Lectures On Algebraic Number Theory

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 = e

2πi 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 degQ K. 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 ∗ for the units in OK i.e. elements in OK later lectures. Let us denote 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 OK is Z/2 × Z. √If K = Q( d) then √ OK = Z( d) if d 6≡ 1(mod4) and OK = Z( 1+2 d ) 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? 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 A1 A2 = λi µi / 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)r2 hR 1 √ . + holomorphic function s−1 w dK 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 Q in OK as pOK = ni=1 pei i 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 irreducible polynomials f¯(x) = f¯i (x)ei in Z/p[x]. The question is whether one can describe a LAW which tells us how many of f¯i ’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 x20 = 5 has solution in p−1 Z/p if and only if 5 2 ≡ 1(mod p) ( which is by defenition denoted as ( p5 ) = 1). This can be seen by noting that (Z/p)∗ is a cyclic group. So if p/5 then f¯(x) = x2 . If ( p5 ) = 1 then f¯(x) = (x − x0 )(x + x0 ), where x20 = 5 in Z/p. And if ( p5 ) = −1 then f¯(x) is irreducible in Z/p. We will later prove the following theorem. There exists a “law” of decomposing 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 −−−→ x  

K x  

Z −−−→ Q

Let p ∈ Z be a prime and let p be a prime ideal lying over p in OK . Let Q pOK = gi=1 pei i . 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. P Lemma 3.1. Let K be a number field. Then gi=1 ei fi = [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 ef g = [K : Q], where e is the ramification index, f is the residual degree and g is the number of primes lying over p.

4

Artin Reciprocity 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 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 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 Q-linear map lx : K −→ K defined by lx (λ) = λx. Define trQK (x) as trace of lx and N mK Q (x) as determinant of lx . If there is no possibility of confusion, we will denote trQK (x), N mK Q (x) simply as tr and N m. 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 0 be another integral basis of Zd . Then there exists an integeral matrix A such that B = AB 0 At . Hence, dB = det B = det(AB 0 At ) = (det A)2 det B 0 = det B 0 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 ANSWERED. 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.

Dirichlet Unit Theorem1

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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 = {x1 v1 + · · · + xm vm | xi ∈ R, 0 ≤ xi ≤ 1} is called a fundamental mesh of the lattice. 1

This 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 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 (r1 p + r3 a + r4 b, r2 p + r3 b − r4 a, 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 2π 2 p2 which is strictly greater than 24 p2 . 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 + B 2 + C 2 + 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 Y j : K −→ KC = C τ

a 7−→ ja = (τ a)

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 and σ ¯ ’s give the 2s pairwise conjugate complex embeddings. We see that the image of K under this Q canonical mapping into Cn = 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 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 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 hjαi , jαk i = τl (αi )¯ τl (αk ) = AA¯t 1≤l≤n

10

Then we get 1

V ol(Γ) = |det(hjαi , jαk i)| 2 = |det(A)| =

p |dK |.[OK : I]



So via the canonical embedding j : K −→ KC , we can identify the ideals of OK with lattices in Rn . But since we are interested in the units of the ring of integers, OK∗ , we pass on to the multiplcative group K∗ by using the standard logarithm map l : C∗ −→ R, z −→ log|z| It induces the surjective homomorphism Y Y C∗ −→ R , (zτ ) −→ (log|zτ | l : K∗C = τ

τ

Note that image of K∗R under the above map lies in the following set " #+ ( ) Y Y R := (xτ ) ∈ R | xρ ∈ R, xσ¯ = xσ τ

τ

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 " #+ Y Tr : R −→ R τ

given by the sum of the coordinates. Also note that the R-vectorspace Q [ τ R]+ is isomorphic to Rr+s . Now conside the following subgroups OK∗ = { ∈ OK | NK|Q = ±1} S = {y ∈ K∗R | N(y) = ±1} Y H = {x ∈ [ R]+ | T r(x) = 0} τ

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

l

OK∗ −→ S −→ H 11

and the composite λ := l o j : OK∗ → H. The image will be denoted by Γ = λ(OK∗ ) ⊆ H and we obtain the Theorem 6.3. Let K be a number field. Then the sequence λ

1 −→ µ(K) −→ OK∗ −→ 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 OK∗ . 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

where A =

sp ( π2 ) |dK |.

Then there exists a ∈ I, a 6= 0, such that

|τ a| < cτ for all τ ∈ Hom(K, C) Proof : We just note that the the set Y = {(zτ ) ∈ KR | |zτ | < cτ } Q is centrally symmetric, convex and has volume 2r+s π s τ cτ . Now, Γ = j(I) p is a lattice with volume |dK |.[OK : I]. So we have 2 sp Vol(Y ) > 2r+s π s ( ) |dK |.[OK : I] = 2n Vol(Γ) π 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 : OK∗ → H arises by restricting the mapping Y Y j l 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∗ | e−c ≤ |zτ | ≤ ec (zτ ) ∈ τ

But it contains only finitely many points of the set j(OK∗ ) 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 Y cτ > A[OK : I] C= τ sp

where A = ( π2 )

|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 } where c˜τ = cτ |yτ | and one has c˜τ = c˜τ¯ , |N(y)| = 1. Then, there exists a point

Q

τ

cτ =

Q

τ

c˜τ = C because

ja = (τ a) ∈ W y, a ∈ OK , a 6= 0 13

Q

τ

|yτ | =

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 [

W (jαi )−1

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 Y cτ = C, |NK|Q (a)| = |N(xy −1 )| = |N(x)| < τ

a is associated to some αi , αi = a,  ∈

OK∗ .

Consequently, we have

y = xja−1 = xj(αi−1 ) Since y, j ∈ S, one finds xjαi−1 ∈ S ∩ W jαi−1 ⊆ 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 OK∗ 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 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} Then A is a ring, called the valuation ring of v. Examples :

S

{x ∈ K ∗ |v(x) ≥ 0}.

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) 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 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

  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 S 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 −→ 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 nontrivial 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

N Z

Q is a field containing Zp such that Qp = Zp [ p1 ].

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|∞ = q deg(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 Q 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 a0 π r + a1 π 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 X f (t) = an tn 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 Q then K Q Qp is a separable algebra which is a product of fields. K

O

Qp =

M

Kv

v|p

Q

Similarly, K

O

R = Rr 1

Q

18

M

Cr2

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. Proposition 8.1. Let K = Q(ζp ), ζp = e is OK = Z[ζp ].

2πi p

. Then the ring of integers of K

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 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 = (x − ζpi ). i=1

Putting x = 1 in this equation,

19

p−1 Y p = (1 − ζpi ) i=1

= (1 − ζp )p−1 .

Y 1 − ζpi 1 − ζp

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

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

p p proving that 1−ζ ∈ Z[ζp ]. Clearly 1−ζpp ∈ Z[ζp ]. Thus 1−ζ ∈ Z[ζp ] is a unit 1−ζpi 1−ζpi 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


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 + b1 ζp + ...... + bp−2 (ζp )p−2 with bi ∈ Q and bi have no p in denominator. Note that if x ∈ OK then tr(xζp−i ) ∈ Z ∀i. Using trζpj = −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. x = b0 + b1 ζp + ...... + bp−2 ζpp−2 = b0 + (b1 − b0 )ζp + ...... + (bp−2 − b0 )ζpp−2 + b0 (ζp + ...... + ζpp−2 ) = (b1 − b0 )ζp + ...... + (bp−2 − b0 )ζpp−2 + b0 (1 + ζp + ...... + ζpp−2 ) X = −b0 ζpp−1 + (bi − b0 )ζpi

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 −→ Q defined by tr(x, y) = trQK (xy). Definition. The set a0 = {x ∈ K|tr(xy) ∈ Z, ∀y ∈ a} is called complementry set. It is not difficult to see that a0 is also a fractional ideal. 0 0 ⊃ OK Definition. OK = {x ∈ K|tr(xy) ∈ Z, ∀y ∈ OK }. Then K ⊇ 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 N m(I) to be the cardinality of OK /I. Theorem 9.1. Let K be a number field. Then N m(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 : N m(IJ) = N m(I)N m(J) N m(α) = |N mK/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 √ √ of Zp [ p]. Then √1p (a + b p) = √ap + b =⇒ tr( √ap + b) = 2b ∈ Zp ∀a, b. That −1 means √1p ∈ δK . √ Let us take case when p = 2 the we claim that 2 ∈ δK . 12 (a + b 2) = 1 a + √b2 =⇒ tr( 12 a + √b2 ) = a ∈ Z2 =⇒ 2 ∈ δK . 2 Lemma 9.1. If L is a finite extension of a local field K, with πK and πL uniformizing parameter such that πLe = πK .u, u a unit in OL . Then πLe−1 /δL/K . In fact δL/K = πLe−1 ⇐⇒ p/e. e−1 e−1 Let us prove that π  . For this we show tr(πL OL ) ⊂ L /δL/K OK . But then tr(πLe−1 OL ) = tr ππKL OL = π1K tr(πL OL ) ⊂ π1K πK OK ⊂ OK . 

Proof:

10

The Riemann Zeta Function

The Riemann zeta function is defined as follows: X 1 Y 1 ζQ (s) = = ns p≥2 1 − p1s n≥1 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 + holomorphic function around s = 1 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). ζQ (s) =

s s ξ(s) = π − 2 Γ( )ζQ (s) 2 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) = 21 . This is one of the most famous unsolved problems in mathematics, 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) =

X I6=0,I⊂OK

Y 1 1 = s (N I) 1 − (N1P)s P

where I ranges over all nonzero ideals of OK , N I is the norm of the ideal I and P is a prime ideal in OK . Both the expressions are valid for re(s) > 1. The function 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)r2 hR 1 √ . + holomorphic function around s = 1 s−1 w dK 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

ζK (s) =

I6=0,I⊂OK

=

X I6=0,I⊂OK

1

Y P

1 (N I)s

1−

1 (N P)s

for Re(s) > 1. Since the equality Y 1 1 = (N I)s 1 − (N1P)s P

is formal, so it suffices to check the convergence of the product. log(ζK (s)) = −

X

log(1 −

P

=

1 ) (N P)s

1 m(N P)ms P,m≥1 X

We have N P = 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)) ≤

N mpf ms N ≥f ≥1

X p, m≥1,

where p is prime. Therefore upto a function which is bounded in a neighborhood of 1, log(ζK (s)) =

X ap ps

ap denotes number of primes above p of degree 1. 24

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

log(ζK (s)) = log

12

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

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. Y 1 ζA1 (s) = 1 (1 − N p(x) )s p(x)

where p(x) runs over irreducible polynomials in Fp [x] with leading term 1 and f(x) is monic polynomial over Fp . And N f (x) = pn where n = deg f (x). But then ζA1 (s) =

p(x)

=

1

Y X f (x)

(1 −

1 (N f (x))s

X pd = pds d

3. Let K = Q(i), then ζK (s) =

=

1 1 − pps

P

1 (m2 +n2 )s

25

1 )s N p(x)

13 13.1

Class Number Formula

2

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. Hence the map If p ≡ 1 mod 4, then −1 p     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 hp √ denote the class number of the quadratic field Q( −p). Then  R − N if p ≡ 7 mod 8, p p hp =  1 (Rp − Np ) if p ≡ 3 mod 8. 3

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 2

This 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) =κ X→∞ X lim

where κ=

 log η 2√

d 2π  √ w |d|

if d > 0, η is the unique fundamental unit > 1, 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, 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 Ld (s) = m−s m m=1 with

d 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.  log η 2√ Ld (1) if d > 0, η is the unique fundamental unit > 1, d h= 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 ( p. ). Using above expression for class number one can √

easily derive that hp = sum.

p L(1, χ). π

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

Lemma 13.2. We have p−1

iπτ1 (χ) X χ(k)k. L(1, χ) = p2 k=1 where τk (χ) =

X

χ(a)ρak .

1≤a 1 we have L(s, χ) =

∞ X χ(n)

ns

n=1

=

X 1≤a 1. We observe that for any α ∈ J, α 6= 0, there exist an integer m such that ω < log η (13.3.1) 0 ≤ log 1/2 |N (ω)| where ω = η m α 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 < log η . 2N (X, C) = ω ∈ J | 0 < |N (ω)| < y, 0 ≤ log |N (ω)|1/2 Let β1 , β2 be an integral basis for J and let 1/2   |ξ1 | 2 0 Ω = (ξ1 , ξ2 ) ∈ R |0 < |ξ1 ||ξ1 | < 1, 0 < log 0 1/2 < log η |ξ1 | where ξ = ξ1 β1 + ξ2 β2 and ξ 0 = ξ1 β10 + ξ2 β20 , β10 , β20 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 0 | = 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 √ NΩ ( y) 2N (X, C) = lim N (J) = N (J)Area of Ω. lim y→∞ X→∞ X y Area of Ω can be computed to be case. Case 2: d < 0 In this case we have

4 log√ η . N (J) d

This completes the proof in this

wN (X, C) = |{ω ∈ J | 0 < N (ω) < y}| . Let  Ω = (ξ1 , ξ2 ) ∈ R2 |0 < |ξ1 β1 + ξ2 β2 |2 < 1 . √ Then wN (X, C) = NΩ ( y), hence √ NΩ ( y) wN (X, C) lim = lim N (J) = N (J)Area of Ω. y→∞ X→∞ X y 2π√ Here area of Ω is calculated to be N (J) , completing the proof of Theorem 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) = hκ. X→∞ X lim

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 lim+

s→1

p∈M (N p)s 1 log s−1

In one of the earlier lecture we had observed that log

1 ∼ log ζK (s) s−1 X 1 ∼ (N p)s p X 1 ∼ (N p)s degp=1

where f1 ∼ f2 if (f1 − f2 )(s − 1) → 0 as s → 1 and N p = 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 = d1 .

33

Proof: We have M = {p ∈ Z | pOK = p1 ...pd , p0 s are completely split}. So using remark preceding theorem we get d

X p∈M

X 1 1 1 = = log s s (N p) (N p) s−1 degp=1

Hence substituting in density formula we get density of completely split  primes ( primes in M ) is d1 . 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 K1 K2 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 Q¯p land inside Qp . Thus if p splits completely in K1 and K2 , it does so in K1 K2 also. Let K1 be degree d1 extension of Q, K2 be degree d2 extension of Q and K1 K2 degree d extension of Q. Then, split primes in K1 are of density d11 , split primes in K2 are of density d12 and split primes in K1 K2 are of density 1 . Since the split primes in K1 , K2 and K1 K2 are the same, d11 = d12 = d1 . d This implies d1 = d2 = d, hence K1 = K2 = K1 K2 .  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 K1 K2 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 = p1 p2 p3 , 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 = p1 p2 in K} = {p ∈ Z|pOL = q1 q2 q3 in L} ≡ {p ⊂ OK | pOL = q1 q2 }. Hence density of M is 21 . 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 − 61 − 12 = 13 .

15

Decomposition of primes in the cyclotomic fields

Let K = Q(ζn ), where ζn = e

2πi 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. 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 Y m−1 m−1 m−1 1 + xp + (x2 )p + ..... + (xp−1 )p = (x − ζ) ζ

where product is over ζ , all pm th roots of unity. Putting x = 1 we get Y p= (1 − ζpi m ). i ∈ Z/pm (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 sends ζpm to ζpi m we get Y p= σ(π). σ∈Gal(K/Q) 1−ζpi m 1−ζpm

∈ Z[ζpm ] is a unit for (i, p) = 1. Hence we get p = Observe that m (1 − ζpm )φ(p ) .u where u is a unit in Z[ζpm ]. By taking p =< 1 − ζpm > we get m pOK = pφ(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 on Q(ζn ) by ζn −→ ζna . 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( mod n). 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 σp x ≡ xp mod p for all x ∈ Z[ζn ]. But σp (ζn ) = ζnp whereas order of σp is the degree of residue field extension of p which is Z[ζn ]/p. Now f

σpf = 1 ⇔ σpf (ζn ) = ζn ⇔ (ζn )p = 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 p−1   X 2πia a e p G= p a=1 36

√ It is easy to see that G2 = ( −1 )p = p hence G = ± p belongs to Q(ζp ). p 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 Let K be a number field which is contained in Q(ζn ). Let n = i pni i . Q Then Gal(Q(ζn )/Q) = (Z/n)∗ = i (Z/pni i )∗ . 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. b the set of all characters of G, forms a group. This group is called Then G, 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 b This correspondence is given by associating to a subgroup A of G, the set G. 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 χ0 : (Z/n0 )∗ −→ C∗ such that following diagram commutes. / (Z/n0 )∗ HH HH HH HH HH χ0 χ HH HH HH HH # 

(Z/n)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 |dK | =

Y

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. 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 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 χ b on Z by ( χ(a) if(a, n) = 1 χ b(a) = 0 otherwise Definition. Let χ be a Dirichlet character. Then L(s, χ) =

Xχ b(n) n≥1

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. s We recall that if ξ(s) = π − 2 Γ( 2s )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 ξK (s) = ( |dK |)s (π 2 Γ( ))r1 (π −s Γ(s))r2 ζK (s) 2 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 Gr¨ossencharacters. 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 b = {ξ : G −→ S1 }. Then G b is a locally compact topological dual of G by G group on which the topology is the compact open topology defined by taking b as W (C, U ) = {ξ ∈ G|ξ(C) b basic open sets of G ⊂ U },where C is a compact 1 set in G and U an open set in S . b is an anti equivaTheorem 19.1 (Pontryagin duality). The map G −→ G 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 R is of the form ψy : x 7−→ ψ(xy) for some y ∈ R. Let us define unitary b = {ξ : R −→ S1 | ξ unitary character of R}. Then the map dual of R by R 40

b One defines for y −→ ψy defines isomorphism of R with its unitary dual R. 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, b fb(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 x= aj (x)pj −∞ 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−

(N p)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(fb, 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 fb(n) f (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 fb(ξ) f (ξ) = ξ∈K

ξ∈K

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

P

ξ∈K

f (aξ) =

P

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 P that if f ∈ L1 (G) then F (x) = γ∈Γ f (x + γ) is a L1 -function on G/Γ R R and G/Γ F (x)dx = G f (x)dx. 55

We continue with the proof of the theorem. Write JK = J1K × R+ (we use (xt) variable). Nm 0 −→ J1K −→ JK −→ R+ Z ζ(f, χ, s) =

f (x)χ(x)|x|s d∗ x

ZJK

dt f (xt)χ(xt)|xt|s d∗ x t ZJK Z dt f (xt)χ(xt)|t|s dx = t R+ J1 Z ∞Z Z 1Z K dt dt s f (xt)χ(xt)|t| dx + f (xt)χ(xt)|t|s dx = t t 1 0 J1K J1K =

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 f (xt)χ(xt)ts dx. ζt (f, χ, s) = J1K

Then Z

"

ζt (f, χ, s) = J1K /K ∗

Z = J1K /K ∗

# X

f (xξt)χ(xξt) ts dx

ξ∈K ∗

"

# X ξ∈K ∗

56

f (xξt) χ(xt)ts dx

since χ is non-trivial, # ξ χ(xt)ts−1 dx ζt (f, χ, s) = fb 1 xt ∗ JK /K ξ∈K ∗   Z 1 = fb χ(xt)ts−1 dx xt J1 ZK   x = fb χ(x−1 t)ts−1 dx 1 t JK "

Z

X



= ζ 1 (fb, χ−1 , 1 − s) t

Thus, Z

1

1

Z

ζ 1 (fb, χ−1 , 1 − s) =

ζt (f, χ, s) = 0

0

t

Z

∞

ζt (fb, χ−1 , 1 − s)

1

Hence, Z

1

dt ζt (f, χ, s) + t

Z

∞

dt ζt (f, χ, s) t 1 Z0 ∞ Z ∞ dt dt = ζ 1 (fb, χ−1 , 1 − s) + ζt (f, χ, s) t t t 1 1

ζ(f, χ, s) =

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(fb, χ−1 , 1 − s).

57

YY

ζ(fp , χp , s) =

p∈S p∈S /

Y p∈S

ζ(fp , χp , s)

Y

YY

ζ(fbp , χ−1 p , 1 − s)

p∈S p∈S /

L(χp , s) =

Y

ζ(fbp , χ−1 p , 1 − s)

p∈S

p∈S /

=

L(χ−1 p , 1 − s)

p∈S /

Y ζ(fbp , χ−1 Y p , 1 − s) p∈S

=

Y

L(χ−1 p , 1 − s)

L(χ−1 p , 1 − s)

p∈S

Y ζ(fbp , χ−1 p , 1 − s) L(χ−1 , 1 −1 L(χp , 1 − s) p∈S

− s)

But using local functional equation (χp , s) =

ζ(fbp , χ−1 ζ(fp , χp , s) p , 1 − 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 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. P D = v nv v 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 nv v 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 (f1 f2 ) = (f1 ) + (f2 ). 3. Let f ∈ K. Then the degree deg((f)) = 0. For a divisor D one defines certain vector space of functions, S 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 P ξ|Ov .πvnv ≡ 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 πvnv 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. Proposition 27.1. The canonical map ρq : Dq −→ Gal following properties.



OK /q Ok /p



has the

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.   OK /q One knows from elementry field theory that Gal Ok /p is cyclic, generated 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 Ou and πv be that  of Ov . Then πu .Ov = πv .Ov and the decomposition group Ov /πv ∼ Du = Gal Ou /πu ∼ = Gal(K/k). 60

This defines the map r :

Y k∗ v −→ Gal(K/k) Ov∗

v ∈S /

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 / Gal(K1 /k) Jk F ∗

FF FF FF FF FF r2 FFF FF FF F#

restriction



Gal(K2 /k) Thus we have a map Jk −→ lim Gal(K/k) = Gal(k ab /k) ←

Theorem 27.2. Let k be a number field. Then, + ∼ ) = Gal(k ab /k) Jk /(k ∗ .k∞

61

i.e. finite abelian extensions of k are in bijective correspondence with open + denotes identity component of subgroups of finite index of Jk /k ∗ . Where k∞ ∗ 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 mod (πvnv )}. Then we take U = k ∗ .U (C) and the corresponding field extension 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 Ik (C) PC .NkK (IK (C))

Jk /(k ∗ .NkK JK ) ∼ =

where C is a cycle in k divisible by ramified primes such that Ov (Cv ) ⊂ NkKvw (Kw∗ ) (called admissible cycles) and Ik (C) is ideals in k which are coprime to C. Proof : We have JC =

Y

Y

Ov∗ (C)

v/C

kv∗

(v,C)=1

≈

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. T

≈

Jk /k ∗   y

−−−→

IC /kC   y Ik (C) PC .NkK (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 −→ F rv . 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 Y

qi

i=1 m

m/d

Then N m(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. (a) Order of the group

Ik (C) PC .NkK (IK (C))

and Gal(K/k) is the same.

(b) The map Ik (C) −→ Gal(K/k) is surjective map. (c) PC belongs to the kernel and this is done via recourse to cyclotomic theory. 2. Deduce reciprocity for general abelian extensions. Below we try to prove some of the steps. Lemma 27.1. The Artin map Ik (C) −→ Gal(K/k) is non-trivial. Proof : If Artin map were trivial then all of the unramified primes are completely split; i.e. ζK (s) = ζkm (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 = K H , 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, I (C) k PC .N K (IK (C)) ≤ [K : k] k Proof : Let H = PC .NkK (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 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) =

Y p

χ(p) 1− (N p)s

−1

This makes sense for Re(s) > 1 log(L(s, χ)) = −

X p

=



χ(p) log 1 − (N p)s



X χ(p) + higher order terms s (N p) p

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

log(L(s, χ)) ∼

p

g∈G

1 (N p)s

#

where p = g ∈ G = I(C)/H. " X χ

log(L(s, χ)) ∼

X

χ(g)

χ,g∈G

= |G|

X p∈H

65

X p

1 (N p)s

1 (N p)s

#

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

1 s−1

So we get h

1−

i 1 m(χ) log s−1 χ6=1

P

|G|

∼

X p∈H

1 (N p)s

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

1−

i 1 m(χ) log s−1 χ6=1

P

|G|

≥

1 [K : k] q

X of degree

1 (N p)s 1

1 ∼ ζK (s) [K : k] 1 1 ∼ log [K : k] s−1 which implies |G| m(χ) ≥ [K : k] I(C) Ik (C) [K : k] ≥ |G| = | P N | i.e. order of PC .N K (IK (C)) ≤ [K : k]. Further we get k m(χ) = 0 ∀χ 6= 1.  1≥1−

X

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 ∼ log s (N p) [I(C) : PC ] s−1 p∈a 0

where a0 ∈ I(C)/PC . 66

Proof : We have X χ(p) (N p)s p X X = χ(a)

log(L(s, χ)) ∼

a

p∈a

1 (N p)s

Then, X

χ(a−1 0 ) log(L(s, χ)) =

XX

χ

χ

χ(aa−1 0 )

a

X p∈a

= [I(C) : PC ]

X p∈a0

1 (N p)s

1 (N p)s

Hence, log ζK (s) = [I(C) : PC ]

X p∈a0

log

1 (N p)s

X 1 1 = [I(C) : PC ] s−1 (N p)s p∈a 0

 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, H 0 (G, A) = AG , the zeroth cohomology AG c0 (G, A) = , the zeroth Tate cohomology H (1 + σ + σ 2 + ... + σ n−1 )A ker(1 + σ + σ 2 + ... + σ n−1 ) : A −→ A 1 H (G, A) = , the first cohomology (1 − σ)A

Definition. The Euler characteristic of A is defined by, χ(A) =

c0 (G, A)) #(H #(H 1 (G, A))

c0 (G, A) and (H 1 (G, A) are if orders make sense, i.e. the orders of both H 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. c0 (Z/n, Z) = Z/n and H 1 (Z/n, Z) = 0. 3. χ(Z) = n as H 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 : N mUK ] = e where e is the ramification index. Proof : 68

1. Since χ(K ∗ ) =

c0 #H , #H 1

by Hilbert Theorem 90, H 1 = 0 (for any cyclic c0 = [k ∗ : N mK ∗ ]. extensions). Hence χ(K ∗ ) = #H

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 H i (H, B) = H i (G, IndG H B). We get χ(Ok ) = χ{e} (Ok ) = 1. 3. Let us look at the following diagram. 0 −−−→ k ∗ −−−→ x  

x7→

x

K ∗ −−−σx → x  

K 1 −−−→ 1 x  

0 −−−→ Uk −−−→ UK −−−→ UK1 −−−→ 1 where K 1 is set of norm 1 elements in K ∗ . Z/eZ ∼ = K ∗ /k ∗ UK ∼ = K 1 /UK1 . χ(UK ) = 1 =

c0 (U ) H K H 1 (UK )

=

Which implies that

[Uk :N mUK ] x 1 :{ [UK }] σx|x∈U K

Observe that = K thus [Uk : N mUK ] = e and [k ∗ : N mK ∗ ] = [K : k] for cyclic extensions of local fields. UK1

1

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 N Gal(K/k). Let M be any Gal(K/k) -invariant lattice in E R. Theorem 29.2. There exists a lattice M 0 ⊂ 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, χ(M ) =

c0 (G, M ) Y #H 0 = χ(M ) = Nv #H 1 (G, M ) v∈S

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

70

Proof : Since M 0 =


IndG Gv Z hence, Y χ(M 0 ) = χ(IndG Gv Z)

L

v

v

=

Y

=

Y

χ(Gv0 Z)

v

Nv 

v

Corollary 29.2. Let KS be the set of S units in K, i.e. w(x) = 0∀w ∈ / SK then Q Nv χ(KS ) = [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 Y χ(KS )χ(Z) = χ(M ) = χ(M 0 ) = Nv Q

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

c0 (Gal, JK ) #H #H 1 (Gal, JK ) c0 (Gal, JK ) (Hilbert 90) = #H

χ(JK ) =

= [Jk : N mJK ] So χ(JK /K ∗ ) = [Jk : k ∗ N mJK ]. Hence the proof of [Jk : k ∗ N mJK ] = [K : k]. 71

JK K∗

=

Q

Q Ov∗ . v∈S kv∗ . KS

This completes

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