PART I

ALGEBRAIC NUMBER THEORY

3

1. Introduction

Let Q ={

ab ; a, b ∈ Z, b 6= 0

}.

Then we can regard Z as

Z = {α ∈ Q; f(α) = 0 for some f(X) = X + b ∈ Z[X]} .

We can associate with Z the Riemann zeta function

ζ(s) =∞∑

n=1

1ns

=∑a6=0

1Nas

where each nonzero ideal a of Z is of the form a = nZ and Na = |n| = #(Z/nZ).

The above series converges for Re(s) > 1, and ζ(s) can be meromorphically contin-

ued to a function which is analytic everywhere except for a simple pole at s = 1,

where it has residue 1, so ζ(s) = 1s−1 + g(s), where g is an entire function of s. The

Riemann zeta function also has a functional equation

π−s2 Γ(s

2

)ζ(s) = h(s) = h(1− s)

and an Euler product

ζ(s) =∏p

(1− p−s

)−1.

By studying ζ(s) as a function of s, one obtains information about primes and

integers, e.g. the existence of infinitely many primes.

Example Let

g(s) =∑

m square-free

1ms

=∏p

(1 +

1ps

)=∏p

1− p−2s

1− p−s=

ζ(s)ζ(2s)

Then

lims→1

(s− 1)g(s) =1π2

6

=6π2

so ≈ 23 of the integers are square-free.

Now let r(n) denote number of ways in which n can be written as a sum of two

squares.

4

Then

r(n) = 4[#divisors of nwhich are ≡ 1 (mod 4)]−[#divisors of nwhich are ≡ 3 (mod 4)]

Define

χ(n) =

1 if n ≡ 1 mod 4−1 if n ≡ 3 mod 40 if n is even.

which is a character mod 4.

Then r(n) = 4∑d|n

χ(d), so

∞∑n=1

r(n)ns

= 4∞∑

n=1

∑d|n

χ(d)

n−s = 4

( ∞∑m=1

m−s

)( ∞∑d=1

χ(d)d−s

)= 4ζ(s)L(s, χ)

And now consider the field Q(i). The solutions of the equation with integer coeffi-

cients

x2 + bx + c = 0

in this field are

x =−b±

√b2 − 4c

2

if b2 − 4c = −m2. If b is odd, the LHS is ≡ 1 mod 4, but the RHS is ≡ 0, 1 mod 4.

Therefore, b has to be even, b = 2b′, and so x = −b′ + m′i.

Hence the ring of integers is O = {a + bi; a, b ∈ Z}. The ideals of this ring are

all of the form a = (a + bi) and Na = #(O/a) = a2 + b2.

Thus

ζQ(i)(s) =∑a6=0

1Nas

=14

∑(a,b) 6=(0,0)

1(a2 + b2)s

=14

∞∑n=1

r(n)ns

= ζ(s)L(s, χ)

and

lims→1

(s− 1)ζQ(i)(s) = L(1, χ) = 1− 13

+15− 1

7+ · · · = π

4

which can be written as

(2π) · 14√

4.

5

For a general imaginary quadratic number field K = Q(√−D), we have a class

number hK and a discriminant dK , and, for D > 3,

Ress=1 ζK(s) =πhk√|dK |

.

As we will see, for an arbitrary number field K, Ress=1 ζK(s) encodes important

arithmetic information about the field K.

2. Facts about fields - review

We assume a basic familiarity with Galois theory, so here we are only reviewing

the main facts that we will need later on.

Let A ⊆ C be an integral domain and let k be its quotient field.

For α ∈ C let B denote the set of elements of k(α) satisfying some monic poly-

nomial with coefficients in A. Then B ⊆ k(α) and B is the integral closure of A in

k(α).

If f ∈ k[X] is an irreducible polynomial over k and f(α) = 0, then k(α) is a

vector space over k of dimension n = deg(f) and {1, α, α2, . . . , αn−1} is a k-basis

of k(α).

If α′ is some other root of the polynomial f , then k(α) ∼= k(α′) via the map that

sends α 7→ α′ and leaves k fixed. In this case α and α′ are called conjugates over k

and k(α), k(α′) are called conjugate fields.

If β =∑

ajαj , a conjugate of β is β′ =

∑ajα

′j .

Theorem 1. (Primitive Element) Suppose the complex numbers α and β are al-

gebraic over k. Then there is an element θ ∈ C, algebraic over k, such that

k(α, β) = k(θ)

Definition 1. An algebraic extension K = k(θ) of k is normal over k if all the

conjugates of θ are in K.

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Theorem 2. An algebraic extension K/k is normal iff for every α ∈ K all the

conjugates of α are also in K.

Theorem 3. Let α1, . . . , αn be the zeros of the polynomial f ∈ k[X] and β1, . . . , βm

be the roots of the polynomial g ∈ k[X]. Then k(α1, . . . , αn, β1, . . . , βm) is normal

over k.

Definition 2. Let K/k be a Galois extension of fields of Galois group G. We say

that an intermediate subfield L of this extension, k ⊆ L ⊆ K is fixed by an element

g ∈ G if g fixes every element of L.We say that L is fixed by H ⊆ G if L is fixed

by every element of H.

The field consisting of all the elements of K that are fixed by H is called the

fixed field corresponding to H and is denoted by KH .

Theorem 4. (Fundamental Theorem of Galois Theory) Suppose K is a normal

algebraic extension of k. Then there is a 1 − 1 correspondence between the inter-

mediate fields k ⊆ L ⊆ K and the subgroups H ⊆ G, given in one direction by

L 7→ Gal(K/L) and in the other direction by H 7→ KH .

Corollary 5. [K : KH ] = #H and [KH : K] =#G

#H.

Definition 3. If L1 and L2 are two subfields of k(θ) containing k, their composite

is the smallest subfield of k(θ) containing both of them and it is denoted by L1L2.

Note that, if Lj = k(αj), j = 1, 2, then L1L2 = k(α1, α2).

Also, if H1, H2 are subgroups of a group G, then will denote by < H1,H2 > the

smallest subgroup of G containing both H1 and H2.

For the foreseeable future, we will consider K/k to be a Galois extension with

Galois group G. (Recall that a Galois extension is by definition normal.)

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Theorem 6. Suppose L1 and L2 are two intermediate fields of the extension K/k

and Hj = Gal(K/Lj), j = 1, 2. Then L1 ⊇ L2 iff H1 ⊆ H2.Also, Gal(K/L1L2) =

H1 ∩H2 and Gal(K/L1 ∩ L2) =< H1,H2 >.

Theorem 7. Suppose k(α) ⊆ K and let α(1), . . . , α(n) denote the algebraic con-

jugates of α, where n = [k(α) : k]. Then the isomorphism k(α) → k(α(i))

can be extended to an automorphism of K. If g ∈ G maps g(α) = α(i), and

H = Gal(K/k(α)), then the [K : k(α)] elements of G that send α 7→ α(i) are

exactly the elements in gH. Finally, Gal(K/k(α(i))) = gHg−1.

Theorem 8. Suppose L is an intermediate field of the extension K/k and H =

Gal(K/L). Then L/k is normal iff H is a normal subgroup of G and in this case

Gal(L/k) = G/H.

Example Let K = Q(√

2,√

3). Then [K : Q] = 4 and K is a normal extension

of Q. Let A =√

2, B = −√

2, C =√

3, and D = −√

3. Let (A,B) denote

the automorphism of K/Q that sends A 7→ B and leaves C and D fixed and let

(C,D) denote the automorphism of K/Q that sends C 7→ D and leaves A and B

fixed. Then G = {1, (A,B), (C,D), (A,B)(C,D)}. The subgroups of G are (1),

H1 = {1, (A,B)}, H2 = {1, (C,D)}, H3 = {1, (A,B)(C,D)}, and G itself.

Then KH1 = Q(√

3), KH2 = Q(√

2), and KH3 = Q(√

6).

We can also write K as K = Q(√

2 +√

3). Let a =√

2 +√

3. The conjugates of

a are a itself, b = −√

2+√

3, c =√

2−√

3, and d = −√

2−√

3. The automorphism

of K that sendsa 7→ b has also the effect of sending c 7→ da 7→ c has also the effect of sending b 7→ da 7→ d has also the effect of sending b 7→ c.

So G = {1, (a, b)(c, d), (a, c)(b, d), (a, d)(b, c)} and

K(a,b)(c,d) = Q(√

3)K(a,c)(b,d) = Q(

√2)

K(a,d)(b,c) = Q(√

6).

8

3. The Ideal Class Group

Let K be an algebraic number field, i.e. a finite extension of Q, of degree n

and OK its ring of integers. To avoid confusion, we will call the elements of Z

rational integers. Note that OK ∩Q = Z. Indeed, if rs is a rational number written

in lowest terms that is a root of some equation with rational integers coefficients

xm + am−1xm−1 + . . . + a0 = 0, then rm + am−1r

m−1s + . . . + a0sm = 0, hence

s | rm. But since gcd(r, s) = 1, this implies that s = ±1, hence rs ∈ Z.

Also note that if amαm + am−1αm−1 + . . . + a0 = 0, aj ∈ C, 0 ≤ j ≤ m, and

α ∈ C, then amα satisfies the monic equation xm + am−1xm−1 + . . . + am−1

m a0 = 0,

so amα is integral over the ring Z[a0, . . . , am].

Lemma 9. Suppose f(X) is a polynomial with algebraic integer coefficients and u

is a root of f . Thenf(X)X − u

is also a polynomial with algebraic integer coefficients.

Proof: We prove this by induction over m = deg f . The polynomial f is of the

form f(X) = θmXm + θm−1Xm−1 + . . . + θ0 with θ0, . . . , θm are algebraic integers.

Therefore Z[θ0, . . . , θm] is integral over Z.

If m = 1, then f(X) = θ1(X − u). Hencef(X)X − u

= θ1, which is an algebraic

integer.

Now assume the lemma holds for all polynomials of degree at most m − 1.

We want to show that it holds for the polynomial f . As we have seen, θmu is

integral over Z[θ0, . . . , θm], which, in turn, is integral over Z. Thus θmu is an

algebraic integer and the polynomial g(X) = f(X)−θm(X−u)Xm−1 has algebraic

integer coefficients. Also, deg g ≤ m − 1 and g(u) = 0. So we can apply the

induction hypothesis to g and, therefore,g(X)X − u

has algebraic integer coefficients.

Butf(X)X − u

=g(X)X − u

+ θmXm−1, so the lemma also holds for f .

Lemma 10. Suppose f(X) = θm(X − u1) . . . (X − um) has algebraic integer coef-

ficients. Then θmu1 . . . uj is an algebraic integer for each j, 1 ≤ j ≤ m.

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Proof: If j = m, the product is actually equal, up to sign, to the constant term of f ,

and therefore it is an algebraic integer. For j < m, the product is equal, up to sign,

to the constant term of the polynomial g(X) =f(X)

(X − uj+1) . . . (X − um). By the

previous lemma, g has algebraic integer coefficients, so θmu1 . . . uj is an algebraic

integer.

Definition 4. If α and β are two elements of some algebraic number field, we say

that α divides β and write α | β if there is an element a of the ring of integers of

the field, such that β = aα.

Theorem 11. Let f(X) =m∑

i=0

αiXi and g(X) =

r∑j=0

βjXj be two polynomials

with algebraic integer coefficients and let h = fg. Suppose that δ 6= 0 divides all the

coefficients of h. Then δ | αiβj for all i and j.

Proof: Let f(X) = αm(X − u1) . . . (X − um) and g(X) = βr(X − v1) . . . (X − vr).

Thenh(X)

δ=

αmβr

δ(X−u1) . . . (X−um)(X−v1) . . . (X−vr) has algebraic integer

coefficients. So δ | αmβr. By the previous lemma,

αmβr

δ· (any product of the ui’s) · (any product of the vj ’s)

is an algebraic integer. Now, by the Viete relations for all 0 ≤ i ≤ m and all

0 ≤ j ≤ r,αi

αmis a sum of products of the ui’s and

βj

βris a sum of products

of the vj ’s. Thus,αiβj

δ=

αmβr

δ

αi

αm

βj

βris equal to a sum of terms of the form

αmβr

δ· (product of some ui’s) · (product of some vj ’s) and, therefore, an algebraic

integer.

Corollary 12. (Gauss) Suppose f, g ∈ Q[X] and fg has integer coefficients. Then

there exists an r ∈ Q s.t. rf(X) and 1r g(X) have integer coefficients.

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Proof: Let h = fg. Choose r ∈ Q s.t. rf has relatively prime integer coefficients.

Denote F (X) = rf(X) =m∑

i=0

aiXi ∈ Z[X] and G(X) =

1rg(X) ∈ Q[X]. Let d ∈ Z,

d 6= 0 for which dG(X) =r∑

j=0

bjXj has relatively prime integer coefficients.

Then d divides all the coefficients of the polynomial dh(X) = F (X) ·dG(X). By

theorem 11 it follows that d | aibj for all i, j. So, for any j, d | gcd(a0bj , a1bj , . . . , ambj) =

bj , since the ai are relatively prime by construction. But the bj ’s are also relatively

prime, so d = ±1. Thus, G has integer coefficients.

Corollary 13. A monic irreducible polynomial in Q[X] that has some algebraic

integer for a root has rational integer coefficients.

Proof: Let f(X) be such a polynomial and let α be its algebraic integer root. Then

there exists some monic polynomial h ∈ Z[X] such that h(α) = 0. Since f is

irreducible, h = fg for some g ∈ Q[X]. By Gauss’s corollary there exists r ∈ Q s.t.

rf(X), 1r g(X) ∈ Z[X] and h(X) = rf(X) 1

r g(X). Since the h is monic, it follows

that the leading coefficients of rf(X)and 1r g(X) are ±1. But, since f is monic it

follows that r = ±1, so f ∈ Z[X].

Recall that K is an algebraic number field of degree n.

Definition 5. If α1, . . . , αm ∈ K, denote by [α1, . . . , αm] =

{m∑

i=1

aiαi; a1, . . . , am ∈ Z

}the Z-submodule of K generated by α1, . . . , αm.

Theorem 14. Any Z-submodule a = [α1, . . . , αm] of K has a Z-basis with at most

n = [K : Q] elements.

Proof: Since a = [α1, . . . , αm] is a subset of the field K, it is a finitely generated

torsion-free abelian group, and thus a free Z-module of finite rank. Therefore it

suffices to show that a can be generated by n elements.

11

Let β1, . . . , βn be a Q-basis of K. Then each αi can be expressed as αi =n∑

j=1

aijβj

with aij ∈ Q. Then any element α of a is of the form

α =m∑

i=1

biαi =n∑

j=1

(m∑

i=1

biaij

)βj =

n∑j=1

cjβj

for some b1, . . . , bm ∈ Z and where cj =m∑

i=1

biaij , for all 1 ≤ j ≤ n.

Let c1,n = gcd(a1n, . . . , amn). Then we can write c1,n as a linear combination of

the ain.

Choose b1,1, . . . , b1,m ∈ Z such that c1,n =m∑

i=1

b1,iain is equal to gcd(a1n, . . . , amn).

Let c1,j =m∑

i=1

b1,iaij , 1 ≤ j ≤ n− 1. Then for each α ∈ a there exists a d ∈ Z such

thatm∑

i=1

biain = d c1,n, where α =m∑

i=1

biαi.

So, α− d

n∑j=1

c1,jβj =n−1∑j=1

ejβj , for some e1, . . . en−1 ∈ Q.

Hence, for each 1 ≤ i ≤ m, there exist di ∈ Z and dij ∈ Q, 1 ≤ j ≤ n − 1 such

that

αi − di

n∑j=1

c1,jβj =n−1∑j=1

dijβj .

Therefore

a = [α1, . . . , αm] =

α1, . . . , αm,n∑

j=1

c1,jβj

=

n−1∑j=1

d1jβj , . . . ,n−1∑j=1

dmjβj ,n∑

j=1

c1,jβj

Now choose b2,1, . . . , b2,m ∈ Z such that

m∑i=1

b2,idi n−1 = gcd(d1 n−1, . . . , dm n−1).

Repeating the procedure, we get

a =

n−2∑j=1

d1jβj , . . . ,n−2∑j=1

dmjβj ,n∑

j=1

c1,jβj ,n−1∑j=1

c2,jβj

.

Keep going and finally obtain that

12

a =

n∑j=1

c1,jβj ,n−1∑j=1

c2,jβj , . . . , cn,nβ1

,

so a can be generated by n elements.

Definition 6. For any finitely generated Z-submodule a of K, the dimension of a

is the number of elements in a Z-basis of a, and it will be denoted by dim a.

Definition 7. If a = [α1, . . . , αr] and b = [β1, . . . , βt] are two Z-submodules of K,

then their product is ab = {αβ;α ∈ a, β ∈ b} = [αiβj ; 1 ≤ i ≤ r, 1 ≤ j ≤ t]

Clearly this definition is independent of the choice of the generators of the two

modules.

By the primitive root theorem, we can find θ ∈ K such that K = Q(θ). Then θ

has n conjugates θ(1), . . . , θ(n), which uniquely determine the n embeddings of K

into Q.

K(i) for 1 ≤ i ≤ n is defined by K(i) = {α(i) : α ∈ K. The K(i) are the conjugate

fields of K.

Definition 8. The discriminant of n elements α1, . . . , αn of K is

D(α1, . . . , αn) =

det

α

(1)1 . . . α

(1)n

.... . .

...α

(n)1 . . . α

(n)n

2

Note that if α1, . . . , αn are linearly dependent over Q, then D(α1, . . . , αn) = 0.

Theorem 15. D(α1, . . . , αn) ∈ Q, and, if α1, . . . , αn ∈ OK , then D(α1, . . . , αn) ∈ Z.

Proof: Let K = Q(θ(1), . . . , θ(n)

), where we recall that K = Q(θ). Then K/Q is a

Galois extension and the effect of the action of the elements of Gal(K/Q) on the

matrix that defines the discriminant is to permute it rows, thus leaving the det2

unchanged. Hence D(α1, . . . , αn) ∈ Q. If all the αj ’s are algebraic integers, then

the discriminant is at the same time an algebraic integer and a rational number.

Therefore it has no choice but to land in Z.

13

Since K = Q(θ), a field basis for K is {1, θ, . . . , θn−1} and

D(1, θ, . . . , θn−1) = det

1 θ(1) . . . θ(1)n−1

.... . .

...1 θ(n) . . . θ(n)n−1

2

=∏

1≤i<j≤n

(θ(i) − θ(j)

)2

6= 0.

Assume that α1, . . . , αn, β1, . . . , βn ∈ K and (β1, . . . , βn) = (α1, . . . , αn)M for

some matrix M ∈ Matn×n(Q). Then for any i, 1 ≤ i ≤ n, applying the ith

embedding of K into C, we get that (β(i)1 , . . . , β

(i)n ) = (α(i)

1 , . . . , α(i)n )M . Hence

β

(1)1 . . . β

(1)n

.... . .

...β

(n)1 . . . β

(n)n

=

α

(1)1 . . . α

(1)n

.... . .

...α

(n)1 . . . α

(n)n

M,

so D(β1, . . . , βn) = D(α1, . . . , αn) (det M)2.

Since K has a field basis whose discriminant is nonzero, it follows that the dis-

criminant of any field basis is nonzero (two field basis differ by an invertible matrix).

On the other hand, if α1, . . . , αn ∈ K, then (α1, . . . , αn) = (1, θ, . . . , θn−1)M for

some n × n matrix M with rational entries. So, if D(α1, . . . , αn) 6= 0, it follows

that detM 6= 0, hence {α1, . . . , αn} is a field basis for K.

Note that by taking a field basis and clearing denominators, one can construct

a field basis consisting entirely of elements of OK and the discriminant of such a

basis is a nonzero integer.

Theorem 16. Let a be an additive subgroup of OK containing n linearly indepen-

dent elements (linear independence over Z is the same as over Q). Then a is an

n-dimensional Z-module.

Proof: Let α1, . . . , αn be the n linearly independent elements of a. They form a

field basis and |D(α1, . . . , αn)| ∈ Z+.

Suppose [α1, . . . , αn] 6= a. Then there is an element αn+1 ∈ a \ [α1, . . . , αn]. By

theorem 14, [α1, . . . , αn+1] has a Z-basis with n elements β1, . . . , βn. Then

14

(α1, . . . , αn) = (β1, . . . , βn)M , for some n × n matrix M with integral entries

and det M 6= 0. Also, |detM | 6= 1, since otherwise αn+1 would be an element

of [α1, . . . , αn]. Thus 0 < |D(β1, . . . , βn)| < |D(α1, . . . , αn)| and they are both

integers. Keep repeating the process. And some point it has to stop, and when it

does we have our n-dimensional basis.

Corollary 17. OK itself is an n-dimensional module.

Proof: It contains a field basis.

Definition 9. A Z-basis for OK is called an integral basis of K.

Note that if {α1, . . . , αn} and {β1, . . . , βn} are two integral bases of K, then

(α1, . . . , αn) = (β1, . . . , βn)M for some invertible matrix M ∈ GL(n, Z). So M ∈

GL(n, Z) and M−1 ∈ GL(n, Z); therefore det M = ±1. Thus D(α1, . . . , αn) =

D(β1, . . . , βn). So we can define

Definition 10. The discriminant of the field K, DK , is the discriminant of some

integral basis of K.

Let {α1, . . . , αn} be an integral basis of K and a = [β1, . . . , βn] be an n-dimensional

subgroup of OK . Then (β1, . . . , βn) = (α1, . . . , αn)M for some M ∈ Matn×n(Z).

Claim |det M | = #OK/a.

To see this, use the Elementary Divisors Theorem to find a Z-basis for a of the

form d1α1, . . . , dnαn with d1, . . . , dn nonnegative integers. Then (d1α1, . . . , dnαn) =

(β1, . . . , βn)M ′ for some M ′ ∈ GL(n, Z). So (d1α1, . . . , dnαn) = (α1, . . . , αn)MM ′.

Then [OK : a] = d1 . . . dn = det M detM ′. But det M ′ = ±1 and d1 . . . dn ≥ 0,

hence [OK : a] = |detM |.

If a is a Z-submodule of K, then it is clear that aOK ⊇ a.

15

Definition 11. A Z-submodule a of K is called a fractional ideal of K if aOK = a.

If in addition a ⊆ OK , then a is called an integral ideal.

Clearly any fractional ideal can be written as 1da, where a is an integral ideal

and d ∈ OK . Also, the index of any integral ideal in OK is finite.

Definition 12. If α1, . . . , αm are elements of K, not all zero, then the ideal gen-

erated by α1, . . . , αm is (α1, . . . , αm) = [α1, . . . , αm]OK . The product of two ideals

a = (α1, . . . , αm) and b = (β1, . . . , βr) is the ideal ab = (αiβj ; 1 ≤ i ≤ m, 1 ≤ j ≤

r).

Principal ideals are of the form (α), with α 6= 0.

Definition 13. If every ideal of OK is principal, then OK is called a principal

ideal domain.

Theorem 18. Given any fractional ideal a of K there exists another fractional

ideal b such that ab = OK .

Proof: Let a = (α0, . . . , αr) and consider the polynomial f(X) =r∑

j=0

αjXj . Let

f (i)(X) =r∑

j=0

α(i)j Xj and set

h(X) = f (1)(X) . . . f (r)(X) and g(X) =h(X)f(X)

=t∑

l=0

βlXl

Let K be the normal closure of K over Q. Then the coefficients of h are fixed

by all the elements of Gal(K/Q), and thus h ∈ Q[X] and the coefficients of g are

fixed by all the elements of Gal(K/K), so g ∈ K[X]. Let N ∈ Q denote the gcd of

the coefficients of h(X).

Claim: (α0, . . . , αr)(β0, . . . βt) = (N)

Suppose the claim holds. If b =(

β0

N, . . . ,

βt

N

), we have ab = (1) = OK and the

theorem is proved.

To prove the claim, apply theorem 11 to N and h(X). It follows that N | αjβl

for all j’s and l’s. So (N) ⊇ (α0, . . . , αr)(β0, . . . βt). On the other hand, the

16

coefficients of h are Z-linear combinations of the αjβl’s, so they are contained

in (α0, . . . , αr)(β0, . . . βt). But N , being the gcd of the coefficients of h, can be

written as a Z-linear combination of these coefficients, so it is also contained in

(α0, . . . , αr)(β0, . . . βt).

Corollary 19. The fractional ideals of K form an abelian group with OK as iden-

tity.

Definition 14. We say that an ideal a divides an ideal b, and write a | b, if there

exists an integral ideal c such that b = ac.

Note that a | b ⇔ a−1b ⊆ OK .

Theorem 20. Let a and b be two ideals of K. Then a | b iff a ⊇ b.

Proof: a | b ⇔ a−1b ⊆ OK ⇔ b ⊆ aOK = a.

Definition 15. If a = (α1, . . . , αr) and b = (β1, . . . , βt) are two ideals of K, then

their greatest common divisor is (a, b) = (α1, . . . , αr, β1, . . . , βt), the smallest ideal

containing both a and b.

Theorem 21. Let d = (a, b). Then d | a and d | b and, if c | a and c | b, then c | d.

Proof: By definition d ⊃ a and d ⊃ b. Now if c | a and c | b, then c ⊃ a and c ⊃ b,

so c ⊃ d, that is c | (α1, . . . , αr, β1, . . . , βt) = d.

Definition 16. A prime ideal p is an integral ideal such that p 6= OK and with the

property that if an integral ideal a divides p, then either a = p, or a = OK

Theorem 22. If p is a prime ideal of K and a, b are two integral ideals with p | ab,

then p | a or p | b.

17

Proof: Since (p, a) is an integral ideal that divides p, it follows that (p, a) = p or

(p, a) = OK .

In the first case, p | a. In the second case b = bOK = b(p, a) = (bp, ab). Clearly

p | bp, and p | ab by hypothesis. So by Theorem 21, p | (bp, ab) = b.

Exercise

(1) Let K/k be a degree n field extension and assume that Ok is a PID. Show

that every ideal of K has a basis of n elements over Ok.

(2) Let K = Q(√−5). Then OK = [1,

√−5]. Which, if any, of the following

three Z-modules are ideals?

• [19 + 7√−5, 43 + 16

√−5]

• [15 + 14√−5, 34 + 32

√−5]

• [−31 + 11√−5,−71 + 25

√−5]

Corollary 23. If p is a prime ideal and α, β ∈ OK with αβ ∈ p, then α ∈ p or

β ∈ p.

Proof: Let a = (α) and b = (β). They are both integral ideals and p | ab. Since p

is prime, it follow that p | a or p | b, so α ∈ p or β ∈ p.

Theorem 24. Every nonzero fractional ideal of K factors uniquely into a product

of the formr∏

i=1

paii with ai ∈ Z. For integral ideals all the ai’s are nonnegative

integers.

Proof: Assume throughout that a is nonzero. It suffices to prove the assertion for

integral ideals. To see this, assume that a is a fractional ideal. Then there is a

d ∈ OK such that da is an integral ideal, b. Both (d) and b factor uniquely into

primes, and therefore so does a =b

(d).

Now consider an integral ideal a. First we are going to prove that a factors as

a product of primes. If a is prime or a = OK , we are done. But suppose a can be

written as the product of two integral ideals: a = b1b2. Both bi’s strictly contain

18

a, and so are of finite index smaller than [OK : a]. The process has to stop at some

point and thus we get a prime factorization of a.

Assume that two such factorizations exist:

a =r∏

i=1

paii =

t∏j=1

qbj

j .

Then for each i there exist a j such that pi | qj . Since they are both nonzero

prime ideals of K, they are maximal ideals of OK and so they must coincide. We

get r = t and, by eventually reordering the factors of the product, pi = qi for all i.

In the same fashion we get that ai = bi for all i.

Theorem 25. The integers of K, OK , form a UFD iff every ideal of K is principal.

Proof: If every ideal of K is principal, then OK is a PID and therefore a UFD.

Conversely, assume that OK is has unique factorization. Since every ideal of K

factors uniquely into product of prime ideals and their inverses, it suffices to show

that every prime ideal is principal. First let π be an arbitrary prime element of OK

and p = (π) the ideal generated by it. For the sake of contradiction, assume that

p is not a prime ideal. Then there exist integral ideals a and b such that p | ab,

but p - a and p - b, i.e. p 6⊃ a and p 6⊃ b. Choose α ∈ a \ p and β ∈ b \ p. Then

αβ ∈ p = (π), so π | αβ. Since π is prime it follows that π | α or π | β. But this

implies α ∈ p or β ∈ p, which contradicts the choice of α and β. Therefore (π) is a

prime ideal.

Now let p be any prime ideal of K. Choose α ∈ p, α 6= 0. Then p | (α). Since

OK is a UFD, α has a unique factorization into primes, α = π1 . . . πr. Therefore

p | (π1) . . . (πr), and, p being prime, this implies that p | (πj) for some 1 ≤ j ≤ r.

By the first part of the argument (πj) is a also prime ideal, so p = (πj).

Definition 17. The ideal class group of K, ClK , is the quotient group of the group

of fractional ideals of K by the group of principal ideals of K. Its order, hK is called

the class number of K.

19

Theorem 26. Given two integral ideals a and b, there exists an integral ideal c

such that ac is principal and (b, c) = (1).

Proof: Let ab =r∏

i=1

plii , where li > 0, for every i, 1 ≤ i ≤ r. Write a =

r∏i=1

paii , with

0 ≤ ai ≤ li for each i.

For each j, 1 ≤ j ≤ r, define

dj =∏i 6=j

pai+1i .

So for each pi | ab, pi | dj iff i 6= j and for i 6= j we have pai+1i | dj .

Now (d1, . . . , dj) = (1), so there are γj ∈ dj such that 1 =r∑

j=1

γj . Choose

αj ∈ paj

j \ paj+1j , so p

aj

j | (αj), but paj+1j - (αj). Define w =

r∑j=1

αjγj . For each

j, we have that paj

j | (αjγj), but paj+1j - (αjγj) and p

aj+1j | (αiγi) for all i 6= j.

This is so because first we have γj ∈ dj , hence di | (γi) and so paj+1j | (αiγi) for

all i 6= j. But if pj | (γj) for any j, then pj would divide all the γ’s, and therefore

would divide their sum, which is 1. This is clearly impossible.

Combining all this we get that for each j, paj

j | (w), but paj+1j - (w), so a | (w).

Let c be the integral ideal for which (w) = ac. Then (ac, ab) =

((w),

r∏i=1

plii

)=

r∏i=1

paii = a, and thus (b, c) = (1).

Theorem 27. Every ideal of K can be generated by two elements.

Proof: It suffices to prove the assertion for integral ideals. So, let a be an integral

ideal. Then a−1 = 1αb for some α ∈ OK and b an integral ideal. Then ab = (α).

By theorem 26 there exists an integral ideal c such that ac = (w) and (b, c) = (1).

Hence (ab, ac) = a, so (α, w) = a

Let a be an integral ideal of K. Then OK/a is a finite ring and we have the

canonical projection OK → OK/a given by α 7→ α. If p happens to be a prime

ideal of K, then OK/p is a finite integral domain, and thus a field. Let p be the

20

characteristic of the finite field OK/p. Then p ≡ 0 (mod p), so p contains a rational

prime, namely p, i.e. there exist a rational prime p such that p | (p) = pOK

The canonical projectionOK → OK/p induces a canonical mapOK [X1, . . . , Xs] →

OK/p[X1, . . . , Xs] given by f(X1, . . . , Xs) 7→ f(X1, . . . , Xs).

Suppose that f, g ∈ OK [X1, . . . , Xs] and that every coefficient of fg is in p. Then

fg = 0 in OK/p[X1, . . . , Xs], so f = 0 or g = 0, i.e. every coefficient of f is in p or

every coefficient of g is in p.

Definition 18. Let f ∈ K[X1, . . . , Xs]. The ideal I(f) of K generated by the

coefficients of f is called the content of f .

Theorem 28. If f, g ∈ K[X1, . . . , Xs], then I(fg) = I(f)I(g).

Proof: Let X = (X1, . . . , Xs). Then f(X) =∑

aiXi and g(X) =

∑bjX

j where i

and j are multi-indices. Without loss of generality, we may assume that f and g

have integral coefficients.

I(f)I(g) ={∑

αiai

∑βjbj ;αi, βj ∈ OK

}⊇{∑

γijaibj ; γij ∈ OK

}⊇ I(fg)

Let a = I(f). This is an integral ideal, and there exists another integral ideal b

such that ab = (α), a principal integral ideal. Also, by theorem 26 there exists

an integral ideal c such that bc = (α′) principal ideal and (c, a) = (1). For any

coefficient a of f we have a ⊇ (a), so a | (a), and(

α′

αa

)=

bc

ab(a) = c

(a)a⊆ OK .

Thereforeα′

αa ∈ OK .

It follows thatα′

αf(X) ∈ OK [X] and

(I

(α′

αf

), I(f)

)=((

α′

α

)a, a

)=(

bc

aba, a

)= (c, a) = (1)

Similarly, there exist β, β′ ∈ OK such thatβ′

βg(X) ∈ OK [X] and(

I

(β′

βg

), I(g)

)= (1).

21

Fix p a prime ideal dividing I(fg). Then all the coefficients of fg are in p, and

therefore, all the coefficients of f are in p or all the coefficients of g are in p, i.e.

p | I(f) or p | I(g). We will discuss the case p | I(f), the other case being similar.

Since(I(

α′

α f)

, I(f))

= (1), it follows that p - I(

α′

α f).

Define γ, γ′ ∈ OK as follows:

• if it also happens that p | I(g), let γ = β and γ′ = β′. Note that in this

case, by the same reason as for f , p - I(

β′

β g)

= I(

γ′

γ g).

• if p - I(g), let γ = γ′ = 1. In this case, we also have that p - I(

γ′

γ g).

So, by the remark we made before, p - I(

α′γ′

αγ fg). Therefore

the power of p in I(fg) = the power of p in(

αγα′γ′

)= the power of p in

(αα′

)+ the power of p in

(γγ′

)= the power of p in I(f) + the power of p in I(g).

Hence I(fg) = I(f)I(g).

4. Extensions of Number Fields

Let K/k be a degree n extension of number fields.

Denote by K(1), . . . ,K(n) the n conjugates of K over k.

Definition 19. For every element α ∈ K, the norm of α relative to k is

NK/k(α) =n∏

i=1

α(i)

Similarly, the norm of a polynomial f ∈ K[X1, . . . , Xs] is NK/k(f) =n∏

i=1

f (i).

Clearly NK/k(α) ∈ k and NK/k(f) ∈ k[X1, . . . , Xs]

Proposition 29. Suppose L is an intermediate field of the extension K/k. Then,

for any α ∈ K and any f ∈ K[X1, . . . , Xs], we have NL/k

(NK/L(α)

)= NK/k(α)

and NL/k

(NK/L(f)

)= NK/k(f).

Proof: Let K be the normal closure of K over k. Then K/k is a Galois extension with

Galois group G. Denote H = Gal(K/K) and H ′ = Gal(K/L). Write G =⊔h

hH,

22

where t denotes a disjoint union. Also write G =⊔h′

h′H ′ and H ′ =⊔h′′

h′′H. Then

we have G =⊔

h′H ′ =⊔

h′h′′H.

For α ∈ K, NK/k(α) =∏h

h(α) and NK/L(α) =∏h′′

h′′(α).

For β ∈ L, NL/k(β) =∏h′

h′(β).

So

NL/k

(NK/L(α)

)=∏h′

h′

(∏h′′

h′′(α)

)=∏

h′ ,h′′

h′h′′(α) =∏h

h(α) = NK/k(α).

We would like to define NK/k for ideals of K. The natural definition would

be NK/k(a) =n∏

i=1

a(i), where, if a = (α1, . . . , αr) is an ideal of K, then a(i) =

(α(i)1 , . . . , α

(i)r ) ⊆ K(i), the image of a under the canonical isomorphism K → K(i).

But we have to precisely define what we mean by the product of ideals of different

fields and we need to know thatn∏

i=1

a(i) ⊆ k. This is obviously an ideal of K that

is fixed by the Galois group, but this does not mean anything. Take, for instance,(√2)

in Q(√

2), which is fixed by the Galois group, but that does not make it

rational.

Let a1, . . . , am ∈ k. We will write ak = (a1, . . . , am)k for the ideal of k generated

by a1, . . . , am and aK = (a1, . . . , am)K = akOK for the ideal of K generated by

a1, . . . , am.

Theorem 30. aK ∩ k = ak

Proof: It is obvious that ak ⊆ aK∩k. To prove the other inclusion, pick any element

b ∈ aK ∩ k. Then b = α1a1 + . . . + αmam for some α1, . . . , αm ∈ OK . Taking norm

of both sides of this equality, we get that

bn =n∏

i=1

(α

(i)1 a1 + . . . + α(i)

m am

).

Consider the polynomial f(X1, . . . , Xm) = α1X1+. . .+αmXm ∈ OK [X1, . . . , Xm]

and let g(X1, . . . , Xm) = NK/k(f) =n∏

i=1

m∑j=1

α(i)j Xj

∈ Ok[X1, . . . , Xm]. This g

23

is a homogeneous polynomial of degree n and bn = g(a1, . . . , am). Therefore bn =

a sum of products of the form αc1 . . . cn with α ∈ Ok and c1, . . . , cn ∈ ak. This

implies that bn ∈ ank , so an

k | (b)n. But, because of the unique factorization of the

ideals this means that ak | (b), i.e. b ∈ ak.

So we can define the equality of ideals in different fields as follows:

Definition 20. If ai is an ideal in the number field Li, i = 1, 2, we say that a1 = a2

if they generate the same ideal in a common extension of the two fields L1 and L2.

Also, the multiplication of ideals of different number fields is done by performing

the multiplication in some larger common extension of the respective fields.

Example Let L1 = Q(√−5)

and L2 = Q(√

2), a1 =

(2, 1 +

√−5)

and a2 =(√2). Then a2

1 =(4, 2 + 2

√−5,−4 + 2

√−5)

=(4, 2 + 2

√−5,−6

)= (2) and a2

2 =

(2). So, in Q(√−5,

√2)

it really does make sense to say that a1 = a2.

In fact(√

2)⊇(2, 1 +

√−5), because 2 =

√2√

2 and 1 +√−5 =

√2

1 +√−5√

2and 1±

√−5√2

are the roots of the polynomial X2 −√

2X + 3.

Also, −2√

2 +1−

√−5√

2

(1 +

√−5)

=√

2, so(√

2)⊆(2, 1 +

√−5).

Going back to our attempt to define the norm of an ideal, let a be an ideal of K

and let g ∈ K[X] be a polynomial with I(g) = a. For instance, if a = (α1, . . . , αr),

take g(X) =∑

αjXj . Then

n∏i=1

a(i) =n∏

i=1

I(g(i))

= I

(n∏

i=1

g(i)

)= I

(NK/k(g)

)⊆ k.

So, now we can actually define the norm of an ideal:

Definition 21. If a is an ideal of K, its norm relative to k is NK/k(a) =n∏

i=1

a(i).

Also if L is an intermediate field of the extension K/k, then NL/k

(NK/L(a)

)=

NL/k

(I(NK/L(g)

))=∏

I(NK/L

(g(i)))

= I(NK/k(g)

)= NK/k(a).

24

And, if a and b are both ideals of K, choose f, g ∈ K[X1, . . . , Xn] such that

I(f) = a and I(g) = b. Then ab = I(f)I(g) = I(fg) and NK/k(ab) = I(NK/k(fg)

)=

I(NK/k(f)NK/k(g)

)= I

(NK/k(f)

)I(NK/k(g)

)= NK/k(a)NK/k(b).

So, if k = Q, then NK/Q(a) = (N) for some N ∈ Z.

Definition 22. By convention, NK/Q(a) = |N | and this is called the absolute norm

of a.

From now on, if K is clear from the context we will write N(a) for the absolute

norm of a.

Example We will see that in Q(√

D), for the prime p ∈ Z, we have

(p) =

(p) if

(Dp

)= −1

p2 if(

Dp

)= 0

p1p2 if(

Dp

)= 1

with Np = Npi = p and N ((p)) = p2.

Note:(

Dp

)is the Legendre symbol, defined by:

(D

p

)=

0 if p | D1 if there exists some a such that a2 ≡ D (mod p)−1 if there is no such a

In general, if a = (l), l ∈ Q and [K : Q] = n, then Na = |l|n.

Also, note that if a is integral, then a | NK/k(a).

Lemma 31. Any ideal a of K may be written as a = (α, β), where(NK/k(α), NK/k(β)

)=

NK/k(a).

Proof: Take c integral ideal with ac = (α). Now take NK/k(c) extended up to K

and choose an integral ideal b of K so that(b, NK/k(c)

)= (1) and ab = (β). Now,

c | NK/k(c) and therefore (c, b) = (1). So (α, β) = (ac, ab) = a(c, b) = a and we

have(NK/k(α), NK/k(β)

)=(NK/k(a)NK/k(c), NK/k(a)NK/k(b)

)= NK/k(a)

(NK/k(c), NK/k(b)

)

25

But(b, NK/k(c)

)= (1), hence

(b(i), NK/k(c)

)= (1) for each i. Thus

(NK/k(b), NK/k(c)

)=

(1).

Note that NK/k ((α)) =(NK/k(α)

).

Assume that Ok is a PID and recall that [K : k] = n.

Lemma 32. Let b = [β1, . . . , βn] and c = [γ1, . . . , γn] be two ideals of K. We

can write (γ1, . . . , γn) = (β1, . . . , βn)M for some M ∈ Matn×n(k). Then detM |

NK/k(c/b).

Proof: Let α ∈ c/b. Then (α)b ⊆ c. So there exists an n× n matrix X with entries

in Ok such that α(β1, . . . , βn) = (γ1, . . . , γn)X = (β1, . . . , βn)MX. Therefore

α(1)β

(1)1 . . . α(1)β

(1)n

.... . .

...α(n)β

(n)1 . . . α(n)β

(n)n

=

β

(1)1 . . . β

(1)n

.... . .

...β

(n)1 . . . β

(n)n

MX

and taking determinants we get NK/k(α) det(β

(i)j

)= det

(β

(i)j

)detM detX. Hence

NK/k(α) = det M detX, and, since det X ∈ Ok, it follows that detM | NK/k(α).

Now, there exist α1, α2 ∈ K such that c/b = (α1, α2) and NK/k(c/b) =(NK/k(α1), NK/k(α2)

).

But this means that detM | NK/k(αi), i = 1, 2, so detM | NK/k(c/b).

Theorem 33. Let k be a number field with Ok a PID. Suppose K/k is a degree

n extension and choose w1, . . . , wn ∈ K such that OK = [w1, . . . , wn]k. Let a =

(α1, . . . , αn) be an ideal of K and write (α1, . . . , αn) = (w1, . . . , wn)M with M ∈

Matn×n(k). Let f(X1, . . . , Xn) = NK/k

n∑j=1

αjXj

.

Then a = [α1, . . . , αn]k iff detM divides every coefficient of f , and in this case

NK/k(a) = (det M).

Proof: Recall that I(f) = NK/k(a).

26

First assume that a = [α1, . . . , αn]k. Then by lemma 32, we have det M |

NK/k (a/Ok). But NK/k (a/Ok) = NK/k (a), so

(1) detM | NK/k(a)

Furthermore, (w1, . . . , wn) = (α1, . . . , αn)M−1, so det M−1 | NK/k(Oka−1)

and therefore (det M−1) | NK/k(a−1). Together with (1) above, this shows that

NK/k(a) | (detM).

Now suppose that a = [β1, . . . , βn]k. Then (α1, . . . , αn) = (β1, . . . , βn)X for some

X ∈ Matn×n(Ok) and (β1, . . . , βn) = (w1, . . . , wn)Y for some Y ∈ Matn×n(k).

Hence (α1, . . . , αn) = (w1, . . . , wn)Y X, and so M = Y X. By the first part of

the proof, it follows that (det Y ) = NK/k(a) = I(f). Therefore (detM) | I(f) iff

(detX) | (1) which is equivalent to [α1, . . . , αn]k = [β1, . . . , βn]k.

Example Let K = Q(√−5)

and a =(2, 1 +

√−5)⊆ K. Then also a =[

2, 1 +√−5].

Here OK =[1,√−5]

and(2, 1 +

√−5)

=(1,√−5)( 2 1

0 1

)and det M = NK/Qa = 2. The associated polynomial is f(X, Y ) = NK/Q

(2X + (1 +

√−5)Y

)=

(2X + Y )2 + 5Y 2 = 4X2 + 4XY + 6Y 2. Note that 2 divides every coefficient of f ,

and in fact I(f) = (4, 6) = (2).

Corollary 34. If k = Q and a = [α1, . . . , αn], then Na = |Ok/a| = |detM |.

Proof: Both quantities are equal to |det M |.

5. Relative Differents and Discriminants

Take K = Q(√

D).

• If D ≡ 2, 3 (mod 4), consider the integral basis{

1,√

D}

. Then

(1

√D

1 −√

D

)−1

=−1

2√

D

(−√

D −√

D−1 1

)=

12

12

12√

D− 1

2√

D

27

So we obtained the dual pair of elements{

12 , 1

2√

D

}. Note that[

12 , 1

2√

D

]=(

12√

D

)as[

12 , 1

2√

D

]= 1

2√

D

[1,√

D]

is a fractional ideal. Note that

any element of this ideal looks like(a + b

√D) 1

2√

D=

a

2√

D+

b

2

We define the trace of any element of K as follows: tr(α) = α(1) + α(2). So

tr(a + b

√D)

= 2a. For α ∈ OK , we clearly have tr(α) ∈ Z. Also notice that for

the elements of(

12√

D

), we have tr

(a

2√

D+ b

2

)= b ∈ Z. What other elements of

K have the property that their trace is an integer? That is, what is tr−1 (Z)?

Consider α = r + s√

D ∈ K. Then tr(α) = 2r. So if the trace is an integer, we

have r ∈ 12Z. Furthermore, tr(

√Dα) = 2Ds. If this is an integer, then s ∈ 1

2D Z.

Thus, we have:

(1

2√

D

)=[12,

12√

D

]= {α ∈ K; tr(λα) ∈ Z for all λ ∈ OK}

Also, note that√

D is a root of the polynomial f(X) = X2 −D and f ′(√

D) =

2√

D, so N(f ′(√

D))

= 4D = DK .

• If D ≡ 1 (mod 4), consider the integral basis{

1, 1+√

D2

}. Then 1 1+

√D

2

1 1−√

D2

−1

=−1√D

1−√

D2 − 1+

√D

2

−1 1

=

− 12√

D+ 1

21

2√

D+ 1

2

1√D

− 1√D

So[− 1

2√

D+ 1

2 , 1√D

]= 1√

D

[1, 1−

√D

2

]=(

1√D

).

Here tr((

a+b√

D2

)(1√D

))= b

2 + b2 = b ∈ Z.

Also, 1+√

D2 is a root of the polynomial f(X) = X2−X + 1−D

2 and f ′(

1+√

D2

)=

√D. So N

(f ′(

1+√

D2

))= D = DK .

Recall that we are going to prove that (p) = p2 iff p | D iff p |(√

D), so primes

that ramify are precisely the primes dividing the different or the discriminant.

28

Now, return to the situation in the previous section. That is, consider a degree

n extension of number fields K/k. By the primitive root theorem, K = k(θ) for

some algebraic integer θ ∈ OK . Let f ∈ Ok[X] be the irreducible polynomial of θ

over k.

We want to write

1 θ(1) . . . θ(1)n−1

.... . .

...1 θ(n) . . . θ(n)n−1

−1

=

β

(1)1 . . . β

(1)n

.... . .

...β

(n)1 . . . β

(n)n

So β(n)j =

signedminor corresponding to θ(n)j−1

det of the θ matrix.

The signed minor corresponding to θ(n)j−1 is (−1)n+j times the determinant of

the minor obtained by deleting the nth row and the jth column of the θ matrix,

which is equal to the coefficient of Xj−1 in the polynomial

h(X) =

1 θ(1) . . . θ(1)n−1

.... . .

...1 θ(n−1) . . . θ(n−1)n−1

1 X . . . Xn−1

=∏

1≤i<j≤n−1

(θ(j) − θ(i)

) n−1∏j=1

(X − θ(j)

)

Hence

β(n)j =

the coefficient of Xj−1 inh(X)h(θ(n)

) =

the coefficient of Xj−1 inn−1∏i=1

(X − θ(i)

)n−1∏i=1

(θ(n) − θ(i)

)

Recall that f is the minimal polynomial of θ, so f(X) =n∏

j=1

(X − θ(j)

)∈ Ok[X].

Thenn−1∏i=1

(X − θ(i)

)=

f(X)X − θ(n)

∈ Ok

[θ(n)

][X] and f ′

(θ(n)

)=

n−1∏i=1

(θ(n) − θ(i)

).

Therefore

β(n)j =

the coefficient of Xj−1 in f(X)X−θ(n)

f ′(θ(n)

) ∈ Ok

[θ(n)

]

29

Let g(X) =f(X)X − θ

=n−1∑l=0

blXl ∈ Ok[θ], where of course bn−1 = 1. Then, for

each 1 ≤ i ≤ n, we havef(X)

X − θ(i)= g(i)(X) =

n−1∑l=0

b(i)l X l ∈ Ok

[θ(i)]. So,

β(n)j =

b(n)j−1

f ′(θ(n)

)and similarly

β(i)j =

b(i)j−1

f ′(θ(i)) , for each i, 1 ≤ i ≤ n.

In general if α1, . . . , αn are k-linearly independent elements of K, then

(α1, . . . , αn) = (θ(1), . . . , θ(n))M for some M ∈ GL(n, k). Hence

α

(1)1 . . . α

(1)n

.... . .

...α

(n)1 . . . α

(n)n

=

1 θ(1) . . . θ(1)n−1

.... . .

...1 θ(n) . . . θ(n)n−1

M

and thus

(α

(j)i

)−1

= M−1(θ(j)i−1

)−1

= M−1(β

(i)j

)=

γ

(1)1 . . . γ

(n)1

.... . .

...γ

(1)n . . . γ

(n)n

.

So we proved the following result:

Theorem 35. If α1, . . . , αn ∈ K are linearly independent over k, thenα

(1)1 . . . α

(1)n

.... . .

...α

(n)1 . . . α

(n)n

−1

=

β

(1)1 . . . β

(n)1

.... . .

...β

(1)n . . . β

(n)n

with βi ∈ K for each 1 ≤ i ≤ n.

In particular if (α1, . . . , αn) = (θ(1), . . . , θ(n)), with K = k(θ) and θ integral with

minimal polynomial f over k, then β(i)j =

b(i)j−1

f ′(θ(i)) , where

f(X)X − θ

=n−1∑l=0

blXl.

Note that (β1, . . . , βn) =(

1f ′(θ)

).

Definition 23. The relative trace of an element α ∈ K is trK/k(α) =n∑

i=1

α(i)

30

Note that for all α ∈ K we have trK/k(α) ∈ k and if L is an intermediate field

of the extension K/k, then trL/k

(trK/L(α)

)= trK/k(α).

In the notation of the theorem we have(β

(i)j

)(α

(j)i

)= (δij), so

β(1)j α

(1)i + +β

(n)j α

(n)i = δij , i.e. trK/k(βjαi) = δij .

Lemma 36. Suppose α1, . . . , αn ∈ K are linearly independent over k and(α

(j)i

)−1

=(β

(i)j

). Then λ ∈ K has the property that trK/k(λαi) ∈ Ok for 1 ≤ i ≤ n iff

λ ∈ [β1, . . . , βn]k.

Proof: Suppose λ has the property that trK/k(λαi) ∈ Ok for all i.

Then(λ(1), . . . , λ(n)

) (α

(j)i

)= (b1, . . . , bn) with bi ∈ Ok.

So(λ(1), . . . , λ(n)

)= (b1, . . . , bn)

(β

(i)j

)and thus each λ(i) ∈

[β

(i)1 , . . . , β

(i)n

]k.

Also note that, in this case, if we write λ =∑

cjβj with cj ∈ k, then bi =

trK/k(λαi) = trK/k (∑

cjβjαi) = ci, so ci ∈ Ok.

The converse is clear.

Definition 24. Let a be an ideal of K. We will denote

tK/k(a) ={λ ∈ K : trK/k(λα) ∈ Ok for all α ∈ a

}Lemma 37. If a is an ideal of K, then tK/k(a) is also an ideal of K.

Proof: First note that if β, γ ∈ OK and λ1, λ2 ∈ t(a), then tr ((λ1β + λ2γ)α) =

tr (λ1(βα)) + tr (λ2(γα)) ∈ Ok for any α ∈ a. So λ1β + λ2γ ∈ t(a).

Therefore we need only show that t(a) is finitely generated, that is to say find

d ∈ OK such that d t(a) ⊆ OK .

Let λ ∈ tK/k(a). Then, for any α ∈ a we have trK/k(λα) ∈ Ok, so trK/Q(λα) ∈

Z. Hence tK/k(a) ⊆ tK/Q(a). Using the fact that Z is a PID, we may write

a = [α1, . . . , αm]Z (where m = [K : Q]), and let(α

(j)i

)−1

1≤i,j≤m=(β

(i)j

)1≤i,j≤m

(here m = [K : Q]). Since λ ∈ tK/Q(a), it follows that tK/Q(λαi) ∈ Z for all

31

1 ≤ i ≤ m and therefore λ ∈ [β1, . . . , βm]Z. Choose d ∈ OK such that dβi ∈ OK

for each i. Then d t(a) ∈ OK .

Definition 25. The ideal DK/k = t(OK)−1 is called the relative different of K/k.

If k = Q, then DK = DK/Q is called the different of K.

Theorem 38. If a is an ideal of K, then t(a) = a−1D−1K/k and DK/k is an integral

ideal of K. If K = k(θ) and OK = Ok[θ], then DK/k = (f ′(θ)) where f is the

minimal polynomial of θ.

Proof: Let a, b be two ideals of K. Then a t(a) = a t(a)b−1b =(a t(a)b−1

)b.

Choose α ∈ a, λ ∈ t(a) and x ∈ b−1 Pick any β ∈ b. Then xβ ∈ OK and

therefore αxβ ∈ a. Hence trK/k(λαxβ) ∈ Ok. Since this holds for any β ∈ b, it

follows that αλx ∈ t(b), that is a t(a)b−1 ⊆ t(b). So a t(a) ⊆ b t(b).

By symmetry, we see that for any two ideals a and b of K, we have a t(a) = b t(b).

In particular take b = OK . We have a t(a) = OK t(OK) = t(OK) = D−1K/k.

Pick any α, λ ∈ OK . Then tr(λα) ∈ Ok, so λ ∈ t(OK) = D−1K/k. Hence OK ⊆

D−1K/k, i.e. DK/k ⊆ O−1

K = OK .

Now assume that K = k(θ) and OK = Ok[θ]. Let(β

(i)j

)=(θ(i)j−1

)−1

. By

theorem 35, (β1, . . . , βn) =(

1f ′(θ)

). Recall that tr(θiβj) = δi+1,j ∈ Ok, so βj ∈

t(OK) for all j. On the other hand, for any λ ∈ t(OK) we have, by lemma 36, that

λ ∈ [β1, . . . , βn]k ⊆ (β1, . . . , βn).

Hence D−1K/k = t(OK) = (β1, . . . , βn) =

(1

f ′(θ)

), so DK/k = (f ′(θ)).

Theorem 39. Let a = [α1, . . . , αn], where we recall that n = [K : k]. Define

β1, . . . , βn as usual by(β

(i)j

)=(α

(j)i

)−1

. Then a−1D−1K/k = [β1, . . . , βn]k.

Proof: Recall that if λ ∈ a−1D−1K/k = t(a), then lemma 36 implies that λ ∈

[β1, . . . , βn]k. On the other hand, for each i and j we have tr(αiβj) = δij ∈ Ok, so

βj ∈ t(a).

32

Definition 26. The relative discriminant of K/k is the ideal DK/k =(det(w

(j)i

))2

,

where OK = [w1, . . . , wn]k.

Corollary 40. If Ok is a PID, then NK/kDK/k = DK/k. In particular, N(DK) =

|DK |.

Proof: Let OK = [w1, . . . , wn]k and(β

(i)j

)=(w

(j)i

)−1

.

By theorem 39 it follows that [β1, . . . , βn]k = D−1K/k. Write (w1, . . . , wn) =

(β1, . . . , βn)M for some n× n matrix with entries in k.

Then, by theorem 33, (det M) = NK/kD−1K/k.

We also have(β

(i)j

)=(w

(j)i

)M , so(

det(β

(i)j

))=(det(w

(j)i

))(detM) =

(det(w

(j)i

))NK/kD−1

K/k

Therefore NK/kDK/k =(det(w

(j)i

))2

= DK/k.

If k = Q, we get that N(DK) =∣∣NK/Q(DK)

∣∣ = |DK |.

Theorem 41. If K ⊇ L ⊇ k are number fields, then DK/k = DK/L · DL/k.

Proof: We will prove that D−1K/k = D−1

K/L · D−1L/k.

Pick some α ∈ D−1K/L = tK/L(OK) and some a ∈ D−1

L/k = tL/k(OL) ⊆ L. Then,

for any w ∈ OK , we have trK/k ((αa)w) = trL/k trK/L (a(αw)) = trL/k

(a trK/L(αw)

)∈

trL/k (aOL) ⊆ Ok. Hence aα ∈ tK/k(Ok) = D−1K/k.

Now we want to prove the other inclusion, which is equivalent to showing that

DL/k · D−1K/k ⊆ D

−1K/L. Choose α ∈ D−1

K/k = tK/k(OK) and a ∈ DL/k ⊆ OL. Choose

w ∈ OK . Then for any b ∈ OL, we have trL/k

(b trK/L(αw)

)= trK/k (α(bw)) ∈ Ok,

since bw ∈ OK and α ∈ tK/k(OK). Thus trK/L(αw) ∈ tL/k(OL) = D−1L/k, and

therefore trK/L(aαw) = a trK/L(αw) ∈ aD−1L/k ⊆ DL/k · D−1

L/k ⊆ OL. Hence aα ∈

tK/L(OK) = D−1K/L.

Corollary 42. Suppose [K : L] = m. Then DmL | DK .

33

Proof: Take k = Q in the previous theorem. We get that

|DK | = NK/Q(DK) = NL/Q(NK/L(DK/L · DL)

)= NK/Q(DK/L)NL/Q(Dm

L ) =

DmL ·NK/Q(DK/L)

and the theorem follows since NK/Q(DK/L) is an integer.

Theorem 43. (Chinese Remainder Theorem) Suppose a and b are integral ideals

of K with (a, b) = (1). If α, β ∈ OK , then the system of equations

{x ≡ α (mod a)x ≡ β (mod b)

has a solution in OK and that solution is unique (mod ab).

Proof:

Theorem 44. Let P be a prime ideal of K. Then p = P ∩ k is a prime ideal of k

and NK/kP = pf for some 1 ≤ f ≤ n = [K : k].

Proof: If a, b ∈ Ok ⊆ OK and ab ∈ p ⊆ P, it follows that a ∈ P or b ∈ P, and since

they are already elements of k, we get a ∈ p or b ∈ p. Therefore p is a prime ideal

of k.

Note that P | p , since pOK ⊆ POK = P. Now let K(i) be a conjugate field of K.

Thenp

P(i)=(

p

P

)(i)

⊆ OK(i) , so P(i) | p. Thus NK/kP | pn, hence NK/kP = pf

for some 1 ≤ f ≤ n.

Lemma 45. Given a prime ideal P of K and an integral ideal a of K with (P, a) =

(1), there exists an element θ ∈ a such that K = k(θ), θ is a primitive root modP

and for any m ∈ Z+ and any w ∈ OK we can find an element α ∈ Ok[θ] with

α ≡ w (mod Pm).

Proof: We know that (OK/P)∗ is a finite cyclic group. Choose θ1 ∈ OK such that

θ1 (mod P) is a generator of this group. Then every element in OK that is not in

P will be congruent to θi (mod P) for some 1 ≤ i ≤ NP− 1.

34

We want to find θ2 ≡ θ1 (mod P) such that π = θNP2 − θ2 ∈ P \P2. To do this,

set θ2 = θ1 + β for some β ∈ P \P2. Then

θNP2 = (θ1 + β)NP = θNP

1 + NPθNP−11 β + β2(. . .) ≡ θNP

1 (mod P2)

since P | NP and β ∈ P.

So θNP2 − θ2 ≡ θNP

1 − θ1 − β (mod P2) and we can choose an appropriate β.

Now, the Chinese Remainder Theorem allows us to pick θ3 ≡ θ2 (mod P2) with

θ3 ≡ 0 (mod a).

Set θ(l) = θ3 + lp2aγ where p is a rational prime such that P | (p), a ∈ a ∩

Z nonzero, γ ∈ OK such that K = k(γ) and l ∈ Z. The determinant of the

matrix(θ(l)(i)

j)

looks like∏i<j

(θ(j)3 − θ

(i)3 + lp2a(γ(j) − γ(i))

), and since γ(j) −

γ(i) 6= 0 we can choose l large enough to this determinant nonzero. In this case

the column vectors are linearly independent, hence {1, θ(l), θ(l)2, . . . , θ(l)n−1} are

linearly independent over k, hence we have K = k(θ(l)). Put θ = θ(l).

Note that θ is an integer.

Change π to π = θNP−θ ∈ P\P2 and set γi = θi for 1 ≤ i ≤ NP−1. Then the

numbersm−1∑j=0

βjπj ∈ Ok[θ] run through all the residue classes mod Pm as the βj ’s

run through {γi; 1 ≤ i ≤ NP− 1} ∪ 0. This is because we have NPm of the formm−1∑j=0

βjπj and if two of them are congruent (mod Pm), then βj ≡ β′j (mod P) for

each j.

Theorem 46. If K = k(θ) and θ ∈ OK , let f ∈ k[X] be the minimal polynomial

of θ. Then DK/k is the gcd of all such f ′(θ). Also, if P is a prime ideal of K,

p = P ∩ k and p = Pea with (a,P) = (1), then, for the θ in lemma 45, the prime

ideal P has the same power in both DK/k and (f ′(θ)).

35

Proof: If θ ∈ OK and K = k(θ) and(β

(i)j

)=(θ(i)j−1

)−1

, then for any λ ∈

D−1K/k = t(OK) we have, by lemma 36, that λ ∈ [β1, . . . , βn]k ⊆ (β1, . . . , βn). So

D−1K/k ⊆ (β1, . . . , βn) = (f ′(θ))−1, by theorem 35.

Thus, (f ′(θ)) ⊆ DK/k, i.e. DK/k | (f ′(θ)).

Hence (f ′(θ)) = DK/kB for some integral ideal B of K.

The theorem will follow if we can show that with the choice of θ from the

lemma 45, we have (B,P) = (1).

Now B | NK/kB = prb for some integral ideal b of k with (p, b) = (1). Given

w ∈ OK , we can find α ∈ Ok[θ] with α ≡ w (mod Per) (see lemma 45).

Then

DK/k(w − α)b(θ)r

(f ′(θ))=

(w − α)Per

· Perarb

B·(

(θ)a

)r

and each of the three terms is integral, so the LHS is an integral ideal.

Now choose b ∈ b ⊆ Ok. Then b | (b), so DK/k(w − α)(b)(θ)r

(f ′(θ))is an integral

ideal. Hence(w − α)b(θ)r

f ′(θ)∈ D−1

K/k and therefore trK/k

((w−α)b(θ)r

f ′(θ) θj)∈ Ok for all

1 ≤ j ≤ n. Again lemma 36 implies that (w−α)b(θ)r

f ′(θ) ∈ [β1, . . . , βn]k ⊆ (β1, . . . , βn) =

(f ′(θ))−1, so (w − α)bθr ∈ Ok[θ]. But αbθr ∈ Ok[θ] and thus wbθr ∈ Ok[θ].

Hence trK/k

(wbθr

f ′(θ)

)∈ Ok for all w ∈ OK and so bθr

f ′(θ) ∈ D−1K/k. Therefore

D−1K/k |

(bθr)(f ′(θ)) , i.e. (bθr)

D−1K/k

(f ′(θ))is an integral ideal. But (f ′(θ)) = DK/kB, so (b)(θr)

B

is integral. This is true for all b ∈ b, hence B | b(θ)r. Since (p, b) = (1) and P | p ,

it follows that (b,P) = (1) But we also have ((θ),P) = (1), so ((θ)rb,P) = (1). As

B | b(θ)r, we get (B,P) = (1), exactly what we wanted.

6. Ramification Theory

Consider the field extensions

K P|k p|Q (p)

36

We have that P | p for each P lying above p, i.e. such that p = P ∩ k. Also, if

P | p, then P ⊇ p, i.e. P lies above p. Thus, we may write any prime ideal of k as

p =∏

Peii , a product of the primes lying above it. Then, taking the norm of both

sides and keeping in mind that NK/kPi = pfi , we get pn =∏(

NK/kPi

)ei = pP

eifi ,

hence n =∑

eifi. In particular, the number of the Pi’s must be ≤ n.

Possibilities are:

(1) p splits completely in K if there are n distinct Pi’s lying above it (each

with fi = ei = 1);

(2) p remains inert in K if there is just one prime ideal P lying above p (here

e = 1, f = n), so p = P;

(3) p ramifies in K if any of the ei > 1. In this case we also say that Pi is a

ramified prime of K relative to k.

Definition 27. The degree of P is the residue class degree, i.e. the degree of the

algebraic extension OK/P over Z/(p).

Therefore NK/QP = pf where f = the degree of P.

Now, OK/P contains the field Ok/p in the following sense: suppose a, b ∈ Ok

with a ≡ b (mod P); then a− b ∈ P ∩ Ok = p, so a ≡ b (mod p).

Claim OK/P is a normal extension of Ok/p.

Proof: Say NK/kP = pf and Nk/Qp = pf0 = q. Then #OK/P = qf . Let θ be a

generator of the multiplicative group of the field OK/P, so θqf−1

= 1 and no

lower power of θ is = 1. For α ∈ OK/P define σ(α) = αq. This defines an

automorphism of OK/P that fixes Ok/p as (α + β)q = αq + βq

and for a ∈ Ok/p

we have aq ≡ a (mod p).

Moreover, σ, σ2, . . . , σf = Id are all distinct automorphisms of OK/P over

Ok/p.

Note that the above field extension is automatically separable, since any finite

field extension is separable.

37

Thus, Gal ((Ok/P)/(Ok/p)) is a cyclic group of degree f with generator σ, given

by σ(θ) = θq, and the f conjugates of θ over Ok/p are θ

qi

, 0 ≤ i ≤ f − 1.

Now assume K is normal over k with Galois group Gal(K/k) = G.

Suppose Pe‖p (exactly divides), and let P1, . . . ,Pr be the r distinct images of P

under the action of G. Then Pei‖p for each i. So, p =

(r∏

i=1

Pi

)e

a, where

(a,Pi) = (1) for all 1 ≤ i ≤ r.

But∏g∈G

gP = NK/kP = pf , so any prime ideal factor of p in K is one of the gP’s,

i.e. one of the Pi’s. Hence a = (1) and∏g∈G

gP = NK/kP = pf =

(r∏

i=1

Pi

)ef

.

Now let GD = GD(P) = {g ∈ G; gP = P}.

Then ∏g∈G

gP =

(r∏

i=1

Pi

)#GD

so #GD = ef .

Note that if g ∈ GD, then g is an automorphism of OK/P over Ok/p.

Let a =p

Peand choose θ ∈ OK so that θ is a primitive root (mod P), θ ∈ a and

K = k(θ), satisfying the conclusions of the lemma 45. Then

π = θ − θNP = θ − θqf ∈ P \P2 and every element of OK is congruent to an

element of Ok[θ] (mod Pm). The irreducible polynomial for θ over k is

f(X) =∏g∈G

(X − gθ) ∈ Ok[X].

Reducing (mod p) we see that θ is a root of f , so (mod p)∏g∈G

(X − gθ) is

divisible byf−1∏i=0

(X − θ

qi)

.

Thus there is a g ∈ G such that gθ ≡ θq (mod P). This g is called theFrobenius automorphism of P and is denoted by σ(P) = σ

(K/kP

).

Let σ be the Frobenius automorphism of P. Then σ ∈ GD. Indeed, suppose this

was not so. Then σ−1P 6= P, hence (σ−1P,P) = (1). So, we can assume that

38

θ ∈ σ−1P. Since σθ ≡ θq (mod P), we have that θ ≡ (σ−1θ)q (mod σ−1P), so

σ−1θ ∈ σ−1P, i.e. θ ∈ P, and we arrived at a contradiction. Thus σ ∈ GD.

Definition 28. GI = {g ∈ G; gw ≡ w (mod P) for all w ∈ OK} is called the

inertia group of P.

Now, if g ∈ GD, then g provides an automorphism of OK/P over Ok/p, so

gθ ≡ θqi

(mod P) for some i. But also σiθ ≡ θqi

(mod P), so g−1σiθ ≡ θ

(mod P), so g−1σiw ≡ w (mod P) for all w ∈ OK . Hence g−1σi ∈ GI , i.e.

g ∈ σiGI .

It follows that GI is a normal subgroup of GD, because

g−1D gIgDθ ≡ g−1

D gIσi(θ) ≡ g−1

D θqi ≡ σ−iθqi ≡ θ (mod P).

Claim GD/GI is a cyclic group of order f and is generated by σGI .

Proof: σGI is a generator as each g ∈ σiGI for some i, as seen above. The only

question is to determine the smallest i such that θqi ≡ θ (mod P). This is i = f

since θ is a primitive root (mod P).

Note that #GD = ef , so #GI = e. Also, GD/GI = Gal(OK

P /Ok

p

).

We now have the field extensions corresponding to G ⊇ GD ⊇ GI :

K P

GI e

kI pI = P ∩ kI

f

kD pD = P ∩ kD

r

k p

Start with an ideal P of K lying above the prime ideal p of k. Denote pI = P∩ kI

and pD = P ∩ kD = pI ∩ kD.

39

Then

NK/kDP =

∏g∈GD

gP = Pef .

At the same time, we know that NK/kDP = pm

D for some m ∈ {1, . . . , ef}. (Recall

that ef = [K : kD].)

Now if OkD/pD were a nontrivial extension of Ok/p, then it would exist a

nontrivial automorphism of OkD/pD that fixes Ok/p. Any such automorphism

would extend to a nontrivial element of Gal(OK

P /Ok

p

)= GD/GI , so it would be

induced by some g ∈ GD. But all the elements of GD fix OkD/pD (contradiction).

Hence OkD/pD = Ok/p.

Therefore NkD/QpD = Nk/Qp = q and NkD/kpD = p.

Also, qf = NK/QP = NkD/QNK/kDP = NkD/Qpm

D = qm, so f = m and

Pef = NK/kDP = pf

D. Hence pD = Pe.

Since the elements of GI fix OK/P, we get just as before that OK/P = OkI/pI

why? and therefore NkI/kDpI = pf

D. It will also follow that NK/kIP = pI and

that pI = Pe.

In conclusion, p splits completely in kD into r different primes. Each of this

primes of kD remains inert in kI , i.e. pD = pI . And each pI lying above p in kI is

totally ramified in K, i.e. pI = Pe and P is the only prime lying above pI in K.

So we have

L Gal(K/L) order of Gal(K/L) e(pL) f(pL) r(pL)

k G n = efr e f r

kD GD ef e f 1

kI GI e e 1 1

or

40

K P

kI pI = Pe NK/kIP = pI

kD pD = Pe NK/kDP = pf

D

k p = Pe NK/k = pf

Now P is ramified in K/k iff #GI(P) > 1.

Recall that K = K(1).

Then

f ′(θ) =#G∏i=2

(θ − θ(i)

)=∏g∈Gg 6=1

(θ − gθ).

Now if P | (θ− gθ), where g 6= 1, then gθ ≡ θ (mod P), so θ ≡ g−1θ (mod g−1P).

Assume that g−1P 6= P. Then we may assume that θ ∈ g−1P, therefore also

g−1θ ∈ g−1P, so θ ∈ P (contradiction). Hence g−1P = P, so P = gP, i.e.

g ∈ GD, and thus g ∈ GI . why?

Thus P | (f ′(θ)) iff #GI > 1 iff e > 1. So we have the following result:

Theorem 47. If K/k is normal, then P | DK/k iff P is ramified in K/k.

Definition 29. GR ={g ∈ G; gw ≡ w (mod P2) for all w ∈ OK

}Gm =

{g ∈ G; gw ≡ w (mod Pm+1) for all w ∈ OK

}, where m = 0, 1, 2 . . .

Note that G0 = GI and G1 = GR.

Theorem 48. Let g ∈ G. Then g ∈ Gm iff gθ ≡ θ (mod Pm+1).

Proof: As before.

Theorem 49. For all m ≥ 0, Gm+1 is a normal subgroup of Gm.

41

Proof: gm+1gmθ ≡ gmθ (mod Pm+2), so g−1m gm+1gmθ ≡ θ (mod g−1

m Pm+2). But

gmP = P as gm ∈ GD, and the assertion follows.

Now, the exact power of P in DK/k = the power of P in∏g∈Gg 6=1

(θ − gθ) =∑g∈Gg 6=1

i(g),

where i(g) = max{i; gθ ≡ θ (mod Pi)} = i such that gθ − θ ∈ Pi \Pi+1.

But this equals

∞∑i=1

∑g∈G\{1}

i(g)=i

1

=∞∑

j=1

i (#Gj−1 −#Gj) =∞∑

j=0

(#Gj − 1)

So we have the following result:

Theorem 50. If K/k is normal, the exact power of P occurring in DK/k is∞∑

i=0

(#Gi(P : K/k)− 1) .

Corollary 51. P | DK/k iff e(p) > 1.

We will now show the following theorem:

Theorem 52. Let L be an extension of k (not necessarily normal), and pL a

prime ideal of L. Then pL | DL/k iff pL is a ramified prime ideal of L relative to k.

This has the following consequence:

Corollary 53. If p = pL ∩ k, then p | DL/k iff p ramifies in L.

Proof of theorem: Let K/k be normal such that K ⊇ L, P a prime ideal of K

dividing pL. Then Gm(P : K/L) = Gm ∩H, where H = Gal(K/L). So, the power

of P in DK/k is∞∑

i=0

(#Gi − 1) and the power of P in DK/L is∞∑

i=0

(#GI ∩H − 1).

Clearly, Gi ∩H = Gi for all i iff G0 ∩H = G0, so the same power of P occurs in

DK/k and DK/L iff e(p) = e(pL).

Now pL occurs in DL/k iff the power of P in DK/k is > the power of P in DK/L.

Therefore pL is ramified in L/k iff e(p) > e(pL) iff P | DK/k iff pL | DL/k.why?

42

FINAL COMMENTS this are just facts, right?

Take K/k to be a normal extension of number fields. For m ≥ 1, Gm/Gm+1 is a

p-group, and every non-identity element has order p. Also, Gm/Gm+1 is abelian.

∞∑i=0

(#Gi − 1) =

{e− 1 , if p - e

≥ e− 1 + p− 1 ≥ e + p− 2 ≥ e , if p | e

p is called tamely ramified if p - e and wildly ramified otherwise.

PART II

ANALYTIC NUMBER THEORY

45

Define

ζ(s) = ζQ(s) =∑a⊆Z

1(Na)s =

∞∑n=1

1ns

=∏p

(1− p−s

)−1

Our aim is to prove that the function

g(s) = π−s2 Γ(s

2

)ζ(s)

satisfies the functional equation

g(s) = g(1− s),

is analytic for all s 6= 0, 1, with simple poles at this points and residue 1 at s = 1

and residue −1 at s = 0.

To do this, we first define

θ(t) =∞∑

n=−∞e−πn2t for t > 0

and prove that

θ(t) =1√t

θ

(1t

).

This is a consequence of the Poisson summation formula: Let f(x) be a function

such that f(x) → 0 sufficiently fast as |x| → ∞. Then F (x) =∞∑

n=−∞f(n + x)

converges for all x and F (x + 1) = F (x). Thus F (x) has a Fourier expansion

F (x) =∑m∈Z

ame2πimx, where

am =∫ 1

0

F (x)e−2πimxdx

=∫ 1

0

∞∑n=−∞

f(n + x)e−2πimxdx

=∞∑

n=−∞

∫ 1

0

f(n + x)e−2πimxdx

Making the change of variables x′ = x + n, we see that

46

am =∞∑

n=−∞

∫ n+1

n

f(x)e−2πimxdx

=∫ ∞

−∞f(x)e−2πimxdx

= f(m) (the Fourier transform of f at m).

Thus∞∑

n=−∞f(n + x) = F (x) =

∞∑m=−∞

f(m)e2πimx

and setting x = 0 we get the Poisson summation formula

∞∑n=−∞

f(n) =∞∑

n=−∞f(n)

Now take ft(x) = e−πx2t. Then

f(m) =∫ ∞

−∞e−πx2t−2πimxdx

=∫ ∞

−∞e−πt

h(x+ im

t )2+ m2

t2

idx

= e−πm2

t

∫ ∞

−∞e−πt(x+ im

t )2

dx

Moving the line of integration we get

f(m) = e−πm2

t

∫ ∞

−∞e−πtx2

dx

=1√t

e−πm2

t

Thus θ(t) =∞∑

n=−∞e−πn2t =

1√t

∞∑n=−∞

e−πn2

t =1√t

θ

(1t

), as desired.

Now, for Re(s) > 1 we have

ζ(s) =∞∑

n=1

1ns

47

and

Γ(s

2

)=∫ ∞

0

ts2 e−t dt

t.

So,

π−s2 Γ(s

2

)ζ(s) = π−

s2

∫ ∞

0

ts2

∞∑n=1

1ns

e−t dt

t

=∫ ∞

0

∞∑n=1

(t

πn2

) s2

e−t dt

t

=∫ ∞

0

∞∑n=1

ts2 e−πn2t dt

t

=∫ ∞

0

ts2

θ(t)− 12

dt

t

=∫ ∞

1

ts2

θ(t)− 12

dt

t+∫ 1

0

ts2

θ(t)− 12

dt

t.

Now, as Re(s) > 1,

−12

∫ 1

0

ts2dt

t= −1

2· 2s· t s

2 |10 = −1s

Also,

12

∫ 1

0

ts2 θ(t)

dt

t=

12

∫ 1

0

ts2−

12 θ

(1t

)dt

t=

12

∫ ∞

1

t1−s2 θ(t)

dt

t

=∫ ∞

1

t1−s2

θ(t)− 12

dt

t+

12

∫ ∞

1

t1−s2

dt

t

and, as Re(s) > 1,

12

∫ ∞

1

t1−s2

dt

t=

t1−s2

1− s|∞1 =

1s− 1

Hence, for Re(s) > 1,

π−s2 Γ(s

2

)ζ(s) =

∫ ∞

1

ts2

θ(t)− 12

dt

t+∫ ∞

1

t1−s2

θ(t)− 12

dt

t−(

1s

+1

1− s

)

48

The two integrals converge for all s ∈ C, so the RHS is analytic on C \ {0, 1}, has

simple poles at s = 0, 1 and reflects into itself as s 7→ 1− s.

Now, let K be a number field with [K : Q] = n. We define the Dedekind zeta

function associated to K to be

ζK(s) =∑

a⊆OK

(Na)−s.

First, does this series converge?

Well, by unique factorization of the ideals, we may write

ζK(s) =∏p

(1−Np−s

)−1

and this converges absolutely iff∑

p

1Nps

converges absolutely. There are at most

n primes p lying above each rational prime ideal (p) ⊇ Z, and Np is smallest when

(p) splits completely, i.e. when (p) =n∏

i=1

p(i) with Np = p.

So∑

p

1Nps

≤ n∑

p

1ps

Therefore, ζK(s) converges for Re(s) > 1 also.

For the functional equation of ζK(s) we will require a theta function that satisfies

itself a suitable function equation, and for this we will need an n-dimensional

Poisson summation formula.

So, let ~x = (x1, . . . , xn). If f(~x) is an infinitely differentiable function such that

f(~x) → 0 as |~x| → ∞ sufficiently fast, then we have the n-dimensional Poisson

summation formula:

∑~m

f(~m) =∑~m

∫Rn

f(~x)e−2πi~m·~xd~x

49

Definition 30. Let P be a real n× n symmetric matrix such that, for

x =

x1

...xn

, txPx is a positive definite quadratic form in n variables.

Then, for u =

u1

...un

and v =

v1

...vn

we define the generalized theta function

Θ(P, u, v) =∑

m=

0@m1

...mn

1Ae−π t(m+u)P (m+u)+2πi tvm

Note that in this case all the eigenvalues of P are positive and there exists an

orthogonal matrix G such that tGPG =

e1

. . .en

, and, from this, there

exists N such that tNPN = I.

First we have to address the question of convergence for the series defining the

theta function.

First note that for any such matrix P , the set {txPx; |x| = 1} has a minimum

d(P ) > 0

Fix P0 a real symmetric positive definite n× n matrix and some y0 > 0. Let

y ≥ y0 and P = yP0.

Clearly d(P ) = yd(P0).

Moreover

∣∣∣∣∣∣∑

m∈Zn\{u}

e−π t(m+u)P (m+u)+2πi tvm

∣∣∣∣∣∣ < c1e−c2y for some constants c1, c2 > 0

To see this, first look at the exponent in the expression of the theta function and

use the following facts:

∣∣tmPm∣∣ ≥ yd(P0) |m|2

50

and

t(m + u)P (m + u) = tmPm +tmPu +tuPm +tuPu.

So, ∣∣∣∣∣∣∑

m∈Zn\{u}

e−π t(m+u)P (m+u)+2πi tvm

∣∣∣∣∣∣ ≤ c

∣∣∣∣∣∣∑

m∈Zn\{u}

e−πyd(P0)|m|2−πc(u,v)|m|

∣∣∣∣∣∣and the assertion follows.

Theorem 54.

Θ(P, u, v) =e−2πiu·v√

detPΘ(P−1, v,−u) =

e−2πiu·v√

detPΘ(P−1,−v, u)

Proof: By Poisson summation

Θ(P, u, v) =∑m

∫Rn

exp[−π t(w + u)P (w + u) + 2πi tvw − 2πi tmw

]dw

=∑m

∫Rn

exp[−π(t(w + u)P (w + u)− 2i t(v −m)w

)]dw

Set x = w + u− iP−1(v −m).

We have thattxPx = t

(w + u− iP−1(v −m)

)P(w + u− iP−1(v −m)

)= t(w + u)P (w + u)− 2i t(v −m)(w + u)−t(v −m)P−1(v −m)

So,

−π(t(w + u)P (w + u)− 2i t(v −m)w

)= −π

(txPx + 2i t(v −m)u +t(v −m)P−1(v −m)

)and

Θ(P, u, v) = e−2πi tuv∑m

exp[−π t(m− v)P (m− v) + 2πi tum

] ∫Rn

exp(−π txPx)dw

Recall that x is a function of u, v, P and w.

Denote Im(P, u, v) =∫

Rn

exp(−π txPx)dw.

This is an analytic function in each of the 2n variables making up u and v.

Setting v = m + iρ, we get x = w + u + P−1ρ, but the integral does not change as

ρ varies.

51

Now Im(P, u, v) is independent of u and ρ, for all real u, ρ, and hence for all

complex u, ρ. Thus Im(P, u, v) is independent of u and v.

So,

Im(P, u, v) =∫

Rn

exp(−π twPw)dw.

To compute its value, choose N such that tNPN = I. Then det N2 det P = 1 and

making the change of variables w = Nz, whose Jacobian is |det N | = detP−1/2,

we get

∫Rn

exp(−π twPw)dw =1√

det P

∫Rn

exp(−π tz z)dz =1√

detPc.

Setting P = I, u = v = 0, we obtain Θ(I, 0, 0) = cΘ(I−1, 0, 0), hence c = 1 and

Θ(P, u, v) =e−2πiu·v√

detPΘ(P−1,−v, u)

Replacing m by −m in the definition of Θ, we get that

Θ(P−1,−v, u) = Θ(P−1, v,−u), and we are done.

Let K be a number field of degree n, r1 the number of real embeddings of K into

C and 2r2 the number of the complex embeddings. By convention

K(r1+r2+j) = K(r1+j), for all j = 1, r2.

Let r = r1 + r2 − 1 and define

ei =

{1 , if 1 ≤ i ≤ r1

2 , if r1 < i ≤ n

e = er+1 =

{1 , if K is totally real, i.e.r2 = 02 , otherwise

and

D = DK

For an ideal b of K let C(b) =(|D|Nb2

)− 1n .

To prove the functional equation of ζK(s) we will look at

52

ϕ(x, t1, . . . , tr, b) =∑δ∈b

exp

[−πxC(b)

(r1∑

i=1

(δ(i))2

ti + 2r1+r2∑

i=r1+1

∣∣∣δ(i)∣∣∣2 ti

)]

where x > 0, ti > 0 for 1 ≤ i ≤ r, and tr+1 is defined byr+1∏i=1

teii = 1.

Note that

ϕ(x, t1, . . . , tr, b) =∑δ∈b

exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 ti

]

By setting tr1+r2+i = tr1+i for i = 1, . . . , r2, we get thatn∏

i=1

ti = 1 and we may

write

ϕ(x, t1, . . . , tr, b) =∑δ∈b

exp

[−πxC(b)

n∑i=1

∣∣∣δ(i)∣∣∣2 ti

]

Claim ϕ(x, t1, . . . , tr, b) = ϕ(x−1, t−11 , . . . , t−1

r ,D−1K b−1)x−

n2 .

Proof: Write b = [α1, . . . , αn]Z. Hence any element δ ∈ b is of the form

δ = (α1, . . . , αn)m for some m ∈ Zn. Thenr+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 ti is a positive definite

quadratic form in m, as

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 ti =

n∑i=1

∣∣∣δ(i)∣∣∣2 ti = tδ

t1. . .

tn

δ

where δ =

δ(1)

...δ(n)

.

Since δ = (α1, . . . , αn)m we get that

(2)r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 ti =tm t

(α

(i)j

) t1. . .

tn

(α(i)j

)m

Let P0 = C(b) t(α

(i)j

) t1. . .

tn

(α(i)j

)

53

Then, clearly, P0 = tP 0 and P0 = P 0, so P0 is real and symmetric. Letting

P = xP0, we have that

ϕ(x, t, b) = Θ(P, 0, 0) =1√

det PΘ(P−1, 0, 0)

Now, det P = xn detP0 and det P0 = C(b)n∣∣∣det

(α

(i)j

)∣∣∣2But

(α

(i)j

)=(w

(i)j

)M , where w1, . . . , wn is a Z-basis for OK and |detM | = Nb,

so∣∣∣det

(α

(i)j

)∣∣∣2 =∣∣∣det

(w

(i)j

)∣∣∣2 Nb2 = |D|Nb2.

Since, C(b)n =(|D|nb2

)−1, it follows that detP0 = 1, and detP = xn.

Now,

P−10 = C(b)−1

(α

(i)j

)−1

t−11

. . .t−1n

t(α

(i)j

) −1

= C(b)−1(α

(i)j

) −1

t−11

. . .t−1n

t(α

(i)j

)−1

But(α

(i)j

)−1

=(β

(j)i

)and t

(α

(i)j

)−1

=(β

(i)j

), where [β1, . . . , βn] = b−1D−1

K .

So

P−10 = C(b)−1 t

(β

(i)j

) t−11

. . .t−1n

(β(i)j

)

Also,

C(b)−1 =(

1|D|Nb2

)− 1n

=

(|D|

|D|2 Nb2

)− 1n

=(

|D|(NDb)2

)− 1n

=(|D| (ND−1b−1)2

)− 1n = C(b−1D−1)

54

So, P−10 = C(b−1D−1) t

(β

(i)j

) t−11

. . .t−1n

(β(i)j

)which is obtained out of x−1, t−1

1 , . . . , t−1r and the ideal b−1D−1 just as P0 was

constructed out of x, t1, . . . , tr and the ideal b.

Also, P−1 = x−1P−10 .

Now, Θ(P−1, 0, 0) =∑m

exp(−πx−1 tmP−10 m),

so

ϕ (x, t1, . . . , tr, b) =1√

det PΘ(P−1, 0, 0)

= x−n2

∑δ∈b−1D−1

K

exp

[−π x−1C(b−1D−1)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 t−1

i

]

= x−n2 ϕ(x−1, t−1

1 , . . . , t−1r , b−1D−1

),

as claimed.

Now, write

ζK(s) =∑

a

(Na)−s =∑C

ζ(s, C),

where

ζ(s, C) =∑

a⊆OKa∈C

(Na)−s

for every ideal class C ∈ ClK .

Fix such a C. Note that the series defining ζ(s, C) converges for Re(s) > 1.

Let b be an ideal in C−1. Then for any ideal a of K we have

a ∈ C ⇔ ab = (δ) for some δ ∈ K.

55

Also, if ab = (δ), then a is an integral ideal of K iff δ ∈ b, in which case

(Na)−s = |Nδ|−s (Nb)s.

So,

ζ(s, C) = (Nb)s∑

(δ)⊆OK

δ∈b

|Nδ|−s.

Then,

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

=[22r2πn |D|−1

Nb−2]− s

2Γ(s

2

)r1

Γ(s)r2∑

(δ)⊆OK

δ∈b

|Nδ|−s =

=([πC(b)]r1 [2πC(b)]2r2

)− s2

Γ(s

2

)r1

Γ(s)r2∑

(δ)⊆OK

δ∈b

|Nδ|−s

For, 1 ≤ i ≤ r1 we have

[πC(b)]−s2 Γ(s

2

)[∣∣∣δ(i)∣∣∣2]− s

2

=∫ ∞

0

xs2i exp

[−πxiC(b)

∣∣∣δ(i)∣∣∣2] dxi

xi

and for r1 + 1 ≤ i ≤ r1 + r2 we have

[2πC(b)]−s Γ(s)[∣∣∣δ(i)

∣∣∣2]−s

=∫ ∞

0

xsi exp

[−2πxiC(b)

∣∣∣δ(i)∣∣∣2] dxi

xi

Here we used the fact that for any real number a and any s ∈ C,

a−sΓ(s) =∫ ∞

0

xse−ax dx

x.

56

Multiplying and summing over (δ) we obtain(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

∫ ∞

0

· · ·∫ ∞

0︸ ︷︷ ︸r1+r2 fold

(r1+r2∏i=1

xeii

) s2 ∑

(δ)δ∈b\{0}

exp

[−πC(b)

r1+r2∑i=1

ei

∣∣∣δ(i)∣∣∣2 xi

]r1+r2∏i=1

dxi

xi

We want to make the change of variables xn =r1+r2∏i=1

xeii and ti = xi

x for

1 ≤ i ≤ r = r1 + r2 − 1.

Also, set tr+1 = tr1+r2 = xr1+r2x .

Note thatr+1∏i=1

teii = x−n

r1+r2∏i=1

xeii = 1.

To find the Jacobian of this transformation, look at

log ti = log xi − log x for 1 ≤ i ≤ r

log x =r+1∑i=1

ei

nlog xi

We are interested in these because our measure was dxi

xi= d log xi.

Now,

∣∣∣∣∂ (log ti, log x)∂(log xi)

∣∣∣∣ =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1− e1n − e2

n · · · − er

n − er+1n

− e1n 1− e2

n · · · − er

n − er+1n

......

......

− e1n − e2

n · · · 1− er

n − er+1n

e1n

e2n · · · er

ner+1

n

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣∣∣∣

11

. . .1

e1n

e2n · · · er

ner+1

n

∣∣∣∣∣∣∣∣∣∣∣=

er+1

n=

e

n

and therefore

57

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

n

e

∫ ∞

0

xns/2

∫ ∞

0

· · ·∫ ∞

0︸ ︷︷ ︸r fold

∑(δ)

exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 ti

]r∏

i=1

dtiti

dx

x

We would like to be able to sum over all the elements δ ∈ b, but for this we need

to be understand what happens when δ is multiplied by some unit of K.

Let w = w(K) denote the number of roots of unity in K. If we count each ideal w

times in the formula above, we get

w

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

n

e

∫ ∞

0

xns/2

∫ ∞

0

· · ·∫ ∞

0︸ ︷︷ ︸r fold

∑(δ)⊆b

′exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 ti

]r∏

i=1

dtiti

dx

x

where∑′ means that each ideal (δ) is counted w times, i.e. if δ is used, then µδ is

used for each root of unity µ ∈ K.

We now begin a detour to deal with other units.

Theorem 55. (Minkowski) If S is a measurable convex body in Rn, symmetric

about 0, and with volume |S| > 2n, then S contains a point of Zn other than 0.

Proof: First assume that S is bounded. We know that x ∈ S ⇒ −x ∈ S, and that

x, y ∈ S ⇒ λx + (1− λ)y ∈ S for 0 ≤ λ ≤ 1.

Define

58

g(x) =

{1 , if x ∈ S

0 , if x /∈ S

and

f(x) =∑

m∈Zn

g(2x− 2m).

As S is bounded, only finitely many terms of the sum are nonzero.

Now use the fact that∫

C

|f(x)|2 dx ≥∣∣∣∣∫

C

f(x)dx

∣∣∣∣2, where C is the unit cube in

Rn.

We have∫C

f(x)dx =∑m

∫C

g(2x− 2m)dxy=x−m

=∑m

∫C−m

g(2y)dy

=∫

Rn

g(2y)dy = 2−n

∫Rn

g(y)dy = 2−n |S|

On the other hand,∫C

|f(x)|2 dx =∑m′

∑m

∫C

g(2x− 2m)g(2x− 2m′)dx

=∑m

∫C

g(2x− 2m)2dx +∑

m6=m′

∫C

g(2x− 2m)g(2x− 2m′)dx

But, just as before,

∑m

∫C

g(2x− 2m)2dx =∑m

∫C

g(2x− 2m)dx = 2−n |S|

and

∣∣∣∣∫C

f(x)dx

∣∣∣∣2 =(2−n |S|

)2> 2−n |S| , since |S| > 2n

Thus,

∑m6=m′

∫C

g(2x− 2m)g(2x− 2m′)dx > 0

59

and so there exist x and m′ 6= m such that g(2x− 2m)g(2x− 2m′) 6= 0, so

2x− 2m ∈ S and 2x− 2m′ ∈ S. Thus, by symmetry, 2m′ − 2x ∈ S, and, by

convexity, m′ −m = 12 (2x− 2m) + 1

2 (2m′ − 2x) ∈ S with m′ −m 6= 0.

If S is not bounded, approximate it by S′ bounded, with S′ ⊆ S and |S′| > 2n.

Corollary 56. If S is compact, |S| ≥ 2n suffices.

Corollary 57. Let txPx be a positive definite quadratic form in n variables (P

symmetric). Then txPx ≤ C has a solution in Zn \ {0} provided that

C ≥ 4(

detP

v2n

)1/n

, where vn =πn

Γ(1 + n

2

) denotes the volume of the

n-dimensional unit sphere.

Proof: Let S = {x ∈ Rn; txPx ≤ C}. This is clearly compact, convex and

symmetric about the origin. If tNPN = I and Ny = x, then∣∣∣∣dx

dy

∣∣∣∣ = |det N | = 1√det P

.

So,

|S| =∫

S

1dx = (det P )−1/2

∫tyy≤C

1dy =√

Cn

√detP

vn

Everything works just fine if |S| ≥ 2n, so we require that C ≥ 4(detP )2/n

v2/nn

.

Corollary 58. (Dirichlet’s theorem on linear forms) Let A = (aij) ∈ GL(n, R). Ifn∏

i=1

bi ≥ |det A|, then the system of inequalities

∣∣∣∣∣∣n∑

j=1

aijxj

∣∣∣∣∣∣ ≤ bi

has a solution in Zn other than 0.

Proof: Let S = {x; |∑

aijxj | ≤ bi for all i}.

This set is clearly compact, convex and symmetric about 0.

Also, |S| =∫

S

1 dx.

60

Let y = Ax. Then∣∣∣ dydx

∣∣∣ = |detA|, so

|S| = 1|detA|

∫|yi|≤bi

dy =2n∏

bi

|det A|≥ 2n

Apply corollary 56 to the set S and we are done.

Corollary 59. Suppose λ1, . . . , λn are real numbers. Given ε > 0, there exist

integers x, y1, . . . , yn not all zero such that |xλi − yi| ≤ ε for 1 ≤ i ≤ n and

|x| ≤ ε−n.

Proof: Consider the (n + 1)× (n + 1) system of inequalities

|yi − xλi| ≤ ε for 1 ≤ i ≤ n

|x| ≤ ε−n

This has determinant

∣∣∣∣∣∣∣∣∣∣∣

1 −λ1

1 −λ2

. . .1 −λn

1

∣∣∣∣∣∣∣∣∣∣∣= 1

and∏

bi = 1, so we may apply the previous corollary.

Theorem 60. (Kronecker) Let m be a positive integer. There exists a constant

δ(m) > 0 such that if α is a nonzero algebraic integer of degree m and

|α(i)| ≤ 1 + δ(m) for 1 ≤ i ≤ m, then α is a root of unity.

Proof: Let α be such an algebraic integer and let k = Q(α).

As each |α(i)| ≤ 1 + δ(m) and N(α) =∣∣∏α(i)

∣∣ ≥ 1, it follows that

|α(i)| ≥ 1(1 + δ(m))m−1 for each i

61

Now consider a ball U of radius ε around 1 in the complex plane and pick l large

enough such that the set V = {z ∈ C; |arg z| ≤ 2πl } intersects U but does not

contain it. Here arg z ∈ (−π, π] for all complex numbers z.

Want to choose integers x, yi such that for all i we have∣∣x · arg(α(i)

)− 2πyi

∣∣ ≤ 2πl . By corollary 59 we can do this with |x| ≤ lm and we

can choose them such that not all x, y1, . . . , yn are zero. To see this take

λi =arg(α(i))

2π for all i in corollary 59.

This means that, for each i, the complex number(α(i)

)x

lies inside V . Since, for

all i,

1(1 + δ(m))m−1 ≤ |α

(i)| ≤ 1 + δ(m)

and

|x| ≤ lm

we may pick δ(m) such that(α(i)

)x

∈ U for all i. Note that δ may be chosen so

that it depends only on m and l, and that l depends only on ε. So δ depends only

on m and ε

Look at the polynomialm∏

i=1

(T −

(α(i)

)x)This has rational integer coefficients, and, if the neighborhood U of 1 we started

with is small enough, these coefficients are within 12 of the coefficients of

m∏i=1

(T − 1), i.e. pick ε small enough such that the polynomialm∏

i=1

(T − λi) has

coefficients within 12 of the coefficients of

m∏i=1

(T − 1) for any λ1, . . . , λm ∈ U .

So T − 1 = T −(α(i)

)xand therefore

(α(i)

)x= 1 for each i. In particular, αx = 1.

To complete the proof, notice that x 6= 0, since otherwise yi = 0 for all i, which

contradicts our choice.

For α 6= 0 in K we can define a map f(α) =(e1 log |α(1)|, . . . , er log |α(r)|

)

62

Theorem 61. If ε is a unit of OK and f(ε) = 0, then ε is a root of unity.

Proof: Since ε is a unit, we have

1 = Nε =n∏

i=1

|ε(i)| =r+1∏i=1

|ε(i)|ei

so,r+1∑i=1

ei log |ε(i)| = 0

.

By hypothesis, log |ε(i)| = 0 for 1 ≤ i ≤ r, so er+1 log |ε(r+1)| = 0.

Hence log |ε(i)| = 0 for all 1 ≤ i ≤ n, and, by Kronecker’s Theorem (theorem 60),

ε is a root of unity.

Proposition 62. Let u ≥ 0 and let η1, . . . , ηu be units of K with f(η1), . . . , f(ηu)

linearly independent over R. Denote by Vu the vector space spanned by

f(η1), . . . , f(ηu). Then there exist ε1, . . . , εu ∈ O×K such that every unit ε of K

with f(ε) ∈ Vu can be written uniquely as

ε = wεa11 · · · εau

u with w a root of unity in K and a1, . . . , au ∈ Z.

Proof: Let Bu =

u∑

j=1

xjf(ηj); xj ∈ R, 0 ≤ xj ≤ 1

.

Suppose, ε is a unit of OK with f(ε) ∈ Vu. Then there exist m1, . . . ,mu ∈ Z such

that

f

(ε

ηm11 · · · ηmu

u

)∈ Bu.

If {εl}l≥1 is a sequence of units of K with f(εl) ∈ Bu for l ≥ 1, then the sequence

{f(εl)}l≥1 has a limit point, so there exists a subsequence with∣∣f(εlj )− f(εlm)∣∣ −→ 0 as j, m →∞.

63

But, by Kronecker’s Theorem, eventually f(εlj ) = f(εlm), and so, by the above,

there are only finitely many distinct f(ε)’s inside Bu. They are given by

f(η1), . . . , f(ηu), f(ηu+1), . . . f(ηu+A), for some A ≥ 0.

If f(ε) ∈ Vu, then there are integers m1, . . . ,mu such that f(

εη

m11 ···ηmu

u

)= f(ηa)

for some 1 ≤ a ≤ u + A, so f(ε) is a Z-combination of f(η1), . . . , f(ηu+A). Hence

Vu is finitely generated over Z.

Now pick ηu+a for any 1 ≤ a ≤ A. For each l ∈ Z, f(ηlu+a) translates by integer

multiples of f(η1), . . . , f(ηu) to some f(εl) ∈ Bu. So for two distinct l1 and l2 will

have the same εl.

Then

f(εl) = f

(ηl1

u+a

ηm1(1)1 · · · ηmu(1)

u

)= f

(ηl2

u+a

ηm1(2)1 · · · ηmu(2)

u

)

So, (l1 − l2)f(ηu+a) =u∑

i=1

(mi(1)−mi(2)) f(ηi).

Thus,{f(ε); ε ∈ O×K , f(ε) ∈ Vu

}is u-dimensional over Q and, at the same time,

finitely generated over Z. This implies that it is actually u-dimensional over Z.

Therefore, it admits a Z-basis f(ε1), . . . , f(εu).

Hence if f(ε) ∈ Vu, then f(

εε

m11 ···εmu

u

)= 0 for some mj ∈ Z, so ε = wεm1

1 · · · εmuu

for some root of unity w.

Moreover, if w1εm1(1)1 · · · εmu(1)

u = w2εm1(2)1 · · · εmu(2)

u , then mi(1) = mi(2) for all i

and w1 = w2.

We will now prove by induction that

For any 0 ≤ u ≤ r there exist units η1, . . . , ηu ∈ OK so that f(η1), . . . , f(ηu) span

a subspace Vu of dimension u in Rr.

This is certainly true for u = 0.

We then know that there exist ε1, . . . , εu s.t. f(ε) ∈ Vu iff ε = wn∏

j=1

εaj

j , aj ∈ Z.

Set ti = exp(

2ei

vi

)for 1 ≤ i ≤ r + 1 . Then dti

ti= 2

eidvi for all 1 ≤ i ≤ r.

64

Notice that, sincer+1∏i=1

teii = 1, we have

r+1∑i=1

vi = 0.

There are two possibilities:

• K is totally real.

Then r = r1 − 1 and e1 = e2 = . . . = er = e = 1, so

n

e

r∏i=1

dtiti

= 2r1−1nr∏

i=1

dvi.

• K is not totally real.

Then r ≥ r1 and e = 2. Also,dtiti

=

{2dvi , for 1 ≤ i ≤ r1

dvi , for i > r1

.

So,

n

e

r∏i=1

dtiti

= 2r1−1nr∏

i=1

dvi.

Hence

w

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

2r1−1n

∫ ∞

0

xns2

∫ ∞

−∞. . .

∫ ∞

−∞︸ ︷︷ ︸r fold

∑(δ)⊆b

′exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 exp

(2vi

ei

)] r∏i=1

dvidx

x

Let f(ε1), . . . , f(εu), ~wu+1, . . . , ~wr be a basis for Rr and let

V ′u =

u∑

j=1

xjf(εj) +r∑

j=u+1

yj ~wj ; yj ∈ R, 0 ≤ xj ≤ 1

.

Claim

w

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

2r1−1n

∫ ∞

0

xns2

∫V ′

u

∑(δ)⊆b

′′exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 exp

(2vi

ei

)] r∏i=1

dvidx

x,

65

where∑′′ means that if δ ∈ b is counted, then δε is also counted, for all ε with

f(ε) ∈ Vu.

Proof: Fix δ ∈ b and look at

∫V ′

u

∑ε

f(ε)∈Vu

exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)ε(i)∣∣∣2 exp

(2vi

ei

)] r∏i=1

dvi.

This may be written as

∫V ′

u

∑ε

f(ε)∈Vu

exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 exp

(2ei

(vi + ei log |ε(i)|

))] r∏i=1

dvi

Let v′i = vi + ei log |ε(i)| for 1 ≤ i ≤ r + 1, so v′ = v + f(ε).

Note that we still haver+1∑i=1

v′i =r+1∑i=1

vi + log (Nε) = 0 + log 1 = 0.

The above integral is therefore equal to

∑ε

∫V ′

u+f(ε)

exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 e

2viei

]r∏

i=1

dvi =

∫ ∞

−∞. . .

∫ ∞

−∞︸ ︷︷ ︸r fold

exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 exp

(2vi

ei

)] r∏i=1

dvi

Now, if every unit ε has f(ε) ∈ Vu, then∑(δ)⊆b

′′=

∑δ∈b\{0}

Suppose this is true and u < r.

Now, let (v1, . . . , vr) be a point in the interior of V ′u and take a neighborhood of

radius ρ of this point lying inside V ′u with ρ < 1

2 |~wr| .

We had, by (2),

C(b)r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 exp

(2vi

ei

)= tmP0m, m ∈ Zn,

66

where P0 = C(b) t(α

(i)j

) t1. . .

tn

(α(i)j

).

Here detP0 is a constant independent of the vi’s.

Thus, by corollary 57, if we pick an appropriate constant C, there is an m 6= 0

with |tmP0m| < C for each point in the neighborhood of radius ρ.

So,

∫. . .

∫︸ ︷︷ ︸

neighborhood

≥ e−cx · ( volume of the neighborhood )

Now, instead of (v1, . . . , vr) use (v1, . . . , vr) + l ~wr, for any l ∈ Z, and the same

radius ρ. Add these up and the integral diverges (because we are summing over

all δ ∈ b \ {0}, so over all m ∈ Zn \ {0}, including the relevant one). This

contradicts the fact that the series defining ζK(s) converges for s real, s > 1.

Thus, if u < r, then there exist some unit ε with f(ε) /∈ Vu. Throw this in, and

eventually get u = r. This concludes the proof by induction.

Therefore f(O×K) is an r-dimensional lattice in Rr.

We have therefore proven the following result:

Theorem 63. (Dirichlet Unit Theorem)

There exist r units ε1, . . . , εr of K such that every unit ε can be expressed

uniquely in the form ε = wεa11 · · · εar

r , where w is some root of unity and ai ∈ Z,

and such that the vectors f(ε1), . . . , f(εr) span Rr.

We now have

67

w

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

2r1n

2

∫ ∞

0

xns2

∫V ′

r

∑δ∈b\{0}

exp

[−πxC(b)

r+1∑i=1

ei

∣∣∣δ(i)∣∣∣2 exp

(2vi

ei

)] r∏i=1

dvidx

x,

where V ′r =

r∑

j=1

xjf(εj); 0 ≤ xj ≤ 1

.

Definition 31. |V ′r | =

∣∣∣det(ei log |ε(i)

j |)∣∣∣ = R(K) is called the regulator of K.

In terms of the function ϕ defined previously,

w

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

2r1n

2

∫ ∞

0

xns2

∫V ′

r

[ϕ(x, t1, . . . , tr, b)− 1]r∏

i=1

dvidx

x

Recall that

ϕ(x, t1, . . . , tr, b) = x−n2 ϕ

(1x

,1t1

, . . . ,1tr

,D−1b−1

)So,

w

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

2r1n

2

∫ 1

0

xns2

∫V ′

r

[ϕ(x, t, b)− 1] +∫ ∞

1

xns2

∫V ′

r

[ϕ(x, t, b)− 1]

Now,

∫ 1

0

xns2

∫V ′

r

[ϕ(x, t, b)− 1]r∏

i=1

dvidx

x=∫ 1

0

xns2

∫V ′

r

ϕ(x, t, b)r∏

i=1

dvidx

x− 2

nsR

as

∫ 1

0

xns2

∫V ′

r

r∏i=1

dvidx

x=

2R

nswhen Re(s) > 1.

Hence,

68

∫ 1

0

xns2

∫V ′

r

[ϕ(x, t, b)− 1]r∏

i=1

dvidx

x=

∫ 1

0

xns2

∫V ′

r

x−n2 ϕ(x−1, t−1, b−1D−1)

r∏i=1

dvidx

x− 2

nsR =

∫ ∞

1

xn2 (1−s)

∫V ′

r

ϕ(x, t−1, b−1D−1)r∏

i=1

dvidx

x− 2

nsR

Now, replacing t by t−1 is the same as replacing vi by −vi, which means

integrating over −V ′r , which is the same as integrating over V ′

r , as we need simply

replace δ by δ(unit).

So, in place of ϕ(x, t−1, b−1D−1) we’ll have ϕ(x, t, b−1D−1).

Also, for Re(s) > 1,

∫ ∞

1

xn2 (1−s)

∫V ′

r

r∏i=1

dvidx

x= − 2R

n(1− s)

We therefore get that

w

(|D|

22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C) =

2r1n

2

[∫ ∞

1

xns2

∫V ′

r

(ϕ(x, t, b)− 1)r∏

i=1

dvidx

x+

+∫ ∞

1

xn(1−s)

2

∫V ′

r

(ϕ(x, t, b−1D−1)− 1

) r∏i=1

dvidx

x− 2R

ns(1− s)

]

Set

g(s, C) =(

|D|22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζ(s, C).

Thus g(s, C) has meromorphic continuation to the entire s-plane, analytic

everywhere except for simple poles at s = 0 and s = 1.

Ress=1 g(s, C) =2r1R

w

69

Also, since(b−1D−1

)−1D−1 = b, we have

g(s, C) = g(1− s, C′),

where b ∈ C−1 and b−1D−1 ∈ C′−1, i.e. C′ = C−1 · (the class of D).

Since g(s, C) is analytic everywhere except for simple poles at s = 0, 1, it follows

that ζ(s, C) is analytic everywhere except for a simple pole at s = 1 and

Ress=1 ζ(s, C) =2r1R

w

(22r2πn

|D|

)1/2

Γ(

12

)−r1

Γ(1)−r2

=2r1+r2Rπ

n2−

r12

w√|D|

=2r1+r2πr2R

w√|D|

Here we used the fact that Γ(

12

)=√

π and Γ(1) = 1.

Since g(s, C) has a simple pole at s = 0 and Γ(

s2

)r1 Γ(s)r2 contributes with a pole

of order r1 + r2, it follows that ζ(s, C) must have a zero of order r at s = 0.

If s is real and s > 1, we have g(s, C) > 2r1Rw , as the integral is positive.

So,2r1R

wh <

∑C

g(s, C) =(

|D|22r2πn

)s/2

Γ(s

2

)r1

Γ(s)r2ζK(s)

which exists for real s > 1, and thus the class number must be finite.

Finally,

g(s) =∑C

g(s, C) =∑C′

g(1− s, C′) = g(1− s)

and

Ress=1 ζK(s) =2r1+r2πr2Rh

w√|D|