Lectures on Algebraic Number Theory

Home , Algebraic number field, Different ideal, Global field, Integral domain, Normal basis, Quadratic field

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.

2πi 3. Cyclotomic ﬁelds Q(ζn), ζn = e n .

1 ∗ 4. Kummer extensions K(a n ) 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 OK for the units in OK i.e. elements in OK ∗ which has inverse in OK . Then OK 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 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 OK is ﬁnite if and only if K = Q or is a quadratic imaginary ﬁeld. √ 2. If K is real quadratic, then O∗ is /2 × . If K = ( d) then √ K Z Z √ Q 1+ d OK = Z( d) if d 6≡ 1(mod4) and OK = Z( 2 ) if d ≡ 1(mod4). Thus existence of units inside OK is equivalent to solving the Pell’s 2 2 equation a − db = 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?

1 1 Exercise: If K = Q[m n ] , then OK contains Z[m n ] as a subgroup of 1 ﬁnite index. Calculate the index of Z[m n ] 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 multiplication with OK as the identity and nX o A1A2 = λiµi / λi ∈ A1, µi ∈ A2 .

Definition. The group of all nonzero fractional ideals Class Group := . 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,

2r1 (2π)r2 hR 1 ζK (s) = √ . + holomorphic function w dK s − 1

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 follows OK −−−→ K x x     Z −−−→ Q

Let p ∈ Z be a prime. Denote the decomposition of the ideal (p) = pOK Qn ei in OK as pOK = i=1 pi 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 ¯ Q ¯ ei irreducible polynomials f(x) = fi(x) 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 2 2 only if there exist x0 ∈ Z/p such that x0 = 5. And x0 = 5 has solution in p−1 5 2 Z/p if and only if 5 ≡ 1(mod p) ( which is by defenition denoted as ( p ) = 1). This can be seen by noting that (Z/p)∗ is a cyclic group. So if p/5 ¯ 2 5 ¯ 2 then f(x) = x . If ( p ) = 1 then f(x) = (x − x0)(x + x0), where x0 = 5 in 5 ¯ Z/p. And if ( p ) = −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 −−−→ K x x     Z −−−→ Q

Let p ∈ Z be a prime and let p be a prime ideal lying over p in OK . Let Qg ei pOK = i=1 pi . 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.

Pg Lemma 3.1. Let K be a number field. Then 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. OK /p 0 −→ Ip −→ Dp −→ Gal −→ 0 Ok/p

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 q field extension Fq ⊂ Fqe , the map σ : Fqe −→ Fqe , x 7→ x 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 theory, 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 -linear map l : K −→ K defined by l (λ) = λx. Define trK (x) Q x x Q as trace of l and NmK (x) as determinant of l . If there is no possibility of x Q x confusion, we will denote trK (x), NmK (x) simply as tr and Nm. Q Q 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 d to be det(B(ei, ej)), where {e1, ...ed} is an integral basis of Z .

Note that dB does not depend on the basis chosen as can be seen as follows. Let B0 be another integral basis of Zd. Then there exists an integeral matrix A such that B = AB0At. Hence, 0 t 2 0 0 dB = det B = det(AB A ) = (det A) det B = det B Here we have used that for A ∈ GL(n, Z), det A = 1 or − 1. This discriminant 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 discriminant. 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 number 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, ··· , vm then 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 euclidean vector space V and let X be a centrally symmetric, convex subset of V. Suppose that n vol(X) > 2 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, 2 2 2 2 A + B + C + D = 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 Y j : K −→ KC = C τ 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, σ¯1, ··· , σs, σ¯s} where the ρ’s give the r real embeddings and σ’s andσ ¯’s give the 2s pairwise conjugate complex embeddings. We see that the image of K under this n Q canonical mapping into C = τ∈X C actually lies inside the following set known as the Minkowski Space. ( ) Y KR = (zτ ) ∈ C | zρ ∈ R, zσ¯ =z ¯σ τ∈X

Q r+2s We note that this space is isomorphic to τ∈X R = R 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

s V olcanonical(Y ) = 2 V olLebesgue(F (Y )) where the canonical volume is the volume on KR induced from the standard 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 p 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

X ¯t hjαi, jαki = τl(αi)¯τl(αk) = AA 1≤l≤n

10 Then we get

1 2 p V ol(Γ) = |det(hjαi, jαki)| = |det(A)| = |dK|.[OK : I]

So via the canonical embedding j : K −→ KC, we can identify the ideals n of OK with lattices in R . But since we are interested in the units of the ring of integers, O∗ , we pass on to the multiplcative group ∗ by using the K K standard logarithm map

∗ l : C −→ R, z −→ log|z|

It induces the surjective homomorphism Y Y l : ∗ = ∗ −→ , (z ) −→ (log|z | KC C R τ τ τ τ Note that image of ∗ under the above map lies in the following set KR " #+ ( ) Y Y R := (xτ ) ∈ R | xρ ∈ R, xσ¯ = xσ τ τ Further the multiplicative group ∗ admits the homomorphism : ∗ −→ KC N KC C∗ given by the product of the coordinates and we have the homomorphism " #+ Y T r : R −→ R τ given by the sum of the coordinates. Also note that the R-vectorspace Q + r+s [ τ R] is isomorphic to R . Now conside the following subgroups

O∗ = { ∈ O | N = ±1} K K K|Q S = {y ∈ ∗ | (y) = ±1} KR N Y + H = {x ∈ [ R] | T r(x) = 0} τ

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

11 and the composite λ := l o j : O∗ → H. The image will be denoted by K Γ = λ(O∗ ) ⊆ H K and we obtain the

Theorem 6.3. Let K be a number ﬁeld. Then the sequence 1 −→ µ( ) −→ O∗ −→λ Γ −→ 0 K K 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() = l(j) = 0 implies |τ| = 1 for each embedding τ of K. Thus

(j) = (τ) lies in a bounded domain of the R-vector space KR. But (j) 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 Y cτ > A[OK : I] τ 2 sp where A = ( π ) |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 } r+s s Q is centrally symmetric, convex and has volume 2 π τ cτ . Now, Γ = j(I) p is a lattice with volume |dK|.[OK : I]. So we have 2 s Vol(Y ) > 2r+sπs( ) p|d |.[O : I] = 2nVol(Γ) π K K 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)- dimensional R-vector space H .

Q + 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∗ → H K arises by restricting the mapping

∗ j Y ∗ l Y K −→ C −→ R τ τ Thus it suffices to show, that for any c > 0, the bounded domain Q {(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 ( ) Y ∗ −c c (zτ ) ∈ C | e ≤ |zτ | ≤ e τ But it contains only finitely many points of the set j(O∗ ) as this is a subset K 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 Y C = cτ > A[OK : I] τ

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

W = {(zτ ) ∈ KR | |zτ | < cτ }

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

W y = {(zτ ) ∈ KR | |zτ | < c˜τ } Q Q Q wherec ˜τ = cτ |yτ | and one hasc ˜τ =c ˜τ¯, τ cτ = τ c˜τ = C because τ |yτ | = |N(y)| = 1. Then, there exists a point

ja = (τa) ∈ W y, 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 M [ −1 T = S ∩ W (jαi) i=1 is a bounded set as W is bounded and we have [ S = T j ∈Θ∗ K

Because for any y ∈ S, we ﬁnd, by the above, an a ∈ OK, a 6= 0, such that ja ∈ W y−1, so ja = xy−1 for some x ∈ W . Since

−1 Y |NK|Q(a)| = |N(xy )| = |N(x)| < cτ = C, τ a is associated to some α , α = a, ∈ O∗ . Consequently, we have i i K −1 −1 y = xja = xj(αi )

−1 −1 Since y, j ∈ S, one ﬁnds xjαi ∈ S ∩ W jαi ⊆ T , and thus y ∈ T j. Now

M = l(T ) is also a bounded set as for any x = (xτ ) ∈ T , the absolute values Q |xτ | are bounded above and are also away from zero (because τ xτ = 1). Now we have [ H = (M + γ) γ∈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∗ of O is K K isomorphic to the ﬁnitely generated abelian group given by

r+s−1 µ(K) × Z where µ(K) is the ﬁnite torsion group consisting of the roots of unity contained 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) ≥ min{v(x), v(y)}

Given a discrete valuation on a ﬁeld K, let A = {0} S{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

1 A = Z(p) = Z q q6=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) n as prime ideal. And for any x ∈ K we get xOK = p 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 valuation 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 f v∞ g = 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} S{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 −→ R such 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 |.|0 on field K are equivalent if there is a positive constant t such that |a|0 = |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 i every i > 0 ; p |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:

n p = lim /p Z ← Z

N 1 Then p = p is a ﬁeld containing p such that p = p[ ]. Q Z Z Q Z Q Z p 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|∞ = deg(f)−deg(g) q 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 Q np(x) ideal xOK and write it as product of prime ideals xOK = p . Define ∗ a discrete valuation on K as vp : K −→ Z as vp(x) = np(x). Using this ≥0 discrete valuation we can define absolute value on K as |.|p : K −→ R by −np(x) |x|p = (#OK /p) . The completion of K with respect to this absolute value is written as Kp. Any element of a discrete valuation ring A looks like

r r+1 a0π + a1π + .... 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 X n f(t) = ant n≥m∈Z where an ∈ C. We have thus defined completion of a number field at various prime ideals. These prime ideals are also called finite places of the number field. There are infinite places (or archmedian absolute values) which correspond to embeddings 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 N a finite field extension of then K p is a separable algebra which is a Q Q Q product of fields.

O M K Qp = Kv Q v|p Similarly, O r M r K R = R 1 C 2 Q

18 8 A sample calculation of the ring of integers of a number ﬁeld

2πi We will calculate the ring of integers of the field K = Q(ζp), ζp = e p . 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.

2πi Proposition 8.1. Let K = Q(ζp), ζp = e p . 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 p−2 x = a0 + a1(1 − ζp) + ...... + ap−2(1 − ζp) with ai ∈ Q. We need to show ai ∈ Z. By using valuation theory we claim that there exists a valuation 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 Y i = (x − ζp). i=1

Putting x = 1 in this equation,

19 p−1 Y i p = (1 − ζp) i=1 i p−1 Y 1 − ζp = (1 − ζp) . 1 − ζp

1−ζp Note that i ∈ [ζp] is unit. For this observe that (i, p) = 1, there existsj 1−ζp Z such that ij = 1 mod p.Therefore

i j 1 − ζp 1 − (ζp) i i = i = 1 + ζp + ..... ∈ Z[ζp] 1 − ζp 1 − ζp

i 1−ζp 1−ζp 1−ζp proving that i ∈ [ζp]. Clearly ∈ [ζp]. Thus i ∈ [ζp] is a unit 1−ζp Z 1−ζp Z 1−ζp Z i in Z[ζp]. Thus the equation p = (1 − ζp) π 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

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

= 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. p−2 Let x = b0 + b1ζp + ...... + bp−2(ζp) with bi ∈ Q and bi have no p in −i j denominator. Note that if x ∈ OK then tr(xζp ) ∈ Z ∀i. Using trζp = −1 ∀j 6= 0 and tr1 = p − 1 we get

(p − 1)bi − (b0 + ....bi−1 + bi+1 + .... + bp−2) ∈ Z p−2 X pbi − bi ∈ Z. i=0

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.

p−2 x = b0 + b1ζp + ...... + bp−2ζp p−2 p−2 = b0 + (b1 − b0)ζp + ...... + (bp−2 − b0)ζp + b0(ζp + ...... + ζp ) p−2 p−2 = (b1 − b0)ζp + ...... + (bp−2 − b0)ζp + b0(1 + ζp + ...... + ζp ) p−1 X i = −b0ζp + (bi − b0)ζp

On taking trace we get b0 ∈ Z. This completes the proof. Remark : Not always the ring of integers OK of a number ﬁeld is generated by one element but it is always generated by atmost two elements.

9 The diﬀerent and the discriminant of a number ﬁeld

Let a be a fractionl ideal in a number ﬁeld K. We look at the map tr : K × K −→ deﬁned by tr(x, y) = trK (xy). Q Q

Deﬁnition. The set a0 = {x ∈ K|tr(xy) ∈ Z, ∀y ∈ a} is called complemen- try set.

It is not diﬃcult to see that a0 is also a fractional ideal.

0 0 Deﬁnition. OK = {x ∈ K|tr(xy) ∈ Z, ∀y ∈ OK }. Then K ⊇ OK ⊃ OK 0 0 and OK is a fractional ideal. The inverse of OK , 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).

p−2 Exercise: disc(Q(ζp)/Q) = p

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 −1 √ δK = {x ∈ K|tr(xy) ∈ Zp, ∀y ∈ OK }. Let a + b p, a, b ∈ Zp is an element √ √ √1 √a √a of Zp[ p]. Then p (a + b p) = p + b =⇒ tr( p + b) = 2b ∈ Zp∀a, b. That −1 means √1 ∈ δ . p K √ 1 Let us take case when p = 2 the we claim that 2 ∈ δK . 2 (a + b 2) = 1 a + √b =⇒ tr( 1 a + √b ) = a ∈ =⇒ 2 ∈ δ . 2 2 2 2 Z2 K

Lemma 9.1. If L is a ﬁnite extension of a local ﬁeld K, with πK and πL uni- e e−1 formizing parameter such that πL = πK .u, u a unit in OL. Then πL /δL/K . e−1 In fact δL/K = πL ⇐⇒ p/e.

e−1 e−1 Proof: Let us prove that π /δL/K . For this we show tr(π OL) ⊂ L L e−1 πL 1 1 OK . But then tr(π OL) = tr OL = tr(πLOL) ⊂ πK OK ⊂ OK . L πK πK πK

10 The Riemann Zeta Function

The Riemann zeta function is deﬁned as follows: X 1 Y 1 ζ (s) = = Q ns 1 − 1 n≥1 p≥2 ps 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

1 ζ (s) = + holomorphic function around s = 1 Q 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).

− s s ξ(s) = π 2 Γ( )ζ (s) 2 Q 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 1 Re(s) = 2 . 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.

X 1 Y 1 ζK (s) = s = 1 (NI) 1 − s I6=0,I⊂OK P (NP) 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,

2r1 (2π)r2 hR 1 ζK (s) = √ . + holomorphic function around s = 1 w dK 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

X 1 ζ (s) = K (NI)s I6=0,I⊂OK Y 1 = 1 − 1 P (NP)s for Re(s) > 1. Since the equality X 1 Y 1 s = 1 (NI) 1 − s I6=0,I⊂OK P (NP) is formal, so it suﬃces to check the convergence of the product.

X 1 log(ζ (s)) = − log(1 − ) K (NP)s P X 1 = m(NP)ms P,m≥1

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, X N log(ζ (s)) ≤ K mpfms p, m≥1,N≥f≥1 where p is prime. Therefore upto a function which is bounded in a neighborhood of 1,

X ap log(ζ (s)) = K ps ap denotes number of primes above p of degree 1.

24 P N Thus log(ζK (s)) for s > 1 is bounded by ps , thus it is convergent for re(s) > 1. Actually we will be proving later that

1 X 1 log(ζ (s)) = log = (up to a bounded function). K s − 1 ps

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.

1 1 2. Define the zeta function ζA (s) of the affine line A over the finite field Fp as follows. Y 1 ζ 1 (s) = A (1 − 1 )s p(x) Np(x)

where p(x) runs over irreducible polynomials in Fp[x] with leading term n 1 and f(x) is monic polynomial over Fp. And Nf(x) = p where n = deg f(x). But then

Y 1 ζ 1 (s) = A (1 − 1 )s p(x) Np(x) X 1 = (Nf(x))s f(x) X pd = pds d 1 = p 1 − ps

P 1 3. Let K = Q(i), then ζK (s) = (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 ?

−1 If p ≡ 1 mod 4, then p = 1. Hence the map

p − 1 p + 1 φ : 1, ··· , −→ , ··· , p − 1 2 2 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 h √ p denote the class number of the quadratic ﬁeld Q( −p). Then  Rp − Np if p ≡ 7 mod 8, hp = 1  3 (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. N(X, C) lim = κ X→∞ X where

 2 log η  √ if d > 0, η is the unique fundamental unit > 1, κ = d √2π if d < 0.  w |d| Now we state the class number formula for the quadratic ﬁeld K = √ Q( d). Theorem 13.3. For a quadratic number ﬁeld K,

lim (s − 1)ζK (s) = hκ s→1+ 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

∞ X d L (s) = m−s d m m=1

d with m 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

lim (s − 1)ζK (s) = lim (s − 1)ζ(s) = Ld(1). s→1+ s→1+ Thus we have the following simplier expression for class number formula.

 2 log η  √ Ld(1) if d > 0, η is the unique fundamental unit > 1, h = d 2π √ Ld(1) if d < 0.  w |d|

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 P∞ χ(n) series is defined to be L(s, χ) = n=1 ns . In this case we define χ to be the quadratic character ( . ). Using above expression for class number one can p √ p easily derive that hp = π L(1, χ). Next Lemma expresses L(1, χ) as a finite sum.

Lemma 13.2. We have

p−1 iπτ1(χ) X L(1, χ) = χ(k)k. p2 k=1 where X ak τk(χ) = χ(a)ρ . 1≤a

proof : For s > 1 we have

∞ X χ(n) X X 1 L(s, χ) = = χ(a) . ns ns n=1 1≤a

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

p−1 1 X c = ρ(a−n)k. n p k=0 Thus

∞ p−1 p−1 ( ) ∞ 1 X X X 1 X X X ρ−nk L(s, χ) = χ(a) ρ(a−n)k = χ(a)ρak . p p ns 1≤a

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

p−1 τ1(χ) X 1 1 L(1, χ) = log . p χ(k) 1 − ρ−k k=1 Let S be the sum in the last formula. Since χ is an even character we have

p−1 X 1 1 1 2S = log − log . χ(k) 1 − ρ−k 1 − ρk k=1

−k π kπ −k Using the expressions log(1 − ρ ) = i 2 − p + log |1 − ρ | we get

p−1 iπ X S = χ(k)k. p k=1

This completes the lemma. √ We recall that τ1(χ) = i p. Thus

p−1 X hpp = − χ(k)k. k=1

29 We split this sum as

(p−1)/2 p−1 X k X k −h p = k + k p p p k=1 k=(p+1)/2 (p−1)/2 X k = (k − (p − k)) p k=1 (p−1)/2 X k = 2 k − p(R − N ). p p p k=1

On the other hand we can write,

(p−1)/2 (p−1)/2 X 2k X p − 2k −h p = 2k + (p − 2k) p p p k=1 k=1 (p−1)/2 2 X k 2 = 4 k − p (R − N ). p p p p p k=1 from above two we get 2 2 2 − 1 h p = −p (R − N ). p p p p p

2 Now Theorem 13.1 follows from the fact that ( p ) = 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 ξ ξ ∆ = (ξ , ξ ) ∈ 2| 1 , 2 ∈ ∆ r 1 2 R r r

30 and N∆(r) = Number of lattice points in ∆r. Then 1 lim N∆(r) = Area of ∆. x→∞ r2 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, −1 IJ = αOK with α ∈ J. Conversely if α ∈ J then J|αOK , hence I = J αOK is 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

ω 0 ≤ log < log η (13.3.1) |N(ω)|1/2 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 = ω2 with = ±1. Hence ω 2N(X,C) = ω ∈ J | 0 < |N(ω)| < y, 0 ≤ log < log η . |N(ω)|1/2

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

1/2 2 0 |ξ1| Ω = (ξ1, ξ2) ∈ R |0 < |ξ1||ξ1| < 1, 0 < log 0 1/2 < log η |ξ1|

0 0 0 0 0 where ξ = ξ1β1 + ξ2β2 and ξ = ξ1β1 + ξ2β2, β1, β2 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 2 0 Ay = (ξ1, ξ2) ∈ Z | |ξ| ≤ 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 √ 2N(X,C) N ( y) lim = lim Ω N(J) = N(J)Area of Ω. X→∞ X y→∞ y

Area of Ω can be computed to be 4 log√η . This completes the proof in this N(J) d case. Case 2: d < 0 In this case we have

wN(X, C) = |{ω ∈ J | 0 < N(ω) < y}| .

Let 2 2 Ω = (ξ1, ξ2) ∈ R |0 < |ξ1β1 + ξ2β2| < 1 . √ Then wN(X, C) = NΩ( y), hence √ wN(X,C) N ( y) lim = lim Ω N(J) = N(J)Area of Ω. X→∞ X y→∞ y

Here area of Ω is calculated to be 2π√ , completing the proof of Theorem N(J) d 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 X A(X) = am. m 1 and we have

lim (s − 1)f(s) = c. s→1+

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

A(X) lim = hκ. X→∞ X

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

lim (s − 1)ζk(s) = hκ. s→1+

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, P 1 p∈M (Np)s lim 1 s→1+ log s−1 In one of the earlier lecture we had observed that 1 log ∼ log ζ (s) s − 1 K X 1 ∼ (Np)s p X 1 ∼ (Np)s degp=1

degp where f1 ∼ f2 if (f1 − f2)(s − 1) → 0 as s → 1 and Np = p ( 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 1 in K has density = d .

33 0 Proof: We have M = {p ∈ Z | pOK = p1...pd, p s are completely split}. So using remark preceding theorem we get X 1 X 1 1 d = = log (Np)s (Np)s s − 1 p∈M degp=1 Hence substituting in density formula we get density of completely split 1 primes ( primes in M ) is d .

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 1 K1K2 degree d extension of . Then, split primes in K1 are of density , Q d1 1 split primes in K2 are of density and split primes in K1K2 are of density d2 1 1 1 1 . Since the split primes in K1,K2 and K1K2 are the same, = = . d d1 d2 d 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 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. 1 1 Let K = Q(2 3 ), and L = Q(2 3 , 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 1 L. Then by the theorem, density of M = 6 . 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 = 1 q1q2q3 in L} ≡ {p ⊂ OK | pOL = q1q2}. Hence density of M is 2 . 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 1 1 1 this case will be 1 − 6 − 2 = 3 .

15 Decomposition of primes in the cyclotomic ﬁelds

2πi Let K = Q(ζn), where ζn = e n 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. m So let us assume n = p 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 d m prime i.e. pOK = p where d = φ(p ).

Observe that the minimal polynomial satisﬁed by ζpm is

m−1 m−1 m−1 Y 1 + xp + (x2)p + ..... + (xp−1)p = (x − ζ) ζ where product is over ζ , all pmth roots of unity. Putting x = 1 we get

Y i m p = (1 − ζpm ). i ∈ Z/p (i,p)=1

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 i sends ζpm to ζpm we get Y p = σ(π).

σ∈Gal(K/Q) i 1−ζpm Observe that ∈ [ζpm ] is a unit for (i, p) = 1. Hence we get p = 1−ζpm Z φ(pm) (1−ζpm ) .u where u is a unit in Z[ζpm ]. By taking p =< 1−ζpm > we get φ(pm) pOK = p . 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 a on Q(ζn) by ζn −→ ζn. Further, for p prime in Z with (p, n) = 1, the ∗ Frobenius element σp ∈ Gal(K/Q) is the element p ∈ (Z/n) . T 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] p p is deﬁned by σpx ≡ x mod p for all x ∈ Z[ζn]. But σp(ζn) = ζn whereas order of σp is the degree of residue ﬁeld extension of p which is Z[ζn]/p. Now

f f pf f σp = 1 ⇔ σp (ζn) = ζn ⇔ (ζn) = ζn ⇔ p ≡ 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 p−1 X a 2πia G = e p p a=1

36 2 −1 √ It is easy to see that G = ( p )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

Q ni Let K be a number field which is contained in Q(ζn). Let n = i pi . ∗ Q ni ∗ Then Gal(Q(ζn)/Q) = (Z/n) = i(Z/pi ) . 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 Gb, 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 Gb. 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 0 ∗ ∗ χ :(Z/n0) −→ C such that following diagram commutes.

∗ ∗ (Z/n) / (Z/n0) HH HH HH HH HH χ0 χ HH HH HH HH H# 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 Y |dK | = cond(χ) χ∈X

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

2. Let p be a prime and consider K = Q(ζp2 ). Then the Galois group 2 ∗ ∼ 2 2 ∗ ∗ G = (Z/p ) = Z/(p − p) and let X = {χ :(Z/p ) −→ 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 2 2 ∼ p . To see this observe that Z/(p − p) = Z/p × Z/(p − 1). Hence we 2p2−3p get |dK | = p .

38 17 L-functions associated to a Dirichlet character

Let χ :(Z/n)∗ −→ C∗ be a Dirichlet character. We can extend χ to deﬁne a function χb on Z by ( χ(a) if(a, n) = 1 χ(a) = b 0 otherwise

Deﬁnition. Let χ be a Dirichlet character. Then X χ(n) L(s, χ) = b ns n≥1 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. s s − 2 We recall that if ξ(s) = π Γ( 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

p s −s s r −s r ξ (s) = ( |d |) (π 2 Γ( )) 1 (π Γ(s)) 2 ζ (s) K K 2 K 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 conductor of χ appears.

2. For an abelian extension K of Q, the zeta function can be written as a product of L-functions as Y 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 Grössencharacters. 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 Gb = {ξ : G −→ S1}. Then Gb is a locally compact topological group on which the topology is the compact open topology deﬁned by taking basic open sets of Gb as W (C,U) = {ξ ∈ Gb|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 −→ Gb is an anti equivalence 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 R is of the form ψy : x 7−→ ψ(xy) for some y ∈ R. Let us deﬁne unitary dual of R by Rb = {ξ : R −→ S1| ξ unitary character of R}. Then the map

40 y −→ ψy defines isomorphism of R with its unitary dual Rb. One defines for an integrable function f its Fourier transform by Z fb(y) = f(x)ψ(xy)dx R 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, fbb(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 x = aj(x)p −∞

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 Qb p is an isomorphism of topological groups.

Proof: (1) If ψ ∈ Qb p, then there is an integer k such that ψ = 1 on −k p Zp. −k Since ψ is continuous, there is an integer k such that ψ maps p Zp into 1 −k {z ∈ S ||z − 1| < 1}. But p Zp is a subgroup of Qp, so its image under ψ is a subgroup of S1; hence equals {1}. j Any ψ ∈ Qb p is completely determined by its values on the numbers p , j ∈ j Z. So if ψ 6= 1 there is an integer j0 such that ψ(p ) = 1 for j ≥ j0, but ψ(pj0−1) 6= 1. −1 (2) Suppose ψ ∈ Qb p, ψ(1) = 1, and ψ(p ) 6= 1. There is a sequence ∞ {cj}0 with c0 ∈ {1, ..., p − 1} and cj ∈ {0, 1, ..., p − 1} for j ≥ 1 such that −k Pk −j ψ(p ) = exp 2πi j=1 ck−jp for k = 1, 2, 3....

41 −k Let ωk = ψ(p ); then

p −k−1 p −k−1 −k ωk+1 = ψ(p ) = ψ(p.p ) = ψ(p ) = ωk

−1 Now ω1 6= 1 and ω0 = 1 so ω1 = exp(2πic0p ) for some c0 ∈ {1, ..., p − 1}. Pk −j Proceeding by induction, suppose ωk = exp 2πi j=1 ck−jp . Since ωk=1 is pth root of ωk, there exists ck ∈ {0, 1, ..., p − 1} such that

k ! k+1 ! X −j−1 −1 X −j ωk+1 = exp 2πi ck−jp exp(2πickp ) = exp 2πi ck+1−jp . j=1 j=1

−1 (3) If ψ ∈ Qb p, ψ(1) = 1, and ψ(p ) 6= 1, there exists y ∈ Qp with |y| = 1 such that ψ = ψy. ∞ P∞ j Let us take {cj}0 as above and set y = 0 cjp . Then |y| = 1 since c0 6= 0, and for k ≥ 1,

k ! −1 ! ∞ −k X −j X j X j ψ(p ) = exp 2πi ck−jp = exp 2πi cj+kp = ψ1( cj+kp ) j=1 j=−k −k

−k −k = ψ1(p y) = ψy(p ) 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

tr ψ ∗ K −→ Qp −→ 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 Kb. Which is K −→ Kb 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 Kb. Also ψy = 1 if and only if y = 0. Since ψ is a non trivial character −1 there exists y0 such that ψ(y0) 6= 1 and if y 6= 0 =⇒ ψ(y0) = ψ(y0y y) = −1 ψy(y0y ) 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 Kb is dense as well as closed. We have Image = {ψy|y ∈ K} ⊂ Kb. 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 −→ φ means that for any compact set X ⊂ K, ψxn (x) −→

φ(x)∀x ∈ K uniformly on K. That is |ψxn − 1| < ∀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 ∼= Kb so fourier analysis of standard kind becomes available on K. For any function f ∈ L1 we deﬁne, Z fb(x) = f(x)ψ(xy)dy K 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 that fbb(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 −1 deﬁned by δ = {x ∈ K|tr(xOK ) ⊂ p} . K|Qp Z

Lemma 19.3. If A is a compact abelian group and χ : A −→ C∗ is a R R character, then A χ(a)da = 0 if χ 6= 1 and A χ(a)da = vol(A) if χ = 1. R R Proof: If χ = 1 then A χ(a)da = A 1.da = vol(A). If χ 6= 1 there R R exists x0 ∈ A such that χ(x0) 6= 1. Put I = A χ(a)da then I = A χ(a)da = R A χ(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 Z Z χ (y) = χ (x)ψ(xy)dx = ψ(xy)dx bOK OK K OK

Then χ (y) = 0 if y∈ / δ−1 and χ (y) = vol(O ) if y ∈ δ−1. Where bOK K bOK K K ψ(x) = exp 2πi tr(x) is a character of K. Then, Z Z χ (y) = χ (x)ψ(xy)dx = vol(O ) ψ(xy)dx bOK OK K b b −1 K δK Z χ (0) = vol(O ) 1.dx = vol(O )N(δ )vol(O ) bOK K K K K b −1 δK But χ (0) = χ(0) = 1. This implies that vol(O ) = √ 1 , where bOK K b N(δK ) −1 N(δK ) = [δK : 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 Z(f, χ, s) = f(x)χ(x)|x|sd∗x K∗

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(OK ) 6≡ 1), where OK is the group of invertible elements of the ring OK .

Z Z(f, χ, s) = f(x)χ(x)|x|sd∗x K∗ Z s ∗ s −s = χ(x)|x| d x + a term belonging to C[q , q ] ∗ OK

Now let us look at the term Z Z χ(x)|x|sd∗x = χ(x)|x|sd∗x ∗ ∞ n n+1 OK ∪n=0[π OK −π OK ] ∞ X Z = χ(x)|x|sd∗x n ∗ n=0 π OK ∞ X Z = χ(πnx)q−ns|x|sd∗x ∗ n=0 OK ∞ X Z = χ(πn) χ(x)q−ns|x|sd∗x ∗ n=0 OK = 0

Therfore for ramiﬁed character Z(f, χ, s) ∈ C[qs, q−s]. If χ| ∗ ≡ 1 i.e. χ is unramiﬁed then by a similar calculation OK

∞ X n −ns ∗ 1 ∗ Z(χOK , χ, s) = χ(π )q vol(OK ) = χ(π) vol(OK ) n=0 1 − qs 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),

−1 −1 Z(f, χ, s)Z(g,b χ , 1 − s) = Z(g, χ, s)Z(f,b χ , 1 − s). Proof: This follows by deﬁnitions as we will see. Z −1 s −1 1−s ∗ ∗ Z(f, χ, s)Z(g,b χ , 1 − s) = f(x)χ(x)|x| gb(y)χ (y)|y| d xd y K∗×K∗ We apply change of variable (x, y) −→ (x, xy). Z s −1 1−s ∗ ∗ = f(x)χ(x)|x| gb(xy)χ (xy)|xy| d xd y K∗×K∗ Z Z ∗ −1 1−s ∗ = f(x)gb(xy)|x|d x χ(y )|y| d y K∗ K∗ Z = f(x)g(z)ψ(xyz)|y|−sχ(y−1)dxdydz K×K×K

This expression is symmetric in f and g; proving the theorem. Corollary 20.1.

Z(f, χ, s) = ρ(χ, s)Z(f,b χ−1, 1 − s) where ρ(χ, s) is independent of f.

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

Note that the value of an unramiﬁed character χ on a uniformizing element π does not depend on the choice of uniformizing element π. Rearranging the functional equation, we get

Z(f,b χ−1, 1 − s) Z(f, χ, s) = (χ, s) L(χ−1, 1 − s) L(χ, s) These epsilons are called local constants. These are monomial function in qs.

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

Proof : Let assume that χ is ramiﬁed. In this case functional equation is Z(f,b χ−1, 1 − s) = (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 (χ, s) ∈ C[qs, q−s]. Similarly (χ−1, 1 − s) ∈ C[qs, q−s]. Using previous proposition we see that (χ, s)(χ−1, 1 − s) = χ(−1), a constant. Hence (χ, s) is a unit in C[qs, q−s] but the only units in s −s C[q , q ] are monomials. Hence the proposition. Proposition 20.2.

(χ, s)(χ−1, 1 − s) = χ(−1)

Proof follows from the functional equation for local zeta functions and the deﬁnition of (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 locally 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 Q Q direct product of the system (Gp,Hp) will be denoted as (Gp,Hp) or Gp. This product is deﬁned as follows.

Y 0 (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: Y u1 × u2 × ..... × ur × Hp p6={1,2,..r} where ui’s are compact open subgroups of Gpi at ﬁnitely many places pi. Q With this topology G = (Gp,Hp) is a locally compact topological group. The corresponding Schwartz space (space of locally constant and compactly supported functions) will be given by

( r r+1 ) O O O S(G) = S(Gp), χHp = lim S(Gpi ) ,→ S(Gpi ) → {1,..,r}⊂p p i=1 i=1 where the maps are deﬁned as

r r+1 O O S(Gpi ) −→ S(Gpi ) i=1 i=1 O O O O O f1 .... fr 7−→ f1 ... fr χHr+1

Q Proposition 21.1. Let G = (Gp,Hp) be the restricted direct product of p 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 Q Q 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 S direct product of {(Qp, Zp)} {R, φ} is called the adeles of Q, denoted as = × where = N . This is a locally compact topological ring. AQ R Qb Qb Q Z Zb Similar notion is associated to the multiplicative group in which we take ∗ ∗ S ∗ {(Qp, Zp)} {R , φ}. Corresponding restricted direct product is called the idele group of denoted by , or ∗ , or . Q JQ AQ Gm

48 For a general number field K one can define adeles and ideles as follows: O Y AK = AQ K = (Kv,Ov) Q !∗ O Y ∗ ∗ JK = AQ K = (Kv ,Ov) Q where Kv denotes completion of K at place v and Ov is corresponding valuation ring. On JK the topology is defined by the system of neighbourhoods ∗ N ∗ coming from O = ( p OK ) in usual way. Note that ,→ is a p Z Z JQ AQ 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 ∈ K it 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 Q of K. In fact let U = U∞ × v Ov where U∞ is a small neighborhood of N r1 r2 0 in K R. We have OK −→ R × C 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. N r r The map K −→ K AQ can be compared to the map Q −→ A to show Q Q AQ Zp×R Q the cocompactness. As we can see = that is = + p × . Q Z AQ Q Z R

∗ Lemma 21.2. The multiplicative group K embedds inside JK as a discrete 1 subgroup. Inside JK there exists a closed subgroup JK = ker{|.| : JK −→ ∗ Q ∗ C ; |x| = v |x|v} which contains K .

1 ∗ Theorem 21.1. The group JK /K is a compact topological group.

49 ∗ Proof : The discreteness of K inside JK is analogous. We prove that ∗ 1 ∗ Q Q ∗ K −→ JK is cocompact. For this observe that K \JK / v Ov ∞ K∞ is nothing but the class group of K. Look at the map M JK −→ Zv = divisors v X xv 7−→ ordv(xv)v v

Q Q ∗ ∗ Then JK / v Ov. ∞ K∞ ≡ divisors, going modulo K we get ∗ Q Q ∗ ∗ K \JK / v Ov. ∞ K∞ ≡ K \divisors, which is the class group of K. Since ∗ Q Q ∗ JK is K . ( v Ov ∞ K∞) up to ﬁnite index, for the purpose of the proof 1 ∗ ∗ ∗ Q Q ∗ 1 of compactness of JK /K we can as well look K −→ K . v Ov ( ∞ K∞) . 1 ∗ Q ∗ 1 Thus JK /K is compact if and only if ( ∞ K∞) /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. Q ni Cf = pi 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) . d1 Proof : Let x = be an element of I(C) with d1, d2 coprime to n. Then d2 ∗ −1 we deﬁne a map I(C) −→ (Z/n) as x 7−→ d1( mod n).(d2( mod n)) . This is a surjective map with kernel PC.

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

∗ Y Y ∗ Y I(C)/PC ≡ JK /K Uv K∞ Uv(C) (v,C)=1 v/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 AcK and give rise to a principal homogeneous structure.

AK × AcK −→ AcK

(a, χ) 7−→ χa

χa(x) = χ(ax)

Q Q ∗ ∗ Lemma 23.2. If G = (Gp,Hp) then Gb = (Gbp,Hp ) where Hp ⊂ Gbp consists of all characters on Gp which are trivial on Hp which is isomporphic to G\p/Hp, a compact group.

Q ∗ 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 Hp = G\p/Hp ⊂ Gp is an open subgroup. To prove this ∗ we look at the continuous map Gp × Gcp −→ C defined by (a, χ) 7−→ χ(a). 1 Define U = {χ ∈ Gcp|χ(Hp) ⊂ a ball of radius 3 around 1} This set of characers are open set in Gcp. But since there are no small subgroups we get ∗ U = 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(2πix)

These mappings can be combined to give a mapping

∼ M ∼ 1 AQ/(R × Zb) = Qp/Zp = Q/Z −→ S p

But = ( + b). =⇒ AQ ∼= R×Zb AQ R Z Q Q Z This gives a character χ(a) = e2πia on R×Zb and hence on AQ . Z Q ∼ Lemma 23.3. For K, an algebraic number ﬁeld we have A\K /K = K as a principal homogeneous space.

Proof Let ψ be a character of AK /K. For any element a ∈ K, one 1 can define another character ψa : AK /K −→ S 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

Y K −→ Kv v6=v0 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 Q places of a global ﬁeld K. Then K is dense in v∈S Kv.

24 Gr¨ossencharacter

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

∗ Definition (Grössencharacters). A Grössencharacter is a character of JK /K ∗ ∗ which is a continuous group homomorphism JK /K −→ C . A continuous ∗ 1 group homomorphism JK /K −→ S is called unitary Grössencharacter.

Gr¨ossencharacters 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

Y 1 L(χ, s) = h χ(πp) i 1 − s (p,c)=1 (Nπp) 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 Q 1 We can also write L(χ, s) = L(χp, s) where L(χp, s) = χ (π ) if χp is p 1− p p (Nπp)s ∗ ∗ an unramified character of Kp i.e. it is trivial on Op 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 Grössencharacter : If χ : JK −→ C is ∗ N ∗ a Grössencharacter, the restriction of χ to K∞ = (K R) is called the infinity type of character. Definition (Algebricity of Grössencharacter). The character χ is called al- n + gebraic if and only if χ∞ is algebraic i.e. χ∞(x) = x if x ∈ R and n −m ∗ χ∞(z) = z z , m, n ∈ Z if z ∈ C . ∗ + Q ∗ Example : For K = Q we have JQ = Q × R × Zp. This means that ∗ ∼ + Q ∗ JQ/Q = R × Zp. 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 Grössencharacter and the corresponding L-function is the Dedekind Zeta function of K. Exercise :

s 1. Prove that any Gr¨ossencharacter can be written as χ = χ0|| where χ0 is a unitary Gr¨ossencharacter.

s0 2. Prove that by using formula L(χ, s) = L(χ.|| , 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, Z ζ(f, χ, s) = f(x)χ(x)|x|sd∗x JK ∗ 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 ∗ Np−1 In our case Hp = OP , we have volume Np . thus to define Haar measure on Np dx JK one multiplies local Haar measure by Np−1 |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 ∈ N Q Q S(AK ) = v S(Kv). So it suffices to assume f = fp, χ = χp, in which Q case if we can prove that p Z(fp, χp, s) converges and defines an analytic function in the region Re(s) > 1, then the integral defining ζ(f, χ, s) will Q also converge to p Z(fp, χp, s) and will be analytic function in the region 1 Re(s) > 1. But we have calculated earlier that Z(fp, χp, s) = χp(πp) is 1− (Np)s 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 Gr¨ossencharacter then ζ(f, χ, s) has analytic continuation to the entire complex-plane and has functional equation

ζ(f, χ, s) = ζ(f,b χ−1, 1 − s)

Proof : It has two basic ingradients.

1. Poisson Summation formula : We recall that Poisson summation formula of R states that X X f(n) = fb(n) n∈Z n∈Z in case of R. where f ∈ S(R). We will be using the analogue of this for AK . X X f(ξ) = fb(ξ) ξ∈K ξ∈K P P Hence for f ∈ S(AK ) and a ∈ JK we get |a| ξ∈K f(aξ) = ξ∈K fb(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 1 P 1 that if f ∈ L (G) then F (x) = γ∈Γ f(x + γ) is a L -function on G/Γ R R and G/Γ F (x)dx = G f(x)dx.

55 1 + We continue with the proof of the theorem. Write JK = JK × R (we use (xt) variable). 1 Nm + 0 −→ JK −→ JK −→ R

Z ζ(f, χ, s) = f(x)χ(x)|x|sd∗x JK Z dt = f(xt)χ(xt)|xt|sd∗x t JK Z Z dt = f(xt)χ(xt)|t|sdx + 1 t R JK Z 1 Z dt Z ∞ Z dt = f(xt)χ(xt)|t|sdx + f(xt)χ(xt)|t|sdx 1 t 1 t 0 JK 1 JK

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, Z s ζt(f, χ, s) = f(xt)χ(xt)t dx. 1 JK Then Z " # X s ζt(f, χ, s) = f(xξt)χ(xξt) t dx 1 ∗ JK /K ξ∈K∗ " # Z X = f(xξt) χ(xt)tsdx 1 ∗ JK /K ξ∈K∗

56 since χ is non-trivial,

Z " # X ξ s−1 ζt(f, χ, s) = fb χ(xt)t dx 1 ∗ xt JK /K ξ∈K∗ Z 1 = fb χ(xt)ts−1dx 1 xt JK Z x = fb χ(x−1t)ts−1dx 1 t JK −1 = ζ 1 (f,b χ , 1 − s) t

Thus,

Z 1 Z 1 Z ∞ −1 −1 ζt(f, χ, s) = ζ 1 (f,b χ , 1 − s) = ζt(f,b χ , 1 − s) t 0 0 1 Hence,

Z 1 dt Z ∞ dt ζ(f, χ, s) = ζt(f, χ, s) + ζt(f, χ, s) 0 t 1 t Z ∞ Z ∞ −1 dt dt = ζ 1 (f,b χ , 1 − s) + ζt(f, χ, s) t 1 t 1 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 functional equation for L-functions associated to Gr¨ossencharacter.

∗ ∗ Proof of Corollary : Let χ : JK /K −→ C be a Gr¨ossencharacter ∗ Q 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,b χ−1, 1 − s).

57 Y Y Y Y −1 ζ(fp, χp, s) = ζ(fbp, χp , 1 − s) p∈S p/∈S p∈S p/∈S Y Y Y −1 Y −1 ζ(fp, χp, s) L(χp, s) = ζ(fbp, χp , 1 − s) L(χp , 1 − s) p∈S p/∈S p∈S p/∈S −1 Y ζ(fbp, χp , 1 − s) Y = L(χ−1, 1 − s) L(χ−1, 1 − s) p p∈S p p∈S −1 Y ζ(fbp, χp , 1 − s) = L(χ−1, 1 − s) L(χ−1, 1 − s) p∈S p

But using local functional equation

−1 ζ(fp, χp, s) ζ(fbp, χp , 1 − s) (χp, s) = = −1 L(χp, s) L(χp , 1 − s) We get L(χ, s)(χ, s) = L(χ−1, 1 − s) Q with (χ, s) = p (χp, s) where (χp, s) = 1 at almost all places and is of s the form ab .

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. P 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 P P 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 P 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} S{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

nv P ξ|Ov.πv ≡ 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 nv of πv 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, OK /q ρq : Dq −→ Gal Ok/p σ 7−→ (x mod q → σ(x) mod q)

This map is clearly well deﬁned homomorphism of groups.

OK /q Proposition 27.1. The canonical map ρq : Dq −→ Gal has the Ok/p 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 /q is cyclic, gener- Ok/p ated by the Frobenius map x → xq where #(k/p) = q. In case of p being unramified the Frobenius element F rp ∈ 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 of O and π be that of O . Then π .O = π .O and the decomposition group u v v u v v v D ∼= Gal Ov/πv ∼= Gal(K/k). u Ou/πu

60 This deﬁnes the map

Y k∗ r : v −→ Gal(K/k) O∗ v∈ /S v πv 7−→ F rv

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

Q ∗ Theorem 27.1 (Artin Reciprocity). The map r : v∈ /S kv −→ 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. ∗ Q ∗ ∗ Q ∗ k ,→ v∈S kv is dense. This implies that k . v∈ /S kv is dense in Jk, hence a ∗ Q ∗ character which is trivial on k . v∈ /S kv 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 number 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. r1 Jk / Gal(K1/k) FF FF FF FF FF r FF restriction 2 FF FF FF F# Gal(K2/k) Thus we have a map

ab k −→ lim Gal(K/k) = Gal(k /k) J ← Theorem 27.2. Let k be a number ﬁeld. Then,

∗ + ∼ ab Jk/(k .k∞) = Gal(k /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

∗ ∗ Y ∗ k k∞. Ov ⊂ Jk v∈kf ∗ More generally if U ⊂ Jk is an open subgroup containing k , then the corresponding 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 define Q ∗ U(C) = v∈C Uv(nv) where C = {v1, v2....vl, v∞} and Uv(nv) = {x ∈ Ov|x ≡ 1 nv ∗ mod (πv )}. 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 subgroups 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 F rp ∈ C |C| is of density |G| . In particular, the set X = {p|F rp ∈ 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 F rp = 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

∗ K ∼ Ik(C) Jk/(k .Nk JK ) = K PC.Nk (IK (C)) where C is a cycle in k divisible by ramiﬁed primes such that O (C ) ⊂ N Kw (K∗ ) (called admissible cycles) and I (C) is ideals in k which v v kv w k are coprime to C. Proof : We have Y ∗ Y ∗ JC = Ov(C) kv v/C (v,C)=1 T ∗ ≈ ∗ Let kC = JC k , then JC/kC −→ Jk/k . This is injective (by deﬁnition) and Q ∗ ∗ Q ∗ surjective by weak approximation theorem ( v/C Ov(C).k = v/C kv). This gives rise to following diagram.

≈ ∗ IC/kC −−−→ Jk/k     y y Ik(C) ≈ ∗ K K −−−→ Jk/(k .Nk JK ) PC.Nk (IK (C))

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 −→ F rv. Observe that K Nk (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.

d Y pOK = qi i=1

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

Ik(C) (a) Order of the group K and Gal(K/k) is the same. PC.Nk (IK (C))

(b) The map Ik(C) −→ Gal(K/k) is surjective map.

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 m completely split; i.e. ζK (s) = ζk (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) K ≤ [K : k] PC.Nk (IK (C)) K Proof : Let H = PC.Nk (IK (C)) and G = I(C)/H. Given group G and χ G −→ 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

m(χ) L(s, χ) = (s − 1) .fχ(s)

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

m(χ) ≥ 0 if χ 6= 1 = −1 if χ = 1

−1 Y χ(p) L(s, χ) = 1 − (Np)s p This makes sense for Re(s) > 1

X χ(p) log(L(s, χ)) = − log 1 − (Np)s p X χ(p) = + higher order terms (Np)s p

The higher order terms are bounded around s = 1. Hence, P χ(p) log(L(s, χ)) = p (Np)s around s = 1 (up to bounded function which we will denote by f ∼ g).

X χ(p) log(L(s, χ)) ∼ (Np)s p " # X X 1 ∼ χ(g) (Np)s g∈G p where p = g ∈ G = I(C)/H. " # X X X 1 log(L(s, χ)) ∼ χ(g) (Np)s χ χ,g∈G p X 1 = |G| (Np)s p∈H

65 m(χ) On the other hand, L(s, χ) = (s − 1) .fχ(s) 1 log(L(s, χ)) ∼ −m(χ) log s − 1 " # X X 1 log(L(s, χ)) ∼ 1 − m(χ) log s − 1 χ6=1

So we get h i 1 − P m(χ) log 1 χ6=1 s−1 X 1 ∼ |G| (Np)s p∈H Note that all primes q in K which are of degree 1 over k have the property T p = q OK belongs to the norm from K to k. Hence,

h i 1 − P m(χ) log 1 χ6=1 s−1 1 X 1 ≥ |G| [K : k] (Np)s q of degree 1 1 ∼ ζ (s) [K : k] K 1 1 ∼ log [K : k] s − 1 which implies X |G| 1 ≥ 1 − m(χ) ≥ [K : k]

I(C) Ik(C) [K : k] ≥ |G| = | | i.e. order of K ≤ [K : k]. Further we get PN PC.Nk (IK (C)) 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. X 1 1 1 s ∼ log (Np) [I(C): PC] s − 1 p∈a0 where a0 ∈ I(C)/PC.

66 Proof : We have X χ(p) log(L(s, χ)) ∼ (Np)s p X X 1 = χ(a) (Np)s a p∈a

Then,

X X X X 1 χ(a−1) log(L(s, χ)) = χ(aa−1) 0 0 (Np)s χ χ a p∈a X 1 = [I(C): P ] C (Np)s p∈a0

Hence,

X 1 log ζ (s) = [I(C): P ] K C (Np)s p∈a0 1 X 1 log = [I(C): P ] s − 1 C (Np)s p∈a0

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

67 cohomology as follows,

H0(G, A) = AG, the zeroth cohomology AG Hc0(G, A) = , the zeroth Tate cohomology (1 + σ + σ2 + ... + σn−1)A ker(1 + σ + σ2 + ... + σn−1): A −→ A H1(G, A) = , the ﬁrst cohomology (1 − σ)A

Deﬁnition. The Euler characteristic of A is deﬁned by,

#(Hc0(G, A)) χ(A) = #(H1(G, A)) if orders make sense, i.e. the orders of both Hc0(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 Hc0(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 ∗ #Hc0 1 1. Since χ(K ) = #H1 , by Hilbert Theorem 90, H = 0 (for any cyclic extensions). Hence χ(K∗) = #Hc0 = [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

Z/n χ(OK ) = χ(Ok[Z/n]) = χ(Ind{e} Ok)

We use Shapiro’s lemma: for a subgroup H of G we have i i G H (H,B) = H (G, IndH B). We get χ(Ok) = χ{e}(Ok) = 1.

3. Let us look at the following diagram.

x7→ x 0 −−−→ k∗ −−−→ K∗ −−−→σx K1 −−−→ 1 x x x       1 0 −−−→ Uk −−−→ UK −−−→ UK −−−→ 1 where K1 is set of norm 1 elements in K∗. Which implies that ∼ ∗ ∗ ∼ 1 1 Z/eZ = K /k UK = K /UK . 0 Hc (UK ) [Uk:NmUK ] χ(UK ) = 1 = 1 = 1 x H (UK ) [U :{ }] K σx|x∈UK 1 1 ∗ ∗ Observe that UK = K thus [Uk : NmUK ] = e and [k : NmK ] = [K : k] for cyclic extensions of local ﬁelds.

69 ∗ K 29 Index calculation [Jk : k Nk JK]

Theorem 29.1. If K is a cyclic extension of degree n of a number ﬁeld k ∗ K then [Jk : k Nk 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 n the places S of k. To each place w in SK deﬁne a symbol Xw. Let E = Q be the free Q module on the symbols Xw. Observe that E is a module for N Gal(K/k). Let M be any Gal(K/k) -invariant lattice in E R.

0 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 Q N ∼ N vector 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 N N N N N 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 polynomial 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,

#Hc0(G, M) Y χ(M) = = χ(M 0) = N #H1(G, M) v v∈S where Nv is the order of the decomposition subgroup at the places v ∈ S.

70 0 L G Proof : Since M = v IndGv Z hence,

0 Y G χ(M ) = χ(IndGv Z) v Y = χ(Gv0 Z) v Y = Nv v

Corollary 29.2. Let KS be the set of S units in K, i.e. w(x) = 0∀w∈ / SK then Q N χ(K ) = v S [K : k] Proof : We look at the map O KS −→ E R X x 7−→ log |x|w.Xw

w∈SK

Dirichlet unit theorem says that image of KS is a lattice in hyperplane in N P E R deﬁned by Xw = 0. Therefore

0 Y χ(KS)χ(Z) = χ(M) = χ(M ) = Nv

Q Nv Thus χ(KS) = [K:k] . We continue the proof of theorem 29.1 as,

0 #Hc (Gal, JK ) χ(JK ) = 1 #H (Gal, JK ) 0 = #Hc (Gal, JK )(Hilbert 90)

= [Jk : NmJK ]

Q O∗. Q k∗ ∗ ∗ JK v v∈S v So χ( K /K ) = [ k : k Nm K ]. Hence ∗ = . This completes J J J K KS ∗ 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).