Lectures On Algebraic Number Theory - …dprasad/ant.pdf · Lectures On Algebraic Number Theory Dipendra Prasad Notes by Anupam 1 Number Fields We begin by recalling that a complex

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 a

subfield of C. Any finite extension of Q 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(ζn), ζn = e2πin .

4. Kummer extensions K(a1n ) 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;

1

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

generated Z 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 conjectured

that there are exactly nine imaginary quadratic fields which are PID, and

this was proved in 60’s 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 O∗K for the units in OK i.e. elements in OK

which has inverse in OK . Then O∗K is a group.

Theorem 1.2 (Dirichlet Unit Theorem). The group O∗K is finitely generated

abelian group. The rank of O∗K is = r1 + r2 − 1 , where r1 is number of real

embeddings of K in R and r2 is number of complex embeddings of K in C up

to complex conjugation.

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

1. The group O∗K is finite if and only if K = Q or is a quadratic imaginary

field.

2. If K is real quadratic, then O∗K is Z/2 × Z. If K = Q(√d) then

OK = Z(√d) if d 6≡ 1(mod4) and OK = Z(1+

√d

2) if d ≡ 1(mod4).

Thus existence of units inside OK is equivalent to solving the Pell’s

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

xx = ±1).

2

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 of Q(√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 of Z[m1n ] in OK .

2 Class Group of A Number Field

Definition. A nonzero additive subgroup A ⊂ K is called a fractional ideal

if there exists λ ∈ K such that λA ⊂ 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 =∑

λiµi / λi ∈ A1, µi ∈ 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). Let K be a number field. Then,

ζK(s) =2r1(2π)r2hR

w√dK

.1

s− 1+ holomorphic function

3

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

embeddings of K in C upto complex conjugate, h is class number of K, R is

the 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 −−−→ Kx xZ −−−→ Q

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

in OK as pOK =∏n

i=1 peii with pi prime ideal in OK . This ei is called

ramification index of pi 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 modulo

p 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 fi’s occur and what are

the possible degrees in terms of p.

Example : Let f(x) = x2 − 5, then quadratic reciprocity law gives the

answer. Let p be an odd prime. Then x2 − 5 = 0 has solution in Z/p if and

only if there exist x0 ∈ Z/p such that x20 = 5. And x2

0 = 5 has solution in

Z/p if and only if 5p−12 ≡ 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/5

then f(x) = x2. If (5p) = 1 then f(x) = (x − x0)(x + x0), where x2

0 = 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

4

the polynomial defines an abelian extension of Q.

Let us look at following field extension.

OK −−−→ Kx xZ −−−→ Q

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

pOK =∏g

i=1 peii . 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. Then∑g

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

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

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

the same, ei’s are the same and efg = [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 Theorem

Let 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 −→ Gal

(OK/p

Ok/p

)−→ 0

The 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 σ : Fqe −→ Fqe , x 7→ xq is called the

Frobenius element. It generates the Galois group of Fqe over Fq. In the case

5

of unramified extension the decomposition group gets a prefered element,

called the Artin symbol σp =< 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

over Q. Then Artin symbol defines a map from I(C) −→ Gal(K/Q), where

C 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(λ) = λx. Define trKQ (x)

as trace of lx and NmKQ (x) as determinant of lx. If there is no possibility of

confusion, we will denote trKQ (x), NmKQ (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 can

restrict 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 deg

of K/Q.

Definition. Let B be a bilinear form B : Zd × Zd −→ Z. One can associate

an 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 of Zd.

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

follows. Let B′ be another integral basis of Zd. Then there exists an integeral

matrix A such that B = AB′At. Hence,

dB = detB = det(AB′At) = (detA)2 detB′ = detB′

Here we have used that for A ∈ GL(n,Z), detA = 1 or−1. This discrim-

inant of the trace form on the ring of integers of a number field is called the

6

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 6= 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 the

realms 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. Let V be an n-dimensional R-vector space. A Lattice in V is a

subgroup of the form

Γ = Zv1 + · · ·+ Zvm

with linearly independent vectors v1, · · · , vm of V . The set

Φ = x1v1 + · · ·+ xmvm| xi ∈ R, 0 ≤ xi ≤ 1

is called a fundamental mesh of the lattice.

1This section is written by Purusottam Rath.

7

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 Γ of V 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 Γ in V is complete if and only if there exists a bounded

subset M ⊆ V such that the collection of all translates M + γ, γ ∈ Γ covers

the whole space V.

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

product

( , ) : V× V→ R

Then we have on V a notion of volume (more precisely a Haar measure).

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

vol Γ := vol (Φ)

Note this is independent of choice of basis vectors of the lattice as any such

two 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 also

belongs to X . It is called convex if for any two points x, y ∈ X the line

segment ty+(1− t)x| 0 ≤ t ≤ 1 joining x and y is contained in X. With

these definitions we have,

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

clidean vector space V and let X be a centrally symmetric, convex subset of

V. Suppose that

vol(X) > 2n vol(Γ)

Then X contains at least one nonzero lattice point γ ∈ Γ.

8

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 has

some 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 Γ 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 Γ has the form (r1p+ r3a+ r4b, r2p+ r3b− r4a, r3, r4). Thus

we 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 a

convex centrally symmetric set with volume 2π2p2 which is strictly greater

than 24p2. Since Γ has volume p2, by Minkowski Theorem, B contains a

non-zero point u = (A,B,C,D), say.

Now, 0 < |u|2 < 2p and |u|2 is a multiple of p. 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 =∏τ

C

a 7−→ ja = (τa)

9

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

given by the ordered set

X = ρ1, · · · , ρr, σ1, σ1, · · · , σs, σs

where the ρ’s give the r real embeddings and σ’s and σ’s give the 2s pairwise

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

canonical mapping into Cn =∏

τ∈X C actually lies inside the following set

known as the Minkowski Space.

KR =

(zτ ) ∈

∏τ∈X

C | zρ ∈ R, zσ = zσ

We note that this space is isomorphic to∏

τ∈X R = Rr+2s given by the map

F : (zτ ) 7→ (xτ ) where xρ = zρ, xσ = Re(zσ), xσ = Im(zσ). However this

map 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 on Cn. A little reflection will show that the mapping

j : K −→ KR identifies the vector space KR with the tensor product K⊗Q R,

K⊗Q R ←→ KR

a⊗ x 7→ (ja)x

Lemma 6.3. If I 6= 0 is an ideal in OK, the ring of integers of K , then

Γ := j(I) is a complete lattice in KR with volume√|dK|.[OK : I], where dK

is the discriminant of K.

Proof : Just note that I has an integral basis i.e. a Z basis α1, · · · , αn,so that Γ = Zjα1 + · · · + Zjαn. Consider the matrix A =(τl(αi)) with τ ’s

running over the embeddings of K in C. Then we have

〈jαi, jαk〉 =∑

1≤l≤n

τl(αi)τl(αk) = AAt

10

Then we get

V ol(Γ) = |det(〈jαi, jαk〉)|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, O∗K, 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 : K∗C =∏τ

C∗ −→∏τ

R , (zτ ) −→ (log|zτ |

Note that image of K∗R under the above map lies in the following set[∏τ

R

]+

:=

(xτ ) ∈

∏τ

R | xρ ∈ R, xσ = xσ

Further the multiplicative group K∗C admits the homomorphism N : K∗C −→C∗ given by the product of the coordinates and we have the homomorphism

Tr :

[∏τ

R

]+

−→ R

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

[∏

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

O∗K = ε ∈ OK | NK|Q = ±1S = y ∈ K∗R | N(y) = ±1H = x ∈ [

∏τ

R]+ | Tr(x) = 0

It is clear that j(O∗K) ⊆ S and l(S) ⊆ H. Thus we obtain the homomorphisms

O∗Kj−→ S

l−→ H

11

and the composite λ := l o j : O∗K → H. The image will be denoted by

Γ = λ(O∗K) ⊆ H

and we obtain the

Theorem 6.3. Let K be a number field. Then the sequence

1 −→ µ(K) −→ O∗Kλ−→ Γ −→ 0

is exact, where µ(K) is the group of roots of unity that lie in K.

Proof : The only non obvious part is to show that ker(λ) lies in O∗K.

Now λ(ε) = l(jε) = 0 implies |τε| = 1 for each embedding τ of K. Thus

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

is a point in the lattice jOK of KR. Therefore the kernel of λ 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 Γ. For this we need

the following lemma which depends on Minkowski Lattice Point Theorem.

Lemma 6.4. Let I 6= 0 be an integral ideal of K, and let cτ > 0, for

τ ∈ Hom(K , C), be real numbers such that cτ = cτ and∏τ

cτ > A[OK : I]

where A = ( 2π)s√|dK|. Then there exists a ∈ I, a 6= 0, such that

|τa| < cτ for all τ ∈ Hom(K,C)

Proof : We just note that the the set

Y = (zτ ) ∈ KR | |zτ | < cτ

is centrally symmetric, convex and has volume 2r+sπs∏

τ cτ . Now, Γ = j(I)

is a lattice with volume√|dK|.[OK : I]. So we have

Vol(Y ) > 2r+sπs(2

π)s√|dK|.[OK : I] = 2nVol(Γ)

So the assertion now follows from Minkowski Lattice Point Theorem.

Finally we determine the structure of Γ.

12

Theorem 6.4. The group Γ is a complete lattice in the (r + s− 1)- dimen-

sional R-vector space H .

Proof : Since [∏

τ R]+ has dimension (r+s),H is of dimension (r+s−1) ,

being the kernel of the linear functional Tr. The mapping λ := l o j : O∗K → H

arises by restricting the mapping

K∗ j−→∏τ

C∗ l−→∏τ

R

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

(xτ ) ∈∏

τ R | |xτ | ≤ c contains only finitely many points of Γ. Since

l((zτ )) = (log|zτ |), the preimage of this domain with respect to l is the

bounded domain (zτ ) ∈

∏τ

C∗ | e−c ≤ |zτ | ≤ ec

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

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

is a complete lattice in H. We chose real numbers cτ > 0, for τ ∈ Hom(K ,

C), satisfying cτ = cτ and

C =∏τ

cτ > A[OK : I]

where A = ( 2π)s√|dK|, and we consider the set

W = (zτ ) ∈ KR | |zτ | < cτ

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

Wy = (zτ ) ∈ KR | |zτ | < cτ

where cτ = cτ |yτ | and one has cτ = cτ ,∏

τ cτ =∏

τ cτ = C because∏

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

ja = (τa) ∈ Wy, a ∈ OK, a 6= 0

13

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

α ∈ OK of a given norm NK|Q = a. Thus we may pick up a system of

α1, · · · , αM ∈ OK, αi 6= 0 such that every a ∈ OK with 0 < |NK|Q(a)| ≤ C is

associated to one of these numbers. The set

T = S ∩M⋃i=1

W (jαi)−1

is a bounded set as W is bounded and we have

S =⋃ε∈Θ∗

K

Tjε

Because for any y ∈ S, we find, by the above, an a ∈ OK, a 6= 0, such that

ja ∈ Wy−1, so ja = xy−1 for some x ∈ W . Since

|NK|Q(a)| = |N(xy−1)| = |N(x)| <∏τ

cτ = C,

a is associated to some αi , αi = εa, ε ∈ O∗K. Consequently, we have

y = xja−1 = xj(α−1i ε)

Since y, jε ∈ S, one finds xjα−1i ∈ S ∩Wjα−1

i ⊆ T , and thus y ∈ Tjε. Now

M = l(T ) is also a bounded set as for any x = (xτ ) ∈ T , the absolute values

|xτ | are bounded above and are also away from zero (because∏

τ xτ = 1).

Now we have

H =⋃γ∈Γ

(M + γ)

Hence the translates (M + γ), γ ∈ Γ , M bounded , covers the whole space

H. Thus Γ is a complete lattice in H(using lemma 6.2).

From previous theorems we immediately deduce,

Theorem 6.5 (Dirichlet’s Unit Theorem). The group of units O∗K of OK is

isomorphic to the finitely generated abelian group given by

µ(K) × Zr+s−1

where µ(K) is the finite torsion group consisting of the roots of unity con-

tained in K.

14

7 Discrete Valuations on a field

Definition. Let K be a field. Then Discerete valuation on the field K is a

map v : K∗ −→ Z such that

• v(xy) = v(x) + v(y)

• v(x+ y) ≥ minv(x), v(y)

Given a discrete valuation on a field K, let A = 0⋃x ∈ K∗|v(x) ≥ 0.

Then A is a ring, called the valuation ring of v.

Examples :

1. Let us consider discrete valuations on Q. Let p be a prime in Z. For

any a ∈ Q write a = pr.b such that neither the numerator nor the

denominator of b has p power. Define vp : Q∗ −→ Z by vp(a) = r. This

is the discrete valuation on Q associated to the prime p, called p-adic

valuation. The valuation ring is

A = Z(p) = Z[1

q

]q 6=p

It can be seen that these are the only discrete valuations on Q.

2. Let K be a number field. Let p be a prime ideal in OK . Then the

localization (OK)p is a discrete valuation ring with p (actually image)

as prime ideal. And for any x ∈ K we get xOK = pn where n ∈ Z. We

define vp : K∗ −→ Z by vp(x) = n. Then this is a valuation on K. It

can be seen that these are all the valuations on K.

3. (Function fields) Let f(x) be an irreducible polynomial in K[x].

Write a general rational function g(x) = f(x)r.b(x) such that neither

the numerator nor the denominator of b(x) is divisible by f(x). We

define vf : K(x)∗ −→ Z by vf (g(x)) = r. This gives a discrete val-

uation on K(x). It can be seen that all the discrete valuations on

K[x] is vf for some irreducible polynomial in K[x], or is v∞ given by

15

v∞

(fg

)= deg(f)−deg(g) for f, g ∈ K[x]. And a function field is finite

extension of the field K(x), or equivalently finitely generated extension

of K of transcendence degree 1. If K is algebraically closed then any

irreducible polynomial is just a liner polynomial, and hence discrete

valuations are in one-one correspondance with points of A1 together

with valuation v∞.

4. (Global Field) A global field is either an algebraic number field or a

function field over finite field Fq.

Proposition 7.1. An integral domain A is a discrete valuation ring (dvr)

if and only if A is a principal ideal domain having a unique nonzero prime

ideal.

One notes that for any Dedekind domain localization at any prime ideal

is a discrete valuation ring.

Given v : K∗ −→ Z, a discrete valuation with valuation ring A, one can

define m ⊆ A to be m = x ∈ A|v(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 its

maximal 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

π ∈ A such that v(π) = 1. Elements π ∈ A such that v(π) = 1 are called

uniformizing 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.

16

Examples :

1. The usual absolute value on Q.

2. For p-adic valuations on Q we can define absolute values as |x|p =

p−vp(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 such

that |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 of Q 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 for

every 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

⊗Z Q is a field containing Zp such that Qp = Zp[

1p].

Theorem 7.1 (Ostrowski’s Theorem). Let K be a prime global field i.e.

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

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

by either the usual absolute value, denoted as |.|∞, or a p-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 irreducible

polynomial p(t) ∈ R.

17

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 fractional

ideal 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 −→ R≥0 by

|x|p = (#OK/p)−np(x). The completion of K with respect to this absolute

value is written as Kp.

Any element of a discrete valuation ring A looks like

a0πr + a1π

r+1 + ....

where ai’s are representations of A/m , a0 6= 0 in A/m and r ∈ Z.

Example : Let us take K = C[|t|][t−1], which is a field. Any element of

this field looks like

f(t) =∑

n≥m∈Z

antn

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 of Q then K⊗

Q Qp is a separable algebra which is a

product of fields.

K⊗

Q

Qp =⊕v|p

Kv

Similarly,

K⊗

Q

R = Rr1⊕

Cr2

18

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(ζp), ζp = e2πip . The

Galois group in this case is Gal(K/Q) = (Z/p)∗. In general Q(ζn) 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 of Q, i.e. K is Galois over Q with abelian Galois group. Then K

is contained in Q(ζn), for some n.

Proposition 8.1. Let K = Q(ζp), ζp = e2πip . Then the ring of integers of K

is OK = Z[ζp].

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[ζp] = Z[1 − ζp] ⊂ OK . Let x ∈ OK we write

x = a0 + a1(1 − ζp) + ...... + ap−2(1 − ζp)p−2 with ai ∈ Q. We need to

show ai ∈ Z. By using valuation theory we claim that there exists a val-

uation with property that v(1 − ζp) = 1 , v(p) = p − 1. To see this we

note,

f(x) = 1 + x+ x2 + .....+ xp−1

=

p−1∏i=1

(x− ζ ip).

Putting x = 1 in this equation,

19

p =

p−1∏i=1

(1− ζ ip)

= (1− ζp)p−1.∏ 1− ζ ip

1− ζp

Note that 1−ζp1−ζi

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

such that ij = 1 mod p.Therefore

1− ζp1− ζ ip

=1− (ζ ip)

j

1− ζ ip= 1 + ζ ip + ..... ∈ Z[ζp]

proving that 1−ζp1−ζi

p∈ Z[ζp]. Clearly

1−ζip

1−ζp ∈ Z[ζp]. Thus 1−ζp1−ζi

p∈ Z[ζp] is a unit

in Z[ζp]. Thus the equation p = (1− ζp)iπ implies that v(p) = p− 1. As the

degree of field is p − 1 which is equal to ramification index of 1 − ζp which

shows that < 1 − ζp > is a prime ideal. There exists a valuation on Q(ζp)

s.t. v(x) ≥ 0 ∀x ∈ OK with v(1− ζp) = 1 , v(p) = p− 1. Also

v(x) = min(v(a0), .......v(ap−2(1− ζp)p−2)

)= min (v(a0), .......v(ap−2)) ≥ 0

This proves that ai’s do not have p in the denominator. The rest of the

argument is simpler.

Let x = b0 + b1ζp + ...... + bp−2(ζp)p−2 with bi ∈ Q and bi have no p in

denominator. Note that if x ∈ OK then tr(xζ−ip ) ∈ Z ∀i. Using trζjp =

−1 ∀j 6= 0 and tr1 = p− 1 we get

(p− 1)bi − (b0 + ....bi−1 + bi+1 + ....+ bp−2) ∈ Z

pbi −p−2∑i=0

bi ∈ Z.

20

This implies that p(b0 − bi) ∈ Z. Since bi’s 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 + b1ζp + ......+ bp−2ζp−2p

= b0 + (b1 − b0)ζp + ......+ (bp−2 − b0)ζp−2p + b0(ζp + ......+ ζp−2

p )

= (b1 − b0)ζp + ......+ (bp−2 − b0)ζp−2p + b0(1 + ζp + ......+ ζp−2

p )

= −b0ζp−1p +

∑(bi − b0)ζ ip

On taking trace we get b0 ∈ Z. 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) = trKQ (xy).

Definition. The set a′ = 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. O′K = x ∈ K|tr(xy) ∈ Z, ∀y ∈ OK. Then K ⊇ O′K ⊃ OK

and O′K is a fractional ideal. The inverse of O′K, denoted as δ, is called the

“different”, which is an ideal in OK.

Definition (Norm of an ideal). Let I ⊂ 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 Nm(δ) = disc(K).

Exercise: disc(Q(ζp)/Q) = pp−2

21

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

ramification index of p over p. Then pe−1 divides δ. Hence, in a number field

the primes which are ramified (ramification index ≥ 2) are exactly primes

dividing the discriminant of the number field.

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

Nm(α) = |NmK/Q(α)|.An example calculation of different :

Let p ∈ Z be a prime. Take K = Qp(√p) then OK = Zp[

√p], for any

prime p. Let us take case p 6= 2. We claim δK =<√p >. Recall that

δ−1K = x ∈ K|tr(xy) ∈ Zp,∀y ∈ OK. Let a + b

√p, a, b ∈ Zp is an element

of Zp[√p]. Then 1√

p(a+ b

√p) = a√

p+ b =⇒ tr( a√

p+ b) = 2b ∈ Zp∀a, b. That

means 1√p∈ δ−1

K .

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

√2) =

12a+ b√

2=⇒ tr(1

2a+ b√

2) = a ∈ Z2 =⇒ 2 ∈ δK .

Lemma 9.1. If L is a finite extension of a local field K, with πK and πL uni-

formizing parameter such that πeL = πK .u, u a unit in OL. Then πe−1L /δL/K.

In fact δL/K = πe−1L ⇐⇒ p/e.

Proof: Let us prove that πe−1L /δL/K . For this we show tr(πe−1

L OL) ⊂OK . But then tr(πe−1

L OL) = tr(πL

πKOL

)= 1

πKtr(πLOL) ⊂ 1

πKπKOK ⊂ OK .

10 The Riemann Zeta Function

The Riemann zeta function is defined as follows:

ζQ(s) =∑n≥1

1

ns=∏p≥2

1

1− 1ps

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

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

equivalent to the fundamental theorem of arithmetic: every positive integer

22

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

The function ζQ(s) has an analytic continuation to a meromorphic function

in s-plane with a simple pole at s = 1 with residue 1. That is

ζQ(s) =1

s− 1+ holomorphic function around s = 1

The Riemann zeta function satisfies a functional equation relating its value

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

ξ(s) = π−s2 Γ(

s

2)ζQ(s)

The function ξ(s) satisfies ξ(s) = ξ(1− s).Riemann conjectured in 1850’s that all the zeros of ζQ(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.

ζK(s) =∑

I 6=0,I⊂OK

1

(NI)s=∏P

1

1− 1(NP)s

where I ranges over all nonzero ideals of OK , NI 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 ζK(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). Let K be a number field. Then,

ζK(s) =2r1(2π)r2hR

w√dK

.1

s− 1+ holomorphic function around s = 1

23

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

complex embeddings of K in C upto complex conjugate, h is class number of

K, 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

ζK(s) =∑

I 6=0,I⊂OK

1

(NI)s

=∏P

1

1− 1(NP)s

for Re(s) > 1. Since the equality∑I 6=0,I⊂OK

1

(NI)s=

∏P

1

1− 1(NP)s

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

log(ζK(s)) = −∑P

log(1− 1

(NP)s)

=∑P,m≥1

1

m(NP)ms

We have NP = pf where P is prime lying over p and number of primes lying

over p of resudial degree f is ≤ N = [K : Q]. Then,

log(ζK(s)) ≤∑

p, m≥1, N≥f≥1

N

mpfms

where p is prime.

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

log(ζK(s)) =∑ ap

ps

ap denotes number of primes above p of degree 1.

24

Thus log(ζK(s)) for s > 1 is bounded by∑

Nps , thus it is convergent for

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

log(ζK(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 ζA1(s) of the affine line A1 over the finite field

Fp as follows.

ζA1(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 Nf(x) = pn where n =

deg f(x). But then

ζA1(s) =∏p(x)

1

(1− 1Np(x)

)s

=∑f(x)

1

(Nf(x))s

=∑d

pd

pds

=1

1− pps

3. Let K = Q(i), then ζK(s) =∑

1(m2+n2)s

25

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 of

quadratic residues in Ip and Np = Number of quadratic non-residues in Ip.

Then the question is

Is Rp = Np ?

If p ≡ 1 mod 4, then(−1p

)= 1. Hence the map

φ :

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 let hp

denote the class number of the quadratic field Q(√−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 ≡ 3

mod 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 of Q. 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 real

2This section is written by Anirban Mukhopadhyay.

26

number X > 0, we define N(X, C) = number of integral ideals I in C such

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

Theorem 13.2.

limX→∞

N(X, C)X

= κ

where

κ =

2 log η√

dif d > 0, η is the unique fundamental unit > 1,

2π

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,

lims→1+

(s− 1)ζK(s) = hκ

where h =Class number of K and ζK 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 ζK .

Lemma 13.1. For s > 1, we have

ζK(s) = ζ(s)Ld(s)

where ζ(s) is the Riemann zeta function and

Ld(s) =∞∑m=1

(d

m

)m−s

with(dm

)denoting the quadratic residue symbol.

27

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

lims→1+(s− 1)ζ(s) = 1 we derive from the above Lemma

lims→1+

(s− 1)ζK(s) = lims→1+

(s− 1)ζ(s) = Ld(1).

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

h =

2 log η√

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

2π

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 number

of roots of unity in K, w = 2. For a Dirichlet character χ, the Dirichlet

series is defined to be L(s, χ) =∑∞

n=1χ(n)ns . In this case we define χ to be the

quadratic character ( .p). Using above expression for class number one can

easily derive that hp =√p

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

sum.

Lemma 13.2. We have

L(1, χ) =iπτ1(χ)

p2

p−1∑k=1

χ(k)k.

where

τk(χ) =∑

1≤a<p

χ(a)ρak.

with ρ denoting the p-th primitive root of unity.

proof : For s > 1 we have

L(s, χ) =∞∑n=1

χ(n)

ns=∑

1≤a<p

χ(a)∑

n≡a mod p

1

ns.

We write ∑n≡a mod p

1

ns=∞∑n=1

cnns

28

where cn = 1 if n ≡ a mod p and cn = 0 otherwise. If ρ denote the p-th

primitive root of unity, we can write

cn =1

p

p−1∑k=0

ρ(a−n)k.

Thus

L(s, χ) =1

p

∑1≤a<p

χ(a)∞∑n=1

p−1∑k=0

ρ(a−n)k =1

p

p−1∑k=0

∑1≤a<p

χ(a)ρak

∞∑n=1

ρ−nk

ns.

We observe that τ0(χ) = 0 and χ(k)τk(χ) = τ1(χ). Using these observations

along with the fact that L(s, χ) is continuous in (0,∞), we get

L(1, χ) =τ1(χ)

p

p−1∑k=1

1

χ(k)log

1

1− ρ−k.

Let S be the sum in the last formula. Since χ is an even character we have

2S =

p−1∑k=1

1

χ(k)

[log

1

1− ρ−k− log

1

1− ρk

].

Using the expressions log(1− ρ−k) = i(π2− kπ

p

)+ log |1− ρ−k| we get

S =iπ

p

p−1∑k=1

χ(k)k.

This completes the lemma.

We recall that τ1(χ) = i√p. Thus

hpp = −p−1∑k=1

χ(k)k.

29

We split this sum as

−hpp =

(p−1)/2∑k=1

(k

p

)k +

p−1∑k=(p+1)/2

(k

p

)k

=

(p−1)/2∑k=1

(k

p

)(k − (p− k))

= 2

(p−1)/2∑k=1

(k

p

)k − p(Rp −Np).

On the other hand we can write,

−hpp =

(p−1)/2∑k=1

(2k

p

)2k +

(p−1)/2∑k=1

(p− 2k

p

)(p− 2k)

= 4

(2

p

) (p−1)/2∑k=1

(k

p

)k − p

(2

p

)(Rp −Np).

from above two we get(2

(2

p

)− 1

)hpp = −p

(2

p

)(Rp −Np).

Now Theorem 13.1 follows from the fact that (2p) = 1 or −1 according as

p ≡ 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 =

(ξ1, ξ2) ∈ R2|

(ξ1r,ξ2r

)∈ ∆

30

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 of

K and J be an integral ideal in C−1. Then for any integral ideal I ∈ C,IJ = αOK with α ∈ J . Conversely if α ∈ J then J |αOK , hence I = J−1αOKis an integral ideal in C. Further |NK/Q(α)| = N(I)N(J). Thus N(I) < X if

and only if |NK/Q(α)| < XN(J) = Y (say). We can conclude that N(X, C) =

Number of pairwise non-associative elements α ∈ J such that |NK/Q(α)| < Y .

We divide the proof in two cases.

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

any α ∈ J, α 6= 0, there exist an integer m such that

0 ≤ log

∣∣∣∣ ω

|N(ω)|1/2

∣∣∣∣ < log η (13.3.1)

where ω = ηmα and we use N to denote norm without mentioning the fields

involved. It is easy to see that if two associates ω1 and ω2 satisfies (13.3.1)

then ω1 = εω2 with ε = ±1. Hence

2N(X,C) =

∣∣∣∣ω ∈ J | 0 < |N(ω)| < y, 0 ≤ log

∣∣∣∣ ω

|N(ω)|1/2

∣∣∣∣ < log η

∣∣∣∣ .Let β1, β2 be an integral basis for J and let

Ω =

(ξ1, ξ2) ∈ R2|0 < |ξ1||ξ′1| < 1, 0 < log

∣∣∣∣ |ξ1|1/2|ξ′1|1/2

∣∣∣∣ < log η

where ξ = ξ1β1 + ξ2β2 and ξ′ = ξ1β

′1 + ξ2β

′2, β

′1, β

′2 denote the conjugates of

β1, β2. We observe that Ω is bounded.

2N(X, C) = Number of lattice points in Ω√y

+Number of lattice points inAy

where

Ay =(ξ1, ξ2) ∈ Z | |ξ|2 ≤ y, |ξ| = |ξ′| 6= 0

31

and ξ 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

y→∞

NΩ(√y)

yN(J) = N(J)Area of Ω.

Area of Ω can be computed to be 4 log η

N(J)√d. This completes the proof in this

case.

Case 2: d < 0

In this case we have

wN(X, C) = |ω ∈ J | 0 < N(ω) < y| .

Let

Ω =(ξ1, ξ2) ∈ R2|0 < |ξ1β1 + ξ2β2|2 < 1

.

Then wN(X, C) = NΩ(√y), hence

limX→∞

wN(X,C)

X= lim

y→∞

NΩ(√y)

yN(J) = N(J)Area of Ω.

Here area of Ω is calculated to be 2πN(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 real

number X, we set

A(X) =∑m<X

am.

Suppose that

limX→∞

A(X)

X= c.

Then the series

f(s) =∞∑m=1

amms

converges for s > 1 and we have

lims→1+

(s− 1)f(s) = c.

32

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= hκ.

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

lims→1+

(s− 1)ζk(s) = hκ.

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,

lims→1+

∑p∈M

1(Np)s

log 1s−1

In one of the earlier lecture we had observed that

log1

s− 1∼ log ζK(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 over Q then the set of primes in Q which split completely

in K has density = 1d.

33

Proof: We have M = p ∈ Z | pOK = p1...pd, p′s are completely split.

So using remark preceding theorem we get

d∑p∈M

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 over Q 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 of K 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 of Q, K2 be degree d2 extension of Q and

K1K2 degree d extension of Q. Then, split primes in K1 are of density 1d1

,

split primes in K2 are of density 1d2

and split primes in K1K2 are of density1d

. 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 in Q which split completely in K are exactly the primes

which split completely in L.

Proof: We have proved earlier that for field extensionsK1, 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 , ω) be Galois clousre of K, where ω 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.

34

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 split

primes 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 primes

in Z falls in one of the above three categories. Hence the density in

this case will be 1− 16− 1

2= 1

3.

15 Decomposition of primes in the cyclotomic

fields

Let K = Q(ζn), where ζn = e2πin and OK = Z[ζn].

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

primes in Z 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(ζpm). 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 = φ(pm).

Observe that the minimal polynomial satisfied by ζpm is

1 + xpm−1

+ (x2)pm−1

+ .....+ (xp−1)pm−1

=∏ζ

(x− ζ)

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

p =∏

(i,p)=1

(1− ζ ipm). i ∈ Z/pm

35

Let us take π = (1− ζpm) then σ(π) = 1− σ(ζpm) for any σ embedding of K

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

sends ζpm to ζ ipm we get

p =∏

σ∈Gal(K/Q)

σ(π).

Observe that1−ζi

pm

1−ζpm∈ Z[ζpm ] is a unit for (i, p) = 1. Hence we get p =

(1−ζpm)φ(pm).u where u is a unit in Z[ζpm ]. By taking p =< 1−ζpm > we get

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

ramifies.

Lemma 15.1. Let K = Q(ζn) be a cyclotomic field. Then the Galois group

Gal(Q(ζn)/Q) is naturally isomorphic to (Z/n)∗ given by a ∈ (Z/n)∗ acting

on Q(ζn) by ζn −→ ζan. Further, for p prime in Z with (p, n) = 1, the

Frobenius element σp ∈ Gal(K/Q) is the element p ∈ (Z/n)∗.

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

Q(ζm) = Q. This

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

Corollary 15.1. Let p be a prime in Z. Then p splits completely in Q(ζn)

if and only if p ≡ 1(modn). In general, for p 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[ζn]. The frobenius automorphism σp : Z[ζn] −→ Z[ζn]

is defined by σpx ≡ xp mod p for all x ∈ Z[ζn]. But σp(ζn) = ζpn whereas

order of σp is the degree of residue field extension of p which is Z[ζn]/p. Now

σfp = 1⇔ σfp (ζn) = ζn ⇔ (ζn)pf

= ζn ⇔ pf ≡ 1 (mod n)

Since p is unramified we get p splits into φ(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(ζp). For this consider

G =

p−1∑a=1

(a

p

)e

2πiap

36

It is easy to see that G2 = (−1p

)p = p hence G = ±√p belongs to Q(ζp).

Let q be a prime. Then σq ∈ gal(Q(ζp)/Q) = (Z/p)∗ and from previous

lemma 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 p

is a quadratic residue mod q. Combining with earlier statement we get p is

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

Remark: One can prove that G =√p if p ≡ 1(mod4) and G = i

√p if

p ≡ 3(mod4).

16 Subfields of Cyclotomic Fields

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

i pnii .

Then Gal(Q(ζn)/Q) = (Z/n)∗ =∏

i(Z/pnii )∗. Using Galois theory there is

a one-one correspondence between the subfields of Q(ζn) and subgroups of

Gal(Q(ζn)/Q) = (Z/n)∗.

Definition. Let G be an abelian group. A group homomorphism χ : 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 group

G. This correspondence is given by associating to a subgroup A of G, the set

of all characters of G which are trivial on A.

Combining this lemma with Galois theory we get that subfields K of Q(ζn)

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

A character of (Z/n)∗ which is a homomorphism χ : (Z/n)∗ −→ C∗ is

called a Dirichlet Character. Let n|m are positive integers. Then we get a

homomorphism (Z/m)∗ → (Z/n)∗.

37

Definition. Let χ : (Z/n)∗ −→ C∗ be a Dirichlet character. Let n0 be the

smallest divisor of n such that χ factors through (Z/n0)∗, i.e. there exist

χ′ : (Z/n0)∗ −→ C∗ such that following diagram commutes.

(Z/n)∗ //

χ

##HHHHHHHHHHHHHHHHHHH(Z/n0)

∗

χ′

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(ζn), for some n. Let this abelian extension K be defined by

a group of characters X = χ : (Z/n)∗ −→ C∗. Then

|dK | =∏χ∈X

cond(χ)

Examples :

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

G = (Z/p)∗ and let X = χ : (Z/p)∗ −→ C∗ be character group.

Then conductor of χ = p, χ ∈ X if and only if χ is not trivial character.

Hence using conductor-discriminant formula we get |dK | = pp−2.

2. Let p be a prime and consider K = Q(ζp2). Then the Galois group

G = (Z/p2)∗ ∼= Z/(p2 − p) and let X = χ : (Z/p2)∗ −→ C∗ be

character 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 we

get |dK | = p2p2−3p.

38

17 L-functions associated to a Dirichlet char-

acter

Let χ : (Z/n)∗ −→ C∗ be a Dirichlet character. We can extend χ to define a

function χ on Z by

χ(a) =

χ(a) if(a, n) = 1

0 otherwise

Definition. Let χ be a Dirichlet character. Then

L(s, χ) =∑n≥1

χ(n)

ns

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

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 χ = 1. These series also have

functional equations like Riemann zeta functions.

We recall that if ξ(s) = π−s2 Γ( s

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

the Riemann zeta function is expressed as ξ(s) = ξ(1− s). Let K be number

field and ζK(s) be the Dedekind zeta function of the number field K. Define

ξK(s) = (√|dK |)s(π

−s2 Γ(

s

2))r1(π−sΓ(s))r2ζK(s)

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

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

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

tor of χ appears.

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

product of L-functions as

ζK(s) =∏

L(s, χ)

39

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 Tate’s thesis

In the following sections we will be discussing the Tate’s 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

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

fields (finite extensions of Q) or function fields (finite extensions of Fp(t)).

19 Harmonic analysis on p-adic fields

Let G be a locally compact abelian topological group. Let us denote unitary

dual of G by G = ξ : G −→ S1. Then G is a locally compact topological

group on which the topology is the compact open topology defined by taking

basic open sets of G as W (C,U) = ξ ∈ G|ξ(C) ⊂ U,where C is a compact

set 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 on R. Let us take the continuous group

homomorphism ψ : R −→ C∗ defined by ψ(x) = exp(2πix). Then any

unitary character (continuous group homomorphisms from R to S1) on Ris of the form ψy : x 7−→ ψ(xy) for some y ∈ R. Let us define unitary

dual of R by R = ξ : R −→ S1| ξ unitary character of R. Then the map

40

y −→ ψy defines isomorphism of R with its unitary dual R. One defines for

an integrable function f its Fourier transform by

f(y) =

∫Rf(x)ψ(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

ψ : Qp −→ C∗ given as follows. For x ∈ Qp write x as

x =∑

−∞<j<∞

aj(x)pj

where aj(x) are integers 0 ≤ aj(x) ≤ p− 1 and aj(x) = 0 for all but finitely

many j < 0. Define

ψ(x) = exp

(2πi

∑−∞<j<0

aj(x)pj

)

This means that ψ(x) = 1 for x ∈ Zp.

Lemma 19.1 (Pontryagin Duality). The map y 7−→ ψy where ψy(x) = ψ(xy)

from Qp to Qp is an isomorphism of topological groups.

Proof: (1) If ψ ∈ Qp, then there is an integer k such that ψ = 1 on

p−kZp.

Since ψ is continuous, there is an integer k such that ψ maps p−kZp into

z ∈ S1||z− 1| < 1. But p−kZp is a subgroup of Qp, so its image under ψ is

a subgroup of S1; hence equals 1.Any ψ ∈ Qp is completely determined by its values on the numbers pj, j ∈

Z. So if ψ 6= 1 there is an integer j0 such that ψ(pj) = 1 for j ≥ j0, but

ψ(pj0−1) 6= 1.

(2) Suppose ψ ∈ Qp, ψ(1) = 1, and ψ(p−1) 6= 1. There is a sequence

cj∞0 with c0 ∈ 1, ..., p − 1 and cj ∈ 0, 1, ..., p − 1 for j ≥ 1 such that

ψ(p−k) = exp(2πi∑k

j=1 ck−jp−j)

for k = 1, 2, 3....

41

Let ωk = ψ(p−k); then

ωpk+1 =(ψ(p−k−1)

)p= ψ(p.p−k−1) = ψ(p−k) = ωk

Now ω1 6= 1 and ω0 = 1 so ω1 = exp(2πic0p−1) for some c0 ∈ 1, ..., p − 1.

Proceeding by induction, suppose ωk = exp(2πi∑k

j=1 ck−jp−j). Since ωk=1

is pth root of ωk, there exists ck ∈ 0, 1, ..., p− 1 such that

ωk+1 = exp

(2πi

k∑j=1

ck−jp−j−1

)exp(2πickp

−1) = exp

(2πi

k+1∑j=1

ck+1−jp−j

).

(3) If ψ ∈ Qp, ψ(1) = 1, and ψ(p−1) 6= 1, there exists y ∈ Qp with |y| = 1

such that ψ = ψy.

Let us take cj∞0 as above and set y =∑∞

0 cjpj. Then |y| = 1 since

c0 6= 0, and for k ≥ 1,

ψ(p−k) = exp

(2πi

k∑j=1

ck−jp−j

)= exp

(2πi

−1∑j=−k

cj+kpj

)= ψ1(

∞∑−k

cj+kpj)

= ψ1(p−ky) = ψy(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 of Qp then we define character

using trace map combining with character ψ as follows

Ktr−→ Qp

ψ−→ C∗

This is a continuous unitary character on K.

Lemma 19.2 (Pontryagin Duality). Let K be a field which is finite extension

of Qp. Let ψ be a non trivial character of K (for example above defined ψ).

Then ψ defines a topological isomorphism of K to its dual K. Which is

K −→ K defined as y 7−→ ψy where ψy(x) = ψ(xy).

42

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 ψy = 1 if and only if y = 0. Since ψ is a non trivial character

there exists y0 such that ψ(y0) 6= 1 and if y 6= 0 =⇒ ψ(y0) = ψ(y0y−1y) =

ψy(y0y−1) 6= 1 =⇒ ψy 6= 1. And if y = 0 =⇒ ψy = 1. This insures

injectivity 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 = ψy|y ∈ K ⊂ K. Let φ be a character in the closure of

the image of K. Thus assume that there exists a sequence xn ∈ K such

that ψxn −→ φ. We can assume that xn’s do not have a convergent sub-

sequence (else if xn −→ x then ψxn −→ ψx = φ) . Thus we assume that

xn −→∞. And ψxn −→ φmeans that for any compact setX ⊂ K,ψxn(x) −→φ(x)∀x ∈ K uniformly on K. That is |ψxn − 1| < ε∀x ∈ K for n sufficiently

large. 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 gives

ψxn(K) = 1, a contradiction to xn −→∞.

From previous proposition K ∼= K so fourier analysis of standard kind

becomes available on K. For any function f ∈ L1 we define,

f(x) =

∫K

f(x)ψ(xy)dy

where dy is a Haar measure on K and ψ : K −→ C∗ is a fixed non-trivial

character.

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

pending on ψ) 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

43

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

defined by δ−1K|Qp

= x ∈ K|tr(xOK) ⊂ Zp .

Lemma 19.3. If A is a compact abelian group and χ : A −→ C∗ is a

character, then∫Aχ(a)da = 0 if χ 6= 1 and

∫Aχ(a)da = vol(A) if χ = 1.

Proof: If χ = 1 then∫Aχ(a)da =

∫A

1.da = vol(A). If χ 6= 1 there

exists x0 ∈ A such that χ(x0) 6= 1. Put I =∫Aχ(a)da then I =

∫Aχ(a)da =∫

Aχ(a+ x0)da = I.χ(x0) =⇒ I = 0.

Let us denote the characteristic function of OK , the valuation ring of K,

by the χOK. The fourier transform will be

χOK(y) =

∫K

χOK(x)ψ(xy)dx =

∫OK

ψ(xy)dx

Then χOK(y) = 0 if y /∈ δ−1

K and χOK(y) = vol(OK) if y ∈ δ−1

K . Where

ψ(x) = exp 2πi tr(x) is a character of K. Then,

χOK(y) =

∫K

χOK(x)ψ(xy)dx = vol(OK)

∫δ−1K

ψ(xy)dx

χOK(0) = vol(OK)

∫δ−1K

1.dx = vol(OK)N(δK)vol(OK)

But χOK(0) = χ(0) = 1. This implies that vol(OK) = 1√

N(δK), where

N(δK) = [δ−1K : OK ] = [OK : δK ].

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 χ be a character on K∗.

Then we define local zeta function by

Z(f, χ, s) =

∫K∗f(x)χ(x)|x|sd∗x

44

for s ∈ C, where d∗x is normalised Haar measure on K∗. Initially Z(f, χ, s)

is defined only for Re(s) > 0 but actually it has meromorphic continuation

to whole of the s-plane. In fact, Z(f, χ, s) ∈ C(q−s) where q = |π|, π is

uniformizing parameter. Let us prove this in case ξ is a ramified character

of K∗ (i.e. χ(O∗K) 6≡ 1), where O∗K is the group of invertible elements of the

ring OK .

Z(f, χ, s) =

∫K∗f(x)χ(x)|x|sd∗x

=

∫O∗K

χ(x)|x|sd∗x+ a term belonging to C[qs, q−s]

Now let us look at the term∫O∗K

χ(x)|x|sd∗x =

∫∪∞n=0[πnOK−πn+1OK ]

χ(x)|x|sd∗x

=∞∑n=0

∫πnO∗K

χ(x)|x|sd∗x

=∞∑n=0

∫O∗K

χ(πnx)q−ns|x|sd∗x

=∞∑n=0

χ(πn)

∫O∗K

χ(x)q−ns|x|sd∗x

= 0

Therfore for ramified character Z(f, χ, s) ∈ C[qs, q−s].

If χ|O∗K ≡ 1 i.e. χ is unramified then by a similar calculation

Z(χOK, χ, s) =

∞∑n=0

χ(πn)q−nsvol(O∗K) =1

1− χ(π)qs

vol(O∗K)

The local zeta functions satisfy a certain functional equation which is

basic to the whole theory.

45

Theorem 20.1. Let K be a local field, χ a non-trivial character and f ∈S(K) then for all g ∈ S(K),

Z(f, χ, s)Z(g, χ−1, 1− s) = Z(g, χ, s)Z(f , χ−1, 1− s).

Proof: This follows by definitions as we will see.

Z(f, χ, s)Z(g, χ−1, 1− s) =

∫K∗×K∗

f(x)χ(x)|x|sg(y)χ−1(y)|y|1−sd∗xd∗y

We apply change of variable (x, y) −→ (x, xy).

=

∫K∗×K∗

f(x)χ(x)|x|sg(xy)χ−1(xy)|xy|1−sd∗xd∗y

=

∫K∗

(∫K∗f(x)g(xy)|x|d∗x

)χ(y−1)|y|1−sd∗y

=

∫K×K×K

f(x)g(z)ψ(xyz)|y|−sχ(y−1)dxdydz

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

Corollary 20.1.

Z(f, χ, s) = ρ(χ, s)Z(f , χ−1, 1− s)

where ρ(χ, s) is independent of f .

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

1−χ(π)qs

if

χ is unramified.

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

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

the functional equation, we get

Z(f , χ−1, 1− s)L(χ−1, 1− s)

= ε(χ, s)Z(f, χ, s)

L(χ, s)

These epsilons are called local constants. These are monomial function

in qs.

46

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

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

is

Z(f , χ−1, 1− s) = ε(χ, s)Z(f, χ, s)

We calculated in the beginning of the section that when χ is ramified Z(f, χ, s) ∈C[qs, q−s]. This implies that ε(χ, s) ∈ C[qs, q−s]. Similarly ε(χ−1, 1 − s) ∈C[qs, q−s]. Using previous proposition we see that ε(χ, s)ε(χ−1, 1 − s) =

χ(−1), a constant. Hence ε(χ, s) is a unit in C[qs, q−s] but the only units in

C[qs, q−s] are monomials. Hence the proposition.

Proposition 20.2.

ε(χ, s)ε(χ−1, 1− s) = χ(−1)

Proof follows from the functional equation for local zeta functions and

the definition of ε(χ, s).

21 Language of Adeles and Ideles

Harmonic analysis on the topological groups is most conveniently done when

the group is locally compact. By Tychonoff’s 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 locally

compact 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 p′s

47

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

of 1 to be:

u1 × u2 × .....× ur ×∏

p6=1,2,..r

Hp

where ui’s are compact open subgroups of Gpiat 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), χHp

)= lim→ 1,..,r⊂p

r⊗i=1

S(Gpi) →

r+1⊗i=1

S(Gpi)

where the maps are defined as

r⊗i=1

S(Gpi) −→

r+1⊗i=1

S(Gpi)

f1

⊗....⊗

fr 7−→ f1

⊗...⊗

fr⊗

χHr+1

Proposition 21.1. Let G =∏

p(Gp, Hp) be the restricted direct product of

Gp, 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 there

is a unique Haar measure dg on G such that for each finite subset of indices

S ⊂ p the restriction of dg on GS =∏

p∈S 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, φ is called the adeles of Q, denoted as

AQ = R× Q where Q = Q⊗

Z Z. This is a locally compact topological ring.

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

(Q∗p,Z∗p)⋃R∗, φ. Corresponding restricted direct product is called the

idele group of Q denoted by JQ, or A∗Q, or Gm.

48

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

AK = AQ⊗

Q

K =∏

(Kv, Ov)

JK =

(AQ⊗

Q

K

)∗=∏

(K∗v , O∗v)

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 O∗p = (Zp

⊗ZOK)∗ in usual way. Note that JQ → AQ is a

continuous map, but JQ is not closed in AQ. Thus the topology on JQ is not

the one it inherits as a subset of AQ.

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 7−→ (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 of

0 in K⊗

R. We have OK −→ Rr1 × Cr2 is discrete, where r1 number of

real 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 mapK −→ K⊗

AQ can be compared to the map Qr −→ ArQ to show

the cocompactness. As we can seeAQQ =

∏Zp×RZ that is AQ = Q +

∏Zp ×R.

Lemma 21.2. The multiplicative group K∗ embedds inside JK as a discrete

subgroup. Inside JK there exists a closed subgroup J1K = ker|.| : JK −→

C∗; |x| =∏

v |x|v which contains K∗.

Theorem 21.1. The group J1K/K

∗ is a compact topological group.

49

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

∗∞ is

nothing but the class group of K. Look at the map

JK −→⊕v

Zv = divisors

xv 7−→∑v

ordv(xv)v

Then JK/∏

v Ov.∏∞K

∗∞ ≡ divisors, going modulo K∗ we get

K∗\JK/∏

v Ov.∏∞K

∗∞ ≡ K∗\divisors, which is the class group of K. Since

JK is K∗. (∏

v Ov

∏∞K

∗∞) up to finite index, for the purpose of the proof

of compactness of J1K/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 set

of embeddings of K −→ R.

Corresponding to a cycle C one can define I(C) as ideals coprime to C and

define PC to be the principal ideals which have generators x such that x > 0

in all embeddings of K corresponding to∞ primes in∞C and x− 1 ∈ Cf i.e.

Cf =∏

pnii 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 = d1d2

be an element of I(C) with d1, d2 coprime to n. Then

we define a map I(C) −→ (Z/n)∗ as x 7−→ d1( mod n).(d2( mod n))−1.

This is a surjective map with kernel PC.

50

Lemma 22.1. Let C = Cf .∞C be a cycle. Then

I(C)/PC ≡ JK/K∗∏

(v,C)=1

Uv∏

K∗∞∏v/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 of AK

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 of AK.

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

give rise to a principal homogeneous structure.

AK × AK −→ AK

(a, χ) 7−→ χa

χa(x) = χ(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 on Hp which is isomporphic

to Gp/Hp, a compact group.

Proof : Since G =∏

(Gp, Hp), then χ : G −→ C∗ gives rise to a charac-

ter χp : Gp −→ C∗. We claim that χp is trivial on Hp for almost all p. This

follows because C∗ has no small subgroups i.e. there exists a neighbourhood

51

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

nuity of χ and the definition of almost direct product that χp = 1 for almost

all p.Then we get H∗p = Gp/Hp ⊂ Gp is an open subgroup. To prove this

we look at the continuous map Gp × Gp −→ C∗ defined by (a, χ) 7−→ χ(a).

Define U = χ ∈ Gp|χ(Hp) ⊂ a ball of radius 13around 1 This set of

characers are open set in Gp. But since there are no small subgroups we get

U = H∗p .

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/QThe local characters we constructed has the required property. We had

Qp −→ Qp/Zp∼= (Q/Z)p −→ exp(2πix)

These mappings can be combined to give a mapping

AQ/(R× Z) ∼=⊕p

Qp/Zp∼= Q/Z −→ S1

But AQ = (R + Z).Q =⇒ AQQ∼= R×Z

Z

This gives a character χ(a) = e2πia on R×ZZ and hence on

AQQ .

Lemma 23.3. For K, an algebraic number field we have AK/K ∼= K as a

principal homogeneous space.

Proof Let ψ be a character of AK/K. For any element a ∈ K, one

can define another character ψa : AK/K −→ S1 using ψa(x) = ψ(ax). We

claim to show that if ψ is non-trivial character of AK/K then all characters

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

group is a discrete subgroup of AK 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.

52

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

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

K −→∏

v 6=v0Kv

is dense.

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

of AK trivial on Kv0 ×K. However all characters of AK , trivial on K are of

the form ψ(ax), a ∈ K and these are non-trivial on Kv0 .

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

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

v∈SKv.

24 Grossencharacter

Characters of idele group JK −→ C∗ which are non-trivial on K∗ play a very

specific role.

Definition (Grossencharacters). A Grossencharacter is a character of JK/K∗

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

group homomorphism JK/K∗ −→ S1 is called unitary Grossencharacter.

Grossencharacters are trivial on Uv, for almost all v.

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

the L-function associated to χ as

L(χ, s) =∏

(p,c)=1

1[1− χ(πp)

(Nπp)s

]where c denotes conductor of χ consists of those primes in K for which χ is

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

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

53

We can also write L(χ, s) =∏

p L(χp, s) where L(χp, s) = 1

1− χp(πp)

(Nπp)s

if χp is

an unramified character of K∗p i.e. it is trivial on O∗p and define L(χp, s) = 1

if χp is ramified. By comparing to ζK(s) one finds that L(χ, s) is holomorphic

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

The infinity-type of a Grossencharacter : If χ : JK −→ C∗ is

a Grossencharacter, the restriction of χ to K∗∞ = (K⊗

R)∗ is called the

infinity type of character.

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

gebraic if and only if χ∞ is algebraic i.e. χ∞(x) = xn if x ∈ R+ and

χ∞(z) = znz−m, m, n ∈ Z if z ∈ C∗.

Example : For K = Q we have JQ = Q∗×R+×∏

Z∗p. This means that

JQ/Q∗ ∼= R+ ×∏

Z∗p. Thus characters are easy to describe in this case. The

characters of JQ/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 χ = χ0||s where χ0

is a unitary Grossencharacter.

2. Prove that by using formula L(χ, s) = L(χ.||s0 , s) = L(χ0, 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 χ :

JK/K∗ −→ C∗ a character,

ζ(f, χ, s) =

∫JK

f(x)χ(x)|x|sd∗x

where d∗x 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.

54

In our case Hp = O∗P , we have volume Np−1Np

. thus to define Haar measure on

JK one multiplies local Haar measure by NpNp−1

dx|x| .

If χ is unitary character then ζ(f, χ, 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, χ =

∏χp, in which

case if we can prove that∏

pZ(fp, χp, s) converges and defines an analytic

function in the region Re(s) > 1, then the integral defining ζ(f, χ, s) will

also converge to∏

pZ(fp, χp, s) and will be analytic function in the region

Re(s) > 1. But we have calculated earlier that Z(fp, χp, s) = 1

1−χp(πp)

(Np)s

is

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

in Re(s) > 1.

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

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

ζ(f, χ, s) = ζ(f , χ−1, 1− s)

Proof : It has two basic ingradients.

1. Poisson Summation formula : We recall that Poisson summation

formula of R states that ∑n∈Z

f(n) =∑n∈Z

f(n)

in case of R. where f ∈ S(R). We will be using the analogue of this

for AK . ∑ξ∈K

f(ξ) =∑ξ∈K

f(ξ)

Hence for f ∈ S(AK) and a ∈ JK we get |a|∑

ξ∈K f(aξ) =∑

ξ∈K f(aξ).

2. Fubini Theorem : If Γ → G is a discrete subgroup and dx is a Haar

measure on G then it descends to give a Haar measure on G/Γ such

that if f ∈ L1(G) then F (x) =∑

γ∈Γ f(x+ γ) is a L1-function on G/Γ

and∫G/Γ

F (x)dx =∫Gf(x)dx.

55

We continue with the proof of the theorem. Write JK = J1K × R+ (we use

(xt) variable).

0 −→ J1K −→ JK

Nm−→ R+

ζ(f, χ, s) =

∫JK

f(x)χ(x)|x|sd∗x

=

∫JK

f(xt)χ(xt)|xt|sd∗xdtt

=

∫R+

∫J1

K

f(xt)χ(xt)|t|sdxdtt

=

∫ 1

0

∫J1

K

f(xt)χ(xt)|t|sdxdtt

+

∫ ∞1

∫J1

K

f(xt)χ(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,

ζt(f, χ, s) =

∫J1

K

f(xt)χ(xt)tsdx.

Then

ζt(f, χ, s) =

∫J1

K/K∗

[∑ξ∈K∗

f(xξt)χ(xξt)

]tsdx

=

∫J1

K/K∗

[∑ξ∈K∗

f(xξt)

]χ(xt)tsdx

56

since χ is non-trivial,

ζt(f, χ, s) =

∫J1

K/K∗

[∑ξ∈K∗

f

(ξ

xt

)]χ(xt)ts−1dx

=

∫J1

K

f

(1

xt

)χ(xt)ts−1dx

=

∫J1

K

f(xt

)χ(x−1t)ts−1dx

= ζ 1t(f , χ−1, 1− s)

Thus, ∫ 1

0

ζt(f, χ, s) =

∫ 1

0

ζ 1t(f , χ−1, 1− s) =

∫ ∞1

ζt(f , χ−1, 1− s)

Hence,

ζ(f, χ, s) =

∫ 1

0

ζt(f, χ, s)dt

t+

∫ ∞1

ζt(f, χ, s)dt

t

=

∫ ∞1

ζ 1t(f , χ−1, 1− s)dt

t+

∫ ∞1

ζt(f, χ, s)dt

t

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

analytic continuation to ζ(f, χ, 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 χ : JK/K∗ −→ C∗ be a Grossencharacter

and f : AK −→ C∗ be factorizable f =∏fp function. We can assume that

ζ(fp, χp, s) = L(χp, s) for almost all place, say outside a finite set S of places

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

57

∏p∈S

∏p/∈S

ζ(fp, χp, s) =∏p∈S

∏p/∈S

ζ(fp, χ−1p , 1− s)

∏p∈S

ζ(fp, χp, s)∏p/∈S

L(χp, s) =∏p∈S

ζ(fp, χ−1p , 1− s)

∏p/∈S

L(χ−1p , 1− s)

=∏p∈S

ζ(fp, χ−1p , 1− s)

L(χ−1p , 1− s)

∏p∈S

L(χ−1p , 1− s)

=∏p∈S

ζ(fp, χ−1p , 1− s)

L(χ−1p , 1− s)

L(χ−1, 1− s)

But using local functional equation

ε(χp, s) =ζ(fp, χp, s)

L(χp, s)=ζ(fp, χ

−1p , 1− s)

L(χ−1p , 1− s)

We get

L(χ, s)ε(χ, s) = L(χ−1, 1− s)

with ε(χ, s) =∏

p ε(χp, s) where ε(χp, s) = 1 at almost all places and is of

the 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 such

that 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 6= 0 the support of the divisor D.

58

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 of f 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. If D1 and D2 are divisors, then deg(D1 + D2) = deg(D1) + deg(D2).

2. If f1, 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

over Fq.

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

$. Let ξ be a non-trivial character on AK/K. This defines character ξv :

Kv −→ C∗. Define the order of ξ at v tobe the minimal integer nv such that

ξ|Ov .πnvv≡ 1. Then the canonical divisor $ =

∑nv.v is a divisor on X, and

is 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 that dimL(D)−dimL(k −D) = deg(D) + (1− g).

Proof : The proof follows from Poisson summation formula applied to

the characteristic function of OD ⊂ AK where OD is defined to be the product

of πnvv Ov for all the places v of K.

27 Artin Reciprocity

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

59

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 let

q be a prime in K lying over a prime ideal p of k. We define decomposition

group of q in G to be Dq = σ ∈ G|σ(q) = q. As residue field OK/q is a

finite field and is a finite extension of Ok/p of degree f . We have canonical

homomorphism,

ρq : Dq −→ Gal

(OK/q

Ok/p

)σ 7−→ (x mod q→ σ(x) mod q)

This map is clearly well defined homomorphism of groups.

Proposition 27.1. The canonical map ρq : Dq −→ Gal(OK/qOk/p

)has the

following properties.

1. The map ρq is surjective.

2. The map ρq 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 σ ∈ Dq extends to an automorphism of the completion Kq that is

trivial on the subfield kp. The induced map jq : Dq −→ Gal(Kq/kp) is

in 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 of p being

unramified the Frobenius element Frp ∈ Dq ⊂ G is called the Artin symbol

attached 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 πu be uniformizing element

ofOu and πv be that ofOv. Then πu.Ov = πv.Ov and the decomposition group

Du∼= Gal

(Ov/πv

Ou/πu

)∼= Gal(K/k).

60

This defines the map

r :∏v/∈S

k∗vO∗v−→ Gal(K/k)

πv 7−→ Frv

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∗ →∏

v∈S k∗v is dense. This implies that k∗.

∏v/∈S k

∗v is dense in Jk, hence a

character 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

##FFFFFFFFFFFFFFFFFFF Gal(K1/k)

restriction

Gal(K2/k)

Thus we have a map

Jk −→ lim←Gal(K/k) = Gal(kab/k)

Theorem 27.2. Let k be a number field. Then,

Jk/(k∗.k+∞) ∼= Gal(kab/k)

61

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 of

k∗∞.

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

k∗k∗∞.∏v∈kf

O∗v ⊂ 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 O∗v ⊂ U .

Ray Class Field : Let C be a cycle in number field k. We define

U(C) =∏

v∈C Uv(nv) where C = v1, v2....vl, v∞ and Uv(nv) = x ∈ O∗v|x ≡ 1

mod (πnvv ). Then we take U = k∗.U(C) and the corresponding field exten-

sion is called Ray class field. Just as k∗v has filtration by congruence sub-

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

Q(ζn)/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 Frp ∈ Cis of density |C|

|G| . In particular, the set X = p|Frp ∈ C is nonempty and

infinite.

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(ζn) of Q. It’s Galois group is

Z/n∗ and Frp = p. The Cebotaraev density theorem implies that there are

infinitely many primes in arithmetic progression.

For the purpose of the proof of Artin’s reciprocity law it’s useful to go

back and forth between ideal theoretic and adelic language.

62

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

Jk/(k∗.NKk JK) ∼=

Ik(C)PC.NK

k (IK(C))

where C is a cycle in k divisible by ramified primes such that

Ov(Cv) ⊂ NKwkv

(K∗w) (called admissible cycles) and Ik(C) is ideals in k which

are coprime to C.

Proof : We have

JC =∏v/C

O∗v(C)∏

(v,C)=1

k∗v

Let kC = JC⋂k∗, then JC/kC

≈−→ Jk/k∗. This is injective (by definition) and

surjective by weak approximation theorem (∏

v/C O∗v(C).k∗ =

∏v/C k

∗v). This

gives rise to following diagram.

IC/kC≈−−−→ Jk/k∗y y

Ik(C)PC .N

Kk (IK(C))

≈−−−→ Jk/(k∗.NKk JK)

Check that the isomorphism in the top horizontol arrow descends to give an

isomorphism 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 −→ Frv. Observe that

NKk (IK(C)) lies in the kernel of the Artin map for K/k (Galois extension of

degree 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

qi

Then Nm(qi) = pmd where m = d.[OK/qi : Ok/p]. Which gives Fr

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

63

1. Prove the reciprocity for cyclic extensions.

(a) Order of the group Ik(C)PC .N

Kk (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. ζK(s) = ζmk (s),m = [K : k]. This contradicts simplicity

of the poles at s = 1 of ζK and ζk.

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.NK

k (IK(C))

∣∣∣∣ ≤ [K : k]

Proof : Let H = PC.NKk (IK(C)) and G = I(C)/H. Given group G and

Gχ−→ C∗, there is the associated L-function L(s, χ). We will look at the

behaviour of this L-function around s = 1. We already know that if χ = 1

then L(s, χ) has simple pole at s = 1 and is holomorphic for χ 6= 1. We write

L(s, χ) = (s− 1)m(χ).fχ(s)

64

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

m(χ) ≥ 0 if χ 6= 1

= −1 if χ = 1

L(s, χ) =∏

p

(1− χ(p)

(Np)s

)−1

This makes sense for Re(s) > 1

log(L(s, χ)) = −∑

p

log

(1− χ(p)

(Np)s

)=

∑p

χ(p)

(Np)s+ higher order terms

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

log(L(s, χ)) =∑

pχ(p)

(Np)s around s = 1 (up to bounded function which we

will denote by f ∼ g).

log(L(s, χ)) ∼∑

p

χ(p)

(Np)s

∼∑g∈G

χ(g)

[∑p

1

(Np)s

]

where p = g ∈ G = I(C)/H.

∑χ

log(L(s, χ)) ∼∑χ,g∈G

χ(g)

[∑p

1

(Np)s

]

= |G|∑p∈H

1

(Np)s

65

On the other hand, L(s, χ) = (s− 1)m(χ).fχ(s)

log(L(s, χ)) ∼ −m(χ) log1

s− 1∑log(L(s, χ)) ∼

[1−

∑χ6=1

m(χ)

]log

1

s− 1

So we get [1−

∑χ6=1m(χ)

]log 1

s−1

|G|∼∑p∈H

1

(Np)s

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

p = q⋂OK belongs to the norm from K to k. Hence,

[1−

∑χ6=1m(χ)

]log 1

s−1

|G|≥ 1

[K : k]

∑q of degree 1

1

(Np)s

∼ 1

[K : k]ζK(s)

∼ 1

[K : k]log

1

s− 1

which implies

1 ≥ 1−∑

m(χ) ≥ |G|[K : k]

[K : k] ≥ |G| = | I(C)PN| i.e. order of

∣∣∣ Ik(C)PC .N

Kk (IK(C))

∣∣∣ ≤ [K : k]. Further we get

m(χ) = 0 ∀χ 6= 1.

Corollary 27.3. For χ 6= 1 the function L(1, χ) 6= 1.

Corollary 27.4 (Density Theorem). There are positive density of primes in

an arithmetic progression. i.e.∑p∈a0

1

(Np)s∼ 1

[I(C) : PC]log

1

s− 1

where a0 ∈ I(C)/PC.

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Proof : We have

log(L(s, χ)) ∼∑

p

χ(p)

(Np)s

=∑

a

χ(a)∑p∈a

1

(Np)s

Then, ∑χ

χ(a−10 ) log(L(s, χ)) =

∑χ

∑a

χ(aa−10 )∑p∈a

1

(Np)s

= [I(C) : PC]∑p∈a0

1

(Np)s

Hence,

log ζK(s) = [I(C) : PC]∑p∈a0

1

(Np)s

log1

s− 1= [I(C) : PC]

∑p∈a0

1

(Np)s

Proof of the index inequality in the other direction needs some algebraic

preleminaries taken up in the next section.

28 Euler characteristic of a group G

Let G = Z/n. Let A be an abelian group which is Z/n-module. Let us

denote the generator of Z/n as σ. Then we define cohomology and Tate

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cohomology as follows,

H0(G,A) = AG, the zeroth cohomology

H0(G,A) =AG

(1 + σ + σ2 + ...+ σn−1)A, the zeroth Tate cohomology

H1(G,A) =ker(1 + σ + σ2 + ...+ σn−1) : A −→ A

(1− σ)A, the first cohomology

Definition. The Euler characteristic of A is defined by,

χ(A) =#(H0(G,A))

#(H1(G,A))

if orders make sense, i.e. the orders of both H0(G,A) and (H1(G,A) are

finite.

Lemma 28.1. 1. If

0 −→ A −→ B −→ C −→ 0

is an exact sequence of Z/n modules, then whenever two of

χ(A), χ(B), χ(C) make sense, so does the third one and

χ(B) = χ(A)χ(C).

2. χ(A) = 1 for A finite.

3. χ(Z) = n as H0(Z/n,Z) = Z/n and H1(Z/n,Z) = 0.

Lemma 28.2. Let K be a degree n cyclic extension of a local field k. Then,

1. χ(K∗) = [K : k].

2. χ(UK) = 1. where UK denotes units in OK.

3. [UK : NmUK ] = e where e is the ramification index.

Proof :

68

1. Since χ(K∗) = #H0

#H1 , by Hilbert Theorem 90, H1 = 0 (for any cyclic

extensions). Hence χ(K∗) = #H0 = [k∗ : NmK∗].

2. Let us look at the following exact sequence of Z/n modules.

1 −→ UK −→ K∗ −→ Z −→ 0

hence, χ(K∗) = χ(UK)χ(Z) = [K : k]χ(UK). Thus it suffices to prove

that χ(UK) = 1.

By the exp map there exists an isomorphism of a subgroup of K of

finite index which is isomorphic to a subgroup of Ok of finite index.

Thus χ(OK) = χ(Ok). By normal basis theorem OK has a subgroup of

finite index which is free of rank 1 as an Ok[Z/n] module. Therefore

χ(OK) = χ(Ok[Z/n]) = χ(IndZ/ne Ok)

We use Shapiro’s lemma: for a subgroup H of G we have

H i(H,B) = H i(G, IndGHB). We get χ(Ok) = χe(Ok) = 1.

3. Let us look at the following diagram.

0 −−−→ k∗ −−−→ K∗x 7→ x

σx−−−→ K1 −−−→ 1x x x0 −−−→ Uk −−−→ UK −−−→ U1

K −−−→ 1

where K1 is set of norm 1 elements in K∗. Which implies that

Z/eZ ∼= K∗/k∗UK ∼= K1/U1K .

χ(UK) = 1 = H0(UK)H1(UK)

= [Uk:NmUK ]

[U1K : x

σx|x∈UK]

Observe that U1K = K1 thus [Uk : NmUK ] = e and [k∗ : NmK∗] = [K :

k] for cyclic extensions of local fields.

69

29 Index calculation [Jk : k∗NKk JK ]

Theorem 29.1. If K is a cyclic extension of degree n of a number field k

then [Jk : k∗NKk JK ] has order n.

The proof depends on the universal norm index inequality proved earlier.

Let S be the finite set of places of k containing all the archmedian places of

k. Let SK denotes the set of places of K (Galois extension of k) lying over

the places S of k. To each place w in SK define a symbol Xw. Let E = Qn

be the free Q module on the symbols Xw. Observe that E is a module for

Gal(K/k). Let M be any Gal(K/k) -invariant lattice in E⊗

R.

Theorem 29.2. There exists a lattice M ′ ⊂ M with basis τw such that

στw = τσw ∀w ∈ SK , σ ∈ Gal(K/k).

This amounts to the following lemma in group theory.

Lemma 29.1. Let G be a group. If V1 and V2 are two finite dimensional Qvector spaces which are G modules and if V1

⊗R ∼= V2

⊗R as G modules

then V1∼= V2 as G modules.

Proof : X = HomQ[G](V1, V2) is a vector space over Q such that

X⊗

R = HomR[G](V1

⊗R, V2

⊗R). Therefore if HomR[G](V1

⊗R, V2

⊗R) 6=

0 then so is HomQ[G](V1, V2). Further one observes that if p(x) is a polyno-

mial on a vector space X defined over Q such that p(x) is trivial on X(Q)

then p(x) ≡ 0.

Corollary 29.1. In the setting of above theorem we get,

χ(M) =#H0(G,M)

#H1(G,M)= χ(M ′) =

∏v∈S

Nv

where Nv is the order of the decomposition subgroup at the places v ∈ S.

70

Proof : Since M ′ =⊕

v

(IndGGv

Z)

hence,

χ(M ′) =∏v

χ(IndGGvZ)

=∏v

χ(Gv′Z)

=∏v

Nv

Corollary 29.2. Let KS be the set of S units in K, i.e. w(x) = 0∀w /∈ SKthen

χ(KS) =

∏Nv

[K : k]

Proof : We look at the map

KS −→ E⊗

R

x 7−→∑w∈SK

log |x|w.Xw

Dirichlet unit theorem says that image of KS is a lattice in hyperplane in

E⊗

R defined by∑Xw = 0. Therefore

χ(KS)χ(Z) = χ(M) = χ(M ′) =∏

Nv

Thus χ(KS) =∏Nv

[K:k].

We continue the proof of theorem 29.1 as,

χ(JK) =#H0(Gal, JK)

#H1(Gal, JK)

= #H0(Gal, JK) (Hilbert 90)

= [Jk : NmJK ]

So χ(JK/K∗) = [Jk : k∗NmJK ]. Hence JK

K∗ =∏O∗v .

∏v∈S k

∗v

KS. This completes

the proof of [Jk : k∗NmJK ] = [K : k].

71

Artin Reciprocity Theorem :

This is true for all cyclotomic extensions of Q i.e. for all extensions K of

Q contained in Q(ζn).

The Artin reciprocity is also true for relative cyclotomic extensions i.e.

for K a extension of k such that K ⊂ k(ζn).

72