Number Fields

Lectured by M. Strauch

Lent Term 2007

Introduction 1

1 The Ring of Integers OK 3

2 Ideals 9

3 Lattices 14

4 Geometry of Numbers 16

5 The Class Number 19

6 Dirichlet’s Unit Theorem 21

7 The Fundamental Unit Of

A Real Quadratic Field 24

8 Factorisation of Prime

Numbers in Number Fields 26

9 Cyclotomic Fields 21

Examples Sheets

Last updated: Wed 2nd Oct, 2013

Please let me know of any corrections: glt1000@cam.ac.uk

Course schedule

NUMBER FIELDS (D) 16 lectures, Lent term

Part IB Groups, Rings and Modules is essential and Part II Galois Theory is desirable.

Definition of algebraic number fields, their integers and units. Norms, bases and discrimi-

nants. [3]

Ideals, principal and prime ideals, unique factorisation. Norms of ideals. [3]

Minkowski’s theorem on convex bodies. Statement of Dirichlet’s unit theorem. Determina-

tion of units in quadratic fields. [2]

Ideal classes, finiteness of the class group. Calculation of class numbers using statement of

the Minkowski bound. [3]

Dedekind’s theorem on the factorisation of primes. Application to quadratic fields. [2]

Discussion of the cyclotomic field and the Fermat equation or some other topic chosen by

the lecturer. [3]

Appropriate books

Z.I. Borevich and I.R. Shafarevich Number Theory. Elsevier 1986 (out of print).

J. Esmonde and M.R. Murty Problems in Algebraic Number Theory. Springer 1999 (£38.50

hardback).

E. Hecke Lectures on the Theory of Algebraic Numbers. Springer 1981 (out of print)

D.A. Marcus Number Fields. Springer 1977 (£30.00 paperback).

I.N. Stewart and D.O. Tall Algebraic Number Theory and Fermat’s Last Theorem. A K

Peters 2002 (£25.50 paperback)

Introduction

Some history

Diophantus of Alexandria, 3rd century AD. Wrote a book on ‘Diophantine equations’ – poly-

nomial equations with integral coefficients. E.g. xn + yn = zn (∗)ax2 + bxy + cy2 = 0, a, b, c ∈ Z (∗∗)

Pierre de Fermat, 1601–1665. Lawyer, but hobby mathematician, famous for his claim that

(∗) does have non-zero integral solutions if n > 3.

L. Euler, 1707–1783. Among others, discovered (instances of) the Law of Quadratic Reci-

procity: for odd primes p and q, one has that x2 − q ≡ 0 (mod p) has an integral solution ⇔x2 − p ≡ 0 (mod q)

{has a solution if p ≡ 1 (mod 4) or q ≡ 1 (mod 4)does not have a solution if p ≡ q ≡ 3 (mod 4)

.

A.M. Legendre, 1752-1833. Worked on QRL, but did not give full proof.

C.F. Gauss, 1777-1855. Among others, studied quadratic forms like (∗∗) and proved QRL

(first publication, ‘Disquisitione Arithmeticae’, 1799). Introduced and studied the arithmetic

of the ‘Gaussian integers’, Z[i] = Z+ Zi (a unique factorisation domain).

P.G.J. Dirichlet, 1805-1859. Introduced ‘Dirichlet L-series’, L(χ, s) =∑∞

n=1

χ(n)ns , where

χ : Z → C. Showed that in every arithmetic progression {an + b}n>1, hcf(a, b) = 1, there

are infinitely many primes. Proved the ‘Dirichlet Unit Theorem’, giving a description of the

group of units of a ring of algebraic integers.

E. Kummer, 1810-1893. Studied Fermat’s equation (∗), and introduced the ring of ‘cyclo-

tomic integers’, Z[ζn] = Z+Zζn + · · ·+Zζnn , where ζn = e2πi/n. Kummer noticed that Z[ζn]

is in general not a UFD, but if it happens to be a UFD then Fermat’s claim is true. Kummer

was led to introduce ‘ideal numbers’, and a ‘measure’ of the failure of Z[ζn] being a UFD,

the so-called ‘class number’.

R. Dedekind, 1831-1916. Proved the finiteness of the class number for general rings of

algebraic integers.

From now on, n = p is an odd prime.

Theorem. Suppose p > 3 is an odd prime and Z[ζp] is a UFD. Then the Fermat equation

(∗) (for n = p) does not have a non-trivial solution (i.e. x, y, z 6= 0).

Sketch (Kummer, 1847). We only treat the case p ∤xyz, the so-called ‘first case’ of Fer-

mat’s Last Theorem. We may assume p > 5 (as p = 3 was known to Euler).

If x ≡ y and y ≡ −z (mod p) then −2zp ≡ zp (mod p), so p|3zp, //\\. So may assume

x 6≡ y (mod p).

Moreover, hcf(x, y) = 1.

1

Write zp = xp + yp =∏p−1

i=0 (x + ζipy) – follows from T p − 1 =∏p−1

i=0 (T − ζjp).

Observation: for all i 6= j, there exist a, b ∈ Z[ζp] such that a(x+ ζipy)+ b(x+ ζjpy) = 1.

Assume Z[ζp] is a UFD. Consider prime factor π of x + ζipy. It appears in zp, so it

divides z, so it appears with pth power on the LHS, so if appears with pth power on the

RHS. As all factors of the RHS are coprime, it appears with pth power in x + ζipy, so

there exists α ∈ Z[ζp] such that x+ ζpy = εαp, for ε ∈ Z[ζp]×. So x+ ζ−1

p y = εαp.

Lemma. (i) Any unit in Z[ζp] is a product ζrpε1, where ε1 = ε1, ε1 ∈ Z[ζp]×.

(ii) There exists c ∈ Z such that αp ≡ c (mod p).

Write ε = ζrpε1, so ζ−1p (x+ ζpy) = ε1α

p ≡ ε1C ≡ ε1C ≡ ε1 αp = ζrp(x + ζ−1

p y) mod p.

So x+ ζpy − ζ2rp x− ζ2r−1p y ≡ 0 mod p.

But any p− 1 elements of {1, ζp, ζ2p , . . ., ζp−1p } form a basis of Z[ζp] over Z as a module.

(a) #{1, ζp, ζ2rp , ζ2r−1p } = 4 ⇒ p | x, p | y. //\\

(b) 1 = ζ2rp then ζpy − ζ−1p y ≡ 0 mod p⇒ p | y. //\\

Last two cases (1 = ζ2r−1 or ζ = ζ2r−1) can be checked similarly.

2

1 : The Ring of Integers OK

Literature. J. Neukirch, Algebraic Number Theory, chapter 1.

Definition. (i) A number field is a finite field extension of Q. The elements of a number

field are called algebraic numbers.

(ii) If K/Q is a number field, an element α ∈ K is called an algebraic integer (or

integral) if α is a root of a monic non-zero polynomial f ∈ Z[X ], i.e. f(α) = 0.

Notation. OK = {α ∈ K : α is integral over Z}.

Definition. Let B be a ring (always commutative and with a 1) and A ⊆ B a subring. Then

b ∈ B is called integral over A if it satisfies an equation bn + a1bn−1 + . . .+ an = 0,

where n > 0 and all ai ∈ A.

B is called integral over A if all elements b ∈ B are integral over A.

Proposition 1.1. Finitely many elements b1, . . ., bn ∈ B are all integral overA iff the subring

A[b1, . . ., bn] (i.e., the smallest subring of B containing A and b1, . . ., bn) is finitely

generated over A as an A-module.

Lemma 1.2. Let S = (aij)16i,j6r be an (r × r)-matrix with coefficients in A. Put S∗ =

(s∗ij)16i,j6r , where s∗ij = (−1)i+j det(Sij), with (Sij) the matrix obtained from S by

deleting the ith column and jth row. Then S∗ S = S S∗ = det(S) Ir .

In particular, for any x =

x1...xr

∈ Br, if Sx = 0 then det(S)x = 0.

Proof. Column-row expansion of the determinant (Laplace’s theorem). 2

Proof of 1.1. Let b ∈ B be integral over A, so that f(b) = 0 for some monic polynomial

f ∈ A[X ], deg(f) = d > 0.

Recall Euclid’s Algorithm in A[X ]. For any g ∈ A[X ], there exists h, r ∈ A[X ] such

that g(X) = h(X)f(X) + r(X) with either r(X) ≡ 0 or deg(r) < d = deg(f). (Here

we use that f is monic.)

From this, we have g(b) = h(b)f(b) + r(b) = r(b) = a0 + a1b+ . . .+ ad−1bd−1, say.

Any element of A[b] is of the form g(b) with g ∈ A[X ], and hence A[b] is (as an A-

module) generated by 1, b, b2, . . ., bd−1.

By induction, R = A[b1, . . .bn−1] is a finitely-generated A-module. The argument above

gives that R[bn] is finitely generated over R, i.e. R[bn] =∑d−1

i=0 Rbin for some d > 0.

Hence A[b1, . . .bn] = R[bn] is finitely generated as an A-module.

3

Conversely, suppose A[b1, . . .bn] is a finitely-generated A-module. Let b ∈ A[b1, . . .bn]

and write A[b1, . . .bn] =∑r

j=1 Acr where c1, . . ., cr ∈ A[b1, . . .bn].

Write bci =∑r

j=1 aijcj , with aij ∈ A. Then(bIr − (aij)

)

c1...cr

=

0...0

.

Lemma 1.2 implies det(bIr − (aij)

)ck = 0 for k = 1, . . ., r, and this determinant is a

monic polynomial in b (of degree r).

Write 1 = α1c1 + . . .+ αrcr with α1, . . ., αr ∈ A.

Then det(bIr − (aij)

)= α1(det . . .)c1 + . . . + αr(det . . .)cr = 0. Hence every b ∈

A[b1, . . ., br] is integral over A. 2

Proposition 1.3. Let A ⊆ B be subrings of a ring C. If B is integral over A, and C is

integral over B, then C is integral over A.

Proof. Let c ∈ C. Then there is an equation cn + bn−1cn−1 + . . .+ bn = 0 with all bi ∈ B.

Put R = A[b1, . . ., bn], then R is a finitely-generated A-module (because B is integral

over A, and Proposition 1.1). So R[C] is a finitely-generated R-module (as C is integral

over R). Hence R[C] is a finitely-generated A-module and so c is integral over A (by

Proposition 1.1). 2

Conclusion. A = {b ∈ B : b is integral over A} is a subring of B, called the integral

closure of A in B. (If A = A, say that A is integrally closed in B.)

Proof. Let b1, b2 ∈ A. Then A[b1, b2] is a finitely-generated A-module.

Now, b1+ b2 and b1b2 ∈ A[b1, b2], so they satisfy an equation of the form xn+a1xn−1+

. . .+ an = 0. Hence they are integral over A. 2

In particular, for a number field K, the set OK is a ring – the integral closure of Z in K.

Proposition 1.4. Let A be a domain (i.e. with zero-divisors) andK = Frac(A), and assume

that A is integrally closed in K. Let L/K be a finite field extension, let β ∈ L, and let

p(X) ∈ K[X ] be the minimal polynomial of β.

Then β is integral over A if and only if p(X) ∈ A[X ].

Proof. (⇐) is by definition.

(⇒). Let g(X) ∈ A[X ] be a monic polynomial such that g(β) = 0. Then Euclid’s

Algorithm implies that p(X) divides g(X) in K[X ], say g(X) = h(X)p(X). Hence all

roots p in L are also roots of g, hence are integral over A. Then p(X) =∏

i(x − βi)

with all βi integral over A.

Hence the coefficients of p are integral over A. As A was assumed to be integrally closed

in K, we get p(X) ∈ A[X ]. 2

4

Example. If K = Q(i) = {a+ ib : a, b ∈ Q} then OK = Z[i].

Proof. Let a+ ib ∈ OK . If b = 0 then clearly a ∈ Z

If b 6= 0 then the minimal polynomial of a+ ib is X2− 2aX+(a2+ b2) which is in Z[X ]

by 1.4. So write a = a′/2 with a′ ∈ Z. And since (2b)2 = 4(a2 + b2)− (2a)2 ∈ Z, write

b = b′/2 with b′ ∈ Z. Then a′2 + b′2 = 4(a2 + b2) ∈ 4Z.

So a′ and b′ are even, and so a, b ∈ Z. 2

In the following, let L/K be a finite field extension.

Definition. The trace and norm of an element x ∈ L are, respectively, the trace and

determinant of the endomorphism Tx : L → L, a 7→ xa, where we consider L as a

K-vector space.

TrL/K(x) = Tr(Tx), NL/K(x) = det(Tx)

If fx(t) = det(t.id − Tx) = tn + a1tn−1 + . . . + an, where n = [L : K], is the characteristic

polynomial of Tx, then TrL/K(x) = −a1 and NL/K(x) = (−1)nan.

Because Tx+y = Tx + Ty and Txy = Tx ◦ Ty, we see that the trace and norm define homo-

morphisms , TrL/K : L→ K and NL/K : L∗ → K∗.

Without proof (see Neukirch, propositions 2.6 & 2.7), we cite the following:

Proposition 1.5. If L/K is separable and Σ = ΣL/K = {σ : L →֒ K field homomorphism

over K} is the set of different K-embeddings of L into an algebraic closure K, then

(i) fx(t) =∏

σ∈Σ(t− σ(x))

(ii) TrL/K(x) =∑

σ∈Σ σ(x)

(iii) NL/K(x) =∏

σ∈Σ σ(x)

If K ⊆ L ⊆ M is a tower of finite extensions (not necessarily separable), then TrM/K =

TrL/K ◦ TrM/L and NM/K = NL/K ◦NM/L.

Definition. The discriminant of a basis {α1, . . ., αn} of a finite separable extension L/K,

where n = [L : K], is defined by

d(α1, . . ., αn) = det((σi(αj))16i,j6n

)2

where {σ1, . . ., σn} = ΣL/K = {σ : L→ K}.

Write S =((σi(αj))16i,j6n

)=

σ1(α1) · · · σ1(αn)...

...σn(α1) · · · σn(αn)

. Then we have

det(S)2 = det(StS) = det((∑n

k=1 σk(αi)σk(αj))

16i,j6n

)

= det((TrL/K(αiαj)

)

16i,j6n

)

5

In particular, d(α1, . . ., αn) ∈ K.

Proposition 1.6. If L/K is separable, then the bilinear form L×L→ K given by (x, y) 7→TrL/K(xy) is non-degenerate, and the discriminant of a basis {α1, . . ., αn} is always

non-zero.

Proof. L/K is separable, so the theorem of the primitive element implies that there is θ ∈ L

such that L = K(θ). Then {1, θ, . . ., θn−1} is a basis for L/K, where n = [L : K].

The bilinear form (x, y) 7→ TrL/K(xy) has the following matrix with respect to this

basis

M =(TrL/K(θi−1θj−1)

)

16i,j6n

This bilinear form is non-degenerate iff det(M) 6= 0. By the calculation above,

det(M) = d(1, θ, . . ., θn−1) = det

1 1 · · · 1θ1 θ2 θn...

......

θn−11 θn−1

2 θn−1n

2

,

where θi = σi(θ), for {σ1, . . ., σn} = ΣL/K .

This Vandermonde determinant is equal to∏

i<j(θi−θj), and this is non-zero since

there are n different K-embeddings Lσi−→ K, and so θi 6= θj for i 6= j.

For an arbitrary basis {α1, . . ., αn}, there is an invertible matrix N ∈ GLn(K) such

that

α1

...αn

= N

θ1...θn

. Thus d(α1, . . ., αn) = det(N)2d(1, θ, . . ., θn−1) 6= 0. 2

Our general setting will be:

• A is an integral domain with field of fractions K, with A integrally closed in K.

• L/K is a finite separable extension, and Σ = ΣL/K = {σ : LK→֒ K}.

• B is the integral closure of A in L.

Obseration. If x ∈ L is integral over A then for all σ ∈ Σ, σ(x) ∈ σ(L) ⊆ K is integral over

A. Hence TrL/K(x) =∑

σ∈Σ σ(x) ∈ A and NL/K(x) =∏

σ∈Σ σ(x) ∈ A. (Here we use that

A is integrally closed.)

Moreover, x ∈ B∗ iff NL/K(x) ∈ A∗. (Where x ∈ B∗ ⇔ ∃y ∈ B : xy = 1).

(⇒) 1 = N(1) = N(xy) = N(x)N(y).

(⇐) There exists a ∈ A such that 1 = NL/K(x).a =(∏

σ∈Σ σ(x))a = x ·

(∏

σ∈Σ,σ 6=id σ(x))a

︸ ︷︷ ︸

∈B

.

Lemma 1.7. Let α1, . . ., αn be a basis of L/K. Assume α1, . . ., αn ∈ B. Then d(α1, . . ., αn)·B ⊆ Aα1 + . . .+Aαn.

6

Proof. Let α = a1α1+ . . .+anαn ∈ B with ai ∈ K. Then Tr(αiα) =∑n

j=1 TrL/K(αiαj)aj .

Hence(TrL/K(αiαj)

)

ij

a1...an

=

Tr(α1α)...

Tr(αnα)

∈ An.

So

a1...an

∈ 1

det(Tr(αiαj))·An =

1

d(α1, . . ., αn)· An.

Hence d(α1, . . ., αn) · ai ∈ A, and so d(α1, . . ., αn) · α ∈ Aα1 + . . .+Aαn. 2

Definition. An integral basis of B over A is a system w1, . . ., wn of elements of B such

that any b ∈ B can be written as b = a1w1 + . . . + anwn with uniquely determined

a1, . . ., an ∈ A. In other words, B =⊕n

i=1 Awi.

Remark. If w1, . . ., wn is an integral basis of B over A, then every element of L can be

written uniquely as a1w1 + . . .+ anwn with ai ∈ K – i.e. w1, . . ., wn is a basis of L/K.

Uniqueness. This follows from K = Frac(A).

Existence. We show that any element x ∈ L is of the form y/a with y ∈ B and

a ∈ A \ {0}. Let x have minimal polynomial td + a1td−1 + . . . + ad ∈ K[t]. There is

a ∈ A\{0} such that aai ∈ A for i = 1, . . ., d. Then (at)d+aa1(at)d−1+. . .+adad ∈ A[t],

and so ax ∈ B.

Proposition 1.8. If L/K is separable and A is a principal ideal domain then every finitely-

generated B-submodule M 6= 0 of L is a free A-module of rank n = [L : K]. In

particular, B has an integral basis over A.

Proof. Let α1, . . ., αn ∈ L be a basis of L/K. By the statement about existence in the

previous remark, there is a ∈ A \ {0} such that aαi ∈ B for i = 1, . . ., n. So we may

assume that α1, . . ., αn ∈ B. Let d = d(α1, . . ., αn). Then by 1.7, dB ⊆ Aα1+. . .+Aαn,

and hence B ⊆ Aα1

d ⊕ · · · ⊕Aαn

d is a submodule of a free A-module of rank n.

Reminder. Over a principal ideal domainA, every finitely-generated torsion-free mod-

ule N is free, i.e. isomorphic to An for a uniquely determined n = rank(N). Ev-

ery submodule of a finitely-generated free module N is itself free, and its rank is

6 rank(N).

So B is free of rank 6 n. So B has a basis over A, and by the preceding remark, any

basis of B over A is a basis of L/K, so rankA(B) = [L : K].

If M 6= 0 is a finitely-generated B-submodule of L, generated by µ1, . . ., µr ∈ L, then

there is a ∈ A \ {0} such that aµi ∈ B for i = 1, . . ., r. So aM ⊆ B, and thus

rankA(M) = rankA(aM) 6 rankA(B). If m ∈ M \ {0} then Bm ⊆ M is a submodule

of M which is free over B of rank 1 and thus free over A of rank [L : K].

Therefore n = rankA(Bm) 6 rankA(M), and hence rankA(M) = [L : K]. 2

7

Let K/Q be a number field. Then by 1.8, every finitely-generated OK-submodule a 6= 0 of

K has a Z-basis α1, . . ., αn – i.e. a = Zα1 ⊕ . . . ⊕ Zαn. The discriminant d(α1, . . ., αn) =

det(σi(αj))2 is independent of the basis α1, . . ., αn. This is because if a = Zβ1 ⊕ . . . ⊕ Zβn

then there is a matrix T ∈ GLn(Z) such that

T

α1

...αn

=

β1...βn

.

Then d(β1, . . ., βn) = det(σi(βj))2 = det(T )2 det(σi(αj))

2 = d(α1, . . ., αn), since detT = ±1.

Definition. For a finitely-generated OK-submodule a 6= 0 of K we put d(a) = d(α1, . . ., αn),

where α1, . . ., αn is a Z-basis of a. We call d(OK) =: dK the discriminant of K.

Example. K = Q(i), OK = Z⊕ Zi, so dQ(i) = d(OK) = det

(1 i1 −i

)2

= (−2i)2 = −4.

Proposition 1.9. Let a, b be finitely-generated non-zero OK-submodules of a number field

K, with a ⊆ b. Then the index [b : a] is finite and d(a) = [b : a]2d(b).

Proof. Let a = Zα1⊕ . . .⊕Zαn and b = Zβ1⊕ . . .⊕Zβn. Then there is a matrixM ∈Mn(Z)

such that

M

α1

...αn

=

β1...βn

.

Then by the definition of d(a) and d(b), we have d(a) = det(M)2d(b).

Exercise. [b : a] = #(Zn/M(Zn)

)= | det(M)|. 2

Corollary 1.10. If 0 6= a ⊆ OK is a finitely-generated OK-submodule of OK such that d(a)

is square-free, then a = OK .

In particular, if θ ∈ OK is such that K = Q(θ) and d(1, θ, . . ., θn−1) is square-free

(n = [K : Q]), then {1, θ, . . ., θn−1} is an integral basis of OK over Z.

Proof. Example sheet 1. 2

8

2 : Ideals

As before, K is a number field and OK ⊆ K its ring of integers.

A non-zero element α ∈ OK , which is not a unit, is called irreducible if for all β, γ ∈ OK

with α = βγ, we have necessarily β or γ a unit, i.e. in O∗K .

Observation. Every non-zero α ∈ OK \ O∗K can be written as a product α = α1. . .αr of

irreducible elements αi ∈ OK .

Proof. If α is not irreducible then there are β, γ such that α = βγ and β, γ /∈ O∗K . Then

we have

|NK/Q(α)| = |NK/Q(β)| |NK/Q(γ)| >{

|NK/Q(β)||NK/Q(γ)|

}

> 1.

By induction on |NK/Q(α)|, we can assume that β and γ can be written as a product

of irreducibles, and so α can be written as a product of irreducible elements.

(Note that if |NK/Q(α)| = 2 then α is automatically irreducible.) 2

In general, the decomposition of an element into irreducible elements is not unique (not even

up to multiplication by units).

Example. K = Q(√−5), OK = Z[

√−5]. (See example sheet 1.)

We have 21 = 3×7 = (1+2√−5)(1−2

√−5), and we can show that 3, 7, 1+2

√−5 and

1−2√−5 are all irreducible elements in OK , and that they are pairwise not associated.

(Elements α, β are called associated if β = εγ with ε ∈ O∗K .)

Conclusion. OK is in general not a unique factorisation domain (i.e. an integral domain

in which the factorisation into irreducible elements is unique, up to order and multiplication

by units).

Ernst Edward Kummer was led by this fact to introduce the concept of ‘ideal numbers’.

These became a forerunner of the concept of an ideal.

Recall that a subset a ⊆ OK is called an ideal if a is an additive subgroup and for all a ∈ a

and x ∈ OK we have xa ∈ a.

An ideal p is called a prime ideal if for all x, y ∈ OK , if xy ∈ p then x ∈ p or y ∈ p.

Equivalently, an ideal p is a prime ideal iff the quotient ring OK/p is an integral domain.

Theorem 2.1. The ring OK is Noetherian, integrally closed in K, and every non-zero prime

ideal is a maximal ideal.

Proof. OK is Noetherian if, by definition, every ideal a ∈ OK is finitely generated as an

OK-module. But this is clear, because by 1.8, OK is a finitely-generated free Z-module

(of rank [K : Q]), and as Z is a principal ideal domain, the submodule a of OK is finitely

generated as a Z-module, and a fortiori it is finitely generated as an OK-module.

9

Let θ ∈ K be integral over OK . By 1.1, OK [θ] is a finitely-generated OK-module. By

1.8, it is a finitely-generated Z-module. Because Z[θ] is a Z-submodule of OK [θ] and

hence finitely generated too, 1.1 implies θ is integral over Z. So by definition it is an

element of OK .

Let p 6= 0 be a prime ideal of OK . Then p∩Z is a non-zero prime ideal. To see that it

is non-zero, let y ∈ p\{0}. Then y satisfies an equation yd+a1yd−1+ . . .+ad = 0 with

all ai ∈ Z, and wlog d is minimal. Then ad ∈ p ∩ Z and ad 6= 0. So p ∩ Z = (p) = pZ

for some prime number p.

Let w1, . . ., wn (n = [K : Q]) be an integral basis of OK over Z, i.e. OK = Zw1 ⊕ · · · ⊕Zwn. Then OK/p is generated by w1 + p, . . ., wn + p over Z/(p) = Fp. So OK/p is

a finite-dimensional Fp-vector space, and thus has only finite many elements. But an

integral domain with only finitely many elements is a field (exercise).

But then p is maximal. 2

Definition. A Noetherian domain which is integrally closed in its field of fractions, and such

that every non-zero prime ideal is maximal, is called a Dedekind domain.

From now on, OK is a Dedekind domain and K = Frac(OK).

For ideals a, b of a ring R, we define

a+ b = {a+ b : a ∈ a, b ∈ b}ab = {∑i aibi : ai ∈ a, bi ∈ b}

Note that (ab)c = a(bc).

Theorem 2.2. Every non-zero, proper ideal a of OK can be written as a = p1. . .pr, with

non-zero prime ideals pi. The pi are uniquely determined up to ordering.

Lemma 2.3. For every non-zero ideal a of OK , there exist non-zero prime ideals p1, . . ., pr

such that p1. . .pr ⊆ a.

Proof. Set M = {a ⊆ OK : a is a non-zero ideal not fulfilling the conditions of the lemma}.We want to show that M = ∅.

Suppose M 6= ∅. Let a1 ⊆ a2 ⊆ . . . be a chain of elements in M . (Remark: M is

ordered by a 6 b iff a ⊆ b.) As OK is a Dedekind domain, if is Noetherian, and hence

this chain becomes stationary, and thus⋃

i ai = an for n ≫ 0. So any chain in M has

an upper bound in M , and by Zorn’s Lemma there is some maximal element a ∈M .

Now, a is not a prime ideal, so there are a1, a2 ∈ OK such that a1, a2 /∈ a but a1a2 ∈ a.

Put a1 = a+ (a1) and a2 = a+ (a2). Then a ( a1 and a ( a2, but a1a2 ⊆ a. And a is

maximal in M , so a1, a2 /∈M .

There are thus non-zero prime ideals p1, . . ., pr and q1, . . ., qs such that p1. . .pr ⊆ a1

and q1. . .qs ⊆ a2, and therefore p1. . .prq1. . .qs ⊆ a. //\\ 2

10

Lemma 2.4. Let p be a non-zero prime ideal of OK . Put p−1 = {x ∈ K : xp ⊆ OK}.

Then for every non-zero ideal a of OK , ap−1 = {∑i aixi : ai ∈ a, xi ∈ p−1} strictly

contains a.

Proof. Let a ∈ p \ {0} and p1. . .pr ⊆ (a) = aOK for non-zero prime ideals pi (by 2.3). We

may assume that r is minimal. Then p1. . .pr ⊆ p and so some pi ⊆ p, since p is prime.

But OK is a Dedekind domain, and hence p1 = p. We may assume that i = 1.

Now r is minimal, so p2. . .pr 6⊆ (a). So there exists b ∈ p2. . .pr such that b /∈ (a), and

then ba /∈ OK . But (b)p = (b)p1 ⊆ p2. . .prp1 ⊆ (a), and hence b

a ∈ p−1. It follows that

p−1 ) OK .

Let a be a non-zero ideal of OK , and α1, . . ., αn be generators of a as an OK -module.

Suppose ap−1 = a. Then, for x ∈ p−1, we have xαi =∑

j aijαj , for (not uniquely

determined) aij ∈ OK .

Put A = (aij)16i,j6n. Then A

α1

...αn

= x

α1

...αn

, so (xI −A)

α1

...αn

= 0.

Hence dα1 = . . . = dαn = 0, where d = det(xI − A), by 1.2. But then, as OK is a

domain, d = 0 and so x is integral over OK . And OK is integrally closed, so x ∈ OK .

But then p−1 ⊆ OK , contradicting what we proved before. 2

Proof of Theorem 2.2.

Existence. Let M = {a ⊆ OK : a is a non-zero, proper ideal, and not a product of

finitely many prime ideals}. We want to show that M = ∅.

Suppose M 6= ∅. By the same reasoning (Noetherian, stationary chain) as in the

proof of 2.3, M has a maximal element a, which is clearly not a prime ideal.

There is thus a prime ideal p such that a ( p, and then OK ⊆ p−1 implies that

a ( ap−1 ⊆ OK (by 2.4).

2.4 implies p ⊆ pp−1 ⊆ OK , and p is a maximal ideal, so the ideal pp−1 is equal

to OK . If ap−1 = OK , then 1 =∑

i aixi, some ai ∈ a, xi ∈ p−1. Then for y ∈ p,

y =∑

i ai(yxi) ∈ a, since yxi ∈ OK . This would imply that p ⊆ a. //\\

Thus ap−1 ( OK . And a maximal in M implies ap−1 /∈ M . Then we can write

ap−1 = p1. . .pr for non-zero prime ideals pi. Since pp−1 = OK , multiplying with

p gives a = p1. . .pr. 2

Uniqueness. For a prime ideal p, for all ideals a, b, if ab ⊆ p then either a ⊆ p or

b ⊆ p. So if p1. . .pr = q1. . .qs, then some qj in contained in p1, wlog j = 1.

As qj is prime and thus maximal, we have q1 = p1. Multiplying by p−11 (using

p1p−11 = OK) gives p2. . .pr = q1. . .qs.

Inductively, we see r = s and qj = pσ(j) for some permutation σ of {1, . . ., r}. 2

11

Convention. When we write a non-zero ideal a as a product of non-zero prime ideals, we

can group the same ideals together and write a = pe11 . . .perr with e1, . . ., er > 1. In this

decomposition, we always assume the pi to be pairwise distinct.

Definition. Two ideals a, b of a ring R are called relatively prime if a+ b = R.

Proposition 2.5. If a1, . . ., an are pairwise relatively prime ideals of OK , then a1. . .an =⋂n

i=1 ai.

Proof. We may assume that all ai are non-zero (otherwise it’s obvious). We may also assume

(by induction) that n = 2. Writing each of a1, a2 as a product of prime ideals, we may

in fact assume that a1 = p and a2 = q are two non-zero prime ideals.

Obviously pq ⊆ p ∩ q. Let a ∈ p ∩ q, a 6= 0. Then (a)p−1 ⊆ OK , so by 2.2 we

can write (a)p−1 = p1. . .pr for prime ideals pi. Then, because pp−1 = OK , we have

(a) = pp1. . .pr. With the same reasoning, we have (a) = qq1. . .qs for prime ideals qj.

As p 6= q we get from the uniqueness assertion in 2.2 that q = pi for some i. And thus

a ∈ (a) = qpb ⊆ pq, for some ideal b. 2

Proposition 2.6. If a1, . . ., an are pairwise relatively prime ideals in OK , then the canonical

map OK/(a1. . .an) −→∏n

i=1 OK/ai is an isomorphism.

Proof. Chinese Remainder Theorem: OK/(⋂n

i=1 ai)∼−→ ∏n

i=1 OK/ai. Now use 2.5. 2

Definition. A fractional ideal of K is a non-zero finitely-generated OK-submodule of K.

For a ∈ K∗ we call (a) = aOK a principal fractional ideal. The non-zero ideals of

OK are called integral ideals.

A non-zero OK-submodule a of K is a fractional ideal iff there exists c ∈ {0} such that

ca ⊆ OK . For if a is a fractional ideals, generated by a1

s1, . . ., an

snas an OK -module, then

ca ⊆ OK for c = s1. . .sn.

Given two fractional ideals a, b, their product ab is again a fractional ideal – for if a1, . . ., an

generate a and b1, . . ., bm generate b, then the aibj generate ab.

Proposition 2.7. The set of fractional ideals of K is an abelian group with respect to the

multiplication of fractional ideals, as defined above. This group is called the ideal

group of K and is denoted by JK . The identity element is OK , and the inverse of

a ∈ JK is given by {x ∈ K : xa ⊆ OK}.

Proof. The map JK×JK → JK , (a, b) 7→ ab is clearly associative, commutative and satisfies

aOK = OKa = a.

If a ⊆ OK is a non-zero ideal, then we can write a = p1. . .pr for prime ideals pi. Because

pip−1i = OK , letting b = p−1

1 . . .p−1r we have ba = OK , and each p−1

i is a fractional

ideal (and thus so is b), because if c ∈ pi \{0} then cp−1i ⊆ OK – now apply the remark

before the Proposition.

12

Since ba = OK , we have b ⊆ {x ∈ K : xa ⊆ OK}. If x ∈ K is such that xa ⊆ OK , then

x ∈ xOK = (xa)b ⊆ OKb = b. Therefore, a−1 is given by {x ∈ K : xa ⊆ OK}.

For a general fractional ideal a, let c ∈ OK \ {0} be such that ca ⊆ OK . Then (ca)−1 =

{x ∈ K : x(ca) ⊆ OK} = c−1{x ∈ K : xa ⊆ OK}. So a−1 = {x ∈ K : xa ⊆ OK}. 2

Corollary 2.8. Every fractional ideal a admits a unique decomposition as a product a =

pe11 . . .perr with e1, . . ., er ∈ Z and pairwise distinct prime ideals p1, . . ., pr.

Proof. Let c ∈ OK \ {0} be such that ca ⊆ OK , and write (c) = p1. . .pr and ca = q1. . .qs.

Then a = (c)−1(ca) = p−11 . . .p−1

r q1. . .qs. The uniquness follws immediately from the

uniqueness of the product decomposition for integral ideals. 2

Definition. The principal fractional ideals aOK = (a) for a ∈ K∗ form a subgroup of JK

denoted by PK , and the quotient group ClK = JK/PK is called the ideal class group

(or class group) of K.

There is a natural exact sequence:

1 −→ O∗K −→ K∗ −→ PK −→ JK −→ ClK −→ 1

Main result of the course. If K is a number field, then

• #ClK <∞• O∗

K∼= {roots of unity in K} ⊕ Zr+s−1,

where r = #{K Q→֒ R} and s = {K Q→֒ C which do not factor through R}.

Main tool. Embed K into an R-vector space.

13

3 : Lattices

Definition. Let V be an R-vector space of finite dimension. A lattice in C is a subgroup

of the form Γ = Zv1 + . . .+ Zvm with linearly independent vectors v1, . . ., vm.

The m-tuple (v1, . . ., vm) is called a basis of Γ, and the set Φ = {∑mi=1 xivi : xi ∈ R

and 0 6 x1 < 1 for all i} is called a fundamental mesh of the lattice.

Γ is called complete if m = dimR V .

Examples. (i) Γ = Z+ Zi ⊆ C = R+ Ri is a lattice

(ii) Γ = Z + Z√2 ⊆ R is not a lattice (it’s a free abelian group of rank 2, but 1 and√

2 are not linearly independent).

From now on, V denotes an n-dimensional R-vector space.

Remark. A lattice Γ ⊆ V is a discrete subgroup of V , i.e. every point γ = a1v1+ . . .+amvm

has an open neighbourhood U ⊆ V such that U ∩ Γ = {γ}.

Proof. Complete v1, . . ., vm to a basis v1, . . ., vn of V and put U = {x1v1+. . .+xnvn : xi ∈ R

and |xi − ai| < 1 for i = 1, . . .,m}. 2

Proposition 3.1. A subgroup Γ ⊆ V is a lattice iff it is a discrete subgroup.

Proof. Let Γ ⊆ V be a discrete subgroup. We show first that Γ is closed in V .

Let U ⊆ V be an open neighbourhood of 0. Then there is an open neighbourhood

U ′ ⊆ U of 0 such that {x−x′ : x, x′ ∈ U ′} is contained in U . If x0 ∈ Γ\Γ, then x0+U ′

contains some γ1 ∈ Γ. We can find another open neighbourhood U ′′ of 0 inside U ′ such

that γ1 /∈ x0 + U ′′ but there is of course some γ2 ∈ Γ in x0 + U ′′. So we have γ1 6= γ2

and γ1 − γ2 = (γ1 − x0)︸ ︷︷ ︸

∈U ′

− (γ2 − x0)︸ ︷︷ ︸

∈U ′′⊆U ′

∈ {x− x′ : x, x′ ∈ U ′} is then contained in U .

Hence we have shown that there exists in any open neighbourhood U of 0 an element

γ = γ1 − γ2 of Γ which is non-zero. This contradicts Γ being discrete, so Γ \Γ = ∅ and

thus Γ = Γ is a closed subset of V .

Let V0 be the subspace of V generated by Γ, and let u1, . . ., um ∈ Γ be a basis of V0.

Consider the lattice Γ0 = Zu1 + . . . + Zum and its fundamental mesh Φ0. Then we

have V0 =⊔

γ∈Γ0(γ + Φ0). For any element γ ∈ Γ there is some γ0 ∈ Γ0 such that

γ − γ0 ∈ Φ0 ∩ Γ ⊆ Φ0 ∩ Γ. Since Φ0 = {x1u1 + . . . + xmum : xi ∈ [0, 1]} ∼= [0, 1]m is

compact and Γ is discrete, we have that Φ0 ∩ Γ is a finite set.

So Γ/Γ0 is finite, say of order d. Then dΓ ⊆ Γ0, or Γ ⊆ 1dΓ0. But then Γ is a subgroup

of a finitely-generated free abelian group, and hence it is free of some rank r 6 m. But

since dimV0 = m and Γ generates V0, we have that r = m.

If v1, . . ., vm is then a basis of Γ, theyare linearly independent (because they span the

m-dimensional subspace V0), and hence Γ = Zv1 + . . .+ Zvm is a lattice. 2

14

Proposition 3.2. A lattice Γ ⊆ V is complete iff there exists a bounded subset M ⊆ V

such that V =⋃

γ∈Γ(γ +M).

Proof. If Γ = Zv1 + . . . + Zvn (where n = dimR V ) is complete, then we can take the

fundamental mesh for M .

For the converse, suppose that M ⊆ V is bounded and V =⋃

γ∈Γ(γ +M). Let V0

be the span of Γ, and let v ∈ V . For j ∈ Z>0, write jv = γj +mj with γj ∈ Γ and

mj ∈M . Since M is bounded, we have limj→∞1jmj = 0 and thus

v = limj→∞

(1j γj +

1jmj

)

=(

limj→∞

1j γj

)

+(

limj→∞

1jmj

)

= limj→∞

1j γj ∈ V0,

since V0 is a closed subspace of V . Hence V0 = V and Γ is complete. 2

Now we suppose that V is a Euclidean vector space, i.e. there is a positive definite symmetric

bilinear form 〈 , 〉 : V × V → R. Let e1, . . ., en be an orthonormal basis of V , then we get an

isomorphism i : V∼−→ Rn. Using i we transfer the Lebesgue measure on Rn to V .

Then vol(Φ0 = {x1e1 + . . . + xnen : xi ∈ [0, 1)}

)= 1. More generally, let v1, . . ., vn be

any basis of V and A = (aij) the change-of-basis matrix from e1, . . ., en to v1, . . ., vn – i.e.

vj =∑aijei. Then vol

(Φ = {x1v1 + . . .+ xnvn : xi ∈ [0, 1)}

)= vol

(A(Φ0)

)= | detA|.

If Γ ⊆ V is a complete lattice with fundamental mesh Φ, we put vol(Γ) = vol(Φ).

Definition. A subset X ⊆ V is said to be centrally symmetric if −x ∈ X for all x ∈ X .

X is called convex if for all x, y ∈ X , the line segment {ty + (1 − t)x : t ∈ [0, 1]} is

contained in X .

Theorem 3.3 (Minkowski’s lattice point theorem). Let Γ be a complete lattice in V

and X a centrally symmetric convex subset of V .

If vol(X) > 2n vol(Γ) (where n = dimR V ), then X contains a non-zero lattice point.

Example. Γ = Z+ Zi ⊆ C.q q q

q q q

q q q

1−1

i

−i

– open cube contains only 0.

Proof. Suppose that the sets 12X + γ, for γ ∈ Γ, are all pairwise disjoint. Then, if Φ is a

fundamental mesh of Γ, we have vol(Φ) >∑

γ∈Γ vol(Φ ∩ (12X + γ)

).

Notice that(Φ ∩ (12X + γ)

)\ {γ} = (Φ \ {γ}) ∩ 1

2X and that all (Φ \ {γ})∩ 12X cover

12X since Γ is a complete lattice. Hence

vol(Φ) >∑

γ∈Γ

vol(

Φ ∩(12X + γ

))

= vol(12X

)= 2−n vol(X).

This is a contradiction, hence there are γ1 6= γ2 satisfying(12X + γ1

)∩(12X + γ2

)6= ∅.

Then there are x, y ∈ X such that 12x+ γ1 = 1

2y+ γ2, and so 0 6= γ1 − γ2 = 12y− 1

2x ∈(Γ \ {0}) ∩X . 2

15

4 : Geometry of Numbers

Motivated by the inclusion OQ(i) = Z+ Zi ⊆ C.

Let K denote a number field. Goal: define an embedding K →֒ KR, where KR is a real vector

space, such that the image of a non-zero ideal in OK is a complete lattice.

Let θ ∈ K be a primitive element, i.e. K = Q(θ). (This always exists by the theorem of the

primitive element.) Let f ∈ Q[X ] be the minimal polynomial of θ, and n = deg(f) = [K : Q].

Let θ1, . . ., θn be the roots of f in C (these are pairwise distinct as f is irreducible). A

field homomorphism τ : K →֒ C (or K → C, as always injective – field) is then uniquely

determined by τ(θ) ∈ {θ1, . . ., θn}. Conversely, for any i ∈ {1, . . ., n} there is a unique field

homomorphism τ : K → C with τ(θ) = θi. Hence there are n distinct (field) embeddings

τ : K →֒ C – let T be the set of these.

Call τ real if τ(K) ⊆ R, and complex otherwise.

Let κ : C → C be complex conjugation. Clearly, τ is real iff κ ◦ τ = τ , and complex iff

κ ◦ τ 6= τ . Write τ = κ ◦ τ , and put KC =∏

τ∈T C = {(zτ )τ∈T : zτ ∈ C}.

Define j : K → KC by j(a) = (τ(a))τ∈T , and define an involution F : KC → KC by

F((zτ )τ∈T

)= (zτ )τ∈T .

Notation. From now on, abbreviate∏

τ∈T to∏

τ , and (zτ )τ∈T to (zτ )τ , etc.

Lemma 4.1. The image of j is contained in the R-vector space KR ⊆ KC consisting of

elements fixed by F . I.e.

im (j) ⊆ KR ={(zτ )τ ∈ KC : zτ = zτ for all τ

}.

Moreover, dimR(KR) = n = [K : Q].

Proof.(F (j(a))

)

τ= j(a)τ = τ (a) = τ(a) = τ(a) = j(a)τ .

Let ρ1, . . ., ρr : K →֒ R be the real embeddings of K, and σ1, σ1, . . ., σs, σs : K →֒ C

the complex embeddings. If (zτ )τ is in KR then we have zρi = zρi= zρi , so zρi is in R.

And we have zσi = zσi , so zσi is determined by zσi .

Hence dimR(KR) = r + s dimR(C) = r + 2s = #{K →֒ C} = n = [K : Q]. 2

Define a scalar product 〈 , 〉 : KR ×KR → R by⟨(zτ )τ , (wτ )τ

⟩=

∑

τ zτwτ .

This is indeed real-valued, since zσiwσi + zσiwσi

= zσiwσi + zσiwσi ∈ R.

Fix an orthonormal basis e1, . . ., en of KR with respect to 〈 , 〉, and define a measure on KR

such that the cube {x1e1 + . . .+ xnen : xi ∈ [0, 1]} has measure 1, as explained in chapter 3.

Proposition 4.2. If a is a non-zero ideal of OK , then Γ := j(a) is a complete lattice in KR.

Its fundamental mesh has volume vol(Γ) = [OK : a]√

|dK |.

16

Proof. Write a = Zα1 + . . .+ Zαn, with αi ∈ OK .

Put A =(j(α1) · · · j(αn)

)=

τ1(α1) · · · τ1(αn)...

...τn(α1) · · · τn(αn)

. Then

d(α1, . . ., αn) = det(τi(αj)

)2(by definition)

= det(A)2

= [OK : a]2 d(OK) (by 1.9)

= [OK : a]2 dK

By 1.6, dK 6= 0, so det(A) 6= 0, and thus the vectors j(α1), . . ., j(αn) are R-linearly

independent (and even C-linearly independent).

Hence j(α1), . . ., j(αn) are a basis, and Γ = Zj(α1) + . . .+Zj(αn) is a complete lattice

(by definition) in KR.

Put B =

〈j(α1), e1〉 · · · 〈j(αn), e1〉...

...〈j(α1), en〉 · · · 〈j(αn), en〉

,

the change-of-basis matrix (e1, . . ., en) 7→(j(α1), . . ., j(αn)

). Then vol(Γ) = | detB|.

Now BtB =(〈j(αi), j(αj)〉

)

i,j= AtA. And hence

vol(Γ) = | detB| = | detAtA|1/2 = | detA|C =√

[OK : a]2|dK | = [OK : a]√

|dK |.

2

Lemma 4.3. With the notation of 4.1, the map f : KR → ∏

τ R∼= Rn given by

(f((zτ )τ )

)

ρ= zρ

(f((zτ )τ )

)

σi= Re(zσi)

(f((zτ )τ )

)

σi= Im(zσi)

is an isomorphism of real vector spaces, and for any measurable set X ⊆ KR, we have

volcan(X) = 2svolLeb(f(X)), where volLeb denotes the Lebesgue measure on Rn.

Proof. f is readily seen to be an isomorphism. The vectors

eρj = (0, . . ., 0,ρj

1 , 0, . . ., 0),

eσj = 1√2(0, . . ., 0,

σj

1 ,σj

1 , 0, . . ., 0),

eσj= 1√

2(0, . . ., 0,

σj

i ,σj

−i, 0, . . ., 0)

form an orthonormal basis with respect to 〈 , 〉, and one immediately concludes that

volLeb(f(X)) = 2−s, where X is the cube in KR spanned by this orthonormal basis.

By definition, volcan(X) = 1, and therefore the assertion follows. 2

17

Theorem 4.4. Let a be a non-zero ideal in OK , and let (cτ )τ be positive real numbers such

that cτ = cτ for all τ . Suppose

∏

τ

cτ >

(2

π

)s √

|dK | [OK : a].

Then there is a ∈ a, a 6= 0, such that |τ(a)| < cτ for all τ .

Proof. The set X ={(zτ )τ ∈ KR : |zτ | < cτ , all τ

}is centrally symmetric and convex, and

f(X) =

{

(xτ )τ ∈∏

τ

R : |xρ| < cρ for ρ real, and x2σ + x2σ < c2σ for σ complex

}

Therefore,

volcan(X) = 2svolLeb(f(X)) by 4.3

= 2s∏

ρ

(2cρ) ·∏

{σ,σ}(πcσ2 )

= 2s+rπs∏

τ

cτ

> 2r+2s√

|dK |[OK : a]

= 2nvolcan(j(a)) by 4.2

By the lattice point theorem (3.3), there is a ∈ a, a 6= 0, such that j(a) ∈ X , i.e.

|τ(a)| < cτ for all τ . 2

Lemma 4.5. For all t > 0, the set Xt ={(zτ )τ ∈ KR :

∑

τ |zτ | < t}is centrally symmetric

and convex, and

volcan(X) = 2n(π

4

)s tn

n!.

Proof. Exercise – or see Neukirch, chapter III, 2.15. 2

Theorem 4.6. If a is a non-zero ideal of OK , then there exists a ∈ a, a 6= 0, such that

|NK/Q(a)| 6M [OK : a], where

M =n!

nn

(4

π

)s √

|dK |

is called the Minkowski bound (or Minkowski constant) of K.

Proof. For ε > 0, put t = n(M [OK : a] + ε

)1/n. Then

volcan(Xt) = 2n(π

4

)s tn

n!

> 2n(π

4

)s nn

n!M [OK : a]

= 2n√

|dK |[OK : a]

= 2nvolcan(j(a))

By 3.3, there is a ∈ a, a 6= 0, such that j(a) ∈ Xt. The AM-GM inequality implies

|NK/Q(a)| =∏

τ

|τ(a)| 6(1

n

∑

τ

|τ(a)|)n

6tn

nn=M [OK : a] + ε.

This holds for all ε, and |NK/Q(a)| is an integer, so we get that |NK/Q(a)| 6M [OK : a]

for some non-zero a ∈ a. 2

18

5 : The Class Number

For a non-zero ideal a of OK , define its norm to be N(a) = [OK : a].

Proposition 5.1. If a = pe11 . . .perr is the factorisation of a, then N(a) = N(p1)

e1 . . .N(pr)er .

Proof. By Proposition 2.6 (an application of the Chinese Remainder Theorem), we have

OK/a ∼=⊕r

i=1 OK/peii . Hence N(a) = N(pe11 ). . .N(perr ).

So let p 6= 0 be a prime ideal of OK , and let m > 1. Consider the chain OK ⊇ p ⊇p2 ⊇ · · · ⊇ pm. Then #(OK : pm) = [OK : p][p : p2]. . .[pm−1 : p].

Claim. pi/pi+1 is a 1-dimensional OK/p-vector space.

Proof. Let a ∈ pi \ pi+1, and write (a) = aOK = piq, where q is relatively prime to p.

Write x+ y = 1 with x ∈ p, y ∈ q.

If b ∈ pi then by = ca for some c ∈ OK , and therefore b = by + bx = ca + bx ≡ca mod pi+1. So a mod pi+1 generates pi/pi+1 as an OK/p-vector space.

Putting all this together: N(pm) = N(p). . .N(p) = N(p)m. 2

By Corollary 2.8, every fractional ideal a can be written uniquely as a = pe11 . . .perr , with the

ei ∈ Z.

If JK denotes the group of fractional ideals of K, then we can define a homomorphism

N : JK → Q>0 by N(pe11 . . .perr ) = N(p1)

e1 . . .N(pr)er .

Lemma 5.2. Let α ∈ K∗, and write [α] for the ideal of OK generated by α.

Then N([α]) = |NK/Q(α)|.

Proof. We may assume that α ∈ OK , α 6= 0. Let w1, . . ., wn be an integral basis of OK .

Then αw1, . . ., αwn is a basis of [α] over Z.

Define the change-of-basis matrix A = (aij) ∈ Mn(Z) by αwj =∑n

i=1 aijwi. Then

| detA| =[OK : [α]

]= N([α]). (See example sheet 1, question 4.)

By definition, detA = detK/Q(K → K : x 7→ αx) = NK/Q(α). 2

Theorem 5.3. Let MK = n!nn

(4π

)s √|dK | be the Minkowski constant. Then any class aPK

of ClK = JK/PK contains an integral ideal a1 ⊆ OK with N(a1) 6MK .

Proof. Let α ∈ a, α 6= 0. Then b := αa−1 is an ideal of OK . By Theorem 4.6, there is some

β ∈ b, β 6= 0, such that |NK/Q(β)| 6MK [OK : b] =MKN(b).

Put a1 = βb−1 = βαa. Then a1 is in the same class as a, and N(a1) = N(β)N(b)−1 6

MKN(b)N(b)−1 =MK . 2

19

Theorem 5.4 (Finiteness of the Class Number). The ideal class group ClK = JK/PK

is finite for every number field K.

The order hK of ClK is called the class number of K.

Proof. By Theorem 5.3, it suffices to show that there are only finitely many integral ideals

a ⊆ OK such that N(a) 6 MK . Writing a as pe11 . . .perr , we see that it suffices to show

that there are only finitely many prime ideals p such that N(p) 6MK .

For a non-zero prime ideal p of OK , the field OK/p is finite and of characteristic p, say,

so it has pf elements, for some f > 1.

Thus it suffices to show that there are only finitely many prime ideals p containing

p. Write [p] = pe11 . . .perr , then the prime ideals containing p are exactly p1, . . ., pr, as

p ∈ p ⇒ pp−1 =: a ⊆ OK ⇒ [p] = pa. 2

Remarks. 1. The proof showsthat for any x > 0 the number πK(x) = #{p ⊆ OK : p a

prime ideal such that N(p) 6 x} is always finite.

Then ‘prime ideal theorem’ asserts that π(x) ∼ x/ log(x) as x→ ∞.

2. For imaginary quadratic number fields K = Q(√d), where d < 0 and square-free,

we have (H. Stark) hk = 1 iff d = −1,−2,−3,−7,−11,−19,−43,−67,−163.

3. The p-part of the class number of Q(ζpn) has known structural behaviour, for

varying n (; Iwasawa theory).

Lemma 5.5. ClK is generated by the classes of prime ideals p of OK such that N(p) 6MK .

In particular, ClK is generated by those prime ideals which contain a prime 6MK .

Proof. Let c1 = a1PK , . . ., cn = anPK be the elements of ClK , with ai ⊆ OK and N(ai) 6

MK by Theorem 5.3. Decomposing ai = pe11 . . .perr , we get N(pj) 6 N(ai) 6MK . 2

Typical example. K = Q(√−17). Since −17 ≡ 3 mod 4, we have dK = 4(−17) = −68.

(Recall example sheet 1, question 2.)

And MK = 2!22

(4π

)√68 < 2

3

√68 < 6. So we look for prime ideals of norm < 6.

Put ω = 1 +√−17. Then [2] = [2, ω]2, [3] = [3, ω][3, ω], and [5] is a prime ideal. Then

we have [ω] = [2, ω][3, ω]2.

These equations show that class([2, ω]

)= class

([2, ω]

)−1, class

([3, ω]

)= class

([3, ω]

)−1,

and class([3, ω]

)2= class

([2, ω]

)−1.

Since class([2, ω]

)2= 1, we have that class

([3, ω]

)4= 1.

So all together, ClK ∼= Z/4Z. 2

20

6 : Dirichlet’s Unit Theorem

Multiplicative version of Minkowski’s theory.

Define j : K∗ → K∗C =

∏

τ C∗ to be the restriction of j : K → KC, j(a)τ = τ(a) for all τ .

Define N : K∗C → C∗ by N

((zτ )τ

)=

∏

τ zτ , so that the composition N ◦ j is the usual norm

NK/Q : K∗ → C∗.

How do lattices enter the picture? By taking logarithms!

Define ℓ : K∗C → ∏

τ R by ℓ((zτ )τ

)=

(log |zτ |

)

τ.

Recall that the image of j is contained in KR ={(zτ )τ ∈ KC : zτ = zτ for all τ

}, the set of

fixed points of F((zτ )τ

)= (zτ )τ . We let F act on

∏

τ R by F((xτ )τ

)= (xτ )τ .

Put[∏

τ R]+

={(xτ )τ ∈ ∏

τ R : F((xτ )τ

)= (xτ )τ

}, and K∗

R = KR ∩K∗C.

Then we have the following commutative diagram (∗)

K∗ j−→ K∗R

ℓ−→[∏

τ R]+

↓ NK/Q ↓ N ↓ Tr

Q∗ →֒ R∗ log | · |−→ R

where Tr((xτ )τ

)=

∑

τ xτ .

Let ρ1, . . ., ρr : K → R be the real embeddings, and σ1, σ1, . . ., σs, σs : K → C be the complex

embeddings. Then[∏

τ

R]+

= R× . . .× R︸ ︷︷ ︸

r

× [R× R]+ × . . .× [R× R]+︸ ︷︷ ︸

s

,

where [R× R]+ = {(x, x) : x ∈ R}.

Sending (x, x) ∈ [R × R]+ to 2x ∈ R gives an isomorphism[∏

τ R]+ ∼−→ Rr+s, and the

composition K∗R

ℓ−→[∏

τ R]+ ∼−→ Rr+s gets identified with

ℓ((xτ )τ

)=

(log |xρ1

|, . . ., log |xρr |, log(|xσ1|2), . . ., log(|xσs |2)

).

Remark. In the additive theory (to prove the finiteness of the ideal class group), we were

working with lattices in KR∼= Rr+2s = Rn, which is an n-dimensional R-vector space, where

n = [K : Q]. For the determination of the structure of O∗K , we work in an (r+s)-dimensional

vector space.

Consider the following subgroups of the groups appearing in the upper row of (∗).

• group of units, O∗K = {a ∈ OK : NK/Q(a) = ±1} ⊆ K∗

• norm-1 surface, S = {y ∈ K∗R : N(y) = ±1} ⊆ K∗

R

• trace-zero hyperplane, H ={x ∈

[∏

τ R]+

: Tr(x) = 0}⊆

[∏

τ R]+

21

Let λ = ℓ ◦ j|O∗

K, so λ : O∗

K

j−→ Sℓ−→ H ⊆

[∏

τ R]+

, and put Γ = λ(O∗K).

Denote by µ(K) the group of roots of unity in K.

Proposition 6.1. The sequence {1} −→ µ(K) →֒ O∗K

λ−→ Γ −→ {0} is exact.

Proof. We only have to show that µ(K) = ker(λ). Clearly, if ζ ∈ µ(K), then |τ(ζ)| = 1 for

all τ (for if ζn = 1 then |τ(ζ)|n = |τ(ζn)| = 1). So ℓ(j(ζ)) = 0, and hence λ(ζ) = 0.

Conversely, if ε ∈ O∗K and λ(ε) = 0, then for all τ , log

(|τ(ε)|

)= 0, hence |τ(ε)| = 1 for

all τ . So j(ε) =(τ(ε)

)

τ∈ K∗

R lies in some bounded subset of KR. But j(ε) is also an

element of j(OK), which is a lattice in KR, by Proposition 4.2.

And a lattice is a discrete subgroup (Proposition 3.1), hence the intersection of j(OK)

with this bounded set (defined by |xτ | = 1 for all τ) is finite. So ker(λ) is a finite group,

and therefore ker(λ) ⊆ µ(K). 2

Lemma 6.2. Up to multiplication by units ε ∈ O∗K , there are only finitely many α ∈ OK of

a given norm a = NK/Q(α) ∈ Z.

Proof. Let a ∈ Z, and wlog a > 1. In each of the finitely many cosets of OK/[a] there exists

(up to multiplication by units) at most one element α such that |NK/Q(α)| = a.

For if we have β = α+ aγ with γ ∈ OK and |NK/Q(β)| = a = |NK/Q(α)|, then

α

β= 1− a

βγ = 1− ±N(β)

βγ ∈ OK , since

N(β)

β=

∏

τ∈T \{id}τ(β) ∈ OK .

The same reasoning shows that βα ∈ OK and so β

α ∈ O∗K , and β = β

αα. 2

Theorem 6.3. The group Γ := λ(O∗K) is a complete lattice in the (r + s − 1)-dimensional

R-vector space H , and is therefore isomorphic to Zr+s−1.

Proof. We show first that Γ is a lattice in H , i.e. a discrete subgroup of H (by 3.1).

It suffices to show that for any c > 0, the intersection of Γ with

Y ={(xτ )τ ∈ ∏

τ R : |xτ | 6 c for all τ}⊆ ∏

τ R

is always finite.

But ℓ−1(Y ) ={(zτ )τ ∈ ∏

τ C∗ : e−c 6 |zτ | 6 ec for all τ

}is compact, and has therefore

finite intersection with the lattice j(OK) ⊆ KR ⊆ ∏

τ C.

Now we show that Γ is a complete lattice. By Proposition 3.2 it suffices to show that

there is a bounded set M ⊆ H such that

H =⋃

γ∈Γ

(γ +M) (∗)

22

We will find a bounded set T ⊆ S ={(zτ )τ ∈ K∗

R :∏

τ |zτ | = 1}such that

S =⋃

ε∈O∗

K

j(ε)T (∗∗)

For (xτ )τ ∈ T we will have that all xτ are bounded from above, and hence (automat-

ically) bounded away from zero, because∏

τ |xτ | = 1, so that M := ℓ(T ) is bounded

too, and (∗∗) implies (∗).

Choose real numbers cτ > 0, cτ = cτ , and C :=∏

τ cτ >(2π

)s √|dK |.

Put X ={(zτ )τ ∈ KR : |zτ | < cτ for all τ

}. For y = (yτ )τ ∈ S, we have that

Xy ={(zτ )τ ∈ KR : |zτ | < c′τ for all τ

},

where c′τ = cτ |yτ |, and we still have c′τ = c′τ and∏

τ c′τ =

∏

τ cτ |yτ | =∏

τ cτ = C.

By Theorem 4.4, there exists a ∈ OK , a 6= 0, such that j(a) ∈ Xy. By Lemma 6.2,

there are α1, . . ., αN ∈ OK , αi 6= 0, such that any α ∈ OK with 0 < |NK/Q(α)| 6 C is

associated to one αi.

Put T = S ∩(⋃N

i=1Xj(αi)−1

)

. Then T is bounded since X is. We show (∗∗).

If y ∈ S, then there is a ∈ OK , a 6= 0, such that j(a) ∈ Xy−1. I.e. j(a) = xy−1 with

x ∈ X . Then

0 < |NK/Q(a)| = |N(xy−1)| = |N(x)| <∏

τ cτ = C.

So a = εαi, for some ε ∈ O∗K and i ∈ {1, . . ., N}, and then

y = xj(a)−1 = j(ε−1)xj(αi)−1

︸ ︷︷ ︸

∈S

∈ j(ε−1)(S ∩Xj(αi)

−1),

and because j(ε−1)(S ∩Xj(αi)

−1)⊆ j(ε−1)T , we get (∗∗). 2

Theorem 6.4 (Dirichlet’s Unit Theorem). The group O∗K is isomorphic to the product

of the finite group µ(K) and a free abelian group ∼= Zr+s−1.

Proof. Let t = r + s − 1, and choose ε1, . . ., εt in O∗K such that λ(ε1), . . ., λ(εt) are a basis

of Γ (or equivalently of H). We claim that O∗K

∼= µ(K)× 〈ε1, . . ., εt〉.

We have µ(K)∩〈ε1, . . ., εt〉 = {1}, since if ζ = εν11 . . .ενtt , for ζ ∈ µ(K) and νi ∈ Z, then

λ(ζ) = 0 implies ν1λ(ε1)+ . . .+νtλ(εt) = 0, which implies that ν1 = . . . = νt = 0, since

the λ(εi) are a basis.

And if ε ∈ O∗K and λ(ε) = ν1λ(ε1) + . . .+ νtλ(εt), then λ

(ε(εν11 . . .ε

νtt )−1

)= 0, and by

Proposition 6.1, this gives ε(εν11 . . .ενtt )−1 = ζ, for some ζ ∈ µ(K).

Therefore, ε = ζεν11 . . .ενtt ∈ µ(K)× 〈ε1, . . ., εt〉. 2

Definition. Let K be a number field of degree n = r + 2s. A set ε1, . . ., εr+s−1 ∈ O∗K of

units is called a set of fundamental units if O∗K

∼= µ(K)× 〈ε1, . . ., εr+s−1〉.

23

7 : The Fundamental Unit Of A Real Quadratic Field

Let D > 1 be a square-free integer and√D ∈ R be the positive square root of D. We

consider the field K = Q(√D) as a subfield of R (consisting of all α+ β

√D, for α, β ∈ Q).

Any such field is called a real quadratic field.

We have r = #{real embeddings} = 2, and s = #{complex embeddings} = 0. By the Unit

Theorem (6.4), O∗K = µ(K)× 〈ε〉 = {±1} × 〈ε〉, for some ε ∈ O∗

K of infinite order.

For x+ y√D ∈ K, we write x+ y

√D = x− y

√D.

Lemma/Definition 7.1. The minimum of {ε ∈ O∗K : ε > 1} does exist, and if ε0 is equal

to this minimum, then O∗K = {±1} × {ε0}. We call ε0 the fundamental unit of K.

Writing ε0 = α+ β√D with α, β ∈ Q, we have α, β > 0.

Proof. By the Unit Theorem (6.4), we have O∗K = {±1}×〈ε0〉 for some ε0 ∈ O∗

K of infinite

order. If ε0 < 0 then we still have O∗K = {±1}× 〈−ε0〉, so we may assume that ε0 > 0.

If 0 < ε0 < 1, then ε−10 > 1 and so we may further assume that ε0 > 1.

Hence there is always some fundamental unit ε0 which is > 1. If ε > 0 is a positive

unit, then ε = εn0 for some n ∈ Z, and ε > 1 iff n > 0. But then ε = εn0 > ε0, and so

ε0 = min{ε ∈ O∗K : ε > 1}.

Write ε0 = α+ β√D. We have

±1 = NK/Q(ε0) = ε0ε0 = (α+ β√D)(α− β

√D).

If N(ε0) = −1, then ε0 = α − β√D < 0, hence −ε0 = −α + β

√D > 0, and then

0 < ε0 + (−ε0) = 2β√D – i.e., β > 0.

But since ε0 > 1, we have ε−10 = −ε0 = −α+ β

√D < ε0 = α+ β

√D, and thus α > 0.

The case N(ε0) = 1 is very similar. 2

Now let ε0 ∈ O∗K , ε0 > 1, be the fundamental unit of K.

Let d = dK =

{4D if D ≡ 2, 3 mod 4D if D ≡ 1 mod 4

}

be the discriminant of K.

Put ε0 = 12 (x0 + y0

√d), where

{x0 ∈ 2Z and y0 ∈ Z if D ≡ 2, 3 mod 4x0, y0 ∈ Z and x0 ≡ y0 mod 2 if D ≡ 1 mod 4

.

Proposition 7.2.

(a) If there is ε ∈ O∗K such that N(ε) = −1 then N(ε0) = −1, and (x0, y0) is the

unique solution of x2 − dy2 = −4 with x, y ∈ Z>0 and x minimal.

(b) If N(ε) = 1 for all ε ∈ O∗K , then N(ε0) = −1, and (x0, y0) is the unique solution

of x2 − dy2 = 4 with x, y ∈ Z>0 and x minimal.

24

Proof. We’ll prove (a), and (b) is very similar. If N(ε) = −1 for some ε ∈ O∗K then write

ε = ±εn0 and get −1 = N(ε) = N(±εn0 ) = N(ε0)n, and so N(ε0) = −1.

Clearly then x20 − dy20 = −4. Let x2 − dy2 be another solution of this equation with

x, y ∈ Z>0. Put ε = 12 (x + y

√d). Then TrK/Q(ε) = x ∈ Z and NK/Q(ε) = εε =

14 (x

2 − dy2) = −1, and hence ε ∈ O∗K .

Since ε > 1 (immediate) we have 12 (x+ y

√d) = ε > ε0 = 1

2 (x0 + y0√d).

And because1

ε6

1

ε0, we have −1

ε= ε =

1

2(x− y

√d) > − 1

ε0= ε0 =

1

2(x0 − y0

√d).

Adding these two inequalities gives x > x0. 2

Theorem 7.3. The fundamental unit ε0 of K = Q(√d) is given as ε0 = 1

2 (x0 + y0√dK),

where x0, y0 ∈ Z>0, and (x0, y0) is the unique (minimal x) solution of x2 − dy2 = −4,

if this equation has an integral solution, or of x2 − dy2 = 4, if the former equation does

not have a solution.

Remark. Let√D = 〈a0, a1, . . ., aℓ〉 be the continued fraction expansion of

√D. Then

x2 − dy2 = −4 has a solution with x, y ∈ Z iff ℓ is odd.

If p/q = 〈a0, a1, . . ., aℓ−1〉 and D 6≡ 5 mod 8 then p + q√D is the fundamental unit of

Q(√D) (for ℓ even or odd).

Algorithm to determine ε0. Check one-by-one for y = 1, 2, 3, . . . whether dy2 ∓ 4 is a

square x2. By the Unit Theorem (6.4) this is bound to happen eventually, with the

plus sign. However, for fixed y, let preference be given to the case where dy2 − 4 is a

square. Then the first case in this order where dy20 − 4 = x20 gives the fundamental unit

ε0 = 12 (x0 + y

√d).

Examples. D 2 3 5 · · · 11 · · · 31 · · ·ε0 1 +

√2 3 +

√3 1+

√5

2 10 + 3√11 1520 + 273

√31

(Online program for ε0: www.numbertheory.org/php/unit.html)

Siegel has shown that log(hQ(

√D) · log(ε0)

)∼ log(

√D) for D → ∞.

(If h = 1, then log ε0 ≈ (√D)2, so ε0 ≈

√D.)

Example: Pell’s equation. We want to find all integral solutions (x, y) of x2−11y2 = −7.

K = Q(√11) has fundamental unit ε0 = 10 + 3

√11, and N(ε0) = 1.

We have x2−11y2 = (x+y√11)(x−y

√11) and −7 = (2+

√11)(2−

√11), and (exercise)

hQ(√11) = 1. So, up to units, we have that x+y

√11 is associated to 2+

√11 or 2−

√11.

Explicitly, x+ y√11 = ±εn0 (2 +

√11) or ±εn0 (2 −

√11), for some n ∈ Z.

And since ±εn0 (2 +√11) = ±ε−n

0 (2 −√11), we get all solutions x, y ∈ Z by writing

±εn0 (2 +√11) = a+ b

√11, i.e. x = ±a and y = ±b.

25

8 : Factorisation of Prime Numbers in Number Fields

Let K be a number field of degree n = [K : Q], and write K = Q(θ) for some θ ∈ OK . Let

f(X) ∈ Z[X ] be the minimal polynomial of θ. Let p ∈ Z be a given prime number.

Question. How do we determine the factorisation of the ideal [p] = pOK into prime ideals?

Theorem 8.1 (Dedekind). Suppose OK is generated over Z by θ – i.e. 1, θ, . . ., θn−1 is

an integral basis of OK . Let f(X) = f(X) mod p, and let f(X) = f1(X)e1 . . .fr(X)er

be the factorisation of f(X) into irreducible monic polynomials fi(X) ∈ Fp[X ] with

fi(X) 6= fj(X) for i 6= j. Let fi(X) ∈ Z[X ] be a monic polynomial of the same degree

as fi(X) such that fi(X) = fi(X) mod p, and put pi = [p, fi(θ)] ⊆ OK .

Then p1, . . ., pr are the distinct prime ideals of OK contained in [p]. The degree of the

finite field extension OK/pi over Fp equals deg(fi), and we have [p] = pOK = pe11 . . .perr

where e1, . . ., er are the same as above.

Proof. The ring homomorphism Z[X ] → OK given by h(X) 7→ h(θ) is surjective, and its

kernel is generated by f(X). So this homomorphism induces an isomorphism

ϕ : Z[X ]/[f(X)]∼−→ OK .

Quotienting both sides by the ideal generated by p induces an isomorphism

ϕ : Fp[X ]/[f(X)]∼−→ OK/[p].

By the Chinese Remainder Theorem,

Fp[X ]/[f(X)] ∼=r∏

i=1

Fp[X ]/[fi(X)ei ].

The prime ideals of this ring are the principal ideals generated by the fi(X). The prime

ideals of OK/[p] are the principal ideals generated by the images ϕ(fi) = fi(θ) mod [p].

Put pi =[ϕ(fi)

]⊆ OK/[p].

Let ψ : OK → OK/[p] be the quotient map. Then the preimages pi := ψ−1(pi) of the

prime ideals in OK/[p] are exactly the prime ideals of OK containing p, and it’s easy

to see that pi = [fi(θ), p].

Since ψ(peii ) ⊆ ψ(pi)ei = pi

ei , we have peii ⊆ ψ−1(pi

ei)and hence

pe11 . . .perr =

r⋂

i=1

peii ⊆r⋂

i=1

ψ−1(pi

ei)= ψ−1

( r⋂

i=1

piei)

= ψ−1((0)

)= [p].

Write [p] = pa1

1 . . .parr with ai 6 ei. Then

deg(f) = dimFp

(Fp[X ]/[f(X)]

)= dimFp

(OK/[p]

)

=

r∑

i=1

ai dimFp

(OK/pi

)6

r∑

i=1

ei deg(fi) = deg(f),

and so ei = ai for all i. 2

26

Definition. Let K be a number field, n = [K : Q], and let [p] = pOK = pe11 . . .perr be the

decomposition of the ideal generated by the prime number p into prime ideals p1, . . ., pr

(with pi 6= pj for i 6= j). We say that p is

(i) ramified if ei > 2 for at least one i (and totally ramified if [p] = pe11 ),

(ii) unramified if ei = 1 for all i,

(iii) completely (totally) split if e1 = . . . = er = 1 and r = n,

(iv) inert if r = 1 and e1 = 1.

Remark. A number K which is Galois over Q (i.e. [K : Q] = #AutQ(K)) is characterised

by the set

SplK = {p a prime number : p is completely split in K}.

More precisely, if SplK1and SplK2

coincide up to finitely many elements, then they

are equal and K1∼= K2. (K1 and K2 are both Galois over Q.)

Let θ ∈ OK be such that K = Q(θ) and the f(X) ∈ Z[X ] be its minimal polynomial. Then

SplK is, up to a finite number of elements, equal to

Spl(f) = {p a prime number : f(X) mod p has deg(f) distinct roots in Fp}.

Question. How can we characterise Spl(f) in terms of the coefficients of f?

Example. If ℓ is an odd prime, then

Spl(X2 − ℓ) ={

p > prime :(ℓp

)= 1

QRL⇐⇒(pℓ

)= (−1)

p−1

2

ℓ−1

2

}

.

Class Field Theory tells us that Spl(f) can be described by congruence conditions if K/Q

is an abelian extension (i.e. AutQ(K) is an abelian group of order [K : Q]).

‘Non-abelian class field theory’ is supposed to give the general answer (but it is a current

subject of research). It is to do with elliptic curves and modular forms.

Theorem 8.3. A prime number p is ramified in K iff it divides the discriminant dK .

Proof. We prove only the case when OK = Z[θ] for some θ ∈ OK .

Let f(X) ∈ Z[X ] be the minimal polynomial of θ. Then p ramifies in K iff f(X) :=

f(X) mod p has an irreducible divisor of multiplicity greater than 1 (iff f(X) has a

root of multiplicity > 1 in Fp, the algebraic closure of Fp).

Then the discriminant of f(X) vanishes. But the discriminant of a polynomial of degree

n is a universal polynomial in the coefficients of f . (I.e., the same polynomial in the

coefficients, for all f . Recall: disc(f) = ±1× the resultant of f and f ′ – example sheet

1, question 6.)

This show that disc(f) is the reduction mod p of the discriminant of f , and the latter

is the discriminant of K. I.e., dK = disc(f) = disc(f) = 0 in Fp iff p|dK . 2

27

9 : Cyclotomic Fields

Definition. A number field K is called a cyclotomic field is K = Q(ζn) for some nth root

of unity ζn.

Below is a summary of some facts.

1. Let ϕ : Z>0 → Z>0 be the Euler ϕ-function, i.e. ϕ(n) = #((Z/nZ)∗

)= #

{d ∈ {1, . . ., n} :

hcf(d, n) = 1}.

Then, if ζn is a primitive nth root of unity (i.e. ζjn 6= 1 for all 1 6 j < n) the polynomial

Φn(X) =∏

16j6n

hcf(j,n)=1

(X − ζjn) ∈ C[X ]

has integral coefficients, and is the minimal polynomial of primitive nth root of unity.

We have∏

d|nΦd(X) = Xn − 1.

2. Let n = ℓν be a power of a prime number ℓ and put λ = 1− ζn. Then the principal ideal

[λ] = λOQ(ζn) generated by λ is a prime ideal with residue field Fℓ, and we have

ℓOQ(ζn) = [λ]µ, where µ = ϕ(n) = [Q(ζn) : Q].

Furthermore, 1, ζn, . . ., ζµ−1n is a basis of Q(ζn)/Q and has discriminant

d(1, ζn, . . ., ζµ−1n ) = ±ℓe, where e = ℓν−1(lν − ν − 1).

3. We use 2 to prove that for any n > 1, OQ(ζn) = Z[ζn].

4. 1 and 3 and Dedekind’s theorem give:

A prime p is completely split in Q(ζn)

⇔ Φn mod p splits into distinct linear factors of Fp[X ]

⇔ Xn − 1 splits into distinct linear factors in Fp[X ]

⇔ F∗p has a subgroup of order n

⇔ n|(p− 1)

⇔ p ≡ 1 mod n

Putting this all together, SplQ(ζn) = {p prime number : p ≡ 1 mod n}.

28

Lent Term 2008 M. Strauch

Number Fields: Example Sheet 1

(1) Which of the following are algebraic integers?

12,

√3 +√

52

,

√3 +√

7√2

,3 + 2

√6

1−√6.

(2) Let D ∈ Z, D 6= 0, D 6= 1, be a square-free integer, and put K = Q(√D).

(a) Show that the ring of integers OK of K is equal to Z[√D] if D ≡ 2 mod 4 or D ≡ 3 mod 4.

Show further that OK = Z[1+√D

2 ] if D ≡ 1 mod 4.

(b) Denote by dK the discriminant of K. Show that dK = 4D if D ≡ 2 mod 4 or D ≡ 3 mod 4,and dK = D if D ≡ 1 mod 4.

(3) (a) Let f(X) = a0Xn + . . . + an ∈ Z[X], a0 6= 0, be a polynomial. Show that, if f(ab ) = 0 for

a, b ∈ Z with gcd(a, b) = 1, then b|a0 and a|an.

(b) Determine which of the following polynomials are irreducible over Q: X3±X+1, X3±X+2,X3 ±X + 3.

(4) (a) Let n be a positive integer and A ∈Mn(Z) be a matrix. By using elementary column and rowoperations, show that there are matrices S, T ∈ GLn(Z) such that SAT is a diagonal matrix.

(b) Let N ⊂ Zn be a submodule of rank n. Show that there is a matrix A ∈ Mn(Z) such thatA(Zn) = N and the index [Zn : N ] of N in Zn is equal to | det(A)|.

(5) Let K be a number field of degree n = [K : Q], and let α1, . . . , αn ∈ OK be a basis of K/Q suchthat d(α1, . . . , αn) is a square-free integer. Show that α1, . . . , αn is an integral basis of OK over Z.

(6) (a) Let f(X) ∈ Q[X] be a monic irreducible polynomial of degree n and θ ∈ C a root of f . PutK = Q(θ). Show that the discriminant of the basis (1, θ, . . . , θn−1) ofK is equal to (−1)

n(n−1)2 R(f, f ′),

where R(f, f ′) denotes the resultant of f and its derivative f ′. The latter is also called the discrim-inant of f .

(b) Show that the discriminant of the polynomial X3 + cX + d is −4c3− 27d2. Show further that(1, θ, θ2) is an integral basis of OK for K = Q(θ), where θ3 + θ + 1 = 0.

(7) Let R be a commutative ring with unit. For a, b ∈ R we say that a divides b (notation a|b) ifb = ac for some c ∈ R. Note that a|1 ⇐⇒ a ∈ R∗. We say that a is associated to b iff a = ubwith u ∈ R∗ (notation a ∼ b). If R is a domain, then a ∼ b ⇐⇒ (a|b ∧ b|a). We call an elementa ∈ R−R∗ irreducible if for any factorization a = bc one of b, c is a unit in R. A non-zero non-unita is called a prime element if a generates a prime ideal. An integral domain R is called a uniquefactorization domain (UFD) if the following two conditions are satisfied:

(i) every element a ∈ R− {0}, which is not a unit can be written as a product of (finitely many)irreducible elements;

(ii) if a = x1 · · ·xr = y1 · · · ys with all xi, yj irreducible, then r = s and there is a permutation σof {1, . . . , r} such that for all i: xi ∼ yσ(i).

(a) Show that in any domain R the prime elements are irreducible, and that in an UFD theirreducible elements are prime elements. Show further that a domain in which (i) holds and in whichthe irreducible elements are prime elements is an UFD.

(b) Recall that a principal ideal domain (PID) is an integral domain in which every ideal isprincipal (that is, generated by a single element). Show that a PID is a Dedekind domain. Showfurther that a PID is an UFD.

Remark. Conversely, we will see later that a Dedekind domain which is an UFD is a PID.

(8) An integral domain R is called a euclidian domain if there is a map N : R−{0} → Z>0 such thatfor all a, b ∈ R, b 6= 0, there are d, r ∈ R with the property that

a = db+ r ,

with either r = 0 or r 6= 0 and N(r) < N(b).

(a) Show that the ring of Gaussian integers Z[i] is an euclidian ring. (Hint: take N = NQ(i)/Q anduse the graphic interpretation of elements of Z[i] as lattice points in C.)

(b) Show that any euclidian domain is a principal ideal domain. Deduce that Z[i] is a UFD.

(c) Show that the group of units of Z[i] is {1,−1, i,−i}.(9) (a) Let p be an odd prime number. Show that the congruence x2 ≡ −1 mod p has a solutionx ∈ Z if and only if p ≡ 1 mod 4. (Hint: use the fact that the multiplicative group F∗p is cyclic.)

(b) Use (a) and the preceding exercise to show that a prime p which is congruent to 1 mod 4 isof the form a2 + b2 with a, b ∈ Z. (Hint: p can not be a prime element in Z[i] because p|(x2 + 1)would then imply p|(x+ i) or p|(x− i). Thus p is not irreducible in Z[i].)

(c) Show that a prime number p which is congruent to 3 mod 4 is a prime element in Z[i].

(10) Prime elements in Z[i]. Use the preceding two exercises two show that the prime elements of Z[i]are, up to associated elements, given as follows:

(1) 1 + i,(2) p, with p ≡ 3 mod 4,(3) a+ ib, with p = a2 + b2 a prime number ≡ 1 mod 4 and a > |b|.

(11) The ring of integersOK of K = Q(√−5) is by exercise 2 equal to Z[

√−5]. Show that 3, 7, 1+2√−5

and 1 − 2√−5 are all irreducible elements in OK . (Hint: use the norm NK/Q.) Deduce from the

equation 3 · 7 = (1 + 2√−5) · (1− 2

√−5) that OK is not a UFD.

(12) Show that the rings of integersOK , whereK = Q(√D), are euclidian domains forD = −3,−2, 2, 3.

These rings are in particular all unique factorization domains. (Hint: proceed as in exercise 8.)

(13) Explain why the equation 2 · 11 = (5 +√

3) · (5−√3) is not inconsistent with the fact that Z[√

3]has unique factorization.

(14) Let G be the Galois group of K = Q(√

2,√

7) over Q. You may assume that G = {1, α, β, αβ}where

α(√

2) =√

2 , α(√

7) = −√

7 , β(√

2) = −√

2 , β(√

7) =√

7 .

By considering the relative traces θ + σ(θ), where σ runs through the elements of G other thanthe identity, show that the integers in K have the form

θ =12

(a+ b√

7 + c√

2 + d√

14) ,

where a, b, c, d are rational integers. By computing the relative norm θσ(θ), where σ ∈ G takes√2 to −√2, or otherwise, show that a and b are even and that c ≡ d mod 2. Hence prove that an

integral basis for OK is 1,√

2,√

7, 12(√

2 +√

14).

(15) Show that an integral domain with finitely many elements is a field.

Comments, corrections and queries can be send to me at M.Strauch@dpmms.cam.ac.uk

Lent Term 2008 Matthias Strauch

Number Fields:

Example Sheet 2

(1) Let O be the ring of integers in a Number Field1 and a ⊂ O a non-zero ideal. Show that any idealof the ring R = O/a can be generated by one element.

Hint: use the Chinese Remainder Theorem to reduce to the case where a = pn is a power ofa prime ideal. Then show that the only non-zero proper ideals of O/pn are p/pn, . . . , pn−1/pn. Ifπ ∈ p− p2 show that pν = [πν ] + pn, ν = 1, . . . , n.

(2) If O be the ring of integers in a Number Field2, show that any ideal can be generated by at mosttwo elements.

Hint: use the preceding exercise.

(3) Let K = Q(√−d), where d is a positive square-free integer. Establish the following facts about the

factorisation of principal ideals in OK .

(a) Suppose d has more than one prime factor. If the odd prime p divides d then [p] = p2, wherep is a non-principal prime ideal of OK .

(b) If d ≡ 1 or d ≡ 2 mod 4, then [2] = p2, with a non-principal prime ideal p ⊂ OK unless d = 1or d = 2.

(c) If d ≡ 7 mod 8 put ω = 1+√−d

2 . Then [2, ω] is not a principal ideal, unless d = 7 in whichcase it is principal. Furthermore, [2] = [2, ω][2, ω̄], where ω̄ = 1−

√−d

2 .

Deduce that if K has class number one, then either d = 1, 2 or 7 or d is prime and d ≡ 3 mod 8.

(4) Let K be a number field of degree n over Q. Assume there is θ ∈ OK such that 1, θ, . . . , θn−1 is anintegral basis of OK . Let f(X) ∈ Z[X] be the minimal polynomial of θ.

(a) Show that the map Z[X] → OK , X 7→ θ, induces an isomorphism Z[X]/[f(X)] '−→ OK .

(b) For any prime number p, show that OK/[p] is isomorphic to Fp[X]/[f̄(X)], where f̄(X) ∈Fp(X) denotes the image of f(X) under the canonical map Z[X] → Fp[X].

(c) Let p be a prime number. Deduce from (b) that the ideal [p] = pOK is a prime ideal if andonly if f̄(X) is an irreducible polynomial in Fp[X].

(5) Let K = Q(√

d), where d 6= 1 is a non-zero square-free integer.

(a) Suppose d ≡ 1 mod 4 and let p be an odd prime number. Show that X2 − X + 1−d4 is

irreducible in Fp[X] if and only if X2 − d is irreducible in Fp[X].

(b) Let p be an odd prime number. Show that the ideal [p] = pOK is a prime ideal if and only ifthe congruence X2 ≡ d mod p does not have a solution.

(c) Suppose d ≡ −3 mod 8. Show that X2 − X + 1−d4 is irreducible in F2[X]. Deduce that

[2] = 2OK is a prime ideal in OK .

(6) We denote by Md = 2π |dQ(

√−d)|

12 the Minkowski constant of K = Q(

√−d). One has M7 ≈ 1.7,

M11 ≈ 2.1, M19 ≈ 2.8, M43 ≈ 4.2, M67 ≈ 5.2 and M163 ≈ 8.1.

Use the fact that the ideal class group is generated by the classes of prime ideals p which appearin the factorisation of the primes p ≤ Md to show that Q(

√−d) has class number one for d =

1, 2, 3, 7, 11, 19, 43, 67 and 163. (For the first three values of d you can cite an exercise on the firstexample sheet.) These values of d are indeed the only positive values for which Q(

√−d) has class

number one.

1this holds more generally if o is a Dedekind domain2this holds more generally if o is a Dedekind domain

(7) Show that the class number of Q(√−5) is two. (Use exercise 3 and that the Minkowski constant of

Q(√−5) is ≈ 2.84.)

(8) Put K = Q(√−6). Show that p = [2,

√−6] and q = [3,

√−6] are prime ideals of OK satisfying

p2 = [2] and q2 = [3] (cf. exercise 3). Find a relation between these two prime ideals and concludethat K has class number two. (You may use that the Minkowski constant of K is ≈ 3.12.)

(9) Prove that the prime 3 generates a prime ideal in the ring of integers of Q(√−10). Show further that

this number field has class number two. (You may use that the Minkowski constant of Q(√−10) is

≈ 4.02.)

(10) Put K = Q(√−17) and ω = 1 +

√−17. Prove that the prime 5 generates a prime ideal in the ring

of integers of K. Show that the following relations hold in the group of fractional ideals of K:

[2] = [2, ω]2 , 3 = [3, ω][3, ω̄] , [ω] = [2, ω][3, ω]2 ,

where ω̄ = 1−√−17. Deduce that the class group of K is cyclic of order four. (You may use that

the Minkowski constant of K is ≈ 5.25.)

(11) Let θ ∈ C be a root of X3 + X + 1, and put K = Q(θ). Show that the Minkowski constant of K is≈ 1.58 (you may use an exercise on the previous example sheet). Deduce that K has class numberone.

(12) (a) Show that if K is a number field of degree n over Q, then

|dK | ≥(

nn

n!

)2 (π

4

)n.

Deduce that |dK | > 1 for every number field K 6= Q.

(b) Show that there are constants A > 1, c > 1, such that for every number field K one has|dK | ≥ 1

cAn, where n is the degree of K over Q. Deduce that for every d > 0 there is some N ∈ Z

such that, if K/Q is a number field whose discriminant is bounded by d, then [K : Q] ≤ N .

(13) Let ζ ∈ C be a primitive fifth root of unity and K = Q(ζ). Use (without proof) that 1, ζ, ζ2, ζ3 isan integral basis of OK to show that the discriminant of K is equal to 125. Compute the Minkowskiconstant and deduce that K has class number one.

(14) Let K be a number field. We define the Dedekind ζ-function ζK(s) by ζK(s) =∑

0 6=a⊂OK

1N(a)s ,

where the summation is over all non-zero ideals a of OK . Show that there is a formal identity

ζK(s) =∏p

11−N(p)−s

,

where the summation on the right is over all non-zero prime ideals of OK . (One can show that bothsides converge for Re(s) > 1 and define holomorphic functions in this domain.) Now let K = Q(i).Use exercise (10) from example sheet 1 to prove that

ζK(s) = ζQ(s) · L(χ, s) with L(χ, s) =∏p

11− χ(p)p−s

,

the product running over all odd prime numbers and χ(p) = (−1)p−12 . Show that

L(χ, s) =∞∑

n=0

(−1)n

(2n + 1)s= 1− 1

3s+

15s− 1

7s± · · ·

Comments, corrections and queries can be send to me at M.Strauch@dpmms.cam.ac.uk

Lent Term 2008 Matthias Strauch

Number Fields: Example Sheet 3

(1) Let D > 1 be a square-free integer and put K = Q(√

D). Recall that the fundamental unit of Kis an element ε0 ∈ O∗

K such that ε0 = min{ε ∈ O∗K | ε > 1}. Use the algorithm explained in the

lectures to determine the fundamental unit of K for D = 13, 17, 26, 29, 35, 37, 53 and 77.

(2) Let m ≥ 1 and D1, . . . , Dm be pairwise co-prime integers, Di /∈ {0, 1} for all i. Put K =Q(

√D1, . . . ,

√Dm). Show by induction over m that [K : Q] = 2m.

(3) For a number field K let as usual r and s denote the number of real and half the number of complexembeddings, respectively. Determine r and s in the following cases:

(a) K = Q(√

D1, . . . ,√

Dm) as in the preceding exercise.

(b) K = Q( m

√D), where D > 1 is a square-free integer and m ≥ 2.

(4) Let K be a number field. Recall that a prime number p is called ramified in K if in the prime ideal

decomposition [p] = pOK = pe1

1· · · per

r at least one of the exponents ei is > 1. Now let K = Q(√

D)for some square-free integer D /∈ {0, 1}. On a previous example sheet we have seen that OK = Z[θ]for some θ ∈ OK . Use the explicit description of θ and Dedekind’s theorem to give a direct proofthat the primes which ramify in K are the prime divisors of the discriminant of K.

(5) Let K = Q(√

26) and let ε = 5 +√

26. Use Dedekind’s theorem to show that the ideal equations

[2] = [2, ε + 1]2 , [5] = [5, ε + 1][5, ε − 1] , [ε + 1] = [2, ε + 1][5, ε + 1]

hold in K. Deduce that K has class number two. (Argue with the Minkowski constant.)

ε is the fundamental unit of K, by a preceding exercise. Use this fact to show that all solutions inintegers x, y of the equation x2 − 26y2 = ±10 are given by

x +√

26y = ±εn(ε ± 1) , n = 0,±1,±2, . . .

(6) Show that ε = 3+√

7

3−√

7is a unit in K = Q(

√7). Show further that [2] is the square of the principal

ideal in OK generated by 3 +√

7. Use the Minkowski constant to show that K has class numberone.

Assuming further that ε is the fundamental unit in K, show that all solutions in integers x, y of theequation x2 − 7y2 = 2 are given by

x +√

26y = ±εn(3 +√

7) , n = 0,±1,±2, . . .

(7) Let K = Q(√

35). By Dedekind’s theorem, or otherwise, show that the ideal equations

[2] = [2, ω]2 , [5] = [5, ω]2 , [ω] = [2, ω][5, ω]

hold in K, where ω = 5 +√

35. Deduce that K has class number two. (Argue with the Minkowskiconstant.)

ω+1 is the fundamental unit of K, by a preceding exercise. Hence show that all solutions in integersx, y of the equation x2 − 35y2 = −10 are given by

x +√

35y = ±ω(ω + 1)n , n = 0,±1,±2, . . .

Calculate the particular solution x, y for n = 1.

(8) Let K = Q(√−34). By Dedekind’s theorem, or otherwise, factorise 2, 3, 5 and 7 into prime ideals

in OK . Show that the ideal equations

[ω] = [5, ω][7, ω] , [ω + 3] = [2, ω + 3][5, ω + 3]2

hold in K, where ω = 1 +√−34. Deduce that the class group of K is cyclic of order four. (Argue

with the Minkowski constant.)

(9) By exercises (6) and (7) of example sheet 2, we know the class groups of the imaginary quadraticfields Q(

√−5) and Q(

√−11). Use this information to find all solutions in integers of the diophantine

equations

y2 + 5 = x3 , y2 + 11 = x3 .

(10) Let K be a number field of degree n = r + 2s. Denote by ρ1, . . . , ρr the real embeddings of K andby σ1, σ̄1, . . . , σs, σ̄s the complex embeddings of K into C. Recall the map λ as introduced in thelectures

λ : O∗K −→ Rr+s , α 7→ (log(|τ1(α)|, . . . , log(|τr(α)|), log(|σ1(α)|2), . . . , log(|σs(α)|2)) .

The image of λ is a complete lattice in the hyperplane

H = {(x1, . . . , xr, ξ1, . . . , ξs) ∈ Rr+s |r

∑

i=1

xi +s

∑

j=1

ξj = 0} .

We consider Rr+s with its standard scalar product and restrict it to H, thereby getting a well-definednotion of volume on H. Show that the volume of a fundamental mesh of the lattice Γ = λ(O∗

K) isequal to

√r + sRK where RK is the absolute value of the determinant of an arbitrary minor of rank

t = r + s − 1 of the following matrix

λ1(ε1) · · · λ1(εt)

......

λt+1(ε1) · · · λt+1(εt)

Here ε1, . . . , εt is a system of fundamental units and (λ1(εi), . . . , λt+1(εi))t = λ(εi), in the standard

coordinates on Rr+s. RK is called the regulator of K. (Hint: The column vector λ0 = 1√r+s

(1, . . . , 1)t

is perpendicular to H and of length one; the volume of a fundamental mesh of Γ is thus given bythe absolute value of the determinant of the matrix (λ0 λ(ε1) · · ·λ(εt)). Then add all rows to a fixedone.)

(11) Let K ⊂ L be number fields and L = K(θ) for some θ ∈ OL. Let f(X) ∈ OK [x] be the minimalpolynomial of θ over K, and put c = {α ∈ OL | α · OL ⊂ OK [θ]}. This is a non-zero ideal ofOL. Generalise Dedekind’s theorem as follows: if the prime ideal p ⊂ OK is co-prime to c (i.e.pOL + c = OL), and f̄(x) = f̄1(x)e1 · · · f̄r(x)er is the decomposition of f̄(x) = f(x) mod p in(OK/p)[x] into irreducible monic polynomials, then P1 = [f1(θ), p], . . . ,Pr = [fr(θ), p] are the rdifferent prime ideals of OL containing pOL and pOL = Pe1

1· · ·Per

r . (Here fi(x) ∈ OK [x] is a monicpolynomial whose reduction modulo p is f̄i.)

(12) Let K = Q(√

D1, . . . ,√

Dm) with D1, . . . , Dm be pairwise co-prime integers, Di /∈ {0, 1} for all i.Use the assertion of the preceding exercise that, up to at most finitely many exceptions, a primenumber p splits completely in OK , i.e. [p] = p1 · · · pn with n = 2m and pairwise different prime idealspi, if and only if all the congruences X2

1 ≡ D1, . . . , X2m ≡ Dm have a solution modulo p.

(13) Use the preceding exercise and the quadratic reciprocity law to show that, up to at most finitelymany exceptions, a prime p splits completely in Q(i,

√3) if and only if p ≡ 1 modulo 12.

Comments, corrections and queries can be send to me at M.Strauch@dpmms.cam.ac.uk