A Course In Algebraic Number Theory

Robert B. Ash

Preface

This is a text for a basic course in algebraic number theory, written in accordance with

the following objectives.

1. Provide reasonable coverage for a one-semester course.

2. Assume as prerequisite a standard graduate course in algebra, but cover integral ex-

tensions and localization before beginning algebraic number theory. For general algebraic

background, see my online text “Abstract Algebra: The Basic Graduate Year”, which

can be downloaded from my web site www.math.uiuc.edu/∼ r-ash/ The abstract algebra

material is referred to in this text as TBGY.

3. Cover the general theory of factorization of ideals in Dedekind domains, as well as the

number ¬eld case.

4. Do some detailed calculations illustrating the use of Kummer™s theorem on lifting of

prime ideals in extension ¬elds.

5. Give enough details so that the reader can navigate through the intricate proofs of the

Dirichlet unit theorem and the Minkowski bounds on element and ideal norms.

6. Cover the factorization of prime ideals in Galois extensions.

7. Cover local as well as global ¬elds, including the Artin-Whaples approximation theorem

and Hensel™s lemma.

Especially helpful to me in preparing this work were the beautiful little book by

Samuel, “Algebraic Theory of Numbers”, Hermann 1971, and the treatment of cyclotomic

¬elds by J. Milne in his online text “Algebraic Number Theory” (www.math.lsa.umich.edu/∼

jmilne/) Some other useful references are:

Esmonde, J., and Murty, M.R., “Problems in Algebraic Number Theory”, Springer 1999

Fr¨lich, A., and Taylor, M.J., “Algebraic Number Theory”, Cambridge 1991

o

Janusz, G.J.,“ Algebraic Number Fields”, AMS 1996

Marcus, D.A., “Number Fields”, Springer 1977

Stewart, I., and Tall, D., “Algebraic Number Theory”, Chapman and Hall 1987

c copyright 2003, by Robert B. Ash. Paper or electronic copies for noncommercial use

may be made freely without explicit permission of the author. All other rights are reserved.

Table of Contents

Chapter 1 Introduction

1.1 Integral Extensions

1.2 Localization

Chapter 2 Norms, Traces and Discriminants

2.1 Norms and traces

2.2 The Basic Setup For Algebraic Number Theory

2.3 The Discriminant

Chapter 3 Dedekind Domains

3.1 The De¬nition and Some Basic Properties

3.2 Fractional Ideals

3.3 Unique Factorization of Ideals

3.4 Some Arithmetic in Dedekind Domains

Chapter 4 Factorization of Prime Ideals in Extensions

4.1 Lifting of Prime Ideals

4.2 Norms of ideals

4.3 A Practical Factorization Theorem

Chapter 5 The Ideal Class Group

5.1 Lattices

5.2 A Volume Calculation

5.3 The Canonical Embedding

Chapter 6 The Dirichlet Unit Theorem

6.1 Preliminary Results

6.2 Statement and Proof of Dirichlet™s Unit Theorem

6.3 Units in Quadratic Fields

1

2

Chapter 7 Cyclotomic Extensions

7.1 Some Preliminary Calculations

7.2 An Integral Basis of a Cyclotomic Field

Chapter 8 Factorization of Prime Ideals in Galois Extensions

8.1 Decomposition and Inertia Groups

8.2 The Frobenius Automorphism

8.3 Applications

Chapter 9 Local Fields

9.1 Absolute Values and Discrete Valuations

9.2 Absolute Values on the Rationals

9.3 Artin-Whaples Approximation Theorem

9.4 Completions

9.5 Hensel™s Lemma

Chapter 1

Introduction

Techniques of abstract algebra have been applied to problems in number theory for a long

time, notably in the e¬ort to prove Fermat™s last theorem. As an introductory example,

we will sketch a problem for which an algebraic approach works very well. If p is an odd

prime and p ≡ 1 mod 4, we will prove that p is the sum of two squares, that is, p can

expressed as x2 + y 2 where x and y are integers. Since p’1 is even, it follows that ’1

2

is a quadratic residue (that is, a square) mod p. To see this, pair each of the numbers

2, 3, . . . , p ’ 2 with its inverse mod p, and pair 1 with p ’ 1 ≡ ’1 mod p. The product of

the numbers 1 through p ’ 1 is, mod p,

p’1 p’1

1 — 2 — ··· — — ’1 — ’2 · · · — ’

2 2

and therefore

p’1 2

)!] ≡ ’1 mod p.

[(

2

If ’1 ≡ x2 mod p, then p divides x2 + 1. Now we enter the ring Z[i] of Gaussian integers

and factor x2 + 1 as (x + i)(x ’ i). Since p can divide neither factor, it follows that p is

not prime in Z[i]. Since the Gaussian integers form a unique factorization domain, p is

not irreducible, and we can write p = ±β where neither ± nor β is a unit.

De¬ne the norm of γ = a + bi as N (γ) = a2 + b2 . Then N (γ) = 1 i¬ γ is 1,-1,i or ’i,

equivalently, i¬ γ is a unit. Thus

p2 = N (p) = N (±)N (β) with N (±) > 1 and N (β) > 1,

so N (±) = N (β) = p. If ± = x + iy, then p = x2 + y 2 .

Conversely, if p is an odd prime and p = x2 + y 2 , then p is congruent to 1 mod 4. [If

x is even, then x2 ≡ 0 mod 4, and if x is odd, then x2 ≡ 1 mod 4. We cannot have x and

y both even or both odd, since p is odd.]

It is natural to conjecture that we can identify those primes that can be represented as

√

x + |m|y 2 , where m is a negative integer, by working in the ring Z[ m]. But the above

2

argument depends critically on unique factorization, which does not hold in general. A

1

2 CHAPTER 1. INTRODUCTION

√ √ √

standard example is 2 — 3 = (1 + ’5)(1 ’ ’5) in Z[ ’5]. Di¬culties of this sort led

Kummer to invent “ideal numbers”, which became ideals at the hands of Dedekind. We

will see that although a ring of algebraic integers need not be a UFD, unique factorization

of ideals will always hold.

1.1 Integral Extensions

If E/F is a ¬eld extension and ± ∈ E, then ± is algebraic over F i¬ ± is a root of

a nonconstant polynomial with coe¬cients in F . We can assume if we like that the

polynomial is monic, and this turns out to be crucial in generalizing the idea to ring

extensions.

1.1.1 De¬nitions and Comments

All rings are assumed commutative. Let A be a subring of the ring R, and let x ∈ R. We

say that x is integral over A if x is a root of a monic polynomial f with coe¬cients in

A. The equation f (X) = 0 is called an equation of integral dependence for x over A. If x

is a real or complex number that is integral over Z, then x is called an algebraic integer.

√

Thus for every integer d, d is an algebraic integer, as is any nth root of unity. (The

monic polynomials are, respectively, X 2 ’ d and X n ’ 1.) The next results gives several

conditions equivalent to integrality.

1.1.2 Theorem

Let A be a subring of R, and let x ∈ R. The following conditions are equivalent:

(i) The element x is integral over A;

(ii) The A-module A[x] is ¬nitely generated;

(iii) The element x belongs to a subring B of R such that A ⊆ B and B is a ¬nitely

generated A-module;

(iv) There is a subring B of R such that B is a ¬nitely generated A-module and x stabilizes

B, that is, xB ⊆ B. (If R is a ¬eld, the assumption that B is a subring can be dropped,

as long as B = 0);

(v) There is a faithful A[x]-module B that is ¬nitely generated as an A-module. (Recall

that a faithful module is one whose annihilator is 0.)

Proof.

(i)implies (ii): If x is a root of a monic polynomial of degree n over A, then xn and all

higher powers of x can be expressed as linear combinations of lower powers of x. Thus

1, x, x2 , . . . , xn’1 generate A[x] over A.

(ii) implies (iii): Take B = A[x].

(iii) implies (i): If β1 , . . . , βn generate B over A, then xβi is a linear combination of the

n

βj , say xβi = j=1 cij βj . Thus if β is a column vector whose components are the βi , I

is an n by n identity matrix, and C = [cij ], then

(xI ’ C)β = 0,

1.1. INTEGRAL EXTENSIONS 3

and if we premultiply by the adjoint matrix of xI ’ C (as in Cramer™s rule), we get

[det(xI ’ C)]Iβ = 0

hence det(xI ’ C)b = 0 for every b ∈ B. Since B is a ring, we may set b = 1 and conclude

that x is a root of the monic polynomial det(XI ’ C) in A[X].

If we replace (iii) by (iv), the same proofs work. If R is a ¬eld, then in (iv)’(i), x is

an eigenvalue of C, so det(xI ’ C) = 0.

If we replace (iii) by (v), the proofs go through as before. [Since B is an A[x]-module,

in (v)’(i) we have xβi ∈ B. When we obtain [det(xI ’ C)]b = 0 for every b ∈ B, the

hypothesis that B is faithful yields det(xI ’ C) = 0.] ™

We are going to prove a transitivity property for integral extensions, and the following

result will be helpful.

1.1.3 Lemma

Let A be a subring of R, with x1 , . . . , xn ∈ R. If x1 is integral over A, x2 is integral

over A[x1 ], . . . , and xn is integral over A[x1 , . . . , xn’1 ], then A[x1 , . . . , xn ] is a ¬nitely

generated A-module.

Proof. The n = 1 case follows from (1.1.2), condition (ii). Going from n ’ 1 to n amounts

to proving that if A, B and C are rings, with C a ¬nitely generated B-module and B a

¬nitely generated A-module, then C is a ¬nitely generated A-module. This follows by a

brief computation:

r s r s

Ayj xk . ™

C= Byj , B = Axk , so C =

j=1 j=1 k=1

k=1

1.1.4 Transitivity of Integral Extensions

Let A, B and C be subrings of R. If C is integral over B, that is, every element of C is

integral over B, and B is integral over A, then C is integral over A.

Proof. Let x ∈ C, with xn + bn’1 xn’1 + · · · + b1 x + b0 = 0, bi ∈ B. Then x is integral

over A[b0 , . . . , bn’1 ]. Each bi is integral over A, hence over A[b0 , . . . , bi’1 ]. By (1.1.3),

A[b0 , . . . , bn’1 , x] is a ¬nitely generated A-module. It follows from condition (iii) of (1.1.2)

that x is integral over A. ™

1.1.5 De¬nitions and Comments

If A is a subring of R, the integral closure of A in R is the set Ac of elements of R that

are integral over A. Note that A ⊆ Ac because each a ∈ A is a root of X ’ a. We say that

A is integrally closed in R if Ac = A. If we simply say that A is integrally closed without

reference to R, we assume that A is an integral domain with fraction ¬eld K, and A is

integrally closed in K.

If x and y are integral over A, then just as in the proof of (1.1.4), it follows from

(1.1.3) that A[x, y] is a ¬nitely generated A-module. Since x + y, x ’ y and xy belong to

4 CHAPTER 1. INTRODUCTION

this module, they are integral over A by (1.1.2), condition (iii). The important conclusion

is that

Ac is a subring of R containing A.

If we take the integral closure of the integral closure, we get nothing new.

1.1.6 Proposition

The integral closure Ac of A in R is integrally closed in R.

Proof. By de¬nition, Ac is integral over A. If x is integral over Ac , then as in the proof

of (1.1.4), x is integral over A, and therefore x ∈ Ac . ™

We can identify a large class of integrally closed rings.

1.1.7 Proposition

If A is a UFD, then A is integrally closed.

Proof. If x belongs to the fraction ¬eld K, then we can write x = a/b where a, b ∈ A,

with a and b relatively prime. If x is integral over A, then there is an equation of the form

(a/b)n + an’1 (a/b)n’1 + · · · + a1 (a/b) + a0 = 0

with all ai belonging to A. Multiplying by bn , we have an + bc = 0, with c ∈ A. Thus b

divides an , which cannot happen for relatively prime a and b unless b has no prime factors

at all, in other words, b is a unit. But then x = ab’1 ∈ A. ™

Problems For Section 1.1

Let A be a subring of the integral domain B, with B integral over A. In Problems 1-3,

we are going to show that A is a ¬eld if and only if B is a ¬eld.

1. Assume that B is a ¬eld, and let a be a nonzero element of A. Then since a’1 ∈ B,

there is an equation of the form

(a’1 )n + cn’1 (a’1 )n’1 + · · · + c1 a’1 + c0 = 0

with all ci belonging to A. Show that a’1 ∈ A, proving that A is a ¬eld.

2. Now assume that A is a ¬eld, and let b be a nonzero element of B. By condition

(ii) of (1.1.2), A[b] is a ¬nite-dimensional vector space over A. Let f be the A-linear

transformation on this vector space given by multiplication by b, in other words, f (z) =

bz, z ∈ A[b]. Show that f is injective.

3. Show that f is surjective as well, and conclude that B is a ¬eld.

In Problems 4-6, let A be a subring of B, with B integral over A. Let Q be a prime

ideal of B and let P = Q © A.

4. Show that P is a prime ideal of A, and that A/P can be regarded as a subring of B/Q.

5. Show that B/Q is integral over A/P .

6. Show that P is a maximal ideal of A if and only if Q is a maximal ideal of B.

1.2. LOCALIZATION 5

1.2 Localization

Let S be a subset of the ring R, and assume that S is multiplicative, in other words,

0 ∈ S, 1 ∈ S, and if a and b belong to S, so does ab. In the case of interest to us, S will

/

be the complement of a prime ideal. We would like to divide elements of R by elements

of S to form the localized ring S ’1 R, also called the ring of fractions of R by S. There is

no di¬culty when R is an integral domain, because in this case all division takes place in

the fraction ¬eld of R. Although we will not need the general construction for arbitrary

rings R, we will give a sketch. For full details, see TBGY, Section 2.8.

1.2.1 Construction of the Localized Ring

If S is a multiplicative subset of the ring R, we de¬ne an equivalence relation on R — S

by (a, b) ∼ (c, d) i¬ for some s ∈ S we have s(ad ’ bc) = 0. If a ∈ R and b ∈ S, we de¬ne

the fraction a/b as the equivalence class of (a, b). We make the set of fractions into a ring

in a natural way. The sum of a/b and c/d is de¬ned as (ad + bc)/bd, and the product of

a/b and c/d is de¬ned as ac/bd. The additive identity is 0/1, which coincides with 0/s for

every s ∈ S. The additive inverse of a/b is ’(a/b) = (’a)/b. The multiplicative identity

is 1/1, which coincides with s/s for every s ∈ S. To summarize:

S ’1 R is a ring. If R is an integral domain, so is S ’1 R. If R is an integral domain and

S = R \ {0}, then S ’1 R is a ¬eld, the fraction ¬eld of R.

There is a natural ring homomorphism h : R ’ S ’1 R given by h(a) = a/1. If S

has no zero-divisors, then h is a monomorphism, so R can be embedded in S ’1 R. In

particular, a ring R can be embedded in its full ring of fractions S ’1 R, where S consists

of all non-divisors of 0 in R. An integral domain can be embedded in its fraction ¬eld.

Our goal is to study the relation between prime ideals of R and prime ideals of S ’1 R.

1.2.2 Lemma

If X is any subset of R, de¬ne S ’1 X = {x/s : x ∈ X, s ∈ S}. If I is an ideal of R, then

S ’1 I is an ideal of S ’1 R. If J is another ideal of R, then

(i) S ’1 (I + J) = S ’1 I + S ’1 J;

(ii) S ’1 (IJ) = (S ’1 I)(S ’1 J);

(iii) S ’1 (I © J) = (S ’1 I) © (S ’1 J);

(iv) S ’1 I is a proper ideal i¬ S © I = ….

Proof. The de¬nitions of addition and multiplication in S ’1 R imply that S ’1 R is an

ideal, and that in (i), (ii) and (iii), the left side is contained in the right side. The reverse

inclusions in (i) and (ii) follow from

ab at + bs a b ab

+= , =.

s t st st st

To prove (iii), let a/s = b/t, where a ∈ I, b ∈ J, s, t ∈ S. There exists u ∈ S such that

u(at ’ bs) = 0. Then a/s = uat/ust = ubs/ust ∈ S ’1 (I © J).

Finally, if s ∈ S © I, then 1/1 = s/s ∈ S ’1 I, so S ’1 I = S ’1 R. Conversely, if

S ’1 I = S ’1 R, then 1/1 = a/s for some a ∈ I, s ∈ S. There exists t ∈ S such that

t(s ’ a) = 0, so at = st ∈ S © I. ™

6 CHAPTER 1. INTRODUCTION

Ideals in S ’1 R must be of a special form.

1.2.3 Lemma

Let h be the natural homomorphism from R to S ’1 R [see (1.2.1)]. If J is an ideal of

S ’1 R and I = h’1 (J), then I is an ideal of R and S ’1 I = J.

Proof. I is an ideal by the basic properties of preimages of sets. Let a/s ∈ S ’1 I, with

a ∈ I and s ∈ S. Then a/1 = h(a) ∈ J, so a/s = (a/1)(1/s) ∈ J. Conversely, let a/s ∈ J,

with a ∈ R, s ∈ S. Then h(a) = a/1 = (a/s)(s/1) ∈ J, so a ∈ I and a/s ∈ S ’1 I. ™

Prime ideals yield sharper results.

1.2.4 Lemma

If I is any ideal of R, then I ⊆ h’1 (S ’1 I). There will be equality if I is prime and disjoint

from S.

Proof. If a ∈ I, then h(a) = a/1 ∈ S ’1 I. Thus assume that I is prime and disjoint from

S, and let a ∈ h’1 (S ’1 I). Then h(a) = a/1 ∈ S ’1 I, so a/1 = b/s for some b ∈ I, s ∈ S.

There exists t ∈ S such that t(as ’ b) = 0. Thus ast = bt ∈ I, with st ∈ I because

/

S © I = …. Since I is prime, we have a ∈ I. ™

1.2.5 Lemma

If I is a prime ideal of R disjoint from S, then S ’1 I is a prime ideal of S ’1 R.

Proof. By part (iv) of (1.2.2), S ’1 I is a proper ideal. Let (a/s)(b/t) = ab/st ∈ S ’1 I,

with a, b ∈ R, s, t ∈ S. Then ab/st = c/u for some c ∈ I, u ∈ S. There exists v ∈ S such

that v(abu ’ cst) = 0. Thus abuv = cstv ∈ I, and uv ∈ I because S © I = …. Since I is

/

prime, ab ∈ I, hence a ∈ I or b ∈ I. Therefore either a/s or b/t belongs to S ’1 I. ™

The sequence of lemmas can be assembled to give a precise conclusion.

1.2.6 Theorem

There is a one-to-one correspondence between prime ideals P of R that are disjoint from

S and prime ideals Q of S ’1 R, given by

P ’ S ’1 P and Q ’ h’1 (Q).

Proof. By (1.2.3), S ’1 (h’1 (Q)) = Q, and by (1.2.4), h’1 (S ’1 P ) = P . By (1.2.5), S ’1 P

is a prime ideal, and h’1 (Q) is a prime ideal by the basic properties of preimages of sets.

If h’1 (Q) meets S, then by (1.2.2) part (iv), Q = S ’1 (h’1 (Q)) = S ’1 R, a contradiction.

Thus the maps P ’ S ’1 P and Q ’ h’1 (Q) are inverses of each other, and the result

follows. ™

1.2. LOCALIZATION 7

1.2.7 De¬nitions and Comments

If P is a prime ideal of R, then S = R \ P is a multiplicative set. In this case, we write

RP for S ’1 R, and call it the localization of R at P . We are going to show that RP is

a local ring, that is, a ring with a unique maximal ideal. First, we give some conditions

equivalent to the de¬nition of a local ring.

1.2.8 Proposition

For a ring R, the following conditions are equivalent.

(i) R is a local ring;

(ii) There is a proper ideal I of R that contains all nonunits of R;

(iii) The set of nonunits of R is an ideal.

Proof.

(i) implies (ii): If a is a nonunit, then (a) is a proper ideal, hence is contained in the

unique maximal ideal I.

(ii) implies (iii): If a and b are nonunits, so are a + b and ra. If not, then I contains a

unit, so I = R, contradicting the hypothesis.

(iii) implies (i): If I is the ideal of nonunits, then I is maximal, because any larger ideal J

would have to contain a unit, so J = R. If H is any proper ideal, then H cannot contain

a unit, so H ⊆ I. Therefore I is the unique maximal ideal. ™

1.2.9 Theorem

RP is a local ring.

Proof. Let Q be a maximal ideal of RP . Then Q is prime, so by (1.2.6), Q = S ’1 I

for some prime ideal I of R that is disjoint from S = R \ P . In other words, I ⊆ P .

Consequently, Q = S ’1 I ⊆ S ’1 P . If S ’1 P = RP = S ’1 R, then by (1.2.2) part (iv), P

is not disjoint from S = R \ P , which is impossible. Therefore S ’1 P is a proper ideal

containing every maximal ideal, so it must be the unique maximal ideal. ™

1.2.10 Remark

It is convenient to write the ideal S ’1 I as IRP . There is no ambiguity, because the

product of an element of I and an arbitrary element of R belongs to I.

1.2.11 Localization of Modules

If M is an R-module and S a multiplicative subset of R, we can essentially repeat the

construction of (1.2.1) to form the localization of M by S, and thereby divide elements

of M by elements of S. If x, y ∈ M and s, t ∈ S, we call (x, s) and (y, t) equivalent if for

some u ∈ S, we have u(tx ’ sy) = 0. The equivalence class of (x, s) is denoted by x/s,

and addition is de¬ned by

xy tx + sy

+= .

s t st

8 CHAPTER 1. INTRODUCTION

If a/s ∈ S ’1 R and x/t ∈ s’1 M , we de¬ne

ax ax

= .

st st

In this way, S ’1 M becomes an S ’1 R-module. Exactly as in (1.2.2), if M and N are

submodules of an R-module L, then

S ’1 (M + N ) = S ’1 M + S ’1 N and S ’1 (M © N ) = (S ’1 M ) © (S ’1 N ).

Problems For Section 1.2

1. Let M be a maximal ideal of R, and assume that for every x ∈ M, 1 + x is a unit.

Show that R is a local ring (with maximal ideal M).

2. Show that if p is prime and n is a positive integer, then Z/pn Z is a local ring with

maximal ideal (p).

3. For any ¬eld k, let R be the ring of rational functions f /g with f, g ∈ k[X1 , . . . , Xn ]

and g(a) = 0, where a is a ¬xed point of k n . Show that R is a local ring, and identify the

unique maximal ideal.

Let S be a multiplicative subset of the ring R. We are going to construct a mapping

from R-modules to S ’1 R-modules, and another mapping from R-module homomorphisms

to S ’1 R-module homomorphisms, as follows. If M is an R-module, we map M to S ’1 M .

If f : M ’ N is an R-module homomorphism, we de¬ne S ’1 f : S ’1 M ’ S ’1 N by

x f (x)

’ .

s s

Since f is a homomorphism, so is S ’1 f . In Problems 4-6, we study these mappings.

4. Let f : M ’ N and g : N ’ L be R-module homomorphisms. Show that S ’1 (g —¦ f ) =

(S ’1 g) —¦ (S ’1 f ). Also, if 1M is the identity mapping on M , show that S ’1 1M = 1S ’1 M .

Thus we have a functor S ’1 , called the localization functor, from the category of R-

modules to the category of S ’1 R-modules.

5. If

f g

M ’’’ N ’’’ L

’’ ’’

is an exact sequence of R-modules, show that

S ’1 f S ’1 g

S ’1 M ’ ’ ’ S ’1 N ’ ’ ’ S ’1 L

’’ ’’

is exact. Thus S ’1 is an exact functor.

6. If M is an R-module and S is a multiplicative subset of R, denote S ’1 M by MS . If

N is a submodule of M , show that (M/N )S ∼ MS /NS .

=

Chapter 2

Norms, Traces and

Discriminants

We continue building our algebraic background to prepare for algebraic number theory.

2.1 Norms and Traces

2.1.1 De¬nitions and Comments

If E/F is a ¬eld extension of ¬nite degree n, then in particular, E is a ¬nite-dimensional

vector space over F , and the machinery of basic linear algebra becomes available. If x is

any element of E, we can study the F -linear transformation m(x) given by multiplication

by x, that is, m(x)y = xy. We de¬ne the norm and the trace of x, relative to the extension

E/F , as

NE/F (x) = det m(x) and TE/F (x) = trace m(x).

We will write N (x) and T (x) if E/F is understood. If the matrix A(x) = [aij (x)] repre-

sents m(x) with respect to some basis for E over F , then the norm of x is the determinant

of A(x) and the trace of x is the trace of A(x), that is, the sum of the main diagonal

entries. The characteristic polynomial of x is de¬ned as the characteristic polynomial of

the matrix A(x), that is,

charE/F (x) = det[XI ’ A(x)]

where I is an n by n identity matrix. It follows from the de¬nitions that the norm, the

trace and the coe¬cients of the characteristic polynomial are elements belonging to the

base ¬eld F .

2.1.2 Example

Let E = C and F = R. A basis for C over R is {1, i} and, with x = a + bi, we have

(a + bi)(1) = a(1) + b(i) and (a + bi)(i) = ’b(1) + a(i).

1

2 CHAPTER 2. NORMS, TRACES AND DISCRIMINANTS

Thus

a ’b

A(a + bi) = .

ba

The norm, trace and characteristic polynomial of a + bi are

N (a + bi) = a2 + b2 , T (a + bi) = 2a, char(a + bi) = X 2 ’ 2aX + a2 + b2 .

The computation is exactly the same if E = Q(i) and F = Q.

2.1.3 Some Basic Properties

Notice that in (2.1.2), the coe¬cient of the second highest power of X in the characteristic

polynomial is minus the trace, and the constant term is the norm. In general, it follows

from the de¬nition of characteristic polynomial that

char(x) = X n ’ T (x)X n’1 + · · · + (’1)n N (x). (1)

[The only terms multiplying X n’1 in the expansion of the determinant de¬ning the char-

acteristic polynomial are ’aii (x), i = 1, . . . , n. Set X = 0 to show that the constant term

of char(x) is (’1)n det A(x).]

If x, y ∈ E and a, b ∈ F , then

T (ax + by) = aT (x) + bT (y) and N (xy) = N (x)N (y). (2)

[This holds because m(ax + by) = am(x) + bm(y) and m(xy) = m(x) —¦ m(y).]

If a ∈ F , then

N (a) = an , T (a) = na, and char(a) = (X ’ a)n . (3)

[Note that the matrix representing multiplication by the element a in F is aI.]

It is natural to look for a connection between the characteristic polynomial of x and

the minimal polynomial min(x, F ) of x over F .

2.1.4 Proposition

charE/F (x) = [min(x, F )]r , where r = [E : F (x)].

Proof. First assume that r = 1, so that E = F (x). By the Cayley-Hamilton theorem,

the linear transformation m(x) satis¬es char(x). Since m(x) is multiplication by x, it

follows that x itself is a root of char(x). Thus min(x, F ) divides char(x), and since both

polynomials are monic of degree n, the result follows. In the general case, let y1 , . . . , ys

be a basis for F (x) over F , and let z1 , . . . , zr be a basis for E over F (x). Then the yi zj

form a basis for E over F . Let A = A(x) be the matrix representing multiplication by x

in the extension F (x)/F , so that xyi = k aki yk and x(yi zj ) = k aki (yk zj ). Order the

2.1. NORMS AND TRACES 3

basis for E/F as y1 z1 , y2 z1 , . . . , ys z1 ; y1 z2 , y2 z2 . . . , ys z2 ; · · · ; y1 zr , y2 zr , . . . , ys zr . Then

m(x) is represented in E/F as

®

···

A 0 0

0 0

···

A

. .

.

°. .»

.

. . .

···

0 0 A

with r blocks, each consisting of the s by s matrix A. Thus charE/F (x) = [det(XI ’ A)]r ,

which by the r = 1 case coincides with [min(x, F )]r . ™

2.1.5 Corollary

Let [E : F ] = n and [F (x) : F ] = d. Let x1 , . . . , xd be the roots of min(x, F ), counting

multiplicity, in a splitting ¬eld. Then

d d d

n

(X ’ xi )]n/d .

n/d

N (x) = ( xi ) , T (x) = xi , char(x) = [

d

i=1 i=1 i=1

Proof. The formula for the characteristic polynomial follows from (2.1.4). By (2.1.3),

the norm is (’1)n times the constant term of char(x). Evaluating the characteristic

polynomial at X = 0 produces another factor of (’1)n , which yields the desired expression

for the norm. Finally, if min(x, F ) = X d + ad’1 X d’1 + · · · + a1 X + a0 , then the coe¬cient

d

of X n’1 in [min(x, F )]n/d is (n/d)ad’1 = ’(n/d) i=1 xi . Since the trace is the negative

of this coe¬cient [see (2.1.3)], the result follows. ™

If E is a separable extension of F , there are very useful alternative expressions for the

trace, norm and characteristic polynomial.

2.1.6 Proposition

Let E/F be a separable extension of degree n, and let σ1 , . . . , σn be the distinct F -

embeddings (that is, F -monomorphisms) of E into an algebraic closure of E, or equally

well into a normal extension L of F containing E. Then

n n n

(X ’ σi (x)).

NE/F (x) = σi (x), TE/F (x) = σi (x), charE/F (x) =

i=1 i=1 i=1

Proof. Each of the d distinct F -embeddings „i of F (x) into L takes x into a unique

conjugate xi , and extends to exactly n/d = [E : F (x)] F -embeddings of E into L, all

of which also take x to xi . Thus the list of elements {σ1 (x), . . . , σn (x)} consists of the

„i (x) = xi , i = 1, . . . , d, each appearing n/d times. The result follows from (2.1.5). ™

We may now prove a basic transitivity property.

4 CHAPTER 2. NORMS, TRACES AND DISCRIMINANTS

2.1.7 Transitivity of Trace and Norm

If F ¤ K ¤ E, where E/F is ¬nite and separable, then

TE/F = TK/F —¦ TE/K and NE/F = NK/F —¦ NE/K .

Proof. Let σ1 , . . . , σn be the distinct F -embeddings of K into L, and let „1 , . . . , „m be

the distinct K-embeddings of E into L, where L is the normal closure of E over F . Then

L/F is Galois, and each mapping σi and „j extends to an automorphism of L. Therefore

it makes sense to allow the mappings to be composed. By (2.1.6),

n m n m

TK/F (TE/K (x)) = σi ( „j (x)) = σi („j (x)).

i=1 j=1 i=1 j=1

Each σi „j = σi —¦ „j is an F -embedding of of E into L, and the number of mappings

is given by mn = [E : K][K : F ] = [E : F ]. Furthermore, the σi „j are distinct when

restricted to E. For if σi „j = σk „l on E, then σi = σk on K, because „j and „k coincide

with the identity on K. Thus i = k, so that „j = „l on E. But then j = l. By (2.1.6),

TK/F (TE/K (x)) = TE/F (x). The norm is handled the same way, with sums replaced by

products. ™

Here is another application of (2.1.6).

2.1.8 Proposition

If E/F is a ¬nite separable extension, then the trace TE/F (x) cannot be 0 for every x ∈ E.

n

Proof. If T (x) = 0 for all x, then by (2.1.6), i=1 σi (x) = 0 for all x. This contradicts

Dedekind™s lemma on linear independence of monomorphisms. ™

2.1.9 Remark

A statement equivalent to (2.1.8) is that if E/F is ¬nite and separable, then the trace

form, that is, the bilinear form (x, y) ’ TE/F (xy), is nondegenerate. In other words, if

T (xy) = 0 for all y, then x = 0. Going from (2.1.9) to (2.1.8) is immediate, so assume

T (xy) = 0 for all y, with x = 0. Let x0 be a nonzero element with zero trace, as provided

by (2.1.8). Choose y so that xy = x0 to produce a contradiction.

2.1.10 Example

√ √

Let x = a + b m be an element of the quadratic extension Q( m)/Q, where m is a

square-free integer. We will ¬nd the trace and norm of x.

The√ Galois group of the extension consists of the identity and the automorphism

√

σ(a + b m) = a ’ b m. Thus by (2.1.6),

T (x) = x + σ(x) = 2a, and N (x) = xσ(x) = a2 ’ mb2 .

2.2. THE BASIC SETUP FOR ALGEBRAIC NUMBER THEORY 5

Problems For Section 2.1

1. If E = Q(θ) where θ is a root of the irreducible cubic X 3 ’ 3X + 1, ¬nd the norm and

trace of θ2 .

2. Find the trace of the primitive 6th root of unity ω in the cyclotomic extension Q6 =

Q(ω). √

3. Let θ be a root of X 4 ’ 2 over Q. √

Find the trace over Q of θ, θ2 , θ3 and 3θ.

4. Continuing Problem 3, show that 3 cannot belong to Q[θ].

2.2 The Basic Setup For Algebraic Number Theory

2.2.1 Assumptions

Let A be an integral domain with fraction ¬eld K, and let L be a ¬nite separable extension

of K. Let B be the set of elements of L that are integral over A, that is, B is the integral

closure of A in L. The diagram below summarizes the information.

L B

K A

In the most important special case, A = Z, K = Q, L is a number ¬eld, that is, a ¬nite

(necessarily separable) extension of Q, and B is the ring of algebraic integers of L. From

now on, we will refer to (2.2.1) as the AKLB setup.

2.2.2 Proposition

If x ∈ B, then the coe¬cients of charL/K (x) and min(x, K) are integral over A. In

particular, TL/K (x) and NL/K (x) are integral over A, by (2.1.3). If A is integrally closed,

then the coe¬cients belong to A.

Proof. The coe¬cients of min(x, K) are sums of products of the roots xi , so by (2.1.4),

it su¬ces to show that the xi are integral over A. Each xi is a conjugate of x over K, so

there is a K-isomorphism „i : K(x) ’ K(xi ) such that „i (x) = xi . If we apply „i to an

equation of integral dependence for x over A, we get an equation of integral dependence

for xi over A. Since the coe¬cients belong to K [see (2.1.1)], they must belong to A if A

is integrally closed. ™

2.2.3 Corollary

Assume A integrally closed, and let x ∈ L. Then x is integral over A, that is, x ∈ B, if

and only if the minimal polynomial of x over K has coe¬cients in A.

Proof. If min(x, K) ∈ A[X], then x is integral over A by de¬nition of integrality. (See

(1.1.1); note also that A need not be integrally closed for this implication.) The converse

follows from (2.2.2). ™

6 CHAPTER 2. NORMS, TRACES AND DISCRIMINANTS

2.2.4 Corollary

An algebraic integer a that belongs to Q must in fact belong to Z.

Proof. The minimal polynomial of a over Q is X ’ a, so by (2.2.3), a ∈ Z. ™

2.2.5 Quadratic Extensions of the Rationals

√

We will determine the algebraic integers of L = Q( m), where m is a square-free integer

(a product of distinct primes). The restriction on m involves no loss of generality, for

√ √

example, Q( 12) = Q( 3).

A remark on notation: To make sure there is no confusion between algebraic integers

and ordinary integers, we will often use the term “rational integer” for a member of Z.

√

Now by (2.1.10) and (2.1.3), the minimal polynomial over Q√ the element a+b m ∈ L

of

(with a, b ∈ Q) is X 2 ’ 2aX + a2 ’ mb2 . By (2.2.3), a + b m is an algebraic integer

if and only if 2a and a2 ’ mb2 are rational integers. In this case, we also have 2b ∈ Z.

For we have (2a)2 ’ m(2b)2 = 4(a2 ’ mb2 ) ∈ Z, so m(2b)2 ∈ Z. If 2b is not a rational

integer, its denominator would included a prime factor p, which would appear as p2 in

the denominator of (2b)2 . Multiplication of (2b)2 by m cannot cancel the p2 because m

is square-free, and the result follows.

Here is a more convenient way to characterize the algebraic integers of a quadratic

¬eld.

2.2.6 Proposition

√

The set B of algebraic integers of Q( m), m square-free, can be described as follows.

√

(i) If m ≡ 1 mod 4, then B consists of all a + b m, a, b ∈ Z;

√

(ii) If m ≡ 1 mod 4, then B consists of all (u/2) + (v/2) m, u, v ∈ Z, where u and v

have the same parity (both even or both odd).

[Note that since m is square-free, it is not divisible by 4, so the condition in (i) can be

written as m ≡ 2 or 3 mod 4.]

√

Proof. By (2.2.5), the algebraic integers are of the form (u/2) + (v/2) m, where u, v ∈ Z

and (u2 ’ mv 2 )/4 ∈ Z, that is, u2 ’ mv 2 ≡ 0 mod 4. It follows that u and v have the

same parity. [The square of an even number is congruent to 0 mod 4, and the square of

an odd number is congruent to 1 mod 4.] Moreover, the “both odd” case can only occur

when m ≡ 1 mod 4. The “both even” case is equivalent to u/2, v/2 ∈ Z, and we have

the desired result. ™

When we introduce integral bases in the next section, we will have an even more

√

convenient way to describe the algebraic integers of Q( m).

If [L : K] = n, then a basis for L/K consists of n elements of L that are linearly

independent over K. In fact we can assemble a basis using only elements of B.

2.2.7 Proposition

There is a basis for L/K consisting entirely of elements of B.

2.2. THE BASIC SETUP FOR ALGEBRAIC NUMBER THEORY 7

Proof. Let x1 , . . . , xn be a basis for L over K. Each xi is algebraic over K, and therefore

satis¬es a polynomial equation of the form

am xm + · · · + a1 xi + a0 = 0

i

with am = 0 and the ai ∈ A. (Initially, we only have ai ∈ K, but then ai is the ratio of

two elements of A, and we can form a common denominator.) Multiply the equation by

am’1 to obtain an equation of integral dependence for yi = am xi over A. The yi form

m

the desired basis. ™

2.2.8 Corollary of the Proof

If x ∈ L, then there is a nonzero element a ∈ A and an element y ∈ B such that x = y/a.

In particular, L is the fraction ¬eld of B.

Proof. In the proof of (2.2.7), take xi = x, am = a, and yi = y. ™

In Section 2.3, we will need a standard result from linear algebra. We state the result

now, and an outline of the proof is given in the exercises.

2.2.9 Theorem

Suppose we have a nondegenerate symmetric bilinear form on an n-dimensional vector

space V , written for convenience using inner product notation (x, y). If x1 , . . . , xn is any

basis for V , then there is a basis y1 , . . . , yn for V , called the dual basis referred to V , such

that

1, i=j

(xi , yj ) = δij =

0, i = j.

Problems For Section 2.2

1. Let L = Q(±), where ± is a root of the irreducible quadratic X 2 + bX + c, with b, c ∈ Q.

√

Show that L = Q( m) for some square-free integer m. Thus the analysis of this section

Q.

covers all possible quadratic extensions of√

2. Show that the quadratic extensions Q( m), m square-free, are all distinct.

3. Continuing Problem 2, show that in fact no two distinct quadratic extensions of Q are

Q-isomorphic.

Cyclotomic ¬elds do not exhibit the same behavior. Let ωn = ei2π/n , a primitive nth

2

root of unity. By a direct computation, we have ω2n = ωn and

’ω2n = ’eiπ(n+1)/n = eiπ eiπ eiπ/n = ω2n .

n+1

4. Show that if n is odd, then Q(ωn ) = Q(ω2n ).

5. Give an example of a quadratic extension of Q that is also a cyclotomic extension.

We now indicate how to prove (2.2.9).

6. For any y in the ¬nite-dimensional vector space V , the mapping x ’ (x, y) is a linear

form l(y) on V , that is, a linear map from V to the ¬eld of scalars. Show that the linear

8 CHAPTER 2. NORMS, TRACES AND DISCRIMINANTS

transformation y ’ l(y) from V to V — (the space of all linear forms on V ) is injective.

7. Show that any linear form on V is l(y) for some y.

8. Let f1 , . . . , fn be the dual basis corresponding to x1 , . . . , xn . Thus each fj belongs to

V — (not V ) and fj (xi ) = δij . If fj = l(yj ), show that y1 , . . . , yn is the required dual basis

referred to V .

n

9. Show that xi = j=1 (xi , yj ). Thus in order to compute the dual basis referred to V ,

we must invert the matrix ((xi , yj )).

2.3 The Discriminant

The discriminant of a polynomial is familiar from basic algebra, and there is also a dis-

criminant in algebraic number theory. The two concepts are unrelated at ¬rst glance, but

there is a connection between them. We assume the basic AKLB setup of (2.2.1), with

n = [L : K].

2.3.1 De¬nition

If n = [L : K], the discriminant of the n-tuple x = (x1 , . . . , xn ) of elements of L is

D(x) = det(TL/K (xi xj )).

Thus we form a matrix whose ij entry is the trace of xi xj , and take the determinant of

the matrix; by (2.1.1), D(x) ∈ K. If x ∈ B, then by (2.2.2), D(x) is integral over A, that

is, D(x) ∈ B. Thus if A is integrally closed amd x ∈ B, then D(x) belongs to A.

The discriminant behaves quite reasonably under linear transformation.

2.3.2 Lemma

If y = Cx, where C is an n by n matrix over K and x and y are n-tuples written as

column vectors, then D(y) = (det C)2 D(x).

Proof. The trace of yr ys is

T( cri csj xi xj ) = cri T (xi xj )csj

i,j i,j

hence

(T (yr ys )) = C(T (xi xj ))C

where C is the transpose of C. The result follows upon taking determinants. ™

Here is an alternative expression for the discriminant.

2.3.3 Lemma

Let σ1 , . . . , σn be the distinct K-embeddings of L into an algebraic closure of L, as in

(2.1.6). Then D(x) = [det(σi (xj ))]2 .

2.3. THE DISCRIMINANT 9

Thus we form the matrix whose ij element is σi (xj ), take the determinant and square

the result.

Proof. By (2.1.6),

T (xi xj ) = σk (xi xj ) = σk (xi )σk (xj )

k k

so if H is the matrix whose ij entry is σi (xj ), then (T (xi xj )) = H H, and again the result

follows upon taking determinants. ™

The discriminant “discriminates” between bases and non-bases, as follows.

2.3.4 Proposition

If x = (x1 , . . . , xn ), then the xi form a basis for L over K if and only if D(x) = 0.

Proof. If j cj xj = 0, with the cj ∈ K and not all 0, then j cj σi (xj ) = 0 for all i, so

the columns of the matrix H = (σi (xj )) are linearly dependent. Thus linear dependence

of the xi implies that D(x) = 0. Conversely, assume that the xi are linearly independent,

and therefore a basis because n = [L : K]. If D(x) = 0, then the rows of H are linearly

dependent, so for some ci ∈ K, not all 0, we have i ci σi (xj ) = 0 for all j. Since the xj

form a basis, it follows that i ci σi (u) = 0 for all u ∈ L, so the monomorphisms σi are

linearly dependent. This contradicts Dedekind™s lemma. ™

We now make the connection between the discriminant de¬ned above and the discrim-

inant of a polynomial.

2.3.5 Proposition

Assume that L = K(x), and let f be the minimal polynomial of x over K. Let D be the

discriminant of the basis 1, x, x2 , . . . , xn’1 over K, and let x1 , . . . , xn be the roots of f

in a splitting ¬eld, with x1 = x. Then D coincides with i<j (xi ’ xj )2 , the discriminant

of the polynomial f .

Proof. Let σi be the K-embedding that takes x to xi , i = 1, . . . , n. Then σi (xj ) =

xj , 0 ¤ j ¤ n ’ 1. By (2.3.3), D is the square of the determinant of the matrix

i

®

··· xn’1

x2

1 x1 1 1

1 xn’1

···

x2

x2

2 2

V = . . .

. . ..

°. .»

. . .

. . . .

···

x2 xn’1

1 xn n n

’ xi ), and the result

But det V is a Vandermonde determinant, whose value is i<j (xj

follows. ™

Proposition (2.3.5) yields a formula that is often useful in computing the discriminant.

10 CHAPTER 2. NORMS, TRACES AND DISCRIMINANTS

2.3.6 Corollary

Under the hypothesis of (2.3.5),

n

D = (’1)( 2 ) NL/K (f (x))

where f is the derivative of f .

n

Proof. Let c = (’1)( 2 ) . By (2.3.5),

(xi ’ xj )2 = c (xi ’ xj ) = c (xi ’ xj ).

D=

i<j i j=i

i=j

But f (X) = (X ’ x1 ) · · · (X ’ xn ), so

(X ’ xj )

f (xi ) =

k j=k

with X replaced by xi . When the substitution X = xi is carried out, only the k = i term

is nonzero, hence

(xi ’ xj ).

f (xi ) =

j=i

Consequently,

n

D=c f (xi ).

i=1

But

f (xi ) = f (σi (x)) = σi (f (x))

so by (2.1.6),

D = cNL/K (f (x)). ™

2.3.7 De¬nitions and Comments

In the AKLB setup with [L : K] = n, suppose that B turns out to be a free A-module

of rank n. A basis for this module is said to be an integral basis of B (or of L). An

integral basis is, in particular, a basis for L over K, because linear independence over A

is equivalent to linear independence over the fraction ¬eld K. We will see shortly that an

integral basis always exists when L is a number ¬eld. In this case, the discriminant is the

same for all integral bases. It is called the ¬eld discriminant.

2.3. THE DISCRIMINANT 11

2.3.8 Theorem

If A is integrally closed, then B is a submodule of a free A-module of rank n. If A is a

PID, then B itself is free of rank n over A, so B has an integral basis.

Proof. By (2.1.9), the trace is a nondegenerate symmetric bilinear form de¬ned on the

n-dimensional vector space L over K. By (2.2.2), the trace of any element of B belongs to

A. Now let x1 , . . . , xn be any basis for L over K consisting of elements of B [see (2.2.7)],

and let y1 , . . . , yn be the dual basis referred to L [see (2.2.9)]. If z ∈ B, then we can write

z = j=1 aj yj with the aj ∈ K. We know that the trace of xi z belongs to A, and we

also have

n n n

T (xi z) = T ( aj xi yj ) = aj T (xi yj ) = aj δij = ai .

j=1 j=1 j=1

Thus each ai belongs to A, so that B is an A-submodule of the free A-module •n Ayj .

j=1

Moreover, B contains the free A-module •n Axj . Consequently, if A is a principal ideal

j=1

domain, then B is free over A of rank exactly n. ™

2.3.9 Corollary

The set B of algebraic integers in any number ¬eld L is a free Z-module of rank n = [L : Q].

Therefore B has an integral basis. The discriminant is the same for every integral basis.

Proof. Take A = Z in (2.3.8) to show that B has an integral basis. The transformation

matrix C between two integral bases [see (2.3.2)] is invertible, and both C and C ’1 have

rational integer coe¬cients. Take determinants in the equation CC ’1 = I to conclude

that det C is a unit in Z. Therefore det C = ±1, so by (2.3.2), all integral bases have the

same discriminant. ™

2.3.10 Remark

A matrix C with coe¬cients in Z is said to be unimodular if C ’1 also has coe¬cients

in Z. We have just seen that a unimodular matrix has determinant ±1. Conversely, a

matrix over Z with determinant ±1 is unimodular, by Cramer™s rule.

2.3.11 Theorem

√

Let B be the algebraic integers of Q( m), where m is a square-free integer.

√

(i) If m ≡ 1 mod 4, then 1 and m form an integral basis, and the ¬eld discriminant is

d = 4m.

√

(ii) If m ≡ 1 mod 4, then 1 and (1 + m)/2 form an integral basis, and the ¬eld discrim-

inant is d = m.

Proof.

√ √

(i) By (2.2.6), 1 and m span B over Z, and they are linearly independent because m

√

is irrational. By (2.1.10), the trace of a + b m is 2a, so by (2.3.1), the ¬eld discriminant

12 CHAPTER 2. NORMS, TRACES AND DISCRIMINANTS

is

20

= 4m.

0 2m

√

(ii) By (2.2.6),√ and (1 + m)/2 are algebraic integers. To show that they span B,

1

consider (u + v m)/2, where u and v have the same parity. Then

√ √

u’v

1 1

(u + v m) = ( )(1) + v [ (1 + m)]

2 2 2

with (u ’ v)/2 and v in Z. To prove linear independence, assume that a, b ∈ Z and

√

1

a + b [ (1 + m)] = 0.

2

√

Then 2a + b + b m = 0, which forces a =√ = 0. Finally, by (2.1.10), (2.3.1), and the

b

√ 2

computation [(1 + m)/2] = (1 + m)/4 + m/2, the ¬eld discriminant is

2 1

= m. ™

1 (1 + m)/2

Problems For Section 2.3

Problems 1-3 outline the proof of Stickelberger™s theorem, which states that the discrimi-

nant of any n-tuple in a number ¬eld is congruent to 0 or 1 mod 4.

1. Let x1 , . . . , xn be arbitrary algebraic integers in a number ¬eld, and consider the

determinant of the matrix (σi (xj )), as in (2.3.3). The direct expansion of the determinant

has n! terms. let P be the sum of those terms in the expansion that have plus signs in front

of them, and N the sum of those terms pre¬xed by minus signs. Thus the discriminant D

of (x1 , . . . , xn ) is (P ’ N )2 . Show that P + N and P N are ¬xed by each σi , and deduce

that P + N and P N are rational numbers.

2. Show that P + N and P N are rational integers.

3. Show that D ≡ 0 or 1 mod 4.

4. Let L be a number ¬eld of degree n over Q, and let y1 , . . . , yn be a basis for L over

Q consisting of algebraic integers. Let x1 , . . . , xn be an integral basis. Show that if

the discriminant D(y1 , . . . , yn ) is square-free, then each xi can be expressed as a linear

combination ot the yj with integer coe¬cients.

5. Continuing Problem 4, show that if D(y1 , . . . , yn ) is square-free, then y1 , . . . , yn is an

integral basis.

6. Is the converse of the result of problem 5 true?

Chapter 3

Dedekind Domains

3.1 The De¬nition and Some Basic Properties

We identify the natural class of integral domains in which unique factorization of ideals

is possible.

3.1.1 De¬nition

A Dedekind domain is an integral domain A satisfying the following three conditions:

(1) A is a Noetherian ring;

(2) A is integrally closed;

(3) Every nonzero prime ideal of A is maximal.

A principal ideal domain satis¬es all three conditions, and is therefore a Dedekind

domain. We are going to show that in the AKLB setup, if A is a Dedekind domain, then

so is B, a result that provides many more examples and already suggests that Dedekind

domains are important in algebraic number theory.

3.1.2 Proposition

In the AKLB setup, B is integrally closed, regardless of A. If A is an integrally closed

Noetherian ring, then B is also a Noetherian ring, as well as a ¬nitely generated A-module.

Proof. By (1.1.6), B is integrally closed in L, which is the fraction ¬eld of B by (2.2.8).

Therefore B is integrally closed. If A is integrally closed, then by (2.3.8), B is a submodule

of a free A-module M of rank n. If A is Noetherian, then M , which is isomorphic to the

direct sum of n copies of A, is a Noetherian A-module, hence so is the submodule B. An

ideal of B is, in particular, an A-submodule of B, hence is ¬nitely generated over A and

therefore over B. It follows that B is a Noetherian ring. ™

3.1.3 Theorem

In the AKLB setup, if A is a Dedekind domain, then so is B. In particular, the ring of

algebraic integers in a number ¬eld is a Dedekind domain.

1

2 CHAPTER 3. DEDEKIND DOMAINS

Proof. In view of (3.1.2), it su¬ces to show that every nonzero prime ideal Q of B is

maximal. Choose any nonzero element x of Q. Since x ∈ B, x satis¬es a polynomial

equation

xm + am’1 xm’1 + · · · + a1 x + a0 = 0

with the ai ∈ A. If we take the positive integer m as small as possible, then a0 = 0 by

minimality of m. Solving for a0 , we see that a0 ∈ Bx © A ⊆ Q © A, so the prime ideal

P = Q © A is nonzero, hence maximal by hypothesis. By Section 1.1, Problem 6, Q is

maximal. ™

Problems For Section 3.1

This problem set will give the proof of a result to be used later. Let P1 , P2 , . . . , Ps , s ≥ 2,

be ideals in a ring R, with P1 and P2 not necessarily prime, but P3 , . . . , Ps prime (if

s ≥ 3). Let I be any ideal of R. The idea is that if we can avoid the Pj individually, in

other words, for each j we can ¬nd an element in I but not in Pj , then we can avoid all

the Pj simultaneously, that is, we can ¬nd a single element in I that is in none of the Pj .

The usual statement is the contrapositive of this assertion.

Prime Avoidance Lemma

With I and the Pi as above, if I ⊆ ∪s Pi , then for some i we have I ⊆ Pi .

i=1

1. Suppose that the result is false. Show that without loss of generality, we can assume

the existence of elements ai ∈ I with ai ∈ Pi but ai ∈ P1 ∪ · · · ∪ Pi’1 ∪ Pi+1 ∪ · · · ∪ Ps .

/

2. Prove the result for s = 2.

3. Now assume s > 2, and observe that a1 a2 · · · as’1 ∈ P1 © · · · © Ps’1 , but as ∈ /

P1 ∪ · · · ∪ Ps’1 . Let a = (a1 · · · as’1 ) + as , which does not belong to P1 ∪ · · · ∪ Ps’1 , else

as would belong to this set. Show that a ∈ I and a ∈ P1 ∪ · · · ∪ Ps , contradicting the

/

hypothesis.

3.2 Fractional Ideals

Our goal is to establish unique factorization of ideals in a Dedekind domain, and to do

this we will need to generalize the notion of ideal. First, some preliminaries.

3.2.1 Products of Ideals

Recall that if I1 , . . . , In are ideals, then their product I1 · · · In is the set of all ¬nite sums

i a1i a2i · · · ani , where aki ∈ Ik , k = 1, . . . , n. It follows from the de¬nition that I1 · · · In

is an ideal contained in each Ij . Moreover, if a prime ideal P contains a product I1 · · · In

of ideals, then P contains Ij for some j.

3.2. FRACTIONAL IDEALS 3

3.2.2 Proposition

If I is a nonzero ideal of the Noetherian integral domain R, then I contains a product of

nonzero prime ideals.

Proof. Assume the contrary. If S is the collection of all nonzero ideals that do not contain

a product of nonzero prime ideals, then, as R is Noetherian, S has a maximal element J,

and J cannot be prime because it belongs to S. Thus there are elements a, b ∈ R such

that a ∈ J, b ∈ J, and ab ∈ J. By maximality of J, the ideals J + Ra and J + Rb each

/ /

contain a product of nonzero prime ideals, hence so does (J +Ra)(J +Rb) ⊆ J +Rab = J.

This is a contradiction. (Notice that we must use the fact that a product of nonzero ideals

is nonzero, and this is where the hypothesis that R is an integral domain comes in.) ™

3.2.3 Corollary

If I is an ideal of the Noetherian ring R (not necessarily an integral domain), then I

contains a product of prime ideals.

Proof. Repeat the proof of (3.2.2), with the word “nonzero” deleted. ™

Ideals in the ring of integers are of the form nZ, the set of multiples of n. A set of

the form (3/2)Z is not an ideal because it is not a subset of Z, yet it behaves in a similar

manner. The set is closed under addition and multiplication by an integer, and it becomes

an ideal of Z if we simply multiply all the elements by 2. It will be pro¬table to study

sets of this type.

3.2.4 De¬nitions

Let R be an integral domain with fraction ¬eld K, and let I be an R-submodule of K.

We say that I is a fractional ideal of R if rI ⊆ R for some nonzero r ∈ R. We call r a

denominator of I. An ordinary ideal of R is a fractional ideal (take r = 1), and will often

be referred to as an integral ideal.

3.2.5 Lemma

(i) If I is a ¬nitely generated R-submodule of K, then I is a fractional ideal.

(ii) If R is Noetherian and I is a fractional ideal of R, then I is a ¬nitely generated

R-submodule of K.

(iii) If I and J are fractional ideals with denominators r and s respectively, then I ©J, I +J

and IJ are fractional ideals with respective denominators r (or s), rs and rs. [The product

of fractional ideals is de¬ned exactly as in (3.2.1).]

Proof.

(i) If x1 = a1 /b1 , . . . , xn = an /bn generate I and b = b1 · · · bn , then bI ⊆ R.

(ii) If rI ⊆ R, then I ⊆ r’1 R. As an R-module, r’1 R is isomorphic to R and is therefore

Noetherian. Consequently, I is ¬nitely generated.

(iii) It follows from the de¬nition (3.2.4) that the intersection, sum and product of frac-

tional ideals are fractional ideals. The assertions about denominators are proved by noting

that r(I © J) ⊆ rI ⊆ R, rs(I + J) ⊆ rI + sJ ⊆ R, and rsIJ = (rI)(sJ) ⊆ R. ™

4 CHAPTER 3. DEDEKIND DOMAINS

The product of two nonzero fractional ideals is a nonzero fractional ideal, and the

multiplication is associative because multiplication in R is associative. There is an identity

element, namely R, since RI ⊆ I = 1I ⊆ RI. We will show that if R is a Dedekind domain,

then every nonzero fractional ideal has a multiplicative inverse, so the nonzero fractional

ideals form a group.

3.2.6 Lemma

Let I be a nonzero prime ideal of the Dedekind domain R, and let J be the set of all

elements x ∈ K such that xI ⊆ R. Then R ‚ J.

Proof. Since RI ⊆ R, it follows that R is a subset of J. Pick a nonzero element a ∈ I,

so that I contains the principal ideal Ra. Let n be the smallest positive integer such

that Ra contains a product P1 · · · Pn of n nonzero prime ideals. Since R is Noetherian,

there is such an n by (3.2.2), and by (3.2.1), I contains one of the Pi , say P1 . But in a

Dedekind domain, every nonzero prime ideal is maximal, so I = P1 . Assuming n ≥ 2, set

I1 = P2 · · · Pn , so that Ra ⊇ I1 by minimality of n. Choose b ∈ I1 with b ∈ Ra. Now

/

’1

II1 = P1 · · · Pn ⊆ Ra, in particular, Ib ⊆ Ra, hence Iba ⊆ R. (Note that a has an

inverse in K but not necessarily in R.) Thus ba’1 ∈ J, but ba’1 ∈ R, for if so, b ∈ Ra,

/

contradicting the choice of b.

The case n = 1 must be handled separately. In this case, P1 = I ⊇ Ra ⊇ P1 , so

I = Ra. Thus Ra is a proper ideal, and we can choose b ∈ R with b ∈ Ra. Then /

’1 ’1 ’1 ’1

ba ∈ R, but ba I = ba Ra = bR ⊆ R, so ba ∈ J. ™

/

We now prove that in (3.2.6), J is the inverse of I.

3.2.7 Proposition

Let I be a nonzero prime ideal of the Dedekind domain R, and let J = {x ∈ K : xI ⊆ R}.

Then J is a fractional ideal and IJ = R.

Proof. If r is a nonzero element of I and x ∈ J, then rx ∈ R, so rJ ⊆ R and J is a

fractional ideal. Now IJ ⊆ R by de¬nition of J, so IJ is an integral ideal. Using (3.2.6),

we have I = IR ⊆ IJ ⊆ R, and maximality of I implies that either IJ = I or IJ = R.

In the latter case, we are ¬nished, so assume IJ = I.

If x ∈ J, then xI ⊆ IJ = I, and by induction, xn I ⊆ I for all n = 1, 2, . . . . Let r be

any nonzero element of I. Then rxn ∈ xn I ⊆ I ⊆ R, so R[x] is a fractional ideal. Since

R is Noetherian, part (ii) of (3.2.5) implies that R[x] is a ¬nitely generated R-submodule

of K. By (1.1.2), x is integral over R. But R, a Dedekind domain, is integrally closed, so

x ∈ R. Therefore J ⊆ R, contradicting (3.2.6). ™

The following basic property of Dedekind domains can be proved directly from the

de¬nition, without waiting for the unique factorization of ideals.

3.2.8 Theorem

If R is a Dedekind domain, then R is a UFD if and only if R is a PID.

Proof. Recall from basic algebra that a (commutative) ring R is a PID i¬ R is a UFD

and every nonzero prime ideal of R is maximal. ™

3.3. UNIQUE FACTORIZATION OF IDEALS 5

Problems For Section 3.2

1. If I and J are relatively prime ideals (I + J = R), show that IJ = I © J. More

generally, if I1 , . . . , In are relatively prime in pairs, show that I1 · · · In = ©n Ii .

i=1

r s

2. Let P1 and P2 be relatively prime ideals in the ring R. Show that P1 and P2 are

relatively prime for arbitrary positive integers r and s.

3. Let R be an integral domain with fraction ¬eld K. If K is a fractional ideal of R, show

that R = K.

3.3 Unique Factorization of Ideals

In the previous section, we inverted nonzero prime ideals in a Dedekind domain. We now

extend this result to nonzero fractional ideals.

3.3.1 Theorem

If I is a nonzero fractional ideal of the Dedekind domain R, then I can be factored uniquely

as P1 1 P2 2 · · · Pr r , where the ni are integers. Consequently, the nonzero fractional ideals

n n n

form a group under multiplication.

Proof. First consider the existence of such a factorization. Without loss of generality, we

can restrict to integral ideals. [Note that if r = 0 and rI ⊆ R, then I = (rR)’1 (rI).] By

convention, we regard R as the product of the empty collection of prime ideals, so let S

be the set of all nonzero proper ideals of R that cannot be factored in the given form, with

all ni positive integers. (This trick will yield the useful result that the factorization of

integral ideals only involves positive exponents.) Since R is Noetherian, S, if nonempty,

has a maximal element I0 , which is contained in a maximal ideal I. By (3.2.7), I has an

inverse fractional ideal J. Thus by (3.2.6) and (3.2.7),

I0 = I0 R ⊆ I0 J ⊆ IJ = R.

Therefore I0 J is an integral ideal, and we claim that I0 ‚ I0 J. For if I0 = I0 J, then the

last paragraph of the proof of (3.2.7) can be reproduced with I replaced by I0 to reach a

contradiction. By maximality of I0 , I0 J is a product of prime ideals, say I0 J = P1 · · · Pr

(with repetition allowed). Multiply both sides by the prime ideal I to conclude that I0 is

a product of prime ideals, contradicting I0 ∈ S. Thus S must be empty, and the existence

of the desired factorization is established.

To prove uniqueness, suppose that we have two prime factorizations

P1 1 · · · Pr r = Qt1 · · · Qts

n n

s

1

where again we may assume without loss of generality that all exponents are positive.

(If P ’n appears, multiply both sides by P n .) Now P1 contains the product of the Pini ,

so by (3.2.1), P1 contains Qj for some j. By maximality of Qj , P1 = Qj , and we may

renumber so that P1 = Q1 . Multiply by the inverse of P1 (a fractional ideal, but there is

no problem), and continue inductively to complete the proof. ™

6 CHAPTER 3. DEDEKIND DOMAINS

3.3.2 Corollary

A nonzero fractional ideal I is an integral ideal if and only if all exponents in the prime

factorization of I are nonnegative.

Proof. The “only if” part was noted in the proof of (3.3.1). The “if” part follows because

a power of an integral ideal is still an integral ideal. ™

3.3.3 Corollary

Denote by nP (I) the exponent of the prime ideal P in the factorization of I. (If P does

not appear, take nP (I) = 0.) If I1 and I2 are nonzero fractional ideals, then I1 ⊇ I2 if

and only if for every prime ideal P of R, nP (I1 ) ¤ nP (I2 ).

’1

Proof. We have I2 ⊆ I1 i¬ I2 I1 ⊆ R, and by (3.3.2), this happens i¬ for every P ,

nP (I2 ) ’ nP (I1 ) ≥ 0. ™

3.3.4 De¬nition

let I1 and I2 be nonzero integral ideals. We say that I1 divides I2 if I2 = JI1 for some

integral ideal J. Just as with integers, an equivalent statement is that each prime factor

of I1 is a factor of I2 .

3.3.5 Corollary

If I1 and I2 are nonzero integral ideals, then I1 divides I2 if and only if I1 ⊇ I2 . In other

words, for these ideals,

DIV IDES M EAN S CON T AIN S.

Proof. By (3.3.4), I1 divides I2 i¬ nP (I1 ) ¤ nP (I2 ) for every prime ideal P . By (3.3.3),

this is equivalent to I1 ⊇ I2 . ™

3.3.6 GCD™s and LCM™s

As a nice application of the principle that divides means contains, we can use the prime

factorization of ideals in a Dedekind domain to compute the greatest common divisor

and least common multiple of two nonzero ideals I and J, exactly as with integers. The

greatest common divisor is the smallest ideal containing both I and J, that is, I + J. The

least common multiple is the largest ideal contained in both I and J, which is I © J.

A Dedekind domain comes close to being a principal ideal domain in the sense that

every nonzero integral ideal, in fact every nonzero fractional ideal, divides some principal

ideal.

3.4. SOME ARITHMETIC IN DEDEKIND DOMAINS 7

3.3.7 Proposition

let I be a nonzero fractional ideal of the Dedekind domain R. Then there is a nonzero

integral ideal J such that IJ is a principal ideal of R.

Proof. By (3.3.1), there is a nonzero fractional ideal I such that II = R. By de¬nition

of fractional ideal, there is a nonzero element r ∈ R such that rI is an integral ideal. If

J = rI , then IJ = Rr, a principal ideal of R. ™

Problems For Section 3.3

√ √

By (2.3.11), the ring B of algebraic integers in Q( ’5) is Z[ ’5]. In Problems 1-3, we will

√

show that Z[ ’5] is not a unique factorization domain by considering the factorization

√ √

(1 + ’5)(1 ’ ’5) = 2 — 3.

1. By computing norms, verify that all four of the above factors are irreducible.

2. Show that the only units of B are ±1.

3. Show that no factor on one side of the above equation is an associate of a factor on

the other side, so unique factorization fails. √

4. Show that the ring of algebraic integers in Q( ’17) is not a unique factorization

domain.√ √

5. In Z[ ’5] and Z ’17], the only algebraic integers √ norm 1 are ±1. Show that this

of

property does not hold for the algebraic integers in Q( ’3).

3.4 Some Arithmetic in Dedekind Domains

Unique factorization of ideals in a Dedekind domain permits calculations that are analo-

gous to familiar manipulations involving ordinary integers. In this section, we illustrate

some of the ideas.

Let P1 , . . . , Pn be distinct nonzero prime ideals of the Dedekind domain R, and let

J = P1 · · · Pn . Let Qi be the product of the Pj with Pi omitted, that is,

Qi = P1 · · · Pi’1 Pi+1 · · · Pn .

(If n = 1, we take Q1 = R.) If I is any nonzero ideal of R, then by unique factorization,

IQi ⊃ IJ. For each i = 1, . . . , n, choose an element ai belonging to IQi but not to IJ,

n

and let a = i=1 ai .

3.4.1 Lemma

The element a belongs to I, but for each i, a ∈ IPi . (In particular, a = 0.)

/

Proof. Since each ai belongs to IQi ⊆ I, we have a ∈ I. Now ai cannot belong to IPi ,

for if so, ai ∈ IPi © IQi , which is the least common multiple of IPi and IQi [see (3.3.6)].

But by de¬nition of Qi , the least common multiple is simply IJ, which contradicts the

choice of ai . We break up the sum de¬ning a as follows:

a = (a1 + · · · + ai’1 ) + ai + (ai+1 + · · · + an ). (1)

8 CHAPTER 3. DEDEKIND DOMAINS

If j = i, then aj ∈ IQj ⊆ IPi , so the ¬rst and third terms of the right side of (1) belong

to IPi . Since ai ∈ IPi , as found above, we have a ∈ IPi . ™

/ /

In (3.3.7), we found that any nonzero ideal is a factor of a principal ideal. We can

sharpen this result as follows.

3.4.2 Proposition

Let I be a nonzero ideal of the Dedekind domain R. Then there is a nonzero ideal I such

that II is a principal ideal (a). Moreover, if J is an arbitrary nonzero ideal of R, then I

can be chosen to be relatively prime to J.

Proof. Let P1 , . . . , Pn be the distinct prime divisors of J, and choose a as in (3.4.1). Then

a ∈ I, so (a) ⊆ I. Since divides means contains [see (3.3.5)], I divides (a), so (a) = II

for some nonzero ideal I . If I is divisible by Pi , then I = Pi I0 for some nonzero ideal

I0 , and (a) = IPi I0 . Consequently, a ∈ IPi , contradicting (3.4.1). ™

3.4.3 Corollary

A Dedekind domain with only ¬nitely many prime ideals is a PID.

Proof. Let J be the product of all the nonzero prime ideals. If I is any nonzero ideal,

then by (3.4.2) there is a nonzero ideal I such that II is a principal ideal (a), with I

relatively prime to J. But then the set of prime factors of I is empty, so I = R. Thus

(a) = II = IR = I. ™

The next result reinforces the idea that a Dedekind domain is not too far away from

a principal ideal domain.

3.4.4 Corollary

Let I be a nonzero ideal of the Dedekind domain R, and let a be any nonzero element of

I. Then I can be generated by two elements, one of which is a.

Proof. Since a ∈ I, we have (a) ⊆ I, so I divides (a), say (a) = IJ. By (3.4.2), there is

a nonzero ideal I such that II is a principal ideal (b) and I is relatively prime to J. If

gcd stands for greatest common divisor, then the ideal generated by a and b is