algebraic number theory notes

8/19/2019 Algebraic Number Theory Notes

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NOTES ON ALGEBRAIC NUMBER THEORY

JEFFREY D. VAALER*

1. Algebraic number fields

Let Q be the field of rational numbers, and let Q denote an algebraic closureof Q. The elements of Q are called algebraic numbers , and Q is the field of allalgebraic numbers. If α is an element of Q then α is algebraic over Q. Thereforeα is the root of a unique, monic, irreducible polynomial mα(x) in the ring Q[x].The polynomial mα(x) is called the minimal polynomial of α over Q, (see TheoremB.4). If the minimal polynomial mα(x) belongs to the subring Z[x] ⊆ Q[x], then wesay that α is an algebraic integer . It is trivial that the elements of Z are algebraicintegers, and we sometimes refer to the elements of Z as rational integers .

Let k be a subfield of Q, so that Q ⊆ k ⊆ Q. We say that k is an algebraic number field if the degree [k : Q] of the extension k/Q is finite. Alternatively, if kis a field of characteristic zero then k is an extension of its prime field Q, and wesay that k is an algebraic number field if the degree [k : Q] of the extension k/Q isfinite. Then an algebraic closure of k is also an algebraic closure of Q and we haveQ ⊆ k ⊆ Q. As an algebraic number field k has characteristic zero, k /Q is a finiteseparable extension. Then it follows from Theorem H.9 that an algebraic numberfield k is a simple extension of Q. That is, there exists an element α in k such thatk = Q(α), and α is said to generate the field extension k/Q.

Yet another point of view is to begin with a field k such that Q ⊆ k ⊆ C. If thedegree [k : Q] of the extension k/Q is finite then k is an algebraic number field, andalso a subset of the complex numbers. As C is algebraically closed, the set

Q = {γ ∈ C : γ is algebraic over Q}

is an algebraic closure of Q (see Corollary C.4) and obviously contains k. Thereforewe have

(1.1) Q ⊆ k ⊆ Q ⊆ C.

As an arbitrary finite extension of Q can always be embedded into a specific al-gebraic closure of Q (see Theorem C.5), the containment (1.1) can always be ar-ranged. It is sometimes useful to work with an algebraic number field embeddedin C because the complete metric topology and the subfield R ⊆ C can be used toadvantage.

Suppose, for example, that [k : Q] = N . Then there exist N distinct embeddingsσn : k → C, where n = 1, 2, . . . , N . If σn(k) ⊆ R we call σn a real embedding,otherwise we call σn a complex embedding. Let z → ρ(z) denote complex conju-gation on elements z in C. Because complex conjugation is an automorphism of C that fixes R, it follows that if α → σn(α) is a complex embedding of k then

Date : 656, February 2, 2016.*Research supported by the National Science Foundation, DMS-06-03282.

1


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2 VAALER

α → ρ

σn(α)

is also a complex embedding of k and is distinct from α → σn(α).Thus the complex embeddings of k into C come in pairs. Let rk be the numberof real embeddings and let 2s

k be the number of complex embeddings of k into

C. Obviously we have rk + 2sk = N . The numbers rk and sk are examples of invariants attached to the number field k. If all the embeddings of k into C arereal embeddings, that is, if rk = N and 2sk = 0, then we say that k is a totally real algebraic number field. At the other extreme, if all the embeddings of k intoC are complex embeddings, that is, if rk = 0 and 2sk = N , then we say that k is atotally complex algebraic number field. Further general results about algebraic fieldextensions are established in Appendix B.

A basic object of investigation in elementary number theory is the integral do-main Z of rational integers in Q. In the theory of algebraic number fields we makea similar investigation of an analogous integral domain contained in an algebraicnumber field k. The analogous integral domain is the set of all algebraic integersin k , which we denote by Ok. In view of our previous remarks, we have

Ok = α ∈ k : mα(x) ∈ Z[x].Our first objective is to show that Ok is a Noetherian domain, that is, Ok is anintegral domain and each ideal in Ok is finitely generated. Later we will prove astronger result: each integral domain Ok is a Dedekind domain.

Theorem 1.1. Let α be an algebraic number. Then the following are equivalent,

(i) the minimal polynomial mα(x) belongs to Z[x],(ii) there exists a monic polynomial f (x) in Z[x] such that f (α) = 0,

(iii) the ring Z[α] is a finitely generated Z-module.

Proof. It is trivial that (i) implies (ii).Assume that (ii) holds with deg f = M . As f is monic we have M ≥ 1. It is

obvious that the Z-module generated by

(1.2) {1, α , α2, . . . , αM −1}

is a subset of Z[α]. Let

β = b0αN + b1α

N −1 + b2αN −2 + · · · + bN

be an element of Z[α], where b0, b1, . . . , bN are integers and b0 = 0. From thedivision algorithm in Z[x] we have

b0xN + b1x

N −1 + b2xN −2 + · · · + bN = q (x)f (x) + r(x),

with q (x) and r(x) in Z[x], and either 0 ≤ deg r < M or r(x) = 0. It follows that

β = q (α)f (α) + r(α) = r(α),

and therefore β belongs to the Z-module generated by the finite set (1.2). Thisverifies (iii).

Assume that (iii) holds, and let g1(x), g2(x), . . . , gL(x) be a finite collection of nonzero polynomials in Z[x] such that

g1(α), g2(α), . . . , gL(α)

generates Z[α] as a Z-module. Let N be an integer larger than the maximum of thedegrees of the polynomials g1(x), g2(x), . . . , gL(x). Then αN belongs to Z[α] andtherefore there exist integers c1, c2, . . . , cL such that

αN = c1g1(α) + c2g2(α) + · · · + cLgL(α).


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ALGEBRAIC NUMBER THEORY 3

it follows that α is a root of the monic polynomial

xN − c1g1(x) − c2g2(x) − · · · − cLgL(x)

which clearly belongs to Z[x]. This establishes (ii).Now assume that (ii) holds. Let f (x) be a monic polynomial in Z[x] such that

f (α) = 0, and let mα(x) be the minimal polynomial for α in Q[x]. By Theorem B.4the principal ideal in Q[x] generated by mα(x) must contain the polynomial f (x).Therefore there exists a monic polynomial g (x) in Q[x] such that

f (x) = mα(x)g(x).

Let d be a positive integer such that dmα(x) has relatively prime integer coefficients.Let e be a positive integer such that eg(x) has relatively prime integer coefficients.Then the identity

def (x) = {dmα(x)}{eg(x)}

holds in Z[x]. Now suppose that p is a prime such that p|d, and let F p denote the

finite field with p elements. Then the image of def (x) in F p[x] is the zero polyno-mial. However, both dmα(x) and eg(x) have relatively prime integer coefficients,and so the image of both dmα(x) and eg(x) in F p(x) is a nonzero polynomial. AsF p[x] is an integral domain this is clearly impossible. It follows that d = 1 andtherefore mα(x) has integer coefficients. This establishes (i).

Theorem 1.2. Let α1, α2, . . . , αN be algebraic integers, then Z[α1, α2, . . . , αN ], the ring of all polynomials in α1, α2, . . . , αN with coefficients in Z, is a finitely generated Z-module.

Proof. We argue by induction on N , the case N = 1 clearly follows from theprevious theorem. Assume then that Z[α1, α2, . . . , αN ] is a finitely generated Z-module and consider the Z-module Z[α1, α2, . . . , αN , β ], where β is an algebraicinteger. Let f 1(x), f 2(x), . . . , f L(x)be a finite collection of polynomials in Z[x1, x2, . . . , xN ], such that

f 1(α), f 2(α), . . . , f L(α)

is a set of generators for Z[α1, α2, . . . , αN ]. Let g1(y), g2(y), . . . , gM (y) be polyno-mials in Z[y] such that

g1(β ), g2(β ), . . . , gM (β )

is a set of generators for Z[β ]. We will show that the finite set

(1.3)

f l(α)gm(β ) : 1 ≤ l ≤ L, and 1 ≤ m ≤ M

is a set of generators for Z[α1, α2, . . . , αN , β ].Let

(1.4) (α)pβ q = α p11 α p22 · · · α

pN N β

q

be a monomial in α1, α2, . . . , αN , β . Then there exist integers a1, a2, . . . , aL, andintegers b1, b2, . . . , bM , such that

(α)p = a1f 1(α) + a2f 2(α) + · · · + aLf L(α),

and

β q = b1g1(β ) + b2g2(β ) + · · · + bM g(β ).


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4 VAALER

Now it is obvious that the monomial (1.4) belongs to the Z-module generated bythe set (1.3). As each element of the Z-module Z[α1, α2, . . . , αN , β ] is a finitesum of such monomials with integer coefficients, it follows that (1.3) generatesZ[α1, α2, . . . , αN , β ]. In particular, the Z-module Z[α1, α2, . . . , αN , β ] is finitelygenerated. The result follows by setting β = αN +1.

Corollary 1.3. The subset of algebraic integers in Q forms an integral domain. If k is an algebraic number field, then the set Ok of algebraic integers in k forms an integral domain.

Proof. Let α and β be algebraic integers in Q. Then Z[α, β ] is finitely generated asa Z-module, and therefore each of the submodules Z[α + β ], Z[α − β ], and Z[αβ ]is a finitely generated Z-module. Then it follows from Theorem 1.1 that α + β ,α − β , and αβ are algebraic integers. Both assertions of the Corollary are nowobvious.

If [k : Q] = N then k is a vector space of degree N over the field Q. Hence thereexists a basis α1, α2, . . . , αN of elements from k , such that

(1.5) k =

b1α1 + b2α2 + · · · + bN αN : bn ∈ Q for each n = 1, 2, . . . , N

.

Moreover, each element β in k has a unique representation of the form

(1.6) β = c1α1 + c2α2 + · · · + cN αN with cn ∈ Q.

If α is an algebraic number and

mα(x) = xN + a1x

N −1 + a2xN −2 + · · · + aN −1x + aN

is its minimal polynomial in Q[x], we may select a positive integer d so that dmα(x)has relatively prime integer coefficients, as in our proof of Theorem 1.1. If we writeeach rational coefficient as

an = rn

sn,

where rn and sn are relatively prime integers and sn is positive, then it is easy tosee that

d = lcm{s1, s2, . . . , sN }.

It follows that dα is a root of the monic polynomial

xN + (a1d)xN −1 + (a2d

2)xN −2 + · · · + (aN −1dN −1)x + (aN d

N ),

and this polynomial has integer coefficients. That is, dα is an algebraic integer. Asα can now be expressed as a ratio of two algebraic integers:

(1.7) α = dα

d ,

it follows easily that k is the field of fractions attached to the integral domain Ok.

Let α1, α2, . . . , αN be a basis for k/Q as in (1.5). Then we can select positiveintegers d1, d2, . . . , dN so that dnαn is in Ok for each n = 1, 2, . . . , N . We find that

{d1α1, d2α2, . . . , dN αN }

is a collection of Z-linearly independent points in Ok. More generally, let I be anonzero ideal in Ok and let β be a nonzero element of I. Then dnαnβ belongs toI for each n = 1, 2, . . . , N , and it follows that

{d1α1β, d2α2β , . . . , dN αN β }


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is a collection of Z-linearly independent points in I. In particular we have

(1.8) spanZ{d1α1β, d2α2β , . . . , dN αN β } ⊆ I.

In order to show that equality can be achieved in ( 1.8), we introduce the discrimi-nant.

Using (L.15) we find that

(1.9) Normk/Q : k× → Q×

is a homomorphism of multiplicative groups, and

(1.10) Tracek/Q : k → Q

is a nonzero linear map from the Q-vector space k into Q. If α is an algebraicinteger in k with minimal polynomial mα(x) in Z[x], then using (L.12) and (L.13)we find that both Normk/Q(α) and Tracek/Q(α) are rational integers. It follows that

if α1, α2, . . . , αN are algebraic integers in Ok, then their discriminant (see (L.17))

∆k/Q(α1, α2, . . . , αN ) = det

Tracek/Q(αmαn)

is a rational integer.

Theorem 1.4. Let k be an algebraic number field with [k : Q] = N , and let I be a nonzero ideal in Ok. Then there exists a Z-linearly independent subset

(1.11) {β 1, β 2, . . . , β N } ⊆ I,

that generates I as a Z-module. That is, every element γ in I has a unique repre-sentation of the form

(1.12) γ = b1β 1 + b2β 2 + · · · + bN β N

with each bn in Z.

Proof. Let A be the collection of all (column) vectors α such that the coordinatesα1, α2, . . . , αN are Z-linearly independent elements of the ideal I. In view of (1.8)it is clear that A is not empty. It follows from Theorem L.4 that for each vector αin A the discriminant ∆k/Q(α) is a nonzero integer. Thus we may select a point βin A such that

(1.13)∆k/Q(β) = min∆k/Q(α) : α ∈ A.

We claim that the coordinates of β form a Z-module basis for I.As the coordinates β 1, β 2, . . . , β N are Q-linearly independent elements of k, they

form a basis for k as a vector space over Q. Thus every element γ in I has a unique

representation as

(1.14) γ = b1β 1 + b2β 2 + · · · + bN β N

with each bn in Q. Assume that γ in I can be selected so that at least one of thecoefficients bn is in Q but not in Z. By reordering the elements β 1, β 2, . . . , β N if necessary, we may assume that b1 is in Q but not in Z. Then by replacing γ with

γ − c1β 1


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6 VAALER

if necessary, where c1 is an integer close to b1, we may assume that −12 ≤ b1 <

12 .

Now observe that we have the matrix identity

(1.15)

γ

β 2β 3...

β N

=

b1 b2 b3 . . . bN 0 1 0 . . . 00 0 1 . . . 0...

......

. . . ...

0 0 0 . . . 1

β 1β 2β 3...

β N

.It follows from (1.15) and (L.21) that

∆k/Q(γ, β 2, β 3, . . . , β N ) = (b1)2∆k/Q(β 1, β 2, . . . , β N ),

and therefore

(1.16) 1 ≤∆k/Q(γ, β 2, β 3, . . . , β N )

≤ 14∆k/Q(β 1, β 2, . . . , β N )

.

As γ belongs to the ideal I, (1.16) contradicts the minimal property (1.13) associ-

ated to the elements β 1, β 2, . . . , β N . Thus every element γ in I has a representationof the form (1.14) with coefficient bn in Z. This proves the theorem.

The collection of elements (1.11), asserted to exist in Theorem 1.4, is called anintegral basis for the nonzero ideal I in the integral domain Ok. By an integral basis for the field k we understand an integral basis for the ring of integers Ok. Plainlyan integral basis for k is also a basis for k as a vector space over Q.

Recall (see Theorem J.1) that an integral domain in which every ideal is finitelygenerated is a Noetherian domain. Thus Theorem 1.4 implies the following result.

Corollary 1.5. Let k be an algebraic number field and Ok the ring of algebraic integers in k . Then Ok is a Noetherian domain.

In our proof of Theorem 1.4 we selected β 1, β 2, . . . , β N in the nonzero ideal I soas to minimize the expression∆k/Q(β 1, β 2, . . . , β N ) = detTracek/Q(β mβ n).We now reconsider this number.

Lemma 1.6. Let k be an algebraic number field with [k : Q] = N , and let Ibe a nonzero ideal in Ok. Let α1, α2, . . . , αN be a Z-module basis for I and let β 1, β 2, . . . , β N be a second Z-module basis for I. Then we have

(1.17) ∆k/Q(α1, α2, . . . , αN ) = ∆k/Q(β 1, β 2, . . . , β N ).

Proof. Let σ1, σ2, . . . , σN be the distinct embeddings of k into an algebraic closureQ. Then define N × N matrices

(1.18) A = σn(αm) and B = σn(β m),where in both matrices m = 1, 2, . . . , N indexes rows and n = 1, 2, . . . , N indexescolumns. From (L.16) and (L.17) we have

(1.19) ∆k/Q(α1, α2, . . . , αN ) = det

AAT

and

(1.20) ∆k/Q(β 1, β 2, . . . , β N ) = det

BBT

.


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By hypothesis there exists an N ×N matrix C = (cmn) with rational integer entriessuch that

(1.21) β l =

N m=1

clmαm for l = 1, 2, . . . , N .

Applying the embedding σn to both sides of (1.19) leads to the system of identities

σn(β l) =

N m=1

clmσn(αm) for l = 1, 2, . . . , N and n = 1, 2, . . . , N .

These identities can also be written as the matrix identity

(1.22) B = CA.

In a similar manner we find that there exists an N × N matrix D = (dmn) withrational integer entries that satisfies the identity

(1.23) A = DB .

It follows from (1.22) and (1.23) that CD = DC = 1N , where 1N is the N × N identity matrix. As C and D have integer entires we conclude that

(1.24) det C = det D = ±1.

Now (1.17) follows immediately from (1.19), (1.20) and (1.24).

If I is a nonzero ideal in the ring Ok, and β 1, β 2, . . . , β N is an integral basis forI, we define the discriminant of the ideal I to be the nonzero integer

(1.25) Disck/Q(I) = ∆k/Q(β 1, β 2, . . . , β N ).

Then it follows from Lemma 1.6 that the discriminant of the nonzero ideal I is welldefined because it does not depend on the choice of an integral basis for I. Thediscriminant of the improper ideal Ok is also called the discriminant of the field k ,

and in this special case we simplify notation by writingDisc(k) = Disck/Q(Ok)

for the discriminant of k. The discriminant of k is a further example of an invariantattached to the number field k .

Lemma 1.7. Let k be an algebraic number field, and let I be a nonzero ideal in Ok. Then there exists an integral basis α1, α2, . . . , αN , for Ok, and positive integers d1, d2, . . . , dN , such that

(i) dn|dn+1 for n = 1, 2, . . . , N − 1, and (ii) d1α1, d2α2, . . . , dN αN is an integral basis for I.

Proof. Let α1, α2, . . . , α

N be an integral basis for Ok, and let β

1, β

2, . . . , β

N be an

integral basis for the ideal I. Write σ1, σ2, . . . , σN for the N distinct embeddings

σn : k → Q. Then letA =

σn(α

m)

and B =

σn(β m)

be the corresponding N × N matrices, where in both cases m = 1, 2, . . . , N indexesrows and n = 1, 2, . . . , N indexes columns. As in our proof of Lemma 1.6, thereexists a nonsingular N × N matrix C = (cmn) with integer entries such that

(1.26) B = CA.

Now, however, we cannot conclude that C is a matrix in GL(N, Z).


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Let U =

umn

be a matrix in the group GL(N, Z), so that

αm =

N n=1

umnαn for m = 1, 2, . . . , N ,

is a second integral basis for Ok. Similarly, let V =

vmn

be a matrix in the groupGL(N, Z), so that

β m =N n=1

vmnβ n for m = 1, 2, . . . , N ,

is a second integral basis for the ideal I. We find that

A =

σn(αm)

= U A,

and

B = σn(β m) = V B.

Therefore we get the matrix identity

(1.27) B = V B =

V CU −1

U A

=

V CU −1

A.

By a basic diagonalization theorem for matrices with integer entries (see [4, Theo-rem 4.3, Chapter 12] or [21, Theorem 4, Chapter 10]) we can select the matrices U and V so that

(1.28) V CU −1 = [dn]

is an N × N diagonal matrix, each diagonal entry dn is a positive integer, anddn|dn+1 for n = 1, 2, . . . , N − 1. Here (1.28) is the Smith normal form of C .Combining (1.27) and (1.28) leads to the identity

B = [dn]A,

which also gives the conclusion (ii). And we have already noted that the positiveintegers dn satisfy (i).

Theorem 1.8. Let k be an algebraic number field, and let I be a nonzero ideal in Ok. Then the index [Ok : I] is finite, and we have

(1.29) Disck/Q(I) = [Ok : I]2 Disc(k).

Proof. By Lemma 1.7 there exists an integral basis α1, α2, . . . , αN for Ok, andpositive integers d1, d2, . . . , dN , such that d1α1, d2α2, . . . , dN αN is an integral basisfor the ideal I. Then every element γ in Ok has a unique representation as

(1.30) γ = b1α1 + b2α2 + · · · + bN αN , where bn ∈ Z.

Define an additive group homomorphism

ψ : Ok → Z/d1Z ⊕ Z/d2Z ⊕ · · · ⊕ Z/dN Z

by

ψ(γ ) = ψ

b1α1 + b2α2 + · · · + bN αN ) = (b1, b2, . . . , bN ).

Obviously the map ψ is surjective. An element γ given by (1.30) is in the kernel of ψ if and only if

bn ≡ 0 (mod dn) for each n = 1, 2, . . . , N .


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That is, γ is in the kernel of ψ if and only if γ belongs to the ideal I. It followsthat

(1.31) [Ok : I] = d1d2 · · · dN = det[dn].As in the proof of Lemma 1.7, we have

B =

σn(β m)

= [dn]

σn(αm)

= [dn]A,

and therefore

Disck/Q(I) = det

BBT

=

det[dn]2

det

AAT )

= (d1d2 · · · dN )2 Disc(k).

(1.32)

The result follows from (1.31) and (1.32).

Let k is an algebraic number field and let α1, α2, . . . , αN be an integral basis for

the ring of algebraic integers Ok. If β is a nonzero element of Ok, then we can formthe nonzero principal ideal

β = βOk

=

β (c1α1 + c2α2 + · · · + cN αN ) : cn ∈ Z

= spanZ{βα1, βα2, . . . , β αN }.

As β α1, βα2, . . . , β αN is clearly an integral basis for the ideal β , we find that

Disck/Q

β

= det

Tracek/Q(β 2αmαn)

= N l=1

σl(β 2)

det

Tracek/Q(αmαn)

= Normk/Q(β )2 Disck/Q(α1, α2, . . . , αN )

= Normk/Q(β )2 Disc(k),

(1.33)

where σ1, σ2, . . . , σN are the distinct embeddings of k into Q. Now Theorem 1.8and (1.33) immediately imply the following result.

Corollary 1.9. Let k be an algebraic number field and β a nonzero element of Ok.Then the discriminant and index of the nonzero principal ideal β satisfy

(1.34) Disck/Q

β

= Normk/Q(β )2 Disc(k),

and

(1.35)

Ok : β

=Normk/Q(β ).

Moreover, β is a unit in Ok if and only if

(1.36) Normk/Q(β ) = ±1.

We recall that a maximal ideal in a commutative ring with an identity element,is also a prime ideal. But in a general commutative ring a prime ideal need not bemaximal. In the ring Ok of algebraic integers in a number field k, we find that eachnonzero prime ideal is also a maximal ideal.

Corollary 1.10. Let k be an algebraic number field and P a nonzero prime ideal in Ok. Then P is a maximal ideal.


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Proof. As P is a nonzero prime ideal, it follows from Theorem I.2 that the quotientring Ok/P is an integral domain. By Theorem 1.8 the integral domain Ok/P isfinite, but it is not the zero ring because P is a proper subset of O

k. From Theorem

I.1 we learn that Ok/P is a field, and then a second application of Theorem I.2shows that P is a maximal ideal.

Let (S, +, ×) be a commutative ring with an identity element and let (R, +, ×)be a subring such that 1R = 1S . We recall that an element s in S is integral overR if there exists a monic polynomial

f (x) = xN + a1xN −1 + a2x

N −2 + · · · + aN −1x + aN

in R[x] such that f (s) = 0. Clearly each element of R is integral over R. We saythat S is integral over R if every element s in S is integral over R. We say that Ris integrally closed in S if the only elements of S which are integral over R are theelements in R. For example, if Ok is the ring of algebraic integers in the algebraicnumber field k, then Ok is integral over Z. The following result shows that Okis integrally closed in the field k, and completes the proof that Ok is a Dedekinddomain.

Theorem 1.11. Let k be an algebraic number field and Ok the ring of algebraic integers in k . Then Ok is a Dedekind domain.

Proof. The integral domain Ok is Noetherian by Corollary 1.5. And each nonzeroprime ideal P in Ok was shown to be a maximal ideal in Corollary 1.10. Thus itremains only to show that Ok is integrally closed in its field of fractions k .

Let α be an element of k which is integral over Ok. As Ok is obviously integralover Z, it follows from Corollary I.9 that α is integral over Z. That is, α belongsto Ok. This shows that Ok is integrally closed in its field of fractions k . Hence Okis a Dedekind domain.

Exercises

1.1. Recall that an integer m = 0 is square free if m is not divisible in Z by thesquare of a prime number. Thus the set of square free integers is

· · · − 10, −7, −6, −5, −3, −2, −1, 1, 2, 3, 5, 6, 7, 10, . . .

.

Let k be an algebraic number field such that [k : Q] = 2. Prove that there exists analgebraic integer α in k and a unique square free integer m = 1, such that α2 = mand k = Q(α).

1.2. Let k = Q(α) and α2 = m be as in Exercise 1.1. Let Ok be the ring of algebraic integers in k . Prove that

Ok = spanZ{1, α}, if m ≡ 1 (mod 4),

and

Ok = spanZ

1,

1 + α

2

, if m ≡ 1 (mod 4).


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1.3. Let k = Q(α) and α2 = m be as in Exercise 1.1. Prove that

Disc(k) = 4m if m ≡ 1 (mod 4),m if m ≡ 1 (mod 4).1.4. Let k be an algebraic number field and Ok the integral domain of algebraic

integers in k. Use the representation (1.7) to show that k is the field of fractions of the integral domain Ok.

1.5. Let k be an algebraic number field and Disc(k) its discriminant. Prove that

(−1)s Disc(k) ≥ 1,

where r is the number of real embeddings of k into C, and 2s is the number of complex (and not real) embeddings of k into C.

1.6. Let k be an algebraic number field and Disc(k) its discriminant. Prove thateitherDisc(k) ≡ 0 mod 4, or Disc(k) ≡ 1 mod 4.

1.7. Let k and l be algebraic number fields such that Q ⊆ k ⊆ l. Prove thatDisc(k)| Disc(l).

1.8. Let α in Q be a root of the polynomial

mα(x) = x3 − x2 − 2x − 8.

Show that mα(x) is in fact irreducible in Q[x], and show that

(1.37) β = α2 + α

2

is an algebraic integer in the field k = Q(α). (Hint: find mβ(x) and observe that itbelongs to Z[x].)

1.9. Let α, β , and k be as in Exercise 1.8. Show that {1, α , β } is an integral basisfor Ok, and conclude that

Disc(k) = −503.

1.10. Let α, β , and k be as in Exercise 1.8 and Exercise 1.9. Prove that if γ is analgebraic integer in Ok then

Disck/Q

γ

≡ 0 mod 4.

Conclude that Ok does not have an integral basis of the form {1, γ , γ 2}.


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12 VAALER

2. Theorems of Blichfeldt and Minkowski

In this section we prove some classical results from the geometry of numbers.

We apply these results to further our understanding of ideals in the ring Ok of algebraic integers in an algebraic number field k. As this subject has many furtherapplications, we take a somewhat general point of view.

We will work in the N dimensional real vector space RN , and write

x =

x1x2...

xN

,for a (column) vector in RN . By a lattice in RN we understand an additive subgroupΛ ⊆ RN generated as a free Z-module by N vectors ξ1, ξ2, . . . ,ξN , which are R-linearly independent. Thus the lattice Λ generated by ξ1, ξ2, . . . ,ξN , is the set

(2.1) Λ = m1ξ1 + m2ξ2 + · · · + mN ξN : m1, m2, . . . , mN in Z.We say that the R-linearly independent vectors ξ1, ξ2, . . . ,ξN , form a basis for thelattice Λ. If λ is a point in Λ then it follows from the R-linear independence of ξ1, ξ2, . . . ,ξN , that there exist unique integers m1, m2, . . . , mN such that

λ = m1ξ1 + m2ξ2 + · · · + mN ξN .

Let

(2.2) e1 =

10...0

, e2 =

01...0

, · · · , eN =

00...1

.

denote the standard basis vectors in RN . Plainly these vectors generate the latticeZN . More generally, write

(2.3) X =ξ1 ξ2 ξ3 · · · ξN

for the N × N real, nonsingular matrix having ξn as its nth column. From (2.1) itis clear that

(2.4) Λ =

X m : m ∈ ZN

.

Thus the lattice Λ is the image of the lattice ZN under the map m → X m.Let GL(N, R) denote the general linear group of all nonsingular N × N matrices

with entries in R. Write GL(N, Z) for the subgroup of N × N matrices with entriesin Z and determinant equal to ±1. If AutZN is the group of all Z-module auto-morphisms of ZN , then each matrix A in GL(N, Z) determines the automorphismϕA : ZN → ZN in Aut

ZN

given by

ϕA(m) = Am.

It is easy to verify that A → ϕA is an isomorphism of the group GL(N, Z) onto thegroup Aut

ZN

. In particular, we have

ϕA

ϕB(m)

= ϕAB(m)


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for all A and B in GL(N, Z), and all points m in ZN . Alternatively, GL(N, Z) isexactly the set of N × N matrices A in GL(N, R) such that

ZN

= {Am : m ∈ ZN

}.If Λ is the lattice in RN determined by (2.4) and ψ : Λ → Λ is a Z-module

automorphism of Λ, then the map

m → X m → ψ

X m

→ X −1ψ

X m

is an automorphism of ZN . It follows from our previous remarks that there existsa unique matrix A in GL(N, Z) such that

X −1ψ

X m

= Am,

and thenψ(λ) = X AX −1λ

for all points λ in Λ.

Lemma 2.1. Let Λ be the lattice in R

N

determined by (2.4). Assume that the columns of a second matrix Y in GL(N, R) also form a basis for Λ, so that

(2.5) Λ = {Y m : m ∈ ZN }.

Then there exists a unique matrix B in GL(N, Z) such that Y = X B.

Proof. The mapm → Y m → X −1Y m

is an automorphism of ZN . Hence there exists a unique matrix B in GL(N, Z) suchthat

(2.6) X −1Y m = Bm

for all m in ZN . In particular, (2.6) holds for each standard basis vector, andtherefore Y = X B.

It follows from Lemma 2.1 that there is a bijection between the collection of alllattices Λ ⊆ RN and the collection of all left cosets of the subgroup GL( N, Z) inGL(N, R). More precisely, if Λ is determined by the matrix X as in (2.4), then theset of all matrices Y in RN such that (2.5) holds, is precisely the left coset

XB : B ∈ GL(N, Z)

.

If E ⊆ RN is a Borel subset, we write VolN (E ) for the N -dimensional Lebesguemeasure of E , but we often refer to this quantity as the N -dimensional volume of E , or simply the volume of E . Each matrix A in GL(N, R) determines a linearhomeomorphism x → Ax from RN onto RN . If E ⊆ RN is a Borel set then theimage

AE = {Ax : x ∈ E }

is also a Borel set, and we have the basic identity

(2.7) VolN (AE ) = | det A| VolN (E ).

If Λ is determined by (2.3) and (2.4), then the corresponding fundamental par-allelepiped is the subset

(2.8) P X =

θ1ξ1 + θ2ξ2 + · · · + θN ξN : 0 ≤ θn < 1 for each n = 1, 2, . . . , N

.

The subset P X ⊆ RN is clearly a Borel set, and is a complete set of distinct cosetrepresentatives for the quotient group RN /Λ. That is, if x is a point in RN , then


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14 VAALER

there exists a unique point λ in Λ such that x − λ belongs to P X . Thus P X is anexample of a fundamental domain for the quotient group RN /Λ. We write

(2.9) C N = θ ∈ RN : 0 ≤ θn < 1for the fundamental parallelepiped associated to the standard basis for ZN . ThenP X is clearly the image of C N under the linear map θ → X θ. It follows from (2.7)that the volume of the fundamental parallelepiped P X is given by

(2.10) VolN (P X) = | det X | VolN (C N ) = | det X |.

If Λ is also determined by a second matrix Y as in (2.5), then Y = X B, where Bis in GL(N, Z), and it is obvious that

(2.11) | det Y | = | det X || det B| = | det X |.

The positive real number | det X | is called the determinant of the lattice . Theidentity (2.11) shows that this number does not depend on the choice of basis, andit is equal to the volume of a fundamental parallelepiped for the lattice. We writed(Λ) = | det X | for the determinant of the lattice Λ.

A basic problem in the geometry of numbers is to identify subsets K ⊆ RN , andlattices Λ ⊆ RN , for which we can draw conclusions about the nature of the setK ∩ Λ. In many cases it is trivial that K ∩ Λ contains the point 0, and we seekresults which assert that K ∩ Λ either does or does not contain further nonzeropoints. For example, K could be the set of points in RN that satisfy a system of inequalities, and we may wish to show that there are nonzero points in a certainlattice Λ that satisfy the system of inequalities. We begin to establish results of this sort by proving the following theorem of Blichfeldt [9].

Theorem 2.2. Let E ⊆ RN

be a Borel set, Λ ⊆ RN

a lattice, and L a positive integer. If

(2.12) Ld(Λ) < VolN (E ),

then there exist L + 1 distinct points y0,y1, . . . ,yL, in E such that each of the differences ym − yl belongs to Λ.

Proof. We may assume that the columns of the N × N matrix X form a basis forΛ, as in (2.4). Let χE (x) denote the characteristic function of the set E , and writeP X for the fundamental parallelepiped (2.8). Then it follows that

(2.13)

λ∈ΛP X + λ

= RN

is a disjoint union of sets. Next we define the Borel measurable function

F : RN → {0, 1, 2, . . . , ∞}

by

F (x) =λ∈Λ

χE (x + λ).


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From (2.13) and the monotone convergence theorem we find that

P X F (x) dx = λ∈Λ P X χE (x + λ) dx=λ∈Λ

P X+λ

χE (x) dx

=

RN

χE (x) dx

= VolN (E ) > L VolN (P X).

(2.14)

It follows that there exists a point p in P X such that F ( p) ≥ L + 1. Hence thereexist L + 1 distinct points λ0,λ1, . . . ,λL, in Λ such that y l = p + λl belongs to E for each l = 0, 1, 2, . . . , L. Then each of the differences ym − yl = λm −λl belongsto Λ. This proves the theorem.

It is often convenient to have the following refinement of Theorem 2.2, in whichthe set E is assumed to be compact.

Corollary 2.3. Let K ⊆ RN be a compact set, Λ ⊆ RN a lattice, and L a positive integer. If

(2.15) Ld(Λ) ≤ VolN (K ),

then there exist L + 1 distinct points y0,y1, . . . ,yL in K such that each of the differences ym − yl belongs to Λ.

Proof. Let 1 ≥ r1 > r2 > · · · > rj > · · · > 0 be a monotone decreasing sequence of positive numbers such that

limj→∞

rj = 0.

Then let Bj = {x ∈ RN : |x|2 ≤ rj}

denote the closed ball in RN centered at 0 and having radius rj . Write

E j = Bj + K = {x+ y : x ∈ Bj and y ∈ K }

for the Minkowski sum of the two compact sets Bj and K . It follows that each of the sets E j is compact,

(2.16) E 1 ⊇ E 2 ⊇ · · · ⊇ E j ⊇ · · · ⊇ K,

and therefore we have a decreasing sequence of nonnegative integers

(2.17)E 1 ∩ Λ

≥E 2 ∩ Λ

≥ · · · ≥E j ∩ Λ

≥ · · · ≥K ∩ Λ

.

Hence there exists a positive integer J such that j → E j ∩ Λ is constant for j ≥ J .As K is compact and(2.18)

∞j=J

E j = K,

we find that E j ∩ Λ = K ∩ Λ for j ≥ J . By the Brunn-Minkowski theorem (see [29,Theorem 3.2.41]) we have

VolN (Bj + K )1/N ≥ VolN (Bj)

1/N + VolN (K )1/N ,


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16 VAALER

and thereforeVolN (Bj + K ) > VolN (K ) ≥ Ld(Λ).

Thus we can apply Theorem 2.2 to the sets E j = Bj + K for j ≥ J . We concludethat for each j ≥ J there exsit L + 1 distinct points x

(j)0 ,x

(j)1 , . . . ,x

(j)L in E j such

that each of the differences x(j)m − x

(j)l belongs to Λ.

As there are finitely many infinite sequences x(j)l and all points of these sequences

are contained in the compact set E J , it follows that there exists a subsequence

J ≤ j1 < j2 < · · · < ji < · · ·

such thatlimi→∞

x(ji)l = yl

exists for each l = 0, 1, 2, . . . , L. In view of (2.18) we conclude that yl belongs to

K for each l. Now each of the differences x(ji)m − x

(ji)l belongs to Λ. Because Λ is

a discrete set, the limiting values

limi→∞

x(ji)m − x(ji)

l = ym − yl

also belong to Λ for each l and m. This is exactly the statement of the corollary.

We say that a subset K ⊆ RN is symmetric if the map x → −x sends K ontoitself. Alternatively, K is symmetric if K = −K . We say that K is convex wheneverit has the following property: if u and v are points in K then the point (1−θ)u+θvis in K for all real numbers θ such that 0 ≤ θ ≤ 1. Thus K is convex if for any pairof points u and v in K , each point of the line segment from u to v is also in K .

The following fundamental theorem was proved by Minkowski [54] in case L = 1.The more general result for L ≥ 1 was first established by van der Corput.

Theorem 2.4. (Minkowski’s convex body Theorem) Let K be a convex,symmetric subset of RN , Λ a lattice in RN , and let L be a positive integer. Assume

that either

(2.19) L2N d(Λ) < VolN (K ),

or that K is compact and

(2.20) L2N d(Λ) ≤ VolN (K ).

Then K contains at least L pairs of distinct, nonzero points ±λ1, ±λ2, . . . , ±λL from the lattice Λ.

Proof. Assume that K satisfies (2.19). Then the volume of 12 K satisfies the in-equality

Ld(Λ) <

12

N VolN (K ) = VolN

12

K

.

By Theorem 2.2 the set 12 K contains L + 1 distinct points y0,y1, . . . ,yL such that

each of the differences ym − yl belongs to Λ. Because 1

2 K is symmetric, both ymand −yl belong to

12

K . Because 12 K is convex, each of the differences

12ym −

12yl =

12ym +

12 (−yl)

belongs to 12 K . Therefore each of the differences ym − yl belongs to K , and so toK ∩ Λ.

Now consider the subspaces

S (l, m) =w ∈ RN : w(ym − yl) = 0

where l = m.


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As RN is not the union of these subspaces, there exists a point z in the complementof their union. Let ϕ : RN → R denote the linear form defined by ϕ(x) = zx.Then the restriction of ϕ to the subset y0,y1, . . . ,yL is injective. By reindexingif necessary, we may assume that

ϕ(y0) < ϕ(y1) < · · · < ϕ(yL).

Set λm = ym − y0 for each m = 1, 2, . . . , L. Then the points λ1,λ2, . . . ,λL aredistinct, hence the points −λ1, −λ2, . . . , −λL are also distinct. And, an identity−λl = λm is impossible because

ϕ(−λl) = −ϕ(λl) < 0 < ϕ(λm).

This proves the theorem under the hypothesis (2.19). If K is compact andsatisfies the inequality (2.20), the argument is essentially the same, but in this casewe appeal to Corollary 2.3.

Corollary 2.5. Let Λ be a lattice in RN . Then there exists a point λ = 0 in Λ

such that (2.21) |λ|∞ = max

λ1, λ2, . . . , λN ≤ d(Λ)1/N .Proof. We apply the theorem to the compact, convex, symmetric set

K =x ∈ RN : |x|∞ ≤ d(Λ)

1/N

.

The result follows because VolN (K ) = 2N d(λ).

Theorem 2.6. (Minkowski’s Theorem on linear forms) Let Λ be a lattice in RN , and let

(2.22) Lm(x) =N n=1

amnxn, where m = 1, 2, . . . , N ,

be a system of N linear forms in N variables with real coefficients. Write A = (amn) for the corresponding N ×N matrix, and let η1, η2, . . . , ηN , be positive real numbers.If

(2.23) d(Λ)| det A| ≤ η1η2 · · · ηN ,

then there exists a point λ = 0 in Λ such that

(2.24)Lm(λ) = N

n=1

amnλn

≤ ηm for each m = 1, 2, . . . , N .Proof. Write H = [ηn] for the N × N diagonal matrix having ηn in the nth rowand nth column. Then define the closed, convex, symmetric set

K =

x ∈ RN :

H −1Ax

∞ ≤ 1

.

If det A = 0 then K is compact, and we find thatVolN (K ) = 2

N

η1η2 · · · ηN

| det A|−1 ≥ 2N d(Λ).

If det A = 0 then the interior of K is unbounded, and in this case we have

VolN (K ) = ∞.

In either case there exists a point λ = 0 such thatH −1Aλ∞

≤ 1.


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18 VAALER

That is,

η−1mN

n=1 amnλn ≤ 1 for each m = 1, 2, . . . , N ,and this proves the theorem.

It is useful to have a variant of Theorem 2.6 in which the linear forms are allowedto have complex coefficients, and the norm | |∞ is replaced by | |1. Suppose thatamong the forms (2.22) there are exactly r forms with real coefficients, there are 2sforms with complex coefficients, and further assume that the complex forms occur incomplex conjugate pairs. Obviously we have r +2s = N . By reindexing if necessary,we may assume that the forms Lm(x) have real coefficients for m = 1, 2, . . . , r, thatthe forms Lm(x) have complex (but not real) coefficients for m = r +1, r+2, . . . , r+s, and that

Ls+m(x) =

N

n=1amnxn

for m = r + 1, r + 2, . . . , r + s. Here z denotes the complex conjugate of the complexnumber z .

Theorem 2.7. Let Lm(x), where m = 1, 2, . . . , N , be a collection of N linear forms in N variables with complex coefficients. Assume that there are r real forms, 2scomplex (but not real) forms, and that the complex forms occur in complex conjugate pairs. Write A = (amn) for the matrix of coefficients. Then there exists a point ξ = 0 in the integer lattice ZN such that

(2.25)N

m=1

Lm(ξ) ≤ 4/πsN !| det A|1/N .Proof. We may assume, as in our previous remarks, that the forms Ll(x) are real

for l = 1, 2, . . . , r, and that Ls+m(x) and Lm(x) form a pair of complex conjugateforms for m = r + 1, r + 2, . . . , r + s. Let W denote the N × N matrix organizedinto blocks as

2W =

21r 0 00 1s 1s0 −i1s i1s

,where 1r and 1s are r × r and s × s identity matrices, respectively. It is easyto verify that det W = (−2i)−s. Because of the way we have arranged the linearforms, the matrix W A has real entries.

Next we define

K =y ∈ RN :

rl=1

|yl| + 2s

m=1

y2m + y

2s+m

1/2≤ 1

.

It follows that K is a compact, convex, symmetric subset of RN , andVolN (K ) = 2

r−sπs

N !−1

.

Hence the linear image

(W A)−1K = {x ∈ K : W Ax ∈ K }

is also a compact, convex, symmetric subset of RN , and

VolN

(W A)−1K

= | det W A|−1 VolN (K ) = 2rπs

N !| det A|−1

.


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Now let t be a positive number. Then we have

tK = y ∈ RN :r

l=1 |yl| + 2s

m=1y2m + y2s+m1/2 ≤ t,and

VolN (tK ) = tN VolN (K ).

We select t so that

(2.26) tN =

4/πs

N !| det A|,

and therefore

VolN

(W A)−1tK

= 2N .

By Minkowski’s Theorem 2.2, there exists a point ξ = 0 in the integer lattice ZN

such that W Aξ belongs to tK . From the definition of W A and tK , we find that

(2.27)N

n=1Ln(ξ) =r

l=1Ll(ξ)+ 2r+s

m=rLm(ξ) ≤ t.The result follows by combining (2.26) and (2.27).

Let

P (x) = a0xN + a1x

N −1 + a2xN −2 + · · · + aN −1x + aN

= a0

N n=1

(x − αn)

= a0

N n=0

(−1)N −nen(α)xN −n

(2.28)

be a polynomial in Z[x] with degree N ≥ 1. Here α1, α2, . . . , αN , are the roots of

P (x), which we regard as elements of C. And we write

e0(α) = 1, and en(α) =

1≤i1


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20 VAALER

As an application of Theorem 2.7, we give a lower bound for the discriminant of an irreducible polynomial in Z[x].

Theorem 2.8. Let P (x) be an irreducible polynomial in Z[x] with degree N ≥ 1.If P (x) has r real roots and 2s complex (but not real) roots in C, then

(2.30)

N N

N !

π4

s2≤ | Disc(P )|.

Proof. Let A denote the N × N complex matrix

(2.31) A =

1 α1 α

21 . . . α

N −11

1 α2 α22 . . . α

N −12

......

... . . .

...

1 αN α2N . . . α

N −1N

.From the van der Monde determinant identity we have

(2.32) det A = 1≤m


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As each monomial in R(x1, x2, . . . , xN ) has total degree at most N − 1, it followsthat

(2.36) aN −10 Re1(α), e2(α), . . . , eN (α) = aN −10 N m=1

β m

is a nonzero integer. Combining (2.32), (2.35), and (2.36), leads to the inequality

1 ≤aN −10 N

m=1

β m

≤

4/πs

N −N N !aN −10 det A

=

4/πs

N −N N !Disc(P )1/2.

(2.37)

Then (2.37) clearly implies (2.30).

Next we give an application of the geometry of numbers to algebraic number

theory. In particular, we prove that a nonzero ideal in an algebraic number fieldcontains a nonzero element with small norm.

Theorem 2.9. Let k be an algebraic number field with r = rk real embeddings and 2s = 2sk complex embeddings. Write [k : Q] = r + 2s = N , and let I be a nonzeroideal in Ok. Then there exists a nonzero point γ in I such that

(2.38)Normk/Q(γ ) ≤ 4πs N !N N Disck/Q(I) 12 .

Proof. Let β 1, β 2, . . . , β N be an integral basis for the nonzero ideal I in Ok. Thenlet σ1, σ2, . . . , σN denote the distinct embeddings σn : k → Q. Here we take Q ⊆ C,and we assume that r = rk embeddings are real and 2s = 2sk embeddings arecomplex. Let

B = σn(β m)be the corresponding N × N matrix, where m = 1, 2, . . . , N indexes rows andn = 1, 2, . . . , N indexes columns. Each column of B determines a linear form

Ln(x) =N

m=1

σn(β m)xm, where n = 1, 2, . . . , N .

Clearly there are r real forms, 2s complex forms, and the complex forms occur incomplex conjugate pairs. By Theorem 2.7 there exists a nonzero point ξ in theinteger lattice ZN such that

(2.39)N n=1

Ln(ξ) ≤ 4/πsN !| det B|1/N .Let

γ = ξ 1β 1 + ξ 2β 2 + · · · + ξ N β N ,so that γ is a nonzero point in I, and

(2.40) Ln(ξ) =N

m=1

σn(β m)ξ m = σn(γ ).

From (1.20) and (1.25) we have

(2.41) det

BBT

= Disck/Q(I).


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By combining (2.39), (2.40), and (2.41), we find that

(2.42) n=1σn(γ ) ≤ 4/πs

N !Disck/Q(I)1

21/N

.

Then from the arithmetic-geometric mean inequality and (2.42) we conclude thatNormk/Q(γ ) = N n=1

σn(γ )≤

N −1N n=1

σn(γ )N ≤ 4

π

s N !N N

Disck/Q(I) 12 .(2.43)

This proves the theorem.

If x is a real number we write

x = min{|x − n| : n ∈ Z}

for the distance from x to the nearest integer. Clearly : R →

0, 12

is acontinuous function with period 1. Hence it is well defined on the cosets of thequotient group R/Z, and we my regard it as a function : R/Z →

0, 12

. Wenote that (x, y) → x − y defines a metric on R/Z, and this metric induces itsquotient topology. The function x → x plays a fundamental role in Diophantineapproximation. As a further application of Minkowski’s convex body theorem, wederive a general form of Dirichlet’s Theorem on rational approximation.

Theorem 2.10. (Dirichlet’s Theorem on linear forms) Let A = (αmn) be an M × N matrix with real entries, and let Q be a positive integer. Then there exist integers q 1, q 2, . . . , q N , not all zero, such that

(2.44) |q n| ≤ Q for each n = 1, 2, . . . , N ,

and

(2.45) αm1q 1 + αm2q 2 + · · · + αmN q N ≤ (Q + 1)−N/M

for each m = 1, 2, . . . , M .

Proof. Let B denote the (M + N ) × (M + N ) real matrix that is organized intoblocks as

B =

1M A

0 1N

.

Let 0 < θ


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If q = 0 then the inequality on the left of (2.46) implies that p = 0, which isimpossible. It follows that q = 0.

Because the coordinates of p are integers, the inequality on the left of (2.46)implies that

(2.47) αm1q 1 + αm2q 2 + · · · + αmN q N ≤ (Q + θ)−N/M

for each m = 1, 2, . . . , M . Because the coordinates of q are integers, the inequalityon the right of (2.46) implies that

(2.48) |q|∞ ≤ Q.

Thus for each θ, 0 < θ < 1, we have established the existence of a point q = 0 inZN that satisfies (2.47) and (2.48). But the set of points q = 0 in ZN that satisfy(2.48) is finite, and it follows that there exists one which satisfies ( 2.47) for valuesof θ that are arbitrarily close to 1. This proves the theorem.

Let M and N be positive integers and write G1 = (R/Z)MN . We write the

elements of the additive group G1 as M × N matrices A = (αmn) with entries αmnin R/Z. Then the group operation in G1 corresponds to addition of matrices. Wenote that in the statement of Theorem 2, we may assume that the matrix A belongsto the group G1, because the conclusion (2.45) depends only on the image of eachentry in R/Z.

Exercises

2.1. Let K ⊆ RN be a convex set such that the interior of K is unbounded. Provethat VolN (K ) = ∞.

2.2. Prove that if 2 ≤ N then GL(N, Z) is not a normal subgroup of GL(N, R).

2.3. Let Γ ⊆ RN and Λ ⊆ RN be lattices such that Γ ⊆ Λ. In this case we saythat Γ is a sublattice of Λ. Prove that the group Γ has finite index in the group Λ.Then show that the ratio d(Γ)/d(Λ) is an integer, and

d(Γ) = d(Λ)[Λ : Γ].

2.4. Let Λ ⊆ RN be a lattice, and suppose that F ⊆ RN is a fundamental domainfor the quotient group RN /Λ. That is, F is a Borel subset of RN , and F is acomplete set of distinct coset representatives for RN /Λ. Prove that VolN (F ) =d(Λ).

2.5. Prove the following more general form of Theorem 2.10. Let A = (αmn) be

an M × N matrix with real entries, let Q1, Q2, . . . , QN , be nonnegative integers,and let 0 < m ≤ 12 be real numbers for m = 1, 2, . . . , M . Prove that if

1 ≤ 12 · · · M (Q1 + 1)(Q2 + 1) · · · (QN + 1),

then there exists an integer vector ξ = 0 in ZN such that

αm1ξ 1 + αm2ξ 2 + · · · + αmN ξ N ≤ m for m = 1, 2, . . . , M ,

and|ξ n| ≤ Qn for n = 1, 2, . . . , N .


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3. The Ideal Class Group

Let k be an algebraic number field and Ok its ring of integers. As Ok is a

Dedekind domain, we may speak of the multiplicative group I (k) of fractional idealsin k and the subgroup P (k) ⊆ I (k) of principal fractional ideals in k. It followsfrom Corollary K.2 that the group I (k) is a free abelian group on the set of all primeideals in Ok. In this section we consider the ideal class group H (k) = I (k)/P (k).The main result is that H (k) is a finite group. Then we define the class number of the field k to be the cardinality

hk = |H (k)|

of the ideal class group. Obviously hk is an invariant of k, but it is not easy tocompute hk, even if k has relatively low degree.

Because I (k) is a free abelian group generated by the set of nonzero prime idealsin Ok, we may define a unique homomorphism

(3.1) normk/Q : I (k) → Q× ∩ (0, ∞)

by specifying the value of normk/Q at each generator. If P is a nonzero prime idealin Ok, we set

(3.2) normk/Q(P) =

Ok : P

.

If A is an integral ideal then

(3.3) normk/Q(A) =

Ok : A

is a positive integer, and equal to the index of A in the ring Ok. If γ is an integralprincipal ideal, then by (1.35) we have

(3.4) normk/Q

γ

= [Ok : γ ] =Normk/Q(γ ).

Lemma 3.1. Let k be an algebraic number field, Ok its ring of integers, and m a positive integer. Then there are at most finitely many integral ideals A in Ok such

that normk/Q(A) = m.

Proof. Let A be an integral ideal in Ok such that normk/Q(A) = m. Then thequotient ring Ok/A has cardinality m, and we find that

m +A = m(1 + A) = 0 + A.

This shows that m belongs to A. We conclude that m ⊆ A, and therefore byTheorem K.3 we have A|m. Let

m = Pe11 Pe22 · · ·P

eN N

be the factorization of m into a product of distinct prime ideals in Ok to positiveinteger powers. Because A|m, the factorization of A must have the form

A = Pd11 Pd22 · · ·P

dN N where 0 ≤ dn ≤ en.

Thus there are only finitely many possible integral ideals with normk/Q(A) = m. Lemma 3.2. Let k be an algebraic number field, Ok its ring of integers, and A a

fractional ideal in k. Then there exists a nonzero element α in k such that

(3.5) αA = αA

is an integral ideal, and

(3.6) normk/Q

αA

≤ 4

π

s N !N N

Disc(k) 12 ,


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26 VAALER

where 2s is the number of complex embeddings of k into C, and N = [k : Q].

Proof. As A−1 is also a fractional ideal in k, there exists a nonzero element β in

Ok so thatβ A−1 = β A−1 = B

is an integral ideal in Ok. Then by Theorem 1.8, Theorem 2.9 and (3.4) there existsa nonzero element γ in B such that

normk/Q

γ

=Normk/Q(γ )

≤ 4

π

s N !N N

Disck/Q(B) 12= 4

π

s N !N N

normk/Q(B)Disc(k) 12 .

(3.7)

Let α = β −1γ . Because γ belongs to B we have

γ ⊆ β A−1,

and thereforeγ β −1A = αA ⊆ Ok.

This shows that αA is an integral ideal. Then from (3.7) we find that

normk/Q

αA

= normk/Q

γ

normk/QB−1

≤ 4

π

s N !N N

Disc(k) 12 .This proves the lemma.

We are now able to prove the following fundamental result about algebraic num-ber fields.

Theorem 3.3. Let k be an algebraic number field and H (k) the ideal class groupof k . Then H (k) is a finite group.

Proof. It follows from Lemma 3.1 that the collection of all integral ideals I in Oksuch that

(3.8) normk/Q(I) ≤ 4

π

s N !N N

Disc(k) 12 ,is a finite set. Then Lemma 3.2 asserts that every element A in I (k) has an equiv-alent coset representative

αA = I

from the quotient group H (k) = I (k)/P (k) such that I is an integral ideal and thenorm of I satisfies the bound (3.8). Hence the number of distinct cosets in H (k) isfinite.


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4. Factorization of primes in an algebraic number field

Let k be an algebraic number field and let I be a proper ideal in the ring Ok of

algebraic integers in k. Because Ok is a Dedekind domain, it follows from TheoremK.1 that I has a factorization

(4.1) I = P1P2 · · ·PM ,

where each Pm is a prime ideal in Ok. Theorem K.1 also asserts that the fac-torization (4.1) is unique up to a permutation of the prime ideal factors. For ourpurposes it will be convenient to reorganize the right hand side of (4.1) as

(4.2) I = Pe11 Pe22 · · ·P

eLL ,

where P1,P2, . . . ,PL are distinct prime ideals in Ok and e1, e2, . . . , eL are positiveintegers. In this section we consider the special case where the ideal I is the principalideal p in Ok generated by a rational prime number p.

Theorem 4.1. Let k be an algebraic number field, and let P be a nonzero prime

ideal in Ok. Then P contains a unique rational prime number p. Moreover, there exists a positive integer f such that

(4.3) normk/Q(P) = [Ok : P] = pf , and 1 ≤ f ≤ N,

where N = [k : Q].

Proof. The prime ideal P is also a maximal ideal. Therefore Ok/P is a finite field of characteristic p for some rational prime number p. The homomorphsim m → m +Pfrom Z into Ok/P, maps Z onto the prime subfield represented by

0 + P, 1 + P, 2 + P, · · · , ( p − 1) + P

.

In particular we have0 + P = p + P,

and therefore p belongs to P. If q is also a rational prime contained in P and q = p,then there exist rational integers a and b so that ap + bq = 1. But this implies that1 belongs to P, which is impossible. Hence p is the unique rational prime thatbelongs to P.

Let p denote the principal ideal in Ok generated by the rational prime number p. Clearly we have p ⊆ P and therefore, by Theorem K.3, we have P| p. Hencethere exists an ideal Q in Ok such that

(4.4) p = PQ.

Then (3.2), (3.4), and (4.4), imply that

pN =Normk/Q( p) = normk/Q p = normk/Q(P) normk/Q(Q).

Using (3.3) we find that

normk/Q(P) = [Ok : P] and normk/Q(Q) = [Ok : Q]are positive integers, and the conclusion (4.3) plainly follows.

If P is a nonzero prime ideal in the ring Ok containing the unique rational primenumber p, then the positive integer f that occurs in the identity (4.3) is calledthe inertial degree (or more simply the degree ) of P. As is clear from the proof of Theorem 4.1, the map

(4.5) P → Char

Ok/P


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28 VAALER

sends each nonzero prime ideal P into the unique rational prime p that belongs toP.

Now suppose that p is a rational prime number and the principal ideal p in Okfactors into prime ideals as

(4.6) p = Pe11 Pe22 · · ·P

eLL ,

where P1,P2, . . . ,PL are distinct prime ideals in Ok, and e1, e2, . . . , eL, are pos-itive integers. Obviously (4.6) implies that Pl| p for each integer l = 1, 2, . . . , L.Therefore p ⊆ Pl, and p is the unique rational prime contained in Pl. In this casethe integer L is called the decomposition number of p in k. The exponent el thatoccurs on the prime ideal Pl in the factorization (4.6), is called the ramification index of the prime ideal Pl.

Alternatively, let P be a nonzero prime ideal in Ok, and let p be the uniquerational prime such that P| p. Then the ramification index of P is the uniquepositive integer e such that

Pe| p and Pe+1 p.

There is a similar terminology that applies to the rational prime p. Assume thatthe principal ideal p in Ok factors into prime ideals as in (4.6). If e1 = e2 = · · · =eL = 1 then we say that the rational prime p is unramified in k . If 2 ≤ el for someindex l, then we say that the rational prime p is ramified in k .

We note that these remarks also show that the map (4.5) from nonzero primeideals in Ok to rational prime numbers is surjective. And the inverse image of eachrational prime p is exactly the finite set of distinct nonzero prime ideals that occuron the right hand side of the factorization (4.6).

We now show that the inertial degrees and ramification indices attached to thenonzero prime ideals in the factorization (4.6) satisfy a simple identity.

Theorem 4.2. Let k be an algebraic number field with N = [k : Q]. Let p be a rational prime and assume that the principal ideal p in Ok has the factorization (4.6). Then the the ramification indices el attached to each prime ideal Pl, and the inertial degrees f l attached to each prime ideal Pl, satisfy

(4.7) N =Ll=1

elf l.

Proof. From (4.6) we obtain the identity

(4.8) pN =Normk/Q( p) = normk/Q p = L

l=1

normk/QPell

.

Then using (4.3) we have

(4.9) normk/QPell = normk/Q(Pl)el = pf lel .The identity (4.7) follows by combining (4.8) and (4.9).

Next we describe a method due to Dedekind [24] for determining the factorization(4.6). To begin with we suppose that k is generated over Q by an algebraic integerβ . Then it follows that

(4.10) Z[β ] =

a0 + a1β + a2β 2 + · · · + aN −1β

N −1 : an ∈ Z

⊆ Ok


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is a subring of Ok. For each rational prime p there exists a unique, surjectivehomomorphism

ψ p : Z → F p,where F p is the finite field with cardinality p. Then ψ p extends to a surjectivehomomorphism

ψ p : Z[x] → F p[x].

In particular, if mβ(x) in Z[x] is the minimal polynomial for β , then its imageψ p(mβ)(x) in F p[x] factors into a product of irreducible factors in F p[x]. Thus wewrite

(4.11) ψ p(mβ)(x) = g1(x)e1g2(x)

e2 · · · gL(x)eL ,

where g1(x), g2(x), . . . , gL(x), are distinct, monic, irreducible polynomials in F p[x],and e1, e2, . . . , eL, are positive integers. The following theorem of Dedekind showsthat for all but finitely many primes p, the factorization (4.11) provides informationabout the factorization of the ideal p into prime ideals in Ok.

Theorem 4.3. (Dedekind) Let k = Q(β ) be an algebraic number field. Assume that β an algebraic integer, mβ(x) in Z[x] is the minimal polynomial for β , and

N = [k : Q] = deg mβ.

Assume that Z[β ] is the subring defined by (4.10), p is a rational prime such that pdoes not divide the index

Ok : Z(β )

, and (4.11) is the factorization of the image

ψ p

mβ

(x) into irreducible polynomials in F p[x]. For each integer l = 1, 2, . . . , L,let Gl(x) be a monic polynomial in Z[x] such that

ψ p

Gl

(x) = gl(x).

Then for each integer l = 1, 2, . . . , L, the ideal

(4.12) Pl = p, Gl(β )is a prime ideal in Ok. Moreover, the prime ideals P1,P2, . . . ,PL are distinct,they satisfy

(4.13) normk/QPl

=

Ok : Pl

= pdeg gl = pdegGl ,

and the ideal p factors into powers of distinct prime ideals as

(4.14) p = Pe11 Pe22 · · ·P

eLL .

Proof. For each integer l = 1, 2, . . . , L the monic polynomial gl(x) is irreducible inF p[x]. Let γ l be a root of gl(x) in an algebraic closure F p. Then

F p(γ l) ∼= F p[x]/gl(x)

is a finite extension of F p of degree deg gl. Hence F p(γ l) is a finite field of cardinalty

pdeg gl . Letϕl : Z[β ] → F p(γ l)

be the surjective ring homomorphism defined by

(4.15) ϕl

f (β )

= ψ p(f )(γ l).

It is obvious that ϕl( p) = 0, and

ϕl

Gl(β )

= ψ p

Gl

(γ l) = gl(γ l) = 0.


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30 VAALER

Therefore both p and Gl(β ) belong to the kernel of ϕl, and we get

(4.16) Pl = p, Gl(β ) ⊆ ker(ϕl).

On the other hand, if f (β ) belongs to ker(ϕl) then ψ p(f )(γ l) = 0, and we findthat ψ p(f )(x) belongs to the principle ideal gl(x) in F p[x] generated by the monicirreducible polynomial gl(x). Hence there exists a polynomial H (x) in Z[x] suchthat

ψ p(f )(x) = ψ p(Gl)(x)ψ p(H )(x) = ψ p

GlH

(x).

That is, each coefficient of the polynomial

f (x) − Gl(x)H (x)

is divisible by p. Therefore we have

f (β ) = f (β ) − Gl(β )H (β ) + Gl(β )H (β )

⊆ p + Gl(β )

= p, Gl(β ).

(4.17)

Now (4.16) and (4.17) imply that

(4.18) Pl = ker(ϕl).

As ϕl induces an isomorphsim

ϕl : Z[β ]/ ker(ϕl) ∼= F p(γ l),

it follows that Pl = ker(ϕl) is a maximal ideal, and therefore a prime ideal in Z[β ].Because Gl(x) in Z[x] is monic and ψ p

Gl

= gl, we also have deg Gl = deg gl.

Exercises

4.1. Let k be an algebraic number field and P a nonzero prime ideal in the ring

Ok. Prove that P ∩ Z is a nonzero prime ideal in the ring Z.


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5. Absolute values

Let K be a field. An absolute value on K is a map

| | : K → [0, ∞)

from K into the nonnegative real numbers that satisfies the following three condi-tions:

(i) |x| = 0 if and only if x = 0,(ii) |xy| = |x||y| for all x and y in K ,

(iii) |x + y| ≤ |x| + |y| for all x and y in K .

The inequality (iii) is called the triangle inequality . For some absolute values definedon a field K it may happen that an inequality stronger than (iii) holds. Moreprecisely, it may happen that an absolute value on a field K satisfies the condition

(iv) |x + y| ≤ max{|x|, |y|} for all x and y in K .

If | | satisfies (iv) then it is obvious that | | must also satisfy (iii). The inequality (iv)is called the strong triangle inequality or the ultrametric inequality . An absolutevalue | | on a field K that satisfies the conditions (i), (ii), and (iii), but does notsatisfy (iv), is said to be an archimedean absolute value. An absolute value | | on K that satisfies the conditions (i), (ii), and (iv), is said to be a non-archimedean abso-lute value or an ultrametric absolute value. Later we will apply these designationsto certain equivalence classes of absolute values.

The first results about a field with an absolute value are immediate consequencesof the definition. Every field K has at least one absolute value which we denotehere by | |0. This is defined by

|x|0 =

0 if x = 0K ,

1 if x = 0K .

The absolute value | |0 is called the trivial or improper absolute value. Any otherabsolute values on K will be called nontrivial or proper . When working in a field K the trivial absolute value is usually not considered. However, there is at least onesituation in which the trivial absolute value may naturally occur. Suppose that Lis an extension of K and | | is an absolute value on L. Then | | restricted to K isobviously an absolute value on K . It may happen that | | is a nontrivial absolutevalue on L, but the restriction of | | to K is trivial.

Let K × denote the multiplicative group of nonzero elements in K . If | | is anabsolute value on K then the restriction of | | to K × is a homomorphism fromthe multiplicative group K × into the multiplicative group of positive real numbers.It follows that |1K | = 1. More generally, the multiplicative group of positive realnumbers is torsion free. It follows that if ζ N is a primitive N -th root of unity in

K ×

, then |ζ N | = 1. We also get| − x| = |x| for all x ∈ K, and |x−1| = |x|−1 for all x ∈ K ×.

If x is in K × and |x| = 1 then x must have infinite order in the multiplicative groupK ×. This shows that the only absolute value on a finite field is the trivial absolutevalue | |0.

When verifying that a function | | : K → [0, ∞) is an absolute value, it is oftenuseful to be able to check an inequality that is weaker than the triangle inequality.


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32 VAALER

Lemma 5.1. Assume that K is a field and | | : K → [0, ∞) satisfies the three conditions

(i) |x| = 0 if and only if x = 0K ,(ii) |xy| = |x||y| for all x and y in K ,(v) |x + y| ≤ 2 max{|x|, |y|} for all x and y in K .

Then | | satisfies the triangle inequality (iii) in the definition of an absolute value,and therefore | | is an absolute value on K .

Proof. If M = 2m for some positive integer m, then it follows easily from (v), byinduction on m, that

(5.1) |x1 + x2 + · · · + xM | ≤ 2m max{|x1|, |x2|, . . . , |xM |}

for all x1, x2, . . . , xM in K . If N is a positive integer and 2m−1 < N ≤ 2m, then by

selecting xN +1 = xN +2 = · · · = xM = 0K in (5.1), we find that

|x1 + x2 + · · · + xN | ≤ 2m max{|x1|, |x2|, . . . , |xN |}

≤ 2N max{|x1|, |x2|, . . . , |xN |}(5.2)

for all x1, x2, . . . , xN in K . In particular, if x1 = x2 = · · · = xN = 1K , then itfollows that

(5.3) |N 1K | ≤ 2N.

Now let x and y be elements of K , and let L be a positive integer. Then using (5.2)and (5.3) we get

|x + y|L = |(x + y)L|

= Ll=0

L

l

xlyL−l

≤ 2(L + 1)maxLl1K |x|l|y|L−l : 0 ≤ l ≤ L≤ 4(L + 1)max

L

l

|x|l|y|L−l : 0 ≤ l ≤ L

≤ 4(L + 1)

Ll=0

L

l

|x|l|y|L−l

= 4(L + 1)

|x| + |y|L

,

and therefore

(5.4) |x + y| ≤

4(L + 1)1/L

|x| + |y|

.

The triangle inequality (iii) follows by letting L → ∞ on the right hand side of (5.4).

If | | is an absolute value on K then the map (x, y) → |x − y| from K × K into[0, ∞) is a metric, and therefore the absolute value induces a metric topology inK . Of course the distance from x to y is |x − y|. Let AK denote the collection of all absolute values on K . We say that two absolute values in AK are equivalent if they induce the same metric topology. It is easy to verify that this is in factan equivalence relation in AK . The trivial absolute value | |0 clearly induces thediscrete topology in K , and it is the unique absolute value in its equivalence class.


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We will show that an equivalence class determined by a nontrivial absolute valuecontains infinitely many distinct, nontrivial absolute values. An equivalence classdetermined by a nontrivial absolute value in A

K is called a place of K . Thus we

may speak of the metric topology in K induced by a place of K , because all theabsolute values in that place induce the same topology.

Equivalent absolute values on K can be characterized in a simple way.

Theorem 5.2. Let | |1 and | |2 be two absolute values on K . Then the following assertions are equivalent:

(i) | |1 and | |2 induce the same metric topology in K ,(ii) {x ∈ K : |x|1 < 1} = {x ∈ K : |x|2 < 1},

(iii) there exists a positive number θ such that |x|θ1 = |x|2 for all x in K .

Proof. Assume that (i) holds and let

U j = {x ∈ K : |x|j


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34 VAALER

we further assume that | | is nontrivial, then |z| = 1 for some point z in K ×, andwe conclude that | |θ and | | are equivalent but not equal. In particular, this showsthat a place determined by a nontrivial absolute value in A

K contains infinitely

many distinct but equivalent absolute values.Let | | be an absolute value on K . Then define

Θ(| |) = {θ > 0 : x → |x|θ is an absolute value in AK }.

It follows from the definition of an absolute value that Θ( | |) is a closed, nonemptysubset of (0, ∞). By our previous remarks, if τ is in Θ(| |) then θτ is in Θ(| |) forall θ such that 0 < θ ≤ 1. This shows that either Θ(| |) = (0, ∞), or Θ(| |) = (0, τ ]for some real number τ ≥ 1. We define a second function on absolute values in AK by setting

Φ(| |) = sup{|x + 1| : x ∈ k and |x| ≤ 1}.

Obviously we have 1 ≤ Φ(| |) ≤ 2. Also, the inequality

(5.5) |x + y| ≤ Φ(| |)max{|x|, |y|}

holds for all x and y in K . To verify (5.5) assume that 0 < |x| ≤ |y|. Then we have

|x + y| = |(xy−1 + 1)y| = |xy−1 + 1||y| ≤ Φ(| |)|y| = Φ(| |)max{|x|, |y|}.

Note that if Φ(| |) = 1 then (5.5) is the strong triangle inequality (iv) in thedefinition of an absolute value.

Theorem 5.3. Let | | be an absolute value on K . Then the following assertions are equivalent:

(i) Θ(| |) = (0, ∞),(ii) |x + y| ≤ max{|x|, |y|} for all x and y in K ,

(iii) Φ(| |) = 1.

Moreover, if 1 < Φ(| |) ≤ 2, then Θ(| |) = (0, τ ] for some real number τ ≥ 1, and

(5.6) Φ(| |)τ = 2.

Proof. Assume that Θ(| |) = (0, ∞). Then the map x → |x|n is an absolute valuefor all positive integers n. It follows that

|x + y| ≤

|x|n + |y|n1/n

for all x and y in K . We let n → ∞ and conclude that

|x + y| ≤ max{|x|, |y|}.

This shows that (i) implies (ii).It is trivial that (ii) implies (iii).Assume that Φ(| |) = 1. In view of (5.5) we find that (ii) is satisfied. Then we

get

|x + y|θ ≤ max{|x|, |y|}θ= max{|x|θ, |y|θ}

≤ |x|θ + |y|θ

(5.7)

for all x and y in K and for all positive θ . This shows that the map x → |x|θ is anabsolute value on K for all positive θ, and so verifies that Θ(| |) = (0, ∞). We haveshown that (iii) implies (i).


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Assume that 1 < Φ(| |) ≤ 2. By our previous remarks we have Θ(| |) = (0, τ ] forsome real number τ ≥ 1. Because | |τ is an absolute value, we find that

Φ(| |)τ = Φ| |τ = sup{|x + 1|τ : x ∈ k and |x|τ ≤ 1}

≤ 2.

(5.8)

If in fact Φ(| |)τ τ such that

Φ

| |θ

= Φ(| |)θ ≤ 2.

It follows from (5.5) that

(5.9) |x + y|θ ≤ 2 max{|x|θ, |y|θ}

for all x and y in K . From Lemma 5.1 and (5.9) we conclude that x → |x|θ is anabsolute value on K . As θ > τ , this contradicts the definition of Θ(| |). We haveshown that if 1 < Φ(| |) ≤ 2, then (5.6) holds.

We are now in position to distinguish between different types of absolute values.If | | is an absolute value on K then by Theorem 5.3, Φ(| |) = 1 if and only if theabsolute value satisfies the inequality

(5.10) |x + y| ≤ max{|x|, |y|}

for all x and y in K . An absolute value that satisfies (5.10) is said to be a non-archimedean absolute value, or to be an ultrametric absolute value. Then (5.10) iscalled the strong triangle inequality, or the ultrametric inequality. If | | is a non-archimedean absolute value then the equivalence class it represents in AK is theset

{| |θ : 0 < θ 1, and |α|n < 1 for 2 ≤ n ≤ N .


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36 VAALER

Proof. Suppose that N = 2. As | |1 and | |2 are inequivalent, by Theorem 5.2 thereexists β in K with

|β |1 < 1 and 1 ≤ |β |2,and there exists γ in K with

1 ≤ |γ |1 and |γ |2 < 1.

In this case we take α = β −1γ .We continue now using induction on N . By the inductive hypothesis there exists

β in K with|β |1 > 1 and |β |n < 1 for 2 ≤ n ≤ N − 1,

and γ in K with1 ≤ |γ |1 and |γ |N 0 there exists a point y in K such that

|xn − y|n < for each n = 1, 2, . . . , N .

Proof. For each integer n with 1 ≤ n ≤ N we apply Lemma 5.4, but with | |n inplace of | |1. In this way we determine a collection of points α1, α2, . . . , αN in K such that

|αm|n > 1 if m = n, and |αm|n < 1 if m = n.

Then we find that

liml→∞

αlm1 + αlm

= 1 in the | |n-metric if m = n,

and

liml→∞

αlm1 + αlm

= 0 in the | |n-metric if m = n.

It follows that for each n, 1 ≤ n ≤ N , we have

liml→∞

N m=1

xmαlm

1 + αlm= xn in the | |n-metric.

Hence we may take

y =N

m=1

xmαlm

1 + αlm

with l a sufficiently large positive integer that depends on .


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Given a field K it is an interesting problem to determine all the nontrivial ab-solute values on K . The following result is useful when the field K is the field of fractions associated to an integral domain.

Lemma 5.6. Let R be a nonzero integral domain, and let K be its field of fractions.Assume that x → x is a map from R into [0, ∞) that satisfies the three conditions

(i) x = 0 if and only if x = 0,(ii) xy = xy for all x and y in R,

(iii) x + y ≤ x + y for all x and y in R.

Then there exists a unique absolute value | | : K → [0, ∞) such that |x| = x for all x in R. If on R satisfies the strong triangle inequality

(iv) x + y ≤ max

x, y

for all x and y in R,

then | | on K is a non-archimedean absolute value. Moreover, if a and b = 0 are in R, and a/b is a point in K , then these maps satisfy the identity

(5.11) ab = ab .Proof. Suppose that a/b = c/d in K . That is, a, b = 0, c and d = 0 belong to Rand ad = bc. It follows that ad = bc and

a

b =

c

d

in [0, ∞). Therefore we define a map | | : K → [0, ∞) byab

= ab

.

Our previous remarks show that this is well defined. Using (i), (ii) and (iii) it iseasy to verify that | | on K satisfies the corresponding conditions required of an

absolute value. For example, if a, b = 0, c and d = 0 belong to R, thenab

+ c

d

= ad + bcbd

= ad + bcbd

≤ ad + bc

bd =

a

b +

c

d

=a

b

+ cd

.(5.12)

If (iv) holds this inequality becomesab

+ c

d

= ad + bcbd

= ad + bcbd

≤ max

ad, bc

bd = maxa

b

, c

d= max

ab

, cd

,(5.13)

and we conclude that the absolute value | | is non-archimedean.We note that (ii) implies that y = 1K y = 1K y and therefore 1K = 1. If

a belongs to R then

|a| = a

1K

= a1K

= a,


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38 VAALER

and this shows that | | on K is an extension of on R. If | |1 and | |2 are bothabsolute values on K that extend the map on R, then we have

ab

1= |a|1

|b|1= a

b = |a|2

|b|2= a

b

2.

This shows that | |1 and | |2 are equal on K . Thus the absolute value defined by(5.11) is the unique extension of on R.

Exercises

5.1. Let | | be an absolute value on the field K . Prove that | | is non-archimedeanif and

algebraic number theory notes

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