TIFR pamphlet on Algebraic Number Theory

8/21/2019 TIFR pamphlet on Algebraic Number Theory

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ALGEBRAIC NUMBER THEORY

RAGHAVAN NARASIMHAN

S. RAGHAVAN

S. S. RANGACHARI

SUNDER LAL

Tata Institute of Fundamental Research, Bombay

School of Mathematics

Tata Institute of Fundamental Research, Bombay

1966


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PREFACE

THIS pamphlet contains the notes of lectures given at a SummerSchool on the Theory of Numbers at the Tata Institute of Fundamen-tal Research in 1965. The audience consisted of teachers and studentsfrom Indian Universities who desired to have a general knowledge of thesubject. The speakers were Raghavan Narasimhan, S. Raghavan, S. S.Rangachari and Sunder Lal.

Chapter 1 sets out the necessary preliminaries from set theory andalgebra; it also contains some elementary number-theoretic material.

Chapter 2 deals with general properties of algebraic number fields; itincludes proofs of the unique factorization theorem for ideals, the finite-ness of class number, and Dirichlets theorem on units. Chapter 3 givesa slightly more detailed analysis of quadratic fields, in particular fromthe analytic aspect; the course ends with Dirichlets theorem on theinfinitude of primes in an arithmetic progression.


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Contents

1 Preliminaries 1

1.1 Sets and maps . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Equivalence Relations . . . . . . . . . . . . . . . . . . . . 21.4 Abelian groups and homomorphisms . . . . . . . . . . . . 31.5 Rings, modules and vector spaces. . . . . . . . . . . . . . 61.6 The Legendre symbol . . . . . . . . . . . . . . . . . . . . 91.7 The quotient field of an integral domain . . . . . . . . . . 101.8 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.9 Ideals and quotient rings . . . . . . . . . . . . . . . . . . . 171.10 Linear mappings and matrices . . . . . . . . . . . . . . . . 191.11 Polynomial rings . . . . . . . . . . . . . . . . . . . . . . . 21

1.12 Factorial rings . . . . . . . . . . . . . . . . . . . . . . . . . 231.13 Characters of a finite abelian group . . . . . . . . . . . . . 26

2 Algebraic Number Fields 29

2.1 Algebraic numbers and algebraic integers . . . . . . . . . 292.2 Unique Factorization Theorem . . . . . . . . . . . . . . . 402.3 The class group ofK . . . . . . . . . . . . . . . . . . . . 432.4 The group of units . . . . . . . . . . . . . . . . . . . . . . 47

3 Quadratic Fields 55

3.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2 Factorization of rational primes in K . . . . . . . . . . . . 583.3 The group of units . . . . . . . . . . . . . . . . . . . . . . 623.4 Laws of quadratic reciprocity . . . . . . . . . . . . . . . . 633.5 The Dirichlet class-number formula . . . . . . . . . . . . . 743.6 Primes in an arithmetic progression . . . . . . . . . . . . 83

i


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Chapter 1

Preliminaries

1.1 Sets and maps

A set is a collection of objects which are called the elementsof the set.We shall suppose that if any object is given, we can decide whether itbelongs to the set or not. The set of all rational integers (i.e. integerspositive, negative and zero) is denoted by Z, the set of all non-negativeintegers by Z+, the set of all rational numbers by Q, the set of all realnumbers by R, and the set of all complex numbers by C.

Ifx is an element of a set A, we write x A. Ifx is not an elementofA, we write x /

A. IfP is a property, the set of all objects with the

property P will be denoted by{x|x possesses the property P}. Thus{x | x Z, x


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2 Chapter 1. Preliminaries

(iii) the cartesian product X

Y ofX and Y as

{(x, y) | x X and y Y}.We say X and Y are disjoint ifX Y = . IfX Y, we define the

complement,Y X, as the set{z| z Y and z / X}.

1.2 Maps

Let X and Y be two sets. A map f: X Y is an assignment to eachx X, of an elementf(x) Y. IfA is a subset ofX, theimagef(A) isthe set {f(x) | x A}. Theinverse imageof a subsetBofY, denoted byf

1

(B), is the set{x | x X, f(x) B}. The map f is said to be ontoor surjective, iff(X) =Y; iff(x) =f(y) impliesx = y, thenfis said tobe one-one or injective. Iff: X Y, g: Y Zare two maps, we definethecomposite (g f): X Zas follows: forx X, (g f)(x) =g(f(x)).The map X Xwhich associates to each x X, the element x itselfcalled the identity map of X and denoted by IX (or by I, if there isno confusion). If f: X Y is both one-one and onto, there is a mapfrom Y X, denoted by f1, such that f f1 = IY, f1 f = IX.The map f1 is called the inverse off. IfA is a subset ofX, the map

j = jA: A Xwhich associates to each a A the same element a inX is called the inclusion map ofA in X. Iff: X Y is any map, themap fjA: A Y is called the restriction off to A and is denoted byf| A.

1.3 Equivalence Relations

Definition 1.1 LetXbe a set. An equivalence relation inX is subsetR ofX X such that

(i) for everyx X, (x, x) R;(ii) if(x, y) R, then(y, x) R; and

(iii) if(x, y) R and(y, z) R then(x, z) R.

We say that x is equivalent to y with respect to R and write xRy if(x, y) R. Then the conditions above simply require that

(i) every element x is equivalent to itself (reflexivity),


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1.4. Abelian groups and homomorphisms 3

(ii) ifx is equivalent to y , y is equivalent to x (symmetry), and

(iii) ifx is equivalent toy and y toz then x is equivalent to z .(transitivity).

Let R be an equivalence relation in a set X. Then for anyx X,the set of all elements ofXequivalent to x with respect to R is calledthe equivalence class of R containing x and is denoted by x. Considerthe family of distinct equivalence classes ofX with respect to R. It iseasy to verify that they are pairwise disjoint and that their union is X.The set of these equivalence classes x is called the quotient ofX by Rand is denoted by X/R.

Example 1.1 The subset R X X consisting of elements (x, x),x X is an equivalence relation. This is called the identity relation.

Example 1.2 Letn Z, n >0. Consider the set in Z Z of pairs ofintegers(a, b)sucha bis divisible byn. This is an equivalence relationin Z and the quotient of Z by this relation is denoted byZ/(n) orZn.

1.4 Abelian groups and homomorphisms

Definition 1.2 LetG be a nonempty set and : G

G

Ga mapping.

Let, forx, y G, ((x, y))be denoted byxyorxy. Then the pair(G, )is said to be an abelian group if the following conditions are satisfied:

(a) x(yz) = (xy)z for every x, y ,z in G(associativity),

(b) there exists an elemente, called the identity element ofG, whichsatisfiesex= xe = x for every x in G.

(c) for everyx G, there exists inGan elementx1, called the inverseofx, such that xx1 =x1x= e, and

(d) for everyx, y

G, xy= yx (commutativity).

Remark 1.1 We often abbreviate(G, )toG when it is clear from thecontext to which map we are referring.

Remark 1.2 The map is called the composition law inG. It is alsocalled the multiplication inG.


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Remark 1.3 The identity element is unique. In fact, if there is an

elemente inG such that condition (b) above is valid for everyx inGwithe replaced bye, we have, in particular, e= ee = e.

Remark 1.4 The inverse of any element is unique.

Remark 1.5 In view of associativity, we define xyz = (xy)z = x(yz)for everyx, y ,z inG.

More generally, the product x1x2 xn is well defined, where x1, x2,. . . , xn G. (Proof by induction). In particular, for any x G, we set

x

m

= xx x(m times) for m >0 inZ

,x0 = e,

xn = (x1)n for n


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1.4. Abelian groups and homomorphisms 5

Definition 1.4 LetG, G be two groups. A homomorphismf fromG toG is a map f: G G such thatf(xy) =f(x)f(y) for everyx, y G.

Letf: G G be a homomorphism. Thenf(e) =e. In fact, f(e) =f(ee) =f(e)f(e) and multiplying both sides byf(e)1, we getf(e) =eIff: GG andg: G G are homomorphisms, theng f: GGis also a homomorphism. For any group G, the identity map IG: G Gis a homomorphism.

Definition 1.5 A homomorphismf: G G is called an isomorphismif there exists a homomorphism g: G G such that f g = IG (theidentity map ofG) andg f=IG(the identity map ofG).

It is easy to see that a homomorphismf: G

G is an isomorphism

if and only if it is both injective and surjective.

Example 1.6 The natural map : Z Z/(n)is a surjective homomor-phism. Forn = 0 it is not one-one and hence not an isomorphism.

Example 1.7 The map f: Z Z given by f(a) = 2a for a Z is aninjective homomorphism. It is not onto and hence not an isomorphism.

Example 1.8 The map g: Q Q given byg(x) = 1/x forx Q isan isomorphism. Furtherg g= IQ.

Definition 1.6 LetG be a group. A non-empty subsetH ofG is called

a subgroup ofG if for everyx, y inH,xy1 also belongs to H.

In particular e Hand for any x H, x1 H. It can be easilychecked that with the composition induced by that ofG, His a groupwith e as the identity element and x1 as the inverse ofx in H.

LetHbe a subgroup ofG. Then the inclusion map j : H Gis aninjective homomorphism.

For any group G, the subsets{e} and G are subgroups of G. LetG1, G2 be two groups and f: G1 G2, homomorphism ofG1 into G2.Then the set of x G1, for which f(x) = e is easily verified to be asubgroup ofG1. It is called thekerneloffand is denoted by ker f. The

homomorphism f is an isomorphism iff is onto and ker f= {e}.LetGbe a group andHa subgroup ofG. The relation x y (x, y

G) if and only if xy1 H is an equivalence relation. If x is theequivalence class containing x, then, on the set of equivalence classes wecan introduce the structure of a group by setting xy= xy. These classesare called cosets ofG modulo H. This group is denoted by G/Hand is


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called the quotient ofG byH. IfG/H is finite, then its order is called

the index ofH in G. There is natural mapping ofG onto G/H takingx Gto xand this mapping is a surjective homomorphism with kernelH.

Letf: G1 G2 be a homomorphism of a group G1 into a group G2with ker f=H. The imagef(G1) ofG1underfis a subgroup ofG2andwe define a homomorphism f: G1/H f(G1) by settingf(x) =f(x) forx G1. Clearlyfis an isomorphism ofG1/H ontof(G1) (fundamentaltheorem of homomorphisms).

Let G be a group and a G such that every element ofG is of theform an, where nZ. Then we say thatG is a cyclic group generatedbya.

Example 1.9 (Z, +)is an infinite cyclic group generated by the integer1. The only subgroups of Z are of the form mZ ={mx| x Z} form 0.

Proposition 1.1 Any cyclic group G is isomorphic to Z orZ/(m) forsome(m > 0).

Proof: LetG be generated by a. Consider the map f: Z Gtakingn Zto an. This is a surjective homomorphism and ker fis a subgroupmZ of Z generated by m

0 in Z. Ifm = 0, G is isomorphic to Z. If

m 0, Gis isomorphic to Z/(m).Definition 1.7 LetG be a group anda G. We say thata is of ordern if the cyclic group generated bya inG is of ordern.

Example 1.10 The element1 in Q is of order2.

Remark 1.6 If G is a group of order h and an element a G is ofordern, thenn dividesh and thereforeah =e.

1.5 Rings, modules and vector spaces.Definition 1.8 Let R be a nonempty set and let , be mappings ofR R into R. Writing ((x, y)) = x+ y, ((x, y)) = xy, the triple(R,,) is said to be a ring if the following conditions are satisfied:

(i) (R, ) is an abelian group,


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1.5. Rings, modules and vector spaces. 7

(ii) x(yz) = (xy)z forx, y ,z inR (associativity).

(iii) x(y+ z) =xy + xz, (y+ z)x= yx + zx inR (distributivity), and

(iv) there exists inR an element 1, called the unit element ofR, suchthatx1 = 1x= x for everyx inR.

Remark 1.7 We call and respectively the addition and the multipli-cation inR; we write+ for and or for.

Remark 1.8 The identity element of (R, ) is called the zero elementofR and is denoted by0.

Remark 1.9 The unit element inR is unique.

Remark 1.10 Associativity is valid for any finite number of elementsinR (in a sense which is obvious).

Remark 1.11 We denote byR the set of non-zero elements ofR.

Definition 1.9 A ring R is commutative if xy = yx for every x, y inR.

Hereafter, by a ring we shall always mean a commutative ring.

Definition 1.10 A subringS of a ringR is a subgroup of(R, +) suchthat1 S and forx, y S, xy S.

We observe that S is a ring under the operations induced from R.

IfR is a ring, Sa subring and Ea subset ofR, the ring S[E] gen-erated by E is the set consisting of 1 and all the elements of the form =

ni=1 siei, si Swhere each ei is a product of finitely many ele-

ments ofE. It is trivial thatS[E] is the smallest subring ofR containingSand E.

Definition 1.11 Let R be a ring and a, b R. Then we say that adividesb (or thata is a divisor ofb, a|b in symbols), if there existsc inR such thatb= ac. Ifa does not divideb, we writea | b.

Definition 1.12 Ifa, b, c R, then a is congruent to b modulo c ( insymbolsa b (mod c)) ifc|(a b).


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Definition 1.13 An elementa

Ris called a zero-divisor if there exists

x = 0 inR such thatax= 0. Trivially, 0 is a zero-divisor.

Definition 1.14 A commutative ring R in which1= 0, is an integraldomain ifR contains no zero-divisors other than0.

Definition 1.15 An elementu R is called a unit inR if there existsv R such thatuv= 1. (R is commutative).

The unitsuinR clearly form a multiplicative group which we denotebyU.

Definition 1.16 We call two elementsa, b inR associates with respectto U if a = ub for some u U. We say, briefly, that a and b areassociates and otherwise we say thata andb are non-associate.

Definition 1.17 A commutative ring R is called a field, if the set R

of elementsa = 0inR forms a group under multiplication (i.e. if everya = 0 inR is a unit ). Clearly a field is an integral domain.

Definition 1.18 A subring R of a fieldK is called a subfield of K, ifR is a field (with respect to the operations which makeR a ring viz. theoperations induced fromK).

Example 1.11 Z, Q, R, and C are commutative rings with the usualaddition and multiplication of complex numbers, and, in fact, integraldomains. C, R, Q are fields and Q, R are subfields ofC; however, Z isnot a field.

Example 1.12 The additive group Z/(m) consisting of residue classes(of Z) modulo (an integer) m(= 0) is a ring, the multiplication beingdefined byrs= rs forr, s Z.

Example 1.13 A finite integral domain A is a field. In fact, let A =

{x1, x2, . . . , xh}, and leta A, a = 0. Then the elementsax1, . . . , a xhare distinct, sinceA has no non-zero zero-divisors. Hence they must beall thexi in some order, so thataxi = 1 for somei.

Definition 1.19 A positive integer p in Z is called a prime if p > 1and its only divisors in Z are1,p.


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1.6. The Legendre symbol 9

Proposition 1.2 Forp >0 in Z, Z/(p) is a field if and only ifp is a

prime.

Proof: In fact, ifp = rs, r, s=1, then rs = 0. But neither r nors is equal to 0. HenceZ/(p) is not even an integral domain, ifp is nota prime. Conversely, if p is a prime, then for any x Z with p x,there exists y Z such that xy 1(mod p). For, consider the set ofr Z such that rx 0(mod p). This is an additive subgroup G of Zand hence, by the example 1.9 on page 6, of the form mZ, m 0. Since

p G, m = 1 or p. But m= 1, since p x. Thus m = p and as aconsequence, the elements,

0, x, 2x , . . . , (p 1)x. are all distinct modulop. Since the order ofZ/(p) is p(see Example 1.4page 4) there exists y Z (1 y p 1) such that yx = 1, i.e.yx 1(mod p). [The last part of the argument is that of Example 1.13above].

1.6 The Legendre symbol

Definition 1.20 Letp Z, p >2 be a prime. An integera withp |ais said to be a quadratic residue modulo p, if there exists x

Z such

thatx2 a(mod p), and a quadratic non-residue modulo p if no suchxexists inZ.

From the definition, it is clear that a is a quadratic residue modulopif and only ifa + mp(for arbitrarym Z) is so. We can thus talk of a(non-zero) residue class modulop being quadratic residue or non-residuemodulop.

Consider now, forp a, the quadratic congruencex2 a(modp) forx Z. Ifa is a quadratic non-residue modulop, this congruence has nosolutionx. Ifais a quadratic residue modulop, saya b2(mod p), thenx2

b2(mod p),i.e. (x

b)(x+b)

0(mod p). Since by Proposition 1.2

Z/(p) is an integral domain, it follows that x b(mod p) are theonly two solutions of the congruence x2 a(modp). Further, since

p >2, b b(mod p). Now consider the mapping x x2 of (Z/(p))into itself. The image of x (= 0) under this mapping is always a quadraticresidue modulo p and by what we have seen, each quadratic residuemodulop in (Z/(p)) is the image of exactly two elements of (Z/(p))


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under this mapping. Thus there are (p

1)/2 quadratic residues modulo

p. It follows that there are (p 1)/2 quadratic non-residues modulo p.The product of two quadratic residues modulo p is again a residue.

For, ifa x2(mod p) and b y2(mod p) then ab (xy)2(mod p).The product of a quadratic residue a modulo p and a non-residueb

modulo p is a quadratic non-residue modulo p. For, there exists x Zfor which x2 a(modp) and ifab z2(mod p) for z Z choose byProposition 1.2, y Z such that xy 1(mod p). Clearly we have thenthe congruenceb (yz)2(mod p), and b would be a residue modulo p.

The product of two quadratic non-residuesa, bmodulopis a quadraticresidue modulo p. For let a1, . . . , aq(q =

12 (p 1)), be the quadratic

residues modulo p. Then since Z/(p) is an integral domain, it follows

from what we have seen above that aa1, . . . , a aq are precisely all theresidue classes modulo p which are non-residues. Since ab is distinctfrom these, it must be a residue.

Definition 1.21 Let p= 2 be a prime and a Z. We define theLegendre symbol(ap ) by

a

p

=

+1 if pa and a is a quadratic residue modulo p1 if pa and a is a non-residue modulo p,0 if p | a.

It is clear from the earlier considerations thatab

p

=

a

p

b

p

for a, b Z (1.1)

1.7 The quotient field of an integral domain

Definition 1.22 LetR, R be two rings. A map f: R R is called ahomomorphism if

(i) f(x + y) =f(x) + f(y),

(ii) f(xy) =f(x)f(y) for everyx, y R, and(iii) f(1) = 1.

For any ringR, the identity mapIR is a homomorphism. The compositeof two homomorphisms is again a homomorphism.


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1.8. Modules 11

Definition 1.23 A homomorphismf: R

R is said to be an isomor-

phism if there exists a homomorphism g: R R such that g f = IRandf g= IR . The ringsR andR are then said to be isomorphic andwe writeR R.

An isomorphism f: R R is called an automorphism of R. Theimage f(R) of a ring R under a homomorphism f: R R is a subringofR. A homomorphism is an isomorphism if and only if it is injectiveand surjective.

Remark 1.12 The natural map q: Z Z/(m) is a homomorphism.

Proposition 1.3 Every integral domainR can be embedded isomorphi-cally in a field.

Proof: Let R be an integral domain and R the set of non-zero el-ements of R. On RR, we define the relation: (a, b) (c, d) ifad= bc. SinceR contains no zero-divisors, it can be verified that this isan equivalence relation. We make the quotient K=R R/ a ring bydefining the ring operations as follows. Ifx/y denotes the equivalenceclass containing (x, y) RR then definea/b + c/d= (ad + bc)/bdand(a/b)(c/d) =ac/bd. These operations are well defined andK is a ring.In fact, K is a field, since b/a is an inverse for a/b, a= 0. The mapi: R Kgiven only byi(a) =a/1 fora Ris a one-one homomorphismofR into K.

We shall identify R with the subring i(R) ofK.

Remark 1.13 K is called the quotient field of R. If f: R L is aone-one homomorphism ofR into a fieldL, thenfcan be extended in aunique way to a one-one homomorphismf ofK into L, by prescribingf(a/b) =f(a)f(b)1, forb = 0. Further, this property characterises thequotient field upto isomorphism. ThusL contains the isomorphic imagef(K) ofK. We see then that ifL contains an isomorphic image ofKand in this sense, K is the smallest field containingR.

Example 1.14 Q is (isomorphic to) the quotient field ofZ.

1.8 Modules

LetR be a (commutative) ring (containing 1)


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Definition 1.24 AnR-moduleM is a triple(M, +, ) where

(i) (M, +) is an (abelian) group

(ii) : R M M is a map such that if we set

((, a)) =a, then for, R, andx, y M, we have

(1) (x + y) =x + y,

(2) ( + )x= x + x

(3) ()x= (x),

(4) 1 x= x.

The elements ofR are called scalars and is called scalar multiplication.

Definition 1.25 IfR is a field, M is called a vector space overR (oranR-vector space) and the elements ofR called vectors.

Example 1.15 Every (abelian) group (G, +) can be regarded as a Z-module (G, +, ) by defining ((n, x)) = nx for x G and n Z.Conversely, everyZ-module is of this type.

Example 1.16 Every (commutative) ringR with 1 is anR-module.

Example 1.17 R andC are vector spaces overQ.

Definition 1.26 Let M be an R-module (resp. vector space over R).Then a subgroup N of M is called an R-submodule (resp. a subspaceoverR) if mapsR N into N.

Definition 1.27 Let M be an R-module. A subset S of M is said togenerate M over R if every x

M can be written in the form x =ni=1 aixi withxi S, ai R andn= n(x) Z+.

The elements ofSare called generators ofM over R. In particular,if there exists a finite subsetSofM overR, thenM is finitely generatedoverR. IfS= , by definition, the generating Mmodule generated byS is{0}.


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1.8. Modules 13

Proposition 1.4 Let 1, 2, . . . , n be a system of generators of a Z-

moduleM, and letNbe a sub-module ofM. Then there exist1, . . . , minN(m n)that generateN overZ and have the formi =

ijkijj

withkij Z, kii 0 and1 i m.

Proof: Let us suppose that the proposition has been proved for allZ-modules with n 1 generators at most, where n 1. (Note that theproposition is trivial when M ={0}.) Let M be a module generatedover Z by n elements 1, 2, . . . , n and N a sub-module ofM. DefineM to be the module generated by 2, 3, . . . , n over Z and N to beN M. If n = 1, M ={0}. If N = N the proposition is trueby the induction hypothesis. If N

= N, then let A be the subgroup

of Z consisting of integers k for which there exist k2, . . . , kn in Z withk1+ k22+ +knn N. Then A is of the form k11Z, k11 0;let 1 = k111 +k122 + +k1nn N. If =

ni=1 hii, with

h1, h2, . . . , hn Z, belongs to N, then h2 A so that h1 = mk11 forsome m Z. Thus m1 N. By the induction hypothesis, thereexist i =

ji kijj (i = 2, 3, . . . , m, kij Z, kii 0), in N, which

generate N. It is clear that1, 2, . . . , m are generators ofN havingthe required form.

Definition 1.28 A subsetSof anR-moduleM is linearly independentover R, if, for any finite set of elements x

1, . . . , x

n in S, a relationn

i=1 aixi = 0, ai R implies necessarily thata1 = = an = 0. Wesay thatSis linearly dependent, if it is not linearly independent.

Definition 1.29 A subsetS ofM is called a base ofM overR (or anR-base) ifS is linearly independent and generatesM overR.

Example 1.18 TheZ-module of even integers has{2} for a base.

Example 1.19 Q is a Z-module but is not finitely generated over Z.Trivially, Q if finitely generated overQ.

Example 1.20 R2 = R R has{(1, 0), (0, 1)} as a base overR.

Proposition 1.5 LetVbe a vector space (over a fieldK) which has aset of generators consisting ofm elements. IfS is any linearly indepen-dent subset ofV, thenScontains at mostm elements.


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Proof: Letx1, . . . , xmbe a set of generators for V overK. If possible,

let S contain more than m elements, say y1, . . . , ym+1. We can writey1 =

mi=1 ixi, i K and i= 0 for at least one i. We may assume,

without loss of generality, that1= 0. Thenx1= 11 y1m

i=2 11 ixi.

Hence y1, x2, . . . , xm generate V. Assume for some i, 1 i m, that(y1, . . . , yi, xi+1, . . . , xm) generate V. (This is true for i = 1.) Then

yi+1=i

j=1

jyj+m

k=i+1

kxk, 1, . . . , m K.

Since y1, . . . , yi+1 are linearly independent, k= 0 for some k i + 1.Without loss of generality, we may suppose that i+1

= 0. Then

xi+1=i+1j=1

jyj+m

k=i+2

kxk

for 1, . . . , i+1, i+2, . . . , m K. Thusy1, . . . , yi+1, xi+2, . . . , xm gen-erate V. By iteration of this process, it follows that y1, . . . , ym generateV. In particular ym+1 =

mi=1 iyi, i K, contradicting the linear

independence ofS.

Corollary 1.1 Ifx1, . . . , xm andy1, . . . , ynare bases ofV, thenm = n.

Remark 1.14 We shall be concerned only with vector spaces, which arefinitely generated (over a field).

Remark 1.15 LetVbe a vector space, finitely generated over a fieldK.From the finitely many generators, we can clearly pick out a maximalset of linearly independent elements which suffice to generate V andconstitute a base ofV overK.

Remark 1.16 LetV be a vector space generated byx1, . . . , xm overKand letx= 0 inV. Then x, x1, . . . , xm again generateV overK and,

from this set, we can always pick out a maximal linearly independent

subset containing x. In other words, any x= 0 in V can be completedto a base ofV.

Definition 1.30 A vector spaceV is said to be of dimensionn over afield K (in symbols dimk V = n) if there exists a base of V over Kcontainingn elements.


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1.8. Modules 15

Observe that, by Corollary 1.1, the dimension is defined indepen-

dently of the base used.We shall write merely dim V for dimkV when it is clear from the

context to which field Kwe are referring.

Corollary 1.2 Let W be a subspace of V with dimkV = n. Then Whas a base consisting of at mostn elements, i.e. dimkW dimkV. IfWis a proper subspace ofV, thendim W


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Definition 1.32 LetV be a vector space over a fieldK anddimkV =

n. By a K-linear form (or briefly, a linear form) on V, we mean ahomomorphism ofV into K(regarded as a vector space over itself).

The set V of linear forms on V forms a vector space over K andis called the dual of V. If 1, . . . , n is a base of V over K, definethe linear forms 1, . . . ,

n by

i (j) = ij (the Kronecker delta)= 1

if i = j and 0 for i= j. We see at once that 1, . . . , n are linearlyindependent over K. For, if

ni=1 ai

i = 0 with a1, . . . , an = K, then

aj =n

i=1 aii (j) = 0 for j = 1, . . . , n. Any linear form

can bewritten in the form =

ni=1 bi

i where

(i) = bi K. ThusdimKV

=n. The base 1

, . . . , n

ofV is called the dual baseof thebase 1, . . . , n ofV .

Definition 1.33 Let V be a vector space over a field K. A bilinearform B on V is a mapping B: VV K such that for any fixedy V the mappings By, By of V into K, defined by By(x) = B(x, y)andBy (x) =B(y, x) respectively, are linear forms onV.

Definition 1.34 A bilinear formB(x, y)onV is non-degenerate if, forany fixed y= 0 inV, the linear formBy, is not zero, i.e. B(x, y)= 0

for at least one x, and for any fixed x

= 0, the linear form Bx is not

zero.

LetV be a vector space of dimensionn over a fieldK. Then we have

Proposition 1.6 Let B(x, y) be a non-degenerate bilinear form on V.Then for any base 1, . . . , n of V, there exists a base1, . . . , n of Vsuch thatB(i, j) =ij (the Kronecker delta) for1 i, j n.

Proof: Consider the mapping ofV to V taking y V to the linearform B y in V. This is clearly a homomorphism ofV into V. Since Bis non-degenerate, this mapping is injective. Since dim V= dim V= n,it follows by Noethers homomorphism theorem and Corollary 1.2 abovethat this mapping is onto V. Let 1, . . . , n be a base of V over Kand 1, . . . ,

n the corresponding dual base of V

. Let 1, . . . , n bethe elements ofVwhich are mapped into 1, . . . ,

n respectively by the

homomorphism above. Then B (i, j) =j (i) =ij .


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1.9. Ideals and quotient rings 17

1.9 Ideals and quotient rings

Let R be a (commutative) ring (with 1). ThenR can be regarded as amodule over itself.

Definition 1.35 By an ideal ofR, we mean anR-submodule ofR.

Clearly an ideal I of R, is a subgroup of (R, +) such that for anyx I and a R, we have ax I.

Example 1.22 R and{0} are ideals of a ringR.

Example 1.23 Subgroups of (Z, +) are ideals of the ring Z with theusual addition and multiplication. Any ideal of Z is clearly of the formmZ, m Z.

Example 1.24 Let R, R be two rings and f: R R be a homomor-phism. Thenker f is an ideal ofR.

An integral domain R={0} is a field if and only ifR and{0} arethe only ideals ofR, as can be easily proved.

Definition 1.36 If a, b are two ideals ofR, the product ab ofa andbis the set of all finite sums of the formi aibi withai a andbi b.

It is easy to check that ab is again an ideal ofR. Clearly ab= ba.

Definition 1.37 An ideala of a ringR divides an idealb ofR, ifa b.

Definition 1.38 By a proper ideal of a ringR, we mean an ideal ofRdifferent fromR and{0}.

Definition 1.39 Let S be a subset of a ring R. An ideal a of R isgenerated by S, if it is generated by S as an R-module. We say a is

finitely generated, if it is finitely generated as anR-module.

Definition 1.40 If an ideal a of a ringR is generated by a single ele-ment R, then a is called a principal ideal of R. (We denote it byR in this case or just by().)

Example 1.25 R and{0} are principal ideals ofR.


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Definition 1.41 An integral domainR all of whose ideals are principal

ideals is called a principal ideal domain.

Example 1.26 Zis a principal ideal domain (see Example1.23, p.17).

Let a be an ideal of a ring R. The additive group ofR/a (in words,R modulo a) is a ring called the the quotient ring of R by a, withmultiplication defined by (x+ a)(y+ a) = xy+ a for x, y R. (Sincea is an ideal ofR, this multiplication is well defined.) The natural mapq: R R/a is a surjective homomorphism with kernel a.

Example 1.27 Z/(m) is the quotient ring ofZ by the ideal(m). This

ring is called the ring of residue classes modulo m. (See Example1.12,page8.) It is a field if and only ifm is a prime, by Proposition1.2.

Letf: R Rbe a homomorphism of a ring R onto a ringRand leta= ker f. The homomorphismfinduces a homomorphism f: R/a R,by setting f(x+ a) = f(x) for x R. Clearlyf is an isomorphism ofrings. (Fundamental theorem of homomorphisms for rings.)

Remark 1.17 Let K be a field. Consider the ring homomorphismf: Z Kgiven byf(n) =n 1(= 1+ +1, ntimes). By the fundamen-tal theorem of homomorphisms referred to above, Z/ ker f

f(Z). We

haveker f = (p) for somep0. in Z. We callp the characteristic ofK. Observe thatp is a prime, ifp > 0. For, ifp = rs, 1 < r, s < pthenf(r)f(s) = f(rs) = 0. But neitherf(r) nor f(s) is zero, contradictingthe fact that K is an integral domain. Ifp = 0, thenK contains f(Z)which is isomorphic to Zand hence contains a subfield isomorphic to Q(see remark on 11). Thus every field contains a subfield isomorphic toeitherQ orZ/(p)(for a primep). The fieldsQ andZ/(p)(pprime) arecalled prime fields.

Definition 1.42 A proper ideal p of an integral domain R is called aprime ideal if, fora, b

R, ab

p implies that eithera orb is inp.

Example 1.28 In Z a primep generates a prime ideal and converselyevery prime ideal ofZ is generated by a primep.

Remark 1.18 A proper idealp is a prime ideal of a ringR if and onlyifR/p is an integral domain.


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1.10. Linear mappings and matrices 19

Remark 1.19 If a prime idealp of a ringR divides the product of two

ideals a andb, then p divides either a orb. For, ifp divides neitheranorb there exista a, b bwhileab ab pwhich is a contradiction.

Definition 1.43 A proper ideala ofR is maximal ifa is not containedin any other proper ideal ofR.

Remark 1.20 An ideala of a ringR is a maximal ideal, if and only ifR/a is a field.

Remark 1.21 A maximal ideal is clearly prime.

1.10 Linear mappings and matrices

Let V be a vector space of dimension n over a field K. Let be alinear mapping (i.e. a homomorphism) of V into V. Taking a fixedbase e1, . . . , en ofV over K, let (ej) =

ni=1 aijei(j = 1, . . . , n) with

aij K. To the linear mapping , we associate the ordered seta11 a1n

a1n ann

of the n2 elements a11, a12, . . . , ann, which will be referred to as thecorresponding matrix(aij). The elements apq(p = 1, 2, . . . , n) are saidto constitute the pth row of (aij) and the elementsapq(p= 1, 2, . . . , n)constitute the qth column of (aij). The matrix (aij) has thus n rowsand n columns and is called an n-rowed square matrix (or an n nmatrix) with elements in K.

Conversely, given ann-rowed square matrix (aij) with elements inK,we can find a unique linear mapping ofVinto itself for which(ej) =n

i=1 aijei. Thus, once a base of V is chosen, the linear mappings ofVinto itself stand in one-one correspondence with the set Mn(K) ofn-rowed square matrices with elements in K. In Mn(K),we can introduce

the structure of a ring as follows. IfA = (aij), B = (bij) are in Mn(K)define A + B to be the n-rowed square matrix (cij) with cij) =aij+ bijand the product AB to be the element (dij) ofMn(K) with

dij =n

p=1

aipbpj.


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The usual laws of addition and multiplication for a ring can be verified

to be true. However, this ring is not, in general, commutative. To theidentity mapping IV: V Vcorresponds the matrix I=In= (ij) andthis serves as the unit element in the ring Mn(K). Note that if, aretwo linear mappings ofV into itself, and ifA, B are the correspondingmatrices, then + corresponds to A+ B and to A B. ForA = (aij) Mn(K), we define the trace Tr (A) of A to be the sumn

i=1 aii. Clearly forA= (aij), B = (bij) in Mn(K),we have Tr (AB) =ni,j=1 aijbji = Tr(BA) and further Tr (I) = n. For A Mn(K), we

denote by det A, the determinant ofA. For A = I , det I= 1.We shall not define the determinant. It has the following properties

which are the only ones we shall use. (See e.g. Halmos [2],)

LetA = (aij), 1 i, j n. Then(a) det A is a homogeneous polynomial in the elements aij of degree

n;

(b) det A is linear considered as a function of any row of ( aij); it islinear also in the columns of (aij);

(c) ifA = (aij) and At = (bij) where bij =aji ,then det A

t = det A;

(d) if A = (aij) and aij = 0 for i > j, then det A = a11 . . . ann; inparticular det I= 1;

(e) for any two n n matrices A and B, we have det(AB) = det A det B.

Letf1, . . . , f n constitute another base ofV over K, and let

fj =ni=1

pijei, ej =ni=1

qijfi, j = 1, . . . , n, pij, qij K.

Then to the linear mapping ofV into V, taking ej to fj correspondsthe matrixP = (pij). Let be the linear mapping ofV takingfj toej .Denote by Q the corresponding matrix (qij). The composite mappings

and are clearly both equal to the identity mapping ofV intoitself. The corresponding matrices are P Q and QP respectively. HenceP Q= QP =In and Q is the inverse P

1 ofP in Mn(K).Suppose now, that instead of the base e1, . . . , enofV, we had referred

everything to another fixed base f1, . . . , f n with fj =n

i=1pijei(j =1, . . . , n). Then the linear mapping taking ej to

ni=1 aijei(j =


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1.11. Polynomial rings 21

1, . . . , n) would take fj toni=1(

nk,l=1 qilalkpkj)fi and the correspond-

ing matrix would be QAP = P1AP Mn(K). Now Tr (P1AP) =Tr (AP P1) = Tr (A). Consequently, if is an endomorphism ofV, andAthe matrix corresponding to it with respect to a base ofV,then Tr (A)is independent of the base chosen, and we set Tr () = Tr (A) and speakof the trace of the endomorphism. Similarly, since

det(P1AP) = det P1 det A det P= det A det P1 det P= det A

we may define det to be det A where A Mn(K) is a matrix corre-sponding to with respect to a base ofV . For = IV, det = 1.

Remark 1.22 A one-one linear mapping ofV into V is necessarilyonto V. For,dim (V) = dim V by the homomorphism theorem. Since(V) is a subspace of V of the same dimension asV, we have (V) =Vi.e. is onto V.

Remark 1.23 A linear mapping ofV into V is one-one if and onlyifdet = 0.

Proof: If is one-one, then is onto V by Remark 1.22 above.Clearly there exists a linear mapping ofV intoVsuch that= IV.Since det( ) = det det = det IV = 1, it follows that det = 0.Conversely, if det

= 0, let, if possible, (e1) = 0 for e1

= 0. But, by

Remark 14 on page 14, e1 can be completed to a base e1, e2, . . . , en ofV. For this base, the corresponding matrixA in Mn(K) has the form(aij) withai1= 0 fori = 1, . . . , nand therefore det = det A= 0,(sincedet Ais linear in the columns ofA). We are thus led to a contradiction.

1.11 Polynomial rings

Let R be a commutative ring (containing 1). Let M be the set of allmappings f: Z+ R such that f(n) = 0 for all but finitely many n.

We introduce on M the structure of an R-module by defining, for

f, g M, a R, f+ g,af by(f+ g)(n) =f(n) + g(n), (af)(n) =a f(n), n Z+.

We make M a ring by defining f g by (f g)(n) =ni=0 f(i)g(n i).The map e for which e(0) = 1, e(n) = 0 for n >0 is the unit ofM. Wehave a map i: R Mdefined by i(a)(0) = a, i(a)(n) = 0 for n > 0, i


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is an isomorphism ofR onto a subring ofM , so that we may identity R

with i(R). Let X Mdenote the map for which X(1) = 1, X(n) = 0if n= 1. Then (with multiplication defined as above) Xk is the mapfor which Xk(k) = 1, Xk(n) = 0 if n= k. Hence any f M can bewritten uniquely in the form

f =

akXk, ak R;

the sum is finite, i.e. ak = 0 for large k .

Definition 1.44 The ringMis denoted byR[X] and is called the poly-nomial ring in one variable over R. The elements of R[X] are calledpolynomials with coefficients inR or polynomials overR.

Definition 1.45 If f =ni=1 aiXi R[X] and f= 0, we define thedegree n of f (in symbols deg f) to be the largest integer i such thatai= 0. We callan the leading coefficient off. If thisan= 1, we sayfis a monic polynomial. Iffis of degree1, we callfa linear polynomial.Iff = 0, we setdeg f = .Remark 1.24 Iff, g R[X] we have

deg(f+ g) max(deg f, deg g).Ifdeg f= deg g, thendeg(f+ g) = max(deg f, deg g).

Remark 1.25 If R is an integral domain and f, g K[X] with f=0, g = 0, then f g = 0, i.e. R[X] is an integral domain. Further,deg(f g) = deg f+ deg g.

Remark 1.26 Let K be a field and let f, g K[X] with deg g > 0.Then there existh, j K[X] such thatf =gh +j wheredegj


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1.12. Factorial rings 23

Let R be a ring with R

Cand let R = C. For any

Cwe have a

uniqueR-linear ring homomorphism: R[X] Csuch that(X) =;in fact we have only to set (

aiX

i) =

aii, ai R. We denote the

image ofR[X] under byR[] and for f =

i aiXi R[X], we write

f() =

i aii.

Definition 1.46 A complex number is a root of f R[X], if fker , i.e. f() = 0.

Example 1.29 TakeR= C. The complex numbers(5) are rootsof the polynomialx2 + 5.

Remark 1.28 If R = K is a field and f K[X], then K is aroot off if and only if (X )|f. In fact, there is K[X] of degree0(or ), i.e. K, such thatf =q (X ) + , q K[X]. Thenf() == 0.

Definition 1.47 IfR= K is a field, K is called a repeated root off K[X] if(X )2|f.

Definition 1.48 If f =

i0 aiXi K[X], then the polynomialf =

i1 iaiXi1 is called the derivative off.

Remark 1.29 It is easily seen that(f+ g)= f + g, (f g) = f g + gfand ifK has characteristic0, thatf = 0 if and only if f K. WhenK has characteristicp >0, f = 0 if and only iff K[Xp].

The quotient field of the polynomial ring K[X] over a field K is de-noted byK(X) and called the field of rational functions in one variableoverK.

1.12 Factorial rings

LetR be an integral domain.

Definition 1.49 An elementa R which is not a unit inR is calledirreducible, if, whenevera = bc forb, c R, eitherb orc is a unit inR.

Definition 1.50 An element p R is said to be prime if it is not aunit and wheneverp dividesab, witha, b R, p divides eithera or b.


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Example 1.30 A prime numberp inZ is prime inZ (Proposition1.2).

Remark 1.30 A prime element is always irreducible but not conversely,(in general). In Z, every irreducible element is (upto sign) a primenumber, by the very definition of a prime number.

Remark 1.31 In a principal ideal domain, an irreducible element gen-erates a maximum ideal.

Definition 1.51 An integral domainR is a factorial ring (or a uniquefactorization domain) if every non-unit a R can be written in theform a = q1 qr where q1, . . . , q r are irreducible elements of R deter-mined uniquely by a upto multiplication by units of R and upto order.the decompositiona = q1 qr is called the factorization of a (into irre-ducible elements).

Remark 1.32 In a factorial ringR, every irreducible elements is prime(and hence, by Remark 1.30 above, the notions of prime and irre-ducible coincide in such a ring). For, if an irreducible elementa dividesbc, thena must occur in the factorization ofbc and hence in that of atleast one ofb, c.

Remark 1.33 Let a = p1 pr be a factorization of an element ainto irreducible elements. Clubbing together the irreducible elementswhich are associated, we can write a in the form upn11 pnss , wheren1, . . . , ns Z, ni > 0, p1, . . . , ps are irreducible elements in R whichare non-associate, andu is a unit.

Definition 1.52 Leta1, . . . , ar be elements of the factorial ringR andai = ui

sj=1p

nj ,ij with nj,i Z+ and irreducible p1, . . . , ps (some of

the nj,i may be zero). The greatest common divisor (briefly g.c.d.) ofa1, . . . , ar is defined to be

sj=1p

jj wherej = min(nj,i, . . . , nj,r), j =

1, . . . , s. Similarly, the least common multiple ofa1, . . . , ar is defined to

be

sj=1p

jj where, forj = 1, . . . , s , j = max(nj,i, . . . , nj,r). These are

uniquely determined upto units.

Proposition 1.7 (Fundamental theorem of arithmetic.) Zis a factorialring.

Proof: The only units in Z are +1 and1. It suffices to prove thatevery positive integer can be written uniquely as a product of positive


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1.12. Factorial rings 25

irreducible elements ofZ, in other words, of prime numbers (in view of

Remark 1.30 above). If two prime numbers are associated, they mustbe the same.

We shall first prove that if a factorization ofa >0 in Z into primesexists it is unique. Let, in fact, fora > 0 in Z, a= p1 pr = q1 qswhere p1, . . . , pr, q1, . . . , q s are primes. Now q1 divides p1 pr andhence divides one of the pi, say p1. Then q1 = p1. Now p2 pr =q2 qs. By repeating the argument above,q2is equal to one ofp2, . . . , pr,say q2 = p2. In finitely many steps, we can thus prove that r = s andthat q1, . . . , q s coincide with p1, . . . , pr order.

We now prove the existence of a factorization for any a >0 in Zbyinduction. For a = 2, this is trivial. Assume that a factorization into

primes exists for all positive integers less than a. Now if a is a prime,we have nothing to prove. ifa is not prime, then a is divisible by b Zwith 1< b < a. In other words,a= bc with b, c Z and 1< b, c < a.By the induction hypothesis, b and c have a factorization into primesand thereforea = bc admits a similar factorization too. The theorem isthus completely proved.

Proposition 1.8 IfKis a field, the polynomial ringK[X] in one vari-able is a factorial ring.

Proof: First, let us observe that the set of units inK[X] is precisely

K. If this were not true, let, if possible, f = anXn

+ +a0 withan= 0, n 1 be a unit inK[X]. Then, there existsg = bmXm+ +b0with bm= 0 such that f g= 1. Then 0 = deg 1 = deg(f g) =n + m >1,which is a contradiction.

We start with the existence of a factorization. Let f be a non-constant polynomial in K[X]. Iffis an irreducible element, then thereis nothing to prove. If not, fhas a non-constant divisor g not associatedwith f, i.e. f=gh with 0< deg g, deg h


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Consider the set a of polynomials of the formup +vg whereu, v

K[X].

Then a is a non-zero ideal ofK[X] and, since K[X] is a principal idealdomain, there is t K[X] with a = (t). Thus t divides p, but since

p is irreducible, it follows that either p = ct with c K, or t K.Ifp = ct, then since t divides every element of a and, in particular, git follows that p| g. This contradicts our assumption. Hence t Kand therefore there existsu1, v1 inK[X] such thatu1p + v1g= 1. Sincegh = pw where w K[X], we have h = h(u1p+v1g) = p(u1h+v1w),i.e. p | h.Remark 1.34 It can be shown, by similar reasoning, that any principalideal domain is a factorial ring.

We now give an example of a ring R which is not a factorial ring.Take R to be the subring of C consisting of numbers of the form a+b

(5) with a, b Z and(5) being a root of the polynomial x2 +5. In R, there are two distinct factorizations of 6, viz. 6 = 2 3 =(1 +

(5))(1 (5)). (It is not hard to see that the four numbers

occurring here are irreducible.) Thus R cannot be a factorial ring andin R, an irreducible element is not prime in general.

We shall see however that R belongs to a general class of rings ad-mitting unique factorization of ideals into prime ideals, which will bethe object of our study in Chapter 2.

1.13 Characters of a finite abelian group

Let G be a finite abelian group, of order h. A character of G is amapping : G C such that 0, and (ab) =(a)(b) for a, b G.

If a G is such that (a) = 0, then for any b G, (a) =(b)(ab1), so that (b)= 0. Hence is a homomorphism of G intoC. Further, since bh =e we have [(b)]h = 1 for any b. Since there areonly finitely manyhth roots of unity in C, it follows thatthere are only

finitely many characters ofG.If we define the product12of two characters1, 2by (12)(a) =

1(a)2(a), then the characters form a finite abelian groupG.Proposition 1.9 LetG be a finite abelian group and leta G, a =e.Then there exists a character ofG such that(a) = 1.Proof: Leta0= e, a1= a, a3, . . . , ah1 be the elements ofG. LetVbe the set of formal linear combinations

iai, i C. Clearly, V is a


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1.13. Characters of a finite abelian group 27

vector space of dimension h over C, and the elements ofG form a base

ofV .We shall use repeatedly the following remark.

Remark 1.35 Let W={0} be a finite dimensional vector space overC, andT: W W an endomorphism. Then there existsx = 0, x Wsuch thatT x= x, C.

In fact, every polynomial overC has a root in C, we can find Csuch that det(t I) = 0. I: W W being the identity map. thenT Icannot be one-one by Remark 1.23 on page 21. Hence there isx = 0 in W with T x x= 0. Such a is called an eigenvalue ofT.

Any element ai (i = 0, 1, . . . , h 1) gives rise to a permutation ofthe elements ofG viz, the permutation given by the mapping x aixof G onto G. There is a uniquely determined linear mapping Ai(i =0, 1, . . . , h 1) of V which maps any element x in G to aix. Further,if the linear mappings Ai, Aj ofVcorrespond in this way to ai, aj in Grespectively, then clearlyAiAj corresponds toaiaj. Moreover, sinceGisabelian, we have AiAj =AjAi. In addition, A0 is the identity mappingI ofV and Ahi =I for i= 0, 1, . . . , h 1. Let us write A for the linearmapping A1 corresponding to a1= a.

We first prove that the linear mapping A corresponding toa Ghasat least one eigenvalue = 1, i.e. there exists = 1 in C and x = 0 inV with Ax= x. Clearly A =I, so that the space W = {Ax x, x V} = {0}. Also A maps Winto itself since A(Ax x) =Ay y wherey= Ax. Hence by the remark above, there existy0= 0 inW and in Csuch thatay0 = y0. Suppose, if possible that = 1. Lety0= Ax0 x0.Then Ak(Ax0 x0) = Ax0 x0 for any k 0 in Z. Since Ah = I, wehave (I+ A + A2 + + Ah1)(A I) = 0 so that

(I+ A + A2 + + Ah1)(Ax0 x0) = 0.But, since Ak(Ax0 x0) = Ax0 x0 this gives us h(Ax0 x0) = 0,contrary to our assumption that y0= Ax0 x0= 0. Hence cannot beequal to 1 and our assertion is proved.

Let now 1= 1 be an eigenvalue of A1 = A and let V0 = V andV1 ={x V0|A1x = 1x}. Then V1={0} and further, the mappingAi of V corresponding to any ai G maps V1 into itself; in fact, ifx V1, thenA1(Aix) =Ai(A1x) =Ai(1x) =1(Aix) so thatAix V1.Hence again by our earlier remark, there exist 2 in Cand x = 0 in V1with A2x = 2x. Let V2 ={x V1 | A2x = 2x}. Again, each


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Ai maps V2 into itself and we may continue the process above. let

Vi+1 ={x Vi| Ai+1x = i+1x} for i = 0, 1, 2, . . . , h 2, where i+1is an eigenvalue of Ai+1|Vi. For any x= 0 in Vh1 we have Aix =ix, i = 0, 1, 2, . . . , h 1. Clearly 0 = 1. If we set (ai) = i fori= 0, 1, 2, . . . , h 1, we obtain a character ofG in fact, since(AiAj)x=Ai(jx) =ijx,we have (aiaj) =(ai)(aj). Further(a) =1= 1.This proves Proposition 1.9.

Proposition 1.10 (Orthogonality relations.) We have

S=

aG1(a) 2(a) =

h if 1= 2,0 otherwise,

S= G

(a)(b) =

k= order ofG ifa= b,0 otherwise.

Here(a) denotes the complex conjugate of(a).

Proof: Since |(a)| = 1, we have (a) = (a)1. If 1 = 2,we clearly have

aG 1(a) 1(a) =

aG1 = h. If 1 = 2, let

b G be such that 1(b) = 2(b). Then we have S (1 2)(b) =aG(12)(ab) = S since ab runs over all the elements of G when a

does so. Since 12(b) = 1, we have S= 0.Similarly, ifa = b, clearlyS=k. Ifa=b, let 1G be such that

1(ab1) = 1. (This exists by Proposition 1.9.) Then

S1(ab1) = G

1(ab1) =S,

so thatS= 0.It can be proved that G andG are isomorphic, so that k = h. This

is, however, unnecessary for our purposes and we do not go into thisquestion.


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Chapter 2

Algebraic Number Fields

2.1 Algebraic numbers and algebraic integers

Definition 2.1 A complex number is said to be algebraic if is aroot of a polynomialanX

n + + a0, a0, . . . , an inQ not all zero.

If is algebraic then the mapping from Q[X] toC takingn

i=0 aiXi

to

aii is a homomorphism with kernel = 0.

Definition 2.2 A complex number is called transcendental if it is notalgebraic.

Let a be the set of all polynomials in Q[X] having a fixed algebraicnumber as a root. Clearly a is an ideal ofQ[X] and, in fact, generatedover Q[X] by a non-constant polynomial, say t in view of Remark 1.27on page 22. Now t is necessarily irreducible. Otherwise, t = uv with0 deg u, deg v < deg t and since 0 = t() = u()v() at least oneofu, v is in a, contradicting the minimality of deg t. The polynomial tis clearly determined uniquely upto a constant factor. We may supposetherefore that the leading coefficient oft is 1; so normalized, it is calledthe minimal polynomial ofx.

Definition 2.3 By the degree of an algebraic number , we mean thedegree of the minimal polynomial of.

Example 2.1 The quadratic irrationality

(5) is of degree2.

Letbe an algebraic number of degreen. Consider the subring Q[]ofCconsisting of numbers of the form

mi=0 ai

i withai Q. We now

29


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30 Chapter 2. Algebraic Number Fields

assert that Q[] is a field. Let, in fact, fbe the minimal polynomial of

. Consider the mappingq: Q[X] C takingmi=0 biXi tomi=0 bii.This is a homomorphism onto Q[] with kernel fQ[X] = (f). By thehomomorphism theorem Q[X]/(f) is isomorphic to Q[]. Let nowgQ[X]/(f) be such that g= 0 i.e. f g . The ideal generated by f andg in Q[X] is a principal ideal b generated by, say h. Since h divides f,we have h = cf, cQ unless hQ. The former case is impossible,sinceh | g and f g. Thusb = Q[X] and consequently there exist k, linQ[X], such thatkf+ lg = 1, so that gl= 1. This proves that Q[X]/(f),and Q[] is a field.

We denote the field Q[] byQ().

Definition 2.4 A subfieldK ofC is called an algebraic number field ifits dimension as a vector space over Q is finite. The dimension of KoverQ is called the degree ofK, and is denoted by [K: Q].

Example 2.2 Q andQ(

(5)) are algebraic number fields of degree1and2 respectively.

Remark 2.1 Any element of an algebraic number field K is alge-braic. (For, if [K : Q] =n then1, , 2, . . . , n are necessarily linearlydependent overQ.)

Remark 2.2 If is an algebraic number of degreen, then Q[] is analgebraic number field of degreen(Q() =Q[]).

Remark 2.3 Any monic irreducible polynomialf inQ[X] is the mini-mal polynomial of any of its roots. For, if is a root off, then is analgebraic number and its minimal polynomial certainly dividesf sincef() = 0. Now, cannot be in Q so that the irreducibility of f givesf =c forc = 0 in Q. Since bothf and are monic, it follows that= f.

Remark 2.4 Let be an algebraic number of degree n(

1) and f =ni=0 aiXi(an = 1) be its minimal polynomial in Q[X]. We now claim

that all the roots offare distinct, i.e. fhas no repeated roots. In fact,suppose thatf= (X )2g. Then, the derivativef off is2(X)g +(X )2g, so thatf() = 0. Sincef is the minimal polynomial of,this implies that f|f. Since deg f < deg f, this is impossible unlessf= 0, so thatf K, which is absurd.


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2.1. Algebraic numbers and algebraic integers 31

Remark 2.5 Let 1 and 2 be two algebraic numbers with the same

minimal polynomial in Q[X]. Then, for any g in Q[X], we see thatg(1) = 0 if and only if g(2) = 0. It is easy now to deduce that themapping: Q[1] Q[2] defined by

m

i=0

aii1

=

mi=0

aii2 (a0, a1, . . . , am Q)

is an isomorphism ofQ(1)onto Q(2). The mapping is the identityon Q and takes 1 to 2. Conversely, let 1 be any algebraic numberwith minimal polynomialf and an isomorphism ofQ(1)into Csuchthat(a) =a, for anya

Q. Then for anyg

Q[X], g(1) = 0 if and

only ifg((1)) = 0. The set of all polynomials in Q[X] having(1)as a root is precisely the idealfQ[X] and therefore(1)is an algebraicnumber withfas its minimal polynomial.

Definition 2.5 Two algebraic numbers 1 and 2 as in Remark 2.5above are said to be conjugates of each other (over Q).

Remark 2.6 Let K be an algebraic number field of degree n. Thenthere exists K such thatK=Q(). (Observe that such a numberis an algebraic number of degreen, by Remarks2.1 and2.2 above).

For proving this, we remark first that the intersection of any familyof subfields of a fixed field is again a field, and denote by Q(1, . . . , p)the intersection of all subfields ofC containingQ and the complex num-bers 1, 2, . . . , p. (This field is also referred to as the subfield of Cgenerated by 1, 2, . . . , p over Q). Since [K : Q] = n, there ex-ist 1, 2, . . . , q(q n) in K such that K = Q(1, . . . , q). By Re-marks 2.1 and 2.4 above, 1, . . . , q. are all separably algebraic overQ i.e. the minimal polynomial of any i has no repeated roots. Ifq = 1, there is nothing to prove. Let first q = 2. We shall prove thatK = Q(1, 2) contains 1 such that K = Q(1). For simplicity ofnotation, let us denote 1, 2 by , respectively and let f and be

their respective minimal polynomials. By the fundamental theorem ofalgebra, we have f= (X1) (Xr) and = (X1) (Xs)where we may assume, without loss of generality, that = 1, = 1.Further, by Remark 2.4 above, 1, 2, . . . , s are pairwise distinct. SinceQis infinite, we can find = 0 inQsuch thati+j=i1 +j1unless

j = j1 and i = i1 (We have just to choose in Q different from 0 and


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(i

i1)/(j

j1) forj

=j1). Set1 = + = 1 + 1and denote

Q(1) byL; obviously, LK. The polynomial(X) =f(1 X)L[X] has 1 as a root, since (1) = f(1 1) = f(1) = 0. Fur-ther, for i= 1, (i)= 0 since, otherwise, f(1 i) = 0 would giveus 1 i = j for some j i.e. 1 + 1 = j +i for i= 1 con-trary to our choice of . Thus and have exactly one root 1 incommon. Let Xbe the greatest common divisor of and in L[X].Then every root of in C is a common root of and . Thus isnecessarily of degree 1 and hence of the form (X1). In other words,,1 L. i.e. 1 =1 1 L(= Q(1)). Therefore 1 and 1 L,i.e. K L. This implies thatK= Q(1). Let now q3. Assume byinduction, that every algebraic number field of the form Q(1, 2 . . . , r)

withr q1 contains a number such thatQ(1, 2, . . . , r) =Q().Then K1 = Q(1, 2, . . . , q1) = Q(1) for some 1 K1. FurtherK=K1(q) =Q(1, q) =Q() for some K(because of the specialcase q= 2 established above).

Remark 2.7 Let K be an algebraic number field of degree n. Thenthere exist preciselyn distinct isomorphisms 1, 2, . . . n of K into Cwhich are the identity on Q. By Remark 2.6 above K = Q() fora number K whose minimal polynomial f in Q[X] is of degreen. Let 1(= ), 2, . . . , n be all the distinct root of f. Then, by Re-mark 2.5, above, there exists, for eachi(i= 1, 2, . . . , n)an isomorphism

i of Q(1) onto Q(i) C defined by i(mj=0 ajj1) =mj=0 ajjifor a0, a1, . . . am Q. By definition i(a) = a for all a Q, 1 isthe identity isomorphism of K. Since i= j for i= j, the isomor-phism 1, 2, . . . , n are all distinct. On the other hand, let be anyisomorphism ofK = Q(1) into C which is the identity on Q. By Re-mark 2.5 above, (1) = i for some i(1 i n) and therefore fora0, a1, . . . , am Q, (

mj=0 aj

j1) =

mj=0 aj

ji . Thus is necessarily

one of then isomorphisms1, 2, . . . , n.

LetKbe an algebraic number field of degree n, and1, 2, n the ndistinct isomorphism ofK into C. We denote the image i(K) ofKby

K(i) and, for K, i() by(i). Let 1 be the identity isomorphismof K; we have then K(1) = K and (1) = for any K. Sinceeach i is an isomorphism which is the identity on Q, it follows thatK(1), . . . , K (n) are again algebraic number fields of degree n. They arereferred to as the conjugates of K. If K(i) R, we call it a realconjugate of K and if K(i) R, then we call it a complex conjugate


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ofK. We now claim that the complex conjugates ofK occur in pairs,

i.e. the distinct isomorphisms iwith i(K) R occur in pairs , with = , where () =() is the complex conjugate of(). For,by Remark 2.6 above, K = Q() for some K. If K(i) = i(K) isa complex conjugate of K, then necessarily (i) = i() is a complexnumber which is not real Now (i) is a root of the minimal polynomialn

i=0 aiXi of and, sincea0, a1, . . . , an Q, it follows that the complex

conjugate (i) of (i) is also a root of

aiXi. Hence by Remark 2.7

above, Q((i)) too occurs among the conjugates K(i), . . . , K (n). Letr1be the number of real conjugates ofKand let s be the number of complexconjugate ofK. By the foregoing,s = 2r2 forr2 Z+. Further we haver1+ 2r2= n.

Remark 2.8 LetKbe an algebraic number field and1, 2, . . . , n bea base of K over Q. With the notation introduced above, let denote

then-rowed complex square matrix((i)j ) with(

(i)1 ,

(i)2 , . . . ,

(i)n ) as its

i-th row. Then has an inverse inMn(C).

In fact,K=Q() for some algebraic number inKof degreen, by Re-mark 2.6, above. Further, if1, 2, . . . , n are the distinct isomorphismsofK into C, then (1) = 1() = ,

(2) = 2(), . . . , (n) = n() are

all conjugates of (by Remark 2.5, above). Moreover, they are distinct,

(by Remark 2.4). Now 1 = 1, 2 = , . . . , n = n

1

form a base ofK over Q. Let A = (

(i)j ) be the matrix for the base 1, 2, . . . , n,

built as was from the base 1, . . . , n. Then, det A is the well-knownVandermonde determinant and is equal to1i


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Remark 2.10 An element ofZ is an algebraic integer.

Remark 2.11 If Q is an algebraic integer, then Z.

Remark 2.12 For any algebraic number , there exists m = 0 inZ such that m is an algebraic integer. (For, if n + an1n1 + + a0 = 0, with a0, a1, . . . , an1 Q then take m Z such thatma0, ma1, . . . , m an1 Z.)

Definition 2.7 A polynomialf=anXn + + a0 Z[X] is said to be

primitive if thegcd(a0, a1, . . . , an) ofa0, a1, . . . , an is1.

In particular, a monic polynomial in Z[X] is primitive. It is clearthat any polynomialf Z[X] can be written in the form cg wherec Zand g is primitive.

Remark 2.13 Any polynomial f Q[X] can be written in the form(a/b)g whereg is primitive anda, bZ are such that (a, b) = 1. (a, b)stands for the g.c.d. ofa andb.

Lemma 2.1 (Gauss.) The product of two primitive polynomials inZ[X] is primitive.

Proof: Let= an

Xn+

+a0

, = bm

Xm+

+b0

be in Z[X] with(a0, a1, . . . , an) = 1 = (b0, b1, . . . , bm). Suppose thatpis a prime dividingall the coefficients off = . Consider the natural map : Z Z/(p);this induces a homomorphism (which we denote again by) : Z[X]Z/(p)[X]. We have (f) = 0, while, since the gcd of the coefficientsof , is 1, we have ()= 0, ()= 0, so that since Z/(p) is afield Z/(p)[X] is an integral domain, and (f) = () ()= 0 , acontradiction. Hence f is primitive.

Proposition 2.1 The following statements are equivalent.

(i) is an algebraic integer.

(ii) The minimal polynomial of is a (monic) polynomial inZ[X].

(iii) Z[] is a finitely generatedZ-module.

(iv) There exists a finitely generatedZ-submoduleM= {0} of C suchthatM M.


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Proof: (i)

(ii). Let n +an

1

n1 +

+a0 = 0 with ai

Z and

= Xn +an1Xn1 + +a0. Let fbe the minimal polynomial of in Q[X]. By definition, = f , where Q[X]. Further, by theRemark 2.13 on page 34, let f= (a/b)f1, = (c/d)1with primitivef1and1anda, b, c, d Z such that (a, b) = (c, d) = 1. Nowbd= acf11.By Gauss Lemma, f11 is primitive. Let us compare the g.c.d. of thecoefficients on both sides. Since is monic and hence primitive, we haveac =bd. Thus =f11; comparison of leading coefficients impliesthat the leading coefficient off1 is1; since f1() = 0, it follows fromthe definition of the minimal polynomial thatf= f1, so thatf Z[X].

(ii) (iii): Let= Xn+an1Xn1 + +a0 Z[X] be a polynomialwith () = 0. Then clearly Z[] is generated by 1, , . . . , n1 over Z.

(iii) (iv): It is obvious that Z[] Z[]. Take M= Z[].(iv) (i): Let M = Zv1+ +Zvn C be a finitely generated

Z-module such that M M. Then vi =n

j=1 aijvj (i= 1, 2, . . . , n)withaij Z. LetA= (aij) andB = InA= (bij). Then

nj=1 bijvj =

0 fori = 1, 2, . . . , n. LetVbe a vector space of dimension n over C ande1, e2, . . . , en a base ofV. The linear mapping ofV into itself takingej to

ni=1 bijei (j = 1, 2, . . . , n) takes the non-zero element

nj=1 vjejto

0. By the remark 1.23 on page 21, det = det B = 0. Expandingdet B = det(In A), we see that n +an1n1 + +a0 = 0 fora0, a1, . . . , an1 Z.

Proposition 2.1 is now completely proved.LetKbe an algebraic number field. Denote by O the set of algebraic

integers in K. If , O, then Z[], Z[] are finitely generated Z-modules by Proposition 2.1. Hence the ring M = Z[, ], is a finitelygenerated Z-module. Since if is one of , we clearly haveM M, it follows from Proposition 2.1 that , are in O sothat O is a ring. By Remark 2.12 on page 34, Kis the quotient field ofO.

If , are algebraic, there exists m Z such that m, m arealgebraic integers, by Remark 2.12. Since m() and m2 arealgebraic integers by the considerations above, it follows that

and are algebraic. Further, if= 0 is algebraic, so is 1. Thus,algebraic numbers form a subfield ofC.

Let Kbe an algebraic number field of degree n and 1, . . . , n bea base of K over Q. For any K,the mapping x x is a linearmapping ofK into itself considered as a vector space overQ. We definethe trace Tr () = Tr K(), and norm N() = NK() of the element


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to respectively, the trace and determinant of this linear mapping. If

wj =ni=1 aijwi, j = 1, 2, . . . , n, and A = (aij), we have Tr K() =Tr (A), NK() = det A; hence Tr K(), NK() Q.

For K, (wj)(k) = (k)w(k)j =n

i=1 aijw(k)i for j = 1, 2, . . . , n.

Denote the n-rowed square matrix (w(k)j ) by , as on page 33, and the

n-rowed square matrix ((i)ij) byA0whereij = 1 fori = j and ij = 0for i =j . Then we have A0 = A. Since, by Remark 2.8 on page 33, has an inverse 1, we have A0 = A1. Hence we have

NK() = det A= det(A1) = det A0= (1) (n). (2.1)

Tr K() = Tr (A) = Tr (A1) = Tr(A0) =(1) + + (n).

Further, if A corresponds to K, B to K, then to + corresponds A+ B and to , AB Hence the mapping A is ahomomorphism ofK into Mn(Q). It is called a regular representationofK (viz. that corresponding to the base w1, . . . , wn ofK.) We verifyimmediately that if, K, we have

Tr K( + ) = Tr K() + Tr K() and NK() =NK()NK().

Let be an algebraic integer in K. By Proposition 2.1, we can sup-pose that the minimal polynomial of is Xm +am1Xm1 + +a0where a0, a1, . . . , am1 Z. Now is of degree m and Q() has1, , . . . , m1 as a base overQ. (Clearlym

n.) LetA

Mn(Q) corre-

spond toin the regular representation ofQ(), with respect to the base1, , . . . , m1 ofQ(). Let1, 2, . . . , l be a base ofKconsidered as avector space over Q(). Then1, 1 , . . . , 1

m1, 2, 2 , . . . , 2m1,. . . , l, . . . , l

m1 constitute a base ofKoverQ. (Incidentallyl m= nand so m|n.) Let A1 correspond to in the regular representation ofKwith respect to this Q-base. Then

A1=

A 0 00 A 00 0 A

Mn(Q) and Tr (A1) =l Tr (A).Since is an algebraic integer, all the elements of A are in Z so that

Tr (A) and Tr(A1) = lTr (A) = lam1 are integers. Thus, for analgebraic integer K, Tr K() is an integer. Similarly, NK() =det A1= (det A)

l Z.The mapping TrK() is clearly a Q-linear mapping ofK into

Q. We define a bilinear formB(x, y) on theQ-vector spaceKby settingB(x, y) = Tr K(xy) for x, y K.


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Proposition 2.2 The bilinear formB(x, y) = Tr K(xy) forx, y

K is

non-degenerate.

Proof: Letx = 0 be inK. Then B x(y) = Tr K(xy) is not identicallyzero in y, since, for y = x1, Bx(x1) = Tr K(1) = n. Similarly fory= 0 in K, By(x) is not identically zero in x.

If we apply Proposition 1.6 to this bilinear form onV =K, we obtainthe

Corollary 2.1 To any Q-base w1, . . . , wn of K, there corresponds abasew1, . . . , w

n such thatTr K(wi, w

j) =ij, 1 i, j n.

If R is a subset ofKand ifa K then, by definition,aR= {ar | r R}.

The following theorem gives more information concerning the struc-ture of the ring of algebraic integers in a given algebraic number field.

Theorem 2.1 LetKbe an algebraic number field of degreen andOthering of algebraic integers in K. Then there exists a Q-base 1, . . . , nofK such thati O andO =Z1+ + Zn.

(Elements1, . . . , n with this property are said to form an integral

base ofO.)Proof: Letw1, . . . , wnbe aQ-base ofK. By Remark 2.12 on page 34there existsm = 0 in Z such that mw1, . . . , m wn are in O. We can thusassume without loss of generality that w1, . . . , wn are already in O. Letw1, . . . , w

n be a base ofKfor which

Tr K(wi, Wj) =ij (1 i, j n). (2.2)

We know that for any z O, z =ni=1 aiwi with a1, . . . , an Q.Sincezwi O for 1 i n, we have, because of (2.2), ai = Tr K(zwi) Z. Thus we obtain

O

Zw1+

+ Zwn.

By Proposition 1.4 (Chapter 1) there exist 1, . . . , m O, m n,such that

O =Z1+ + Zm.We claim that, necessarily, m = n. In fact, if m < n, then, the Q-subspace ofKgenerated by 1, . . . , m, which is clearly K itself, would


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have dimension

m < n over Q, contrary to our assumption that K is

of degreen. Further, we see that1, . . . , n are Q-independent, so that,necessarily, the sum above is direct. This proves Theorem 2.1.

Remark 2.14 Any set of elements1, . . . , n as above forms aQ-baseofK. We have only to repeat the argument in the last few lines above.

Remark 2.15 Leta be any non-zero ideal inO. Thena Z = {0}. Infact if = 0is an algebraic integer ina and ifr+ar1r1+ +a0= 0where ai Z, a0= 0, then a0 =(a1+ +ar1) Z. It followsthat for any K, there isa Z, a = 0 such thata a.

If1, . . . , n are as in Theorem 2.1, then a O =Z1+ + Zn.By Proposition 1.4 (Chapter 1), there exist1, . . . , m a, m n, suchthat

a= Z1+ + Zm.As in Remark 2.14,we must have m= n.Theiare said to constitute

an integral base ofa.Further, we may choose the i so that i =

jipijj, pij Z.

Remark 2.16 As in Remark 2.14, any elements 1, . . . , n such thata = Z1+ + Zn form a Q-base ofK. In particular, any non-zeroideal ofa containsn elements which are linearly independent overQ. It

follows that ifi = jipijj ,thenpii= 0.Remark 2.17 If a={0} is an ideal ofO, then by Remark2.15, thereexists0= aZ such that aOa O. Now ifO = Z1+ +Zn,then aO = Za1+ +Zan so thatO/aO is of order an. ThereforeO/a is also finite.

Remark 2.18 If p is a prime ideal in O then p contains exactly oneprime numberp > 0 of Z.

In fact, let a = p1

pk in p

Z with primes p1, . . . , pk

Z. Since p

is prime, at least one pi p. Ifp, qare distinct primes in p, we can findintegersx, y with xp + yq= 1; then 1 p and p= O, a contradiction.

Definition 2.8 For any ideala = {0} ofO, the number of elements inthe residue class ringO/a is called the norm ofa and denoted byN(a);ifa= {0} we setN(a) = 0.


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Clearly N(O) = 1. For a proper ideal a ofO, N(a)> 1.

Proposition 2.3 Let a={0} be an ideal of O = Z1 + + Zn.Then there exist i =

nj=1pijj , pii > 0, pij Z such that a =

Z1+ + Zn. FurtherN(a) =p11p22 pnn.

Proof: By Remark 2.15 above, a has an integral base 1, . . . , n ofthe required form.

We claim that the p11p22 pnn numbers

= (x1, . . . , xn) =n

i=1xii, 0 xi < pii, x Z

form a complete system of residue ofO modulo a, In fact, if

=ni=1

cii O, ci Z,

and we set Oi = O(Zi+1+ +Zn), then there existx1, 0 x1< p11and m1 Zwith

1= x11 m11 O1;further, x1, m1 are determined by the relation

c1= m1p11+ x1.

We can, in the same way, find m2 Z, 0 x2< p22 with

2= 1 x22 m22 O2;

continuing in this way, we find that

=ni=1

(mii+ xii), mi Z, 0 xi < pii,

so that the generate O modulo a.

Now, ifni=1 xii =ni=1 mii, where mi Z, 0 x < pii, wenote that

ni=1 mii =

ni=1 cii, where c1 = mip11. Since thei are

linearly independent, we have x1 = m1p11 and since 0 x1 < p11, weconclude that m1 = x1 = 0. This implies that c2 = m2p22, which inturn implies that x2 = m2 = 0, and so on; hence xi = 0. This provesthat the s are distinct modulo a and with it, the proposition.


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2.2 Unique Factorization Theorem

LetKbe an algebraic number field of degree n and Othe ring algebraicintegers inK . Letpbe a prime ideal in O. Then O/pis finite and indeeda commutative integral domain with 1, by Remark 1.18 on page 18. ByExample 1.13, page 8, O/p is a field. Thus:

(D1) Every prime ideal ofO is maximal.

We say that an element C is integral over O if there exists amonic polynomial f with coefficients in O such that f() = 0. As inProposition 2.1, one can prove that C is integral over O if andonly if there is a non-zero finitely generated O-module M C withM

M (alternatively, if and only if O[] is finitely generated over

O.) By Theorem 2.1, such a module M is finitely generated over Z, andconsequently an element Cintegral overO is an algebraic integer.

It follows at once that, ifK is an algebraic number field and O thering of algebraic integers in K, then any Kwhich is integral overObelongs to O. If we define, for any integral domain R the integral closureofR in its quotient field as the set of all elements of the quotient fieldofR which are roots of monic polynomials with coefficients in R, we cantherefore assert the following:

(D2) The integral closure ofO inK isO itself.

To each ideal a={0}, associate its norm N(a) > 0 in Z. ThemappingN: a

N(a) is, in general, not one-one. However, ifa

band

a = b, then N(a)> N(b). [For, letf: (O/a, +) (O/b, +) be the mapdefined by f(x+a) =x+b. Then fis well-defined, onto (O/b, +) butis not one-onesince there exists y b, y a. We deduce at once thefollowing statements.

(N1) If a1 a2 an an+1 is an increasing sequenceof ideals in O then am= am+1 for m m0 Z+.

(N2) Any non-empty set S of ideals in O contains a maximal ele-ment, i.e. an ideal a such that a b for any b S, b =a. (For, any setof positive integers contains a least element.)

(N3) Any ideal a in O with a = O is contained in a maximal idealofO.

(Take for S in (N2) the set of all proper ideals b with a b O.)

Remark 2.19 A ringR is noetherian if the statement (N1) is true ofideals inR. One can show thatR is noetherian if and only if statement(N2) is true of ideals inR. This is further equivalent to the statementthat any ideal of R is finitely generated (as an R-module). Statement


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2.2. Unique Factorization Theorem 41

(N3)is true in arbitrary (even non-commutative) rings with unity, and,

in this generality, is due to Krull.

We have thus proved that(D3) O is noetherian.Any commutative integral domain R satisfying the conditions (D1),

(D2), (D3) is known as a Dedekind domain and these conditions arethemselves known as axioms of classical ideal-theory. The ring of al-gebraic integers in an algebraic number field is therefore a Dedekinddomain.

Definition 2.9 By a fractional ideal inK, we mean anO submodulea

ofK for which there existsm = 0 inZ such thatma O.Any ideal in O is trivially a fractional ideal. By analogy withZ we

may call an ideal in Oan integral ideal in K. Any fractional ideal acanbe written as a1bfor a = 0 in Z and an integral idealb. Ifcis an integralideal, then for any b = 0 in Z, b1c is clearly a fractional ideal in K. Ifc, d, are fractional ideals inK, then for a suitable c Z, c = 0, cc, careboth integral ideals and the sum c+ d = c1(cc+cd) and the productcd= c2(cc cd) are again fractional ideals in K.

We now prove the important

Theorem 2.2 (Dedekind.) Any proper ideal of the ringO

of algebraicintegers in an algebraic number fieldKcan be written as the product ofprime ideals inO determined uniquely upto order.

For the proof of the theorem, we need two lemmas.

Lemma 2.2 Any proper ideala O contains a product of prime idealsinO.

Proof: Let S be the set of proper integral ideals not containing aproduct of prime ideals. If s=, then by statement (N2), S containsa maximal element, say a. Clearly a cannot be prime. Thus there

exist x1, x2 O, x1x2 a but x1, x2 / a. Let ai (i = 1, 2) be theideal generated by a and xi. Then a1 and a2 contain a properly. Bythe maximality of a in s, a1 / S, a2 / S. Hence a1 p1p2 pr anda2 q1 qs where p1, . . . pr, q1, . . . , qs are prime ideals in O. Sincea1a2 a, we have p1 prq1 q3 a giving us a contradiction. HenceS = .


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Lemma 2.3 Any prime ideal p

O is invertible, i.e. there exists a

fractional idealp1 such thatpp1 = O.

Proof: Let p1 be the set ofx K such that xp O. Clearly p1 isan O-module containing O. Sincepcontainsn = 0 inZ(see Remark 2.15on page 38), we have np1 p1p O. Hence p1 is a fractional ideal.Now p pp1 O. Since p is maximal, either pp1 = O in which caseour lemma will be proved or p= pp1.

Ifp= pp1, then every x p1 satisfiesxp p. Since pis a finitelygenerated Z-module (vide Remark 2.15 on page 38), we see that x Oin view of Proposition 2.1. Hence p1 O i.e. p1 = O. This, as wenow show, is impossible. Takex

p such that xO

= O. ThenxO is a

proper integral ideal and by Lemma 2.2, p1 pr xO for prime idealsp1, . . . ,pr. Assumer so chosen that xO does not contain a product ofr 1 prime ideals in O. Now p xO p1 pr. By Remark 1.19 onpage 19 p divides one ofp1, . . . , pr, say p1. But by Property (D1), p =p1. Now p2 pr, is not contained in xO, by minimality of r. Hencethere exists b p2 pr, b / xO, i.e. x1b / O. But since bx1pp2 pr(x1O)p= x1O p1 pr x1O xO = O, we havebx1 p1.But x1b O, i.e. p1 = O. Thus pp1 = O.

We may now give the following

Proof of Theorem 2.2. As in Proposition 1.7, the proof is splitinto two parts.

(i) Existence of a factorization. Let S be the set of proper ideals ofOwhich cannot be factorized into prime ideals. We have to show thatS =. Suppose then that S=. Then by (N2) S contains a maximalelement say a O. Now obviously, a cannot be prime. Hence by (N3)a p, a =p where pis prime. By Lemma 2.3, there exists an ideal p1such that pp1 = O. Thus ap1 is a proper ideal in O and contains aproperly sincep1 contains Oproperly. But ifap1 =p1 pr for primeideals p1, . . . , pr then a = pp1 pr contradicting the assumption thata S. Hence ap1 S but this contradicts the maximality of a in S.Thus S =, i.e. every proper ideal of O can be factorized into primeideals.

(ii)Uniqueness of factorization. If possible, let a proper idealain O havetwo factorizationsa = p1 pr = q1 qrwherep1, . . . , pr, q1, . . . , qs, areprime ideals. This means thatq1dividesp1 prand by Remark 1.19, q1divides one of the ideals p1, . . . , pr sayp1. But since p1 is maximal, q1=p1. Now by Lemma 2.3, q

11 a = q

11 q1 qs = q2 qs and q11 a =

p11 p1 pr = p2 pr. Thus p2 pr = q2 qs. By repeating the


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2.3. The class group ofK 43

argument above, q2 is equal to one of the prime idealsp2, . . . , pr, sayp2.

In finitely many steps, we can thus prove that r = s and that p1, . . . , prcoincide with q1, . . . , qr upto order.

Corollary 2.2 Any fractional ideal a can be written uniquely in theform

a= q1 qsp1 pr

1p

stands for p1

where theqi, pj are prime; and no qi is apj.

This follows immediately if we choosec = 0, c Zwith b = ca O,and write (c) = p

1 pr

, b = q1

qs and cancel equal q

i and p

j in

pairs.

Corollary 2.3 Given any fractional ideala = {0} inK, there exists afractional ideala1 such thataa1 = O.

For proving this, it suffice to show that every integral ideal is in-vertible. But this is an immediate consequence of Theorem 2.2 andLemma 2.3.

Remark 2.20 Let a = pa11 parr , b = pb11 pbrr be integral idealsp1, . . . ,pr being prime ideals a1, . . . , ar, b1, . . . , br Z

+

. (We definep0i = O, i= 1, 2, . . . , r.) The greatest common divisor (a, b) ofa andbis defined to be pc11 pcrr whereci = min(ai, bi) i = 1, 2, . . . , r. Clearlyci is the largest integerc such thatp

ci divides botha andb. But now if

c, d are integral ideals thenc dividesd if and only ifd= cc1 forc1 O.(For, ifc d, thenc1 = dc1 O and conversely ifd= cc1 withc1 Othen c d.) Thus the greatest common divisor of a and b is actuallythe smallest ideal dividinga andb which is nothing buta+b. For anyset of s ideals a1, . . . , as the greatest common divisor of a1, . . . , as maybe similarly defined.

2.3 The class group ofK

Let K be an algebraic number field of degree n. By Corollary 2.3 toTheorem 2.2, the non-zero fractional ideals in Kform a multiplicativegroup which we denote by . The ring O of algebraic integers is theidentity element of .


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A fractional ideal in Kis said to be principal if it is of the form O

with K. Clearly the principal (fractional) ideals a={0} form asubgroup of . The quotient group h = / is called the group ofideal classes in Kor briefly the class group ofK.

Definition 2.10 Two idealsa, b = {0} inKare in the same ideal class,or are equivalent if and only ifa= ()b for some K.

We shall prove in this section that h is a finite group. The orderdenoted by h ( =h(K)) is called the class number ofK.

Ifh = 1, then O is a principal ideal domain.For proving that h is finite, we need a few lemmas.

Let a be an integral ideal in Kand let

a= Z1 + + Zn= Z1 + + Zn with 1, . . . , n, 1, . . . , n a.Clearly i =

nj=1 qijj , qij Z and i =

nj=1 rijj, rij Z

for i = 1, 2, . . . , n. Denoting the n-rowed square matrices (qij), (rij)by Q, R respectively, we see that ifQR = (sij), then i =

nj=1 sijj .

By Remark 2.16 after Theorem 2.1, 1, . . . , n are linearly independentover Q. Thus sij = ij (the Kronecker delta) for 1 i, j n i.e.QR = In. Taking determinants, it follows that det Q det R = 1 andsince det Q det R Z, we have det Q= det R= 1.

We now use the foregoing remarks to identity the norm N(()) of

the principal ideal () generated by = 0 in O with the absolute value|NK()| of the norm NK().

Lemma 2.4 For = 0 inO, N(()) = |NK()|.

Proof: IfO= Z1+ = Zn, then by Proposition 2.3, there existi =

nj=1pjij, i = 1, 2, . . . , n;pji Z, p11, . . . , pnn > 0 such that

() =Z1+ + Zn and N(()) =p11p22 pnn. LetQ stand for then-rowed square matrix (qij) whereqij = 0 for 1 i < j nandqij =pijotherwise. Since () = Z1+ +Zn as well, we have in view ofthe remark immediately preceding this lemma, i =

nj=1 rjij, i =

1, . . . , n, and ifR denotes then-rowed square matrix (rij), then det R=1. Let Sbe the matrix QR. Taking the regular representation withrespect to the base 1, . . . , n, we have NK() = det S. But

det S= det Q det R= det Q= p11 pnn= N(()).Thus N(()) = |NK()|.


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2.3. The class group ofK 45

Corollary 2.4 Fort

Z, N(tO) =

|N(t)

|=

|tn

|.

Lemma 2.5 For any two integral ideals a, b there exists O, suchthatgcd(ab, ()) isa.

Proof: Let a = pa11 parr , b = pb11 pbrr , ai, bi Z+ and p1, . . . , prare all the prime ideals dividing a and b. We can find an element ipa1+11 pai1+1i1 paii pai+1+1i+1 par+1r but i / pa1+11 pai+1i pan+1n(since pi= O). Take = 1+ +n. Clearly pai+1i divides j(for

j=i) and paii is the highest power ofpi dividing i. Hencepaii and nohigher power ofpi divides . Consequently (ab, ()) =

ri=1p

aii =a.

Remark 2.21 Given any integral ideal a there exists t= 0 in Z suchthatb

TIFR pamphlet on Algebraic Number Theory

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