teoria_divizibilitatii

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Ioan TOFAN Aurelian Claudiu VOLF

Ring Arithmetic, Field Extensions

and Applications


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Table of Contents

Foreword..............................................................................................7 I. Arithmetic in integral domains ....................................................10I.1 Divisibility...................................................................................10

I.2 Euclidian domains.......................................................................22

I.3 Euclidian rings of quadratic integers...........................................26

I.4 Principal ideal domains ...............................................................31

I.5 Unique factorization domains .....................................................36

I.6 Polynomial ring arithmetic..........................................................43

Exercises...........................................................................................50

II. Modules.........................................................................................58II.1 Modules, submodules, homomorphisms....................................58

Exercises...........................................................................................72

II.2 Factor modules and the isomorphism theorems.........................74

Exercises...........................................................................................80

II.3 Direct sums and products. Exact sequences ..............................81

Exercises.........................................................................................102

II.4 Free modules............................................................................104

Exercises.........................................................................................116

III. Finitely generated modules over principal ideal domains....118III.1 The submodules of a free module...........................................118

Exercises.........................................................................................128

III.2 Finitely generated modules over a principal ideal domain.....130

Exercises.........................................................................................138

III.3 Indecomposable finitely generated modules ..........................140

Exercises.........................................................................................149


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III.4 The endomorphisms of a finite dimensional vector space ..... 150

Exercises ........................................................................................ 167

IV. Field extensions ........................................................................ 169

IV.1. Algebraic extensions............................................................. 169Exercises ........................................................................................ 190

IV.2 Roots of polynomials. Algebraically closed fields ................ 193

Exercises ........................................................................................ 207

IV.3 Finite fields ............................................................................ 209

Exercises ........................................................................................ 221IV.4 Transcendental extensions ..................................................... 223

Exercises ........................................................................................ 230

V. Galois Theory............................................................................. 232

V.1 Automorphisms ....................................................................... 233V.2 Normal extensions................................................................... 238

Exercises ........................................................................................ 242

V.3 Separability.............................................................................. 244

Exercises ........................................................................................ 259

V.4 The Fundamental Theorem of Galois Theory......................... 262

Exercises ........................................................................................ 271

VI. Applications of Galois Theory ................................................ 273VI.1 Ruler and compass constructions........................................... 273

VI.2 Trace and norm ...................................................................... 289

Exercises ........................................................................................ 296

VI.3 Cyclic extensions and Kummer extensions ........................... 297

Exercises ........................................................................................ 312

VI.4 Solvability by radicals............................................................ 313

VI.5 Discriminants, resultants........................................................ 320

Exercises ........................................................................................ 326

Appendices ...................................................................................... 331

1. Prime ideals and maximal ideals................................................ 3312. Algebras. Polynomial and monoid algebras............................... 333

3. Symmetric polynomials ............................................................. 348

4. Rings and modules of fractions.................................................. 354

5. Categories, functors.................................................................... 362

6. Solvable groups.......................................................................... 373

Index ................................................................................................ 379

Bibliography.................................................................................... 387


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7

Foreword

The book is aimed at undergraduate students in Mathematics, Com-

puter Science or technical universities having a background of stan-

dard courses of Abstract Algebra and Linear Algebra as well as at

general Mathematics readers with an interest in Algebra.

The topics covered by the book are mandatory algebraic back-

ground of any mathematics graduate: arithmetic in integral domains,

module theory basics, the structure of the finitely generated modules

over a principal ideal domain with applications in abelian groups and

Jordan forms of matrices, field extensions and Galois theory with

applications.Throughout the book the reader is motivated by concrete applica-

tions and exercises.

The book is reasonably self contained, in the sense that the reader

is assumed to be familiar with general notions on algebraic structures

(monoids, groups, rings, fields), factor rings and isomorphism theo-

rems, vector spaces and bases, matrices, polynomials, permutation

groups basics, elementary arithmetic of cardinals. Some topics less

likely to appear in standard general Algebra courses are presendted in

the Appendix.

The language of Category Theory is used beginning with the chap-

ter on Modules, the reader being invited to refer to the appendix (or to

a book on Categories) when categorical notions appear in the text. We


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think that the Categories are particularly useful for a better understan-

ding and for unifying many algebraic concepts and proofs.

The chapters on Modules (II, III) can be read independently of therest of the book. The section VI.1, Ruler and compass constructions, is

not a prerequisite for the other sections in chapter VI.

Some notations used in the text:

- |A | denotes the cardinal of the setA (the number of elements ofA, ifA is finite).

-

x :=y means x is equal by definition to y (wherey is alreadydefined)or we denotey withx"

- ! marks the end or the absence of a proof.- N is the set of natural numbers, {0, 1, 2, }- N* is the set of positive natural numbers, {1, 2, }- Z is the set of integers


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10

I. Arithmetic in integral domains

The set Z of integers, endowed with the operations of addition and

multiplication, is the prototype for the familiar concept of ring. The

classical divisibility theory in Z (the arithmetic ofZ) can be extended

with outstanding results to a large class of rings, the integral domains.

Such a generalization is interesting by itself and it also illuminates and

yields nontrivial results on the divisibility in Z.

After a general study of the divisibility in integral domains, three

classical and important classes of domains are studied (Euclidian do-mains, principal ideal domains and unique factorization domains).

The definitions of these classes of rings originate in fundamentalarithmetic properties ofZ. The material in this chapter is at the very

basis of all Algebra, and vital in Algebraic Number Theory, Field

Extensions and Galois Theory.

I.1 Divisibility

The classical definition for the relation of divisibility in the ring of

integers Z generalizes easily to an arbitrary ringR:


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I.1 Divisibility 11

1.1Definition. Let R be a ring and let a, bR. We say that adi-vides bin R (and write a | b) if there exists cR such that b=ac.

The fact that a | b can be also expressed by writing b !a (read b isdivisible by a). Other ways of reading a | b are: a is a divisor of borb is a multiple of a.

One says a divides binRbecause the ringR plays an essential

role here. For instance, 2 | 3 in Q, but of course not in Z! We shall

omit any reference toR in the notation a | b if the ringR is clear from

the context. Write a-b ifa does not divide b.

If the ringR lacks some natural properties, the theory of divisibility

inR can be very poor or very peculiar when compared to the classical

theory in Z. For instance, in a ring without identity element, an ele-

ment may not divide itself; other difficulties arise ifR is not commuta-

tive or ifR has zero divisors.

This motivates the following definition:

1.2Definition. A ringR is called an integral domain(a domain, forshort) if it has identity (denoted by 1), it is commutative and has nozero divisors: for anyx, yR,xy= 0 impliesx= 0 ory= 0. This canalso be said: for any nonzero x, yR, we havexy 0.

In what follows, all rings we consider are domains, 0 denotes the

zero element of the domain and 1 its identity element. All subrings

considered are unitary (they contain the identity element of the ring).

1.3Examples. a) Any field F(for instance Q, R, C, ...) is a do-

main. Indeed, if x, yF are nonzero, then xy= 0 implies (bymultiplying with x

1, which exists in Fsince F is a field and x 0),

that y= 0, contradiction. The theory of divisibility in fields is trivial,though (cf. 1.7).

b) Every subring of a domain is itself a domain. In particular, everysubring of a field is a domain. So, ifdZ is squarefree (i.e.: d 0,


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I. Arithmetic in integral domains12

d 1 and dis not divisible by the square of any integer greater than 1),the subring ofC generated by 1 and d , denoted by [ ]d" , is a do-main. One easily checks that [ ]d" consists of the complex numbersof the form dba + , with a, b Z. The ring [ ]1" is called thering ofGauss integers.

1

c) IfR is a domain and n N*, then the polynomial ring in nindeterminates with coefficients inR,R[X1,...,Xn], is a domain.

In a domain,one can simplify the nonzero factors:

1.4 Proposition.Let R be a domain and let a, b, cR, with c 0.

If ac=bc, then a=b.Proof. We have ac=bcacbc= 0 (ab)c= 0. SinceR is a

domain, ab= 0 orc= 0. But c 0, so ab= 0. !

We will develop a theory of divisibility in R, with Z as a model

(and a particular case). The proof of the following properties is an easyexercise:

1.5Proposition.Let R be a domain. Then:a) For any a R, a | a.b) For any a, b, c R such that a | b and b | c, we have a | c.c) For any a R, we have a | 0 and1 | a.d) For any x, y R and a, b, c R such that a | b and a | c, we

have a | (bx+cy). !

Properties a) and b) say that the divisibility relation on R is reflex-

ive and transitive.

1.6 Definition. The elements a, b in R are called associated indivisibility (or, simply, associated) ifa | b and b | a. Notation: ab.

1 Carl Friedrich Gauss (1777 1855), famous German mathematician.


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I.1 Divisibility 13

Ford, a R, dis called aproper divisor of a ifd | a and d is neitherinvertible inR, nor associated with a.

The relation "" defined above is an equivalence relation on R(exercise!) and it is very important when studying the arithmetic ofR:two elements associated in divisibility have exactly the same divisors

and the same multiples. One can say they are indistinguishable as far

as divisibility is concerned.

An invertible element uR is called a unitofR, because u1 (sou behaves just like 1 from the divisibility standpoint). Let U(R) denote

the set of all invertible elements ofR:U(R)= {xR | ()yR such thatxy =1}.

U(R) is a group with respect to the ring multiplication (as astraightforward checking shows) and is called thegroup of units of R.

1.7 Proposition.Let R be a domain. Then :

a) For any u R, we have: u U(R) u 1 u | a, () a R uR=R.

b)For any a, b R, we have: a b there exists u R such thata = bu. !

For a given domainR, knowing the group U(R) is a first step, veryimportant, in the study of divisibility inR.

1.8 Examples.a)U(Z)= {1, 1}.b) IfKis a field, U(K[X])= {fK[X] | degf= 0} =K*(we iden-

tify nonzero elements inKwith the polynomials of degree 0).

c) IfdZ is squarefree, then:

U{Z[ d]}= { dba + | a, bZ, a2db 2=1}

Proof. LetR=Z[ d ]. It is useful to define the norm N :R Z,

by N()=(), where () is the conjugate of, defined as :

( ) dbadba =+ , for any a,bZ.


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So ( ) 22N dbadba =+ , a,b Z. An easy computation

shows:

N()N()= N()N(), , R.This implies: if, R with |inR, then N() | N() in Z.Let u = dba + U(R). Then N(u)= 22 dba divides 1 in Z, so

N(u)=1. Conversely, if N(u)=1, then ( )( ) 1=+ dbadba ,so ( )dba is the inverse ofu. Thus:

U{Z[ d]}= { dba + | a, bZ, a2db 2=1} == {Z[ d] | N()=1}.

We define the central concepts of greatest common divisorandleast common multiple. Since an order relation like the one on N is not

available, we use the divisibility relation itself to order the common

divisors.

1.9 Definition. Assume R is a domain, nN* and a1, ..., anR.We call the element dR a greatest common divisor(abbreviatedGCD)of the elements a1, ..., an if:

i) d | a1, ..., d | an.(dis a common divisor ofa1, ..., an)ii) For any e R such that e | a1, ..., e | an, it follows that e | d(dis

the greatest among the common divisors ofa1, ..., an).If 1 is a GCD ofa1and a2, we call a1 and a2coprime ormutually

prime (orrelatively prime).We call mR a least common multiple (LCM for short)ofa1, ...,

anif:

i') a1 | m, ..., an | m.ii') For any e R such that a1 | e, ..., an | e, it follows that m | e.We denote by (a1, ..., an) or GCD(a1, ..., an) a GCD ofa1, ..., an, if it

exists.

Similarly, [a1, ..., an] or LCM(a1, ..., an)denotes a LCM ofa1, ...,an, if it exists.


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I.1 Divisibility 15

1.10 Remarks. a) Given a1, ..., anR, if there exists a GCD fora1,..., an, say d R, then d is uniquely determinedup toassociation indivisibility: ife is also a GCD of a1, ..., an, then e d. Moreover, ife d, then e is a GCD ofa1, ..., an.

The same remark applies to the LCM.b) When the domainR in which we work is not clear from the con-

text, we use occasionally a subscript, like in the notation (a1, ..., an)R.For a given domainR and givenx, y R, a GCD(x, y) may not ex-

ist (see the Exercises for some examples). A domainR with the prop-erty that any two elements x, y R possess a GCD is called a GCD

domain. For instance, Z is a GCD domain.Writing d= (a1, ..., an) means that d is associated with a GCD of

a1, ..., an. This can lead to some oddities: in Z, we can write 1=(1, 2)=1, but this does not imply that 1=1(of course, it implies that 11).

c) Note that a1, a2are coprime if and only if all their common divi-

sors are units in R.

d) For any domainR and any aR, there exists GCD(a, 0)=a. Ifuis a unit, then there exists GCD(a, u)=u. What can you say about theLCM in these cases?

Ifd 0 and d|a, a/ddenotes the unique elementxR with a=dx.

1.11Proposition.Let R be a domain and let a1, ..., an, rR \{0}.a) If there exists d = (a1 , ..., an) , then a1/d,..., an/d have a GCD,

equal to 1.

b) If there exists (a1, ..., an)=: d and exists (ra1, ..., ran)=: e, thene=rd. Thus:

(ra1, ..., ran)=r(a1, ..., an).c) If there exists [a1, ..., an] =m and exists [ra1, ..., ran] =: , then

=rm. Thus:[ra1, ..., ran] =r[a1, ..., an].


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Proof.a) Let xiR such that ai=dxi, ni ,1= . IfeR is a com-mon divisor ofx1, ..., xn, then de is a common divisor ofa1, ..., an, so

de|d. This implies e | 1.b) Since rd | rai, ni ,1= , we get rd | e. Let uR with e=rdu. It is

enough to prove that u | 1. Letxi,yiR such that ai=dxi and rai=eyi,ni ,1= . For any ni ,1= , rai=rdxi=rduyi. It follows that u is a com-

mon divisor of the elementsxi, whose GCD is 1, by a). So, u | 1.c) Because rm is a common multiple of rai, ni ,1= , we have

| rm, so rm=t, for some tR. We can write m=aibi, =raixi,for somexi, bi R, ni ,1= . We have rm=raibi=t=raixit, ni ,1= .

Simplifying, bi=xit. Also, a1x1= ... =anxn is a common multiple ofa1,..., an. Thus, m | aixi , ni ,1= . Since =raixi, we also get mr | . !

1.12 Corollary. Let R be a GCD-domain and let K be the field of

quotients of R. Then every element in K can be written as a quotient

b

a, with a, bR, b 0 and(a, b) = 1.

Proof. If d

c

K, with c, dR, d 0, then let e= GCD(c, d). Thenc=ea, d=eb, for some a, bR, with (a, b)= 1 (use the statement a)

above). Moreover,b

a

d

c= . !

The next result is ubiquitous in divisibility arguments.

1.13 Corollary.Let R be a GCD domain and a, b, cR such that

a | bc and(a, b)= 1. Then a | c.Proof. (a, b) = 1 and the preceding result, part b), imply that

(ac, bc)=c. Since a | ac and a | bc, the definition of the GCD ensuresthat a | (ac, bc)=c. !

Although the definitions of the GCD and the LCM are dual to

each other, the situation is not entirely symmetric (the existence of the


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I.1 Divisibility 17

LCM implies the existence of the GCD, but not conversely, in gen-

eral).

1.14 Proposition. Let R be a domain and x, yR. Then:a) If a LCM of x, y exists, [x, y] =mR, than a GCD of x, y exists,

(x, y), andxy [x, y](x, y).

b) If any two elements in R have a GCD, then any two elements in

R have a LCM.

c) If any two elements in R have a GCD, then, for any nN, n >1,

any n elements a1, ..., an in R have a GCD and a LCM.Proof.a) Whenx= 0, [0,y] exists and it is 0. Similarly, (0,y)=y.

Suppose now thatx andy are nonzero. The definition of the LCM im-

plies m | xy. Let d, a, bR withxy=mdand m=xa, m=yb. We needonly prove that d=(x, y). We havexy=xad, soy=add | y. Like-wise, d | x. Take eR with e | x, e | y and pickr, sR such thatx=erand y=es. Then ers is a common multiple ofx and y, so m | ers.Let tR such that mt=ers. We have dm=xy= temrse =2 . Simplify-

ing by m, d=te, so e | d.b) Let a, bR \ {0} and let d=(a, b). There existx, yR with a=

dx, b=dy. The element m=dxy is obviously a common multiple ofaand b. Let be another common multiple ofa and b. There existz, tR such that =az=dxzand =bt=dyt. So m divides y=dxyzandx = dxyt, which means m divides also (x, y) = (x, y) = . Thisshows that m is a LCM ofa and b.

c) Induction on n. (Exercise!).!

1.15 Example. In R = Z [ ]5 , x = 51 + and y = 2 have aGCD, but no LCM. Indeed, let d= 5+ ba (a, b Z) be a common

divisor ofx andy. Using the properties of the norm N (see 1.8.c.), we

get N(d) | N(x)= 6 and N(d) | N(y)= 4 in Z. So,N(d) | 2 in Z. Since


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N(d)=22 5ba + ,a case-by-case inspection leads to the conclusion that

a=1and b =0. Thus, d is invertible. We proved that any common

divisor ofx andy is a unit, sox andy have GCD equal to 1.Suppose a LCM R ofx andy exists. Then 6 | N() and 4 | N()

in Z, so 12 | N() in Z. On the other hand, 6 = 23= ( )( )5151 + and ( )512 + are common multiples ofx andy, so they are commonmultiples of . Thus, N() divides N(6)= 36 and

( ) ( ) 24512 =+NN in Z, so N() | 12. Combining with 12 | N(),we get N()=12, which is impossible (the equation 125 22 =+ ba has

no solutions in Z).

IfR is a domain, letR designate the set of nonzero and non-invert-

ible elements inR:

R :=R \{0}\ U(R)

In Z, prime numbersplay a central role in divisibility questions.

Usually, the (elementary) definition for the notion of prime number is

the natural numberp> 1 is prime if its only divisors in N are 1 andp. The generalization to the case of a domain of this definition leads

to the notion ofirreducible element (also compare with the notion ofprime elementbelow).

1.16 Definition. LetR be a domain.

The element p R is called irreducible (in R) if it has no properdivisors. In other words, any divisor ofp is either a unit or is associ-

ated top: dR, d | pd1 ordp.The elementp R is calledprime (in R)if, for any a, b R, p | ab

p | a orp | b.We emphasize that a prime element or an irreducible element is by

definition nonzero and non-invertible.


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I.1 Divisibility 19

A quick argument shows that, for any mN*, ifp is prime and pdivides a product ofm factors inR, thenp divides one of the factors.

1.17 Proposition.Every prime element is also irreducible.

Proof. LetpRbeprime. IfdR is a divisor ofp, there existsxR (nonzero) such thatp=dx. So p |dx, which impliesp | d(and weare finished) orp | x. Butp | x means thatp x (sincex | p), sop=ux,with u a unit. So, ux=dx=p and thus u=dis a unit. !

The notions of prime element and irreducible element (which coin-cide forZ, as we will see)are not the same in general.

1.18 Example.In [ ]5" ,2is irreducible and it is not prime. In-

deed, 2 divides ( ) ( )1 5 1 5 6 + = , but 2 divides neither factor.

On the other hand, ifdis a divisor of 2, then N(d) can only be 1, 2 or

4. An examination of the possible cases shows that dis 1 or2.

Thus, the notion of prime element depends heavily on the ring in

which it is considered: 2 is prime in Z, but not in [ ]5" . The same

remark applies to the notion of irreducible element.

The GCD domains do not have the peculiarity described in the

example above:

1.19 Proposition.Let R be GCD domain. Then any irreducible ele-

ment in R is prime in R.

Proof. LetpR, irreducible andx, yR such thatp | xy. Ifp-x,

then the GCD ofp and x(which exists!) is 1. Indeed, ifd | x and d | p,we cannot have dp(we would get p | x),so d 1. Thus, p |xy and(p, x)= 1. Corollary 1.13 guarantees thatp | y. !

The notion of divisibility can be translated in the language of ide-

als. This approach allows extending classical results on the divisibility


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in Z to much more general classes of rings (for instance, the primarydecomposition theory).

Recall that a subsetIof the commutative ringR is called an idealofR if:

a) (I,+) is a subgroup in the additive group (R,+): x,yIx+yI.

b) xI, rRrxI.Write IR ifI is an ideal in the ring R. The ideal I is proper if

IR.ForaR, the ideal generated by a is the set {ra | rR}, denoted

byRa oraR and is called theprincipal ideal generated by a. Thesum

of two idealsIandJofR is the ideal:

I+J:= {i+j | iI,jJ}.

1.20 Proposition. Let R be a domain, nN* and a, b,x1, ..., xnR. Then:

a) a | b if and only if RaRb.b) a b if and only if Ra=Rb.c) a U(R) if and only if Ra=R.d) a is prime in R if and only if Ra is a prime ideal.

e) a is irreducible in R if and only if Ra is a maximal ideal among

the principal proper ideals of R (more precisely: xR such thatRa Rx, we have Ra=Rx or Rx=R).

f ) a is a common divisor of x1, ..., xn if and only if Rx1+...+Rxn isincluded inRa.

g) If Rx1+...+Rxn=Ra, then a=(x1, ..., xn).2h) a is a common multiple of x1, ..., xn if and only if Rx1 Rxn

includesRa.

2 The converse is false in general. For a counterexample, see the section

Principal Ideal domains.


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I.1 Divisibility 21

i) a= [x1, ..., xn] if and only if Rx1 Rxn= Ra.Proof.a)a | b cR with b=cabRaRbRa.b) Obvious, by a).

c) Ifa is invertible, then cR with ca=1. So 1Ra Ra=R.Conversely, ifRa=R, then 1Ra, so there exists cR such that 1=ca.

d) Letx, yR. We havexyRaa | xy. Ifa is prime, then a | xora | y, i. e.xRa oryRa, which shows thatRa is prime. IfRa isa prime ideal and a | xy, thenxyRa, soxRa oryRaa | x ora | y.

e) Suppose a is irreducible. IfRx is a proper principal ideal ofRwith RaRx, then x | a. Since a has no proper divisors, x is associ-ated with a or it is a unit. But x cannot be a unit because RxR. So,xa, i.e. Rx =Ra. Suppose now Rx is maximal among principalproper ideals, and dR is a divisor ofa.ThenRa Rd, soRd=RaorRd=R. This means da ord1.

f ) Ifa is a common divisor ofx1, ..., xn, then a | nnxrxr ++ ...11 , forany r1, ..., rnR, so any element in the idealRx1+...+ Rxn is divisibleby a. The other implication is left to the reader.

g) From f ), a is a common divisor ofx1, ,xn. Let dR be an-other common divisor. Because aRx1+ +Rxn, c1, , cn Rwith a=c1x1+ +cnxn. Since d | x1, , d | xn, we obtain d | a.

h), i) are left to the reader. !

An essential role in the arithmetic ofZ is played by the theorem of

division with remainder:

For any a, bZ, b 0, there exist q, rZ,such that a=bq + rand |r|


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Any nonzero non-invertible integer can be written as a finite prod-

uct of prime integers and this writing is unique up to the order of fac-

tors and an association of the factors in divisibility. This result is

called the fundamental theorem of integer arithmetic or the unique

integer factorization theorem.

Abstracting these properties of Z, one obtains the notions of

Euclidian domain (a domain in which a property analogous to thetheorem of division with remainder holds), principal ideal domain (adomain whose every ideal is principal) and, respectively, uniquefactorization domain (a domain in which every nonzero non-invertibleelement can be written as a product of primes).

The following sections are devoted to the study of these classes of

rings.

I.2 Euclidian domains

2.1 Definition. A domain R is called an Euclidiandomain if there

exists a mapping :R \ {0} N satisfying:For anya, bR,b 0, there existq, rRsuch that:

a=bq+rand (r= 0 or(r)


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I.2 Euclidian domains 23

degree function deg : K[X] \ {0} N). These rings are also the mostimportant examples of Euclidian domains.

2.2 Remark. Sometimes, in the definition of an Euclidian domain,

the following condition on is also required:

For any a, bR \ {0}, a | bimplies (a)(b).This extra condition is not essential and in fact defines the same

class of rings as the definition above (see 4.9).

2.3 Remark. IfbR, b 0, and (b)= 0, then bU(R). Indeed,the remainderrof the division of 1 to b satisfies 1 =bq +r, for someqR,and (r= 0 or(r)


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Moreover, there exist (and can be algorithmically determined) u, v

R such thatd=au+bv.

Proof. The algorithm3 above implies the following sequence of

divisions with remainder performed inR:

(1) a=bq1+r1 with r1= 0 or(r1)


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I.2 Euclidian domains 25

r2=b r1q2. By induction, e | ri for any i


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Given a, bZ, the quotient q and remainderrof the division ofaby b are not unique: for instance, 3 = 21 + 1 = 22 + (1). If we re-quire that the remainder should be positive, then q and rare unique.

b) The ring K[X] of polynomials in one indeterminate X withcoefficients in the field K is Euclidian with respect to

deg :K[X] \ {0}N. The proof originates in the procedure of long

polynomial division taught in school. Letf=a0+ + anXn

and

g=b0+ + bmXm K[X], with g0 (i.e. bm0). We prove by

induction on n = deg f. If n


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I.3 Euclidian rings of quadratic integers 27

A somewhat surprising fact is that the ring

Z[ ( ) 21 d+ ] = {a+b ( ) 21 d+ | a, bZ} has sometimes betterarithmetic properties than Z[ d ]. This fact is closely connected to

the theory of the rings of quadratic integers, an important topic in

(algebraic) number theory. In the following, some elementary facts on

these rings are presented. The unproven statements below are pro-

posed as exercises (some of them in the next chapters). For a system-atic treatment of the theory of algebraic integers, see for instance

LANG [1964].

3.1. Definition. A subfield ofC that has dimension 2 viewed as avector space overQ is called a quadratic number field.

Using elementary tools of field extensions theory, one can readily

show that any quadratic number field is of the form

Q[ d ] = {a+b d | a, bQ}, where dZis squarefree.

3.2. Definition. A complex number that is the root of a monic4

polynomial in Z[X] is called integral over Z (or algebraic integer).

For instance, 2 is integral overZ, but 1/2 is not. Sometimes, foravoiding confusions, the numbers in Z are called rational integers.5

An element of a quadratic number field that is integral overZ is

called a quadratic integer.One can prove that: a quadratic integer is a

root of a monic polynomial in Z[X] of degree 2.

Fix dZ, squarefree. If =a+b d Q[ d ], the elementa b d= is called the conjugate of . Define the norm

N : [ ]dQ Q and the trace Tr : [ ]dQ Q,N() :==a2db2

4 A polynomial is called monic if the coefficient of its monomial of maximum

degree is 1.5 Because every algebraic integer overZ which is rational (in Q) must be in Z.

Prove!


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Tr() :=+= 2afor any =a+b dQ[ d] (a, bQ).

The norm N is multiplicative and the trace Tr is additive: for any

, Q[ d],N()= N()N(),

Tr(+)= Tr()+ Tr().One can prove that, for any x [ ]dQ :xis integral overZ(x is a

quadratic integer) Tr(x)Z andN(x)Z.

The quadratic integers in [ ]dQ form a ring, called the ring ofintegers of [ ]dQ . This ring is traditionally called a ring of quadraticintegers (imaginary ifd< 0, respectively real ifd> 0). We have thefollowing (for the proof, see the Exercises):

3.3 Proposition. The ring of integers of [ ]dQ is Z[] , whereZ[] = {a+b |a, bZ} and

=

d ifd 2 or 3 (mod 4)

( ) 21 d+ if d 1 (mod 4). !

In what follows, for a fixed dZ, squarefree, denotes the num-ber above.

We remark also that

Q[ d] =Q[] = {a+b | a, bQ}.Ifd 1(mod 4), then Z[] = {a+b ( ) 21 d+ |a, bZ} can also

be described as the set of complex numbers of the form (u+v d)/2,with u, vZ having the same parity.

According to the above, Z[i], [ ] ( )[ ] ( )[ ]251,231,2 ++ ZZZ i are examples of rings of quadratic integers.

The norm N : [ ]dQ Q has the property that N()Z,Z[], as we saw above. We obtain a mapping:

|N| : Z[] N, |N|()=|N()|, Z[].A natural problem arises:


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I.3 Euclidian rings of quadratic integers 29

For whichdZ, squarefree,Z[] is Euclidian with respect to |N|?

We remark first that, since the norm N : [ ]dQ Q is multiplica-

tive (N(xy) = N(x)N(y), x, y [ ]dQ ), |N| : Z[] Nis alsomultiplicative.

Ifd< 0, the representation of the complex numbers in Z[] in theplane yields agrid (rectangular if d 2 or 3 (mod 4) and oblique ifd 1(mod 4)); moreover, for any x, y [ ]dQ , |N|(x y) is theEuclidian distance between the points x and y.

Ifd> 0, Z[] is a (dense) subset ofR and no geometric interpreta-tion is available.

3.4 Lemma. Let dZ , squarefree. Then R :=Z[] is Euclidianwith respect to |N| if and only if for any x [ ]dQ , there exists Rsuch that |N|(x)< 1.

Proof. [ ]dQ is the field of quotients of R, i.e.: x [ ]dQ , , R,0 such thatx=/. SinceR is Euclidian with

respect to |N|, , R such that =+, with |N|()< |N|() or= 0. So,

x=/=+/,with |N|(x )= |N|(/)= |N|()/()< 1.

d 2 or 3 (mod 4) d 1 (mod 4)


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Let , R. Forx=/ [ ]dQ , R with |N|(x )3,i.e. d=1ord=2.

b) In this case, the grid forms isosceles triangles with base 1 and

height 2d . The points inside the triangle have distance less than1 to some vertex if and only if the circles of radius 1 centered in base

vertices intersect in a point Psituated at a distance less than 1 to the

third vertex. An easy geometric reasoning shows that the distancefrom P to the third vertex is ( ) 23 d . We must have then( ) 23 d < 1 d>7 34 d {3, 7, 11}. !

6 For anyzC and any R, > 0, there existsx [ ]dQ such that |zx|


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I.4 Principal ideal domains 31

We obtain that the following rings are Euclidian with respect to |N|:

Z[i], [ ] ( ) ( ) ( )2 , 1 3 2 , 1 7 2 , 1 11 2i i i i + + + " " " " .

One can prove that these are all the imaginary Euclidian rings of

quadratic integers (not necessarily with respect to |N|). The real cased> 0 has no geometric interpretation and is considerably more diffi-cult.

I.4 Principal ideal domains

4.1 Definition. A domain R is called a principal ideal domain(PID) if any ideal ofR is principal. In other words, for every idealIofR, there exists aR such thatI=Ra.

Any field is a PID8. The most important examples of PIDs are

given by the following result.

4.2 Theorem. Any Euclidian domain is a principal ideal domain.

Proof. LetRbe Euclidian with respect to and letIbe a nonzero

ideal ofR. The set of natural numbers {(x)| xI, x 0} contains aminimum element: aI, a 0, such that (a)= min{(x)| xI,x 0} (a may not be unique). We claim that a is a generator of theideal I. Obviously, RaI. In order to prove the opposite inclusion,

suppose by contradiction that there exists b I \ Ra. Applying thedivision with remainder inR,we obtain elements q, rR such that b=aq+r, r0 (since bRa) and (r)


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For instance, ifK is a field, K[X] is a PID; ifI 0 is an ideal inK[X], a generator ofI is a polynomial gI of minimum degreeamong the degrees of nonzero polynomials inI.

The principal ideal domains are GCD-domains; for any a, bR,there exists their GCD, namely any generator of the ideal aR+bR:

4.3 Proposition. Let R be a PID and let a, bR. Then:a) A GCD d of a and b exists and, for some u, vR, d=au+bv .

Moreover, the element d R is a GCD of a and b if and only ifdR = aR+bR.

b) The element m R is a LCM of a and b if and only if mR= aR bR.

Proof.a) Since R is a PID, dR such that the ideal aR+bR={ax+by |x, yR} is generated by d. Since daR+bR, there existu, vR such that d=au+bv.We have a, b dR, so d | a, d | b. IfeR is such that e | a, e | b, then e | ax+by, x, yR. In particular,e | d. Thus, any generatordofaR+bR is a GCD ofa and b.

Conversely, ifdis a GCD ofa and b, ddivides a and b, so dRaR and dRbR. Thus, dRaR+bR. Ife is a generator ofaR+bR,this means d| e. But e is also a common divisor ofa and b, so d is

associated to e. Thus, dR=eR=aR+bR.b) The proof is similar to the one above and is proposed as an exer-

cise. !

Recall that, fora, bR, the notation (a, b) is used to designate theideal generated by a and b, aR+bR. The proposition above shows thatthis notation is consistent to the fact that it designates also the GCD of

a and b(which is a generator of the ideal aR+bR).

4.4 Example. LetR be a domain that is not a field and take rRnonzero and non invertible. Then the ideal (r,X) inR[X] is not princi-


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pal, soR[X] is not a PID. In particular, the rings Z[X],K[X, Y], with K

a field, are not PID's.

Indeed, suppose that for somefR[X], we have (f)=(r, X). Thenf | r. We obtain degf= 0, so fR. Since f | X, there exists gR[X]withX=fg, sofis invertible inR. Thus, the GCD ofrand Xis 1, sof= 1. But the ideal generated by randXdoes not contain 1: ifh, qR[X] are such that 1 = hr + qX, setting X = 0 in this equality ofpolynomials, it follows that 1 =h(0)r, which means r is invertible, acontradiction.

Using Proposition 1.19 and the fact that any PID is a GCD-domain,

we obtain:

4.5 Proposition.In a PID, an element is irreducible if and only if it

is a prime element. !

Thus, the domains containing irreducibles that are not primes are

not principal ideal domains. such a ring is Z[ 5 ], as Example 1.18shows.

4.6 Corollary. In a PID R, the prime nonzero ideals are maximal

ideals. Any maximal ideal is of the form pR, where p is irreducible in

R. Moreover, p R is irreducible if and only if pR is a maximal ideal.Proof. It is sufficient to remark that any nonzero prime ideal is

principal, generated by a primep (Prop. 1.20.d)). The element p isirreducible, so (Prop. 1.20.e)) the idealpR is maximal. The other state-ments are obvious. !

The case R=Z of the following proposition is known as the thefundamental theorem of integer arithmetic. Recall that

R= {x R | xis nonzero, non-invertible}.


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4.7 Theorem. Let R be a PID. Then every nonzero non-invertible

element rinR is a finite product of prime elements.9

Proof. SinceR is a PID, the primes inR coincide with the irreduci-

bles inR. Suppose by contradiction that there exists r0R such thatr0 cannot be written as a finite product of irreducibles. In particular, r0

is reducible, so r0=r1s1, with r1,s1R, not associated with r0. Ifr1and s1 are finite products of irreducibles, then r0 is too, false. So, at

least one of them (say r1) is not a product of irreducibles. We obtainthus r1R with r1 | r, r1rand r1 is not a product of irreducibles.This reasoning applies to r1, so we get r2R, r2 | r1, r2r1, and r2 is

not a product of irreducibles. By induction, there exists a sequence(rn)n 0 of elements inR, with the property that for any nN, rn+ 1 is aproper divisor ofrn. In other words, we obtain a strictly increasing se-

quence of ideals r0Rr1R rnRrn+1R . This is impossi-ble in a PID, as the following lemma shows.

4.8 Lemma. Let R be a PID and let(In)n 0 be a sequence of idealsin R such that In In +1, for any nN. Then there exists mN such

that Im = Im + i , for any i N. (In other words, every ascending se-quence of ideals is stationary).

Proof. LetIbe the union of all idealsIn, nN. Since the sequence(In)n 0 is ascending,Iis an ideal inR: ifx, yI, then, for some i, jN,xIi,yIj. Sox, y It, where t= max(i, j), hence x+yItI.IfrR, rxIi I. ButR is a PID, so there exists aR such thatI=aR. Since aI, there exists mN such that aIm, so aRIm

I=aR. So aR=I=Im +i, iN.!

A ringR satisfying the ascending chain condition (ACC): every as-

cending sequence of ideals of R,I0I1 , isstationary (there ex-

9 The products may contain a single factor (i.e., the element itself is a prime).


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ists mNsuch that Im=Im + i, iN) is called aNoetherian ring.10

We have shown that every PID is Noetherian. Note that the proof

above can be slightly modified to obtain the theorem: A ring in which

every ideal is finitely generated is Noetherian. The converse is also

true. Noetherian rings are an important topic in Commutative Algebra.

We have remarked that sometimes in the definition of the Euclidian

domain R with respect to :R \ {0}N, an extra condition is re-quired:for any a, bR*, a | bimplies (a)(b). We show that thiscondition is not essential.

4.9 Proposition. Let R be a Euclidian domain with respect to: R*N. Then there exists : R*N such thatR is Euclidian with

respect to and, for any a, bR*, a | bimplies (a)(b).Proof. Let : R

* N, defined by (x)= min{(y) | yx},xR. Clearly,xy implies (x)=(y). We claim thatR is Euclid-ian with respect to . Let a, bR, b 0 and let b0 be associated withb such that (b)=(b0). There exist q, rR such that a=b0q+randr = 0 or (r)


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I.5 Unique factorization domains

5.1 Definition. A domain R with the property that every nonzero

and non-invertible element is a finite product11

of prime elements is

called a unique factorization domain(UFD).Theorem 4.7 shows that any principal ideal domain (thus also any

Euclidian domain) is a UFD. Every field is a UFD, because it has nononzero and non-invertible elements.

5.2 Proposition. Every irreducible element in a UFD is also a

prime.

Proof. LetR be a UFD and letpbe irreducible inR. SincepR,p is a product of primes. But this product can have only one factor,

otherwisep would not be irreducible. So,p is itself a prime. !

The next Proposition justifies and explains the epithet unique in

the name of a UFD.

5.3 Proposition.Let R be a domain and rR. If r has a prime de-composition, then this decomposition is unique up to an ordering of

the factors and an association in divisibility. This means that: if r=p1pn=q1qm are two prime decompositions of r, then m=n andthere exists a permutation of the set{1, , n}such that piq(i), i {1, , n}.

Proof. Let r= p1 pn be a prime decomposition of r. Call the

natural numbern the length of the decomposition. Define l(r) as thesmallest nN* such that there exists a prime decomposition of roflength n.

11 Such a product is also called a prime decomposition of the element. Products

may have a single factor.


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I.5 Unique factorization domains 37

We prove the claim by induction on l(r).Ifl(r)= 1, then let r=p1=q1 qm, withp1, q1,, qm prime. Since

r is a prime and rdivides the product q1 qm, rdivides one of the

factors, say q1(relabel if necessary). But q1 is irreducible, so we haverq1. Thus r= q1u, with u invertible. We must prove that m= 1. Ifm 2, simplifying by q1 in q1u = q1 qm, we obtain q2 qm=u, soq2,, qm are invertible, contradiction.

Suppose the claim is true for any x R with l(x)


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c)d) Let a, bR (ifa, b {0} U(R), it is trivial to exhibit aGCD ofa andb). A GCD ofa and b can be determined by taking thecommon prime factors, at the smallest power. To be precise, letPbe

a system of representatives of the equivalence classes of the irreduci-

ble elements inR(with respect to the association in divisibility). Thismeans that every irreducible inR is associated to exactly one element

in P. Property c) assures that there exist and are unique distinct p1,

, pnP,s1,, sn, t1,, tnN, u, vU(R) such thata= upp nsn

s 11 and b= vpp ntn

t 11 .

The uniqueness claim follows the uniqueness assertion in c). Let ri

= min(si, ti) and define d=n

rn

r

pp 11 . We have d | a, d | b. If e | a,e | b, then every irreducible cPthat divides e divides also a and b.This implies c {p1, ,pn}. Indeed, otherwise a(orb) would possesstwo decompositions in irreducible factors, one containing c and the

other one not, contradicting the uniqueness property. So

e= qpp nwnw 11 , with w1,, wnN, qU(R). From e | a it follows

that wisi, and e | b implies witi, i= n,1 . So wiri and e | d.d)a) Prop. 1.19 makes sure that any irreducible is a prime, since

R is a GCD domain. The implication is now obvious. !

Note that in a UFD any two elements a and b have a LCM. Given

two prime decompositions ofa and b, their LCM can be determined

by taking all prime factors in the decompositions, at the greatest

exponent; with the notations in the proof above, define

qi= max(si, ti); the element m= nqn

q pp 11 is then a LCM ofa and b.

5.5 Example.Z [ ]5 is not a UFD, since 2 is irreducible and nota prime (cf. Ex. 1.18). The reader can also check that 6 has twodecompositions as a product of irreducibles in Z [ ]5 :

6 = 23 = ( )( )5151 + ,and 2 is not associated with 51 + or with 51 .


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5.6Proposition. Let R be a UFD, nN* and a, b1, , bnR. If(a, bi)= 1, for any 1 in, then (a, b1bn)= 1.

Proof. It is enough to show that there is no primep that simultane-

ously divides a and b1bn. Ifp is such an element, then there existsj,

1 jn such thatp | bj. Sincep | a, we havep | (a, bj)= 1. Therefore,p is invertible, contradiction. !

The UFDs have many characterizations. The following, due to

Kaplansky, is inspired from Commutative Algebra techniques.

5.7 Theorem. Let R be a domain. Then R is a UFD if and only if

any prime nonzero ideal of R contains a prime element.Proof. Suppose R is a UFD andP 0 is a prime ideal. Take aP,

a 0. SincePR, a is not a unit, so a decomposes in prime factors: a=p1 pn. BecausePis a prime ideal, there exists i such thatpiP.

Suppose now that any nonzero prime ideal in R contains a prime

element. Consider

S= {aR | aU(R) ora is a product of prime elements}.IfS=R{0}, then R is a UFD. Suppose, by contradiction, that

there exists aR, nonzero, with aS. We note that Sis a multiplica-tive system (1 Sand a, bS, abS). Then there exists a primeidealPinR containing a, with PS=. This fact, together with thehypothesis, implies the existence of a prime elementpP. ButpS,contradictingPS=. So we must have S=R{0}.

We have to prove the existence ofP. We will prove that any idealI,

maximal with the properties IS= and aR I, is prime. Let us

prove first that such an ideal exists. The idea is to use Zorn's Lemma.Consider the set of ideals (ordered by inclusion):

J= {Iideal inR | IS= and aR I}.First,Jis not empty, since aRJ. Indeed, ifarSfor some r

R, then aris a unit or is a product of primes. Ifaris a unit, then a is

also a unit, contradiction. Ifar is a product of primes, we show that


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aS, by induction on the number of factors. Ifar=p, withp a primeimpliesp r(so a is a unit) orp a(so aS). Ifar= p1 pn withp1,, pn primes, thenp1| a orp1| r. Ifp1 | a, then a=bp1; simplifying, br=p2 pn. The induction hypothesis shows that b S, so a S. Ifp1| r, let r = cp1, for some c, so ac =p2 pn. Again by induction,aS.

Jis inductively ordered: every chain in J, indexed by some set L,

(Il)lL, is bounded above in Jby lLIl. The checking is straightfor-ward and is left to the reader.

Zorn's Lemma guarantees now that some maximal element PJ

exists.The idealPis prime: letx, yR, withx Pandy P. If, by ab-

surd,xyP, consider the idealsP+Rx and P+Ry, which strictly in-clude P. Since P is maximal, there exist some elements sS(P+Rx) and tS(P+Ry); thenstSP, contradiction. !

The following important result shows the property ofR being a

factorization domain is inherited by the polynomial ring R[X]. This

has far from trivial consequences: for instance, it is not obvious thatevery polynomial in n indeterminates with coefficients in Z decom-

poses uniquely in irreducible factors.

5.8 Theorem. If R is a UFD, then the polynomial ring in one

indeterminate R[X] is a UFD.

The proof of this theorem needs some preparations, which are also

interesting on their own. First, we determine the units ofR[X].

5.9 Proposition. Let R be a domain. Then

U(R[X])=U(R)and R[X] =R {fR[X] | degf 1}.

In particular, if K is a field, U(K[X])=K* andK[X] = {fK[X] | degf 1}.


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Proof.U(R)U(R[X]) is evident. Iffis a unitR[X], there is somegsuch that fg= 1. Since R is a domain, degf+ degg= 0, sodegf= 0 = degg, which means that f, gR. From fg= 1 we deducethatfU(R).

IfKis a field, then U(K)=K* and the other claims follow. !

5.10 Definition. LetRbe a UFD and letf=a0+a1X++anXn

R[X]. The GCD of the coefficients a0, a1, , an is called the contentof

f, denoted c{f}. A polynomial with content (associated with) 1 iscalled a primitive polynomial. Note that f is primitive if and only if

there exists no primep inR such thatp divides all the coefficients off.AnyfR[X] can be written

f=c{f}g, for some primitivegR[X].Also, iff=ag, with aR andgprimitive, then c{f}=a.

5.11 Proposition. a) Let R be a domain. If p is prime in R, then p is

prime in R[X], too.

b) Let R be a UFD and let f, gbe primitive in R[X]. Then the prod-

uct fg is also primitive.12

c) Let R be a UFD and let f, gR[X]. Then c{f g}=c{f}c(g).Proof.a) Note first that:p divides a polynomial inR[X] if and only

ifp divides all the coefficients of the polynomial. Letf=a0+a1X++anX

n, g=b0+b1X++bmX

mR[X] such thatp-fandp-g. Let

us prove thatp-fg. Fromp-fwe deduce that there exists i, 0 in,such thatp-ai . Take i minimal withp-ai. Similarly, letj be minimal

such thatp-bj. Then the coefficient ofXi+j

infgis

+=+ jilk

lkba

In this sum, aibj is not divisible byp and the other terms are divisi-

ble byp(as products of two factors, at least one of which is divisible

12 This is called Gauss' Lemma.


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byp). So, the coefficient ofXi+j

is not divisible byp and neither is the

polynomialfg.

b) Iffgis not primitive, there existspR, prime, such thatp | fg.The previous paragraph impliesp | forp | g, contradiction.

c) Letf=c{f}f1,g=c(g)g1, wheref1 andg1 are primitive. Thenfg=c{f}c(g)f1g1,

withf1g1primitive by b). It is clear now that c{fg}=c{f}c(g). !

5.12Proposition. Let R be a UFD and let K be its field of quo-

tients. Then a polynomial finR[X] of degree 1 is irreducible in R[X]

if and only if f is primitive and irreducible in K[X].Proof. Letfbe irreducible inR[X]. Thenfis clearly primitive. Let

us prove that f is irreducible in K[X]. Iff=gh, with g, hK[X], bymultiplying with the LCM of the denominators of the coefficients ofg

and h, we get something like af=g1h1, for some polynomials g1,h1R[X] and some a R. Computing the contents, we have a =c(g1)c(h1), since c{f}= 1. We have g1=c(g1)g2, h1=c(h1)h2, withprimitive g2, h2R[X]. So af=c(g1)c(h1)g2h2; simplifying by a(=

c(g1)c(h1)), we obtainf=g2h2. Sincefis irreducible inR[X], degg2= 0or deg h2= 0. But degg= degg1= degg2 and similarly forh, so degg= 0 or deg h= 0.

Conversely, iffis irreducible inK[X], it has no proper divisors (ofdegree 1) inK[X]; so it does not have divisors of degree 1 inR[X].Sincefis primitive,fhas no non-invertible factors of degree 0. !

This result is also important from a practical point of view: in order

to prove that some polynomial with coefficients in R is irreducible inK[X] , it is sufficient to prove it is irreducible inR[X], which is often

easier to accomplish.

We can give now the Proof of the Theorem5.8. We will use the

characterization in 5.4.b). LeR be a UFD,Kits field of quotients andfR[X], irreducible. Let us prove thatfis a prime. Iff | gh, withg, h


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I.6 Polynomial ring arithmetic 43

R[X], becausefis irreducible inK[X] (and thus also prime inK[X]), wehave f | gorf | h inK[X]. Sayf | ginK[X]; so there exist aR, qR[X], such that ag =fq. This implies ac(g) = c(q). We can writeq=c(q)q1, g=c(g)g1, with q1, g1 primitive in R[X]. Thus ac(g)g1=fc(q)q1; simplifying by c(q), we haveg1=fq1, sof | ginR[X].

Next, we must show that any nonzero non-invertible polynomial

fR[X] is a product of irreducibles. We prove this by induction onthe degree off. If degf= 0 andfis not a unit inR[X], thenfR andit has a decomposition in irreducible factors inR, factors that are also

irreducible inR[X]. If degf> 0, writef=c{f}f1, withf1primitive. It is

sufficient to prove the existence of a decomposition forf1. Iff1 isirreducible, we are finished; if not, f1 has a proper divisor in R[X],

which must be a polynomial of degree strictly less than degf (f1hasno proper divisors in R, being primitive). Thus, f1=gh, with g, hR[X], having degrees smaller thanf. Applying the induction forgand

h, we infer thatf1is a product of irreducibles inR[X]. !

Thus, the following rings are unique factorization domains:

Z[X], Z[X1, ,Xn],K[X1, ,Xn], whereKis a field.

An analogous result holds forNoetherian rings: if R is a commuta-

tive Noetherian ring, then R[X] is Noetherian.(David Hilbert'sBasis-satz).

I.6 Polynomial ring arithmetic

In this section, R designates a domain and Kits field of quotients.

Recall thatR denotes the set of nonzero and non-invertible elements

inR.


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The problem of deciding the irreducibility of a polynomial inR[X]

is important and often nontrivial. A rich collection of irreducibility

criteria is thus very useful in such problems.

6.1 Remark. GivenfR[X], the following simple facts are worthremembering:

- if degf= 0, thenfR. In this case, fis irreducible inR[X] if andonly if itis irreducible inR. IfR=K(R is a field), thenfis invertibleand thus reducible.

- if degf=n > 0, thenfis irreducible inR[X] if and only iffhas no

non invertible divisors of degree 0 and there are no decompositionsf=gh, withg, hR[X] and 1 degg, deg h


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a) a is a root of f.

b) the polynomial Xa divides f in R[X].Proof. There exist q, rR[X] such that f=(Xa)q+r, where

deg r= 0 orr= 0. Note that Xa divides f if and only if r= 0. Butf(a)=(aa)q(a)+r(a)=r, sof(a)= 0 is equivalent to r= 0. !

This theorem is the basis of the notion ofmultiple root:

6.4Definition. IffR[X] and aR, a is called a multiple rootofmultiplicity m forfif(Xa) m |fand (Xa) m+ 1-f; the natural num-berm is called the multiplicity of the root a. A root of multiplicity 1 is

called asimpleroot.

6.5 Corollary. Let fR[X], degf> 1. If fhas a root a R, then fis reducible in R[X] (it is divisible with Xa). !

The converse of this statement is false: (X2+ 1)2 has no roots in Q,

yet it is obviously reducible in Q[X]. Nevertheless, we have:

6.6 Proposition. Let K be a field. Then a polynomial f of degree 2

or3 in K[X] is irreducibleif and only if f has no roots in K. In particu-

lar, if R is a UFD, a primitive polynomial of degree 2 or3 inR[X] is

irreducible in R[X] if and only if it has no roots in K.

Proof. Sincefis irreducible inK[X] and degf> 1,fhas no roots inK. Conversely, iffis reducible and has degree 2 or 3, then, by looking

at the degrees of the factors in a decomposition off, one concludes

thatfhas a divisor of degree 1, which has a root inK. The remaining

claims follow from fis irreducible inR[X] if and only iffis primitiveand irreducible inK[X]. !

If the ringR is not a UFD, then the criterion above may not work:

6.7 Examples. a)f=(2X+ 1)2is reducible in Z[X], but has no rootsin Z. Of course,fhas roots in Q, the field of quotients ofZ.


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b) Let R= {a+ 2bi | a, bZ}. A quick check shows that R is asubring in Z[i], so it is a domain. The polynomialX

2+ 1 is irreducible

inR[X] (prove!), but has the roots i, i in Q[i], the field of quotients ofR. This means that a polynomial of degree 2 or 3 inR[X] that has roots

in Q(R) is not necessarily reducible inR[X].

The following criterion is widely used to find all rational roots of a

polynomial in Z[X] (also see Exercise 26).

6.8 Proposition. Let R be a UFD and f=a0+ a1X+ +anX

nR[X]. Ifp/q K is a root of f, with p, qR, (p, q)= 1 , then

p | a0 and q | an.

Proof. We remark first that every element ofKcan be written as

p/q, with p and q coprime. Writing that f(p/q)= 0 and multiplying

with qn, we have:

a0qn=a1pq

n1++anp

n,

sop | a0qn. Since (p, q)= 1, we also have (p, q

n)= 1 (R is a UFD), sop | a0. A similar proof can be given forq | an. !

6.9 Example. Let f=X3X+ 2 Z[X]. Ifp/qQ is a root off,(p,q)= 1, thenp | 2 and q | 1. So, the rational roots offbelong to theset {1, 1, 2, 2}. By direct testing, we get that none of these numbersis a root. So,fhas no rational roots. Sincefhas degree 3,fis irreduci-

ble in Q[X] (also in Z[X], being primitive).

Here are some general tricks that may prove useful in irreducibility

problems.

6.10 Proposition.Let f=a0+a1X+ + anXnR[X],f 0.

a) Let c, dR, where c is a unit in R. Then f is irreducible if andonly if f(cX+d) is irreducible.

b) Suppose f(0)=a0 0. Then f is irreducible if and only if

r{f}=an+an 1X+ + a0Xn


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(the reciprocal of f) is irreducible.c) Suppose f has no divisors of degree 0 other than units. If S is a

commutative ring and: RSis a unitary ring homomorphism suchthat (an) 0 and (a0)+ (a1)X+ + (an)X

n is an irreducible

polynomial in S[X], then f is irreducible in R[X].

d) (Eisenstein criterion) Let R be a UFD.If there exists a prime ele-ment pR such that p | ai, i


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deg(h)= deg h. But{f} is irreducible, so (g)(to make a choice) isa unit, of degree 0. So, 0 = deg(g)= degg. Since fhas no 0 degreenon-invertible divisors,gU(R).

d) Write f=c{f}f1, with f1 primitive. We have that f and f1 areassociated in K[X]. By replacing fwith f1, we can assume f is primi-

tive. It is sufficient now to prove thatfis irreducible inR[X]. Iffwere

reducible, then:

f=a0+a1X++anXn=(b0+b1X++bmX

m)(c0+c1X++cpXp),

where m> 0,p> 0, b0, b1, , bm, c0, c1, , cpR, bm 0, cp 0. Wehave b0c0=a0,so p | b0c0 and p

2-b0c0 ; thus p divides exactly one of

b0and c0. Suppose p | b0 and p-c0. Because p-an, p does not divideall the bi's; thus there exists some i, 1 im, such thatp-bi andp | bj,j 1. The proof is proposed as an exercise. This sim-ple remark is useful in reducibility issues.

A few instances of using the above criteria on, concrete cases will

give an idea on the strategies of approaching the problem of irreduci-

bility of a polynomial. The exist algorithms that decide if a given

polynomial inZ

[X] is irreducible. Such an algorithm(due to

Kronecker) that also outputs a factor of the polynomial if it is not irre-ducible is described in Exercise 31. This algorithm, applied repeat-

edly, yields a factorization algorithm (producing a decomposition inirreducible factors) for any polynomial in Z[X] orQ[X]. The modernsymbolic computation software (Maple, Mathematica, Macaulay,Axiom, etc.) have powerful routines that decide polynomial irreduci-


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bility, including polynomials in several indeterminates, polynomials

with coefficients in algebraic extensions ofQ or in a finite field. One

can prove that, if there exists a factorization algorithm forK[X], with

Ka field, then there exists one orL[X], whereL is a finitely generated

extension ofK. For details and developments, see SPINDLER [1994],

WINKLER[1996].

6.12 Examples. a) The polynomial 6X9+ 13X2+ 26 is irreduciblein Q[X] (and in Z[X], it is primitive), by the Eisenstein criterion withp= 13.

b) For any prime p and any nN*

, X

n

p is irreducible in Q[X]and in Z[X] (use Eisenstein again).c) Letpbe a prime number and let

f=Xp1

+Xp2

+ +X+ 1 Z[X].

The Eisenstein criterion cannot be applied directly to f. Consider

the polynomial

g= ( )( )

( )1

1

1 11

1 1

pp

i

i

p

i

Xf X X

X

=

+ + = =

+

Remark that the Eisenstein criterion can be applied tog, sincep di-

vides all the binomial coefficients ( )pi , with 1 i


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with a, b, cZ2. Identifying the coefficients, we obtain a system ofequations in a, b, c, which is readily seen to have no solutions in Z2.

Hence,fis irreducible in Z2[X].

f) A typical use of 6.10.c) for a polynomial f with integer coeffi-cients is to reduce the coefficients modulo n. More precisely, for

some nN conveniently chosen, consider the unique ringhomomorphism : ZZn and investigate the polynomialfreduced

modulo n, (denoted with {f} in the Proof). Take, for instance,f= 7X5+ 4X3X2+6X+ 9 Z[X]. The polynomial f reducedmodulo 2 isX

5+X2+ 1 Z2[X], which is irreducible. The conditions

in 6.10.c) are satisfied, sofis irreducible in Z[X] (also in Q[X]).g) 10X7+ 5X2+ 2 is irreducible in Z[X]: its reciprocal is

2X7+ 5X5+ 10, irreducible by Eisenstein withp= 5.

Exercises

Throughout the exercises, R is a domain and Kis its field of quo-

tients (unless specified otherwise).

1. Let R be a commutative unitary ring that has zero divisors and let

aR be a zero divisor. IfbR \ {0} is not a zero divisor, prove thatax+b= 0 has no solutions in any ring Sthat includesR as a subring.

2. LetRbe a finite domain. Prove thatR is a field.

3. Let R be an infinite unitary ring. Prove that the setR of the non-

zero non-invertible elements inR is infinite. (Hint: IfR is finite, thenU(R) is infinite. Let S(R) be the set of all bijections from R to R.The mapping : U(R) S(R),x&x, x(a)=xa, aR, is injec-tive, contradiction.)


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Exercises 51

4. Let R be a GCD domain. Then any element in Kcan be written as

a/b, (b 0), with a, b R, coprime. What can you say about theuniqueness of such a representation?

5. Show that a commutative ringR is a domain if and only ifR[X] is a

domain.

6. Let pR. Prove that the ideal generated by p in R[X], pR[X], isprime if and only ifp is a prime element in R. (Hint: R[X]/(pR[X])(R/pR)[X]). Deduce a new proof for 5.11.a).

7. Let dZ, d< 0 be squarefree. Determine U(Z[ d]).

8. The elements 6 and 2 + 5 have no GCD (and thus no LCM) inZ[ 5 ].

9. Show that Z[ 2 ] is Euclidian. (Hint. Use 3.4.)

10. Let = ( ) 251+ . Show that the norm N : Z[]Z isN(a+b)=a2+abb2, a, bZ. Prove that ( )[ ]251+Z isEuclidian. (Hint. Use 3.4 and Q[ 5 ] =Q[]).

11. Let = ( ) 231 i+ . Write a formula for the norm N : Z[]Z

and determine U(Z[]).12. Let dZ be squarefree.

a) Any element in [ ]dQ can be uniquely written as a+b d ,with a, bQ.

b) Suppose known that any quadratic integer is a root of a monic

polynomial of degree 2 with integer coefficients. Show that:

x=a+b d [ ]dQ (a, bQ) is a quadratic integer Tr(x)= 2aZ and N(x)=a2db2Z.

c) Show thatx=a+b d (with a, bQ) is a quadratic integer2aZ, 2bZ and 4a 2 4b 2d 0 (mod 4).

d) Show that R := { [ ]dQ | is a quadratic integer} is asubring of [ ]dQ andR=Z[], where is given by Prop. 3.3.

e) Ford< 0, determine explicitly U(R).


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13. Let dZ be squarefree and , Z[ d ] such that Z.Show that there is some Z[ d] and a, bZ such that =aand=b

(

is the conjugate of).

14. The purpose of the exercise is to determine all primes in Z[i]. As a

bonus, we find the primes in Z that can be written as a sum of two

squares.

a) Prove: for any dZ and any a+biZ[i], d | a+bi in Z[i] d | a and d | b in Z.

b) Suppose thatpZ is a prime in Z and in Z[i]. Show that theequationx

2+ 1 = 0 has no solutions in Zp (=Z/pZ, the field of inte-

gers modp).c) Let p be prime in Z. Then: ()a, bZ such that p=a2+b2(p

can be written as a sum of two squares) the equationx2+ 1 = 0 hassolutions in Zp.

d) Ifp is a prime in Z andp 3(mod 4), thenp cannot be written asa sum of two squares.

e) Ifp is a prime in Z andp 1(mod 4), then ( )pp

mod1!2

1

(Hint. Use Wilson's Theorem: (p 1)! 1(modp)).

f) Prove that a prime pZ is also prime in Z[i] if and only if

p 3(mod 4).

g) Prove that all the prime elements in Z[i] (up to association in

divisibility) are: 1 +i, 1 i and the primesp in Z withp 3(mod 4).

h) Prove that a primepZ can be written as a sum of two squares

p 1(mod 4).

15. LetR be a unitary subring of the commutative ring S. An element

ofSis called integral overR if it is a root of a nonzero monic polyno-

mial inR[X]. Prove that, ifR is a GCD-domain, Kis its field of quo-

tients andxKis integral overR, thenxR. (A domainR with thisproperty is called integrally closed).


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Exercises 53

16. Let R be a PID and let Sbe multiplicatively closed system in R.

Then the ring of quotients S1

R is a PID.

17. Let R be an Euclidian domain with respect to the function andlet Sbe a multiplicatively closed system inR. Then S

1R is an Euclid-

ian domain. (Hint. Take Ssaturated. Use exercise 4.)

18. LetR be a UFD and let Sbe a multiplicatively closed system inR.

Then S1

R is a UFD.

19. Does the property of a domainR of being Euclidian (respectively aPID, a UFD) is inherited by the unitary subrings ofR?

20. LetR be Euclidian with respect to . Show that there exists uRnonzero and not a unit with the property: xR, qR such thatxqu is a unit or 0. Find such a u forR=Z, K[X]. (Ind. min {(v) |vR} =(u) for some uR.)

21. Let d13, dsquarefree and denote by R the ring of integers of[ ]dQ (so,R =Z[], with as in Prop. 3.3). Show that U(R)= { 1,

1}. Show that R is not Euclidian. Is this result a particular case of

Proposition 3.5? (Ind. IfR is Euclidian, let uR given by the preced-ing exercise. Then, xR, we have u | x oru |x 1. Takex= 2 anddeduce that uZ and u=2 or3. Then findyR such that u doesnot divide any ofy ory 1).

22. Let dZ be squarefree, d 1 (mod 4). Then Z[ d ] is not aGCD-domain. (Hint. 2 is irreducible and not a prime.)

23. Show that Z contains an infinity of prime elements not associated

in divisibility to each other.

24. Let R be a UFD that is not a field, such that the group of units

U(R) is finite. Then R contains an infinity of prime elements notassociated in divisibility to each other. (Hint. Ifp1, ,pn are all the primes up to association in divisibility , then there exists m 1such that 1 +(p1pn)

mR.)


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25. Let R be a UFD and let pR be a prime. Using the canonicalhomomorphism :R R/pR and its extension to a homomorphism:R[X](R

/pR)[X], give a new proof for the Eisenstein criterion.

(Hint. Iff=a0+a1X+ + anXn

satisfies the hypothesis of the crite-

rion and f=gh, then{f}=(an)Xn=(g)(h). If degg, deg h 1,

theng(0)and h(0) are divisible byp.)

26. LetR be a UFD andf=a0+a1X+ + anXnR[X].

a) Ifp/qK is a root off, where p, qR and (p,q)= 1, thenp | a0, q | an and (pqr) | f(r), rR. Write down explicitly theconclusions foran= 1.

b) Letg= nnnnnnn XXaXaaaa ++++ 111201 . Then ( )1nna f X = g(anX). What connection is between the roots ofgand the roots off?

c) Find the rational roots of 2X3+ 5X2+ 9X 15 and4X

3 7X2 7X+ 15.

27. LetKbe a field. Prove that any nonzero polynomialfK[X] hasat most degfroots inK(every root is counted with its multiplicity).

28. LetR be a commutative ring. Prove that the following statements

are equivalent:a) Any nonzero polynomialfR[X] has at most degfroots inR.b) Any polynomial of degree 1 has at most one root inR.c)R is a domain.(Hint. Consider the field of quotients ofR and use the previous

problem).

29. Let R be a commutative ring. IffR[X], define the polynomial

function f

~

:R R, defined by: xR, f

~

(x)=f(x)(the value offinx). Prove that, if R is an infinite domain, then the mapping:R[X] R

R, {f}= f

~, fR[X], is injective. Is the conclusion

still valid if one does not assume thatR is infinite?

30.(The Lagrange interpolation polynomial) LetKbe a field, n 1 aninteger, fix n+ 1 distinct elements x0, ,xnKand (not necessarily


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Exercises 55

distinct)y0, ,ynK. Prove that there exists a unique polynomialLK[X] satisfying: degLn andL(xi)=yi, 0 in.

31. LetpZ[X], primitive, degp=n and let m= the largest integern/2.a) Show thatp is reducible in Z[X] p has a divisor of degree be-

tween 1 and m.

b) Choose m+ 1 distinct integers, (x0, ,xm)Zm+ 1

. Show that

the following algorithm terminates in a finite number of steps and out-

puts a nontrivial factor ofp of degree m or proves thatp is irreduci-ble:

1. Ifi withp(xi)= 0, thenXxi is a factor ofp and STOP. If not,go to 2.

2. Let D= {d=(d0, , dm)Zm+ 1

| di|p(xi), i}. D is a finiteset.

For any dD, let LdQ[X] be the Lagrange interpolation

polynomial withLd(xi)=di, i, and degLdm. If there exists

dD with LdZ[X] and Ld |p, then Ld is a factor ofp and

STOP. If not, thenp is irreducible.c) Deduce an algorithm of deciding the irreducibility of polynomi-

als in Q[X].

d) Suppose m= 2. Propose a choice for(x0, ,xm).e) Suppose there is available a factorization algorithm forR (an

algorithm that produces a decomposition of any element in R in

prime factors). What properties should R have in order to adapt the

algorithm above toR[X]?f) SupposeR is a UFD and there exists a factorization algorithm for

R[X]. Then there exists a factorization algorithm forK[X].

32. Decide the irreducibility ofX4+X2+ 2X 1 Z[X].

33. Show that a0+a1X+ + anXn Z2[X] (an 0) has no roots in

Z2 if and only ifa0(a0+a1+ + an) 0.


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34. Using the equality (in Z2[X]):X

5+X+ 1 =(X

3+X2+ 1)(X

2+X+ 1),

prove thatX5X 1 Q[X] is irreducible.

35. a) Let fR[X], degf=m. Iffhas at least m+ 1 roots in R, thenf= 0.

b) LetgR[X1,,Xn], with R infinite. If g(a1, , an)= 0,(a1, , an) R

n, then g= 0. Deduce that two polynomials in

R[X1,,Xn] are equal if and only if the associated polynomial func-

tions are equal.

c) Give an example of a finite field Kand distinct polynomials in

K[X] that have the same associated polynomial function.

d) (Generalization of b) Let R be infinite and let gR[X1,,Xn],with deg(g,Xi)=mi. Suppose that ()SR with |S|>m1; ()a1S,()S(a1)R with |S(a1)|>m2; ()a1S, ()a2S(a1), ()S(a1, a2)R with |S(a1, a2)| >m3 and so on. If g(a1, , an)= 0,a1S, a2S(a1), a3S(a1, a2), , anS(a1, , an), theng= 0.

36. Let R be a domain. A polynomial in n indeterminatespR[X1,,Xn] =:R[X] is called homogeneous of degree q if all themonomials inp have total degree q(see the Appendix). Show that:

a) AnypR[X] can be written uniquely as:p=p0+p1+ +pm, withpi R[X], homogeneous of degree i. b) The product of two homogeneous polynomials of degrees a,

respectively b, is homogeneous of degree a+b.c)p is homogeneous of degree qp(TX1,, TXn)=T

qp(X1,,Xn)

(equality inR[X1,,Xn, T] =R[X1,,Xn][ T]).d) LetpR[X] be homogeneous. Then any divisor ofp inR[X]

is homogeneous. (Hint. Forp=gh, writeg=ga+ +gm, wheregi ishomogeneous of degree i, 0 am and ga 0 gm. It suffices toprove that a=m.)


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Exercises 57

e) IfR is infinite andpR[X], thenp is homogeneous of degree qp(tx1,, txn)=t

qp(x1,,xn), t,x1,,xnR.

f) Is it true that any symmetric polynomial in R[X] has all its divi-

sors symmetric polynomials inR[X]?

37. LetKbe a field of characteristic not equal to 2 (1 + 1 0 inK) andp a homogeneous polynomial of degree 2 in K[X,Y], i.e.

p=aX2+bXY+cY2, with a, b, cK. Prove that p is reducible inK[X,Y] b2 4ac is a square in K b2 4ac= 0 or there exist, K, (,)(0, 0), withp(,)= 0.

38. Assume K is a field and p=a0Yn+a1 Y

n 1X+ + anX

nis a

homogeneous polynomial of degree n in K[X, Y]. Let

p(X, 1)=a0+a1X+ + anXnK[X]. Prove that, for any

gK[X, Y], g |p in K[X, Y] if and only ifg is homogeneous andg(X, 1) |p(X, 1) inK[X].

39. LetKbe a field and letpK[X, Y] be homogeneous. Prove thatpis irreducible inK[X, Y] p(X, 1) is irreducible inK[X].

40. Write a decomposition in irreducible factors for

X13+X23K[X1,X2]. (Hint. The cases charK= 3 and charK 3should be treated separately).

41. LetKbe a field, charK 3 andf=X13+ +Xn

3K[X1,,Xn].

Show that f is irreducible if and only if n 3. Generalization. (Hint.Forn= 3, apply Eisenstein tofK[X1,X2][X3]. Use then an inductionon n.)

42. Considern2

indeterminates Xij, 1 i, jn, and consider the nn

matrixA=(Xij)1 i, jn Mn(Z[Xij;1 i, jn]). Then the polynomial:detA={X1(1)Xn(n) | Sn}

is irreducible in Z[Xij;1 i, jn].

43. Prove that a commutative ringR is Noetherian if and only if every

ideal inR is finitely generated.


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58

II. Modules

Module theory can be seen as a generalization of the classic linear

algebra (which studies vector spaces over an arbitrary field1). The the-ory is fundamental in many areas of mathematics: commutative alge-

bra, algebraic number theory, group representation theory, algebra

structure theorems, homological algebra etc. Also, module theory

illustrates and uses concepts of category theory (we use some elemen-tary notions from category theory in this chapter, notions that can be

found in the Appendix). Module theory language and results areindispensable throughout most of modern algebra.

II.1 Modules, submodules, homomorphisms

The notion of a module over a ring can be obtained by replacing

the word field with the word ring in the definition of the vector

space.

1 Although the volume Algbre linaire (1961) of the famous Bourbaki series

Elments de Mathmatique begins with the definition of the module.


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II.1 Modules, submodules, homomorphisms 59

1.1 Definition. Let R be a ring with identity (not necessarilycommutative) and (M, +) an Abelian group. We say that M is a leftR-module (orleft module over R) if there exists an external operationon Mwith operators inR2, i.e. a function

:RM M(notation: (r, x)=: rx, rR, xM),satisfying, for any r, sR andx, yM:

i)r(x+y)=rx+ry;

ii)(r+s)x=rx+sx;

iii)(rs)x=r(sx);

iv) 1x=x,

1.2 Remark. The addition inR is denoted by + , as the addition inM. Also, the zero element in R is denoted by 0, like the zero element

in M. For instance, in axiom ii), the + in the LHS denotes the additionin R, whereas the + in the RHS denotes the addition in R. This nota-tional abuse (which is widely used) should not confuse the reader.

If the axiom iii) is replaced by:

iii')(rs)x=s(rx), r, sR, xM,we say that Mis a rightR-module. The usual notation for the scalar

multiplication in the case of right R-modules is with the scalars on

the right, i.e. the scalar multiplication is a function : MR M,with the notation (x, r)=xr, rR, xM. The axioms for therightR-module become in this case:

i')(x+y)r=xr+yr;

ii')x(r+s)=xr+xs;iii')x(rs)=(xr)s;

iv')x1 =x,

2 Also called multiplication of elements in Mwith scalars inR.


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II. Modules60

for any r, sR andx, yM.There is a handy notation for the fact that M is a left R-module,

namely R M.

Mis a rightR-module is denoted by MR.

IfR is commutative, then the notions of left R-module and right

R-module are the same (look at axiom iii').IfR is an arbitrary ring and Mis a rightR-module, then Mbecomes

a left Rop

-module, where (Rop

, +, *) is the opposite of the ring R (Rop

andR have the same underlying Abelian group R, but the multiplica-

tion * inRop

is defined by r*s=sr, r, sR).

The construction above shows that a result that holds forany ring Rand anyright R-module is valid also forany left R-module, and con-

versely. In the same way, all definitions for left modules have a natu-

ral correspondent for right modules.

1.3 Examples. a) IfKis a field, a K-module is exactly a K-vectorspace.

b) IfR is a ring with identity, R has a (canonical)structure of left

R-module, denoted RR. Indeed, (R, +) is an Abelian group; the exter-nal operation RR R is the ring multiplication: (r,s)&rs, r,sR. Similarly,R is canonically a rightR-module, denotedRR.

c) Any Abelian group (A, +) is canonically a Z-module. FornZand aA, na is defined as the multiple ofa in the additive groupA(ifnN, na=a++a (n terms); ifn< 0, na= (a)+ +(a) (nterms)). This is the only external operation that endows A with a

Z-module structure (exercise!). The theory of Abelian groups is thus aparticular case of module theory.d) Let R be a ring and let nN*. The n-fold Cartesian product

Rn= {(x1, ,xn) |xiR, 1 in} becomes anR-module if the addi-

tion and the scalar multiplication are defined component-wise:

(x1, ,xn)+(y1, ,yn)=(x1+y1, ,xn+yn), (x1, ,xn),(y1, ,yn)R

n.


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r(x1, ,xn)=(rx1, , rxn), rR, (x1, ,xn)Rn.

e) IfR is a ring and m, nN*, the set Mm,n(R) ofmn matrices withentries inR is an Abelian group endowed with usual matrix multipli-

cation and becomes an R-module by defining the multiplication of

matrices with scalars: for rR, A=(aij) Mm,n(R), r(aij) :=(raij)(multiply every entry of the matrix with r).

f) Let R := M2(Z) (the ring of 22 matrices with entries in Z) andM:= M2, 1(Z)(the Abelian group of 21 matrices with entries in Z). Mhas a natural structure of left R-module: A M2(Z)=R,U M2, 1(Z)=M,AUMis the usual matrix product. Checking the

module axioms is straightforward and it boils down to the knownproperties of matrix operations. Can you generalize this example? Can

Mbe endowed with a natural structure of a rightR-module ?

g) Let :R S be a unitary ring homomorphism. If M is a leftS-module, then Mhas a structure of left R-module by restriction of

scalars:rR, xM, rx :=(r)x. In particular, Sbecomes a leftR-module (and also a right R-module). This example generalizes asituation often encountered in field extensions: any field Sis a vector

space over any subfieldR.

1.4 Remark. For a ringR and an Abelian group M, defining a left

R-module structure on M amounts to defining a unitary ring

homomorphism :R End(M), where (End(M), + ,) is theendomorphism ringof the Abelian group M, defined as follows: + is

homomorphism addition: u, v End(M), (u+v)(x) :=u(x)+v(x),

xM, and is the usual map composition: u, v End(M),(uv)(x)=u(v(x)), xM. 3

3Sometimes this ring is called the ring ofleftendomorphisms ofM, emphasizing

that the functions are written at the left of the argument, like u(x); this forces the

definition of the composition of functions in the usual manner defined above. But


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II. Modules62

Indeed, if M is a left R-module, define :R End(M) by(r)(x)=rx, rR, xM. Conversely, given :R End(M), de-fine scalar multiplicationRM Mby (r,x)&rx := (r)(x), rR,xM. The reader should check the details. What corresponds to astructure of rightR-module ofM?

Throughout this section,R denotes a ring with identity. All modules

are leftR-modules (unless specified otherwise).

1.5 Proposition. Let M be an R-module. The, for any xM andrR, we have:

a) 0x=r0 = 0.b) r(x)=(r)x=(rx).Proof.a) 0x=(0 + 0)x= 0x+ 0x. Since Mis a group, by simplifica-

tion we deduce that 0x= 0. The same method shows r0 = 0.b) 0 =r0 =r(x+(x))=rx+r(x). So (rx), the opposite of rx in

(M, +), is r(x). !

We define the natural notion ofsubmodule:

1.6 Definition. a) Let Mbe a leftR-module. A non-empty subsetLofMis called leftR-submodule of Mif

i)L is a subgroup in (M,+): x, yLxyL;

ii)rR, xLrxL.Usually one saysL is asubmoduleofMif no confusions can occur.

Notation: LR M (or, simpler, LM ). The fact that L is a rightR-submodule ofMis writtenLMR.

if one writes (x)u for the value ofu atx, then the composition ofu and v is defined as

(x)(uv)=((x)u)v. With this multiplication, End(M) is called the ring of rightendomorphisms ofMand it is the opposite of the ring of left endomorphisms ofM.


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II. Modules64

1.9 Proposition. Let(Mi)iI be a family of submodules ofRM. Thenthe intersection of this family, iIMi , is a submodule of M. !

This simple result allows us to define the notion ofsubmodulegenerated by a subset:

1.10 Definition. Let Mbe anR-module and letXbe a subset ofM.

a) The intersection of all submodules of M that include X is a

submodule ofM, called thesubmodule generated by Xand denoted by

R (or simply ). IfLR Mand =L, one says also thatXis asystem of generators forL(or thatXgeneratesL).

b) Define the set of linear combinations of elements in X withcoefficients in R as the setRX, where

RX:= {r1x1++rnxn|nN, r1, , rnR, x1, , xnX}.IfX=, defineR= {0}.

1.11 Proposition.Let M be an R-module and X M. Then:a) < X> is the smallest (inclusion-wise) submodule of M that in-

cludes X.

b) < X> =RX, that is,the submodule generated by X is the same asthe set of linear combinations of elements in X with coefficients in R.

Proof.a) Evidently, is a submodule and includesX. IfL is a

submodule in Mthat includesX, then LbecauseL is a mem-ber of the family of submodules the intersection of which isL.

b) We show thatXRXand thatRXis the smallest submodule thatincludesX. The case X= is trivial. IfX and xX, then x is a

linear combination, x= 1xRX. So XRX. The difference of twolinear combinations in RXand the product of any rR with a linearcombination is still inRX. So,RXis a submodule. IfL is a submodule

that includesX, 1.7.c) impliesRXL. !

1.12Definition. For any aRM, the submodule generated by {a}is Ra= {ra | rR} and it is also called the cyclic submodule


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generated by a. The R-module M is called finitely generatedif there

exists a finite system of generators for M, i.e. a finite subset FofM

such that =M.

While the intersection of a family of submodules is a submodule,

the union of a family of submodules is not a submodule in general.

1.13Definition. Let Mbe anR-module and letE, Fbe submodules

in M. The submodule generated by EF is called the sum of the

submodules EandF, and is denotedE+F. Thus,E+Fis just anothernotation for .

For an arbitrary family (Ei)iI of submodules ofM, the submodulegenerated by iIEi is called the sum of the family of submodules

(Ei)iI, denoted iIEi orIEi.The sum of the submodules E1, , En is denoted E1+ +En

or=

n

iiE

1

.

The sum of the family of submodules (Ei)iIis the smallest submod-

ule of M including all submodules Ei.

1.14 Proposition. a) If E, F are submodules ofRM, then the sum of

E and F is

E+F= {e+f | eE, fF},b) If E1, , En are submodules of M, then

E1++En= {e1++en | e1E1, , enEn}.Proof.a) Let S:= {e+f | eE, fF}. A straightforward verifica-

tion shows that Sis a submodule. IfL is a submodule containingEand

F, then e+fL, eE, fF. Thus, SL and S is the smallestsubmodule includingEF. !

In order to formulate a similar result for the case of the sum of an

arbitrary ( possibly infinite) family of submodules, we introduce thefollowing notion: for a setI(seen as a set of indexes),anR-module M


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II. Modules66

and a family of elements4(ei)iI, with eiM, iI, define the sup-

portof the family (denoted Supp((ei)iI)):Supp((ei)iI) := {iI| ei 0}

For any family (ei)iI having finite supportJI, itssum is definedas

iIei :=iJej.

1.15 Proposition. If(Ei)iIis a family of submodules of M, then

( ){ }, , having finite supporti i i i i Iii Ii I

e e E i I eE = =

={ }1 1 11, , , , , ,n n ni i n i i i ie e n i i I e E e E + + ' .

Proof. Let S= ( ){ }, , having finite supporti i i i i Ii I

e e E i I e

.As above, we show that S is a subm

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