegree m). Let p be a prime in k and q be a prime in K over p, and suppose

this is an unramified prime.

pOK =d

i=1 qiThen N m(qi) = p

md where m = d.[OK/qi : Ok/p]. Which gives F r

m/dp = 1,

thus all the norms are in the kernel of the Artin map. Proving PC is in kernel

of Artin map is non-trivial part.

Steps in the proof of Artin Reciprocity Theorem

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1. Prove the reciprocity for cyclic extensions.

(a) Order of the group Ik(C)PC .NKk (IK(C))

and Gal(K/k) is the same.

(b) The map Ik(C) Gal(K/k) is surjective map.(c) PC belongs to the kernel and this is done via recourse to cyclotomic

theory.

2. Deduce reciprocity for general abelian extensions.

Below we try to prove some of the steps.

Lemma 27.1. The Artin map Ik(C) Gal(K/k) is non-trivial.Proof : If Artin map were trivial then all of the unramified primes are

completely split; i.e. K(s) = mk (s), m = [K : k]. This contradicts simplicity

of the poles at s = 1 ofK and k.

Corollary 27.2. Artin map Ik(C) Gal(K/k) is surjective.Proof : Let G = Gal(K/k), an abelian group. Let H be the image of

Artin map. Let L = KH, the subfield of K fixed by H. Then L = K H = G. But the Artin map for L/k is trivial. Hence by the previous lemma,L = K = G = H.

We will in fact prove that the Artin map is surjective restricted to the

primes in k.

Theorem 27.4 (Universal Norm Inequality). With notations all above and

Artin map,

Ik(C)PC.NKk (IK(C))

[K : k]

Proof : Let H = PC.NK

k (IK(C)) and G = I(C)/H. Given group G andG

C, there is the associated L-function L(s, ). We will look at thebehaviour of this L-function around s = 1. We already know that if = 1

then L(s, ) has simple pole at s = 1 and is holomorphic for = 1. We write

L(s, ) = (s 1)m().f(s)

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where f(s) at 1 is holomorphic and nonzero and

m() 0 if = 1= 1 if = 1

L(s, ) =p

1 (p)

(Np)s

1This makes sense for Re(s) > 1

log(L(s, )) = p

log1 (p)(Np)s

=p

(p)

(Np)s+ higher order terms

The higher order terms are bounded around s = 1. Hence,

log(L(s, )) =

p

(p)(Np)s

around s = 1 (up to bounded function which we

will denote by f g).

log(L(s, )) p

(p)(Np)s

gG

(g)

p

1

(Np)s

where p = g G = I(C)/H.

log(L(s, )) ,gG (g)p1

(Np)s= |G|

pH

1

(Np)s

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On the other hand, L(s, ) = (s

1)m().f(s)

log(L(s, )) m()log 1s 1

log(L(s, ))

1 =1

m()

log

1

s 1

So we get 1 =1 m() log 1s1

|G| pH

1

(Np)s

Note that all primes q in K which are of degree 1 over k have the propertyp = q

OK belongs to the norm from K to k. Hence,

1 =1 m() log 1s1

|G| 1

[K : k]

q of degree 1

1

(Np)s

1[K : k]

K(s)

1

[K : k]

log1