A Course in Constructive Algebra (Universitext)

Unive rsitext Editors

F.W. Gehring P.R. Halmas

Universitext Editors: J. Ewing. F.W. Gehring. and P.R. Halmos Booss/Bleecker: Topology and Analysis Charlap: Bieberbach Groups and Flat Manifolds Chern: Complex Manifolds Without Potential Theory Chorin/Marsden: A Mathematical Introduction to Fluid Mechanics Cohn: A Classical lnvitation to Algebraic Numbers and CIass Fields Curtis: Matrix Groups. 2nd ed. van Dalen: Logic and Structurc Devlin: Fundamentals of Contemporary Set Theory Edwards: A Formal Background to Mathematics I alb Edwards: A Formal Background to Mathematics II alb Endler: Valuation Theory FrauenthaI: Mathematical Modeling in Epidemiology Gardiner: A First Course in Group Theory Godbillon: Dynamical Systems on Surfaces Greub: Multilinear Algebra Hermes: lntroduction to Mathematical Logic Humi/Miller: Second Order Ordinary Differential Equations Hurwitz/Kritikos: Lecturcs on Number Theory Kelly/Matthews: The Non-Euclidean. The Hyperbolic Plane Kostrikin: lntroduction to Algebra Luecking/Rubel: Complex Analysis: A Functional Analysis Approach Lu: Singularity Theory and an Introduction to Catastrophe Theory Marcus: Number Fields McCarthy: lntroduction to Arithmetical Functions Mines/Richman/Ruitenburg: A Course in Constructive Algebra Meyer: Essential Mathematics for Applied Fields Moise: Introductory Problem Course in Analysis and Topology 0ksendal: Stochastic Differential Equations Porter/Woods: Extensions of Hausdorff Spaces Rees: Notes on Geomctry ReiseI: Elcmentary Theory of Metric Spaces Rey: lntroduction to Robust and Quasi-Robust Statistical Methods Rickart: Natural Function Algcbras Smith: Power Series From a Computational Point of View Smorynski: Self-Reference and Modal Logic Stanisic: The Mathematical Theory of Turbulence Stroock: An lntroduction to the Theory of Large Deviations Sunder: An lnvitation to von Neumann Algebras Tolle: Optimization Methods

Ray Mines Fred Richman Wim Ruitenburg

A Course in Constructive Algebra

Springer-Verlag New York Berlin Heidelberg London Paris Tokyo

Ray Mincs Fred Richman Department 01 Malhcmatical SCiCIllTS New Mexicll Stale University Las Cruccs, NM XX003 U.S.A.

Wim Ruitenburg Department of Mathematics. Statistics, anel Computer Science Marljuette University Milwaukcc, Wl 5323,

U.S.A.

Llbrary "I Congrl~" Cataloging-in-Publicrtion Ibta Mines. Ra). A coup,c in CD!1')lructivc algehra.

(Universitc:xt i H ibliograpln' P Incluues Illlkx. I. Algebra I. Kidllllan. hcu. 11. Kultenbur,~. Willl. 111. Titk. IV. Tit!c: Construclivc algebra. QA I:;:i. 1"1).\ 1l)~X 512 X7-2h65X

'( ILJXX by Springer- Verlag New York Ine All right> rcserveu. This wnrk may not be transla\cu ur (opied in whole ur in pari withou\ the written perlllission 01' the puhlisher (Springer-Verlag, 175 Fifth Avenue. New York. N"( 10010. USA). except I,n brief excerp\s in eonnection with reviews or scholarly analysis. Lse in c()nneLtion with any form "I inlorlllation storage and retrievaL e1ectronic adaptation. Clll1lputer software. or by slinilar or di"imilar lllcthodlliogy IHlW known or hereaftcr developeu is forbiuuen. 'fhe liSt' 01' general descriptive names, tralk names. trademarks, ete. in Ihis publicatioll, cyen iI Ihc former are not especially identified. is not 10 oe taken as a sign Ihat such names, as lInderstood hy the Trade Marks allu Merehulldise Marks Act. nray accordingly be lIscd fredy by anyonc. Text prcparcd by thc authms using T" software output on a LN03 printer. Printed and bound by Quinn-Woodbine (ne. Woodbinc. Ncw Jersey Printed in the Uniled States ur Ameriea. LJ X 7 h 5

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3 2 I

ISHN O-JX7-%64ll-4 Springer-Verlag Ncw York Berlin Heidelberg ISBN 3-540-lJ6640-4 Springer-Verlag Berlin Heidelberg New York

Dedicaled

10

Errett Bishop

Preface

The constructive approach to mathematics has enjoyed a renaissance, caused in large part by the appearance of Errett Bishop's book Foundations of constructiue analysis in 1967, and by the subtle influences of the proliferation of powerful computers. Bishop demonstrated that pure mathematics can be developed from a constructive point of view while maintaining a continuity with classical terminology and spirit; much more of classical mathematics was preserved than had been thought possible, and no classically false theorems resulted, as had been the ca se in other constructive schools such as intuitionism and Russian constructivism.

The

computers created a widespread awareness of the intui ti ve notion of an effective procedure, and of computation in principle, in addition to stimulating the study of constructive algebra for actual implementation, and from the point of view of recursive function theory. In analysis, constructive problems arise instantly because we must start with the real numbers, and there is no finite procedure for deciding whether two given real numbers are equal or not (the real numbers are not discrete) . The main thrust of constructi ve mathematics was in the direction of analysis, although several mathematicians, including Kronecker and van der waerden, made important contributions to constructive algebra. Heyting, working in intuitionistic algebra, concentrated on issues raised by considering algebraic structures over the real numbers, and so developed a handmaiden'of analysis rather than a theory of discrete algebraic structures. paradoxically, it is in algebra where we are most likely to meet up with wildly nonconstructive arguments such as those that establish the existence of maximal ideals, and the existence of more than two automorphisms of the field of complex numbers. In this book we present the basic notions of modern algebra from a constructive point of view. The more advanced topics have been dictated by our preferences and limitations, and by the availability of constructive treatments in the literature. Although the book is, of vii

viii

Preface

necessity, somewhat self-contained, it is not meant as a first introduction to modern algebra; the reader is presumed to have some familiarity with the classical subject. It is important to keep in mind that constructive algebra is algebra; in fact it is a generalization of algebra in that we do not assume the law of excluded middle, just as group theory is a generalization of abelian group theory in that the commutative law is not assumed. A constructive proof of a theorem is, in particular, a proof of that theorem. Every theorem in this book can be understood as referring to the conventional universe of mathematical discourse, and the proofs are acceptable within that universe (barring mistakes). We do not limit ourselves to a restricted class of 'constructive objects', as recursive function theorists do, nor do we introduce classically false principles, as the intuitionists do. We wish to express our appreciation to A. Seidenberg, Gabriel Stolzenberg, Larry Hughes, Bill Julian, and steve Merrin for their suggestions.

Ray Mines Fred Richman New Mex(co State Uniuersity

Wim Ruitenburg Marquette Univer'sity

Contents

0IAP1'ER I. SETS

1. 2. 3. 4. 5. 6.

Construetive vs. e1assiea1 mathematies Sets, subsets and funetions Choiee Categories partia11y ordered sets and 1attiees We11-founded sets and ordinals Notes

1

7

14 16 20

24

30

0IAP1'ER II. BASIC ALGEBRA

1. 2. 3. 4. 5. 6. 7. 8.

Groups Rings and fields Real numbers Modules polynomial rings Matriees and veetor spaees Determinants Symmetrie polynomials Notes

0IAP1'ER 111. RIIGS

1. 2. 3. 4. 5. 6. 7.

AN!)

35

41 48

52

60

65

69 73 77

l'DIXJLES

Quasi-regular ideals Coherent and Noetherian modules Loealization Tensor produets Flat modules Loeal rings Commutative loeal rings Notes

78 80

85

88 92 96 102 107

0IAP1'ER IV. DIVISIBILITY IN DISCRETE DOMAINS

1. 2. 3. 4.

Caneellation monoids UFD's and Bezout domains Dedekind-Hasse rings and Euelidean domains Polynomial rings Notes

ix

108

114 117 123 126

Contents

x CHAPI'ER V. PRINCIPAL IDEAL lXI'IAINS

1. 2. 3. 4.

Diagonalizing matrices Finitely presented modules Torsion modules, p-components, elementary divisors Linear transformations Notes

128 130 133 135 138

CHAPI'ER VI. FIELD THEORY

1. 2. 3. 4. 5. 6. 7. 8.

Integral extensions and impotent rings Algebraic independence and transcendence bases Splitting fields and algebraic c10sures Separability and diagonalizability Primitive elements Separability and characteristic p Perfect fields Galois theory Notes

139 145 150 154 158 161 164 167 175

CHAPI'ER VII. FAC'IDRING POLYNaoUALS

1. 2. 3. 4.

Factorial and separably factorial fields Extensions of (separably) factorial fields condition P The fundamental theorem of algebra Notes

176 182 186 189 192

CHAPI'ER VIII. CCtIlMUTATIVE NJE'IHERIAN RINGS

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

The Hilbert basis theorem Noether normalization and the Artin-Rees lemma The Nullstellensatz Tennenbaum's approach to the Hilbert basis theorem Primary ideals Localization Primary decomposition Lasker-Noether rings Fully Lasker-Noether rings The principal ideal theorem Notes

193 197 201 204 208 211 216 220 224 228 231

CHAPI'ER IX. FINITE DIMENSICNAL ALGEBRAS

1. 2. 3. 4. 5.

Representations The density theorem The radical and summands Wedderburn's theorem, part one Matrix rings and division algebras Notes

232 235 237 242 245 248

xi

Contents CBAPl'ER X. FREE GRaJPS

1. Existence and uniqueness 2. Nielsen sets 3. Finitely generated subgroups 4. Detachable subgroups of finite-rank free groups 5. Conjugate subgroups Notes

249 253 255 257 261 263

CBAPl'ER XI. ABELIAN GRaJPS

1. Finite-rank torsion-free groups 2. Divisible groups 3. Height functions on p-groups 4. Ulm's theorem 5. Construction of Ulm groups Notes

265 269 273 277

281 285

CBAPl'ER XII. VALUATICl'i TBEDRY

1. Valuations

2. 3. 4. 5. 6. 7. 8.

Locally precompact valuations Pseudofactorial fields Normed vector spaces Real and complex fields Hensel's lemma Extensions of valuations e and f Notes

287 292 295 299 302 306 315 319 324

CBAPl'ER XIII. DEDEKIND IXEAINS

1. Dedekind sets of valuations 2. Ideal theory 3. Finite extensions

326 329 332

BIBLIOGRAPHY

335

INDEX

339

Chapter I. Sets

1. crnsTRUCTIVE VS. CLASSlCAL MATHEMATICS

The classical view of mathematics is essentially descriptive: we try to describe the facts about a static mathematical universe. Thus, for exarople, we report that every polynomial of odd degree has a root, and that there is a digit that occurs infinitely often in the decimal expansion of

In opposition to this is the constructive view of

Ir.

mathematics, which focuses attention on the dynamic interaction of the individual with the mathematical universe; in the words of Hao wang, it is a mathematics of doing, rather than a mathematics of being. The constructive mathematician must show how to construct a root of a polynomial of odd degree, and how to find a digit that occurs infinitely often in the decimal expansion of Ir. We

picture

an

idealized

mathematician

U

interacting

wi th

the

mathematical universe; this is the "you" who finds the 6, and to whom the is given, when we say "given E you can find 0." The phrases "there exists" and "you can find" mean that U can carry out the indicated

E

constructions. The disjunction of two statements "P 1 or P2" means that either Pi is true or P 2 is true, and that U can determine which of these alternatives holds. As "Pi or P2 " means that there exists i in {l,2} such that Pi is true, the meaning of "or" can be derived from the meaning of "there exists", and it is the interpretation of this latter phrase that is fundamental to constructive mathematics. Classical mathematics can also be encompassed by this picture; the difference lies in what powers we ascribe to U. An omniscient U can decide whether any given mathematical statement is true or false; so U can, for exarople, survey the decimal expansion of Ir and determine which digits appear infinitely often. With an omniscient U, our picture is just a more dynamic, anthropomorphic portrayal of classical mathematics. In constructive mathematics we assume only that U can carry out 1

2

Chapter I. Sets

constmctions that are finite in nature. As Errett Bishop put it, "The only way to show that an object exists is to give a finite routine for finding it."

In this setting, we are not entitled to say that some digit

appears infinitely often in the decimal expansion of

11"

until we are

prepared to exhibit such a digit, or at least produce an algorithm that will compute such a digit. We consider U to be capable of carrying out any finite construction that is specified by an algorithm, but we do not rule out the possibility that U can do other things--even that U might be omniscient. The picture that

results

when

we

restrict

U to

finite

computational interpretation of mathematics.

constructions

is

the

Because any statement that

admits a constructive proof is true under the computational interpretation, we say that constructive mathematics has numerical meaning; because any statement that admits a constructive proof is true under the classical interpretation, we say that constructive mathematics is a generalization of classical mathematics. Constructive mathematics is pure mathematics done algorithmically in order to respect the computational interpretation.

The central notion of

a finite routine, or algorithm, is taken as primitive. My attempt to define what an algorithm is ultimately involves an appeal to the notion of existence--for example, we might demand that there exist a step at which a certain computer program produces an output. If the term "exist" is used here in the classical sense, then we have failed to capture the constructive notion of an algorithm; if it is used in the constructive sense, then the definition is circular. Consider the difference between the constructive and the classical use of the connective "or". In order to prove "Pt or Pz " constmctively, we must construct an algorithm that will either prove Pt or prove P2 , and by executing that algori thm we (the idealized mathematicianl can determine which is true. To prove "Pt or P2 " classically, it suffices to show that Pt and P 2 cannot both be false. For example, let Pt be the statement: there exist positive integer-s x, !J, z, ami n such that x n + 2 + !Jn+2 = zn+2,

and let P 2 , the famous unproved theorem of Fermat, be the denial of Pt. If Pt is false then P 2 is tme, so "Pt or P z " is classically provable; but

1. Constructive vs. c1assica1 mathematics

3

we don't know, at this time, how to determine which of P, or P2 ho1ds, so we do not yet have a constructi ve proof of "P, or P2" • A constructive proof of a theorem proves more than a c1assica1 one: a constructive proof that a sequence of real numbers converges imp1ies that we can compute the rate of convergence; a constructive proof that a vector space is finite dimensional implies that we can construct a basis; a constructive

proof

that

a

po1ynomia1

is

a

product

of

irreducib1e

polynomials imp1ies that we can construct those irreducib1e polynomials. 'IWo statements P and

Q

may be c1assically equiva1ent without being

constructive1y equiva1ent. Let P be the statement that every subgroup of the additive group ~ of integers is cyc1ic. This means that we can a1ways produce a generator from the da ta specifying the subgroup. Let Q be the statement that no subgroup e of ~ can have the property that for each m in

e, there is an integer in e that is not a multiple of m. The statements P and Q are

immediate1y

equiva1ent

c1assica11y,

but

quite

different

constructive1y. The statement Q is true: as 0 is in e, there must be a nonzero integer n in e; as n is in e, there must be an integer proper1y dividing n in e, and so on unti1 we arrive at a contradiction. But P is un1ike1y to be true, as we can see by considering the subgroup

e generated

by the perfect numbers: to construct a generator of e we must construct an odd perfect number or show that all perfect numbers are even. On the other hand, any two constructive1y equiva1ent statements are c1assica11y equiva1ent; indeed any theorem in constructive mathematics is also a theorem in c1assica1 mathematics: a constructive proof is a proof. Suppose we are trying to find a constructive proof of a statement P which is c1assica11y true. After many unsuccessfu1 attempts to prove P, we may be inclined to look for a counterexamp1e. But we cannot hope to prove the denia1 ,p of P, which is what a bonn fide counterexamp1e wou1d entai1, because ,p is c1assica11y fa1se. As this avenue is closed to us, we need some other alternative to persisting in trying to prove P. One approach is to fix a formal language in which P is expressible, specify precise1y what sequences of words in that language consti tute proofs of P, and show that no such proof can be constructed (possibly by giving an unintended interpretation of the formal language, and showing that P is fa1se in that interpretation). Such a program can be il1uminating, but it is often doubtful that the formal system adequate1y reflects the informal mathematics. A more serious objection is that

Chapter I. Sets

4

engaging in such independence arguments requires a drastic shift in point of view; a procedure that stays closer to the subject at hand would be preferable. To this end we introduce the idea of an omniscience principle and of a Brouwerian example. A rule a that assigns to each positive integer n an element an of {O,l) is called a binary sequence. An omniscience principle is a classically true statement of the form "P (o) for all binary sequences 0" which is considered not to have a constructive proof. For example, from the classical point of view, for each binary sequence a either {P}

or

(Q)

there exists n such that on = 1, an = 0 for each posi ti ve integer n.

The assertion that either P or Q holds for any binary sequence a is called the limited principle of omniscience (LPO). As Q is the negation of P, the limited principle of omniscience is a form of the law of excluded middle: the assertion that for any statement P, the statement "p or not P" holds. The limited principle of omniscience and, a forti01'i, the law of excluded middle, is rejected in the constructive approach as no one seriously believes that we can construct an algori thm that, gi yen 0, chooses the correct alternative, P or Q. Another argument against LPO is that i f we restrict our idealized mathematician to specific kinds of algorithms, as is done in the Russian school of constructive mathematics, we can show that LPO is false. Fix a computer programming language capable of expressing the usual number theoretic functions and symbol manipulations. It can be shown that there is no computer program that accepts computer programs as inputs, and when applied to a program that computes a binary sequence, returns 1 if the sequence contains a 1, and returns 0 otherwise. So if we require our rules to be given by computer programs, then LPO is demonstrably false. This argues against accepting LPO because any informal algorithm that we use will undoubtedly be programmable, so our theorems will be true in this computer-program intepretation; but we do not restrict ourselves to computer programs lest we rule out the possibility of intepreting our theorems classically: indeed, LPO is classically true. If a statement P can be shown to imply LPO, then we abandon the search for a constructive proof of P. We do not assert that P is false, as we denied the existence of the computer program in the preceding paragraph;

1. Constructive vs. classical mathematics

5

after all, P may admit a classical proof. Rather, statements like LPO are thought of as independent in the sense that neither they nor their denials are valid. For exarnple, consider the classically true statement P that every subset of the integers is either empty or contains an element. Given a binary sequence a, let A = {1} and let B = {an: nE rn}. Then A n B is a subset of the integers. If A n B contains an element, then that element must be 1, so, from the definition of B, there exists Tl such that an = 1; if A nB is empty, then an = 0 for all n. Thus if P ho1ds, then so does LPO. A weaker omniscience principle is the lesser limited principle of omniscience (LLPO), which states that for any binary sequence a that contains at most one 1, either an = 0 for all odd n, or an = 0 for all even Tl. This implies that, given any binary sequence, we can tell whether the first occurrence of a 1, if any, occurs at an even or at an odd index n. As in the case of LPO, if we restrict ourselves to rules given by computer programs, then we can refute LLPO. If you think of a as a b1ack box into which you put n and get out an' then it is fairly clear that you cannot hope to establish LPO or LLPO. Or consider the sequence a defined by: if, and only if, there are 100 consecutive 6's in the first n places of the decimal expansion of ~.

1

I

As

~/4

if, and only if, there are 100 consecutive 7's in the first n places of the decimal expansion of ~.

= 1 - 1/3

+ 1/5 - 1;7 + 1/9 - ••• ,

there

is

an

algorithm

for

computing the terms an. But unless we chance upon 100 consecutive 6's or 7's, we are hard put to find an algorithm that will tell us the parity of the index n for which an = 1 for the fi rst time (if ever). A Brouwerian example E is a construction E (a) based on an arbi trary binary sequence a. We say that a Brouwerian example E satisfies a condi tion C if E (a) satisfies C for each ai we say that E does not satisfy a condition C if there is an omniscience principle "P(a) for all a" such that whenever E(a) satisfies C, then P(a) holds. A Brouwerian counterexample to a statement of the form "C, implies C2 " is a Brouwerian exarnple that satisfies C, but does not satisfy C2 • OUr construction A n B above is a Brouwerian exarnple of a subset of the

6

Chapter I. Sets

integers that neither contains an element nor is empty.

We now construct

a Brouwerian example of a bounded increasing sequence of real numbers that does not have aleast upper bound.

For each binary sequence 0' let E(a) be

the sequence ß of real numbers such that (in = sup'; =la i bounded increasing sequence of real numbers.

.

Then E (0') is a

Let C be the condition, on a

sequence of real numbers, timt it has aleast upper b01.md.

We shall show

Let P(O') be the property that eUher an = 0

that E does not satisfy C.

for all ", or there exists " such that

0'"

= 1, and suppose E (0') satisfies

C.

If the least upper bound of E (0') is less than 1, then an = 0 for all

n.

If the least upper bound of E(a) is positive, then there exists n such

that an > 0, and hence an = 1.

Thus P (,,) holds.

EXERCISES 1. Show that LPO imp1ies LLPO. 2. Each subset of

{O, I}

has 0,

1 or

2 elements.

Construct a

Brouwerian counterexample to that statement. 3. Construct a Brouwerian example of a nonempty set of positive integers that does not contain a smallest element. 4. Construct a Brouwerian example of a subgroup of the additive group of int.egers that is not cyclic. 5. Construct a Brouwerian example of t.wo binary sequences whose sum contains

infinitely

many

l' s,

yet

neither

of

the

original

sequences does. 6. Call a st.at.ement simply existential i f it is of the form "there exists

Tl

such that an = 1" for some binary sequence o.

Show that

LLPO is equivalent to "",(A

and

B) i f

and only if

~A

or

~B

for each pair of simply existential statements A and B. 7. The weak limited principle of OIIlI1iscience (WLPO) is the statement that for each binary sequence

(l

ei t.her

impossible that a" = 0 for a11 ".

u'"' =

0 for all n or i t is

Show that LPO implies WLPO,

and that WLPO implies LLPO. 8. Let S be the set of a11 fini te sequences of posi ti ve int.egers.

By a finitary tree we mean a subset T of S such that

1. Constructive vs. classical mathematics (i) For each

7

S, either sET or s (

s €

T,

(U) If (xl'''''x n ) E T, then (x1"",x n _1) E T, (iii) For each (x1""'x n ) E T, there is m Ern such that if

(x1"'''x n ,z) An infinite path in T

such that (x l' ... ,x n )

E T,

then z S; m.

is a sequence {xi) of positive integers

E T

for each n.

I!Wnig' s lellllla states that

i f T is infinite (has arbitrarily large finite subsets) , then T

Show that König's lemma implies LLPO.

has an infinite path.

2. SETS, SUBSETS AND FUNCTICNS

We deal with two sorts of collections of mathematical objects: sets and eategories.

OUr notion of what constitutes a set is a rather liberal one.

2.1 DEFINITlOO.

A set S is deftned when we descrLbe how to construct

L ts member's from objecLs timt have been,

01'

could iwve been, constn!cted

prior' to S, and descr'tbe what it means for' two membETs oF S to be equaL.

Following

Bishop

we

regard

the

equality

relation

on

a

set

as

conventional: something to be deterrnined when the set is defined, subject only to the requirement that it be an equivalence relation, that is reflexive:

a = a.

synmetric:

If

transitive:

If o.



B to indicate

Two functions fand 9 from A to Bare The identity function f : A -> A is

for each a in A.

10

Chapter I. Sets To constcuct a function from A to H it suffices to construct a subset S

of the Cartesian product A (i) for each a

B with the properties

x

A there exists b ERsuch that (a ,b) ES,

E

= b2

(ii) if (a,b , ) and (a,/),,) are elements of S, then b,

•

In the computational interpretation, the algorithm for the function comes from (i), which specifies the construction of an element b depending on the parameter

(1.

wi thout (ii), however, the algori.t.hm implici t

need not be extensional.

in (i)

The fact that a subset S ofAx B satisfying (i)

and (ii) determines a function f such that (o,f

(a)) E

S for each

0

in A is

known as the axiom of unique choice. Let

r

We say that f is

be a function from A to B. one-to-one if 0,

=

a 2 whenever ((u,)

= f(02)'

onto i f for each h in B there exists CI in A such that f (1) strongly extensional if

(1,

t

Qc

whenever F (ll,) t f (Cl 2

b,

) •

Note that any function between sets with denial inequalities is strongly extensional.

If S

~

F(S)

A, then the image of S under F is the set

= {b

E

B : b = F(a) for some"

Thus f is onto if and only i f

r (A)

=

B.

If S

~

E A).

13, then the preimage of S

under f is the set F···1(S)

=

{Cl (

A : ((a) ES).

Two sets A and 13 have the same cardinality if there are functions f from A to B, and 9 from B to A, such that Fg is the identity function on B and gf is the identity function on A; we say that the functions fand 9 are inverses of each other, and that each is a bijection. the same cardinali ty,

we wri te #A = #B.

If A and B have

The axiom of unique choice

implies that a one-to-one onto flillction is a bijection

(Exercise 6).

Classically we think of sets of the same cardinality as simply having the same size; constructively it is more accurate to think of them as having the same s t ruc t twe . When we refer to the cardinality of a set we mean the set itself, ignoring any structure other than equali ty that

it

might

have.

The

distinction between referring to a set and referring to i ts cardinali ty is primarily one of intent: when we refer to the cardinality of a set, we do not plan on paying attention to any characteristics of the set that are not shared by aU sets with the same cardinali.ty.

For example, if x is an

2. Sets, subsets and functions

11

element of a group, then the set S = {I, x, x 2 , x " , ••• } is the submonoid generated by x, while the cardinality of S is the order of x. It is much the same distinction as that between a fraction and a rational number. A common device for dealing with this kind situation is to introduce equivaLence cLasses but, following Bishop (1967), we prefer to deal with the equivalence directly and not introduce cumbersome new entities. If a set A has the same cardinality as {l, ... ,n} (empty if n = 0) for some nonnegative integer n, then we say that A is an n-element set, or A has cardinality Tl, and write #A = n. A finite set A is a discrete set which has cardinali ty n for some nonnegati ve integer n. Recall that a discrete set must be discrete in its specified inequality, if any, so a set can have finite cardinality without being finite; such sets are somewhat pathological, which is why we give them the longer name. A set A is finitely enumerable if it is empty or there is a function from {l, ••. ,n} onto A. Note that a finitely enumerable set is discrete if and only i f it is finite. We say that A has at most n elements i f whenever aO, •.• ,an E A, then there exist 0 ~ i < j ~ n such that a i = a j . A set is bounded in number, or bounded, if it has at most n elements for some n. A set is infinite i f it contains arbitrarily large finite subsets. A set A is countable if there is a function from a detachable subset of the positive integers onto A. Thus the empty set is countable, as is the set of odd perfect numbers. Nonempty countable sets are the ranges of functions from the positive integers, so their elements can be enumerated (possibly with repetitions) as a"a 2 , • • • • A seqUence of elements of a set A, or a sequence in A, is a function from the nonnegative integers IN to A. We shall also speak of functions from the positive integers as being sequences. A family of elements of A, indexed by a set I, is a function F from I to A; the image of i in A under f is usually denoted f i rather than f (i ) . Thus a sequence is a family indexed by the nonnegative integers IN. A finite family of elements of A is a family of elements of A indexed by {l, ... ,n} for some positive integer n. If {Ai}iEI is a family of subsets of S, then its union is defined by UiEIA i {x E S there exists E I such that x E Ai}' and its intersection is defined by niEIA i = {x E S x E Ai for all i EI}. If S is a set with an inequality, and X is a set, then the set SX of

12

Chapter I. Sets

functions from X to S inherits an inequality from there exists x in X such that f(x) ~ g(x). 2.2 'lHOOREM.

S

by setting f

if

~ g

If

Le t S be u se t with i ne qua li ty und Let X be u se t .

the inequality on S is consistent, symmetric, cotrunsitive, or ti.ght. then so, respectiveLy, is the inequality on SX.

PROOF.

Consistency

and

symmetry

are

clear.

Suppose

that

the

inequality on S is tight. If f, i f 2 is impossible, then there cannot exist x such that f, (x) ~ f 2 (x). Thus , given x I it is impossible that f,(x) ~ f2(X), whence f,(x) = f 2 (x) for each x, 50 f, = f 2 • Now suppose the inequality on S is cotransitive. If f, ~ f 3 1 then for some x we have f,(x) i f 3 (x) so either F,(x) ~ f2(X) or f 2 (x) ~ F3 (x) whereupon either

f,

~

f 2 or f2

F3

~

•

0

As an example of Theorem 2.2, take S to be the discrete set {O,l} and X to be the nonnegative integers. Then SX is the set of binary sequences. As {O,l} is discrete, the inequality on {O,l} is a consistent tight apartness, so the inequality on the set of binary sequences is also a consistent tight apartness. However if the set of binary sequences were discrete , then we could establish LPO. EXERCISES 1. Give an example (not Brouwerian) of a consistent apartness that is not tight. 2. Show that the set of binary sequences is discrete if and only if LPO holds. 3. A denial inequality that is not an apartness. Let A be the set of binary sequences. For x and y in Adefine x = y i f there exists N such that x n = Yn for all n ~ N, and let x ~ y be the denial inequality. Show that if this inequality is an apartness, then WLPO holds; show that if it is a tight apartness, then LPO holds. 4. A notty problem. A difference relation is a symmetric inequality such that any of the following three conditions holds: (i) x

~

(ii) • x

z implies .(. x ~ y and· y ~ y and • y ~ z implies • x

~

z

~

z

13

2. Sets, subsets and functions (iii) x

~

z and

~

x

y implies

~

~~

y

z.

~

Show that these conditions are equivalent, and that an apartness is a difference relation. 5. Define a natural tight apartness on the set of detachable subsets of a set.

Show that a subset A of a set S is detachable if and

only if it has a characteristic function, that is, a function f from S to {O,l} such that

A=

{s

ES:

= I).

{(sl

6. Show that a function is a bijection i f and only i f it is one-·toone and onto. 7. show

that

a

fini tely

enumerable

discrete

set

is

finite.

Construct a Brouwerian example of a finitely enumerable set, with a tight apartness, that is not finite. enumerable set is bounded in number.

Show that a finitely Construct a Brouwerian

example of a set that is bounded in number but is not finitely enumerable. 8. Show that a nonempty set A is countable if and only if there exists a function from IN onto A.

Show that a discrete set is

countable i f and only i f it has the same cardinali ty as a detachable subset of IN. 9. Show that a subset A of IN is countable if and only if there is a detachable subset

S

of IN x IN such that A

= lTS,

where

lT

is the

projection of IN x IN on its first component. 10. Show that the set of functions from a discrete bounded set A to (O,l) need not be discrete.

(Hint:

Let A be the range of a

binary sequence1 11. Show that if a set S is bounded in number, then any one-to-one function from S to S is onto. 12. Give a Brouwerian example of a subset A of IN such that it is

contradictory for A to be finite, but A is not infinite. 13. Let S be a nonempty set with an apartness,

integer.

and n a positive

Show that the following are equivalent. (i) There exist elements xO"'. ,x n in S such that xi for

i

-je

.i.

~

xj

14

Chapter I. Sets Given Y1""'Yn in S, there exists z in S such that

(ü)

z

Yi

je

for i = 1, ... ,n.

14. A set with inequa1ity is Dedekind infinite if it is isomorphie, as a set wi th inequa1ity, to a proper subset .

Show that any

Dedekind-infini te set satisfies (i) of Exercise 13 for each n. 15. call a set S w-bounded i f for each sequence (si) exists m t n such that sm

=

s,,'

in S, there

Show that i f S is a discrete

ul-bounded set, and (si) is a sequence in S, and m is a positive integer, then there is a finite set I of m positive integers such that si

=

S

j

if i

and j

are in 1.

Show that i f A and Bare

discrete ld-bounded sets, then so is A x B. 3. crIOICE

The axiom of choice asserts the existence of a function with a certain property, so we might weIl suspect that its validity wou1d be more dubious in

a

constructive

a1gorithms.

AXIOM OF CHOlCE.

Fo,- each a

setting

where

functions

may

be

interpreted

as

We phrase the axiom of choice as folIows: Le t A mill B

be se t s, (md S

in A U,e,-e is a" element b

[n

Cl

subse t ofAxB SllCh tha t

B such that

(a,b) E S.

TI",n

Ulere is a Function F:A ->B such that (a.((a)) (S For- eaell a in A. The axiom of choice can be criticized from the computational point of view on two grounds.

The first concerns whether we can find an aLgor-i thm

F (not necessarily extensional) with the

required property.

We have

a1ready come across this issue with the axiom of unique choice, and we take the position that the algorithm is inherent in the interpretation of the phrase "for each a in A there is an element b in B".

A more serious criticism is that, although we can find an algorithm we

cannot

find

a

In

Fun,tion.

fact

we

can construct a

F,

Brouwerian

counterexample to the axiom of choice. 3 .1 EXAMPLE.

Let a be

Cl

binor!)

sequence,

l"t

A = (x,!))

with the

equaUt!) on A deFined b!) setting x = 1J iF and onl!) iF the,-e exists n SUdl that an AxB.

=

1, ond let B

Suppose F

F (Y), then an

A ->

=

{O,l}.

Cansid'T tll" s11hset S

=

{(x,O). (!),1J) oF

B satis(les (a,F(a)) ES For each a E A.

1 For- same n; If f(xl t f(y),

then

an

=

IF f(x)

0 for an n.

15

3. Choice

There are two restricted versions of the axiom of choice which are commonly accepted in constructive mathematics. CCUNTABLE AXICI'l OF CHOlCE.

The weaker of these is:

This is the axiom of ehoiee wi th A being

the set of post t ive tnteger-s.

If A is the set of positive integers, then there is no real distinction between an algorithm and a function in the computational interpretation, as each integer has a canonical representation. from the interpretation of the phrase "for all

So this axiom follows a

there exists b"

as

entailing the existence of an algorithm for transforming elements of A to elements of B. Even stronger than the countable axiom of choice is: AXICI'l OF DEPENDENl' CHOlCES.

A be a nonempty set and R a subset of

Let

AxA such that for- eaeh a in A there ls an eLement a' in A wUh (a,a') ER. Then

there

a

is

eLements

of

of

A

such

that

(al,ai+l) ER for each i.

The axiom of dependent choices implies the countable axiom of choice as folIows.

Suppose S is a subset of

is an element b

in B with

~

x B such that for each n in

(n,b) ES.

Let A consist of all

~

there finite

sequences b O,b l , ... ,bm in B such that (i ,bi) E S for each i, and let R consist of all paiI:s (a,a') of elements of A such that deleting the last element of

a'

gives

a,

Applying the axiom of dependent choices to R

yields a sequence in A whose last elements form the required sequence in

B. The argument for the axiom of dependent choices is l\Ulch the same as that for the countable axiom of choice.

We shall freely employ these

axioms, although we will often point out when they are used. We shall have occasion to refer to the following axiom of choice for which we have no Brouwerian counterexarnple,

yet which we

be l i eve

is

unprovable within the context of constructive mathematics. WORLD'S SIMPLEST AXICI'l OF CHOlCE. such that if a, E A and o? E A, fr-am A to

Ix :

then

Let A be 0,

=

O2 ,

0

set of

two-eLement sets

Then ther-e is a function f

xE a far" some 0 E Al such thot ((0) E 0 for- eaeh a E A.

16

Chapter 1. Sets EXERCISES 1. Modify (3.1) to show that the axiom of choice implies the law of excluded middle. 2. Show that LLPO, together with dependent choice, implies König's lemma (see Exercise 1.7). 3. Show that the axiom of choice implies the world's simplest axiom of choice. 4. A set P is projective if whenever

lT

:

A ..... B

f : P ..... B, then there exists 9 : P .... A such that

is onto,

and

Show that finite sets are projective. Show that countable discrete sets are projective if and only if the countable axiom of choice holds. Show that if discrete sets are projective, then the lTg

=

f.

world's simplest axiom of choice holds. 4. CATEmRIES.

The collection of binary sequences forms a set because we know what it means for two binary sequences to be equa1. Given two groups, or sets, on the other hand, it is generally incorrect to ask if they are equal; the proper question is whether or not they are isomorphie, or, more generally, what are the homomorphisms between them. A category, like a set, is a collection of objects.

An equality relation on a set constructs, given any two objects a and b in the set, a proposition 'a = b'. To specify a category 'C, we must show how to construct, given any two objects A and B in~, a set ~(A,B). In concrete

categories, the objects of ~ are mathematical structures of some kind, and the set ~(A,B) is the set of maps from A to B that respect that structure: if

to

~

is the category of sets, then 'C(A,B) is the set of functions from A if ~ is the category of groups, then ~(A,B) is the set of

B;

homomorphisms from A to B. When we abstract the notion of composition of maps from these concrete situations, we see that a category must have, for any three objects A,B,C in ~, a function from ~(A,B) x 'C(B,C) to 'C(A,C), called composition and denoted by juxtaposition, and an element 1B E 'C(B,B), such that i f f E ~(C,D), 9 E ~(B,C), and h E ~(A,B), then

4. Categories

17

(i) I Bh = hand gIB (ii) (fg)h = f(gh)

g.

Every set S can be considered a category by setting S(a,b)

{x

E

{Ol : a

=

bl.

The reflexive law for equality gives the element IB' and the transitive law gives (ii). The sets and functions introduced in the previous section constitute a category: the objects of this category are sets, and the maps are the We may also consider the category whose functions between the sets. objects are sets with inequality, and whose maps are strongly extensional functions. The idea of category theory is to forget about the internal structure of the objects and to concentrate on the way the maps combine under composition. For example, a function f from A to B is one-to-one if a l = a 2 whenever f(a t ) = f(a 2 ) . This definition relies on the internal structure of the sets A and B, that is to say, on the elements of A and B and the equality relations on A and B. The categorical property corresponding to a function f being one-to-one is that if 9 and h are maps from any set C to A, and fg = fh, then 9 = hi that is, F is left cancellable. It is routine to show that F is one-to-one if and only if it is left cancellable. A map F from A to B is onto if for each b in B there exists a in A such that F(a) = b. The corresponding categorical property is that f be right cancellable, that is, i f 9 and h are maps from B to any set C, and gF = hf, then 9 = h. The proof that a function F is right cancellable i f and only if it is onto is less routine than the proof of the corresponding result for left cancellable maps. 4.1 'lHOOREM.

A function is r-ight canceUable in the category oF sets

iF and only iF it is onto.

PROOF. Suppose f:A --> B is onto and gf = hf. If bEB, then there exists a in A such that f(a) = b. Thus g(b) = g(f(a)) = h(f(a)) = h(b), so 9 = h. Conversely suppose f:A --> B is right cancellable, and let n be the set of all subsets of {Ol. Define g:B --> n by g(b) = {Ol for all b, and define h:B --> n by h(b)

Thus

h(b)

= {x E

{Ol : b

=

f(a)

for some al.

is the subset of {Ol such that 0

E

h(b) i f and only i f there

18

Chapter I. Sets

exists a such that b

f (a ) .

=

C1early gf

I Ol.

element of A to the subset that b

=

for some a.

f(a)

So 9

=

hf is the map that takes every

=

h, whence 0

E

h (b), which means

0

isomorphism between two objects A and B of a category 'C is an

An

element f E 'C(A,B) such that there is 9 E 'G(B,A) with fg = IB and gf = 1A. The element 9 is called the inverse of fi it is easily shown to be unique.

A bijection between sets is an isomorphism in the category of sets.

We

say that A and B are

an

isomorphie,

and write A

~

B,

there

if

is

isomorphism between A and B. We will be interested mainly in categories of sets with algebraic structures,

in which the maps are

structures.

In this case,

the

functions

that preserve

the maps are called homomorphisms.

those If a

homomorphism is one-to-one, it is called a monomorphismi if it is onto, it is called an epimorphism.

A homomorphism from an object

to itself is

called an endomorphism, and an endomorphism that is an isomorphism is called an automorphism.

A funetor T from a category

to a category

~

~

is a rule that assigns to

each object A E r.4 an object T(A) E ,'11, and to each map f T(F)

E

E

d(A"A 2

)

a map

such that

~(T(A, ),T(A 2 ))

(i) T : "i(A"A 2

)

-+~7!(T(A,),T(A2))

is a function

(ii) T(Fg) = T(F)T(g)

(iii) T(lA)

=

ly(A)'

A functor between two sets, viewed as categories,

is simply a function.

Note that if f is an isomorphism, then so is T(F). Using the notion of a funetor, we can extend our definition of a family of elements of a set to a family of objects in a eategory 'Co set.

A family A of objeets of

as a category,

IAiliEI'

If i

'(I

to the category =

Let I be a

indexed by I is a functor trom I, viewed 'C.

We often denote such a family by

j, then the map from Ai to Aj is denoted by

A), and is an

i somorphi sm. An element of the disjoint union of a family (i ,x)

such that i

E

I and x

disjoint union are equal if subset {(i,x)

:

xEA i

}

E

i

Ai'

=

j

Two

{AiJ iEI of sets is a pair

elements

and A~(x)

= y.

of the disjoint union.

(i ,x)

and (j ,~J) of the

We identify Ai with the Thus after constructing

the disjoint union, we may consider the family IAiltEI to be a family of elements of the power set of the disjoint union.

4. categories

19

Let (Ai }U:I be a family of sets, and let P be a set. to Ai may be identified with a map f (At}tEI such that f(P)

~

Let F denote the set of maps f from P to the

Ai'

disjoint union of {Ai} tEL such that f family of maps that

1T i

(P)

~

for some

(P) !;;; Ai

from P to Ai we mean a family

1T t

1T

in I.

By a

of elements of F such

Ai for each i in I.

Let IAlliET be a family of objects in a category product of the objects At (projection) maps maps f i

Then a map from P

from P to the disjoint union of

ce.

A categodcal

is an object P together with a

family of

from P to Ai such that for any object Sand family of

1T t

from S to Ai' there exists a unique map f from S to P such that

rr i f = f i

fOl:

each i

i.n

r.

categorical

A

isomorphism in the sense that if (P'

an isomorphism e from P to p' such that

product

is

unique

up

to

is another one, then there exists

,1T')

rri8

=

1f i

for each i.

If

'e 1.S the

category of sets, then it is easy to verify that the set of all functions Ä from I to the disjoint union of (AiJ iEI such that Ä(i) E Ai for each i, and rr i (Ä) defined to be A(i), is a categorical product of the At, which we refer to as the product and denote by lIiE! Ai' If I = {l, ... ,nJ,

then the product of the sets Ai

product Al )( ••• )( An'

If Al = S

for each i

is the Cartesian

in I, then we wri te the

product, which is the set of functions from I

to S,

as SI,

or Sn if

I = {l, ... ,n}.

EXERCISES 1. Show that a function is one-to-one i f and only i f it is left

cancellable, 2. Show that the categorical product of a

family of objects

is

unique up to isomorphism. 3. Show that, in the category of sets, the set of all functions Ä from I to the dis joint union of (At I tEl' such that

Ä (i.)

E At for

each L, is a categorical product of lAI}iEI' 4. Let I be the set of binary sequences, and, for each i in I, let Ai be {x E {O,l} : x ~ i j

for all j}.

Show that the natural map

from ITiA i to AO is onto if and only if WLPO holds, 5. Consider

the

extensional

category functions.

of

sets

with

Show that

inequality

the

and

product ITiA i

strongly in

this

20

Chapter I. Sets category is the product in the category of sets equipped with the inequality ~ t ~ if there exists such that ~(i) t ~(i). Generalize Theorem 2.2 to this setting. 6. Let a be an object in a

category~.

Show how

functor from cl to the category of sets. to be representable. 7. If

~

is a category, then the dual

= ~(a,b) is a

T(b)

Such a functor T is said

category~'

is defined to have

the same objects as 'C, but ~'(a,b) = 'Il(b,a). The coproduct of a family of objects of 'Il is the product in the category 'C'. Describe the coproduct directly. What is the coproduct in the category of sets? 8. Direct limits. A direct system is a sequence of objects An and maps f n : An .... An + 1 . An upper bound of a direct system is an object B together with maps b" : An .... B such that b n +1 f n = bn for each n. A direct limit of a direct system is an upper bound L so that, for any upper bound B, there is a unique map ~ : L .... B such that ~en = bn for each n. (i) Show that any two direct limits are isomorphie. (ii) Show that the direct limit in the category of sets is the disjoint union of the An with the equality generated by requiring a = fn(a) for each a E An' (iii) Show that the direct limit of discrete sets need not be discrete, but it is discrete if all the maps are one-to-one. 5. PARTIALLY ORDERE!) SETS AND IATTICES

A partially ordered set is a set P with a relation a (i)

( ii ) (i ii )

a

~

~

b satisfying:

a,

if a

~

if a

~

band b band b

~ ~

c, then a a, then a

~

=

c, b.

A map between two partially ordered sets Pt and P 2 is a function f from Pt to P2 such that i f a

~

b, then f

(a)

~

f (b).

For the most part we will be

interested in discrete partially ordered sets; in this ca se we write a < b for a ~ band a t b. Let a, band c be elements of a partially ordered set P.

We say that c

21

5. partially ordered sets and lattices is the greatest lower bound, or infimum, of a and b, and write c i f for each x E L we have x

~

c i f and only if x e n ) is isomorphie to {x C TJ

C

(iii ) There is a permutation a

of

x d n }

so that [ci ,ci+l]

is

< Tl.

We leave the proof that pieeewise isomorphism is a transitive relation as Exercise 4.

If C and D are pieeewise isomorphie discrete linearly ordered

sets, then C and D have the same eardinali ty (Exereise 5). 5.2 THEOREl'i maximal.

(Jordan-li:jlder-Dedekind).

finite!u erturnet-able chain X,

If

a

modu!ar-

lat tice

/1OS

a

then each finite1!) enumer-nble chatn

5. partially ordered sets and lattiees is

in u

eontuined

maximuL

isomorphie to X.

...

Let Xo ~ xl ~ as the formal length of PROQF.

m

finitel y

xm = 1. eontained contained piecewise

~

enumer'ubLe

23 ehain

timt

is

pieeewise

xm be the maximal chain X; we will refer to

x. I t is readily verified that Xo = 0 and If xl 11 Ul = xl' then Y is Let Yl ~ ~ Un be a ehain Y. in the lattiee [xl,l]. By induetion on m, the ehain Y is in a maximal finitely enumerable ehain in [xl,l] that is isomorphie to xl S ... ~ x m ' and therefore in a maximal finitely

...

enumerable chain that is piecewise isomorphie to X. and we have the following pieture

Otherwise xl 11 Yl = 0

. / xl V YI "'xl YI

~O/

where

[~JI,xl

V YI] is isomorphie to [O,xl], and [xllxI V LIl] is isomorphie

to [0 ,UI]. By induetion on m the chain xl S xl V Yl is contained in a maximal finitely enumerable chain in [xI,ll, of formal length m-l, consisting of a maximal finitely enumerable chain C in [xI,xI V YI! of formal length

e,

of formal length

and a maximal finitely enumerable chain D in [xl V UI' 1] m-I'-l.

of formal length at most

The ehain hJI} UD is a maximal chain in [YI,I] m-I'

so, by induction on m, Y is contained in a

maximal chain in [Ul,l] that is pieeewise isomorphie to {Yl} U D.

Lemma

5.1 shows that the ehain C is isomorphie to a maximal ehain in [O,LIlJ. Thus Y is contained in a maximal ehain that 1S piecewise isomorphie to X. o It follows from (5.2) that if a diserete modular lattiee has a maximal finite ehain of length n, then any finite ehain is eontained in a maximal chain of length Tl. A partially ordered set P satisfies the ascending chain condi ti on

if

of elements of P, there is Tl such for eaeh sequenee P, ~ P2 ~ P3 0; LLPO is equivalent to the statement that, for each real number x, either x ~ 0 or x ~ O. Karkov's principle states that if a is a binary sequence, and an = 0 for a11 n is impossible, then there exists n such that an = 1. The idea is that we can construct the number n by successively computing a" a2, a3' •.• unti1 we get a 1. Markov's princip1e is emp10yed by the Russian constructivist school, which is a constructive form of recursive mathematics; for details see [Bridges-Richman 1987]. An argument against Markov's principle is that we have no prior bound, in any sense, on the length of the computation required to construct Tl. We regard Markov' s principle as an omniscience principle. The position that "for a11 0 in A there exists b in B" entails the existence of an a1gorithm (not necessarily extensional) from A to B is suggested by the assertion in [Bishop 1967, page 9]

that "A choice

function exists in constructive mathematics, because a choice is impLied by the ver'y meoning of existence." Bishop defines a function f from B to A to be surjective i f there is an a1gorithm g from A to B such that f (g (0)) = 0 for a11 0 in A. OUr Brouwerian counterexample to the axiom of choice, and Exercise 3.1 that the axiom of choice implies the law of excluded middle, is due to Myhill and Goodman (1978). An earlier proof , in a topos-theoretic setting, was given by Diaconescu (1975) who shows that the axiom of choice implies that every subset A of a set B has a complement in the sense that there is a subset A' such that B = A U A' and A n A' =~. Fourman and Scedrov (1982) have shown, using topos-theoretic methods, that the world' s simp1est axiom of choice is not provab1e in intuitionistic set theory with dependent choice. The arguments against countab1e choice, and dependent choices, must be based on more fundamental grounds than we used for the axiom of choice. That is, we must question the interpretation of the phrase "for all 0 there exists b" as entailing the existence of an algorithm for transforming elements of A to elements of B. One reason for rejecting

Chapter I. Sets

34

this interpretation is that by so doing our theorems will also be theorems in other (unintended) models, such as arise in topos theory, which are of interest in classical mathematics (see, for example, [Scedrov 1986]). A more relevant reason, perhaps, is that arguments that depend essentially on this interpretation often have an unsatisfying feel to them which seems to be connected wi th the gratuitous "completion of an infini ty" that occurs when

we

subsume

a

potentially

infinite

information into a single algorithm. A rank function on a well-founded set W is a map

number ~

of

items

of

from W to an ordinal

< 'P(y) whenever x < y. It seems unlikely that we can always construct a rank function although, classically, every well-founded set admits a unique smallest rank function. Induction on rank is a common technique in the classical theory (see Exercise 6.11).

A such that 'P(x)

Chapter H. Basic Algebra

1.. GROUPS

A monoid is a set G together with a function written as

=

~(a,b)

~

from GxG to G, usually

ab, and a distinguished element of G, usually denoted

by 1, such that for all a, b, c in (ab)c

(i)

(ii)

e

a(bc),

=

la = al = a.

(associative law) (identity)

The function 'fi is called lIIUltiplication and the element 1 the identity. The associative law allows us to ignore parentheses in products 0la2" The monoid is said to be abelian, or cOIIIIIUtative, elements a and b.

aa"'a

the

identity element

is then

In this case we speak of an additive monoid, as opposed to

a multiplicative monoid. product

ba for all

=

In an abelian monoid the function 'fi is often called

addition and written as 'fi(o,b) = a + b; denoted by O.

if ab

'a,,'

In a multiplicative monoid we write the ,,-fold

as an for each posi ti ve integer

Tl,

and set a 0

=

1; in an

addi ti ve monoid we wri te the n-fold sum a + a + ••• + a as na, and set

Oa = O. rf x is a set, then the set of all functions from X to X forrns a monoid:

rnultiplication

is

cornposition of

element is the identity function.

The set

functions,

m of

and the

identity

nonnegative integers is a

cornrnutative monoid under addition, with identity element O. A homomorphism of monoids is a function f horn a monoid G to a monoid H such that f(l)

=

1, and f(ab)

rnultiplicative monoid, and

,p is a homomorphism. subset of im F.

(J

= f(a)f(b)

for all

Cl

and b in G.

E G, then the map horn IN to

e

If

e

is a

that takes n to

A homomorphism f is nontrivial if {l} is a proper

A subset H of a monoid G is a submonoid i f 1

closed under rnultiplication.

If S is a subset of the monoid

E

Hand H is

e,

then the

set consisting of 1 together with all finite products of elements of S is a submonoid of G called the submonoid generated by S. generated by {3,4) is m\(1,2,5}. 35

The submonoid of m

36

Chapter 11. Basic algebra Let X be a set.

elements of X,

Define X* to be the set of all finite sequenees of

including the empty sequenee.

The elements of X* are

'!'Wo words u =' x1x2" 'x m and v = Y1 Y2" 'Y n are equal if m = Yi for i = 1,2,···,m. If u and v are equal in X* we will write

ealled words. n and xi

=

u

Define a multiplieation on X* by setting uv

= v.

= x1"'xmY1'''Y,,'

This multiplieation is assoeiative and the empty word is the identity element, so X* is a monoid, ealled the free monoid on the set X. then x*

a one-element set,

is isomorphie

to the additive

If X is

monoid of

nonnegative integers. If a and b are elements of a monoid, and ab

1, then we say that a is

=

a left inverse of band that b is a right inverse of a. inverse a and a right inverse e, then a

=

a (be)

we say that a is the inverse of band write a

If b has a left

= (ab)c = e; in this ease

= b- 1 •

If b has an inverse

we say that b is a unit, or that b is invertible. A group is a monoid G in whieh every element is invertible. additive group, the inverse of a is denoted by -a rather than a- 1 a

positive

integer,

exponents hold.

we define a-fl

to be

(a-

1 )Il

i

the usual

In an •

For n

laws of

In an additive group this definition takes the form

(-11)a = n(-a), and the appropriate assoeiative and distributive laws hold

(see the definition of an R-module in Seetion 3). group is the group

of integers under addition.

~

The order of an element {an: nEIN}. for m

The prototype abelian

II

of a group is the eardinali ty of the set = 1 and a m t- 1

The order of a is nEIN if and only if lln

= 1, ... ,n-1.

In

~

the element 0 has order 1, as does the identity

in any group, and eaeh nonzero element has infinite order.

In a diserete

group the order of an element a is the eardinality of the set {n EIN: a m t- 1 whenever 0 < m ~ n} (which eontains 0), henee is an ordinal ß ~ w. If G and H are groups, and f is a monoid homomorphism from G to H, then f(a-

1 )

=

f(a)-1

and f(a-')F(a)

homomorphism between

two

=

groups

f(a-'a)

=

preserves

multiplieation, identity, and inverse.

F(l)

all

=

the

1.

Thus a monoid

group

If G is a group, and

struetures: CI

E

G, then

the map that takes x E G to axa-' is easily seen to be an automorphism of Gi such an automorphism is ealled inner.

If G and H are abelian groups, then the set Hom(G,H) of homomorphisms from G to H has a natural strueture as an abelian group by setting

(F, + Fz)(x) = F,(x) + F2 (x). The group Hom(~,H) is naturally isomorphie to H under the map taking f to f(1). If h is a homomorphism from H to H' ,

1. Groups

37

then hinduces a homomorphism horn Hom(C,H) to Hom(C,H") by taking f to hf; that is, h(f, + f 2 ) = hF, + hf 2 • Simi1arlya homomorphism 9 : C· ~ C induces a homomorphism horn Hom(C,H) to Hom(C· ,H) by taking f to fg. A category ~, like the category of abelian groups, such that ~(G,H) is an abelian group for each pair of objects C and H of ~, and such that the functions induced from ~(G,H) to ~(G,H') by a map H ~ H', and from ~(G,H) to 'fl(C' ,H) by a map C' ~ C, are homomorphisms of groups, is called a pre-additive cateqory. A permutation of a set X is a one-to-one map of X onto itself.

The set

of permutations of X is a group called the S}'IIIIIetric group on X. If xl' ... ,xn are distinct elements of a discrete set X, then we denote by (xl' ... ,xn ) the pernrutation Tr of X such that TrX i TrXn TrX

= x i +1 =

for

i

= 1, •.• ,n-1

xl

= x otheLWise.

Such a pernrutation is called an n-cycle with support (x1' •.. 'x n }, and two cycles are called disjoint i f their supports are disjoint. If X is a finite set, then every pernrutation is a product of disjoint cycles. As (xl' ••. 'x n ) = (x1,xn )···(xl,x3)(xl,x2)' every pernrutation of a finite set is a product of 2-cycles (not necessarily disjoint). A pernrutation that can be written as a product of an even number of 2-cycles is said to be even, otheLWise odeL If Tr is a pernrutation of a finite set, then we define sgn

Tr

={

I if

Tr

i s even

-1 i f Tr is odd. The product of an odd number of 2-cycles is odd (Exercise 7), so sgn Tr , Tr 2 (sgn Tr, )(sgn Tr 2 ) (Exercise 8). A subgroup of a group is a submonoid that is closed under inverse. If C is a group, then C and (I} are subgroups of G; we often denote the subgroup (I} by 1. If 8 is a subset of a group C, then the set

(I} U (sls2"'sh :

si E 8

U 8-1 ,

h l

I}

of all finite products of elements that are in 8, or are inverses of elements of S, is a subgroup of G called the subgroup generated by S. If = G, then 8 is called a set of generators for C. A group is finitely generated if it has a finitely enumerable set of generators, cyclic if has a one-element set of generators. The additive group ~ of rational numbers

Chapter 11. Basic algebra

38

is not finitely generated; in fact, each finitely generated subgroup of

~

is cyclic (it is easy to show that each finitely generated subgroup of ~ is contained in a cyclic subgroup). A subgroup H of G is normal i f ghg - 1 E H for each 9 in G and h in H. Every subgroup of an abelian group is normal.

rf f is a homomorphism from

G to H, then the kernel of f,

ker f

=

{x E G : f(x)

=

I}

is easily seen to be anormal subgroup of G. measures how badly f

fails

=

f- 1 (1)

The size of the kernel of f

to be one-to-one,

as the following are

equivalent: f(a) = f(b)

f

(ab-

1 )

f(a)f(bab -

1

1 )

1

= f(a)f(b)-1

E ker' f,

so f is a monomorphism if and only i f ker' f = 1. We will be studying algebraic structures that are abelian groups with additional structure. In these cases, the kernel of a homomorphism f means the kernel of f as a homomorphism of groups; if the group is written additively, as will normally be the case for the more complex structures, ker' f = f- 1 (0). Each normal subgroup H of a group G is the kernel of a homomorphism that is constructed as follows.

Let G,IH be the set whose elements are

precisely the elements of G, but equality is defined by setting a

=

b if

When it is necessary to distinguish between the equalities on G and on G,IH we will wri te a = b (mod H) to denote the equali ty in G,IH.

ab- 1 E H.

Multiplication and inverse remain functions with respect to the equality on G,IH, so G,IH is a group, called the quotient group of G by H. The prototype of a quotient group is obtained by taking G to be the group ~ of integers, and letting H be the subgroup of ~ consisting of all multiples of a fixed integer Tl; the resulting group G,IH is the group module n. The fundamental facts relating normal subgroups, homomorphisms are contained in the following theorem. 1.1 THEOREM.

Le t

N be a

nor'ma!

sllbgr'oup

homomor-phism From G to a gr'oup L wilh F(N) = from GIN to L.

IF

F is onto.

isomorphism From GIN to L.

and

1.

the ker-nd

of

a

~n

of integers

quotients,

gr'oup (;,

Then F is of F is N.

and

and

F "

a homomorphism then F is an

1.

Groups

39

PROOF. If a = b (mod N), then ab- t € N, so f(a) = F(b); therefore F is a function on G/N, which is clearly a homomorphism. Conversely, if F(a) = F(b), then F(ab- t ) = 1, so ab- 1 E N and a = b (mod N). Therefore F is a one-to-one map from G/N to L; so if F is onto, then F:G/N inverse g.

Clearly 9 is also a homomorphism.

~

L has an

0

Let N be anormal subgroup of the group G. A (normal) subgroup of G/N is a (normal) subgroup H of G that is a subset of G/N, that is, if a E H and a = b (mod N), then b € H. It is easily seen that a subgroup H of G is subset of G/N just in case N ~ H. The difference between a subgroup H of G containing N, and a subgroup H of G/N, is the equality relation on H. We distinguish between H as a subgroup of G, and H as a subgroup of G/N, by writing H/N for the latter.

If H is normal subgroup of G, containing then (G/N)/(H/N) is isomorphie to G/H; in fact, the elements of both groups are simply the elements of G, and the equalities are the same.

N,

1.2 THEDREM.

Let H and K be subgroups of the group G.

IF K is normal,

then

(i) The set HK

(ii) The

=

subgroup H

(iii) The quotient PROOF.

{hk: h € H and k E K} is a subgroup,

Exercise.

n

K is

nor-mal in H. and

groups HK;K and

H/(H n

K) are

isomorphie.

0

In an additive group, the subgroup HK is written as H + K. If a E G, and H is a subgroup of G, then Ha = {ha : h E H} is called a right coset of H, while aH = rah : h EH} is called a left coset of H. The inverse function induces a bijection between left and right cosets of H

that takes

aH

to

Ha- t

,

so we can speak unambiguously of the cardinality

of the set of cosets of H in G. This cardinality is called the index of H in G and is denoted by [G:H]. If H is normal, then Ha = aH for each a in G. EXERCISES

1. Show that in a finite monoid, if a has a left inverse or a right inverse, then an = 1 for some positive integer n; so an element can have at most one left or right inverse. Give an example of an element of a monoid that has two distinct left inverses; two

40

Chapter 11. Basic algebra distinct right inverses. 2. Show that a monoid may be identified with a one-object category, and that homomorphisms of monoids are categories.

functors between such

Which of these categories correspond to groups?

3. Show that the set of units of a monoid is a group.

4. Show that the set S of binary sequences forms an abelian group G under coordinatewise addition module 2.

Let a

€

Sand define

b,c € S by b n = 1 if and only if Gn = 1 and Gm = 0 for all m

and

cn = 1

if and only if bn - 1

=

1.

< n,

Show that if the subgroup of

G generated by band c is generated by a finite subset of G, then

ei ther G = 0 or a "" O. 5. Let G be

a multiplicative

group with an

inequality.

The

called translation invariant if x "" y implies Assuming the inequali ty is translation

inequality is

and xz "" yz. invariant, show that

zx l' zy

(i) the function taking x to

x-t

is strongly extensional if and

only if the inequality is symmetrie. (E) the inequality is cotransitive if and only i f multiplication is strongly extensional (the inequality on G x G is given by Finally, show

,X2) "" (YI,Y2) if Xl "" Yt or X2 "" Y2)' that if the inequality is consistent,

(Xi

and

multiplication is strongly extensional, then the inequality is translation invariant. 6. Considering an inequality on G to be a subset of G x G, show that the union of a family of inequalities on a group G, under which the group operation is strongly extensional, is again such an inequality.

Show that there is a unique inequality on

GjN

that

makes Theorem 1.1 true in the category of groups with inequality, and strongly extensional homomorphisms.

Show that if G is the

set of binary sequences under coordinatewise addition module 2, and N = {x x "" 0 in

€ G :

GjN if

there is m such that xn = 0 for all n 2 m}, then = 1 infinitely often and LPO.

and only if xn

7. Let xl"" ,xmdJl"" 'Y n be distinct elements of a finite set X, let G be the symmetrie group on X, and let 1 ~ < j ~ m. verify the following two equalities in G.

41

1. Groups (i) (xi ,xj)(xl"" ,xm )

=

(xl"" ,xi_l'x j"" ,xm ) (xi"" ,xj_l)

('!i.i) (xl'Yl)(xl,,,,,xm)(Yl'''''Y n )

=

(Yl' .... 'Yn'xl' ... 'xm ).

For lr in G, we can write lr in an essentially unique way as a product of disjoint cycles whose supports exhaust X. Let Nlr be the number of cycles in such a product. Use (i) and (ii) to show that if T is a 2-cycle, then NTlr = Nlr ± 1. Conclude that an even permutation cannot be written as a product of an odd number of 2-cycles. 8. Show that sgn is a homomorphism from the symmetrie group on a

finite set to the group {-l,l} under multiplication. 9. Give a Brouwerian example of a countably generated subgroup of a

finite abelian group that is not finitely generated. 10. Show that the set of normal subgroups of a group is a modular

lattice. 2. RINGS AND FIELDS A ring is an additive abelian group R which is also a multiplicative monoid, the two structures being related by the distributive laws:

(i)

(ii)

a(b

+ e)

(a +

b)e

(ab) + (ac), (ac)

+ (be).

If the monoid structure is A ring is said to be trivial i f 0 = 1. commutative, then R is a cammutative ring. A unit of R is a unit of the multiplicative monoid of R. A ring is said to have recognizable units if its units form a detachable subset. If A is an abelian group, then the set of endomorphisms E(A) = Hom(A,A) is a ring (multiplication is composition) called the endomorphism ring of A. More generally, if ~ is apre-additive category, then ~(A,A) is a ring

for each object A

in~.

A ring k is a division ring if, for each a and b in k, a #- b if and only i f a - b is a unit.

We remind the reader that the interpretation of the symbol a #- b depends on the context: if k comes with an inequality, then a #- b refers to that inequality, otherwise a #- b refers to the denial inequality. An immediate consequence of the definition is that if k is a division ring, then the

Chapter 11. Basic algebra

42

inequality on k is symmetrie, and translation invariant: if a -;! b, then a+c-;!b+c. Note that the denial inequality is automatically translation invariant because addition is a function. Of course you could define the inequali ty a -;! b on any ring to mean that a - b is a unit and, technically, you would have a division ring; thus the general theory of division rings will include the theory of rings. However the idea is to use the symbol ° -;! b to represent relations that can reasonably be called inequalities: if you pick a silly inequality and get a silly division ring, don't blame uso As a rule of thumb, you should use a standard inequality. For the most part we will be interested only in discrete division rings, and, to a lesser extent, in division rings with a tight apartness. The nonzero elements of a division ring are exactly the units; in the discrete case, this characterizes division rings (Exercise 3). A field is a commutative division ring. A Heyting field is a field with a tight apartness. In a Heyting field, or more gene rally in a field wi th a cotransitive inequality, the arithmetic operations are strongly extensional (Exercise 5).

The rational numbers

the real numbers (next section)

~

form a discrete field;

form a Heyting field.

The rational

quaternions (Exercise 4) form a noncommutative discrete division ring. A subset of a ring is a subring i f it is an additive subgroup and a multiplicative submonoid.

Let 8 be a subring of a commutative ring R, and

let 01"" ,an be elements of R. the form

Then the set of sums of elements of R of

sam (1 )am (2) •• • am (n)

1

2

n'

with s E 8 and m(i) E IN, is a subring of R denoted by 8[a1"" ,an] which is contained in each subring of R that contains 8 U {a1, ..• ,an }. If 8 and Rare fields, then 8(01"" ,an) denotes the set of quotients F/g with F,g E 8[01, ... ,an ] and g"# O. It is easily seen that 8(a1, ... ,on) is a field which is contained in each subfield of R that contains S U (a1, ... ,an ). An integral domain, or simply a domain, is a ring that admi ts an inequality preserving isomorphism with a subring of a field; more

informally, an integral domain is simply a subring of a field. If R is a subring of a field, then (ab- 1 : o,b E Rand b -;! O) is a field containing R called the field of quotients of R.

The field of quotients of an

43

2. Rings and fields integral domain is essentially unique (Exercise 6).

If R is a nontrivial discrete integral domain, then, for each a and b in R, i f a f. 0 and b f. 0, then ab f. O.

( *)

Conversely, i f R is a discrete commutati ve ring satisfying (*), then we can embed R in a discrete Held I< by imitating the construction of the rational nUl11bers with

(a,b) =

Let I, = {(a,b) € R x R : b f. 0 l

frorn t.he int.egers 71.

(Q

(e,d)

if

ad = be.

DeHne

multiplication

on

h

(a,b)'{e,d) = (ae,bd) and addition by (a,b) + (e,d) = (ud + bc,bd).

by

It. is

routine t.o verify that this makes I< a ring with additive identity (O,l) and multiplicative identity (1,1).

If (a,b) f. (0,1),

then a f. 0 so the

element (b,a) is in hand (a,b)(b,a)

(1,1); conversely, if (a,b)(c,d) =

(1,1), then ae = bd f. 0, hence

so (b,al Eh.

(1.

f. 0,

Thus I, is a Held.

We embed R into h by taking a to (a, 1) . An

intrinsie characterization of an arbitrary integral domain is given

in Exercise 7.

To prove that this characterization is correct,

you

construct a field of quotients as in the discrete case. If h is a field, then the subfield of h consisting of a11 elements of

the

form

(n·1)/(m·1)

where m and n

aLe

integers and m'l f. 0 is the

smallest subfield contained in I o.

(iii) a is invertibLe. Suppose (i) holds. We may assume that laN - an I < fej2 whenever If aN i fe, then an > fe/2 whenever ni N, so a > 0; if aN ~ - t then

PROOF. n l N.

a

< O. Now suppose (ii) holds. If a > 0, then there are positive fe E III and such that an > fe whenever n iN. We may assume that an > fe for at t

NEIN

Then the sequence {I/an} is Cauchy and is the inverse of a in ffi. Finally suppose (iii) holds. Then there exists a Cauchy sequence Ibn} such that anbn - 1 converges to O. Choose positive fe € III such that

nEIN.

Ibnl

< I/fe for all n.

Then I~ll is eventually greater than fe.

0

We get an inequality on ffi by defining a ~ b to mean that b - a is invertible, so (3.1) says that a ~ b if and only if a < b or b < a. The cotransitivity of a < b is the constructive substitute for the classical trichotomy . 3.2 'IHFDREM (cotransitivity). a

< c,

then ei ther a

PROOF.

Choose m

0 so that < c m - 6E lan - a m I < E !e n - c m I < E Ibn - b m I < E

am

whenever n l m. Either b m < c m - 3fe, in which case bn < c n - fe for all im, so b < c, or bm > a m + 3E, in which ca se bn > an + E for all n im,

n

50

Chapter 11. Basic algebra

so b > Cl.

0

We wri te

0, either

!

Heljtlng field.

(1

assume that a + b > 0.

Tl

b.

=

Cl

defines areal number

0

If

and bare

n

that is the

C

The infimum of n and b is

= Sllp(o,b).

The absolute value of areal number n may be defined as

Inl = sup(o,-a).

The Held ([: of complex numbers is obtained from the veetor space 1R 2 by setting (a,b) (r ,d) = (ar - bd, od +

It is readily verified that ([ is

hel).

a Heyting field with multiplicative identity (1,0). see that i

2

Setting

i

=

(0,1) we

= -1 and ([: = IR + [Ri.

Ametrie space is a set S together wi th a function ci, called a metric, from S x S to [R such that

(i) ci(x,y) - d(y,x) (ii) d (x,u) = (iii) d(x,z)

~

°

~

0,

if and only if

x

'I,

d(x,y) + d(lj,?).

The real numbers form ametrie space under the met.ric d(x,U)

=

Ix - yl.

A

Cauchy sequence in ametrie space is a sequence {x n } in S such t.hat for

51

3. Real numbers each positive

t

E

~,

there exists N E d(xn,x m )

A

~ t

such that

~

whenever m,n ? N.

sequence in (xnJ in S converges to y E S if for each positive

there

exists NEIN

such

that d (xn,Y)

whenever

~ t

t

If

n ~ N.

E (Q (x n

I

converges to y, we say that y is the limit of (xn ).

It is easy to verify

that each convergent sequence is a Cauchy sequence.

If, conversely, each

Cauchy sequence converges to some element of S, we say that S is complete. The space ffi is complete. By imitating the construction of ffi from ID, we ean embed any metrie space S in i ts completion S, whose elements are Cauchy sequences in 8, wUh d(a,b) d(a,b) =

O.

equal

to

the

limit

of d(an,bn ), and a = b defined by The space 8 is dense in S, that is, for each positive E, and

s E 8, there is a E S such that d(a,s)


b is impossible if and only if a

b.

~

2. Show that the following are equivalent. (i) For all a E

rn,

either a > 0 or a

~

O.

(U) LPO.

3. Show that IR is a distributive laUiee under ~.

4. Show that lai i 0, and that lai> 0 if and only if a ~ O.

Show

that la + bl ~ lai + Ibl.

5. Show that the following are equivalent. (i) LLPO, (ii) For all a E IR, either a (iii) For all a,b

(iv) If a,b E (v)

rn,

For all a,b

~

0 or a i O.

E IR, if ab = 0, then a

o or

b = O.

then there is c E IR such that a E ffi,

= eh

or b

=

ca.

if sup(a,b) = 1, then a = 1 or b = 1.

6. Show that Markov's principle is equivalent to IR being a denial field.

7. The field of p-adic numbers. on ~ is defined, for x; "

p7l(x, - x 2

)

X2'

If

p

is a prime, the p-adic metric = pn such that

by sett.ing cl (x, ,xz)

can be written with numerator and denominator not

divisible by p.

Show that d is ametrie, and that the completion

52

Chapter 11. Basic algebra of

m with

respect to this rnetric is a Heyting fjeld.

8. Show that each metric space is dense in its eornpletion, and that its cornpletion is complete. 4. MODULES If R is a ring,

then a left R-module is an additive abelian group fr1

together with a function

trom R

jJ

x M

t.o M, written 11(,-,a) ""

and called

1'0

scalar multiplication, satisfying (i)

+ b)

I" (0

(ii) (iii )

+ d,

r~a

'-0 +

(I- + s)o

Set

(T"S )0 = ,- (so)

( iv)

1'0 = a

for all ", s in Rand a, b in M.

Right R-modules are defined similarly,

except the sealar multiplieation is on the right.: the only real differenee is

in

(iE)

whieh by ,-s

multiplying

will is

read

the

N is a fami ty of homomorphisms, then there is a unique homomorphism f from M to N sueh timt

f

=

f i on Ai for eaeh i E I.

then x ~ L~=I a i (m)' with a i (m) E Ai (m). Therefore fIx) must equal L~=I fi(m)(a i (m))' so f is unique. If we define f (x) to be L~=1 f i (m) (ai (m))' we must show that f (x) is weH defined; i t suffices to show that if x = 0, then fIx) ~ o. Suppose x = L~~1 ai(m) = o. As the AL are independent, either aL(m) ~ 0 for each m, or there exist m ~ m' such that i(m) = L(m'). In the latter case, we can add aL(m) and PROOF.

If x E M = LiEIAi'

Chapter 11. Basic algebra

56

ai(m') within Ai (m)'

o by

F(x)

50

that f is a homomorphism.

induction on n.

0

An R-module F is free on a family of elements

function F mapping I

It is readily seen

into an R-module M,

homomorphism f* from F to M such that f*(x i

(xi)iEI of F if for each

there is a unique R-module )

= F(i).

We say that {xi)iEI

The uniqueness of f * implies that free modules on

is a basis for F.

{xi} iEI and {Yi} i EI are isomorphic under an isomorphism taking xi to Yi ' so

any

two

free

modules

essentially the same.

whose

bases

have

takes T' to rX i is an isomorphism for each is free on (xi }iEI' i

E I, so if Xi

=

xj

the

index

set

are

in I, then (4.2) shows that F

i

If R is a nontrivial ring, ,

same

If F = Gl i EI Rx i , and the map from R to Rx i that

then

i

= j;

then Xi t 0 for each

so the basis elements Xi are in one-to-

one correspondence with the elements i of I.

Thus if we were to restrict

ourselves to nontrivial rings, we could define a basis to be a set rather than a family.

If R is a trivial ring, then any family of elements of any

R-module M is a basis for M. Let I be a discrete set, and for each i E I let 0i E R(I) be such that 0i(i)=1,

and 0iU)=O

for

jti.

Then R(I)

is free

on

Wil iEI •

Similarly, suppose I is an arbitrary index set, and (R i }iEI is a family such that each Ri is a copy of R. If for each i E I we let Xi be the sequence of length one whose term is the identity element of Ri , then R(I) is free on (xi liEI' By abuse of language we say that R(I) is the free module on I. If I = (1, ... ,n), then we write Rn instead of R(I).

A module M has a

basis of n elements i f and only i f it is isomorphic to Rn, in which case we say that M is a free module of rank n.

In Section 6 we shall see that,

for nontrivial commutative R, the rank of M is an invariant.

Exercise 3

contains an example of a nontrivial noncommutative ring R such that the left R-modules Rand R2 are isomorphic. An R-module M is finitely generated if there is a map from Rn onto M

for some positive integer n; that is, if there exist elements x1, ..• ,xn in M such that each element of M can be written in the form

27=1

l'iXi'

An

R-module M is cyclic if there is a map from R onto M; that is, if there exists x E M such that each element of M is a scalar multiple of x. 4.3 THEDREM.

Let R be a subT'ing oF the f'ing E.

IF M is a

finitety

generated (FT'ee of rank n) E-module, und E is a finitely generated (free

4. Modules

57

of ronl< m) R-moduLe,

then M ;8 0 firtitdy geneT-nted

(free of nlnl, mn) [{-

moduLe.

PROOF.

Let ~ : ~ ~ E be an epimorphism (isomorphism) of R-modules, n and t : E .... M be an epimorphism (isomorphism) of E-modules. Then ~n : R"'n -+ En is an epimorphism (isomorphism) of R-modules, and t~n R"'n

M is an epimorphism (isomorphism) of R-modules.

-+

0

An R--·module P is said to be projective if whenever 9 maps an R-module A

onto an R-module B, and such that gh

=

f.

r maps

P into B, then there exists a map h : F

Finite-rank free modules are projective: if xl'"

is a basis for P, then there exist (11"" ,an in A such that 9 (ni) for each

i.,

and there is a map h such that h (xi)

gh and

are equal because they agree on the basis.

Let M be a finitely generated module.

If

f (xi)

maps a finite-rank free

TI

The following

theorem shows what happens when we do this for different F's and is similar to the

A

fo[" each i; the maps

= ai

module F onto M with kernel K, then M is isomorphie to PIK. resul t

=

->

"xn

TI'S;

the

rule for determining when two fractions are

equal.

4.4

(Schanuel's trick) .

THEDREM

ff i

, then K i ffi

PROOF. 4'2

: Pi

P2

/Je

R-moduLe,

an

an map from Pi. onto M. i.s i somor-pll i c to K2 ffi Pi'

projecli.ve R-moduLes, and of

M

Let

?TL

IF Ki

As the Pi are projective, we can find maps

...., P21

such that

lTt'!',

=

?Tz

and

?T2~2

~

?Ti'

~1

P,

P"

the kernet

is

:

and

P2

...,

P, and

Map 1\, Gl P 2 to K2 ffi Pt

by taking (1

Construct x ERsuch that xA

V is an i somo rphi sm, and

y

ERsuch that!jB

0 and 0 and

y A --> V is an isomorphism. Show that Rx ~ Ry ~ R as R-modules, and that R = Rx al Ry. What is the point of this exercise?

4. Let

be an R-module and {Ai liEI a family of R-modules. Let be a family of R-module homomorphisms such that F i maps M to Ai. Let 1f i be the projection of lI iEI Ai to Ai. Show that there exists a unique R-module homomorphism F taking M to lI iEl Ai M

{FiliEl

such that

1f

i

F = Fi for each

i

in I.

5. Let F be the external direct sum of IAiliEl' as constructed for an arbitrary index set I. Show that the map from Ai into F given

4. Modules

59

by taking a

Ai to the sequence (a) is a monomorphism. Show that if we identify Ai with its image under this monomorphism, then F

E

= ~iEIAi .

6. SuDmands need not be SUIIIIIaIlds. Let a be a binary sequence with at most one 1, and let S = {O, s, 2s, t, 2t} with s = t and 2s = 2t if an = 1 for an even n, s = 2t and 2s = t if an = 1 for an odd n. Let I = {x,y} with x = y if an - 1 for some n. Finally let Ax = {O,s,2s} and Ay = {O,t,2t} with the obvious three-element group structures. Show that this is a Brouwerian example (LLPO) with Ax not a summand of ~iEIAi. 7. Let {Ai }iEI be a family of modules and f i : Ai ... A a family of isomorphisms. Show that the kernel of the map f : ~iEIAi ... A induced by the ismorphisms f i is a complementary summand of each submodule Ai • 8. Show that A ~ B is projective if and only i f A and Bare projective. Construct a two-element projective module over the the ring ~/( 6) • 9. Show that the free module on a projective set (see Exercise I.3.4) is projective. What is the free module on an empty set? 10. Free modules need not be projective. Construct a Brouwerian example of a map a from a rank-2 free module F, onto a free module Fz such that there is no map ~ from Fz to F, with a~ equa1 to the identity on F,. Hint: Let Fz and F, be free k-modules on the sets A and B of Example I. 3.1, where k is the ring of integers modulo 2. 11. Show that i f free modules with discrete bases are projective, then the world's simplest axiom of choice holds. 12. Let I be a discrete set and ~ a nontrivial ~-module map from ~(I) to~. Show that her ~ is a summand of ~(I) if and only if im ~ is cyclic.

Construct a Brouwerian example of a (not necessarily

nontrivial) cyclic.

~

such that

ker- ~

is a summand but im

~

is not

60

Chapter 11. Basic algebra

5. POLYl'DIIAL Rn«iS If M is a monoid, and R is a ring, let R(M) denote the free R-module on the set M.

We may think of the elements of R (M) as formal finite sums

r1m1 + ••• + rnm n with each mi in M and each of two elements of R(M) by

(2~ =1

r imi

)(2']~1

rjm

j)

=

in R.

"i

2~ =1 2'J~1

("i

DeHne the product

"j )(m i mj ) .

The product mim] is the product in the monoid M while r t ,'] is the product in the ring R. This makes R(M) into a ring, called a monoid ring, with identity element 1, the identity of M. called a group ring.

If M is a group, then R(M)

is

The map taking " to "I maps R isomorphically onto a

subring of R(M), and we shall consider R to be a subset of R(M) via this embedding, that is, the element "I will be denoted by r. Let M be the free monoid on the one-element set {X}. {,'O + "IX + ••• +

R(M)

",ln:

"i

E

Then

Rand nE w}.

The element X is called an indeterminate and the elements of R (M) are called p:>lynomials.

The monoid ring R(M)

is denoted by R[X), and is

called the p:>lynoudal ring in the indeterminate X over R. The polynomial

ring

in n

indeterminates,

R [X I' .•• ,X n ),

is defined

inductively to be R[X 1 , ... ,X n _ 1 )[X n ). form Xl (1) •.• X~ (n) discrete, R[X 1 , .. .

then

An element of R[X 1 , •.• ,X n ) of the is called a monomial of degree 2'~ =1 e( i) . If R is

the

(total)

degree

of

a

nonzero

polynomia1

f

in

,X n ) is the maximum of the degrees of the monomials that appear in

f with nonzero coefHcients.

the monomials;

in fact

The ring R[X 1 , ... ,Xnl is a free R-module on the monomials form a commutative monoid, and

R[X 1 , ... ,X n ) is the monoid ring, over R, on that monoid. If R is commutative, then each polynomial f in R[X 1 , ••• ,X n ) defines a function from Rn to R: i f 01"" ,on are in R, or in any commutative ring

containing R, we let F (01' ... ,on) be the resul t of substi tuting in the formal expression for operation in R.

We

F,

°i

for Xi

and interpreting the formal operations as

require commutativi ty because the

commute with each other, and with the elements of R.

indeterminates

As the 0i commute

wi th each other, and wi th the elements of R, the map that takes f

to

F(ol, ... ,on) is a homomorphism of rings. For

II

E IN, a polynomial F in R[Xl that can be written as 2~:Ö "iXi is

said to have degree at most n-1, written deg f

~

n-1, or deg f < l l .

A

61

5. polynomial rings If deg f

palynomial is zero if and only if it has degree at most -1. and rd

1, then we say that f is monic.

=

no reference to an inequality on R. that f

Note that these definitions make

If '-i fc 0 for same

;: d, then we say

and wri te deg f 2: d.

has degree at least d,

deg f i 13., then we say that f has degree d rd the leading coefficient of f.

d,

~

f

I f deg f ~ d and

written deg f

and we call

= 13.,

If R is not discrete, then F need not

have a degree even if f has a nonzero coefficient. If fand gare polynomials, then we write deg f ~ deg 9 i f deg 9

implies

deg f

REf!l (UIlique interpolation).

of

any

a fietd k. and tet vO •... ,vTl be

a unique potynomiat

fE

k[X)

of

Let aO'" .. an be Tl

distinct etements Then there is most n such tllUt f(a i ) = vi for

+ 1 elements of k.

degr'ee at

aLl i.

PROOF. We prove existence by induction on n. If n = 0 take f = vo. If Tl > 0, then, by induction, there is a polynomial 9 of degree at most 1 such that g(a i ) (X - aO)g(X) + vo.

Tl

-

(vi - vO)/(a i

=

- aO)

for 1

~

~

n.

Take

f(X) =

To prove uniqueness it suffices to show that if f is a polynomial of degree at most

Tl,

and f

induction on n. If n Suppose Tl > 1.

O.

0 for all i, then F

(ai) =

= 0, then By the

= O.

We proceed by

50 f = 0 because f(aO) = remainder theorem we can write f (X) =

f is a constant,

(X-o.n)g(X) where deg 9 ~ n-1. As follows that g(a j ) = 0 for j < n,

aj

t- an for j

50

9

< Tl, and

k

is a field, it

= 0 by induction. Therefore f = O.

o The proof of Theorem 5.5 gives coefficients

Äi

a

recursive

construction of

the

of the Newton Interpolation Formula

F =Ä O +Ä 1 (X -aO) +Ä 2 (X -aO)(X

•••

-al) + + A,,(X -

...

aO)(X -

al)"'(X - (ln-l)'

For the Lagrange Interpolation Formula, see Exercise 5. We state the Euclidean algorithm fOl: discrete commutative rings with recognizable units, rather than just for discrete fields. The algorithm

5. polynomial rings

63

constructs either the desired common factor or a nonzero nonunit. The model application is to the ring k[X]/(f) where k is a discrete field: the construction of a nonzero nonunit results in a factorization of f. 5.6 '.lm!XlREM (Euclidean algorithm).

Let R be a discrete commutative ring with recognizabte units, and I a finitety generated ideaL of R[X]. Then either I is principat or R has a nonzero nonunit. Either I = 0, so I is principal, or there is n € IN and a nonzero polynomial f in I with deg f = n. We may assume that f is monic (otherwise we have a nonzero nonunit), and we proceed by induction on n. PROOF.

If n = 0, then f = 1, so I = k[X] = (f). If n > 0, then each generator 9 of I can be written as 9 = qf + r with deg r < n. Note that r € I. Either each r = 0, or some r # O. If each r = 0, then I = (f). If some r # 0, then we have a nonzero polynomial in I of degree less than n, and we are done by induction. 0 If c

= ab in a commutative ring, then we say that

a

divides c; i f a

divides c we say that a is a divisor, or a factor, of c. 5.7 COROLIARY. Let k be a discr"ete field, and a,b € k[X]. Then there exist s,t € k[X] such that sa + tb divides both a ruill b. Hence sa + tb is the greatest common divisor of a and b in the sense that every common divisor of a will b divides sa + tb. Let I be the ideal of k [X] gene ra ted by a and b. As k is a discrete field, Theorem 5.6 says that I is principal; that is, there exist s and t such that sa + tb divides a and b. 0 PROOF.

We say that two polynomials a and b over a commutative ring are strongly relatively prime i f there exist polynomials s and t such that sa + tb = 1. Thus (5.7) implies that if two polynomials over a discrete field have no common factors of posi ti ve degree, then they are strongly relatively prime. It is easy to see that if a and bare strongly relatively prime, and a and c are strongly relatively prime, then a and bc are strongly relatively prime (multip1y the two equations). EXERCISES 1. Let R and S be commutative rings,

~

S, and sI' ... ,sn elements of S.

a ring homomorphism from R to Show that

~

has a unique

Chapter 11. Basic algebra

64

extension to a homomorphism from RIX I' ... ,Xn I to S mapping Xi to 2. An apartness domain is an integral domain whose inequality is an apartness. Show that the Held of quotients of an apartness domain is a Heyting field. Let fand 9 be polynomials over an apartness domain R. Show that if deg f ~ t and deg 9 ~ j, then Use this result to show that if R is an deg Fg ~ t + j. apartness domain, then so is R[XI. 3. Let R be a ring. The formal power series ring R[[XII is defined to be the set of sequences {an} in R, written ao + a,X + a 2 X2 + ••• ,

with addition and multiplication as suggested by the notation. Show that R[[XII is an apartness domain if R iso Hint: Suppose R is an apartness domain and fg = h in R[[X)) with f 0 + F,X + go + g,X +

f 9

h

Assume that f i

~

= ho +

0 and gj

~

h,X +

0 for some t,j; show that hk

~

0 for

somekS;i+j. 4. Lagrange Interpolation. satisfies Theorem 5.5. f(X)

5. Let

k

r

= t=O

Show that the v. l

rr

.~.

J l

following polynomial

(X - a j )

(a i - a j )

be a Heyting Held and let f E k[XI be nonzero and have

Show that if aO ,al' ••. ,am are distinct degree at most m. elements of k, then there is i such that f(a t ) ~ O. 6. Let k be a Heyting Held and let F E k [X l' ... ,X n J be nozero and have degree at most m in each of its variables separately. Show that i f k contains m + 1 distinct elements, then F(al' ••. ,an) for some a i E R. 7. Let k be a discrete Held.

~

0

Show that any nonzero proper prime

ideal in k[XI is maximal. 8. Let f be a nonzero polynomial over a discrete field k, and a E k. Show that there is a unique nonnegative integer n, and a unique

65

5. Polynomial rings polynomial u € '{[Xl, such that f(X) = (X - a)TlIl(X), and u(a) t- O. If Tl = 1, then a is said to be a simple root of fi if Tl > 1, then a is said to be a root of multiplicity n. 9. Find a polynomial of degree 2 over the ring of integers module 6

that has 3 distinct reots. quaternions.

Do the same for

the

rational

10. Give a Brouwerian counterexample to the statement that either ~ deg g, or deg 9 a commutative ring.

deg f

~

deg f, for all polynomials fand 9 over

6. MATRICES AND VECTOR SPACES

Let a be an R-module homomorphism from a free right R-module N to a free right R-module M. If el, ... ,e n is a basis for N, and FI, ... ,f m is a basis for M, then adetermines, and is determined by, the m x n matrix A {a i j } such that

a(e.)=~ J

Li=l

F.a ... t

tJ

If ß is a homomorphism from a free R-module L, with basis dl, ..• ,d e , to N, then we get an Tl x e matrix B = {b jk) such that ß(dk)=f e].b]·k· ]=l Thus aß(dk )

~ afj=l e.b· =f ]] k j=l

a(e.)b· k ) J

=f

j=l

r

i=l

F·a .. b." t

t

J J{

So the matrix corresponding to aß is the matrix product AB, which is an If we consider only maps m x I! matrix whose ik th entry is :2:J=laijb jk. from N to N, then we get an isomorphism between the ring of endomorphisms of the free right R-module N and the ring Matn(R) of n x Tl matrices over the ring R. The matrix corresponding to the identity endomorphism is called an identity matrix and is denoted by I.

If ei, ••. ,e~ is a new basis for N, and fi, ... ,F~ is a new basis for M, then let a and T be the automorphisms of N and M defined by a(e j ) = ej, and T(f i ) = fi, and let Sand T be the matrices of a and T with respect to the old bases. Then the matrix of a with respect to the new bases is computed by a(e]'.)

=

a(ae).)

=

.. ) = \'LiI~ f,aL.s .. = L \' Li e.s t t) ~.REK. Then

M is

PROOF.

,

so A is finitely presented by (2.3.i).

0

Let N be a finitely gener'ated submodule of a module M.

coherent iF and only if N and

MjN are

coherent.

Let rr be the natural map from M to MjN.

N is clearly coherent.

If M is eoherent, then

If B is a finitely generated submodule of MjN,

then B = rr(A) for same finitely generated submodule A of M. But A n N is finitely generated by (2.4) so B is finitely presented by (2.3.i). Conversely suppose N and MjN are eoherent, and A is a finitely generated submodule of M. Then A n N is finitely generated by (2.2), and therefore finitely presented because N is coherent. Also rrA is finitely presented because MjN is coherent. Therefore A is finitely presented by

83

2. Coherent and Noetherian modules (2.3.H).

0

2.6 COROLIARY.

A finitely presented module ouer a

coherent ring is

coherent.

PROOF.

Repeated application of

(2.5)

shows that finite

One more application of

modules are coherent.

(2.5),

rank free

in the other

direction, shows that finitely presented modules are coherent.

0

A module has detachable submodules if each finitely generated submodule is detachable.

A ring R has detachable left ideals i f R has detachable

submodules when viewed as a left module over itself.

Note that R has

detachable left ideals if and only if R/I is discrete for each finitely generated left ideal I of R. Polynomial

rings

over

discrete

fields

Noetherian rings with detachable ideals.

are

examples

of

coherent

In classical mathematics euery

Noetherian ring is coherent with detachable ideals (but see Exercises 3 and 4). 2.7 'l'BEDRE2t. module M.

PROOF.

Let N be a

firli tely gener'ated submodule of a

rhen M has detachable submodules

iF

und only

iF

coherent

N and MIN da.

If M has detachable submodules then clearly N and MIN do.

Conversely, suppose that N and MIN have detachable submodules and let the natural map from M to MjN.

let P be a finitely generated submodule of M and let x E M. then

x ( P.

If

~x

be

~

To show that M has detachable submodules,

= ~p for

pEP,

then x

E

P

if

If and

~x

(

~P,

only

if

x - pEP n N, which is a finitely generated submodule of N because M is coherent.

0

2.8 COROLIARY.

If

M is

a

coherent left Noetherian ring R. If,

in addition,

finitely

presented

left

module

ouer

a

then M is a coherent Noetherian module.

R has detacllabLe

Left

ideaLs,

then M llas detacllable

submoduLes.

PROOF.

Let '€ be the class cf coherent Noetherian R-modules

(wi th

detachable submodules).

As R E '-C, induction on n and (2.1) and (2.5) (and

(2.7)) gives Rn E '€.

M is finitely presented, ME 'e by (2.1) and (2.5)

(and (2.7)).

0

As

84

Chapter 111. Rings and modules EXERCISES 1. Show that a commutative ring R has detachable ideals if and only if R/I is discrete for eaeh finitely generated ideal I of R. 2. Show that each nonzero fini tely genera ted ideal in the ring 1L of integers

has

finite

depth

in

finitely generated ideals of 1L. generated ideal in the ring has bounded depth. 3. Let

0

be a binary sequence.

k [X 1,

partially ordered

set

of

where

k

is a diserete field,

Let k be the two-element Held and R

the subring of the finite ring and lar/l.

the

Show that eaeh nonzero finitely

k[X,Yl/(X,Yl 2

generated by 1, X,

Show that R is a Brouwerian example of a Noetherian

ring with detaehable ideals that i s not coherent.

Do the same

for the ring 71/1 where I is genera ted by the elements 0n ll r • 4. Construet a Brouwerian example of a diserete eoherent Noetherian ring R, and a fini tely generated ideal I of R, such that

J

is not

(Hint: Let R lie between 1L and mJ

detaehable from R.

5. Let 1 be a finitely genera ted ideal of a eommutati ve eoherent

Noetherian ring R with detaehable ideals. of

J

is detaehable.

(Bint: Given x

E

Show that the radieal

R, eonsider the aseending

chain of ideals T:x m ) 6. Show that finitely generated eoherent modules over rings wi th detachable ideals are discrete. 7. Show that if S is a subring of R, and M is an R-module that is a Noetherian S-module, then M is a Noetherian R-module. 8. Show

that

if

a

module

is

bounded

in

number,

then

it

is

Noetherian. 9. Show that any discrete field is coherent.

Show that the ring l

is coherent. 10. A submodule A of an R-module R is called pure if ",henever a finite family of equations

with

r'ij E

Hand

solution in A.

Gi (

A,

has a solution in B,

then it has a

Shmv that A \: 13 i5 pure if and only i f every map

2. Coherent and Noetherian modules

85

of a finitely presented module to B/A lifts to a map to B. Show that any nonzero pure subgroup of the infinite cyclic abelian group ~ is equal to ~. 11. Show that every Noetherian module M is Hopfian (Exercise 1. 9) •

(If f(z) = 0, construct a sequence such that

Xo = z

and f(x i

)

=

xi-I· ) 3. LQCALlZATIrn

We construct rings of fractions of Let R be a commutative ring. elements of R in much the same way that we construct the rational numbers from the integers. First we decide on a subset S of R whose elements we will allow as denominators; in the construction of the rational numbers, this set consists of the positive integers. In general we want S to be a multip1icative submonoid of R, by which we mean that I € Sand, if S, and S2

are in S, then so is S,S2. The elements of the ring of fractions S-IR consist of pairs r/s

consisting of an element r in Rand an element s of S. multiplication are defined by the usual formulas: r,/s, + r 2 /s2 (r,/s,) ("2/S2)

Addition and

(r,s2 + r2s,)/(S,S2) (r,r2 )/(S,S2)

but we have to be a little careful about equality of fractions because any element s of S is invertible in S-IR I so any element r of R such that sr = 0 must be set equal to zero in S-IR. With this in mind we say that two fractions ",/s, and r 2 /s 2 are equal if there exists an element s in S such that S(",S2 - r 2 s,) = O. There is a natural map R .... s-IR defined by taking the element " to "/1. We leave the verification that S-lR is a ring to the reader. More gene rally, if M is an R-module, then we can form the module of fractions S-IM consisting of fractions m/s with m in M and s in s. Equality and addition are defined as for s-IR , while multiplication by elements of R, or of s-IR, is defined in the obvious way. Thus S-IM is an s-IR-module. If N is a submodule of M, then s-IN is a submodule of s-IM I f P is a proper prime ideal of R, then S = R\f is a multiplicative submonoid of R. In this ca se we denote the module s-lM by Mp • If P is detachab1e, then the ring Rp is loca1 in the sense that for each x € R, either x or I-x is invertible; indeed an element r/s of Rp is invertible

86

Chapter 111. Rings and modules

if I' 'l P, while if I' € P, then s-" 'l P, so 1 - I' /s (s-r )/s is invertible. When R is a discrete integral domain, then P = 0 is a proper detachable prime ideal and the ring Rp is the field of quotients of R. Let I' be an element of a commutative ring R, and S the multiplicative submonoid of R generated by r. It follows immediately from the defini tions that " is nilpotent i f and only i f S-1R = O. We illustrate the use of this fact in the proofs of (3.1), (3.2) and (3.4) below. OUr first result generalizes (11.7.5) which showed, using determinants, that the rank of a finite-rank free module over a nontrivial commutative ring is an invariant. 3.1 THEDREM.

If R is a commutatlve ring, and ther'e is a monomorphism

from ~ to Rn where m

> n,

then R is trivial.

Let A be the m x n matrix of a one-to-one map 'P from Rm into Rn. We first show that the elements in the first column of Aare nilpotent. If,' is an element in the first column, then let S = (l,r',r 2 , ••• } and pass to the ring T = S-1R • I t is easily seen that A is the matrix of a one-to-one map from 1~ into Tn , Apply elementary row and column operations to A, which amounts to changing the bases of r n and Tm, so that the first column and row of Aare 0 except for a 1 in the upper PROOF.

left corner. Let e l' ... ,e m be the new basis for Tm. If n = 1 then 0, so 0 = 1 in T because 'P is one-to-one. If n > 1, then 0 = 1 in T by induction applied to the matrix that results from A by deleting the first row and column. Thus r' is nil potent in R. Let r be the ideal in R generated by the elements of the first column of A, and let el, •.• ,e m be the natural basis of~. Then r k = 0 for some k. But i f Ikel = 0 for k > 1, then 'P(Ik-1el) = 0, so rh-lel = 0 as 'P is 'P(e m ) =

one-to-one.

Thus el

=

0 so R is trivial.

0

Theorem 3.1 generalizes (Ir. 7.5) because i f Rn maps onto Rm with m > n, then, as ~ is projective, ~ maps one-to-one into Rn. The next theorem will be used in Chapter VIII. 3.2 LEMMA.

Let

R C;; T be commutatiue T'ings, and A a matl'ix ouer' R.

Suppose that (i) if (",0, .•. ,0) is in the r'ow space of A over R, then r' (ii) (1,0, ••• ,0) is in the row space of A over T.

Then R

O.

0,

87

3. Localization PROOF.

Designate a finite number of rows of A as good, the rest being

designated bad, so that if a ww of A is goOO, then it contains a 1, called a good 1, in a column whose other entries are O. designate aB rows as bad.

To start we may

Ne induct on the number of bad rows. Let S = {1,r,r 2

Suppose ,. 1S in a bad row. matrix over s-1R I:;; s-lr.

, •••

1 and consider

A as a

Ne can reduce the number of bad rows of A by

elementary row operations that replace r by a good 1 and leave the good rows goOO, so s-lR

=

0 by induction; thus

I'

is nilpotent in R.

Ne have

shown that aB elements in bad rows are nilpotent. If Pi denotes ww i of A, then by hypothesis (ii) we have

(1,0, ... ,0) for some elements t 1 , ... ,t m and we are done. then

t j

=

E

T.

rf all rows are bad, then 1 is nilpotent

rf Pj has a good 1 in a colurnn other than the first,

So we may assume that some Pi has a good 1 in the first

O.

column, and the remaining t j P j consist of nilpotent elements. entry of

Pi

except the first 1S nilpotent.

generated by these nilpotent elements. Therefore I

(i), so R = O.

0, so Pi

Let

=

such tim t I n R

O.

R I:;;

r

IF 1

(1,0, ... ,0) whereupon 1

0 by hypothesis

o by

hypothesis

be commutatiue hngs, and I an idea~ E

Tl,

t hen

R

=

in R[X]

O.

Nrite 1 this way, and apply (3.2) to the matrix of

coefficients of the Pi' 3.4 COROLLARY.

that

so,·

Each element in TI can be wri tten as (lP1 + ••• + tmpm, wi th

Pi € land t i € 1'.

stich

=

I',

0

3.3 THEDREM.

PROOF.

=

Then every

be the ideal of R

If Th = 0, and r € I h - 1 , then f'Pi

has zero entr les except for the fi rst, which 1S (i).

Let I

1 E TI.

0

Let R !;:

r

be commutatiue rings, und I an ideal in R[Xj

Then Uw annihi lotor· of Rn I

consists of l1i Ipotent

e!.ements. PROOF.

Suppose ,. (R n 1) = O.

R generated by r s-1R = 0, so

T'

1

Let S be the multiplicative submonoid of

and pass to S-lR.

is nilpotent .

0

Now (3.3) applies and tells us that

88

Chapter 111. Rings and modules

EXERCISES 1. Let R be a coherent commutative ring with detachable ideals, and

let P be a finitely generated proper prime ideal of R.

Show that

RI' is discrete.

2. Let S be a multiplicative submonoid of a commutative ring R, let M

be an R-module, and let

and

N1

N2

be submodules of M.

Show

that S-l(N , f1 N,,) ~ s-1N, n s-1Nz S-1 U'I , + N,,) ~ s-1 N, + S-1 N2 •

and that

3. Let S be a multiplicative submonoid of a commutative ring R, and let

'f!

:

R ... s-1R be the natural map.

i f and only i f her

discrete

Show that s-1R is discrete

is detachable from R.

'f!

Show that i f

ring wi th no nilpotent elements,

and S

is

I~

is a

fini te1y

genera ted, then s-1R is discrete. 4. Let S be a multiplicative submonoid of a commutative ring R. Show that if M is a coherent R-module, then S-1M is a coherent S-1g.··module. 5. Let P be a detachable proper prime ideal of a commutative ring R. Show that the Jacobson radical of Rp is detachable, and that its complement consists of the units of Rp • 6. Let /, be a discrete field, and G,b ,c and s indeterminates. R ~ h.[a,b,cl/(ca,cb,c 2

and let S ~ R[sJl(sa + (1-s)b).

)

the ideal of R[Xj generated by 1+aX and l+bX. generated

by c

and b-v ,and

that

Let

Let I be

Show that I n R is

1 : Gi

from

bi

one

,

We

and define two such formal sums to be equal to

the

other

by

applying

left

and

right

89

4. Tensor products

distributive laws, the associative law (ar)11 = a (r-b) , and computations that lie totally within A or B. The ring structure of A is used only to get the A-module structure on the formal sums; in general, all we need is for A to be a right R-module, and the result 1s an abelian group. Here is apreeise speeifieation of the construetion. 4.1 DEFINITICN.

R be a

1'ing,

The tensor product A ® B

R-module,

nbeUnn gnJllp elements

Let

or

F(AxB) on AxB

es

A

R-module,

right

0

and B a

left

defLned as the quol fent of fhe FreI'

bll [he subgnJup

K(AxB)

of

F(AxB) generated

b,}

the form (i)

(a 1

+ n2. b) - (n "

(ii) (a, b , +b z ) (Hi) (m', b) -

b) -

(n, b , ) -

(a 2

,

b)

(n, b z )

(0, 1'b).

We denote the image in A ® B of the element (a,b) E AxB by a

@

b.

A

map f

from AxB to an abelian group C is bi linear if (i) f (a,

+

(ii) f(n,

b,

n z , b)

(iil) F(ar, b)

+

f(o"

+ [(° 2

,

b)

f(n, b , ) + f(a, b 2

bz )

=

b)

)

f(n, rb),

that is, if the induced map from F(AxB) to G takes K(AxB) to zero.

Henee

each bilinear map f from AxB to Ginduces a unique abelian group homomorphism from A ® B to C that takes a ® b to f(a,b). Because of this we ean define a map from A ® B by specifying its values on the elements ® b, and cheeking to see if this assignment is bilinear.

a

The tensor produet A ® B depends on the ring R; if we want the notation to indicate that ring, we write A ®R B.

If A is an S-R-bimodule, and B is

a left R-module, then the abelian group A ®R B admi ts a natural left S-module structure by setting S (0 ® b) = so ® b. In particular, if R is commutative, then A ® B is an R-module. The tensor product of cyclic modules has a simple description. 4.2 'lHEXlREM. lefi R-iIlodule,

Let R be n r'(ng, 1 a 1-ight ideal, ] a Left idea!, nnd B n Then (i)

(ii)

(R/I) ® B

(R/I) ® (R/J)

B/IB '"

R/(I+J),

If I is an idenl, thert these are isomaqJhisms of left R-modules,

PR'OOF.

To prove (i) we deHne a map

~)

from (R/I)

® B

to B/1B by

90

Chapter 111. Rings and modules

'f(r- ® b) = rb,

if

r' -

and '# horn B/lB to (R/I) ® B by t(b) then dJ - "OIJ E IB,

Er,

",

then 1 ® b - 1 ®

b - b' E IB,

/)0

We note that

is well-defined.

'f

in (R/I) ® B,

(b-bO) = 0

@

As ,. ® b = 1 ® "b, the two maps

well-defined.

'f

and t

Also,

if

so t

is

are inverses of

Part (ii) follows from (i) because I(R/J) = (1+J)/J.

each other.

an ideal, then R/r and E/rE are left on (R/l)

so

= ].

= 1 ® b.

® 13

homomorphisms

is given by

R ..modules,

so

= " ® b,

"(1 ® b)

If I is

and the R-module structure and '# are R-module

'f

0

o

In particular, taking r

in (4.2.i), if B is a left R-module, then

R ® B 20 B. 4.3 THEDREM.

LI' t R !JE> (] r-ing,

(ffiiE1A i PROQF.

(Ai} i EI

of LeFt R-moduLes.

(BjljE] a fmnUy

Exercise 1.

4.4 COROLIJ\RY.

®

)

(ffijE]B j '

-

famiLy of r-igh t R-mod.uLes, and.

(1

is a nattu'al isomoqJ/üsm

tllfTE'

ffi j ,jElxJ(A j ®

Bj )

0

The

tensor

pr'oc!ucl

commutat tue ring R is a F,'ee module.

PROOF.

Tlten

Theorems 4.2 and 4.3.

of

fnee

two

In particII!ar,

modules

over

a

Jtl ® Rn 20 Jtln.

0

The tensor product is a bifunctor in the sense that given R....module maps

FA ..·, A ° and 9 : B

A ® B -;

A°

@

there is a natural abelian group map f®g

--> B' ,

defined

B'

commutative, then

to the kernel of B

-->

->

(f@g)(a®b)

If

F(a)®g(b).

R

is

is an R-module homomorphism.

f®g

A pair of maps A

by

B

it is exact at Al for

-->

C is exact at B i f the image of A

A sequence of maps Al

C. i

2, ... ,n-1.

-->

B is equal

A2 --> ••• -> An is exact i f The sequence 0 .... A .... B is exact if -->

and only if A .... B is one-to-one; the sequence A .... B .... 0 is exact if and only i f A -; B is ünto. 4. 5 THEDREM.

co . .,

LE't

1< /Je

a

r' i nfl '

and

A -; B --> C --> 0 and

Let

A' --> 13' ....

0 be exact sequences of r-lght and Zert R-modules ,-especUveiy.

Then

the sequenc:e

(A ® B' ) ffi (B

@

A')

->

B 0 13'

->

C ® C'

->

0

is exac:t.

PROOF.

Let K be the image of (A ® BO ) ffi (B

from B ® B' to C

@

@

A') in B @ BO.

The map

CO takes K to 0, so i t induces 'f : (B ® BO )jK .... C

@

Co.

91

4. Tensor products We shall show that Define a map

'I'

is an isomorphism by constructing its inverse.

: C ® C' .... (B ® B')/1
B ® M is onp-to-one. (i)

holds and I

= 0, so there exist elements = 2 j G i j Yj and Iiria ij = O. Thus

'<jXi

I

E

"j®X j llj

® M

goes to zero in M.

in M, and

G

ij

Then

in R sueh that

o so (ii) holds. Now suppose (ii) holds. We shall prove (iii) by induction on the rank of B. HB is rank one, then B is isomorphie to R and (ii) applies. Let B = BI aJ B2 where the Bi are free of rank less than the rank of B. Let Ai = A n Bi, let A2 be the projection of A into Bz , and consider the commutative diagram. 0

->

1 ®M 1

->

A2 ®

->

B2

Ai ® M

->

A

Bi ® M

->

B ®M

1

0

0

0

!

1

M 1 ®M

The first row is exact by (4.6), and Bi ® M is a summand of B ® M because

B, 1s a summand of B. The first and third columns are exact by induction. easy diagram ehase shows that the seeond column is also exact, so (iii) holds. Suppose ( i i i ) holds. It suffices to verify (iv) for B a fini tely generated right R-module, because if an element is zero in B ® M, then it is al ready zero in B ® N for some finitely generated submodule N of M. Map a finitely generated free right R-module F onto B with kernel K, and let F' be the preirnage of A in F. Consider the commutative diagram An

0

1 0

->

K®M

->

0

->

K®

M

F'® M

->

1

11 -->

F ®M

o

A® M 1

->

B ® Iif

-->

The rows and columns are exact because of (iii).

o An

easy diagram chase

shows that the map A ® M ~ B ® M is also one-to-one. Finally suppose that (iv) holds, and L;'

B®K

->

C ® K

->

1

->

->

1l ® F

->

1l ® M

®F

->

C ® M

1

1

0

A® M

A ® F

1

C

!

The middle column is exact because F is free; because C 1.5 flat.

0

->

1

the last row is exact

Let x E A ® M go to zero in B ® M.

!d in A ® F, which goes to z in B ® F, which comes from

goes to zero in C ® F, hence to zero in C 181 K.

Then x comes from Now

10

Therefore w comes from

11

10

in B ® K.

in A 181 K, which goes to y in A 181 F because it goes to z in B ® F. x = O.

0

EXERCISES 1. Let S be a multiplicative submonoid of a commutative ring R.

Show that S-lR is a flat R··module. 2. Let A be a Hat right module over a ring R,

such that Al is

detachable from A for all finitely generated left ideals I of R. Let B be a finitely generated coherent left R-module.

Show that

A 181 B is discrete. 3. Show that the following conditions on a ring Rare equivalent. (i) 17 iEINMi is a flat left R-module whenever each Mi iso (ii) RIN is a flat left R-module. (iii ) If 'f' : Rn ->R is a map of right R-modules, then countable

of

I,el' 'f'

finitely generated submodule of

/zer' n, kills the vector

j

I f d is

(bO, ... ,bm)t.

the determinant of the matrix, then dbj = 0 for all j.

Either d is

invertible, in whieh ease we are done, or d - 1 is a unit, in whieh ease

< i, and we are done by induction on

a j is invertible for some j

To prove (iii), first suppose

Fg

E 8, so ch is invertible for some k.

As R is loeal, Gib) is invertible for some i , j with i + j = k. invertible I so

r E 8.

Next suppose f

9 ES.

-I-

for some i, so either 0i or b i i5 invertible.

have on

invel'tible coefFiclent.

PROOF. t(as'

the ring R[Xl s wher'e S ts

Ir R is

R[X].

0

Heytfng fleld,

If als +

0'

18'

+ a's) es fm: some

thus R (X) i5 local. R (X) is one-to-one.

0

invertible,

Then

(os' + a'

= tC

IIlI' set oF eLements of R[X]

R(X)

8

is

0

loeal

r'ing

that

containing

R(X).

LI,en so is

)/ss'

S, so either

0

is invertible in R (X), then or

0'

is in S by (7.1.iii);

From (7.1. i.i) we see the natural map from R [X 1 to Suppose that the inequality on R is tight.

cannot be a unit in R(X), then can be

Thus ai is

Then a i + b i is invertible

Let R be a commutative Locot r'ülg ond X an indeter'minate.

7.2 '1'HroREM. Let R(X) be

i.

whence

all

0

If als

carmot be in S, so no coefficient of n coefficients

of a

are

zero.

If

the

inequality on R is con5istent, then 1 cannot equal 0 in R or in R[Xli but the rnap from R[X] to R(X) is one-to-one,

0

104

Chapter III. Rings and modules 7.3 THEOREM.

Let

endomonJ/üsm of

0

R be

conunutolivE'

0

locol

finite-r"anl, fr"ee R--;nodule F.

,-(no ond

e on

Then the

I~enlel

(dempotpnt

of e is

(l

f(ni te-'"Lmk f,"ee R--;nodule.

PROOF.

The endomorphism ring of the R-module R is isomorphie to the

ring R, so the Azumaya theorem (6.2) applies.

0

rf e is an idempotent in a local ring, then either e is a unit, in which case e

1, or 1 - e is a unit, i.n vlhich ca se

=

I"

O.

=

By an impotent

ring we mean a commutative ring with no nilpotent elements, such that any idempotent

is

either

0

or

Any

l.

commutative

local

ring with

no

nilpotents is impotent. 7.4 'rHElOREM.

Let R be

PROOF.

gl1 = Xm + em_lX

Let

9

and

h

=

Suppose G(b j

+ j,

(

m-l

...

+ ClO' + b O'

=

0

we note

if

>

i + j

that

Llib j

I, +l.

Mu1tip1ying this equation bY"ibh_i we get

=

0 trivially

If "

> m, then

Multiplying by

Gjb m_(

is impotent.

If they are all 0,

"

is a unit.

= Gi'

we find that

> i , then

i

+ j > 2n.

,= 0, so Gjbh _ i Now consider

0 as

(Ctib h _ i )"

is idempotent, hence 0 or 1 as R

Gjb m_ i

then R is trivial and the theorem is

Otherwise there exist If h

if

O.

R has no nilpotents, completing the induction.

G,b j

oF Cl monic

factm-

Proceeding by backwar-ds

m.

GoI'h + G1hh_l + ••• + GkbO

trivially true.

G

+ cO, where

+ (ln_I'Xn - 1 +

0 whenever i + j

=

+

b n Xn + b n-1 Xn - l +

= o"xn

We shall show that Clib j induction on

Ir 0 is

impotent dng.

CUl

then ther-e ls a lmit 1\ of R so that ,,-lg is monie.

polynornial in R[XJ,

0l,b j

=

i

and

such that ( + .i

j

0, so Gh

=

=

m and

0; thus we can choose

0

7.5 COROLLlIRY.

fr

0

und

h orT

l'oLynomials

impotent ,'ing, und gh = Xel, lhen then' is u unit

with

r..

cuefficicnts

(llld 0

in

(U)

~ l ~ cl such t!J"1

9 = /\XL.

PROOF. Let" be as in (7.4), and choose i so that Xl g(ljX) and Xd - i h(ljX) are polynomials. Then 1 = Xclg(ljX)h(ljX) = X i g(ljX)X d - i h(ljX),

7. Commutative loeal rings

105

= AX i

so deg Xig(l;X) ~ 0, by (7.4), whenee g(X)

7.6 LEI'IMA. cm iJnpotel!t

of

.

0

Let a be on endomorphism of a Fini te-rank free module over"

If a is

r'irlf].

ni lpotent,

then the ehw'aetcr'ist ie polynomiol

Let A be an

Tl

is apower of X.

0

PROOF.

the n x

Tl

Suppose om

O.

=

identity matrix.

x

matrix representing

Tl

0

and I

I

Then

Xml = (XI - A) (XJn····1I + Xm- 2A + •.. + XA m- 2 + Am-I).

Taking determinants of both sides yields XJnn = f(X)g(X)

F is the eharaeteristic polynomial of o.

where

(7.5).

7.7 'l.'HOOREi'1.

over-

Let

eornrnutativc

0

bc on cndomOl'phism of

0

toeot

ri.ng

wi th

no

H

Then V

•

K wller'e

(j)

autoJnorphisJn on K, and (o-i\)nH PROOF.

H is

0

fini te-r'onl< Free moduLe V

0

=

f (X)

Le t

nUpotents.

dw.r'oetcristie pol!}nomiol, ond suppose f(X) uni t

Then f is apower of X by

0

be

(X-i\)Tlg(X) wherc g(i\)

,-onlF.

For each j = l, ... ,m consider the sequence x n = o;CCD(o,p'j). By the divisor chain condition, there exists n such that xnlxn+l; set

= n and 0j = o/x n = GCD(a,Pjl relatively prime, so are the 0j.

= GCD(o,Pj+ll.

there

00a1·· ·a m •

e

exists 00

GCD(aO,Pj)

=

such

that a

1 for each j. 0j

=

As the Pj are pairwise

Repeated application of (l.l.iv) shows It

remains

But

= GCD(0,Pj+1) = GCD(oOOj,Pjpj)

to

show that

1. Cance11ation monoids

1.1 LEMMA.

Let

111

M be

condition. und Let a.b E M. (i) a

= a+cb-

a

GCD-monoid

satisFying

the

divisor

chain

rhen there are a+.a-,c.b+,b- E M such that

und b = a-cb+.

(ii) a-Ia+ and b-Ib+. (iii) a+, c, and b+ar'e pairwise reLativeLy prime.

PHOOF. Let x = a/d and y = b/d where d = GCD(a,b). Then x and y relatively prime, so by Lemma 1.6 we can write d = a-cb- where a-Ixn , b-Iy'll, and c is relatively prime to both x and!}. Let a+ = xab+ = yb-, so a = dx = a+cb- and b = d!} = b+ca-. As xn +1 , ym+1 and c pairwise relatively prime, so are a+, b+ and c.

are and and are

0

As an example of the decomposition of Lemma 1.7 consider the monoid m+ of positive integers. If a and b are in m+, then a+ and a- are the

largest factors of a and b respectively, that consist of primes that occur more often in a than in b. Similarly b+ and b- are the largest factors of b and a respectively, that consist of primes that occur more often in b than in a, while c is the largest factor of a (or b) that consists of primes that occur equally often in a and in b. If a = 560 = 23 '3.5'7 and b = 300 = 22 '3'5 2 , then a+ = 56, a- = 4, c = 3, b+ = 25, and b- = 5. Note that GCD(a,b) = a-cb-. 1.8

~

(Quasi-factorization).

Let xl, •.• ,xk be eLements oF a GCD-

monoid M satisFying the divisor chain condition.

Then there is a famHy P

oF pairwise reLativeLy pl'ime eLements oF M such

that

each xi

is

an

associate oF a product oF eLements oF P.

PHOOF.

k = 2. We construct sequences r n = = b 1 (n)"'bm (n)(n) as follows. Let m(O) = 1 and = b 1 (0) = x2' We shall suppress the dependence of

Consider first the case

al(n)'''a m(n)(n) and sn

rO = al(O) = xl and So m, a i and b i on n for cleaner notation. Suppose we have constructed r n = al"'am and sn = b 1 "'bm with GCD(ai,aj) = GCD(bi,b j ) = GCD(ai,b j ) = 1 if i t- j. Then we construct r l1 +1 and sn+l as follows. For each iwrite

ai

= atcibi and b i = aicibt as in Lemma 1.7. Then set (a!/ai)"'(a:/a;)bi"' b;

r n +1

=

sn+1

= ai"'a;(b!fbl)'" (b~fb;).

We easily see that, except for the pairs (at/ai, ail and (btfbi, bi l , the

112 2m

Chapter IV. Divisibility in discrete domains factors of r"n+1 and of sn+1 are pairwise relatively prime.

principal ideals Mr'OsO' Mr"lsl' Mr"2s2""

form an ascending chain.

divisor

such

chain

condition

there

is

n

that

r"nSn

The By the

r"n+1sn+1

at"'a~t"'b~, so the elements ai, ci and bi are all units.

Thus the

elements a1"" ,a m,b 1 ,.·. ,bm are conjugates of at, ... ,a~,bt, ... ,b~ so are pairwise relatively prime.

1t now suffices to show that if we can write

the elements in the family

E

=

{ai, (at/ai), bi, (bt/bi) : i.

=

l, .. ,m}

as products of pairwise relatively prime elements, then we can do the same for a1, ... ,Gm,b l , ... ,bm.

Suppose that

Q

is a finite family of pairwise

relatively prime elements, and that each element of E is an associate of a product of elements of Q. some

element

of

E.

We may assume that each element of Q divides Then

by

Lemma

1.7

each

of

the

elements

a1' ... a m,b 1 , ... bm is an associate of a product of elements in Q U {c1""'c m }, and the elements in the latter family are pairwise

relatively prime. If k

> 2 we proceed

by induction on k.

Let P = {P1"" ,Pm} be a family

of pairwise relatively prime elements such that each of xl"" ,xh.-1 is an associate xn i

of

a

product of elements

of P.

By

(1.6)

we

can wri te

= Goal" ·am where Gi. divides apower of Pi.' and CCD(oO'pi.) = 1 for = 1, ... ,m. From the case k = 2 we can construct a finite family Si of

pairwise relatively prime elements such that 0i and Pi are associates of products elements of Si' and each element of Si. divides apower of Pi' Then {ao} U sl U ••. U Sm forms a family of pairwise relatively prime elements for x1""'x m ,

0

EXERCISES 1. Show that the set of positive even integers, together with 1,

form a (multiplicative) discrete cancellation monoid M. elements

°

Find

and b in M that do not have a GCD.

2. Construct elements

G,

b, and c in a discrete cancellation monoid

such that albc, and CCD(a,b)

=

1, but a does not divide c.

3. Show that the set of positive integers that are congruent to I module 3 form a (multiplicative) discrete cancellation monoid M. 1s M a GCD-monoid?

1. Cancellation monoids

113

4. The least cammon multiple LCM(a,b) of two elements a and b in a cancellation monoid M is an element m € M such that alm and b Im and, if ale and ble, then mle. Show that, if LCM(a,b) exists, then GCD(a,b) exists and is equal to abjLCM(a,b). Show that if M is a GCD-monoid, then LCM(a,b) always exists. Construct elements a and b of a discrete cancellation monoid exists, but LCM(a,b) does not.

M

such that GCD (a,b )

5. Let M be a cancellation monoid. Define what it means to be a greatest common divisor GCD(a1, ..• ,an ) or aleast commen multiple LCM (al' •.• ,an) of the finite family a1' .•• ,an of elements of M.

Show that if M is a GCD-monoid, then these always exist. 6. Let R be a GCD-monoid. Show GCD(a,bne) = GCD(a,e) for all n.

that

if GCD(a,b)

= 1,

then

7. Let M, = {2n : n € m}, and let M2 = M,\{2}. Use these monoids to construct a Brouwerian example of a discrete cancel1ation monoid with recognizab1e units in which divisibility is not decidable (you can't tell if 418). 8. Let M a submonoid of a multip1icative abelian group G. y

in G, we say that

x

divides

y

For x and (relative to M), if yx- 1 € M.

Define GCD in G using this notion of divides. Show that i f M gene rates G as a group, and M is a GCD-monoid, then GCD(a,b) exists for all a,b in G, and (1.1) holds. Show that every cancellation monoid can be embedded as a submonoid in an essentially unique abelian group that it generates as a group. 9. Let P be the set of principal ideals of a GCD-domain R, partially ordered by inclusion.

Show that P is a distributive lattice.

10. Let a, band e be elements of a cancellation monoid. Show that if GCD(ca,cb) exists, then GCD(a,b) exists, and GCD(ea,eb) = c-GCD(a,b) . 11. Let R = l[Y,X l ,X 2 , ... I/I, where I

is the ideal generated by {X i +1Y - Xi : i ~ I}. Show that R is a GCD-domain that does not satisfy the divisor chain condition. Show that there is no finite family Q of pairwise relatively prime elements such that Y and Xl are associates of products of elements of Q.

114

Chapter IV. Oivisibility in discrete domains 12. Show that the elements a+,a-,c,b+, and b- of Lemma 1.7 are unique up to units.

2.

UFO' S AND BEZCl1T

~NS

Questions involving factoring are touchier in constructive algebra than they are in classical algebra because we may be unable to tell whether a given element has a nontrivial factorization. OUr definition of a unique factorization domain is straightforward. 2.1 DEFINITION.

A discrete domain R is called a unique factorization

domain, or UFO, if each nonzero element r' in R is either a unit or has an essentia11y unique factorization into irreducible elements, that is, if and r- = Q1'" qn are two factorizations of r- into irreducible elements, then m = n and we can reindex so that Pi - qi for each i. We say that R is factorial if R[X) is a UFO.

r-

= P1'"

Pm

Note that a UFO has recognizable units, that is, the relation u 11 is decidable.

oiscrete fields are trivial examples of UFO's; the ring

well known to be a UFO.

~

is

The somewhat peculiar looking definition of

Factor'ia!. agrees with our usage of the term as applied to discrete fields,

and allows us to show that R[X] is factorial if R iso The following is a Brouwerian counterexample to the classical theorem that if R is a unique factorization domain, then 50 is R[X). 2.2 EXAMPLE. numbers, i unique

2

Let a be a binary sequence, «l the field of rational

= -1, and k = Un«l(ia n ).

factorization domain.

Then k is a discrete field, hence a

However we cannot factor X2 + 1

into

irreducibles over k[X). The notions quasi-UFO, bounded GCD-domain, UFO, and factorial domain are classically equivalent, but the ring k[X) of Example 2.2 is a Brouwerian example of a bounded GCD-domain that is not a UFO, while the field k of Example 2.2 is a Brouwerian example of a UFO that is not a factorial domain. In (3.5) we shall give a Brouwerian example of a quasiUFO that is not a bounded GCD-domain. I t is easily verified that the other implications hold. 2.3 'lHEOBEM.

Let R be a discrete domain. Then

(i) If R is Factorial , then R is a UFD.

115

2. UFD's and Bezout domains (ii) (iii)

rf

R is a UFD, then R is a bounded GCD-domain.

IF R ts a bounded GCD-domain, then R is a quasi-UFO.

A multiplicative

submonoid S of a

commutative

ring R is

0

called

saturated if xy E S implies x E S. 2.4 TiIEORF.:M..

Let

R be

a

discr'ete

submonoid of R not containi.ng O.

mui

domain

S

is

a

muttipUcative

S-lR •

(i) If R is a GCD-domain, then so is (ii) If

S

Then

saturated

arui

detachab!e,

recognizable units.

To show (i) we observe that GC:rJ(a/s,b/t) = GCD(a,b)/l. To show (ii) we shall show that o/s is invertible i f and only if a ES. If als is invertible, then ab/SI = 1/1 for some bit ES-IR, so ab = st E S, whence PROOF.

a E S.

Conversely, if a

2.5 THElOREl'I.

Let

E

S, then a(l/a)

=

1/1.

0

R be a UFD (md tet S be a

muLtipU.catilie submonoid of R not conlaining O.

satur'ated detachoble

Then S-l R 1S also a UFD.

If r/s ES-IR, then rand s can be written as products of

PROOF.

irreducibles in R. By (2.4) we can decide for each irreducible factor of whether it is invertible in S-lR or not. Those irreducible factors of r

r

that are not invertible constitute the unique irreducible factorization of r/s in S-lR . 0 2.6 COROLIARY.

Ir

R ts

factor-ial.,

(md

S

is

a

detachable

saturated

muLtipUcative submonoid not contni.ning 0, then S-lR is also factoriol.

0

The assumption of Theorem 2.5 that S be saturated is essential. Consider the following Brouwerian example. Let 0 be a binary sequence containing at most one 1. Let R = S-l~ with S

=

{q ; q = 1 or q =

znn

for some m,n such that on

Then S is a detachable multiplicative submonoid of

~

=

1).

not containing O.

But we can't tell whether or not 2 is a unit in R. 2.7 DEFINITION. A Bezout damain is a discrete domain such that for each pair of elements o,b there is a pair s,t such that S

R such that

2ß/a -

e =

a'

+ b'v'-19

with Ib' I ~ 1/8 and la' I ~ 1/2, so N(2ß/a - 8') < 1. The only problem is that a might divide 2ß but not ß. However it is easy to show that 2 divides '( E R i f and only i f Nh) is even, so 2 is a prime in R. Therefore if aö = 2ß, then either 21ö, in which case alß, or 21a. But if 2 is a common factor of a and ß, then we are done by induction because our condition depends only on ß/a.

120

Chapter IV. Divisibility in discrete domains

the other hand, N is not a Euclidean norm because there is no e in R such that N(ß/a - e) < 1 if ß = (1+vCI9)/2 and a = 2. In fact, R does not admit a Euclidean map (see Exercise 11). 0 On

Classically any bounded domain R admits a pseudonorm:

set v(x) equal

to the least n such that x is bounded by n. If R is a principal ideal domain, then v is a multiplicative Dedekind-Hasse norm; constructively we must require more.

The following theorem gives the construction of such a

norm if the principal ideal domain is also a UFD. 3.5 'l'HEDREM.

An!}

UFD admits

a

muttiplieative

pseudonorm on a diser-ete domain R.

nar-m.

IF R is a Bezout domain,

Let

v

be

a

then v is a

Dedelünd-Hasse map.

Let R be a UFD. For nonzero a in R define v (a) = 2n where n is the number of primes, including multiplicities, in a prime factorization of a. Clearly II is a multiplicative norm. PROOF.

Suppose R is a Bezout domain and a,b ER. There exists d in R such that (a,b) = (d). If a ~ 0, then either a ~ d, so a divides b, or there exists d' ~ d such that II (d') < v (a). Therefore v is a Dedekind-Hasse map.

0

Absolute value is a multiplicative Euclidean norm on the ring of If F is a discrete field, then the degree function is a integers. Euclidean map on the ring F [X]. Note that the degree function is not multiplicative, but the Euclidean norm zdeg fis. A multiplicative Euclidean norm on the ring of algebraic integers ~[J2] is provided by v(a + bJ2) = la 2 - 2b 2 1 = I(a + bJ2)(a - bJ2)I. We easily see that v is multiplicative. To show that v is Euclidean, let a + bJ2 and e + dJ2 ~ 0 be elements of ~[J2]. As J2 is irrational, we can find rational numbers integers

p

and

q

such that

mand n with Ip - ml



is either finite or infinite.

0m = I} is bounded by n.

is equivalent to 2IN pdnciple 21N = UnC n ?

j

is in Crl ) .

°

be a

binary

sequence,

Let I, be the two and define

= R/J l' Sn

lL

Then :>:1]=1 R/(Ji+ J 1 )

M/(JIM)

-

--

27=1 R/(T;+ J1 )

We can map ~~ onto 27=1 R/(I i + J 1 ), so S = 0 by (11.7.5). As (a) holds we may assume that m = n. To prove (bl it suffices, by

as S-modu1es.

symmetry, to show that I k t;: J" for I,

xM where

(K:x)

r'x C

1, ... ,n.

=

-

Let xE I k ,

Then

2:~"~,+1 R/(I; :x)

Applying

K).

(al

to

xl
we

get

EXERCISES 1. Show that a finitely presented abelian group is a direct sum of a

finite number of infinite cyclic and finite cyc1ic groups. 2. Give a Brouwerian examp1e of a cyclic group that is nei ther finite nor infinite. 3. Show that if an m x n matrix inverse, and m < n, then k

=

OV8r

a commutati ve ring R has a 1eft

O.

4. Let H be a detachab1e subgroup of a free abelian group (free Z'-module) on a discrete set.

Show that for each h in H there

exists x in H such that h ( <x>

and H/<x>

is

subgroup of a free abe1ian group on a discrete set.

a

detachab1e

2. Finitely presented modules

133

5. Use Exereise 4 to show that a detaehable subgroup of a free abelian group on a eountable diserete set is a free abelian group on a eountable diserete set. (Construet generators xi for H induetively so that H/<xl"",xn_l> is a detaehable subgroup of a free abelian group on a eountable diserete set for eaeh n L 1.) 3. TORSICfi KlIXJLES, p-aJn'(lIIENTS, ELPl'mNTARY DIVISORS

Let M be a module over a diserete integral domain R. The torsion suI::IIIodule T (M) of M is defined to be {m E M : um = 0 for some a f- O}. We easily see that T(M) is a submodule of M. If T(M) = M, then we say that M is a torsion module. If d is a nonzero element of R, and say that M i s bowlded by d. 3.1 'l'IfI!DREK.

Let M be a

dM

= 0,

finitety p,-esented module ouer a

then

we

prineipal

ideal domain R.

Then T(M) is a finitely presented detaehable submodule of

M,

61

and

M

=T(M)

~

for

some

n.

Mor-eover

{d

ER:

dT(M)

=

O}

is

a

nonze,-o prineipal ideal.

PROOF. By Theorem 2.3 there exist prineipal ideals 1 1 d 1 2 d '" d Im such that M is isomorphie to the direct sum R/I 1 61 R/I 2 6l ••• 6l R/I m. If I k = 0 for all k, then the eonelusions are elear. Otherwise ehoose k sueh that I k f- 0 and I i = 0 for > IL Then T(M) R/I 1 61 ... 61 R/I k and M T(M) 6l ~-k. Moreover I k {d ER: dT(M) = O}. 0

=

=

Let M be a module over a eommutative ring R, and let dER. d-caaponent of M is defined by Md = {m easily verify that all

a,b E

Rand

3.2 LEMMA.

Md

M(an) Let

is a submodule of

E

M.

M :

dkm

Observe that

Ma + Mb ~ Mab

We for

= Ma for all n > O.

M be

R. Let a and b Ma EIl Mb' and, if M

a module ove,- a eommutatiue r-ing

be strongly relatively p,-ime elements of R.

Then Mab =

is

Mab onto Ma

finitely

Then the

= 0 for some k}.

generated,

the

pr-ojeetion

multiplieation by an etement of R.

of

is

,-ealized

by

The module Mab is eyelie if and only

if the submodules Ma and Mb ar-e eyel ie.

PROOF. To show that Mab = Mo Gl Mb' it suffiees to show that K = Ka EIl Kb for eaeh fini tely generated submodule of Mab I so we may assume that M is finitely generated. Then al~b'~Mab = 0 for some positive integer k.

Chapter V. Prineipal ideal domains

134

There exist sand t in 1I"b = saR • Then

so

Mab = Ma

R

sueh that

sa k

+

tb k

= 1.

Let 1I"a =

(i) 1I"aMab ~ Ma and 1I"bMab ~ Mb' (ii) 1I"aMb = 0 and 1I"bMa = 0, (iii) 1I"ax = x for x E Mb and 1I"bx = x for x E Ma , e Mb and multiplieation by 1I"a gives the projeetion of

and

tb k

Mab

onto

and Mb is generated by

1I"bx.

Ma •

lf

Mab

= Rx, then

Conversely suppose

= 1I"bx

Rx and z

E

is generated by

Ma

1I"aX,

= Ry and Mb = Rz, and let = Rx' 0

Ma

=y

x

+ z.

Then

= 1I"ax

y

E

Rx, so Mab

Let 3.3 'lBEDREM. Let M be a module over a eommutative ring. a = p(!)e(!)"'p(m)e(m). where the p(i) are pairwise strongLy reLatively prime. Then Ma = Mp (!) e ... e Mp(m)' and, if M is finUeLy generated, the

projection of

element

of

PROOF.

onto

Ma

Mp(i)

is realized

by

multiplieation

by

on

ring.

the

Apply Lemma 3.2 repeatedly.

0

lf p is a prime in a diserete integral domain R, then an R-module M is said to be p-primary if Mp = M. If dM = 0 for some nonzero element of R whieh is a produet of powers of strongly relatively prime primes, then we ean deeompose M into a direet sum of primary submodules by (3.3). 3.4 'lBEDREM.

Let

presented p-primary of

R_odules,

PROOF.

R

be a

PID.

R_odule,

and p a prime in

then

M is

R.

If

M is

a

finitely

isomorphie to a finite direet

Stlln

eaeh of the form R/(pn) for some n > O.

By the strueture theorem (2.3), M is isomorphie to a finite

direet sum of R-modules, each of the form R/I for some prineipal ideal I. As M is p-primary, eaeh I eontains a positive power of p. If pm E I = (a), then pm = ab. Beeause p is a prime, we ean write a = UIP where u is a unit; so I = ( pn ) .

0

The powers of p oceurring in (3.5) are ealled the elementary divisors of M. If M ean be written as a direct sum of primary submodules, then the elementary divisors of Mare the elementary divisors of the various primary submodules of M.

135

3. Torsion modules, p-eomponents, elementary divisors EXERCISES

1. Find the primary eomponents of the abelian group Z/l2l. 2. Let R be a Bezout domain and p a prime in R.

Show that R/(pm) is

Prove Theorem 3.4 for R a Bezout domain.

a valuation ring. 4. LINEAR TRANSPORMATlCHl

Let V be a finite dimensional veetor spaee over a diserete field k, and let T : V

~

V be a linear transformation.

We ean make the veetor spaee V

into a module over k[X) by defining Xu =T(u) for eaeh u Cayley-Hamilton

theorem,

the

eharaeteristie polynomial of T).

k[X)-module

V

is

E

By the

V.

bounded

(by

the

We shall show that the k [X I-module V is

finitely presented. 4.1

LEMMA.

u1"" ,un'

and

Let V be a uector space ouer a discrete field k with basis T : V -+

V a

tinear"

transformation

such

that

Let el, ••• ,e n be a basis For k[X)n, und let ~ : k[X)n 2 F i (X)e i to 2 Fi(T)u i • Define d t E k[X)n by 2 ajiu j •

T(u i ) -+

=

V take

d i = Xe t - 2J=la ji e j • Then

ker

~

PROOF.

is a free F [X )-module wi th basis d 1 , .. · ,d n . Obviously d 1 , ..• ,dn

ker" ~, where gi

E

ker~.

Suppose gle1 + ••• + gnen E

Using the relations Xe i = d i + 2J=lajiej' we ean

E k[X].

write = h 1d 1 + ••• + hnd n + bleI + ••• + bne n ,

gle 1 + ••• + gnen

So bleI + ••• + bne n E ker ~. Sinee ul"" ,un is a basis of V as veetor spaee over k, this implies that eaeh b i = O. Henee

where b i E k.

d 1 , ••• ,dn generate ker

~.

If h 1d 1 + ••• + hnd n = 0, then 27=lh t Xe i = 27=12J=lhiajiej" h i t 0,

If some

then we may assume that the degree of h 1 is maximal among the

degrees of h l , .•. ,h n • linearly independent.

But hlX = 27=lhiali' so h 1 = O.

Thus d l , ... ,dn are

0

By (2.3) the k[X)-module V ean be written as V = Cl $ ••• $ Cs ' where the Ci are eyelie k[X)-modules, isomorphie to k[X)/(F i ) for nonzero monie polynomials f i , with f i dividing Fi +1 for i = 1, ••. ,s-1. The polynomial f s generates the ideal {g E k[X] : gV = O} = {g E k[X) : g(T) = O}, and is ealled the minimal polynomial of T.

By the Cayley-Hamilton theorem, the

Chapter V. Principal ideal domains

136

minimal polynomial of T divides the characteristic polynomial of Ti the two polynomials are equal if and only if V is a cyclic k[X]-module. If A is a root of the characteristic polynomial of T, then we say that A is an eigenvalue of T. If A is an eigenvalue of T, then there exists a nonzero u € V, called an eigenvector of T, such that (T - A)U = 0, that is, Tu = AU. So X - A must divide the minimal polynomial of T, whence A is also a root of the minimal polynomial of T. The decomposition of V into a direct sum of cyclic k[X]-modules provides a basis of V as vector space over F relative to which T has a canonical form. Let ci be a generator for Ci over h [X 1, and suppose Fi has degree m. Then ci,Xci, ••• ,xm-1ci is a basis of Ci as a vector space over , O.

=

0,

then a

0

4. Show that any Heyting field, and any denial field. is an impotent ring. 5. Let I< be the integers modulo 2. I< [X ]/(X 2

X) ,

-

Let E =k[Xl/(X 2

)

and F

=

Show that E and F are algebraic over h. but that

neither is a field.

Why doesn't (1.9) apply?

6. Let h be a discrete subfield of a commutative ring E.

If a E E

satisfies an irreducible polynomial over k of degree n, show that 1,er, ... ,er n- 1 is a basis for kral over k.

m2 denote the ring of integers localized at 2. Use the pair m ~ ~ = m[1/2] to show that the phrase 'or R contains a nonzero

7. Let 2

2

nonunit' cannot be removed from the conclusion of the weak Nullstellensatz. Use the same pair to construct a Brouwerian example where the weak Nullstellensatz fails because R does not have recognizable units.

145

2. Algebraic independence and transcendence bases 2. ALGEBRAIC INDEPENDmcE AND TRANSCENDmcE BASES

Let k ~ K be commutative rings. The elements xl, •.. ,x n of Kare called algebraically independent over k i f whenever f € k[X1, ••• ,Xnl, and If k is discrete, xl' ... ,xn are called f (xl' ... ,xn ) = 0, then f = O. algebraically dependent over k i f there exists nonzero f in k [X I' .•. 'X n 1 such that f(xl' .•• 'xn ) = 0; in this case, xl, ... ,xn are algebraically independent if and only if they are not algebraically dependent. If K is discrete, and S is a finite subset of K, then we say that S is algebraically dependent, or algebraically independent, if its elements in some (any) order are. We may reduce the notion of algebraic dependence to that of algebraic elements as follows. If

2.1'lHEDREM.

k~K

are discr'ete fietds,

then xl, •.. ,xn in Kare

algebraicaUy dependent ouer k if and only if ther'e exists i such that xi is algebraic ouer k(xI, ••• ,xi_I).

PROOF.

Suppose xi satisfies a nonzero polynomial f (Xi) with coefficients in k(xl, .•. ,xi_l). Then there are nonzero polynomials such that f (Xi.) = 9 € k[X l ,··· ,Xi 1 and h € k[X I ,··· ,Xi_I] g(Xl, •.• ,Xi._I,Xi.)!h(Xl, •.. ,Xi._l) and g(xI' ••. 'xi.) = O. hence xl' ••. ,xn ' are algebraically dependent over k.

So xl' ... 'xi.' and

Conversely, suppose xl' ..• ,xn are algebraically dependent over k, so there exists nonzero 9 € k[Xl, ... ,Xnl such that g(xl' ... 'xn ) = O. Then g(xl' .•. ,xn_I'Xn ) is a polynomial in Xn with coefficients in k[xI, ... ,xn_l], satisfied by xn . If g(xI, ... ,xn_I'X n ) = 0, then any nonzero coefficient of g, where 9 is thought of as a polynomial in Xn with coefficients in k[X I , ••• ,Xn _l ], gives an algebraic dependence among xI, ... ,xn_l' and we are done by induction on n. then x n is algebraic over k(xl' .. "xrt-l).

If g(xI, ... ,xn_l'X n ) # 0,

0

As a corollary we have the exchange property for algebraic dependence. 2.2 COROLIARY. algebraic

ouer

Let k ~ K be discr-ete

k(y),

then

either

y

is

fietds, and x,y € K. algebraic

Oller'

k(x)

If or'

X

is

x

is

algebraic ouer k.

PROOF. By (2.1) we have y,x are algebraically dependent over k, so x,y are algebraically dependent over k. The conclusion follows from (2.1). 0 Corollary 2.2 parallels the exchange property for linear dependence of

146

Chapter VI. Fields

vectors: if x is a linear combination Yl""'Y n ' then either x is a linear combination of Y1"" 'Y n -1' or Yn is a linear combination of Y1"" 'Y n -1 and x (see Exercise 2). Let h!;; K be discrete Fields.

2.3 'l'HEXlRDl. subsets

oF

K such

that

K is algebr-aic

algebraically dependent over8 0 and 8, such that #So = ItT,

PROOF. 8 0 = T.

I~,

Let 8 and T be Fini te

over- Id8).

Then

either

T

is

or- we can par-tition 8 into finite subsets

and

K is algebr-aic over- I«T

Proceed by induction on m Otherwise let x E T\S.

=

If m

It(T\8).

=

U 8,).

0, then we can take

Let 8 = {s1, ... ,s,J, with the elements of

n S listed first. Repeated application of (2.2) to x and l«s1"" ,s j)' for j 1 1. Note that m is a power of p since m divides q. As h(O)m = -a. we can set b = -h(O)m/p , and a = bq . 0

h(X)m

Let E be an impotent ring and Iz a discrete subfield of E.

The

separable closure of Iz in E is the subfield of E consisting of those elements that are separable over k. i f the separable closure of

I~

The field k is separably closed in E

in E is k.

The following generalizes

Coro11ary 5.5 and shows that the separable closure is separably closed. 6.7 THEOREM. Let k ~ E be diserete fieids. Suppose a E E is algebraie (separable) over k, und ß E E is separable over kral. Then k[a,ßI = k[ßI

164

Chapter VI. Fields

Fm- same

e

(und

PROOF.

Choose q by (6.4), ei ther 1 or apower of a prime p that is

zero in k, (6.2),

ß is sepora.bie aver k). Then k [a,ß] = h [0 ,(lq] by

so that ßq is separable over I

= O.

0

S is one-to-one,

There is an analogous

result in the eontext of fields. 8.2 LEMl'lA.

I