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
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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, then b = a.
band b
c, then a
c.
An n-ary relation on a set S i5 a propez:ty P that 1S applicable to n-tuples of elements of S, Xi
~
Ui' for
i.
and is extensional in the sense that
if
= 1, ... ,n, then P(xl""'x n ) if and only if P(Ul""'Yn )'
Note that equality is a binaz:y r'elation in this sense.
The relation f' is
decidable if for eaeh n-tuple xl"" ,xn either f'(xl"" ,xn ) holds or it doesn't hold. A unary relation P on S defines a subset A
=
(x ES: f'(xl) of S:
an
element of A is an element of S that satisfies P, and two elements of A are equal i f and only i f they are equal as elements of S.
rf A and Bare
subsets of S, and if every element of A is an element of B, then we say that A is contained in B, and wri te A B and B
~
B.
Two subsets A and B of a set S
are equa1 if A
~
subsets of S.
We have described ho", to eonstruct a subset of S, and ",hat
~ Ai
this is clearly an equivalence z:elation on
it means for two subsets of S to be equal. of all subsets, or the power set, of S.
Thus we have defined the set A subset of S is nonempty i f
8
Chapter I. Sets
there exists an element in it. The union of two subsets A and B of S is the subset of S defined by AU B =
{x
Ei
~
x
:
~
A or x
interpreted constructively, subset
The y
'or'
in this
definition
is
in A U 13 we can determine which
is in (although we may not be able to tell whether it is in
x
both) . {1,2}
B}.
~
so given
In terms of existence, x E A U 13 means that there exists
=
so that i f i
1,
then x
E
A,
and if
interseetion of A and B is the subset A n B We
regard
the
relation
of
=
i
=
2,
then x
in The
E B.
{x e s : x E A and x E B}.
inequality
as
conventional
and
not
necessarily the denial of equality; the interpretation of the symbol a T iJ will depend on the context. defined by a T b if a natural
inequalities:
b
=
if
(1
Every set admits the denial
is impossible.
inequality
Some sets admit other more
and bare binary sequences,
then the good
interpretation of a T b is that there exists " such that an T b".
If a
set has no speeified inequality, we interpret n T b as being the denial inequality.
We employ the usual terminology involving inequality: to say
a and bare distinet means
(l
Tb; to say
is nonzero means
(l
(1
T O.
An inequality on a set may be one or more of the following:
eonsistent:
a T n is impossible.
syn:metrie:
if
(1
T b, then b T a.
eotransitive: if a T c, then for any b ei ther tight:
ifa T b is impossible, then a
(1
=
T b or b TC. b.
We almost always want an inequali ty to be symmetrie because n T b
is
supposed to embody the idea that n and bare distinet, which should be a symmetrie relation.
It is also natural
to demand eonsisteney,
praetice this property is usually unneeessary.
but in
A eonsistent, symmetrie,
eotransitive inequality is called an apartness; the inequality specified above for the set of binary sequences is a tight apartness, as is the standard inequality on real numbers (see 11.3).
The denial inequality
need not be an apartness, nor need it be tight. to
be
equivalent to the denial inequality using the law of excluded middle.
A
An
inequali ty
is
said
to
be
standard
if
tight consistent inequality is standard beeause ~(a
one
= b).
it ~~(a
can
be
shown
T b) is equivalent to
The denial inequality is trivially a standard inequality.
important exeeption
standard inequalities.
(loeal
rings),
With
we will be interested only in
It should be noted, however, that the requirement
2. Sets, subsets and functions
9
that an inequality be standard has very little constructive content: one cannot even prove that every standard inequality on a one-element set is consistent
(a
statement
that
can
be
refuted using LEM need not
be
refutable) . A set S with a
consistent inequality is discrete if
elements a and b of S, inequality,
then S
inequality.
either a
b or a t b;
=
is discrete
if
it
is
fOl:
any two
if S has no specified
discrete
under
the
denial
The inequality of a discrete set is the denial inequality,
and is a tight apartness.
However the assertion that a set is discrete
does not a prior-i refer to the denial inequality, but rather to whatever inequali ty comes wi th S:
to say that a set S of binary sequences i s
discrete means that for all a and b in S, ei ther a such that an t bn . The set 71 of integers is discrete.
=
b or there exists n
The set Cl of rational numbers is
also discrete: a rational number is a pair of integers m/n with n t 0, two rational numbers m,/n, and m2 /n 2
being considered equal if m, n 2
=
m2n, .
Another example of a discrete set is the ring 71 12 of integers module 12: the elements of 71 12 are integers, and two elements of 71 12 are equal if their difference is divisible by 12. As we can decide whether or not an integer is divisible by 12, the set 71 12 is discrete. If x E Sand A is a subset of S, then we define x
~
A to mean that
x t a for each a E A; if the inequality on S is the denial inequality, or if S has no specified inequality, then x A.
The complement of A in S is S\A
~
A if and only if x cannot be in
{x ES: x 'l Al.
=
proper if there exists x in S such that x 'l A.
A subset A of S is
A subset A of a set S is
detachable if for each x in Seither x E A or x 'l A. Sn
we
define
their
cartesian
product
SI x S2 x •.• x Sn to consist of the n-tuples (x1,x2'." ,x n ) with xi in Si for each i . Two such n-tuples (x1,x2"."x n ) and (Yl'Y2'."'Y n ) are equal if xi
=
Yi
for each
Relations may be identified with subsets of
i.
Cartesian products: a binary relation on S is a subset of S x
s.
If A and B are sets, then a function from A to B is a rule that assigns to each element a of A an element ((a) of B, and is extensional in the sense that f(a,}
= f(a 2
}
whenever
that F is a function from A to B. equal if f(a)
(l,
=
= g(a} for each a in A.
defined by setting f(a}
=a
a?.
We write f : A
->
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 <; a and x
easily seen that such c is unique, if i t exists.
~
Similarly c
the least upper- bound, or supr-emum, of a and b if for each x c " x if and only if
b.
a 1\ b, I t is
a V b 1.S
=
we have
E L
x and b " x.
~
(1
=
A lattice is a partially ordered set in which any two elements have an If 5 is a set, then the set of all subsets of 5,
infimum and a supremum.
ordered by inclusion, forms a lattice: the supremum of A and B is A U B and the infimum is A
n
B.
The set of positive integers,
setting a " b i f b is a multiple of
(1,
ordered by
is a lattice: the supremum of a and
b is thei r least common multiple, the infimum is their greatest common divisor.
Note that
(1
~
equivalent to a 1\ b
=
a.
b is decidable in a discrete lattice because it is
If a lattice has a least element, then we denote that element by 0; if the 1attice has greatest element, we denote it by 1. A lattice is distributive i f i t satisfies the identity a /\ (b V c)
The lattice of all subsets of a
(a 1\ b) V (a 1\ c).
set is distributive.
The following
S-element discrete lattice is not distributive.
A lattice 1S modular- i f a V (b 1\ c) = b 1\ (a V c) whenever a is easily seen that any distributive lattice is modular i nondistributive lattice shown above is also modular.
~
b.
It
the 5-element
If G is a finite
abelian gwup, then the set of finite subgmups of C is a modular lattice, which is distributive only if G is cyclic.
More gene rally, the set of
submodules of an R-module form a modular lattice.
The simplest nonmodular
lattice is the following 5-element discrete lattice.
*
If a
~
b are elements of a partially ordered set P, then we use the
interval notation [a,b 1 to denote the set {x
E
P : n
~ x "
b}.
If P is a
lattice, then [o,b J is a lattice wi th the same suprema and infima as in f'.
22 A
Chapter 1. Sets key
fact
about
modular
lattiees
is
that
isomorphie lattiees for any elements a and d.
and
[al\d ,d]
[a ,aVd
1 are
We prove this in a slightly
disguised form. 5.1 LEMMA.
Let
IJ
b anLl e <; d be elements of a modular-
,;
laU tee L.
TJefine fIx) = a
V (b 1\ x)
=
=c
V (d 1\ y)
=d
g(y)
b 1\ (a V x) 1\ (c V y).
Then 9 maps [f(c).f(d)] isofnorphicoUy onto [g(a),g(b)] wUh inver-se f.
It suffices to show that if c <; x
PROOF.
then fgf Ix)
~ cl,
=
f (x) .
We
ean write fgf(x) as
fgf(x)
('" )
b 1\
(IJ
V
V (d 1\ b 1\ (a V x)))
C
or as (** )
fgf(x)
Ta show that fIx) fgf(x)
~
a V (b 1\ d 1\ (c V IJ V (b 1\ x))).
fg{(x), use (*) and fIx)
= IJ V (b
use (**) and fIx) = b 1\ (a V x).
fIx),
Taking b
~
=
= IJVd
and c
=
(Ll\d
1\ x).
Ta show that
0
in (5.1) we see that [IJl\d,d] and [a,aVdj are
isomorphie.
A subset C of a partially ordered set P is a chain if for eaeh
IJ
and b
in C, either a <; b or b S ai if P itself is a ehain, we say that P is linearly ordered. sueh that C U {al
A maximal ehain in a partially ordered set is a ehain C
is a ehain only if
(l
The simplest nonmodular
C C.
lattiee above has two maximal finite ehains, one of length 2 and one of length 3.
For modular lattiees this ean't happen.
linearly ordered sets
C
and
D
We say that two
are pieeewise iSOIIIOrphie if there exist
elements el" .. c n and d 1 , .•. ,dn such that (i) {x (ii) (x
{x
C C
x S cl) is isomorphie to
c
x :> e n ) is isomorphie to {x C TJ
C
(iii ) There is a permutation a
of
x <; cl l
C D
[1, ... ,n-I)
isomorphie to [da; ,d ai +1] for eaeh i
}
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 <; that Pn = Pn+l' the descending ehain eondition is defined similarly. Classieally, if P satisfies the ascending ehain eondi tion, then we can find Tl sueh that Pm = P" for each m ;> 11. FLom a construetive point of view, even the two element set {O,l} fails to satisfy this form of the aseending chain condition. We say that an element p of a partially ordered set P has depth at IIIOst
24
Chapter I. Sets if
n i
s:
whenever
p =
Po
s: ...
<: PI ~ [12
S Pn+l'
then
Pi
=
l'i+1
If P is diserete, we say that P has dept.h at least
n"
exists a ehain
p =
has depth at most and at least
11,
Po 11
< PI < ••• < for some
for some
Pfl
"
for n
same
if there
An element has bounded depth if it
finite depth if it has depth at most
11,
11,
Similar definitions apply to !tel gilt instead
fl.
of depth.
EXERCISES 1. Show that a lattiee is diserete if and only if the relation a
s:
b
is deeidable. 2. Show that a lattiee is distributive if and only it satisfies the identity (l V (b /\ c)
=
(n V b) /\
V cl.
((t
3. Let L be a modular lattice eontaining a maximal chain that is finite (denial inequality).
Show that L is diserete.
4. Show that if two linearly ordered sets are pieeewise isomorphie to a third, then they are pieeewise isomorphie to eaeh other. 5. Show that two pieeewise isomorphie diserete linearly ordered sets have the same eardinality. 6. Two intervals A and B of a modular lattice are called transposes if they are of the form [(l,(lVd] and projective if there is sequenee A
(in either order),
[a/\d,d]
1 1 ,,, .,I n = B of intervals such that r land I i +1 are transposes for l = l, ... ,n-l. Show that any two maximal
=
finitely enumerable chains in a modular
lattiee are piecewise projective. 7. Show that a partially ordered set may be eonsidered a eategory 'C in which the
set
'G(a,b)
{x E
n <: öl"
{Ol
What
is
the
categorical description of the infimum of two elements? 8. Suppose that for each binary sequence a , find m such that a"
=
a m whenever
rt
~
~
a2
~
m.
(1"
~
•••
we eould
Conclude that LPO
holds. 6. WELL-roJND!ID SETS AND ORDlNALS
Let W be a set wi th a relation a < b. hereditary if w
C
S whenever
UJ'
C
A
S for eaeh
subset S of W is said to be UJ'
< UJ.
The set W (or the
6. Well-founded sets and ordinals
25
relation a < b) is weIl founded if each hereditary subset of W equals W. A discrete partially ordered set is well founded i f the relation a < b (that is, a ~ band a ~ b) on it is well-founded. An ordinal, or a wellordered set, is a discrete, linearly ordered, well-founded set. Well-founded sets provide the environment for arguments by induction. The prototype well-founded set is the set ~ of nonnegative integers, with the usual order. 6.1~.
The set
~
of nonnegative integers 1S weH founded.
PROOF. Let S be an heredi tary subset of ~. Then 0 E S because the hypothesis w' E S for each w' < 0 holds vacuously. From 0 E S we conclude 1 E S, from 0 E Sand 1 E S we conclude 2 E S, etc. 0 The set ~ of nonnegative integers, viewed as an ordinal, is denoted by We shall see how to construct other well-founded sets from w. First we observe that any subset of a well-founded set is weIl founded. More w.
generally we have: 6.2~.
tImt W is weH Founded. whenever a
< b.
< b, such ",(a) < ",(b)
Let P and W be sets, each wUh a relation a Let", be a map from P to W such tImt
Then P is welt Founded.
PROOF. Let S' be an heredi tary subset of P, and let S = {w E W : ",-l(w) ~ S'}. We shall show that S is hereditary, so S = W and therefore S' = P. Suppose v E S whenever v < w. If xE ",-l(w) and y < x, then
< w so ",(y) E S, whence y E S'. As S' is hereditary, this implies that x E S' for each x in ",-l(w), so wES. Thus S is hereditary. 0
",(y)
In particular, any subset sequence provides an example exhibit a first element. It well-founded relation is weIl
of w is an ordinal. The range of a binary of an ordinal for which we may be unable to follows from (6.2) that any subrelation of a founded.
We say that a well-founded set, or relation, is transitive if a < band example of a well-founded relation on rn that is not transitive is constructed by taking defining a < b to mean a + 1 = b. This relation is well-founded by (6.2). An i nduc ti on argument with respect to this relation is proof by induction as ordinarily defined; an induction argument with respect to the usual relation a < b is sometimes called proof by complete induction.
b
< c implies a < c.
An
26
Chapter I. Sets A discrete,
transitive well-founded set admits a natural partial ordering by defining a S b to mean a < b or a = b; the only nontrivial thing to check is that a = b if a S band b S a, but this follows from the fact
that a < a
is impossible in any well-founded set (Exercise 1).
Conversely, the relation a < b on a discrete partially ordered set is transi ti ve. One
way
to
construct
a
well-founded
set
is
by adding
together
previously constructed well-founded sets: we get the ordinal w + placing two copies of the nonnegati ve integers side by side.
(,1
by
More
gene rally, let I be a well-founded set, and let {Ai }iEI be a family of well-founded sets indexed by I. Then the disjoint union LiEI Ai = {(a,i) : a
E Ai and (a,i)
E I} admits a relation:
<
(b,j) if i
< j or if
i
=
j and a
If I {l, ... ,n}, with the usual order, we write Al + ••• + An' Note that if I and each Ai is discrete and transitive (and linearly orderedl, then so is LiEI Ai' 6.3 '.mEX>REM.
If I
is a weU-founded set. and {Ai }iEI is a famUy of
weU-founded sets indexed by
PROOF.
r.
then W = 2: iEI Ai is a weU-founded set.
Suppose S is an hereditary subset of W.
{a E Ai : (a,i) ES}, and let I' = {i E I : Ai that I' is hereditary, so I' = I, which says S = W.
Ai
For each i in I, let =
We shall show Suppose i' E I' for Ai}'
each i' < i. We shal1 show that Ai = Ai by showing that Ai is hereditary. Then w' ES for each w' < (a,i), so Suppose a' E Ai for each a' < o. (o,i) ES whence 0 E Ai:. Therefore Ai = Ai as Ai is well-founded, so E
1'.
0
Let {Ai}iEI be a family of well-founded sets indexed by a discrete set We say that an element f of IT iEI Ai has finite support if there is a finite subset 1 of I so that for each i in I, either i E 1, or a < f i is
I.
impossible for any 0 in Ai (that is, f i is a minimal element of Ai)' Note that if I is finite, then every element of IT i EI Ai has finite support, while if some element has finite support, then all but finitely many of the Ai have minimal elements. If I is a well-founded set, then the lastdifference relation on the elements of finite support in the product set IT iE1 Ai is defined by f < 9 if:
6. Well-founded sets and ordinals
27
(i) there exists i E I such that fi. < 9i' and (ii) for each i E I, either f i = 9i' or f j < 9j for some j
~
1.
If I, and each Ai' is an ordinal, this may be described as ordering distinct elements according to the last place where they differ (reverse lexicographic order). 6.4~.
Let I be an ordinaL, and (AiliEI a famiLy of weLL-founded
Then the set F of eLements of finite support in sets indexed by I. UiEI Ai is weLL founded under the Last-difference reLation.
PROOF. We first note that the theorem is true when I = {1,2}; in this case F = A1 xA 2 = 2bEB A(b) where B = A2 and A(b) = A1 for each b in B. So by Theorem 6.3 the last-difference product A1 xA 2 is well-founded if A1 and A2 are. Adjoining a greatest element 00 to I and setting Aoo = {Ol does not change anything, so we may assume that I has a greatest element 00. Let Fi
be the set of elements of finite support in Uj~i Aj and let I' = {i EI: Fi is well-founded}. We shall show that I' is hereditary, hence equal to I, so F = Foo is weIl founded. Suppose k EI' for each k < i. Let Fi* be the elements of finite support in Uj
F;
for each k < i, so soiEI. 0
~
Fi* = S.
F;,
Thus Fi* , and therefore Fi , is weIl founded,
If I is an arbitrary set, and Ai is a partially ordered set for each i in I, then the cateqorical product of the Ai is the set UA i equipped with the partial order: f
~
9 if f i
~ 9i
for each i in I.
If I = {l, ... ,n}, and each Ai is discrete and well-founded, then the identity map of the categorical product into the last-difference product preserves order (but not nonorder) , so the categorical product is wellfounded by (6.2).
Chapter I. Sets
28
If a is an ordinal, and ß is a well-founded set, then the well-founded set of functions with finite support from a to ß is denoted If A and from A to a
~
are ordinals, then an injection of A into
then p
PROOF.
p
We shall show that there is at most one injection
~.
IF A and
6.5 'lHEXlREM. into~,
by~.
is a function
such that if a < b then pa < pb, and if c < pb, then there is
~
E A such that pa = c.
from A to
~
=
Let S
~
ar'e or'dirwls, and p and
0
ar'e injections of A
a. =
{a E A : pa
= aal,
and suppose a E S for all a < b.
If
= ab < pb, so a < b, so pa = aa; > aa, a contradiction. Similarly we cannot have pb < ab. Therefore pb = ab, so b E Si thus S is hereditary, so S = A. 0
ab < pb, then there is a E A such that pa but pa
= ab
If there is an injection from the ordinal A to the ordina1
A
C1early compositions
~ ~.
relation is transi ti ve.
By
of
(6.5)
injections are
i t follows that i f
~
injections, A ~ ~
and 11
we write so
this
S A,
then
A and ~ are isomorphie, that is, there is an invertible order preserving function from A to~. It is natural to say that two isomorphie ordinals are equal. EXERCISES 1. Show that a < a is impossible in any well-founded set. 2. If a < b is a relation on a set W, define the transitive closure relation a <* b to mean a < b or there exists xl"" ,x n such that a < xl < x2 < ... < xn < b. Show that a <* b is well-founded i f a < b is we1l-founded (mimic the proof that ordinary induction on ~
implies complete induction on rn).
3. A relation a < b is acyclic if a <* a is impossible for any a (see Exercise 2).
Show that any acyclic relation on a two-
element set is well-founded.
show that any acyc1ic relation on a
set that is bounded in number is well-founded. 4. Show that each discrete,
well-founded,
partially ordered set
satisfies the descending chain condition. 5. Let W be a well-founded set.
Let S be a subset of W such that,
given w in W, either wES or there exists w' < w such that wES
6. Well-founded sets and ordinals i f w' ES.
29
Show that S = w.
6. Show
that a discrete partially ordered set that satisfies Exercise 5 for each subset W has the descending chain condition.
7. Show the converse of Exercise 6 holds choice) •
(this uses dependent
< b defined to mean band b - a is odd. Show that the relation a < b is wellfounded but not transitive.
8. Let W be the nonnegative integers with a a ~
9. Show that the categorical product of partially ordered sets as
defined in this section is the categorical product. 10. Show that if every nonempty ordinal has a least element, then LPO holds. 11. A rank relation on a discrete partially ordered set W is an
ordinal A together with a subset R of W x A such that (i) For each w in W there is a in A such that (w,a) E R. (ii) If v < wand (w,a) E R, then there is b < a such that (u,b) E R. Prove Theorem 6.4 for I a discrete, well-founded, partially ordered set with a rank relation. 12. A Grayson ordinal is a set W with a well-founded relation a < b satisfying: (transitivity) (i) If a < band b < c, then a < c (ii) If c < a is equivalent to c < b for each c, then a = b (extensionality). Define a ~ b in a Grayson ordinal to mean c < a implies c < b for all c. Show that if the relation a < b is decidable, then W is a Grayson ordinal if and only if W is an ordinal. (Hint: prove that a < b or a = b or b < a in a decidable Grayson ordinal) 13. Let a be a binary sequence, and let S = {x,y,z} with x = y if a is identically zero. Define a relation 11 < II on S by setting (i) y < z (ii) x < y if an = 1 for some n (iii) x < z i f x = y or x < y. Show that this turns Sinto a Brouwerian example of Grayson ordinal with elements such that x
~
y
< z, but not x < z.
Chapter I. Sets
30
Bishop's definition of when an object may be said to exist can be found in [Bishop 1967, page 8].
The idea that the notion of an a1gorithm is
primitive has also been advanced by the Russian mathematicians Uspenskii and Semenov (1981); "The concept oF algorithm I Uze such
a
fundamental
concept
that
that it
of set and of natur-al mnnber is
cannot
be
explained
through
other-
concepts and should be r'egarded os [an) undeFinable one."
Attempts to explain constructive existence in classical terms are always somewhat unsatisfactory, but the classical notion of existence is no less mysterious than the constructive one, we are just more familiar with its use. What does it mean for a well-ordering of the real numbers to exist? or a basis of the rea1s as a rational vector space? or an automorphism of the field of complex numbers that takes e to 1T? or a noncomputab1e function? Formal systems that specify the COHect use of the phrase "there exists" are available to the constructive mathematician as well as to his classical counterpart. OUr definition of a set is a combination of the formu1ations in [Bridges 1979, page 2] and [Heyting 1971, 3.2.1].
Many authors use the
positive term "inhabited" to describe nonempty sets; this avoids possible confusion with the notion of a set that cannot be empty. The notion of a (tight) apartness is found in [Heyting 1971, 4.1.1). The terminology "tight" is due to Scott (1979). Troelstra and van Dalen use the term "pre apartness" to denote what we call an apartness. The part of Theorem 2.2 that states that i f the inequality on S is tight, then so is the inequality on
SN,
is essentially [Bishop 1967, Lemma 5, page 241, which
says that the natural inequality on the real numbers is tight. A standard inequality on the set (0] is gotten by setting 0 ~ 0 if LPO Ag LPO is refutable in two main branches of constructive is false. mathematics--intuitionism and Russian constructivism--we cannot show that For more on intuitionism and Russian this inequali ty is consistent. constructivism from the point of view of constructive mathematics see [Bridges-Richman 1987). Difference relations are symmetrie inequalities that satisfy , x
~
y
and,
y ~
z implies ' x
~
z.
Notes
31
They were studied by van Rootselaar (1960) and by Olson (1977). We could demand that every set come with an inequality, putting inequality on the same footing as equality; it would then be natural to demand that all functions be strongly extensional. with such an approach, whenever we construct a set we must put an inequality on it, and we must check that our functions are strongly extensiona1. This is cumbersome and easily forgotten, resulting in incornplete constructions and incorrect proofs. An exarnple of the cornplications: if H is a subgroup of an abelian group G, then the inequality on G/H as a group may differ from the inequality on G/H as a set because the group operation on G/H need not be strongly extensional with respect to the latter inequality (unless the inequality on G is decidable)--see Exercise 11.1.6. Dur defini ti on of a subset agrees wi th the informal treatment in [Bishop 1967, page 32). A categorical definition of a subset is found in [Bishop 1967, page 63) where a subset of a set S is defined to be a set A together with a one-to-one map from A to S.
The categorical definition is
attractive for constructive mathematics, where it is irnportant to keep in mind that an element of a subset, by virtue of belonging to the subset, carries irnplicitly the additional data that establishes its mernbership in the subset. The categorical approach allows us to make this additional data explicit. For example, if S is the set of binary sequences, and A is the subset of S consisting of those sequences a such that a m = 1 for some m, then to specify a mernber of A we must not only construct a sequence a in S, but also an integer m for which a m = 1. Thus an element of A may be viewed as a pair (a,m), two pairs (a,m) and (a' ,m') being equal i f a = a·. The map from A to S that takes (a,m) to a is one-to-one but is not really an inc1usion map as i t forgets the additional datum m. We find the informal approach more natural. The axiom of unique choice allows us to identify a function with its graph; Myhill has called this the axiom of nonchoice. It is easy to see that the axiom of unique choice is equivalent to either of the following. (i) Every one-to-one onto function has an inverse. (ii) If S is a set, and S* is the set of one-element subsets of s, then s* has a choice function: that is, a function F from s* to S such that f(xl E x for each x in s*. Bishop ernploys the notion of a nonextensional function or operation. In almost all applications, one can consider an operation from a set A to
32
Chapter
I.
Sets
a set B to be a function from A to the set of nonempty subsets of B. OUr definition of a finite set differs from that in [Bishop 1967], [Bridges 1979] and [Bishop-Bridges 1985] in that we consider the empty set to be fini te. Another, more subtle, difference is that we require that finite sets be discrete with respect to their given inequality. Thus a set S of functions between two discrete sets is finite only if for each f and 9 in S, either f
= 9 or there exists x such that f(x)
# g(x).
The functions that we call onto are called surjective in [Bishop 1967] where the word onto is reserved for a map f from A to B that has a crosssection, that is, for which there exists a function 9 from B to A such that fg is the identity map on B. In [Bishop 1967] a finitely enumerable set is called subfinite, and a set is said to have at most Tl elements if it can be written in the form {xl' •.. ,x n }. The term 'subfinite' suggests a subset of a finite set to us, while the [Bishop 1967J usage of 'at most' precludes saying that each subset of
{l, ... ,n}
contains at most
Tl
elements. Greenleaf (1981) examines cardinality of sets, and related questions, from a constructive point of view. Our definition of a countable set differs from that of [Bishop 1967], [Bridges 1979] and [Bishop-Bridges 1985) in that we do not require that a countable set be nonempty (or even that we can decide whether or not it is emptyl; for discrete sets, it is equivalent to the definition in [Brouwer 1981]. We are unlikely to be able to construct a Brouwerian example of a subset of
m that
is not countable.
However, any acceptable proof of the
theorem T that every subset of m is countable could probab1y be converted into a proof that every subset of m is recursively enumerab1e, which is false. A well-known variant of T is Kripke's schema, which states that for each proposition P there exists a binary sequence 0 such that P holds if and only i f on
=
1 for some
Tl.
Kripke' s schema has some plausibility
within Brouwer's framework of the creating subject, where we imagine the idealized mathematician as effecting a nonpredeterminate sequence of attempts at proving P, and we keep in mind the intuitionistic criterion that to prove ,p is to show how to convert any proof of P into a contradiction . Bishop used an ita1icized flOt to indicate the existence of a Brouwerian counterexample. For example, he would say "there is an inequality that is
Notes
33
not an apartness" to mean "if every inequality were an apartness, then LPO
holds". We shall not employ this convention. The lesser limited princip1e of omniscience (LLPO) was introduced in [Bishop 1973]. Both LPO and LLPO have simple interpretations in terms of LPO is equivalent to the statement that, for each real real numbers: number x, either X ~ 0 or x > 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 <8>
(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 <8> = 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<, and is called the prime field of h. This field is the field of quotients of the subring {n'l : n
of h.
E 7l)
If a1"" ,an Eh, and I, 0 Is the prime field of h, then we say that h is generated by a1' ••• ,an i f RO (al' •.• ,an) = I<. A function F from a ring R to a ring S is a ring homomorphism if it is a
homomorphism
of
the
multiplicative monoids. takes
11
additive
groups, An I'
in
I.
f
her'
F
It
r-
1
is
a
homomorphism
of
the
The map horn the ring of integers 71 to S that
to n·l is a ring homomorphism.
is an ideal if for each x in land each =
and
easily
seen
(0) is an ideal.
that An
if
additive subgroup I of a ring R in R the elements rx and i5
a
ring
Xl'
homornorphism,
ideal I i5 proper if 1
are then
A left (re5p.
right) ideal I of a ring R is an additive subgroup of R such that for each I"
in Rand
X
in I the element .-x (resp.
ring i5 nonzero if I
is in I.
A left ideal I of a
contains a nonzero element.
Ta avoid confusion
XI')
between left ideals, right ideals and ideals, an ideal i5 often referred to as a two-sided ideal.
rf I is a two-sided ideal of the ring R, then
multiplication is a function on the quotient group R/I, so R/I is a ring.
44
Chapter 11. Basic algebra
Let X and Y be subsets of the ring R. Define XY to be the addi ti ve subgroup of R generated by (xy : x E X and y E Y). If X, Y, and Z are subsets of R, then (XY)Z = X(YZ). A subset I is an ideal if and only if RIR
= I, while I is a left (resp. right) ideal if RI = I (resp. IR = I). If 8 is a subset of a ring R then (8) = RSR is the smallest ideal of R
containing 8, called the ideal generated by 8.
If 8 is the finite family
(s l' ... ,sn)' then the ideal genera ted by 8 is denoted by (s l' ... , sn) . The left ideal generated by 8 is RS, while the right ideal generated by 8 is 8R; if 8 is a one-element set (s), then the corresponding left or right ideal is called principal, and is denoted Rs or sR.
If land J are ideals, then IJ and I
n J are ideals.
need not be an ideal; the ideal genera ted by I U ] is j
E J} and is denoted by I + ].
{i
The set I U J + j i E land
More gene rally , if (I i) is a family of
ideals, then the ideal generated by
Uil i
is denoted by 2 i l i
.
The quotient
of a left ideal I by a set 8 is the left ideal 1:8 = (x ER: x8!;; I). The radical of an ideal I in a commutative ring is the ideal JI = (x ER: x" € I for some n). The fundamental theorem of homomorphisms of rings follows immediately from the corresponding theorem for groups. 2.1 'lHEDREM.
Let I be an ideal in the ring R.
from R to a ring 8 wUh f(I)
IF f
=
0,
maps onto 8 and the I~er"nel of
R/I to 8.
IF f is a homomor"phism
then f is a homomorphism from R/I to 8.
F is I, then F is an isomorphism fr"om
0
Let I be an ideal in the ring R. ideal ] of R containing I.
A (left) ideal of R/I is a (left)
If J is an ideal in R containing I,
then
R/] ::= (R/I )/U/I).
2.2 'lHEDREM.
Let R be a dng, 8 a subring and I an ideal of R.
8 + I is a subring oF R containing I as an ideal, 8 and (8 + I)/I ::= 8/(8
n
n
I
Then
is an ideal in 8,
I).
PROOF. Clearly 8 + I is a subring and I is an ideal in 8 + I. Define a function f from 8/(8 n I) to (8 + 1)/1 by setting f(s) = s. Note that r is indeed a function because if s, = S3 in 8/(8 n I), then s, - S2 E I so s, = S2 in (8 + 1)/1. Clearly F is a homomorphism. Now define a function 9 from (8 + 1)/1 to 8/(8 n I) by setting 9 (s+i) = s. Ta see that 9 is a function we note that if 5, + L , = S2 + i 2 in (8 + I)/I, then s, - S2 EI,
45
2. Rings and fields so
S 1
= S2
n
in 8/(S
It follows that f is an i somorphi sm.
I).
0
If P is an ideal in a commutative ring R, then we say that P is a prime ideal
if whenever xy E P,
then either
or y EI'.
x E P
rf P
is
a
detachable proper ideal of R, then it is easy to see that P is prime i f and only if R/p is an integral domain.
rf p is a prime number, then the
ideal (p) in E is a detachable proper prime ideal, as is the ideal O. 2.3 'I'HEOREM.
R such af
t.hal
Let PI""
Pi is pr-ime for- i
R, then either I PROOF.
'Pn be detachable ideals af a cammutative ring
Ir I i.s a (ini teLy gener'ated ideal
n-2,
~
Pi. Far same i, ar there exist.s z E
~
Let xl, ••• ,xm generate I,
#F,
on
x j E I \ U j l' { #S <; 1,
there exists a i
ES,
number
for some
then I
then set x m+l
the
~
so xm+l
=
~ U jE8P j"
F.
in
Otherwise #S
i.
E
ai. + IT jE8\{ i. Ja j
2.4 THOOREM. commulatiue
R.
Then I i
PROOF.
then If
E
In ei ther ca se we can
I \ UiESP i .
~
F, and we are done
0
Let
,'i.ng
= 0,
#F
2 and for each i E S,
~
enlarge (xl"" ,xm ), without enlarging I, so that S by induction on #F.
We proceed by
If
'''rn J such that a i E Pi \ u8 \{ i )p j . If #8 2, r \ UiE8p i . If #8 > 2, then Pi is prime for some
E {xl""
j
= 2iE~~i
elements
Otherwise choose 8 in F minimizing #8.
j.
for some
Pi
of
•
and let F be the set of finite
subsets 8 of {1, 2, .•• ,n} such that {xl"" ,x m} induction
I \ UiP f
I 1 , •.. ,I"
and ~
P a
be
pr'ime
(i,nUel!) ideo.l
or
gener'ated R
such
of
ideals
that
the
a
product
P far some f.
By induction on
Tl
it suffices to consider the case Aß
o.i E P, or h j E P for all j.
Dib j E 1',
n
= 2.
for each i
Let
either
0
A denial field is a commutative ring that i5 a field under the denial inequali ty, and such that 0 i5 a prime ideal. denial field.
Any discrete field is a
A maximal ideal in a commutative ring R is an ideal M such
that R/M is a denial held; so an ideal M of R is maximal if and only if it is a prime ideal, and A detachable maximal
x E R\)I
if and only if
r'x -
1 E M for some r' E R.
ideal is an ideal M such that R/M is a discrete
field. The characterlstic of a ring h is the order of the element I in the addi ti ve group of Iz.
It is a standard convention to say that h has
46
Chapter 11. Basic algebra
characteristic 0 if the order of 1 is infinite.
Thus the characteristic
of a discrete ring k is the least positive integer n such that n·1 = 0, if such a positive integer exists, and is 0 i f no such positive integer exists. ~P
The field of rational numbers has characteristic 0, and the field
= ~/(p) has characteristic
need not be in
~
p. The characteristic of a discrete field as the following Brouwerian exarnple shows.
2.5 EXAMPLE. Let a be a binary seguence with at most one 1, and define Pn by
[
0
the
if
0
an =
n th
prime i f an = 1
Let P be the ideal of the integers ~ generated by the numbers Pn ' and let = ~/p. Then R is a discrete integral domain; let k be its field of guotients. The characteristic of 1~ is not in~. 0
R
EXERCISES 1. Show that in any ring the following identities hold. (i) aO
= Da
(ii) a (-b) =
2. Use the rings
~
0,
(-a)(b)
=
-ab.
and
discrete integral domain that does not have recognizable units. 3. Show that a discrete ring is a division ring if and only if the nonzero elements form a multiplicative group. 4. 'l'he rational quatemions.
Let the elements of (a,b,c ,d ) in
R =
Note that
(a + bi + cj +
dk)
(a - bi - cj -
dk)
= a 2 + b2 + c 2 + d 2
and show that R is a noncollUllUtative discrete division ring. 5. show that in a cotl:ansitive field the operations of addition, subtraction, multiplication and division (restricted to units) are strongly extensional (see Exercise 1.5). 6. Let R be a subring of a field K. k
{ab -, : a,b
Show that E
R and b
"# O}
2. Rings and fields
47
is a field eontaining R. Show that if R is a subring of another field K', and the inequalities on K and K' agree on R, then k is isomorphie to k·. Show that the inequality on R is eonsistent, eotransitive, tight or discrete if and only if the inequality on k iso
7. Show that a commutative ring is an integral domain if and only if the following conditions hold: (i) 1
0,
~
(ii) a ~ b if and only if a - b ~ 0, (iii) if a ~ 0 and ab = 0, then b = 0, (iv) a ~ 0 and b ~ 0 if and only if ab ~ O. Show that a necessary and sufficient condition for a commutative ring with the denial inequa1ity to be an integral domain is that a ~ 0 if and only if a is caneellable (ab = 0 implies b = 0). 8. Show that the fol1owing hold for ideals in a commutative ring.
(i)I]!;;InJ (ii) IJ !;; K if and only if I (iii) If I ~ J then K:J !;; K:I
~
K:J
(iv) (niIi):J = ni(Ii:J) (v) I:2 i J i = ni(I:J i ) (vi) I:JK = (I:J):K 9. Show that the following hold for ideals in a eommutative ring.
(i) JIj = JInY = .jf n .Ji. (ii) If In !;; J for some n, then.jf!;;.Ji. (iii)
JI+J = JJI
+ JJ.
(iv) NI = ';1. 10. Let ~ : R ~ R' be a map of eommutative rings, and let I and J be ideals of R'. Show that ~-1 (I) n ~-1 (J) = ~-1 (I n J), that ~-l(1)~-l(J) !;; ~-l(IJ), and that ';~-l(I) = ~-l(,;I). Show that if ~
is onto, then
~-l(1:J)
= ~-lI:~-lJ.
11. Show that (12) U (45) is not an ideal in the ring ~ of integers. Show that (12) + (45) and (12):(45) are principal ideals. 12. In (2.3) we don't have to know which n-2 ideals are prime. Prove (2.3) under the weaker hypothesis that if aib i E Pi for i = 1, ..• ,n, then for at least n-2 indices i, ei ther a i E Pi or
48
Chapter 11. Basic algebra
l3. Change (2.3) so that none of the Pi are assumed prime, and the conclusion is that I t;;; Pi for some i, or there exist three distinct indices sense) .
j
such that Pj is nonprime (in a suitably strong
Can you prove that the three Pj are distinct (and not
just their indices)? 14. Let Band C be two detachable subgroups of a group G. Show that if A is a finitely-generated subgroup of G, then either A t;;; B or At;;; C or there exists x E A\(B U Cl. Give a counterexample (not Brouwerian) to show that the result is false for three subgroups. 15. Show that a nontrivial discrete ring R is a division ring if and only if each finitely generated ideal is either R or O. Brouwerian example of an ideal in
mthat
is neither
mnor
Give a O.
16. Show that an ideal of a commutative ring R is a proper prime ideal if and only if it is the kernel of a homomorphism of R into a denial field. 17. Show that a finite integral domain is a field. Show that a detachable ideal I in the ring 7L is maximal if and only i f I = (p) for some prime number p. 3. REAL NllMBERS
The prototype Heyting field is the field ~ of real numbers. The set ~ of sequences of rational numbers forms a commutative ring under coordinatewise addition and multiplication. A sequence {qrJ of rational numbers is a cauchy sequence if for each positive r= E m, there exists NEIN such that I qn - qm I ~ r=
whenever m,n ~
N.
It is readily seen that the set C of Cauchy sequences of rational numbers forms a subring of ~. A sequence {qn} of rational numbers converges to 0 if for each positive r= E m, there exists NEIN such that Iqn l ~ r= for all n ;:> N. The set r of rational sequences that converge to 0 is easily seen to be an ideal in the ring C of Cauchy sequences. The set ~ of real numbers is the quotient ring C/I.
By associating with each
q
E
mthe
all equal to q, we get a natural embedding of
sequence whose elements are
min ~.
3. Real numbers
49
The set ffi of real numbers admits a natural order. Define a E m to be positive, i f there exist positive fe E III and NEIN, such that an l fe whenever n l N. It is easily checked that this definition respects the equality on m = C/l, and that the set of positive real numbers is closed under addition and multiplication. We write a < b, or b > a, if b - a is positive; in particular, a > 0 means a is positive. 3.1 'IHFDREM.
The
fot towing
condi t ions
on
a
reat
munber
a
Q7"e
equivatent. (i) There exist positive
an ( ii) a
fe
E
III and
lanl l
NEIN such that
fe for
n l N.
< 0 or
a
> 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
or b
and
E
Le t
a, b and c be
r"eat
number"s.
IF
< c. > 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
<; b if
(l
<
0
transitive and reflexive.
h + ~
for all
> O.
L
This relation is clearly
'ro show that it gives a partial order, we need
the following. If
3.3 THroREM.
PRCQF. an -
on
bn
<; b ami b <; Cl, [hell n
0
Suppose c E ([) is positive.
< e for all
< r for all
n ~ N.
As b S
Cl
I
As
U
b.
<; b, we can find N ( IN so that
we can find such N so that also b" -
But this says that 10" - b n
n ~ N.
I
for all
we have Sh01
PRCQF.
If
0
IR is
former ease
n
+ b t 0, then eithe.
11
Then either n >
° or
n
b, by (3.2), in the
Thus the inequali ty
0.
on [R is cotransitive. To show that the inequality on so
n
< c.
Thus
n
~
> 0 or
n
[R
is tight, suppose
n
l'
° is impossible.
< IC by (3.2); the former is impossible,
n
Similarly
D.
n
2 0, so n
=
0 by (3.3).
Not only is IR a partially ordered set, it is a lattice. real numbers, then
c n = max(nn ,b n )
supremum of n and b, written c -sup(-n,-b).
N, so 0
+ /' ) 0 or n + b < 0, and we may
t 0, in the latter ease 0 < b so b
For each (' > 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)
<
The space 8 is complete.
t.
EXERCISES 1. Show that a > 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
<1('"s)
same
as
(ar-)s,
first
so
for
Light
multip1ying
by ,-
modules, and
then
multip1ying by s, while for 1eft modules multiplying by ,-s is the same as first multip1ying by sand then multip1ying by ".
Any abelian group is a
The set R" of n-tuples of element.s of R is an R-module undee
Z'-module.
eoordinatewise addition and scalar multiplieation.
is a division
If R
ring, then an R-module is ealled a vector space over R. A group homomoephism F trom M t.o N is
Let. M and N be 1eft R-modules. an R-module homomorphism if f kernel of (F (Cl )
r Ct
is trer F Ec M}.
=
The
(1'0) = I f (u)
(n ( M : f (n) eategory
of
for all ,- in Rand
in M.
0
The
0), and the image of f is im f
=
R-modules
is
easily
seen
to
=
be
pre-additive. The set E(M) of endomorphisms of an abelian group M forms a ring, where
r
is a
then M ean be endowed with a
1eft
mul tiplication is eomposition of funetions. ring homomorphism from R to E(M) , R-module
structure
by
setting
nn =
f (r)m
If R is a ring and
for
,-
Conversely if M is an R-module, then we ean define from R to E(M) by setting 'I'(,")(m)
in Cl
R and
The homomorphism
= ,mo
representation of R as a ring of endomorphisms of M. modules are two ways of looking at. the same thing.
m
in
ring homomorphism 'I'
nn
'p
is called a
Representations and If the kernel of the
representation is 0 then the representation is said to be faithful. g-module M is faithful if ,. = 0 whenever
N.
= 0 for eaeh
In
in M.
An
4. Modules
53
A subgroup N of an R-module M is an R-slltmrxrule if ra E N for each a in N and r in R. The submodule of M generated by a subset X is the additive subgroup generated (rx : r E R and x E X}. The quotient group MiN is an R-module because scalar multiplication is a function on it. Theorems 1.1 and 1.2 hold for R-modules i f we replace the term 'group' by 'module' throughout, and consider all submodules to be normal (which, as subgroups, they are). If Rand S are subrings of a ring A, then Ais, among other things, a left R-module and a right S-module. This kind of situation occurs often enough to deserve a name. Let Rand S be rings and let M be a left R-module and a right S-module. Then M is an R-8-bimodule i f (ra)s = r(as) for all each r E R, a E M, and sES. Thus the ring R is an R-R-bimodule, and every left R-module is an R-l-bimodule. If R is a commutative ring, then there is no distinction between left modules and right modules, and each is an R-R-bimodule. Ideals in R can be described in terms of the module structures on R: a left ideal is a submodule of the left module R, a right ideal is a submodule of the right module R, and a two-sided ideal is a submodule of the R-R-bimodule R. Let M be an R-module and (Ai) iEI a family of submodules of M. The submodule of M generated by UiE1A i is denoted by 2 iE I Ai' The Ai are said to be independent if whenever i(l), ..• ,i(n) and j are elements of I, and x E Aj n (A t (l) + ••• + At(n))' then either x = 0 or tIm) - j for same m. We say that M is the (internal) direct sum of the submodules Ai' and write O\EIAi' if M = 2iEIAi and the Ai are independent. If I = (1, ... ,n}, then we write M = Al (j) ••• (j) An' If I = (l,2}, then M = A, (j) A2 i f and only if M = A, + A2 and A, n A2 = O. As an example let M = l/(6), let A, = (O,2,4) and let A2 = (O,3).
M =
The product of a family of R-modules (Ai}iEI admits a natural R-module structure under which it is the categorical product, or the direct product, of the modules Ai. If fand gare functions in the product, define f + 9 by (f + g)(i) = f(i) + g(i), and direct product is denoted by IT iE1 Ai .
'-f
by
('-F)(i)
= '-f(i).
The
Let I be a discrete set and {Ai }iEI a family of R-modules. We can construct a categorical coproduct (external direct sum) as follows. An element f E IT iEI Ai has finite support if there is a finite subset J of I so that f(i) = 0 if i EI\]. For I a discrete set, the external direct
54
Chapter 11. Basic algebra
sum of the family {Ai ljEI is the set of elements in the direct product that have finite support.
It is easy to see that if fand 9 have finite
then so do F + 9
support,
and r'f,
submodule of the direct product.
so the external direct sum is
Note that if I
is finite,
direct product and the external direct sum are the same. the module Aj
: F(j)
wi th the submodule {r
then the
If we identify
0 for j t i
=
a
I, then we see
that the external direct sum is a direct sumo The above construction runs into trouble if the index set I
is not
discrete, because if Ai is discrete and (f : f(j) = 0 for j t il contains a
nonzero element,
then
(i)
is
detachable
horn I.
To construct an
external direct sum when the index set 1 is not necessarily discrete, let
F be the set of finite sequences of elements of the disjoint union of the Let the equality on F be generated by
Ai'
( i)
(a l' ... ,on)
{1, ... ,n}.
(ii) (iii)
(al""
,u n __ 1'u n )
(ClI'"
(u l' ... ,Cl n _1,Cl n )
the same
.,on_l) if on = O.
if u n _1 and Cl n are in
(Cl1, .•• ,on_1+on)
f\.
More precisely, call two sequences in F adjacent if some pernrutation of one is gotten by applying (ii) or (iii) to some pernrutation of the other. Then a and
are equal in F if there is a chain
T
of sequences in F such that si is adjacent to si +1 for i We want to identify {(Cl) show that i f (Cl) =
(b),
E
F :
0
Ai) with Ai'
(
then u = IJ.
=
1, ... ,rn-I.
To da this we must
This follows horn the Church-Rosser
property of F: 4.1 LEMMA (Church-Rosser property). Clnd let
f' (s) denate
the !ength of
ehut" a = sl,s2"" ,sm = s[+l Fm'
i
=
T
(/le
Suppase
und
0
sequenee s.
T
Cll-e
T/Jen then
equal
in F,
thef'f' i
S
(j
of sequenccs in F s((ch thut si is adj((eenl ta
l, ... ,m-l, Clnd fcw i
"' 2, ... ,m-l,
if P(si_l) < I'(si)' !hen
l'(si) < ['(s'+1)' PROJF.
Let er
=
sl,s2"" ,sm
sl is adjacent to si+l for i
N
=
2 i e(si)'
=
=
T
be a chain of sequences in F such that
1, ... ,rn-I.
showing that if l'(si_l)
i, then we can reduce N.
<
We will pLOceed by induction on
P(si)
and P(si)
>
P(si+l)
for some
4. Modules
55
Suppose we get from si to si+l by deleting a zero z. If z appears in Si_I' then we can replace si by si_l with z deleted. Otherwise si_l comes from si by deleting z, in which case we can omit si. The same argument applies if we get from si to si_l by deleting a zero. Suppose we get from si to si+l' and from si to Si-I' by applying (iii). Depending on how many distinct positions in si are involved, we can represent the various cases by si-I (a+b)
si
si+l
(a,b)
(a+b)
(a+b,e)
(a,b,e)
(a,b+c)
(a+b,e,d)
(a,b,e,d)
(a,b ,eid)
In the first ca se we can omit si and si+l. In the second we can replace si by (a+b+c), and in the third we can replace si by (a+b ,eid). 0 '!'wo sequences in F are added by concatenating them, scalar multiplication is done coordinatewise, and the empty sequence serves as an identity. Because of the equality relation, we can safely, and
unambiguously, write an element
(al,a2' •••
,an) of F as a formal sum
al + a2 + ••• + an
If the module Ai is identified wi th {(a) : a E Ai}' then F is an internal direct sum of the Ai; the verification of this rash claim is left as Exercise 5. If I is discrete, and (al' .•. ,an) E F, then we may assume that a m E Ai (m) with i (m) ~ i (m') i f m ~ m', and we can identify F with the set of elements of ITiEIA i of finite support, as above. If each Ai is a fixed module M, and I is an arbitrary index set, we denote the external direct sum by M(I).
The following theorem says that direct sums are categorical coproducts. 4.2 THEX>REM. IF M ~ QliEIAi' and f i : Ai --> 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<, ,pz) to (hz ,Pt) where
Pi
h, +
~,P2
and map K2 Gl F, to K, ffi Fz by setting
Tt. is readily checked that these maps are inverses of each ot.her. An
a
element e of a ring is idempotent if
(direct)
e'/
= evA
submodule A of M is
sUllllliIDd of M if t.here exists a submodule B of M,
complementary sUllllliIDd of A, so that M = A Mare always summands.
$
B.
0
called a
The submodules 0 and M of
58
Chapter II. Basic algebra 4.5 THEDREM.
Let
A be a submodute
dir-ect swnmand oF M if and of M such timt A = eM. compLementary summand
oF
onLy iF
then'
this
In
an
oF is
case
R_odute
an the
M.
Ais
a
idempotent endomor-phism
e
submoduLe
Then
(l-e)M
is
a
A.
PROOF. Suppose M = A al B. If x E M, then we can write x uniquely as a + b for some a in A and b in B. Define an endomorphism e of M by setting e(x) = u. It is easy to see that e is idempotent and that eM = A. Conversely suppose e is an idempotent endomorphism of M, and A = eM. Set B = (l-e)M. Aß x = ex + (l-e)x we have A + B = M. If xE A n B, then x = (l-e)y and x = ez. Thus ex = e(l-e)y = (e-e 2 )y = 0 and ex = e 2 z = ez = x, so x = o. So A n B = o. 0
The idempotent e in (4.5) is called the projection of M on A (along B). EXERCISES 1. The opposite ring ROP of a ring R consists of the additive group R with the product ab in ROP defined to be the product ba of R. Show that every left R-module is a right ROP-module in a natural way. If R is a commutative ring, the opposite ring is isomorphie to R, so each left R-module is also a right R-module, and we need not distinguish between left and right R-modules. 2. Let R be a ring and M an R-module. R-module mapping onto M.
Show that there exists a free
3. Let A and B be vector spaces over a discrete field k, each having Let V = A al Band let R be the ring
a countably infinite basis. of endomorphisms of V. x
B
-->
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
deg 9
< Tl
for
each
Tl
and
EIN;
< Tl
deg f
implies
for
we
each
wri te
< deg
deg F
if
9
if 9 = Xn + rn_1Xn-1 + ••• + '-0' then deg f ~ deg 9 if and only if deg f ~ n, Tl
Note
< Tl
EIN.
that
even if the ring might be trivial.
5.1 deg 9 =
Let
'l'mlOREM.
Let a
PROOF.
f + deg g.
=
fie/d,
(,g
und
Ir deg f
k[X].
E
and b
om-n+1 f
coefftcient
=
PROOF.
the 1eading
Then ab is the leading coefficient of Fg, and deg fg
=
Let R be a commutative ring, artd let
f.g E R[Xj be potynomials such timt deg f S m and deg 9 the
m and
0
5.2 THOOREM (Division algorithm).
is
=
m + n.
the 1eading coefficient of f,
be
coefficient of g. deg
be a
h
rt, then deg fg
of XIl in g.
then
exist
ther-e
q,r
~
n
~
E RIXj
lf
m+1.
Cl
timt
such
qg + rund deg r ~ n-1. We proceed by induction on m-n.
If m-n
=
-1, choose q
=
0 and
= b O + b 1X + ••• + bmXm , and set fl = (If - bmXm-ng . Then deg f1 S m-1, so by induction, a m-rl f1 = qlg + r for same q1,r E R[X] with deg r ~ n-1. Thus a m-rl+l f = (q1 + am-nbmXm-n)g + r-. If m-n-" 0,
r=f.
let f
o If the divisor 9 is monie, whieh, i f R is a discrete Held, we may assume it to be, the division algorithm has a roueh tidier form. 5.3 COROLLARY.
R, wUh 9 monic, '1g + r-. und deg r PROOF.
1
f.g
Tlten (her'" is
<
be polynomi(1!s
E R[X] [1
uni'1ue
pai,'
001'"
(1 rommutatioe dng
q." E R[X) such tlwt
= qg
< deg g.
+ r-.
Then
F=
g.
By Theorem 5.2 there exist q and r in R[XJ, with deg r-
such that f deg r
Let
To prove uniqueness, (q - q,)'1
= '-, -
< deg g,
suppose f = '1, 9 + r- 1
rand deg(r, - r)
wi t.h
< deg g, so
62
Chapter II. Basic algebra
q - q, = 0, because 9 is manic (induction on n such that and so r, - r- = O. 0 5.4 COROLLARY (Remainder theorem).
ring R.
conunutative
such
that f(X) = q(X)(X-o.)
PROOF.
f E
Let
E R.
and tet a
Then
deg(q - q,) ~
n),
R[X) be a potynomiaL over- a
ther'e
exists
unique q E
a
R[X)
+ f(a).
By Theorem 5.2 there exist unique q E R[X) and + r'. But then F(a) = q(a)(a-o.) + r = r'.
f(X) = q(X)(X-o.)
r'
ERsuch that
0
We can construct a polynomial of degree at most n over a field taking on prescribed values on n + 1 distinct points. The remainder theorem shows that this polynomial is unique, 50 a nonzero polynomial cannot have n + 1 distinct roots. This is one of the few results in the general theory of fields (as opposed to discrete fields, or Heyting fields). 5.5 THEX>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) ~.<1 t] ik
a(\'
T- '
(f,'~ )a,u .s t] ..
66
Chapter 11. Basic algebra
So Ta(ej) = IiI~ fkahisij' whence the new matrix of matrix of a is r- 1 AS, where r- 1 is the matrix of T- 1
Ta
is AS,
SO
the new
•
The i th row (ai l' ... ,a in ) of A may be considered as an element of the left R-module Rn. The row space of A is the submodule of Rn generated by the rows of A. An elementary row operation on A consists of either (i) Interchanging two rows, (ii) Multiplying a row by a unit of R, (iii) Adding a multiple of a row to a different row. The matrix resulting from applying an elementary row operation to A is the matrix of a with respect to some other basis of M. The matrix obtained by interchanging rows sand t of A is the matrix for a if we interchange the basis elements f s and Ft • The matrix obtained by multiplying row s of A by the unit u is the matrix for a if we replace the basis element Fs by FSU -
'.
The matrix obtained by adding " times row s to row
t
is the matrix
for a i f we replace the basis element Fs by Fs - Ft'" The row space of A is unchanged by elementary row operations. The jth column of (alj, ... ,amj ) of A may be considered as an element of the right R-module either
~.
An
elementary column operation on A consists of
(i) Interchanging two columns, (ii) Multiplying a column by a unit of R, (iii) Adding a multiple of a column to a different column. The matrix resulting from applying an elementary column operation to A is the matrix of a wi th respect to some other basis of N. The matrix obtained by interchanging columns sand t of A is the matrix for a if we interchange the basis elements es and e t • The matrix obtained by multiplying column s of A by the unit u (on the right) is the matrix for a if we replace the basis element es byesu.
The matrix obtained by adding
r times column s to column t is the matrix for a if we replace the basis element e t by e t + es'" By an elementary matrix we mean the result of applying an elementary row operation to the identity matrix. If E is the elementary matrix obtained by applying the elementary row operation p to the identi ty matrix, then E may be obtained by applying an elementary column operation p' to the identity matrix. Furthermore, if A and Bare matrices of the appropriate shapes, then EA is gotten by applying
p
to A, and BE is gotten
67
6. Matrices and vector spaces
by applying p' to B. A matrix of O's and l's, with exactly one 1 in each column and row is called a permutation matrix, and is a product of elernentary matrices. It is readily seen that elementary matrices have two sided inverses, which are also elernentary. If k is a division ring, then a 1<-rnodule is called a vector space over Classically every vectot space over a division ring is free. This is not the case constructively, even for finitely generated vector spaces
k.
over discrete fieIds, as the following Brouwerian exarnple shows. 6.1 EXAMPLE. Let a be a binary sequence, let sequence of subfields
of the field of Gaussian nurnbers (]) (i
i
2
-1 and consider the
=
U h n , then k is a discrete field, and Q(i) is a discrete 1<-module generated by the two elements 1 and L But we cannot construct a basis for Q(i) over IL 0 ).
If
=
I~
If the vector space V is a free k-module of rank n, then n is referred to as the dimension of V and written dim1< V, or simply dim Vi the space V is then called a finite-dimensional vector space over k. The following theorem shows, arnong other things, that dim V is well defined if k is discrete. 6.2 THEDREI'I. r'ing
k,
or
Let V and W be l'ectm' spaces ouer a discTete division
di.menslons
tr-onsfor'mnt i on from V to
or w.
f1""'( m T(e t ) = 0 fm' i
PR(X)F.
Let
n
W.
1.1I1I1
m
r'espectively,
Ilten tlter'e exist
Ir
T
a
linear
bases elf .... ",e n of
such tlwt T(e t ) = f i '
amI an index I< ~ n.
ls
for
V, and
<; k, and
> h. A = {a i j } be
bases for V and W.
the matrix of
T
with respect to the given
By elementary row and column operations we can arrange
that a ij = 0 for i ;t j, that Uu E {O,l}, and that a it ;: a j j i f But this amounts to constructing the desired new bases for V and W.
i
~ j.
0
Taking f to be the identity map in (6.2) shows that the dimension of a finite-dimensional vector space is well defined. It follows irnrnediately from (6.2) that her T and im T are finite-dimensional summands, and dirn ker T + dirn im T - dim V. The trouble in Examp1e 6.1 is that the h-subspace generated by 1 is not detachable: we cannot tell whether
i
is in it or not.
A
sun~nd
A of a
68
Chapter II. Basic algebra
discrete vector space is detachable because x
E
projection of x onto A is equal to x.
A if and only if the
Thus the following corollary
implies that fini tely genera ted subspaces of finite-dimensional
vector
spaces are detachable. 6. 3 COROLLARY.
Lel
V be
discrete division ring k.
fini le-dimensionül
a
Let W be
°
PROOF,
Since
W is a
finitely
spnce
olJer'
a
v.
oF
Then W is a fini te dimeflsiorml summünd
vectDr'
fini tel!) generated subspoce of V.
generated,
there
exists
a
finite
dimensional vector space F over k and a linear transformation T from F onto W.
It follows from (6.2) that W is a finite-dimensional summand.
6.4 COROLLARY. discrete
Let
division
V l::,le
I<.
r'ing
generated suIJspoces of V is
PROOF.
Fini te-cUmensional
a
the
Then
°
an\!
oF
(Jue,-
spo_ce
uectCH"'
inter'secUon
lwo
D
a
fini let!)
finUe-dimensional summand of V.
Let A and B be finitely generated subspaces of V.
can find a complementary summand C of B in V.
By (6.3) we
The projection on C,
restricted to A, is a linear transformation from A to C whose kernel is A
n B.
Hence A n B is a finite dimensional summand of V.
Let V be a vector space over a division ring
0
We say that lJ1, ... ,l'n
/<,
in V are dependent if there exist Cl i in h such that 2 i O i v i = 0, and a i t 0 for some i. If k and V are discrete, then we say that vl' ~ .. ,un are independent if they are not dependent; in this case it is easily seen that lIl""v"
form
a
basis
for
Let
V be
V if
only
and
if
they
generate
V
and
independent, 6.5 COROLLARY.
Cl
fini te-dimensionnl
space
veclor·
over'
(l
discn,'e division rlrlg h. are dependent,
Let T map 1<"
PROOF, The
Ure!) (JT'e
01'
kernel
of T
is
independent.
into V by taking the natural basis to I' l' ... ,I'n . finite-dimensional,
by
(6.2),
dependent if and on1y if the kernel of T is nonzero. Let h
6.6 THFDREM,
finUe-dimensiono( K.
Then
V
is
~
!>ector·
fini te
K be SPOCf'
cUscr~pte Olle I'
dimensional
dilllSioi1
k, ond
K i(
(wel·
dimensional over' h, in which CClse dim k V
/I't
=
and
"1""
fU"
r'ings
sr1.ch
that
K
V he Cl "ectell' spnce nnd
are
0
on!\!
di.m" K dim K E.
if
V is
fs
(l
Ollel
rini te
6. Matrices and vector spaces
69
PROOF. If V is finite dimensional over K, then it follows from (4.3) that V is finite dimensional over 1<, and the product formula for the dimensions holds. Conversely suppose that V is finite dimensional over 1<,
and that we have constructed a K-independent set Xl"" ,x m in V. Then + KX m is a summand of V as a vector space over I< by (6.3). If Kx l + + Kx m = V, then we are done; otherwise anyelement in V that is Kx l +
not in KXl + ••• + Kxm will extend the K-independent set xl' ... ,xm,and we are done by induction on the dimension (over 1<) of a complementary summand of Kx 1 + ••• + Kx m . 0 It is also true, in (6.6), that if V is finite dimensional over both I< and K, and nonzero, then K is finite dimensional over 1<. We will not need this result, which follows immediately hom the Azumaya theorem in the next chapter. EXERCISES 1. Construct a Brouwerian example of a vector space V over (Q that
contains two finite-dimensional subspaces whose intersection is not finite dimensional. (Hint: Let Y = (Q2jS for an appropriate subspace S of (Q2) You can arrange for your example to be discrete. 2. Generalize Corollaries 6.3 and 6.4 to the case where Y is a free module on a discrete set over a discrete division ring k. 3. Show that any type (i)
row operation can be achieved by row
operations of types (ii) and (iii). 4. A ring R is von Neumann regular if for each
a E
R there is
X E
R
such that axa = a. Show that the ring of n x n matrices over a discrete division ring is von Neumann regular. Show that a ring is von Neumann regular if and only if every principal left ideal is generated by an idempotent. 7. DETERMINANTS
Let Ma t n (R) be the ring of n x and let A = {ai j) be an element of defined to be
Tl
rnatrices over a cOIlUllUtative ring R, Then the determinant of A is
Ma t n (R) •
70
Chapter II. Basic algebra det A
=
la
sgn(a)ala(1)a2a(2)"'a ner (n)
where er ranges over Sn' the set of permutations of {I, 2, ••• ,ft) • 7.1 THEOREM.
Let A and B oe
x
Tl
matr'ices over' a commutati ve r'ing.
Tl
(i) det A = det At (ii) det Ais a Linear function of each
(iii) If Iwo
rOIDS
(i v) de t AB
PROOF.
of A are
['OlV
or
A.
then det A = O.
~qual.
de t. A de t B.
To verify claim
(i)
note
that i f
La sgn(o)a T (l)l"'a T (n)n = dei At as sgn(o)
=
= o-
T
sgn(T).
1
then det A
=
Claim (ii) is clear
from the definition of de I A, as each term in the defining sum contains exactly one element from each row.
As for (iii), i f rows i
and j of Aare
equal, and i t j, then the permutations may be partitioned into pairs {a, The tel:ffiS corresponding to the elements of each pair in the
a·(i,j)}.
defining sum are equal but of opposite sign,
[la
a
sgn(a) Gl (l)"'G
Lo,T
na (n)]' [lT
=
sgn(aT) Gla(l)' "G no (n)b h (l)' "bnT(n)
1T
is
a
funetion
from
{1, 2, ... ,n}
permutation, anda(i) =a(j) for
I
1T
add to O.
(i,j)
i ; t j,
to
{I, 2, ... ,n},
1T
rather
than
a
then
SfJn(rr) (!la(l)b o (l)1r(l)"'G n (J(n/'a(n)1T(n) .
because for each permutation 1T'
sgn(T) b h (l)"'b nT (n)]
l a,T sgn(aT) ala(l)bo(l)To(l)"'ana(n)ba(n)Ta(n) La l sgn(1T) Ula(l)ba(l)1T(l)"'CLlw(n)ba(n)1T(n)'
(7.2) If a
so t.he sum is zero whieh
Finally eonsider det A det B =
establishes (iii).
= 0
the t.erms in the sum indexed by
1T
and by
Therefore we may let (J range over all functions from
{1,2, ... ,n} to (1,2 ••••• n) in (7.2) so de t A de t B 'I'he cofactor Aij of the element
=
G
ij
det AB.
o
in the matrix A is (_1);+] times
the determinant of the Tl-I x 11-1 lTIatrix obtained by deleting row column
j
from A.
i
and
It follows easily from the definition of det A that, fOl:
each i, dei A
7. Determinants whence the
nan~
71 Prom (7.1.iii) we also have
cofactor.
a i 1A j1 + a i 2 A j2 + ••• + ainA jn
0
If we deHne the adjoint of A to be the matrix B whose i j th
i f i t- j.
entry is A jil then (7.3)
AB = (det All = (det At)I
=
(AtBt)t = BA.
Thus we can construct an inverse for A if we can construct an inverse for det
A. 7.4 THE()REM.
Let R be a commutatille ring und A E Mutn(R).
unH in Mat r1 (R) if and onLy
if
det A is a unU in R.
If AB = BA = I, then (det A)(det B)
PRODF.
1, so det A is a unit in R. (7.3)
Then A is a
=
(det B)(det A) = det I
=
Conversely, if det A is a unit in R, then
shows that A is a unit in
Matn(R).
0
We can now show that the rank of a finite-rank free module over a nontrivial commutative ring is an invariant; in fact, we show a litt1e more. 7.5 THEORE!'!.
Let
f<
commutative
be a
hng.
Let
m
< 11
i.ntegers and Let 'P : ~ .... Rn be an epimor'phism of R-moduLes.
There is an R-modu1e map ~ : Rn
PRODF. 'P
to Rn
=
~ EIl Rn -i71 by setting
<{J
(Rn-m)
ry
~
be
such that 'Pt = 1.
= 0 ,and view t as
equa1~.
Let R be a commutative ,-ing. M an R-moduLe, and A an n x n
matrix wUh enth.es in R. AU = 0,
Extend
a map into Rn.
As (det 'f')(det t) = 1, the map~' has a (left) inverse, which nrust As .p(Rn - m ) = 0, we have Rn -i71 = >J;(O) = 0 so R = O. 0
7.6 LEMMA.
positive
Then R = O.
Ir U is an n x 1 matr'lx with entT-ies in M, and
rhen (det A)U = O.
PRODF.
Let B be the adjoint of A.
'rhen BAU = 0, so (det AlU = O.
0
If A is an n x n matrix with entries in a commutative ring R, then XI - A is a matrix with entries in R[Xl.
The determinant of XI - A is
called the characteristic polynomial of A.
The characteristic po1ynomia1
of an endomorphism a of a free R-module F of rank n is the characteristic po1ynomial of the matrix of characteristic polynomial of of
Cl
wi th
respeet
Cl
Cl
wi th respect to a basis
is monie of degree n.
to another
basis
of F,
for F.
The
If B is the matrix
then B = S" 1 AS
for
some
Chapter 11. Basic algebra
72
invertible S in Mo t n (R) • determinant
of
xr -
determinant of
Thus the characteristic polynomial of B is the
XI - S- 1 AS
A,
S- 1 (XI - A)S /
which
is
equal
so the characteristic polynomial of
depend on the choice of basis of F.
to
the
does not
0
The Cayley-Hamilton theorem says that
o satisfies its characteristic polynomial. 7.7 THElOREM (cayley-Hamilton).
be
polynomia[
the cll(tr'acte,-i.stic
Tl1e/1 f (Cl)
f,'ee R-modlll f' F.
R be a COffiffiutattUf' dng, and f(Xl
Let
of an endoma,-phi.sm
0
of a
finUe-r-ank
= O.
If S is the (commutative) subring of the endomorphism ring of F
PROOF.
generated by a and R, then F i5 an S-module via the multiplication in F. Let A = {Ct; j} be the matrix of a with respect to a basis ltl""
,tl
n of F,
so (JU
= '\ o .. {1 . • Li 1...1 t
J.
Let U = (ul""'un)t. and let C be the n x n matrix aI - A with entries in S.
Then
CtU
But de t Cl
= 0 whence (det
= de t.
C
= r' (a).
Ct)U
= 0 by (7.6).
Therefore de(
Ct
= O.
D
EXERCISES 1. Let f : Mat n (R) .... R satisfy
(i) frA) is linear in each row of A. (ii) F(A) = 0 if two rows of Aare equal. Show that there exists r
E. R
t.o show that det AB = det
such that f (A)
2. Let M be a free module of rank
let a det A
over a commutative ring R, and
n
be an endomorphism of M.
matrices
=
of
wi th
a
Use thi s
" de t A.
A det B.
respect
If A and B in to
bases
of
Matn(R)
are
show
that
M,
det B.
3. Another proof of the Cayley-Hamilton theorem. (i) Show that Mat n (R [X l) is isomorphie to Mat n (R) [X 1 for any ring R. (iil Let S be
a
ring,
[l
f (X) = g (X) (X - a),
true that i f (iii) Prove
(7.7)
fIX)
E S,
and f,g E S[Xl.
then f (a) =
= (X -
a)g(X),
by letting S =
o. then
Mr,ltn(R)
Show that i f
(Warning: f(a)
it is not
= 0.)
and using
(7.3)
to
73
7. Deterrninants the characteristic polynornial F (X)
faetor
=
9 (X)
(X - 0),
viewed as an element of S(X], and apply (ii).
4. Consider the determinant det M of the Vandermonde matrix M 1
Xl Xz
x1Z xZZ
Xm- l I Xm- l
1
Xm
xmZ
Xm'm
1
Z
l
wi th entries in the commutati ve ring k (X l' ... ,Xm]. det
f
=
M = ITl(j(X j
Fa
Xi)'
-
Use
+ fIX + ••• + fm_lX
this
rn-l
t 0
formula
to
prove
Show that. that
if
is a polynornial over a Heyting
Held h, and 01' ... 'Dm are distinet. elements of k, then F(ai) 1'- 0 for some i • 8. SYMMETRIe l'OLYNCJIlIALS
Let R be a eommut.ative ring and fE R[XI, ... ,XnJ a pclynomial with We say that f is invariant under apermutation w of
coefficients in R.
the set of indeterminates (Xl"" ,Xnl i f f(Xl, .. ·,Xn
) =
f(w(Xl),···,w(X n ))·
If f is invariant. under eaeh permutation of its indeterminates, then we say that F is asymmetrie pclynomial.
If we consider
as a polynomial in the indeterminates Y ,Xl' ... ,Xn with coeffieients in R, then i t is readily seen that f is invariant under any permutation of the indete rminates Xl" .. ,X n
So i f we wri te f as a pol ynomial
•
yn
+ °lyn-l + ••• +
an
in Y with coeffieients in R[Xl,o .. ,Xn
),
symmetrie pclynomials in R[XI, ... ,Xn
Expanding f to a sum of monomials
].
then the eoefHcients u j
are
we find al
=
an =
2 i Xi'
u2
= 2 l (J
XiX},
03
=
2 i (j<1< XiXjX h '
XI X2 " 'X n '
The polynomials 01"" ,un are ealled the elementary symmetrie polynamials in n indeterminates. Clearly eaeh pclynomial in the subring R[ol,oo.,unl of R[Xl, .. o,X n ] is symmetric; the fact that every symmetrie pclynomial has a unique
74
Chapter II. Basic algebra
representation as a member of R[ol, ••• ,u"l is the fundamental theorem on symmet.rie polynomials . B.l THEXlREM.
f be a summeU-ic
Let
pO~!Jrwmial
there i.s a llniquE' polynomia/. h in R[Y l' .... Yn
1
We eonstruct 11 by induetion on n.
PROOF.
in R[X 1 , ..• ,X"l_
Then
SUel1 that f = h (ul' - ... on)'
Replacing R by
;~
[1-1' ... "-m
1,
where the r a r e indeterminates eorresponding to the eoeffieients of the monomials in the expression for (', eonstructing h, that
we
ean
polynomia1
for the purpose of
that R is discrete; this is a teehnieal convenience so
talk
is
we rnay assume ,
about
degrees.
symmetrie,
so we
If
n =
1,
then 01
=
ean ehoose h=F(Y 1 ).
and
Xl
every
If n>l,
let
Tl"" ,Tn-l be the elementary symmet.rie polynomials in the indeterminates Note
Xl"" ,X n _ l ·
that Ti
= 0i
(X:lt
'"
By induetion we
,Xn_I,O).
ean
eonstruet 9 E R[Y I , ... ,Y n - 1 1 such that f(X I ,· .. ,X n __ 1 ,O)
We may assume
1
:;:'(1:
that
9
=
9(Tl,···,7,,_1)·
eontains no monomial
1e (i) exeeeds the (total) degree of
f1
=
yE' (1) ••• yE' (n-l)
1 n-l F(X l' .•. ,Xn _ 1 , 0). Then
sueh that
F - 9(01" .. ,on_l)
is symmetrie and f 1 (X 1 , ••• ,X n _ 1 ,O) = 0.
Thus fl is divisible by Xn , so by
symmetry is divisible by Xi for eaeh i.
This implies that fl is divisible
byo n = X1X2 "'X fl lower degree eonstruet
E,O
,
than
rI
vle earl writ.e fl = unf2 where f2 1S symmetrie and of
or f.
r
By induction on the degree of sueh
l' E R[Yl""'Ynl
that
f2
=
P(ul, ... ,on)'
and
we
ean
we
set.
h = YnP + g.
'1'0 show that h 1S unique, i t suffices to show that if 9 (ul' .•• ,un ) =
for 9 in R [Y I' ... , Yn 1, then 9 = 0.
We proceed by induetion on n.
n
may
=
1
this
m
is m-I
0mYn + 9 m-I Yn
R[Y 1 ,···,Y n - 1 ]· 90(Tl"" ,Tn-l)
trivial,
so
we
assume
Tl
>1
and
°
When
write
9 =
+ ••• + 90 as a polynomial in Yn with eoefficients 9i in Substituting 0i for Y i and then setting X" = 0 we get =
0, '",he re the Ti are the elementary symmetrie polynomia.1s
in X1 , ... ,X n _ 1 . By induetion on 11 we eonclude that 90 = O. 'I'hen 9 = 1',,1' for some p in R[YI, ... ,Y n ]. As 0 = 0(01""'u n ) 0n P (ul'''''u n ), we conelude that p(al"" ,un ) 9 = O.
0
=
0,
and by induetion on m that
[J
=
0,
so
75
8. Symmetrie polynomials
independent
atgebrai.eaHy 9 = O.
elementar-y
The
8.2 co:ROLL!\RY.
symmetrie
ouer- R;
that
is,
poLynomi.ats
if
,un are
01""
g(ol"" ,on) = 0,
then
0
The
ring of polynomials R[X l , ... ,Xn ] is a module over the subring R[ol,.",unl of symmetrie polynomials. We shall show that it is a finiterank free module. The subring R[ol"" ,0n,Xj"" ,XnJ o( R[X l , .. · ,Xnl consists
8.3 LEMMA.
those
uf
polynomials
that
invoriant
are
under-
permutat ion
eo"ch
uF
Xl' ... ,X j-l' Let 8=R[Xj
PROOF.
, .. "Xn],
and let Tl, .. "Tj_l be the elementary
symmetrie polynomials in X1 ' ... 'X j _ 1 ' The polynomials in R[X 1 , ••• ,X n ] = 8[X 1 , ... ,X j _ 1 ] that are invariant under each permut.a.tion of X1 ""'X j _ 1 eonstitute the subring 8fTI" '. ,T j _1] by Theorem 8.1. Obviously 0i is invariant
under
permutations Let
8[T1,· .. ,T j _l]'
of
X1 , ... ,X j _ 1 ,
8[01, ... ,un J c
so
(Y + X 1 )(Y + X 2 )"'(Y + Xn )
f(Y)
and
g(Y) =
Then fand gare monie polynomials of (Y + X j) (Y + X j+1)'" (Y + Xn ). S[ol, .. "un l!Yj, so there are polynomials q and r in S[ol, ... ,onJ[Y] such that
deg r <;
polynomials
Tl
q
j,
-
and
f = qg + '-.
and rare
unique
yj-1 + Tl yj-2 + .•• + Tj_l
is
8[01, ... ,on] =S[Tl, ... ,Tj_11. 8.4 'HJElDREM.
'file monomial" s
Sinee
9
is
in R [X l' ••• 'Xn 1, in
caneellable, so
r
=
the
0 and q = Thus
8[01"" ,on][Y]'
0
xi (1 )X~ (2) ••• X~ (n)
wUh i. (h) <;
I~ - 1 form a
set of nl free generators uf R[X 1 , .. · ,Xnl as a moduLe ovel- Rlo 1 ,··· ,anl. PROOF.
Let Rj
=
R[al, ... ,un,Xj, ... ,Xnl.
We
shall
show
that
the
form a. family of free generators of Rj over Rj +1. By Lemma 8.3 the polynomial fj(y) = (Y - X 1 )(Y - X 2 )"'(Y - Xj) has its elements
1,X j ,XJ. .. ,x{-1
eoefficients in Rj +1 . fj(X j ) = O. Thus 1,X j
The polynomial f j
remains to show that 1,X j
,xJ' ...
j,
and
generate Rj = Rj+1[Xjl over Rj +1 . It ,xJ"'l are independent over Rj + 1 • Suppose
such that deg 9 < j-I.
g (Xi) = 0 for some 9 € Rj+1 [Y]
As g is invariant
we have g(X i ) = 0 for 1 <; i S j, so 9 has distinct roots; thus g = 0 as Xi - Xj is caneellable by the uniqueness
under permutations of X1, .. ,Xj j
is monie of degree
,XJ.,.,xj-1
part of the remainder theorem.
,
0
76
Chapter 11. Basic algebra EXERCISES
1. Let K be a diserete field, and let 01"" 'an be the elementary
symmetrie polynomials in K[X l' ... ,Xn ] •
Show that the set of Il!
monomials xi(1)x~(2) ••• x~(n) with iOd ~ I~ - 1 form a basis of the field of quotients of K[X1, ... ,XIl ] as a veetor spaee over the field of quotients of K[ol, ••• ,on]' 2. Show that the following algori thm exhibi ts a gi yen symmetrie
f
polynomial
as
a
polynomial
in
the
elementary
symmetrie
Order the monomials xi(1)x~(2) ••• x~(n) lexieograph-
polynomials.
ieally by setting xi(1)x~(2) ••• x~(Il) ~ xi(1)x~(2) ••• x~(Il)
i f for eaeh k
i (k)
j (k) or there is m < k
~
such that
Let Qxi(1)x~(2)"'Xr~(Il) be the 1eading term of f
i(m) < j(m). with
ei ther i (k)
respeet
to
this
lexieographie
2 i (k+l) for k = 1, ... ,n-1.
ordering.
Show
that
Subtraet
aaf (l)-i (2)oi (2)-i (3) •• 'oi (n)
1
from
f.
2
11
Show that the differenee has a lower degree with respeet
to the 1exieographie ordering.
Replaee f by the differenee and
repeat. 3. Express 27=1 X~ in terms of the elementary symmetrie polynomials
4. Let R be a eommutative ring and E = ITi<j(X i - Xj) (a)
Show that E
is
altemating,
E R[X 1 , ... ,Xn
].
E(lTX 1 , ... ,lTXn ) = sgn(lT)E(X 1 , ... ,X n ) for eaeh pernrutation lT, and that E2 is symmetrie. Show that i f 2 is a unit in R, then any alternating polynomial
that
is,
is of the form FE where F is a
symmetrie polynomial. (b)
Show that there is a polynomial d E Z[Y 1'''' 'Y n J such that if rr'1=1 (X - r- i ) = Xn + a1xn-1 + ••• + an' where the r-i and a j are in a fie1d K, then the if d(a1, ... ,a n )
(e)
#-
r- i
are distinet if and only
0 in K.
Show that E is invariant under even pernrutations of the Xi' Show that if 2 is a unit in R, then R[E,ol, ... ,on J is the ring of all polynomials invariant under even pernrutations of the
Xi'
Notes
77
We leave open what a reasonable inequality on a division ring should be. In order that a division ring be a division ring in the classical sense, the inequality must be standard; but this requirement is too weak to provide any useful consequences. Many important nondiscrete fields, such as the real and complex numbers, are Heyting fields. Although discrete fields are also Heyting fields, the temptation to identify the notion of a field with that of a Heyting field is lessened by the existence of a naturally arising denial field (the residue class field of a nonarchimedean valuation) with an inequality that is neither cotransitive nor tight. If an ideal P in a commutative ring is not detachable, then it is not clear just what i t should mean for P to be prime. In the ring of real numbers, the ideal 0 is not prime according to our definition, because it is possible to construct two real numbers whose product is 0 yet we cannot tell which of the two is O. Indeed LLPO is equivalent to the statement about real numbers that if ab = 0, then a = 0 or b = 0 (Exercise 3.5). In rings with a positive notion of inequality, like the real numbers, it is natural to define an ideal P to be prime if whenever a,b ( P, then ab ( P. Our definition of a prime ideal has the virtue of not referring to an inequali ty . Ag P is a prime ideal if and only if P is the kernel of a homomorphism into a denial field, i t is natural to define a maximal ideal to be the kernel of a homomorphism onto a denial field. There may be a better definition, but we won't know until we find some interesting theory involving nondetachable maximal ideals. Free modules on nondiscrete sets are not just curiosities. They are used, for example, to construct singular homology groups and to construct tensor products of arbitrary modules.
Chapter BI. Rings and Modules
1.
THE JACOBSCN RADICAL
~I-RmJLAR
EI.FJIIfBl'S
An element r
of a ring R is called left (right) quasi-regular i f l-r'
AN!)
has a left (right) inverse.
If r is both left and right quasi-regular,
then we say that r is quasi-regular. An element r in R is nilpotent if r n = 0 for some positive integer 11; we say that R has no nilpotent elements i f r = 0 whenever r' is nilpotent. Observe that i f r n = 0, and s
=
1 + r + r 2 + ••• + r n - 1 , then (l-r)s
are quasi-regular.
= s(l-r) =
1, so nilpotent elements
A left ideal L of R is called quasi-regular if each
element of L is left quasi-regular. Let L be a quasi-r'egul.ar' l.eft ideaL
1.1 '1HI!XJRm.
Then each eLement
of L is quasi-regular. PRCOF.
If r'
element s (l-r)s
=
ts
=
1.
1.2 LEMMA. PRCOF. If (1-Qb)c
L,
E
then s(l-r)
=
1 for some s
1 + sr has a left inverse t.
=
E
R.
Thus 1-r
AB -sr
=
E
ts(l-r)
L,
the
= t,
so
0
If ab is l.eft (right) quasi-regular. then so is ba.
If c(l-Qb) = 1, then (1+bca) (1-ba) = 1 - ba + bc(1-Qb)a
= 1,
1.3 'lHEXlREl'l.
then (1-ba)(1+bca)
= 1.
1.
0
The set of aU elements r such that Rr is a quasi-regular'
left ideal is a two-sided ideal. PRCOF.
In view of (1.2), i t suffices to show that i f Ra and Rb are
quasi-regular, then so is a + b. such that t(l - sb)
= 1.
Choose s such that s(1 - a)
Then ts(l - (a+b))
= 1.
=
1, and t
0
The ideal of Theorem 1.3 is called the Jacobson radical of R. 1.4 LEMMA (Nakayama).
Let M be a flni teLy gener'ated left R-illodule and
L a quasi-regular left ideal of R. PRCOF.
IF LM
=
M. therr M = O.
We may assume that L is the Jacobson radical of R, hence a two78
79
1. Quasi-regular ideals
sided ideal. Let x1, ••• ,xn generate M. 'Ihen xl E LM, so we can write xl = u1x1 + ••• + unxn where each u i is in L, as L is a two-sided ideal. Thus xl = (1 - u1)-' (u2x2 + ••• + unxn ), so M is generated by x2"" ,xn ' and we are done by induction on n. 0 EXERC1SES 1. Use the identity in the proof of (1.2), with U = 1, to show that if an element of a ring has a unique left inverse, then it has a right inverse. [Rudin 19851 2. Show that the following rings have 0 as their Jacobson radical. (i) The ring ~ of integers. (ii) Any discrete division ring. (iii) 'Ihe polynomial ring R[X 1 for R a discrete integral domain. (iv) A von Neumann regular ring (see Exercise 11.6.4). 3. Find the Jacobson radical of the subring of ~ consisting of those rational numbers that can be written with odd denominator. construct a Brouwerian example of a countable discrete integral domain whose Jacobson radical is not detachable. 4. An element X of an R-module M is called a nongenerator if whenever A is a submodule of M such that Rx + A = M, then A = M. Show that an element x of R is in the Jacobson radical of R if and only if x is a nongenerator of the left R-module R. 5. A detachable maximal left ideal of a ring R is a proper left ideal L such that i f x E R, then either x E L or Rx + L = R. Show that each detachable maximal left ideal contains the Jacobson radical. Show that if each element of R that is not contained in any detachable maximal left ideal has a left inverse (Zorn' s lemma establishes this c1assically), then the Jacobson radical is equal to the intersection of the detachable maximal left ideals. 6. Let A be a square matrix with entries in the Jacobson radical of a ring. Show that I - A is invertible, where I is the appropriate identity matrix.
80
Chapter III. Rings and modules 7. If 1 is the Jacobson r.adica1 of R, show that the Jacobson radica1 of R/J is zero. 8. Use Lemma II. 7.6 to show that if M is a fini tely generated module over a commutative ring R, and 1 is an ideal in R, then IM = M if and only i f (1- r')M
=
0 for some
i'
R.
E
Use this to prove the
Nakayama lemma for commutative rings. 9. Use Exercise 8 to show that any finitely generated module M over a commutative ring
I{
is Hopfian: each epimorphism from M onto M
is an automorphism. (For F:M
~
M, consider N as an R[Fl-module.)
2. COHERENT AND OOETHERIAN l'()OOLES
Let R be a ring, and M an R-module.
'I'hen M is Noetherian i f the set of
finitely generated submodules of M, ordered by inclusion, satisfies the ascending chain condition.
A ring R is 1eft Noetherian if R is Noetherian
as a left module over itself.
Thus R is 1eft Noetherian if and only if
for each sequence 1 1 I:;; 1 2 I:;; I Cl I:;; ••• of finitely generated 1eft ideals of such that In = I n +1 . It is easy to check that the ring of integers is Noetherian, as is IdX J
R there is if k
Tl
is a discrete field.
contains a discrete Held
I~,
Finite modules are Noetherian,
and if R
then R-modules that are finite dimensional
lz-spaces are Noetherian--more gene rally, if S is a subring of R, and M is an R-module
that
is
a
Noetherian S-module,
then M is
R-module.
'I'he Hilbert basis theorem (see Chapter VI II)
polynomial
rings
in
several
variables
over
the
a
Noetherian
implies that
integers,
or
over
a
discroete field, are Noetherian.
2.1 THElOREM.
iF
(md
Let N be a subrnodu!e of
0
modutE' Mo
PROOF.
Let
rr
denote
the
natural
map
from M to
Noetherian, then clearly N is also Noetherian. chain of finitely generated submodules of MIN, chain I , I:;; I 2 rrl rn
ThE'rl M is NOE'thE'dan
GnLy i.r both N and M/N arE' NoethE'ri.an.
= 1m •
~
of
There exists
fini tely Tl
generated
If
J,
NIN. ~
If
M
is
is a
.1 2 I:;;
then we can construct a
submodules of M such
that
such that 1,,+1
= I",
so 1n +1 = Jn , whence MIN is
both
M/N
are
Noetherian. Conversely, 1 1 «;;; 1 2 «;;; •••
suppose
N
and
Noetherian,
and
be a chain of finitely generat.ed submodules of M.
let Then
81
2. Coherent and Noetherian modules rrI ,
~
rrI z
~
•••
is a chain of
finitely gene ra ted submodules of
Noetherian module MIN, so there exists n such that rrI n = rrI n +1 .
the
Therefore
I n +1 = In + K for some finitely generated submodule K ~ N. Iterating this construction, we construct a sequence n(1) < n(2) < ••• of positive integers, and an ascending chain of finitely generated submodules Ki of N, such that In (i )+1 = In (i) + Ki • that Ki An
=
Ki - 1
~ In(i)'
so I n (i)+1
As =
N is Noetherian there exists i such In(i)'
0
R-module M is finitely presented i f there is a map from a finite
rank free R-module onto M with a finitely generated kernei.
The following
theorem shows that i f M is finitely presented, then an!} map !rom a finitely generated R-module onto M has a finitely gene ra ted kernei. 2.2 THEOREM. a
firlitei!}
If f is a map from a finitei!} generated R-moduie M, onto
presented R-module Mz ,
then
the
kernel
of f
is
finitel!}
generated.
PROOF. Let Fi be a finite rank free R-module, and rr i a map from Fi onto Mi such that the ker rr z is finitely generated. By Schanuel's trick (rr.3.3) we have Therefore ker frr ,
is
finitely
generated.
rr,(ker frr,) is also finitely generated.
2.3 THEOREM. (i)
As rr l
is
onto,
ker f
0
Let N be a submoduie of the R-moduLe M.
Then
lf M is finitel!} presented, and N is finiteLy generated, then MIN is finitely presented.
(ii) If N and
MIN are finiteL!} presented, then M is finitel!}
presented.
PROOF. Let rr be the natural map from M to MIN. To prove (i) let «J map a finite rank free module F onto M with finitely generated kernel K, and let N' C F be finitely generated with «J(N') = N. Then K + N' = ker rr«J is finitely generated, so MIN is finitely presented. To prove (ii) let «J, and «J2 map finite rank free modules F, and F2 onto N and MIN with finitely generated kerneis K, and Kz respectively. L.et «J : F, ~ Fz ~ M be such that «J restricted to FI is «JI and rr«J restricted to F2 is «Jz. Then «J maps F, ~ Fz onto M. If K = ker «J and rr z is the projection of F, ~ Fz on Fz ' then rr 2 K = Kz and K n F, = K" so K is
82
Chapter 111. Rings and modules
finitely generated. An
R-module
0
is coherent i f every finitely generated submodule is
finitely presented. A ring R is left coherent if it is coherent as a left R-module. Classically every left Noetherian ring is left coherent. 2.4 THEX>REK.
An R-illodule M is coher'ent
if and only if
the
following
two conditions hold. (i) The
inter'section oF any
two Finitely presented submodules
oF M is Fini tely generated . (ii)
IF x E M.
then
{r ER:
rx
0)
is a
finitely
gener'ated
left ideal of R.
PROOF. Suppose that M is coherent, N, and N2 are finitely generated submodules of M, and x E M. Let 'Pi be a map of a fini te-rank free R-module Fi onto Ni' and let K be the (finitely generated) kernel of the indueed map from F, ffi F2 into M. If rr, is the projeetion of F, ffi F2 onto then N, n N2 = 'P, (rr,K) is finitely generated establishing (i). The eyelie submodule Rx of M is finitely presented, so the kernel of the map
F"
taking R onto Rx is finitely generated, which shows (ii). Conversely, suppose (i) and (ii) hold, and A is a finitely generated submodule of M. If A is cyclie, then A is finitely presented by (ii). Otherwise A = N, + N2 where eaeh Ni is generated by fewer elements than A iso
By induetion,
each Ni
is finitely presented,
and by
(i)
the
interseetion N, n N2 is finitely generated. By (2.3.ii) the direet sum N, ffi N2 is finitely presented. But N, ffi N2 maps onto A = N, + N2 with kernel isomorphie to N, n N2 2.5 THEX>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
Thus we
cannot
strengthen (3.3) to conclude that the annihilator of I n R is zero. 7. Show that an m x n matrix A over a commutati ve ring R has a left inverse i f and only if the deterrninants of the set of square matrices whose rows are rows of A generate R as an ideal. 4. TflilSOR PRODUCTS Let R be a ring and B a left R-module.
If R is a subring of a ring A,
then there is a natural way to construct a left A-rnodule horn B. consider formal sums ifwe
can
get
::>: 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<
~
as follows.
choose b in B mapping onto c, and b'
Given c ® c' in C ® C',
mapping onto c'.
Define
in (B ® B' )/1<; if b l and b 2 map to c,
® c') to be the image of b ® b'
~(c
in B'
and bi and b 2 map to c', then (b l - b 2 ) ® bi + b 2 ® (bi - b 2)
b l ® bi - b 2 ® b 2 = is in K, so
is well defined.
~
It is readily seen that the bilinearity
requirement is met so that we have indeed defined a map from C ® C' • Clearly'l' and
~
are inverses of each other.
0
Let M be a ieft moduLe ouer a 4.6 COROLLARY. A .... B ... C .... 0 be an exact sequence of ,-igIlt R_oduLes.
A ® M .... B ® M .... C ® M .... 0 PBOOF.
i
R,
and
iet
Then the sequence
s exac t .
Take A' = 0 and B' = C' = M in Theorem 4.5.
4.7 COROLLARY.
ring
If C and C'
are
finitely
0
presented modules ouer a
commutatiue '-ing R, then so is C ® C' . PBOOF.
We can choose A', B', A, and B to be finitely generated free
R-modules in the hypothesis of Theorem 4.5.
0
The equality relation on A ® B admits the following description.
4.8 'lH&lRDl.
Let R be a
,-ing, A a right R_odule generated by the
elements al"" ,am' and B a Left R_odule.
The element tr .. l a t ® b i is 0
in A ® B if and only if there exist elements r ij in R and c j
in B such
that (i) (ii)
PBOOF.
Clearly the condition implies 2 a i ® b i = O.
Conversely, let F
be a rank_ free right R-module, and map F onto A with kernel K by taking
a basis el"" ,e m of F onto al"" ,am'
The element tr=l e i ® b i € F ® B
goes to zero in A ® B, so it comes from an element of K ® B which we can write as
r:
i=l
But the e i
e· ® b. t
t
are a basis of a free module, so we have b i .. 2 r i jC j '
2 i e i r ij € K, we have 2 i a i r i j = O.
0
As
92
Chapter 111. Rings and modules EXERCISES 1. Let {Bi} iEI be a family of left R-modules.
If
A is a right
R-module, show that A®
(~iEIBi)
~
~iEI(A
® Bi)'
2. If S is a multiplieative submonoid of a eommutative ring R, and M is an R-module, show that S-lM ~ (S-lR) ® M as S-lR-modules.
3. Let a be a binary sequenee, let A be the group of integers module 5, let C be the group of integers module 25, and let B be the
subgroup of C generated by 5 and the set {an : Tl = 1,2, ... }. Show that A and Bare diserete abelian groups, and that A ® B is diserete i f and only if an = 1 for some n, or an = 0 for all n. 4. Let a be a binary sequenee.
Let F be the ring of integers modulo
2, and let R be the subring of F[x,U,s,t]/(sx+tu-1) generated by
x,u and the set {ans,a"t B
: n = 1,2, ... }. Let A = R/(x) and Show that A and Bare fini tely presented diserete
= R/(y) •
R-modules, and that A ® B is diserete if and only i f an some n, or an
=
=
1 for
0 for all n.
5. Show that A ~ B is pure if and only if the map M ® A ~ M ® B is one-to-one for eaeh finitely presented right module M. Show that the expression 'finitely presented' ean be dropped from the preeeding statement. 5. FIAT lOXJLES
Let R be a ring and M a left R-module.
We say that M is flat i f
whenever xl"" ,xm are elements of M, and 1'1"" ,rm are elements of R such that Ll'ixi = 0, then there exist elements u1"" 'Y n in M, and elements a ij ERsuch that xi = L.i0ijYj and 2i1'iaij = O. Flatness is clearly a loeal property in the sense that an R-module is flat if and only if eaeh finitely generated submodule is eontained in a flat submodule. 5.1 THEX>REM.
IF
(Mi
}iEI
is
a
Fami ly aF R--modules,
then ~iEIMi is Flot
iF
and
we
PROOF. Suppose eaeh Mi is flat. Beeause flatness is a loeal property, may assume that I is finitely enumerable, say I = (s(l), ..• ,s(n)).
antu if each Mi is Flot.
Suppose t2=lr"x" = 0, where
I'" E Rand x" E ~iEIMi'
Write
xl'
= 21~=lUl'i
93
5. Flat modules with Yl'i E MS(i)' Then 27=1 O:e=l r "UI' i i = 1, ... ,n, or S (i) = s (j) for some
0, so either ~=lr'l'YI'i = 0 for
)
"# j .
In the latter ca se we are
done by induction on n.
In the former case, as Ms (i) is flat, t.here exist elements zl'ij E Ms(i) and elements al'i.j ERsuch that Yei = 2] aetjZeij "le r'"apij = O.
so
xe
UeiFpij and "le r'e"ei.j = 0, which shows that 6\EIMi is flat.
= 2 ij
Conversely, suppose GliOM{ is flat and 2er eXe
0 with xe E Mi'
=
Then
we can wdte xe = "l/'l'jYj' with Yj E GliE1M i and Iereaej = O. Let. s(l), ••. ,s(n) be an enumeration of the indices in I used in expressing t.he elements Yj '
Then Yj
=
.2k=l
Z
with
jl<
Z
jl< E Ms (k)'
and
xe - .2k=l(.2 j a pj z j k)· So
either s(l')
=
for some Iz t 1',
s(k)
in which ca se we are done by
induction on n, or xe - 2 j a ej Z j e and we have shown that.
Mi
is flat.
0
The 'only i f ' part of (5.1) says more than just that a summand of a flat module is flat, because the Mi need not be summands. 5.2 COROLLARY.
PROOF.
Free modules and pr"ojecttve modules ar"c flat.
That. the left R-module R is flat follows by taking
Yl = 1 in the definition of flatness. are flat.
If P is projecti ve,
n
= 1 and
Then (5.1) shows that free modules
then the natural map of the free module
R (1') onto P has a right inverse, so l' is a summand of R (1'), whence flat.
l'
is
0
A diagram of modules and maps, such as the square
01 ~rl
C~D
is said to be commutative if any two compositions of maps, beginning at the same place and ending at the same place, are equal.
The square is
cornmutative i f and only if w = ö(l. 5.3 THroRElt!:.
Let M
be
(l
(eft
R-modlile
Then
t Iw fo LI om; ng rond; ti ons
m-e equi.valerlt. (i) M i,s flat. (ii) For each T"ight
i.den!
I
oF U thp map I ® M .... M i.s one-to-
Chapter III. Rings and modules
94
one (iii) For
A
each
~
fini te-rank free ";ght R-modllLe Band submoduLe
B the müp A ® M -> B ® M 1s one-to-one.
(iv) FOT" euch right A ®
PROOF. I xi
Suppose
R-modllie
Band
submodllLe
A ~ B the map
M -> 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;'<,X;
=
O.
Let 1 be the right
95
5. Flat modules
ideal of R generated by the r' t and consider the map horn I ® M to R ® M. The element Li l' t ililx i goes to zero under this map, so by (iv) we have Ljri®x i = O. That (i) holds now follows from Theorem 4.8. 0 5.4 COROLLARY. submodule of B.
Let M be a
Left R-moduLe. B a r-ight R-moduLe. (md A a
IF B/A is nat.
then the map from A ® M to B ® M is one-
to-one,
PRQOF.
Map a free module F onto M with kernel K, and consider the
following commutative diagram where C
=
B/A.
0
1 A ® K
->
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'
Show that (iii)
set
of
elements
is
contained
each
in a
is equivalent to right coherence of R i f R is
So
96
Chapter 111. Rings and modules eountable and diserete. 4. Finitely presented Hat modules are projective. left
R-module
of
submodule of F.
finite
rank,
and
K a
Let F be a free
fini tely
generated
For x ( Flet 1 x denote the right ideal of R
generated by the coordinates of x in F.
Show that the following
are equi valent. (i) FjK is Hat. (ii) x
E
IxK for eaeh x in K.
(iii) for eaeh x ( K there is ( iv) K is a swmnand of F.
r
f (x ) •
F ... K such that x
Use your proof to show that if H. is eommutative, and finitely generated submodules of F are detaehable,
then we can decide
\,hether or not FjK is projeetive. 5. Show that any module over a discrete division ring is flat.
Show
that if k t;; Kare discrete fields, and V is a vector spaee over I<, then lII"",un in V are dependent over /< if and only if they
are dependent over K in K ®h V. 6. Use (5.3.ii) to show that if .A t;; B, and.A and Bill are Hat, then B is flat.
Conclude that i f every eyclie module is flat, then
every module is flat.
Show that a ring R is von Neumann regular
(see Exercise II.6.4)
if and only i f every R-module is flat.
Show that
if R is von Neumann
regular,
then every
finite1y
generated 1eft ideal of R is a swmnand whenee R is eoherent. 6. LOCAL RINGS
A ring R is ealled loeal if for each unit. I',
or
An equiva1ent condition is that if 1'2
is a unit.
l'
ER, ei ther
1'1
+
1'2
l'
or 1 -
l'
is a
is a unit, then either
Any Heyting field is a loeal ring, and many of the
results in this seetion, and the next, are of interest because of what they
say
about
Heyting
fields.
In
fact
a
Heyting
field
ean
be
eharacter:ized as a eommutative loeal ring in whieh 1 earmot be equal to 0 and any element which cannot be a unit is equal to O.
A virtue of vlOrking
with loeal rings rather than Heyting fields, in addition to the gain in generality, is that we do not eoneern ourselves with negative notions 1ike the ones in the preeeding sentence. An endomorphism f of a ring R as a 1eft R-module is given by f{x)
6. Local rings
97
xf(l), so taking f to f(l) is an isomorphism of the endomorphism ring of R as a left R-module with the opposite ring of R. As the opposite ring of a loeal ring is loeal, if R is loeal then the left module R has a loeal endomorphism ring.
If e is an idempotent in a loeal ring, then either e
is a unit, so e = I, or 1 - e is a unit, so e
= 0 (or both).
Thus if a
module M has a loeal endomorphism ring, then any summand of M is either 0 or
M,
that is,
is indeeomposable.
M
The next few theor·ems, dealing with
direet deeompositions involving modules with loeal endomorphism rings, are ealled Azumaya theorems. Keep in mind that if R is a Heyting field, or even just a Ioeal ring, then Rn is a direet sum of R-modules with Ioeal endomorphism rings. 6.1 LEl'IMA.
Al Gl .•• Gl An be a moduLe.
Let B Gl C
endomorphism ring, then B 6l C
= B Gl
Ir C Ims a loeal
D, wller-e D!;; Ai fei("" same 1..
PROOF. Let lT i , I TB , and 17C be the projeetions on Ai' Band C respeetively. Then 1rC(lTl + ••• + lTn ) is the identity map on C, so some 17ClT I. is an automorphism of C. Let D = 17 jC. Then 17C maps D isomorphieally
onto C, so B Gl C
=
B Gl D.
0
The first Azumaya theorem shows that summands of direet sums of modules wi th loeal endomorphism rings are again direct sums of modules vlith Ioeal endomorphism rings. 6.2
THEDRFl'I.
Let
AGlB
en.domo,phi.sm ring of each Ci
Cl
(j)
•••
is IDeal."
Gl Cn be modules such
lha t
'fllen tlter"e exist modules Dj
the such
that
and ther-e is apermutation a SUdl timt Ci == Da(i) for eaeh i. PROOF. By (6.1) we can find Dl , contained in A or contained in B, such that D1 al C2 Gl ••• al Cn = Cl Gl C2 Gl .•. Gl CH • We may assume that Dl t;;: B. Then B = B' (j) Dl and A Gl B' '" C2 Gl ••• (j) so we are done by induetion on Tl.
0
lt follows from (6.2) that a summand of a finite-rank free module over a loeal ring is finite-rank free.
The next Azumaya theorem shows that we
ean cancel modules with Ioeal endomorphism rings.
98
Chapter III. Rings and modules 6.3
If
THEOREM.
then A
endomaqJhism ,';ng, PROOF.
~
~
B ffl C
C'
Bor
~
C
0
B' fIl C', so A ffl C'
L:'
~
modu(('s,
and
B fIl C wi th C'
~ C'
(1/-('
C.
If the latter, then C'
a
/lOS
(ocnt
r C' = 0, whence A B' ffl C' ffl C.
co: B.
Then A
co: B'
If C'
~ B,
ffl C
B.
~
By (6. 1) we may
1S a summand of C,
From (6.3) we see that if R is a loeal ring, and R!" R is trivial.
C
B.
We may assume tha t A ffl C'
assume that C' so either:
A ffl C
then write B = 0
2'
Rn, then
m
=
11
or
Reeall that this is also true for R eommutati ve (11. 7 . 5)
but is not true in general (Exereise II.4.3).
"2
Define an inequality on a loeal ring by setting ", t unit.
i f ", - '-"
is a
Using this nonstandard inequality we ean develop, in a natural way,
mueh of the theory of finite-dimensional veetor spaces over Heyting fields in the more general setting of finite-Lank free modules over a loeal ring. This inequal.i ty is symmetrie and translation invariant for any ring and, for a loeal ring, it is also eotransitive. addition
and
subtraetion
are
strongly
Exercise Ir .1. 5 shows that
extensional.
To
show
tha t
and ab
is a
multiplieation is strongly extensional we need the following. 6. 4 LEMMA.
IF
un i t, t hen a and b PROOF.
G Gml b (1/-e (l/"'C
elements af a
,-i.ng I<,
l.acal
url i t s ,
",e may assume that ab = 1.
It suffiees to show that either -n
or b is a unit, so we may assume that 1+0 and 1-!J are units. (l+n)(l-b) is a unit, so either a or -b is a unit. If
Then a-b =
0
b, c and d are elements of a loeal ring R, and ab cf cd, then
(l,
either ab t ad or a.d cf cd by eotransitivity; so either b 't cl or (6.4) .
Thus multiplication is strongly extensional.
ct -je
c by
I t follows that i f
f (Xl"" ,Xn ) is a funetion built up fLOm elements of Rand the variables X1 "",Xn using only multiplication and addition, and if f(O, ••.• ,O) = 0, then fLOm fkl, ... ,r n ) cf A standard
° it follows that
elassieal
"i f
eharaeterization of
0 for some loeal
i.
rings
is
that
Exel:eise 4 outlines why we do not use
nonurtits form an ideal.
the this
ehal:aeterization, but nontdvial local rings do have the property.
6.5 'lHEOREM. then R is t,-ivi.nt
PROOF.
Let R be
l
ct
toeat r-i.ng. Thefl M = {r' ER: if I' is a uni t,
is (he ]aco!Json radienl of R,
Suppose Rr is a quasi-l:egular left ideal.
If ,- is a unit, then
99
6. Loeal rings 1 E Rr,
suppose
°
so
m E M
is a unit, whence R is trivial; thus and
r' E R.
Either
rm
or 1 -
rom
Conversely,
r' E M.
is a unit.
If
nn
is a unit
then m is a unit by (6.4), so R is trivial whence 1-rm is also unit. Therefore Rm is a quasi,-regular left ideal. Let M be a module over a loeal ring R.
0
The strong inequality on M is
defined by setting x t y if there exists a homomorphism f:M
~
R such that
The strang equality on M is the srnallest inequality that
((xl t f(y).
rnakes all the homomorphisms from M to R strongly extensional.
On Rn
the
strong inequali ty is the natural one t.o impose; it can be described in terms of coordinates as foliows. 6.6 THIDREM.
Let R be a LocaL ri.ng, and x und y elements of Rn.
Tlten
u t v in the str'ong ineqlw.U t!J on Rn i fand onL y i F x - !J /Jas a coordinate that is a unH in R.
PROOF.
Let el"" ,e n be the nat.ural basis for Rn and let. x -
U
=
If a i is a unit, then f(xl t f(u) where F is the prajection of Rn onto its i th factor. Conversely, suppose f(xl t f(y) for some R-module
2 i a ie i .
rnap f : Rn -;. R. i.
Then "\'P(rlie i ) t 0, whence ai'P(ei) = ~)(aiei) t 0 for some
Therefore a i t O.
0
We say that ul"" ,um in Rn are linearly independent, over the loeal
2'i=l
0 whenever r'l"" ,rm are in R, and r· i. t 0 for some Clearly any basis is linearly independent. We will show in (6.10)
ring R, if t .
ri.ut t
that, conversely, if ul"" ,um generate Rn, and are linearly independent, then they form a basis.
We will also show that any linearly independent
set in Rn can be extended to a basis. Lemma 6.4, that ab = 1 implies a and bare units, extends from loeal rings to rnatriees over loeal rings.
Moreover, the invertible rnatrices are
all products of elementary matrices. 6.7 THIDREM.
Let A be an
Jl
x
Jl
matr'lx over' a tocat ,-ing R.
IF A l1O.s a
left or ,-i.ght 1Iluer'se, then A is a product of eLementar-y matrices, so A has
D
two-sided inlJer-se.
PROOF.
It suffiees, by considering transposes, to assume that A has a
left inverse B. unit.
Then 2]=1 b 1j a j l = 1, so there is j such that b 1j D.!1 is a By ( 6.4) thi s irnplies that a j 1 isa unH. Then we ean find a
product E of elementary rnatriees such that the first eolumn of EA is all
100
Chapter II!. Rings and modules
0'5 except for a 1 at the top.
As (BE-')(EA) = I, and E has a tWQ-sided
inverse, we may assume that a i 1 = 0 for i t 1, and all = 1; note that this implies that b i1 = 0 for i t 1, and b 11 = 1. If M* denotes the matrix M without its first colurnn and row, then B*A * I* 50 by induction on n we can find a product E of elernentary matrices such that EA is the identity matrix except for a12, ... ,a u1 ' But these latter entries are easily made equa1 to 0 by elementary row operations. 0 The next two lemmas concern linearly independent elements of fini terank free modules. 6.8 LEMMA.
of R.
Let R be a iocal r-ing and M be an m x n matrix oF elements
Let A be an invertible m x m matrix. and B be an inver"tible n x n
matrix.
Then
the rows oF M ar"e linearly independent iF and only if the
rows of AMB are.
PROOF. Because A and Bare invertible, it suffices to prove that if the rows of Mare linearly independent, then the rows of AMB are. The rows of Mare linearly independent if and only if whenever X is a 1 x m matrix such that X t 0, then XM t O. If X t 0, then XA t 0 by strong extensionality of the ring operations, because (XA)A- 1 = X. We then have XAM
t 0 because the rows of Mare linearly independent, and XAMB t 0 by
strong extensionality. 6.9 LEMMA.
R.
0
Let R be a local r"ing and M an m x n matr"ix oF elements of
If the rows oF M are I inear"ly independent. then ei ther R is trivial or"
ther"e exists an invertible squar"e matr"ix A. and apermutation maU"ix B. such that the first m columns oF AMB for"m an m x m identity matrix.
PROOF.
The matrix A is constructed by composing elementary row
operations, the matrix B by permuting colurnns. As the first row of M is nonzero, we can permute colurnns and rnultiply by a unit so that the first entry in the first row is 1. We can then clear the rest of the first colurnn by elementary row operations. The rows of the resulting matrix are still linearly independent by Lemma 6.4. If m > n = 1, then R is trivial; otherwise, by induction on the nurnber of rows, either R is trivial or we can perform elementary operations on rows 2 through m, and permute colurnns 2 through n, to get an (m-1) x (m-I) identity matrix in rows 2 through m and colurnns 2 through m. We can then use elementary row operations to
6. Loeal rings
101
eonvert the first m eolurnns into an identity matrix.
0
We ean now show that any linearly independent set in a finite-rank free module over a loeal ring ean be extended to a basis. 6.10 THEDREM.
Let
R be a
Ioeal
"lft9 ami
let
VI"" ,vm be Linear-l!}
independent e tements of a fini te-rank f,-ee R-module F. basis Fot' F.
If m
< n,
then
LI" t e l ' ... ,e n be a
such that vl t
,vm ,e j
are
lineOl-ly independent; if m = n, then vl"",vm is a basis for- F; if m
> n,
tf,el-e exists j
•••
then R is trivial.
PROOF.
We may assume that F is the set of 1 x n matriees of elements
of R, that Vi is the jth row of an
m x
Tl
matrix M, and that
1 x n matrix with a 1 in the jth eolurnn and 0 elsewhere. by (6.9) we ean find an inverti.ble square matrix A, matrix B, such that the first
is the
If m ~ n, then
and apermutation
colurnns of AMB form an
m
ej
m x m
identi ty
If m = n, then M is invertible, henee i ts rows are a basis for F. < n, then, as B is a permutation matrix, there exists j such that
matrix. If m
e jB = em+l'
independent. independent.
The rows of AMB, together with the row e m+l' are linearly Thus
the
rows
of
AM,
As A is invertible,
together
with
ej
are
linearly
the rows of M together with e j
are
linearly independent by (6.8). If m
> n, then
01" .• , v ll are a basis for F, so v n +l ean be wri tten as a
linear eombination of them.
o~
But VI"" ,v n +l are linearly independent, so
0 in F, whence R is trivial.
0
EXERCISES 1. Show that i f Cl
D1 Gl '" (jj Dn , and each Ci and Di has a loeal endomorphism ring, then there is apermutation a of {I, ... ,n I
50
{j)
•••
that Ci
(jj
C"
~ Da (l)
~
for each
i .
2. Use the Azumaya theorems to show that if A and B are square matriees over a loeal ring such that AB
=
I, then A and Bare
invertible. 3. Show that the inequality defined on a local ring is symmetrie and cotransi ti ve. 4. Show that the following are equivalent.
(i) Markov's prineiple. (ii) If R is a countable diserete commutative ring such that the
102
Chapter III. Rings and modules nonunits of R form an ideal, then R is loeal. Bint: To show that (i) follows from (ii) let S
m
1,
5. Gi ve an example of elements a and b of a ring sueh that ab
1,
or
0 and
m ;t.
on =
1 for some
11
l, and let R
=
{m E E
=
s-lz,:.
but neither a nor b is a unit.
6. Show that Rn is Hopfian (Exereise 1.9) if R 15 loeal. 7. Let R be a loeal ring and M an R-module equipped wi th the strong inequali ty .
Show that addition and sealar multiplieation in M
are strongly extensional. 8. Generalize Theorem 11. 6.5 to loeal rings I<
s
K, assurning that the
K-module V comes with an inequality such that sealar multiplieation is strongly extensional, and that each of the three oeeurrences of the
term 'finite-dimensional'
implies an inequali ty
preserving isomorphism with Rn. 9. Let I, be the field of integers modulo 2.
Show that K
= "[X l!(X 2
)
is a loeal ring isomorphie to 1<2, but that the inequality on K as a loeal ring differs from the inequality on K as a I<-module.
For
V = k [X, Y l!(X, Y)2 show that Exercise 8 fails beeause V is not a
free K-module. 10. Show that if R is a loeal ring, and u1"'. ,um is a basis for Rn, then either m = n, or R = O. 11. Show that if R is a loeal ring, e1"" '''n is a basis for a free R-module F, and "1"." v m generate F, then there exists i
such
that Vi ,e2"" ,e" is a basis for F. 7.
~TIVE
LOCI\.L RIN:iS
If [,: i5 a eomrnutative loeal ring we define R(X), in analogy with the
rational funetion field over a field, by inverting all those elements in R[X 1 whieh have an invertible coeffieient. 7.1 LEMMA. Let
S
Let R be a commUloliliC
the
!N
coeffiöent.
set
of
polurwmiuls
[oclll iTi
hng and X
R[X]
1110/
0"
haue
an
ThE'll
(i) S is mu!tipUcotivelu closed. (ii)
If fg
=
0 few fES and 9 E R[XJ, Ihert 9
inciettTmi.note.
O.
irwertible
7. Commutative local rings
103
If f + 9 E 8, tlten f E 8 or 9 E 8.
(iii) IF fg E 8, then f E 8.
=
Let f
PROOF.
.•. + b O' invertib1e.
anXn + an_1xn-1 + ... + 00 and 9
Let fg
=
cm+nX m+n
+
...
+ cO,
=
bmXm + b m_ 1Xm- 1 +
Suppose 0i
and b j
are
Then 0tbj is invertib1e, so either c[+j or c i +j - 0ibj is In the former case we have established (.i); in the latter we
invertible.
have 1 apb q is invertible, where the sum runs over indices p,q such that p + q = i + j and P"Fe i. As R 15 loeal, there exists p < i or q < j such that apb q is invertible, induetion on
i
so either apb j
Ta prove (ii) suppose a i a i = 1.
or aib q is invertible,
so by
+ j we have proved (i).
is invertible and
Fg
=
O.
We may assume that
Then the (m+1)x(m+1) matrix
1
where a j = 0 i f j
°i_2
°i-m
(11_1
°i-m+l
(1[+1
°i-1 1
a t +fll
a i +m-1
< 0 or
1
> 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\)
,-onl<-n fr'ce
module,
a-i\
it S
is is
0
an
O.
=
The I:emainder theorem shows that X - i\ and g (X) are strongly
relati vely prime,
so
(X - i\)n and 9 (X)
al:e strongly relati vely prime I
whenee thel:e exist polynomials s(X) and t(X) such that s(X)(X-i\)n + t(X)g(X) Let e
=
s (a) (o-i\)n.
Then e 2
= e(
1. c, because f (0) = O.
I-t (o)g (0) )
= ca
be the kernel of e, and K the kernel of l-e; as oe nH ~ Hand oK ~ K.
As e
on K, and aK !: K, we have
= S (0) a-),
fra)
=
g(a) (a_i\)Tl.
it follows that
(n_i\)Tl = (o-i\)s (IX) (o-i\ ),,-1 is the identHy
is an automorphism of K.
rf the finite-rank free module K is zero, then e is one-to-one because
Let H
t (lX)g (IX)
=
=
0 and H
1. In that case (n_i\)n
By (7.6) we have f(X)
=
=
=
V, so g(o)
0 because 0
is apower of X-i\, so 9
1
and the theorem holds. If the rank of K is greater than 0, then we can write f
f H and FK are the eharacteristic polynomials of respectively. fK(i\)
As fK(i\)
is a unH.
O.
0
=
fHf K, where
restricted to Hand K
is the determinant of 0-:\ acting on K, we have
Thus Fjj
=
(X-/\)Tlg*(X)
where g*(i\)
induction on the dimension of V we have H is a (a-/\ )Hjj =
0
is a unit.
By
rank-n free module and
106
Chapter 111. Rings and modules EXERCISES
1. Show that i f o,L> and e are polynomials over a Heyting field, and dcg (]
~
dcg b, then deg oe
2. Units in R[X].
~
deg /Je.
Let R be a commutative ring.
Show that if
the constant term of 9 1S a unit, and all the other coefficients of gare nilpotent, then 9 is a unH in R[X].
Show
that, eonversely, if 9 is a tmit in R[X], then the eonstant term of 9 is a unit in R, and all the other eoefficients of 9 are nilpotent (Let S ~ {l,o,o~, ... } where
is the highest
Cl
coefficient of [] not known to be nilpotent, and show that S-lR is trivial). 3. Show that Theorem 7.4 characterizes impotent rings (you need consider only polynomials of the form oX + /J). 4. Gi ve an example of a
commuta ti ve
local
ring whe ce
( 7 .6)
fails. 5. Give a Brouwerian example of an endomorphism a of ffi2, where ffi is the real numbers, such that the characteristic polynomial
f of a has " as a root, but cannot be wri tten in the for:m given in (7.7). 6. Jordan canonical fo~ 1.
Let a
be an endomorphism of a
finite-rankfree module V over a eommutative loeal ring R with no nilpotents. a
is
a
produet of
Suppose the charaeteristic polynomial of polynomials of
the
distinet " (in the inequalHy on R). sum of submodules VA such that aV7\
form
(X - ,,)m
for
Show that V is a direet ~
VA and the character-
istic polynomial of a r:estricted to V" is (X - A)m. 7. Jordan canonical form 11.
Let R be a commutative loeal ring,
and a an endomorphism of Rn sueh that a m =
a
and im a i
is
finite-dimensional for i = 1, ... ,m. (i) Show that Hj = ("er a) n (im a i ) is finite-dimensional, hence a summand of Rn. (ii)
show that her a = Va m '"
(iii) Choose a basis e iJ
e ij
Rn.
m Vm- 1 where Hi = Vi (jJ Hj +1 for each Vi' henee for ker· a. Let
= ai.xi.j and show that {alzx ij
: 1< <:; i}
is a basis for
107
7. Commutative local rings 8. Give a Brouwerian example of a nilpotent endomorphism a of m2 such that im a is not finite-dimensional. NOTES
OUr theory of Noetherian rings and modules makes extensive use of the axiom of dependent choices. To attempt a theory that does not avail itself of this axiom seems too ambitious for the purposes of this book, and it appears likely that the classical theory would be significantly distorted at best. Classically, the ascending chain condition on submodules is equivalent to the ascending chain condition on finitely generated submodules, and the latter condition admits interesting constructive examples (the former does not). The descending chain condition, on the other hand, does not seem to lend itself to a constructive treatment. A test case would be to formulate adescending chain condition that was satisfied by the abelian group 7L (pCIJ) , the p-primary component of IQß. The definition of coherent comes from [Bourbaki 1961, §2, Exercise 11] where it is called pseudo-coherent, the term coherent being reserved for finitely generated, pseudo-coherent modules. The classical theorem that a ring is right coherent i f and only i f products of flat left modules are flat involves the full axiom of choice. Exercise 5.3 is the countable version. The proof that finitely presented flats are projective in Exercise 5.4 is from [Bourbaki 1961, §2, Exercise 23(a)]. The app1ication to deciding whether a finite1y presented module is projective is from [Baumslag et a1 1981, Lemma 5.1) where the commutativity hypothesis is not stated but seems to be usedi the problem is that IxK need not be a submodu1e.
The
Baumslag paper is in the eontext of reeursive funetion theory and uses Markov's princip1e. For the relationship between constructive algebra and recursive algebra see [Bridges-Richman 1987]. In [Julian, Mines and Richman, 1978] a field is defined to be a commutative loeal ring, with no nilpotent elements, in which 0 cannot equa1 1.
Chapter IV. Divisibility in Discrete Domains
1. DIVISIBILITY IN CANCELIATlal KN)IOS
A commutative monoid is called a cancellation monoid if ab
=
ac implies
b = c. If R is a discrete integral domain, then the set of nonzero elements of R forms a discrete cancellation monoid. This example motivates our study of general cancellation monoids. Te rminology introduced for cancellation monoids transfers to discrete domains by applying it to the monoid of nonzero elements. rf a and b are elements of a cancellation monoid M, then we say that a divides b, and write alb, i f there exists c in M such that b = ca. The divisors of 1 are the units of M and form a submonoid U of M which is a group. If U is detachable, we say that M has recognizable units. '!Wo elements a and b of Mare associates, written a ~ b, if each divides the other. As M has cancellation, a ~ b if and only if there is a unit u such that b = ua. The monoid MjU is constructed from M by declaring two elements equal if they are associates. We operate in MjU when we ca re what the elements are only "up to a unit": in particular, the relation alb may be viewed in MjU where it constitutes a partial order. We say that d is a greatest common divisor, or GCD, of a and b, if dia and dlb, and for each c such that cla and clb, we have cld. Observe that GCD's are unique in MjU, and that the GCD of a and b is the infimum of a and b in the partially ordered set MjU. If d is a GCD of a and b, then we let OCD(a,b) denote d as an element of MjU. '!Wo elements a and bare relatively prime i f CCD (a,b) = 1. A GCD-monoid is a cancellation monoid in which each pair of elements has a greatest common divisor. A GCD-domain is a discrete domain whose nonzero elements form a GCD-monoid. Let a, band c be elements of a CCD--monoid M. (i) OCD(CCD(a,b).c) = CCD(a,CCD(b,c)). (ii) c'OCD(a,b) = CCD(ca,cb).
1.1 'lHEDREM.
108
Then
109
1. Cance11ation monoids (iii)· (iv)
= OCD(a,b),
x
IF
albe, and OCD(a,b) = 1, then ale.
then
OCD(a,be)
= OCD(a,xe).
IF
Claim (i) is easi1y verified. For (ii), let d = OCD(a,b) and Then edle, so e = edx. It remains to show that x is a Now ea = ea' = edxa', so a = dxa'. In the same way, b = dxb' •
P9OOF. e
= OCD(ea,eb).
unit.
Thus dxld, so x is a unit. For (iii) we have OCD(a,bc) = OCD(OCD(a,ac),bc) = OCD(a,OCD(ae,be» = OCD(a,c·OCD(a,b» = OCD(a,xc). Claim (iv) follows immediately from (iii) upon taking x = 1 0 nonunit p of a cancellation monoid is said to be irreducible i f whenever p = ab, then either a or b is a unit. We say that a nonunit p is prime if whenever plab, then pla or plb. Clearly every prime is irreducible. A
1.2 LEMMA.
Eaeh irredueible
element in a GCD-monoid is prime.
P9OOF. Let p be irreducible and suppose plab. We shall show that pla or plb. Let d = OCD(p,a) and let p = ed. As p is irreducible either e is a unit or d is a unit. If e is a unit, then pld, so pla. If d is a unit then OCD(p,a) = 1 so p Ib by (1.1.iv). 0 1.3 LEMMA. divisibility
A OCD-monoid M has
reeognizable
tmits
iF
and only
iF
is deeidable in M.
P9OOF. If divisibility is decidable in M, then we can decide whether 11, so uni ts are recognizable. Conversely suppose that M has recognizable units. Let a and b be elements of M, and let d = OCD(a,b).
u
There is s such that a = sd. is a unit. 0 1.4 DEFINITION.
Then alb if and only if ald if and only if s
Let M be a cancellation monoid.
An
element a E M is
said to be bounded by n if whenever a = aO"'a n with a i E M, then a i is a uni t for some i . An element of M is bounded if it is bounded by n for some n E m; the monoid M is bounded if each of its elements is bounded. A discrete domain is bounded if its nonzero elements form a bounded monoid. The units of Mare exactly those elements of M that are bounded by 0; an element of M is irreducible if and only if it is bounded by 1 but not by O.
110
Chapter IV. Oivisibility in discrete domains
A principal ideal of a commutative monoid M is a subset I of M such that I
=
Mo
=
{ma : m E
M} for some 0 in M.
We say that M satisfies the
divisor chain condition if for each ascending chain I, principal ideals, there is n such that In
= I n +l .
~
12
~
13
~ •••
of
A discrete domain is
said to satisfy the divisor chain condition if its monoid of nonzero elements does. A cancellation monoid M is bounded if and only if for each 0 in M there ~ I l I:::: ••• k In of principal ideals, with 1 0 = Mo, there exists j < n with I j = 1 j +l ; thus any bounded monoid satisfies the divisor chain condition. The ring of integers is bounded as
exists n such that for any chain I 0
each nonzero integer n is bounded by Inl.
The polynomial ring F[X] over a
discrete field F is bounded as each nonzero polynomial f deg f.
is bounded by
A GCD-domain satisfying the divisor chain condition is called a
quasi-UFO. 1.5 EXAl'IPLE.
Let 0 be a binary sequence.
Let R = Unlf[on/2].
Then R
is a Brouwerian example of a quasi-UFO without recognizable units, as we cannot tell whether 2 is invertible.
Clearly we cannot write the elements
6 and 10 of R as products irreducible elements.
However we shall show
that, given a finite set S of nonzero elements of a quasi-UFO, we can find a set P of pairwise relatively prime elements such that every element of S is an associate of a product of elements of P (Corollary 1.9). 1.6 LE21MA.
Let
M
be
0
GCD-monoid
sotisFying
the
divisor
choin
condition,ond tet Pl, ... ,Pm be pairwlse retotiuely pdme elements oF M. If 0 E M, then we con construct elements 00.01 ..... om in M such that (i) a
= aOOl···o m
(ii) For j = l, ... ,m there exists e such that ojlpj (iii) 00 ond Pj ore ,'elotiuely prime fo,' j
= l, ... ,m.
PR<X>F.
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<1 + tb divides a and b. A principal ideal damain is a Bezout domain which satisfies the divisor chain condition.
116
Chapter IV. Divisibility in discrete domains
Observe that if sa + tb divides a and b, then sa + tb principal
ideal
I, !:;; I 2 !:;;
domain
of
is
finitely
Noetherian,
that
=
is,
given
generated
ideals,
there
is
discrete
domain,
then
A
GCD(a,b).
a n
sequence such
that
I n +1 ·
In
IF
2.8 'HJEXJREM.
R
is
a
the
FoUowing
are
equiuaLent.
(i) R is a Bezout domain. (ii) Euery FiniteLy generated ideaL in R is principaL. (iii) Euer'y FiniteLy generated ideaL in R is principaL, and R is a
GCD~omain.
2.9 COROLIARY.
K is
If
0 a
discrete
FieLd,
then
K[X)
is
a
bounded
principaL ideaL domain.
PROOF. An
Combine Theorem 2.8 and Theorem 11.4.7.
exarnple of a Bezout domain that is not a principal ideal domain is
constructed as fallows.
Let I. be a discrete field, and M the monoid of
nonnegative rational numbers under addition. I. (M);
0
Let R be the monoid ring
the elements of R may be thought of as polynomials in X wi th
coefficients
in k
and exponents
decreasing sequence of positive
in M. rationals,
If m, ,m2"" is a strictly then (X m, ),(Xm2), •.. is a
strictly increasing sequence of finitely generated ideals of R.
On the
other hand, given a finite number of elements of R, there exists m
E
M
such that they are all contained in k[X m ), so R is a Bezout domain. If k is a discrete field,
then k [X) is a principal ideal domain, but
IdX] but not necessarily a UFD -
see Exarnple 2.2.
A related Brouwerian
exarnple of a principal ideal domain which is not a UFD is constructed as fallows. principal
Let a be a binary sequence, and let R ideal
domain,
irreducible factors.
but
we
cannot
= Un~[ian)'
factor
the
Then R is a
element
2
into
Note that R is a bounded GCD-domain. In general a
principal ideal domain is a quasi-UFD. In Section 4 we show that ©[X] is a UFD, that is, © is factorial.
In
Chapter 6 we shall show that a discrete field" is factorial i f and only i f it has a root test, that is, each polynomial in I<[X] has a root in I< or
has no raot in Iz.
That will provide us with more exarnples of discrete
fields k such that k[X) is factorial.
2. UFO's and Bezout domains
117
EXERCISES 1. Let R be
a
discrete
domain.
Show that
the
following
are
equivalent. R is a UFD. (ii) Each nonzero element of R is a unit, or is a product of primes. (iii) R is a bounded ~omain, and each nonzero element is either a unit, or is irreducible, or has a proper factor. (iv) R is a quasi-UFO, and each nonzero element is either a unit, or is irreducible, or has a proper factor. Remark: A proof of (iv) implies (i) requires the use of the axiom of dependent choice. (i)
2. Eisenstein criterion. Let R be a discrete domain, and let f aO + ••• + anXn E R[X] such that any common factor of the a i is a unit. Let pER be a prime element such that p does not divide an' and p2 does not divide aO' but pla i for i < n. Prove that f is irreducible in R[X]. 3. Let R be a principal ideal domain, and let S be a multiplicative submonoid of R not containing O. Show that s-1R is a principal ideal domain. 4. Show that any Bezout domain is coherent. Show that if a Bezout domain has recognizable units, then it has detachable ideals. 5. Show that if R is a Bezout domain, then each finitely generated submodule of Rn is free of rank at most n. 3. DEDEKINI)-HASSE RIlIGS
AN!)
EUCLIDEAN
IJaItAINS
Oiscrete domains often admit maps to the nonnegative integers that can be used to study questions of divisibili ty. Examples are the absolute value function for the integers, the degree function for polynomials over a discrete field, and norms on rings of algebraic integers. If the map fits in with a division algorithm, or satisfies the Oedekind-Hasse condition, then the domain under consideration is a principal ideal domain.
118
Chapter IV. Divisibility in discrete domains Let
3.1 DEFINITlOO.
be a function from the nonzero elements of a
]l
discrete domain R to the nonnegative integers. pseudonorm if whenever
(i)
CI
such that IJ 10, then either suchthatu(b')
Then
is a
IJ
and bare nonzero elements of R (J
",
IJ
or there exists b ' " b
(ii) Dedekind-Hasse map if for any nonzero elements a
R, either 11(1')
<
alb,
or there exists nonzero " in (a,b) such that
v(a).
(iii) Euclidean map if for any nonzero elements either olb,
al (b We say that
or
- ,,) and
IJ
there exists nonzero r
Ü')
that
LI
((1b)
>
v(o)ll(b)
for all nonzero
0
(1
A multiplicative pseudonorm is called a multiplicative norm.
The notion of a pseudonorm is a technical convenience. norm
and b of R
Cl
in R such
< "(0).
is multiplicative i f
!l
and b in R.
and b of
should
probably
]je
somewhere
between
a
The notion of a
pseudonorm
and
a
multiplicative norm, possibly at one end or the other. AnlJ Euel idean map is c, Dedekind-Ho,sse map.
3.2 THEOREM. Hosse
map
is a
If
pseudonot'm.
element a is bounded b\J u(o). un i t i
f (md on I \I
PROOF.
f
i
lJ
v
i
S
If v is
psetldonor'm,
L1
0
An!) Dedekind-
The first claim is obviously true.
such that u(,-)
< v{n).
we can pick b'
nlb
Either
v(a).
Suppose a
Tl •
'~b
where b
= 01' "on
< u(r) < ll(a).
such that lJ(b')
is a pseudonorm and
l'
We shall show that a is bounded by n = aOh
We pmceed by in (n,b) = (b)
l'
In the latter case, by induction, either r ' b, so
To prove the third claim suppose that on
0
To prove the second,
or there exists nonzero
'-, or there exists b'
nonzero element.
is
0
= 1.
(0)
suppose that v is a Dedekind-Hasse map and b divides a. induction on
each ,wnzen)
then
mtllliplicotilJe nm'm, then
if
Tl
>
= 11 ((1)
0, and b I} ,
=
(l
is a
by induction 1 if n
=
O.
I, such that so 0.0 is a unit, or there is By induction /J' , emd hence 1>, i s bounded by Tl - 1, so (li is
Then
either olb,
u(b' )
<
u(o) .
~
a unit for some i . Finally suppose that 0,
we have
lJ(n)v(c) II
=
I. (
l)
=
1.
1 whence v(n)
is a fiul tiplicati ve norm.
II
If " =
is a pseudonorm, either
1. (l
'"
is a unit,
As
Conversely, suppose u(n) 1, so
(l
t'( 1) =
then oe = 1 for =
1.
I'
(l)" (1) >
some c,
so
As 11([, and
is a unit, or there exists b'
such
3. Dedekind-Hasse rings and Euclidean domains that v(b') < vIa)
~
1, which is impossible.
119 0
The Dedekind-Hasse condi tion provides a cri terion for a ring to be a bounded principal ideal domain. 3.3 THEOREM.
A diserete domain with a Dedekind-Hasse map ts a bounded
prineipal ideal domain. PROOF. Suppose v is a Dedekind-Hasse map for R. As R is bounded, by (3.2), it suffices to show that R is a Bezout domain. Given nonzero a,b € R, and a nonzero element e in the ideal (a,b), we shall show by induction on v(e) that there is a common divisor of a and b in (a,b). As v is a Dedekind-Hasse map, either ela or there exists nonzero r in (e,a) such that v(r) < v(e). Similarly either elb or there exists nonzero r in (e,b) such that ll(r) < v(c). Hence either e is the desired common divisor
of a and b, or there exists nonzero r in (a,b) such that v(r) < v(e), and we are done by induction. 0 3.4 EXAJIIJPLE. A multiplicative Dedekind-Hasse norm that is not a Euclidean map, Let R = H:[ (1+F-I9)/2], which is easily seen to be a free H:-module with basis 1 and (1+F-I9)/2. The function N(a+vJ:19) ~ a 2 + 19b 2 is a multiplicative function on ~(vCI9) which restricts to a norm on R. If a and ß are nonzero elements of R, we shall show how to construct e in R such that N(ß/a - e) < 1
or
0 < N(2ß/a - e) < 1.
This will show that N is a Dedekind-Hasse norm on R. ß/a -
e =
can easily find
Ibl
~ 3/16, then N(ß/a - 8) ~ 235/256
3/16, then we can find 8'
€
such
a + bF-I9
we
8 € R
Writing
that
Ibl ~ 1/4 and lai ~ 1/2. If < 1, and we are done. If Ibl >
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
<:;
(p + qJ2) (c + dJ2) = (0 +
so v(s + tJ2) is Euclidean. 3.6 EXAMPLE.
I (p
-
m)2 -
(a
1/2 and
b/2)
=
+ bJ2)/(c + dJ2) = p + q~2,
Iq - Tl 1 S
1/2.
and
Then
(m + nh) (c + d~2) + (s + t~2)
2(q - n)"lv(e + dJ2) <:; v(c + dJ2l/2.
Thus
v
0
A principal ideal domain that is not bounded.
binary sequence with at most one 1, and let
Let a be a
121
3. Dedekind-Hasse rings and Euclidean domains R =
([l[X,a,Y, ,azY z , .. . 1I(a, (X-Y,) ,az(X-Y~), •.• ).
Then R is a principal ideal domain, but we can't find abound for X. 3.7 'lHEDREM.
Let R be a discrete domain and v a Dedekind-Hasse map on
R such that v(a) = v(b) if a
not containing O. Euclidean
if
PROOF. v(a)
=
v(b)
properties.
~
b.
Let S be a multipUcative subset of R
Then v extends to a Dedelünd-Hasse map on S-IR that is
v iso
Set v(r/s) if a
~
b.
=
This is weIl defined because routine to check that v has the desired
v(r-;CCD(r,s».
It is
0
3.8 EXAMPLE. A domain with a Euclidean map that does not have recognizable units. Let R be the ring of integers, and S the nrultiplicative subset generated by {2a n : nEIN} for some binary sequence a. Extend the absolute value function on R to a Euclidean map on S-IR by (3.7) . EXERCISES
1. Call a bounded discrete domain R a DH-ring if for all nonzero a and b in R, with a bounded by Tl, either alb or there exists c in (a,b) that is bounded by n-l. Show that a DH-ring is a principal ideal domain. 2. CaIl a bounded domain R strictly bounded if whenever a divides an element b that is bounded by n, then either b divides a, or a is bounded by n-l. Show that a bounded domain with recognizable units is strictly bounded. Show that a DB-ring (see Exercise 1) is strictly bounded. Show that a strictly bounded principal ideal domain is a DH-ring. 3. Let a be a binary sequence, and let R = ([l[y Z ,a,Y,a,/(y Z _l), azY, a z /(y Z -l), ... ]. Show that R is a Brouwerian example of a bounded principal ideal domain that is not strictly bounded (see Exercise 2) by considering the elements a = y 2 and b = y 2 (y 2 _1). Show that R admits a Euclidean map, but not a multiplicative one. 4. Show that the ring of Example 3.6 is a principal ideal domain. 5. Let R be a Euclidean domain. Show that for each x unit u such that v(x) " v(u).
#
0 there is a
Chapter IV. Divisibility in discrete domains
122
6. Show that the ring ?[i 1 of Gaussian integers has a multiplicative Euclidean norm.
7. show that
any
pseudonorrn on
a
principa1
ideal
domain
is
a
Dedekind-Hasse map. 8. A peculiar omniscience pünciple.
Consider the subsets Cn of the
set 2IN of binary sequences defined inductively as folIows.
Co
{Ol·
=
if (ll = 1 then either
{o
or °;+1,° 1 +2"" Show that °
E
Cl if {m : 0m
Show that u E Ca if {m
=
I}
= 1 for some
j
I, (X n )
-> Ic
let
i,
Show that LPO
What do you make of the omniscience
9. A bounded PID that does not admi t a pseudononn.
element field,
>
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
:
(X n + 1 ) by 'i'n (X n ) = u n
(X'~+l +
n) + Xn +1 ·
Let R,l be a subring of k(X n ) such that R1 = "[Xl]' and R,Hl = if u n = 0, and R,Hl = S-l<1' (~), where S is the
multiplicative set generated by X;~+l +- X'l+l + I, if an that the direct limit R of the rings
r
1.
Show
is a bounded PID.
Show
=
that the omniscience principle of Exercise 8 would hold i f I< adrnitted a pseudonorm. 10. Show that ll'[ (1+-J-19)/2) is the integral closure of 7L in il)[-J-19]. 11. Show that the units of R in Example 3.4 are ±1. (3
=
(1+-J-19)/2.
Let
°
=
2 and
Using N, show that a, a+1, tl-1, (3, (3+1, and (3-1
are pairwise relatively prime. v, and 1 is a nonunit of R.
that there is a nonullit
l'
Suppose R admits a Euclidean map By dividing
1
into a and (3, show
of R such that nh')
< "h).
12. Construct a Brouwerian example of a principal ideal domain with a Dedekind-Hasse map that does not have recognizable uni ts or a Euclidean map.
(Look at U' [ (1+';-19 )j2] (;;
([l
[-J--19] )
123
4. Polynomial rings 4. POLYlQoJIAL RIlG"l
In this section we consider properties of a discrete domain R that are inherited by the polynomial ring R[XI. 4.1 DEFINITIrn.
Let R be a GCD-domain, and let f E R[XI.
the coefficients of f cont(f).
If cont(f)
4.2 LEMMA.
=
The GCD of
is called the content of fand is denoted by 1, then f is said to be priudtive.
Let R be a GCD-domain with Fietd of quotients K.
lf
f E K[X], then we can find c E K and primitive 9 E R[XI such tllat f = eg. IF f = c'g' u
For c' E K und primitive g' E R[X], then c
Write
PROOF.
go/cont (gO) and c
F
=
uc'
for some unit
E R.
= c'g'
f = cogo
for
some
= cO·cont (gO)'
so f
where c' E K and g'
Co E K and go E R[X I.
= eg
and 9 is primitive.
= lIdc'
=
Choose
Thus c
= lIC'.
0
Let R be a GCD-domain, und Let f,g E R[XI.
cont (f )cont (g) •
Let m
PROOF.
As cont (ah)
= deg
= a'cont (h)
fand n
= deg
g.
We proceed by induction on m + n.
for constants a and polynomials h, we may divide f
and 9 by their contents and prove the claim for primitive fand g. c
=
eont(fg) and d
= GCD(e,f m ), where Fm If f = fmXm the
Then dl(f - fmXm)g. induction,
d !cont (f - f mXm )cont (g).
dl(f - fmXm), so dlf. the
same
way we
GCD(c,fmgn )
=
Then dc and dc' are each the
for some uni tuE R.
4.3 LEMMA (Gauss's leDllla). riten cont (fg)
9
Supppose
is a primitive po1ynomial in R[X).
nonzero dER such that dc and dc' are in R. content of dF, so dc
Let
= 1,
4.4 COROLLARY.
As f
is the leading coefficient of f. lemma is clear. Since
9
is
is primitive, we get d
prove GCD(c,gn)
so fg is primitive.
=
1.
Thus
by
Otherwise,
primitive GCD(C,f m )
Lemma
we
=
get
1.
1.1.iii,
by
In c
0
Let R be a GCD-domain with fietd of quotients h.
and 9 be polynomials in R[XI.
Let
Let F
rhen F divides 9 in R[XI if und only iF f
divides 9 in h[XI and cont(F) divides cont(g).
PROOF.
The "only if" is immediate from Gauss's lemma.
"if" we may assume that f is primitive.
where h is a primitive polynomial in R[Xj and a E h. is a primitive polynomial, so a
=
To prove the
By (4.2) we can write 9 = ahf By GaU55'S lemma Fh
cont(g) E R by (4.2).
Thus f divides 9
Chapter IV. Divisibility in discrete domains
124 inR[X].
0
4.5 THEOREM.
Let R be
0
CCD--domoin mUh quotient (ietd k.
be polynomiols in 1,[Xl, eaeh
lf f9 is
of R.
=
and b-
bg, .
1
(s
unit
(1
r
=
in R.
By Gauss's
lermna f,g,
=
o-'b-'fg
is a
ar, o
-
,
primitive
So a -, b -, is a uni t of R, so u and b are in Rand thus fand
are in /, IX
J.
0
4.6 TUEDREM. be
thnt
Since Fand 9 have units of R among theiL coefficients,
are
polynomial. 9
wllLeh has some eoeffieient
phmitiue pol!Jnomied of R[X). llten fand 9 an, in R[X].
0
There are a,1:! E Rand pdmitive f,g, E R[X] such that
PROOF.
and 9
01'
Let fond 9
Let
1, lJE' u C;CD--
in k[X] such lhat fg E R[X J.
pot~j,wmüt!s
TI1('n tfleTe
(s
Let
F und 9
b E k such lha t
bf and 9;1) al-e in R[XJ. PROOF.
There are cl,b E k and primitive f"g, E R[X] such that f =
and 9 = bg,.
By Gauss's 1ermna f,9,
(4.1) whence bf
and 9/b
= abf,
4.7 THEOREM.
= g,
is primitive so ab are in R[X J.
=
ar,
cont(fg) E. R by
0
Let R be a disen'te domoin.
Ir R is a CCD--{lomn i Il , then so is R[X) • (ii) If R tS bounded, thell so lS R[X) • (i)
(iii) (iv) (v)
If
R
If R
If
,'ecogn i zab I e units, thell so does R(X] •
11[IS
lLLls deci.tlüble diuisihi.lil\},
R snl i sft es
tlien so does R[X
the eLi uL sor chuin condL t i on,
J.
then so docs
R[X ].
(vi) PROOF.
Ir
I~
is
(!
CjIJC1si-{]FD,
then sn is R[X].
Let /, be the field of quotients of R.
Ta prove (i) let rand 9
be in R[X), and let h be primitive in R[X) such that h is a GCD of Fand 9 inl<[X).
Letd =CCJJ(cont(fJ,cunt(g)).
By (4.4) d/i divides rand gin R[X]. and 9 in R[X]. eont(g),
We shall showthatdh =GCD(f,g). Conversely, suppose Ci divides f
Then q divides h in ldXj, and cant ('1) divides con! (f) and
hence divides d.
Therefore Cl divides rlh by (4.4) and we hav(,
shown that cl/i is a GCD of fand 9. To prove (ii) coefficient of f.
let
r E R [X j
have degree
fi,
If" is bounded by m, then
r
and let
(!
be the leading
is bmmded by
In
+ n.
Claim (iii) is trivial, as the ring R[X] has the same units as R. Ta prove (iv) we show by induction on n
=
deg 9 that we can decide
125
4. Polynomial rings whether f di vides g.
Let a be the leading coefficient of f, and b the We may assume that deg f
leading coefficient of g. then f
not divide b,
does not divide g.
~
deg g.
If alb,
polynomials q and h such that 9 = qf + hand deg h $ n - 1. and only if flh.
If a does
then there are Then f Ig i f
By induction we can decide whether f divides h, so we
can decide whether f divides g. Suppose R satisfies the divisor chain condition. be an ascending chain of principal ideals in R[X).
such that (fm)
Let (f1)
~
(f 2 ) ~
•••
We shall construct m
= (Fm+l) by induction on
coefficient of f i •
deg Fl' Let a i be the leading Then ai+1lai for each positive i, so there is m such
If deg f m
=
Otherwise deg f m+1 < deg Fm
~
that amlam+1'
starting at f m+l .
deg Fm+1'
then Fm1Fm+1 and we are done. deg fl' and we consider the sequence of F's
By induction we
find
the desired pause
in this
sequence. Claim (vi) follows from (v) and (i). 4.8'l'HIDREM (Kronecker 1). unUs, then so is R[X).
Let F
PROOF.
0
If R is an infinite UFD wUh finitet!} man!}
Thus R is factorial.
be a polynomial of degree n.
€ R [X)
It suffices to
construct a finite co1lection of polynomials that contains all divisors of f of degree at most n/2. n
~
2m.
Let aO, ... ,am be distinct elements of R, where
Since R is a UFD with finitely many units, each nonzero f(ai) has
a finite set of divisors.
If some f(ai)
=
0, then f
=
are done by induction on n.
So we may assume that F(a i
Note that i f gl F, then
IF(ai) for each i.
9 (ai)
(X - ai)g and we ) #
0 for each i.
There are only fini tely
many sequences b O"" ,bm such that b i divides F (ai) for each i. By the unique interpolation theorem (11.5.5) there is, for each sequence bO, ... ,bm ,
9 (ai)
= bi
a unique polynomial g€h[X) for all i, where h
of degree at most m such that
is the field of quotients of R.
The
collection of polynomials 9 forms a finite set of polynomials in h [X) which contains all divisors of F in R [X 1 of degree at most n/2. polynomial 9 of degree at most n/2 is a factor of f 9
€ R[X)
and F/g
from IdX).
€ R[X),
Then a
if and only i f
and this is decidable since R[X) is detachable
0
Kronecker 1 shows that TL [X I' ... ,X n ) is factorial for each n. Using Corollary 2.6 we see that !ll[X l , ... ,X n ) is also factorial. If h is a
126
Chapter IV. Divisibility in discrete domains
fini te field, then k [X I' .•• ,X n ) is factorial , since each F E k [X l' ... ,Xn ) has only finitely many divisors. However, not all UFD's are factorial, as (2.2) shows. Kronecker I requires that the ring R be infinite. certainly factorial.
But finite UFD's are
In Theorem VI.6.ß we shall eliminate the requirement
that R be either finite or infinite. The algebraic closure k of III -
see
(VI. 2.5) -
polynomial ring k [X) is a UFD, so k is factorial.
as weH as the
But Kronecker 1 is not
enough to show that k[X) is factorial, as was the ca se for k
= Ill.
Another
trick of Kronecker's shows that R[X,Y) is a UFD if R[X) iso 4.9 THEOREM (Kronecker 2).
IF R is a
dornain,
Factol"iat
then so
i.s
R[X). PRODF. For m > 0, let ~m : R[X,Y] ~ R[X] be the ring homomorphism that is the identity on R[X] and takes Y to Xm. Let ~ : R[X) ~ R[X,Y] be the R-module homomorphism that takes m.
xl1
to
yqx r-
where n
=
qm + r-, and 0
~ r-
<
Let R[X'Y)m be those polynomials in R[X] of X-degree less than m.
Then ~m~
is the identity on R[X]
~~m is the identity on R[X,Yl m R[X,Y] is closed under taking factors.
To factor a polynomial F in R[X, Y), of degree less than m, look at the ~m(f) = ab. Check to see if = f. Any factorization of F must have this form. 0
finite number of factorizations (up to units) ~(a)~(b)
EXERCISES 1. Show that X4 + 1 is irreducible over Ill[X) using the techniques of Kronecker 1 and (2.6). 2. Factor X' +
4y 4
Use Kronecker 1 to factor X' + 4.
into primes over the ring
~[X,Y).
3. Use Kronecker 2 on the polynomial X2 + Y to see that even though the image
~m(F)
factors in I,[XI, the original polynomial need not
factor.
The divisor chain condition is defined in terms of ascending chains of principal ideals I 1
~
12
~
•••
instead of descending chains of elements
Notes
127
a, ,a2' ••• with a i lai_I.
The two versions are equivalent in the presence of the axiom of dependent choices. The version with ideals allows us to derive in this chapter and in Chapter VI - the basic properties of principal ideal dornains without the use of the axiom of dependent choices.
In quasi-UFD's, principal ideal dornains and Euclidean dornains we need not be able to factor nonzero elements into irreducible factors. Instead we can use the quasi-factorization theorem to wri te nonzero elements al, ••• ,an of a quasi-UFD as products of pairwise relatively prime elements PI'·· .,Pm • Kronecker 1 and 2 are found in [Kronecker 1882].
These results provide algori thrns for factoring polynornials in several indeterminates over the integers as products of primes. We are not concerned here with the efficiency of these algorithrns.
Chapter V. Principal Ideal Domains
1. DIJ\GCHU.IZll«> MATRICES
The theory of modules over a principa1 ideal domain is closely related to the theory of vector spaces over a field and is almost identical to the theory of abelian groups, which are modules over the integers. The analogue of a finite-dimensional vector space is a finitely presented module over a principal ideal domain. A finitely presented module is given by matrix. In this section we prove some basic facts about matrices over a principal ideal domain. An m x n matrix A = (aij) is diagonal if a ij = 0 whenever i F j. Two m x n matrices A and Bare equivalent if there is an invertible m x m
matrix C, and an invertib1e n x n matrix D, such that A 1.1 LEMMA.
=
CBD.
Euch matrix over' a principat ideat domain is equivaLent to
a diagonaL matrix.
PROOF.
matrices.
The key is the construction of certain invertible 2 x 2 If sa + tb
=d
f.
0 is the GCD of a and b, then for all u,v
there are w,x such that
[~
b ). [ s -b/d ) v t a/d
[~ 2)·
Moreover, the right hand factor is invertible since its determinant is 1. Similarly, if a and b are in a column, we multiply on the left as follows.
[ -b/ds
t
a/d
].[ab u] v
[i
~].
Thus if a and bare entries in the same row (column) of a matrix B, we can multiply B on the right (left) by an invertible matrix leaving d in the posi tion occupied by a, and 0 in the position occupied by b, and fixing the entries not in the rows or columns of a or b. We shall operate on a given matrix A by left and right multiplication by invertible matrices to obtain a diagonal matrix. If A = 0 we are done. Otherwise, by row and column interchanges, we can bring a nonzero element a, to the upper left hand corner. By a sequence of right multiplications 128
1. Diagona1izing matrices
129
by invertib1e matrices we can replace a, bya 2 , the GeD of all elements in
the first row, 1eaving zeros in the remaining places in the first row. Similarly, by 1eft mu1tiplications, we can replace a2 by a", the GCD of all elements now in the first column, leaving zeros in the rest of the first column. Continue by replacing a" by a4, the GeD of all elements now in the first row, and so on. In this manner we generate a sequence (a, ), (a 2
), •••
of principal ideals such that ai+llai for each
i.
Hence,
for some n, we have an lan+l. But that means that an is a GCD of the elements in the first row (or column), while the remaining elements in the first column (or row) are zero. Hence, by elementary row or column operations, we can clear both the first row and the first column, making all elements in the first row and column zero, except for the corner. By induction on the size of the matrix, we can diagonalize A. 0 A
matrix A =
(a i j )
is in Smith normal form i f it is diagonal and
aiilai+1,i+1 for each i.
1.2 'HiEDREM. Each matrix ouer a pdncipaL ideaL domain is equiuaLent to a maU-ix in Smi th nor-maL form.
By Lemma 1.1 we may assume that we are given a diagonal matrix. Let a,b be diagonal elements and d sa + tb be the GeD of a and b. Then PROOF.
-tb/d ] sa/d Repeated application of this observation allows us to convert our diagonal matrix to a diagonal matrix wUh the property that the corner element divides all the remaining elements. Then, by induction on the size of A, we obtain a matrix in Smith normal form. 0 We want to show that each matrix is equivalent to an essentially unique matrix in Smith normal form. Given a matrix A, define Ai (A) to be the ideal generated by the determinants of all i x i submatrices of A. Let A und B be equivaLent m x n matrices oue'- a GCD-domain Then Ai (Al = Ai (Bl for aLL i.
1.3 LEMMA.
R.
PROOF. It suffices to show that if C is an invertible m x m matrix, then Ai (CA) = Ai (A) . The rows of CA are linear combinations of the rows of A. Hence the determinants of i x submatrices of CA are linear combinations of the determinants of x i submatrices of A. So Ai (A) :2 Ai (CA). Since C is invertible, we also have Ai (CA) :2
Chapter V. Principal ideal domains
130
Two m x n matr-ices in Smi th nor-mat For-m ovel- a GCD-domoin
1.4 THEOREM.
if and ont9 if corTesponding etements aroe ossociates.
ar-e equivatent
PROOF.
Let D
=
(d ij ) be a matrix in Smith normal form.
We easily
verify that A1 (D) = (d n ), and Ai (D).(d i +1 ,i+l) = Ai +1 (D) for each i ~ m-l. So the diagonal elements of D are determined, up to a unit, by the ideals Ai (D) • By Lenuna 1. 3 this implies that if two matrices in Smi th normal form are equivalent, then their elements are associates.
0
EXERCISES
1. Find a matrix in Smith normal form, the matrix
over~,
that is equivalent to
1 2 3
2.
A
4 5 6. 789 valuation ring is a commutative ring R such that for any two
elements 0 and b in R, either (0)
~
(b) or (b)
~
(0).
Show that
each matrix over a valuation ring is equivalent to a matrix in Smith normal form. 3. Show that a square matrix over
~
is invertible if and only if it
is a product of elementary matrices. 4. Show that a square matrix over it
is
a
product
of
~
has determinant 1 if and only if
elementary
matrices
corresponding
to
elementary operations of type (iii). Hint:
(-1° -10]
-1] [121' 0] (1°-1]l' [1210] [101'
2. FINITELY PRESENTED !DXJLES
Let M be a module over a commutative ring R.
A finite presentation of
is a tripie (M, (xl"'.'x n ), A) where xl, ... ,xn is a finite set of generators of M, and A is an m x n matrix of elements of R whose rows
M
generate the module of relations among the x' s. (ol, .•. ,on)
Rn
Thus for each element
°
we have 0lxl + ••• + 0l/n if and only if (al' .•• 'an) = vA for some II € R"'. We may identify a fini tely presented module with its finite presentation. The matrix A contains all the information about the structure of the Given M, the matrix A depends on the choice of generators module M. €
131
2. Finitely presented modules x1, .•• ,xn and on the choice of generators for the module of relations. M
If
'= R/(r-1) ES ••• ES R/(r n)' then we can get a diagonal matrix A for M; if we can get a diagonal matrix A for M, then we have
conversely,
exhibited M as a direct sum of cyclic modules.
Thus it behooves us to
examine how to pass from one matrix for M to another. Let
(M, (x1""'x n ), A)
(M, (x1""'x n ), Bl
be
a
finite1y
presented R-module.
Then
is a finitely presented R-module i f and only i f the
rows of B generate the same submodule of Rn as the rows of A; happens if B
this
= FA for some invertib1e m x m matrix F. What happens to the
matrix A of relations of a fini tely presented module (M, (xl"" ,xn ), A) when we change the generators x1""'xn ? 2.1 THEDREM.
Let
(M.
(xl"" ,xn ), A) be a
ouer a commutatiue ring R, and let E be Then (M, (x1 ..... xn)' A)
PROOF.
M.
=
(M.
Wl
fini tel!} presented module
n x n inuertible matrix ouer R.
(x1 ..... xn)Et, AE- t
Let (!}l'''''!}n)t =E(x1, ... ,xn )t.
).
clearly !}l""'!}n generate
Moreover the following are equivalent.
o (r1, ... ,rnl
(sl, ... ,sm)A
(r1,· .. ,rn )E-
for some
(Sl'''''Sm)
(Sl"",sm)(AE-').
t
Hence the module of relations among !}l""'!}n is generated by the rows of AE- 1
•
0
2.2 COROLLARY.
Let A be an m x n matrix ouer a commutatiue ring R, und
let (M, (xl'''' ,xn ), Al be a FinUd!} presented R--module. Let E be an n x n inuertible matrix ouer R, and F an m x m inuertible matrix ouer R. Then (M, (x1""'Xn )' A) = (M, (x1 ....• xn)Et, FAE-
PROOF.
As the rows of FAE- 1 have the same span as those of AE- t
corollary is an immediate consequence of 2.1.
2.3 THEDREM (structure Theorem). ouer a
principal
ideal
1 1 :2 1 2 :2 ••• :2 In such R/I 1 ES R/I 2 ES '" ES R/I n •
PROOF.
1 ).
M
the
Let M be a Finitet!} presented module
domain R. that
,
0
Then is
ther-e
isomor-phic
exist to
pr-incipal the
ideals
dir-ect
Let (M, (x1, ... ,xnl, Al be a finite presentation of M.
Theorem 1.4 there exist invertible matrices E and F such that D
Sllm By
= FAE- 1 is
132 a
Chaptee V. Principal ideal domains
diagonal matrix (M,
If (Yl""'Y n )
where I i = The
=
in Smith nOemal
form.
(d i i ) .
then M
,
have
(xl, ••• ,xn )E t , D).
( x l " " ' x n ), A) = (M,
(xl, ... ,xn)E t
By Corollary 2.2 we
= Ru1!ll '"
(j)
Ry n , and RU; - R/l t
0
decomposition
in
Theorem
2.3
is
essentially
unique
over
an
arbitrary conunutati ve ring. Let R be a eommutative ring, m ~ n positive integer's, and
2.4 TflEX)REM.
1 1 :;;J 1 2 :;;J ••• :;;J Im and 11:;;J J2 :;;J ••• :;;J J n ideals of R. R-module that i.s isomorphie 10 L7=1 R/I t and to Lr]~1 R/J (a)
h
J 2 = •••
(b)
Ii
In-m+i
rOl~
;
=
r
CUt
Then
R.
=l, .... m.
To prove (a) it suffices to show that if m < n, then 1 1
PRCOF. Let S
1n -;n
=
Suppose l
= 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 ,<. rf Fi = Xm - b1Xm- 1 - ••• - bm , then the matrix of T restricted to Ci' relative to the basis ci' Xci"'"
o
-1. Xm ci' 1S
o o
1
o
1
o o o
o
The matrix Bi is called the companion matrix of Fi • Note that Fi is the characteristic polynomial of Bi. By choosing such a basis for each Ci' we obtain an F-basis relative to which the matrix of T has the form
This matrix is called the where Bi is the companion matrix of Fi . rational canonical form for T. The characteristic polynomial of this matrix is readily seen to be F1F2···Fs. 4.2 '.lHOOREM (Jordan canonical form).
Let T be an endomorphism oF a
finite-dimensionaI uector- space V ouer a discrete fietd k, such timt character-istic potynomiat oF T is a pr-oduct of Hnear- factars.
can find a basis for V r-e tati ue to which the maU-ix for
J1 A
J2 Jr
1
where each matrix J i is a rn-by-m maU-ix of the For-rn
the
Then we
T has the form
4. Linear transformations
J(m,A)
for uarious m and A.
137
A
0
1 0
A 1
0
0
0
0
0 0
0 0
0
0
A 1
A
0
The diagonal of A, that is, the matrix A wi th the
off-diagonal entries set to 0, ean be wI'Uten as a polynomiaL in A.
PROOF.
By (3.3) we ean write the k[X]-module V as a direet sum of
primary modules VX- A' and (3.4) says that VX- A is a direet sum of modules, This latter eaeh of whieh is isomorphie to R [X 1/( (X-A)m) for some m. module has an k-basis 1,(X - A), ... ,(X - A)m-l, and relative to this basis the matrix of T restrieted to VX- A has the form J(m,A). Performing this eonstruetion on eaeh VX- A' we obtain a basis relative to whieh the matrix of T has the form A. The diagonal of A is a polynomial in A beeause, by (3.3), the projeetions of V on the various VX_A are given by polynomials in A. 0 We say that T is diagonalizable if V admits a basis of eigenveetors of Thus T is diagonalizable if and only i f there is a basis for V relative to whieh the matrix of T is diagonal. We ean express the faet that T is diagonalizable in terms of the minimal polynomial of T.
T.
4.3 'l'HEXJREM.
Let T be a linear transfor'mation of a finite-dimensionaL
ueetor spaee V ouer a diserete field k.
Then T is diagonalizable
iF
and
only if the minimal polynomial of T is a produet of distinct monie linear factors.
PROOF. If Al' ••. '~ are the distinet entries on the diagonal of a diagonal matrix representing T, then elearly the minimal polynomial of T is (X - Al)(X - A2)···(X - ~). conversely, if the minimal polynomial of T is (X - Al)(X - A2)···(X -~) where the Ai are distinet, then by (3.3) V admits a basis of eigenvectors of T. 0
Any two diagonal matriees eommute, 50 there is no hope of getting a basis relative to whieh linear transformations T, and T2 both have diagonal matriees, unless T, and T" commute. This eondition turns out to suffieient.
138
Chapter 4.4 'l'HEDREM.
Let T,
and T 2
v.
Principal ideal domains
be commuting linear"
tr"ansfor'mation of a
fini te-
,
If V admi ts a
then V admi ts a basis whose
•
PROOF. Let V~ = her Ti -A be the subspace of V whose nonzero elements are the eigenvectors of Ti wi th eigenvalue A. Then, for i = 1,2, the space V is a direct sum of the subspaces V~ as A ranges over the eigenvalues of Ti' As T2 commutes with T,-A, the subspace V~ is invariant under T 2 ' hence also under the projection of V onto V~, which is a polynomial in T 2'
Therefore V~
= 211
V~
n
V~, so V
=
2A,~ V~
n
V~.
0
EXERcrSES 1. Show that the Jordan canonical form of a linear transformation is unique in the sense that, for each pair m and A, the number of Jordan blocks J(m,A) that appear is invariant.
Smith normal form for matrices over the integers goes back to [Smith 1861]. The diagonal elements are called invariant factors; the proof of (1.4) shows why. The ascending chain condition is used in Lemma 1.1 to diagonalize a The question as to whether it is really needed, or whether matrix. matrices over Bezout domains are diagonalizable, has tantalized people for many years. The standard proofs of the uniqueness of the decornposition of a module over a pro into a direct sum of cyclics involve factoring into primes, despi te the fact that Kaplansky proved the more general Theorem 2.4 in 1949 (using two proofs by contradiction) . As we need not be able to factor in a PIO, we were led to rediscover the better theorem. The quasifactorization theorem (IV.1.8) was forrnulated in order to constructivize the inferior theorem.
Chapter VI. Field Theory
Let R be a subring of a eommutative ring E. An element of E is integral over R if it satisfies a monie polynomial in R[X]. The integral closure of R in E is the set of elements in E that are integral over R. 1f every element of E is integral over R, then we say that E is an integral extension of R. 1f R is equal to the integral elosure of R in E, then we say that R is integrally closed in E. 1f R is a field, the word integraL in the above definitions may be replaeed by the word algebraic. We shall show that the integral elosure of R in E is a subring of E. First a lemma. 1.1 LEMMA. Let R!;; E be commutative r"ings. If a E E satisfies a monie poLynomiat f of degree n ouer" R, then the ring R[a] is generated by 1,a, ... ,an - 1 as an R-module. PROOF. algorithm
1f ß E R[a], then ß = g(a) for some 9 E R[X]. (11.4.3) there exist polynomials q and r
r ~ n-1, so that 9 = qf + r". eombination of 1,a, .•• ,an - 1 . 0
deg
m
Then ß =
Reeall that an R-module M is faithful if r in M.
g(a)
By the division in R[X], with
= r(a) is an R-linear
o wheneve r
r"m
o for
1.2 '1lIEXJRE2Il. Let E be a eommutative ring, R a subring of E, and a The foUowing are equivaLent.
all
E
E.
(i) a satisfies a monie poLynomiat of degree n ouer R. (ii) R[a] is generated by n elements as an R-moduLe. (iii) E has a faithFul R-submodule M. genemted by n elements, sueh that aM !;; M. If (i) holds, then R[a] is generated by 1,a, ••• ,an - 1 . 1f (ii) holds, then M = R[a] satisfies (iii). Suppose (iii) h01ds, and m1, •.• ,mn PROOF.
139
140
Chapter VI. Fields
generate
Then
M.
2: "i.imi.
amj
with
by (II.6.6), so {(al = 0 as M is faithful. 1.3 COROLLARY. If a
Le t
1.4 COROLLARY.
Ir
a
is
of deg,'ee
moni.e IJoLYT1omiaL or degree
11
Tl
Olle"
R.
aue,' R.
0
Lntegr'aL over' R, ond {3 is i.ntegroL ouer' Rla],
R[u,(3] is i.ntegr'al. ouer'
(/le11
the
0
R[ß] and apply Theorem 1.2.
Take M
PROOF.
0
F be
Let
Then f(a)m j = 0 for each j
(3 sn ti sFy u mord c po Lynomi 01
RIß], then a soUsfies
E
R.
r ij E
characteristic polynomial of the matrix ["i J}'
Ilenee the elements in E tllat ar'e ifltegr'al
I~.
over R for'm a subdng of E.
From (1.2) we have R[ul is a finitely generated R-module, and
PROOF.
RlaJl] is a finitely generat.ed R[u]-modl.lle.
Fwm (I1.3.2) we have R[a,ßl
is a (faithful) finitely generated R-module, so (1.2) says that Rla,{31 is integral over R.
0
1.5 COROLLARY.
If
Let R (;; E (;; F be commutative r-ings wilh E integr'oL ave"
Then every element of F tlwt is
R.
intt~QraL
OUtT
E is [nteQnIl ouer R.
E i.s a Fini. teLy gent'l'oted ,'ing extension of R,
gene
"0 t ecl
then f
is Cl f'irü tel,)
R-modu I e •
PROOF.
We prove the seeond statement first.
Let E = R[al, ••• ,an ].
Then R[al] is a finitely generated R-modl.lle by (1.1), and E is a finitely generated R [al ]-rnodule by induetion on n. genera ted R-rnodule by (II. 4.3) .
assume that. E is finitely generated. t.here
is a
aM (;; M.
faithful
Therefore E is a
If a E F is integral over E, t.hen
finitely generat.ed E-submodule M of F such that
But M is a fini tely generated R-module by (II. 4.3)
integral over R.
f
so a
is
0
Corollary 1.5 shows that the integral elosure of R in f closed in E.
fini tely
To prove t.he fi rst st.at.ement we may
is integrally
A discrete integral dornain is said to be integrally closed
if it is integrally closed in i ts Held of quotients. 1.6 THOOREM.
PROOF. u/v where Xn + a
Ir
R is Cl GCD-;:jomn;n,
th"n R is integr'nlly cICJsPc/.
Each element of the field of quotients of R ean be written as 11
and v are relatively prime elements of R.
x,,-1 + ••• + n-l
(u/u) = O.
alX + 00
Suppose f
(X)
=
is a monie polynomial in R[X J such that vlu n . As u and II are relatively prime,
Then ull f(u/u} = 0 so
141
1. Integral extensions and impotent rings
vlu
so u/v E R.
0
Reca1l that a commutative ring E is impotent if (i) an = 0 for some n E ( ii)
~
imp1ies a
= 0,
and
a 2 = a implies a = 0 or a = 1.
Condition (i), which is equivalent to a 2 = 0 imp1ies a = 0, says that E has no nilpotent elements. Condition (ii) says that E has no idempotent elements other than 0 and 1. Any Heyting field is an impotent ring. We shall show that if E is an impotent extension of a discrete field k, then the elements of E that are algebraic over k form a discrete field. As the complex numbers are a Heyting field, this shows that the algebraic numbers are a discrete field. 1.7~.
Let a und b be elements of an impotent ring such that a + b
1 und ab = O.
Then ei.ther a = 0 and b = 1, or a = 1 and b = O.
PROOF. Multiplying the equation a + b = 1 by b, and making use of the fact that ab = 0, we obtain b2 = b. As the ring is impotent, it follows that b = 0 or b = 1.
0
Let E be an impotent ring, k a subring oF E, and a € E. IF k[X] are strongly ,-elatively prime, and F (a)g(a) = 0, then ei.ther
1.8~.
F,g
€
F(a) or g(a) is a unit, so either g(a)
PROOF.
= 0 or F(a) = O.
As Fand gare strongly relatively prime, there exist sand t
in k[X] such that s(a)f(a) + t(a)g(a) = 1. From (1.7) it follows that s(a)F(a) = 1 or t(a)g(a) = 1. Therefore either F(a) or g(a) is a unit, so the other is O. 0 1.9 'lHOOREM. Let E be an impotent r-ing and k a discrete subfield oF E. If a € E is algebraic over k, then kral is a discrete field. Thus the set oF elements in E algebraic over h [s a dlscrete Fleld. PROOF.
Let,
€ k [a].
We may assume that h [a]
=
E, and we shall show
that , is either a unit, or that , = O. From Corollary 1.4 it follows that , is algebraic over I~. Thus there is a monic polynomial 9 in k [X] so that g(""() = O. Wüte g(X) = Xmh(X) wUh 11(0) ~ O. By Lemma 1.8, either ,m is a unit, and so ""( is a unit, or ,m = O. In the 1atter case ""( = 0, as E is impotent. 0
142
Chapter VI. Fields 1.10 COROLIARY.
Let k !: K
= k(x1""'xn )
be discrete fieLds.
Then the
following are equivalent.
(i) K is algebraic over k,
(ii) xl"" ,xn are algebraic over k, (iii) K is a finitely generated vector space over k. In addition, each of (i), (ii), and (iii) implies
(iv) PROOF.
K = k[x1""'x n ],
Clearly (i) implies (ii).
Suppose
(ii)
holds.
To prove
(iii),
it
suffices
to
show that
k(x1"",xi) is a finitely generated vector space over k(x1"",x i _1) for
each
i
i 1.
As
is
xi
algebraic
over
it
k,
is
integral
over
k(x1"",x i _1); the result then follows fram (1.2).
Theorem 1.2 shows that (i) follows from (iii).
Final1y, if (i) ho1ds,
then k[x1""'x n ] is a field by (1.9), so (iv) holds.
D.
In fact, if condition (iv) holds in (1.10) then so do the others; this result is sometimes called the weak Nullstellensatz. We prove the following somewhat stronger version. 1.11 'l'HOOREM. recognizable units.
R!: 8
Let
be
If 8 = R[x1'"
discrete
.,xn ],
commutative
r-ings
with
then 8 is integr-al. over R, or 8
contains a nonzer-o nonuni t, or R contains a nonzer-o nonuni t •
PROOF.
It suffices to show, for each i, that xi is integral over R, or
8 contains a nonzero nonunit, or R contains a nonzero nonunit. R(x i ) = {fg-
Then R(x i
)
1
:
f,g E R[x i ]
Let
and 9 is a unit in 8).
has recognizable units, so by induction on n we may assume that
8 is integral over R(xi)'
By taking apower of the product of the
denominators appearing in the monic polynomials satisfied by the x j over R(x i ) we construct a polynomial
and ß(xi)x j is monic, or else assume that xi assume that ß power of ß(x i )
ß E R[X] such that
ß(x i ) is a unit in 8,
integral over R[x i ] for each j. We mayassume that ß is its leading coefficient is a nonzero nonunit in R. We may is a unit of 8 and, multiplying by X if necessary, we may is nonconstant. Any element of 8 can be multiplied by a to make i t integral over R[x i ]. In particular 1 - ß(x i )
equals 0, in which ca se xi is integral over R, or 1 - ß(x i ) is a nonzero nonunit of 8, or there exists m such that a(x i ) = ß(x i )m(l - ß(X i ))-l is integral over R [xi]' and thus a root of a monic polynomial f (X) E
143
1. Integral extensions and impotent rings R[xt)[X]. of
F,
Multiplying F(a(xi)) = 0 by (1 - ß(xi))d, where d is the degree
we get
ß(Xt)md = g(x i )(l - ß(x i ))· Since ß(X)md and 1 - ß(X) are strongly relatively prime in R[X], implies that (1 - ß (xi) )h (xi)
this
= 1 for some hER [X], whence xi satisfies
the polynomial (1 - ß(X))h(X) - 1.
Either the leading coefficient of this
polynomial is a nonzero nonunit of R, or xi is integral over R.
0
The more familiar form of the weak Nullstellensatz is as follows. 1.12 ODBOLLARY.
Let k
~
K
=
k[x1' ••• 'x n ] be discrete fields.
rhen K
is algebrate ouer k. The discrete fields K and k have no nonzero nonunits, so K is
PROOF.
algebraic over k by (1.11).
0
If E is a conunutative ring containing a field k,
then E is a vector
space over le k[a],
1f a in E satisfies a polynomial of degree n over k, then being generated by 1,a, ... ,an - 1 , is a finitely generated vector
space over k. 1.13
When is k[a] finite dimensional?
THEOREM.
E, and a E E.
Let E be a commutatiue ring, k a discrete subField oF
If a satisfies an ir-reducible polynomial ouer k, then k[a]
is a finite-dimensional uec tor space ouer- k.
IF E is impotent, and k [a]
is contained in a finite-dimensional k-subspace of E, thell a satisFies an irr-educible polYllomial ouer k. PROOF.
1f a satisfies an irreducible polynomial of degree n over k,
then 1,a, ••• ,an - 1 is a basis for kral over k.
Conversely, suppose E is
impotent and V is an n-dimensional k-subspace of E containing a.
Then
(1.2) says that a satisfies a monic polynomial f of degree n over k.
Thus
kral
is generated by 1,a, ••• ,an - 1
dimensional by (11.5.3).
as a k-subspace,
We shall show that f is irreducible. field, so i f f
so kral
is finite
By induction on n, we may assume that k[a]
= gh, then either
By (1.9) the ring kral is a discrete
g(a)
= 0 or
f(a)
= o.
But a cannot
satisfy a monic polynomial of degree less than n because k [a) n-dimensional.
= V.
= V is
0
Let E be a conunutative ring that is finite dimensional over a discrete subfield k.
Each a E E is integral over k by (1.2).
Because we can
144
Chapter VI. Fie1ds
decide whether or not the elements 1,er, ••• ,erm are linearly independent, we can construct a monic polynomial f E
I~
[X]
of smallest degree such that
I f 9 (er) = 0 for some 9 E I
f (er) = O.
Euclidean algorithm construct a polynomial satisfied by er of smaller degree than f. The polynomial f is called the minimal polynomial of er over 1<. I f E is impotent, then (1.13) shows that f is irreducible. EXERCISES 1. Let R ~ E be commutative rings with Eintegral over R, and let I be an ideal in R. Show that if er EIE, then there exist elements r- t, E Ii. such that an + r a n - 1 + ... + r= 0 Conclude that 1 n'
n
R
IE ~
Fr.
2. Let K be a finite-dimensional extension field of~. Show that an element of K is integral over 71. i f and only i f i ts minimal polynomial over
~
has integer coefficients.
integral closure of 71. in the integral closure of
71.
~(i)
Show that the
is 7l.[i] (the Gaussian integers are
in the Gaussian numbers).
3. In any ring show that if Cl = 0 whenever a 2 whenever an = 0 for some n > 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<-1, ... , results either in showing that x is algebraic over I<, hence T is algebraically dependent, or in finding an element si that is
T
algebraic over Ids1"" ,si_l'x), with x algebraic over l«s1"" ,si)' If then T is algebraically dependent. If si E T, replace si in 8 by x, and we are done by induction on m. 0 si E T,
Let I< !;; K be discrete fields, and let B be a finite subset of K that is algebraically independent over k. The field extension I< !;; I«B) is said to be purely transcendental, and if transcendence basis of K over h.
K
An
is algebraic over k (B), we call B a immediate consequence of (2.3) is the
following. 2.4 COROLLARY.
Any two tr-anscendence bases oF a discrete Field K over-
a discrete field I< have the same car-dinaUty.
0
If B is a transcendence basis for K over 1<, then the cardinality of B is called the transcendence degree of K over k, and is written as tr .degh K. The transcendence degree of an algebraic extension is zero, the empty set being a transcendence basis. A purely constructive consequence of (2.3) is that we can decide algebraic dependence if we have a transcendence basis. 2.5 COROLLARY. a
subField
1<.
If the discr-ete Field K has a tr-anscendence basis over-
then
any
fini te
subset
oF
K
is
ei ther-
algebraically
dependent. or algebraically independent, ouer- k.
PROOF. Let 8 be a transcendence basis for K of ca~dinality n, and T a finite subset of K. By (2.3) either T is algebraically dependent, o~ we can enlarge T to a set T' of cardinality n such that K is algebraic over
147
2. Algebraic independence and transcendence bases
Were T' algebraically dependent, then we could construct a set T n of cardinality n-1 with K a1gebraic over k(T n ) ; but this would contradict (2.3) as 8 is algebraically independent. 0
k(T' ).
2.6 LEMMA.
subFieid k.
Let K be a discr-ete Fieid oF transcendence degree n ouer a IF K is aigebraic ouer k(8)
contains a transcendence basis oF K ouer k.
that
is aigebraicaUy independent
ouer k,
For same subset 8 !;; K,
then 8
IF 8 is a finite subset oF K then 8
can be extended
to a
transcendence basis.
Choose a finite subset B of 8 such that each element of the transcendence basis, and hence K itself, is algebraic over kIB). By (2.5) we can decide whether or not B is algebraically dependent. If B is algebraically independent, then B is the desired transcendence basis. If B algebraically dependent, then there exists bEB such that K is algebraic over k(B\{b}), in which ca se we are done by induction on iB. Now suppose 8 is a finite algebraically independent set. By (2.3) we PROOF.
can enlarge 8 to a finite subset S' of K of cardinality n such that K is a1gebraic over k(S'). By the first part of this theorem, 8' contains a transcendence basis; so 8' is a transcendence basis by (2.4). 0 2.7 'IHEDREM.
Le t R !;; K !;; L be discr-e te
fieids.
If two of
the
thr-ee
extensions k !;; L, K !;; Land R !;; K haue fini te transcendence bases, then so does the third, and tr.degh L = tr-.degK L + tr.degk K.
If BQ is a transcendence basis for K over k, and Bi is a transcendence basis for Lover K, then one easily verifies that BQ U Bi is a transcendence basis for Lover h. If Bo is a transcendence basis for K over R, and Bi is a transcendence basis for Lover R, then by (2.3) we may assume that Bo !;; Bi, whereupon B,\ßo is a transcendence basis for Lover PROOF.
K. Finally, suppose that BQ is a transcendence basis for Lover K, and B, is a transcendence basis for Lover R. We can find a finite subset 8 of K so that each element of B" and hence L, is algebraic over R(S U Bo ). By (2.3) we mayassume that S is algebraically independent over R. If x E K, then by repeated application of (2.2) we find that x is algebraic over R(8), because Bo is algebraically independent over K. 0
148
Chapter VI. Fields Let R be a discrete field and X an indeterminate.
states that every field between
Lüroth's theorem
and !?(X) is of the form
I,
for same
k(z)
We can't hope to prove Lüroth's theorem even for detachable
z E I«X).
subfields of I«X). subfields,
However Lüroth' s theorem holds for finitely generated
and a trivial classical argument then yields the classical
Lüroth's theorem.
First we show that if
E k(X)'\.r,
t
then k(X) i5 finite
dimensional over /, (l ) •
2.8 IDIMA. tE h(X)\j,.
h[X],
be
~~al,
t =
If p(Y)
then
k
Let
tv(Y)
discn'lc
a
WUll u
-
u(Y)
amI v 1S
X an
field,
r'e!ativeliJ [wime in'educible
L1Il
indeterminate,
and
poLiJnomials
polynomiai.
ouer
i.n
h(t)
sa I i sfied by X.
Note that p (Y) t 0, because
PRO:>F.
is algebraic over Iz (t By Gauss's
t
Clearly p(X)
q /{.
with 9 and h in I<[t ][Y].
= gh
divides p(}')
=
that gE k.
Thus p(Y) i5 irreducible over k(t).
is irreducible in So gEI, [Y J.
If K i5 a subfield of IdX), and
over
t
then Lemma 2.8 says that h (X)
E K\j"
As,
1< (t ) •
0
classically,
implies that K is finite-dimensional ovet: 1< (t ), theorem is
cl
the
1.e t
Let K
=
field k(X)
is
finite = k (al)
polynomial I' in
K [Y J •
of f is not in h. let p(}') I«z) • I«z) =
=
= uJX)
iiTXT
dimensional [cx2' • •• ,an J • X
11
zv(Y) - u(}').
and
1«z) ~
OUCT
k, and
somf' z E Ie(X).
over Idol)'
By Lemma 2.8
so k(X)
is
finite
Thus X satisfies an irreducible k,
some coefficient
.7
= K.
relatively prime polynomials in /, [X J, and
II
Lemma 2.8 says that p(Y)
We shall show that deg
K, as
rar'
is transcendental over
We will show that /0' (/) with
l'
=
elf'Cf
r,
is irreducible over
fram which it follows
that
K.
For any polynomial '1 k[X][y]
As
Id/)
=
We may assume that al q h.
Idu1, ... ,on)'
dimensional over K
write z
immediately
k be a d (seTe t e f i e tel, X an Lnde ter'mLnn t e
'Flten Idu1""'''n)
E Ie(X).
PRO:>F.
this
the classical Lüroth' s
diLect consequence of Lemma 2.8 and the following.
2.9 THEOREM.
"l".""n
As 9
and u and v are relatively prime, it follows
tv(}') - u(Y),
finite-dimensional
so I, (X)
As the t-degree of p is
1 i t follows that ei ther !! or Il has t -degree 0, say g.
is
= 0,
is tran5cendental over /,.
l
lemma it suffices ta show that p(Y)
Suppose p
I<[t )[Y].
Therefore
).
such that q/q*
E idX)[Yj,
(ldX);
let '1* denote a primitive polynomial in
in pat:ticulat: let p*
= u(X),,(Y) -
,,(l()1I(Y).
149
2. Algebraie independenee and transcendenee bases Let m '" degX
F*,
the maximum of the X-degrees of the eoeffieients of
r*.
S m.
As
As f is a monie po1ynomial, degx p* = degy p* '" max(deg u, deg f(X) = 0,
and f
is irredueible in K[Y], we ean write p
K[Y].
Then p*(Y) = dF*(Y)g*O')
degx g*
=
irreducible deg p
=
Therefore g*(Y)
O.
over
deg f.
so
k(z),
E
for some d k[y];
in 1<,
=
v)
fg with 9 in
so degx p* = m and
but g*(Y) divides p(Y), which is Thus
g*(Y) E: h.
degy p* = degy
f*,
so
0
We elose this seetion with the construction of an example that will provide a Brouwerian example of a field that is separably factorial but not factorial, and show the necessity of the separability assurnptions in Theorems VII.1.2 and VII.2.3. Let F be n diso"ete fieLd of clWTacter-istlc 2, nnd Let
2.10 THIDREM. K
F(b,s,t), where b, s und t are indetermina!es. Then I<.
I< = F(a,b).
Let a = bs 2 +
[2
and
a!.gebrai.cally closed i.n K.
is
Note that K has transcendence degree 3 over F, and that t 2 E
PROOF.
F(a,b,s) = h(s), so K is algebraic over I«s).
I t follows that a and bare
algebraically independent over F, and that s is transcendental over h. (1.6) and (IV.4.7),
By
the üng F[a,b] is integrally closed in k, hence in
k(s) . Now let 8 E K be algebraic over I<, and let f polynomial with F(8)
=
The polynomial ,.n-I f
(~)
1S monie and has coefficients in F[a,bl. so ,"e is
integral over F[a,b] whence
IV '" (dl)2 E
p/q + (u/u)t
IV = p2/q 2
+
Write m
(U 2 /V 2 )
+
(a
F[a,b,X] be a nonzero
Let r E F[a,b] be the leading coefficient of f.
O.
so
IV E F[a,b].
E
=
bs 2 ),
k(s) is also integral over F[a,b], where p,q,u,v E F[a,b,sj.
Then
as our fields have characteristic 2, so
(* )
If u = 0,
then rS
=
p/q is integral over F[a,bj and in h(s),
F[a,bJ, and hence 8 E k as desired,
of s dividing
U2
q2 .
so 1"8 E
Otherwise, let n be the largest power
The coefficient. of sn on the left hand side of
equation (*) contains only even powers of band is nonzero as it. contains an odd pcwer of a.
Hence w (which is in F[a,bj) and so wq2v2, contains
only even powers of b.
But the coeffieient of sn+2 on the left hand side
of (*) contains an odd power of b.
Thus u fc 0 is impossible.
0
150
Chapter VI. Fields
EXERCISES 1. Let /,
that
F be discrete fields,
~
a
if
and S a finite subset of E.
subfield K of F
is
algebraic over h,
Show
then S
is
algebraically dependent over K if and only if S is algebraically dependent over I,. 2. A
span operation on a
discrete set S is a map s frorn finite
subsets of S to subsets of S such that (i) if A ~ B, then sA (i i) A
c::
~
sB,
sA,
(iii) if B is a finite subset of sA, then sB c:: sA, (iv) if Let h
~
Ix
either
or
xE sA
{x)).
K be discrete fields,
define sA
then
, C s(A LJ {y}),
U E s(A LJ
and for A a finite subset of K,
is algebraic over /, (A)} •
Let V be a
discrete vector space over the discrete field /',
and for A a
~
E K
:
subset of V, define
xc
sA
to be the subspace of V generated by
Show that we get a span operation in ei ther case.
A.
Develop the
theory of this section in this more general setting.
3. Give a Brouwerian example of a discrete field K with a detachable subfield /, such that K
= /, (8)
but K does not have a transcendence
basis over /, .
4. Construct a Brouwerian example of a detachable subfield of that is not of the form
~(X),
(j)(z).
5. Let F be a discrete field of characteristic 2, let hand t be
indeterminates, and let P be a statement. /, =
and
K
{x ( F(h)
{xEk(t
2
+bt)
xE F(/)2)
Let
or Pl,
x E " ( t 4 + ,,2 t 2)
or P}.
Show that K is a detachable subfield of "( t ), properly containing h, and that K = kC,.) for some
7,
if and only if P or not P.
3. SPLITrIlIKi FIELDS AND ALGEBRAIC CLOSURES
Let f be a nonconstant polynomial over
il
discrete held /,; we want to
construct a discrete field K containing I! such that f has a root in K.
If
we are content that K be a discrete r-i nq, then we let K be the quotient ring 1,[Xl/(f): if adenotes the image of X in K, then K
=
h[u], and er is a
151
3. Splitting fields and algebraic closures root of F.
However K is a Fietd only when f is irreducible. Let F be a noneonstant potynomiat ouer a diserete Fietd
3.1 THEOREM.
k.
Let K
= k[X1!(Fl.
and tet adenote the image of X in K.
is a diserete eommutatiue ring eontaining k, and F(a)
=
O.
rhen K
= kral
Moreouer K is
a diserete Field iF und onty iF F is irredueible. PROOF. The division algorithm (II.5.2) allows us to decide whether or not a polynomial in k[X] is divisible by F, so K is discrete. As F is nonconstant, k maps onto an isomorphie copy of itself in K, which we may identify with k. Clearly K = kral and f(a) = o. Suppose K is a field. If F = gh, then gh = 0 in K so, as K is a field, either 9 = 0 in K or h = 0 in K. Thus either Fig or flh, whence fis irreducible. Conversely, if F is irreducible, and 9 is an arbitrary element of K, then by (II.5.7) there exist sand t in k[X] such that sf + tg divides both fand g. As F is irreducible, we may assume that sF + tg is either F or 1. If sF + tg = F, then Flg, so 9 = 0 in K; if sF + tg = I, then t is the inverse of 9 in K. 0
If f is not irreducible, then it is more difficult to construct an extension field in which F has a root: the c1assical technique is to work wi th an i rreducible factor of f, but we may be unable to find one (see Example IV.2.2). In fact, we cannot always construct such an extension field (Exercise 1). If k is countable, however, then we can construct such a field by constructing a detachable maximal ideal M in k [X I, containing f, without constructing a generator for M. 3.2 LEMMA.
Let R be a eountable eommutatiue ring in mhieh Finitely
generated ideals are detaehable. ideal oF R, then
r
If
I
is a
finitely
generated proper
is eontained in a detaehable maximal ideat.
PROOF. Let r"l,r"2' •• ' be an enumeration of R. We shall construct an ascending sequence of finitely generated ideals I j starting with I l = r. If I j has been constructed, construct 1 j+l as follows: if 1 E r j + Rr j' then set I j +1 = Ij' otherwise set Ij+1 = Ij + Rr j . Let M be the union of the ideals I j . As r' j E M if and only if r"j E I j+1' the ideal M is detachab1e. If r"j ( M, then r"j ( I j+1 so 1 E: I j + Rr' j t;; M + Rr"j' so M is maximal. 0
152
Chapter VI. Fields 3.3 'l'HEDREI'I.
Let k be a countabie discrete fieid and f a nonconstant
poLynomiai in k[X]. k, and a E
E
PROOF.
Then there is a countabie discrete Fieid E containing
= O.
such that F(a)
Fini tely generated ideals of k [X]
are prineipal,
by the
Euelidean algorithm, henee detaehable by the division algorithm.
Thus
(3.2) applies and we ean eonstruet a detaehable maximal ideal M of k[X] eontaining F.
Let E
= h
[X]IM, and a the image of X in E.
Let f be a monie polynomial over a field I<.
0
An extension field K of h
is called a splitting field for F over h if FIX) = (X - al)···(X - an)
and K
=
h [al' ... ,an] .
Note that K is eountable i f h is eountable.
3.4'l'HEDRE1'1 Let f(X) be a nonzero poiynomiai ouer a countabie discr-ete heid h.
PROOF.
Then we can construct a discr-ete spiitting fieid for f ouer IL
Repeated applieation of Theorem 3.3.
0
Spli tting field over eountable fields may be used when working wi th arbitrary diserete fields
h
as follows.
If f
is a polynomial in
consider the subfield h o of R generated by the eoefficients of f. is countable, f has a splitting field over Ro •
h[X],
As Ro
We use the roots of f in
this splitting field to obtain information about F as a polynomial in h[X].
Classieally the splitting field for a polynomial over a field h is unique in the sense that any two splitting fields are isomorphie over k. Construetively we may be unable to eonstruet the isomorphism (see Exereise 2).
There are problems even if f is irreducible beeause after we adjoin a
root a of f to R we may not be able to find irredueible faetors of f in k[a] [X] (see Exercise 3).
A field a is algebraieally elosed if eaeh monie polynomial of degree at least I in a[X] has a root in
o.
This implies that eaeh monie polynomial
in 0 [X] factors into linear faetors in
a.
If an algebraically closed
field a is algebraic over a subfield h, then 0 is ealled an algebraie elosure of k.
Constructively we may not be able to embed a diserete field
in an algebraically closed field;
or we may construet two algebraic
closures without being able to construct an isomorphism between them.
For
countable discrete fields, however, we can construct an algebraie closure.
3. Splitting fields and algebraic closures 3.5 THEDREJ!Il.
Let I< be a countabte discr-ete fieLd.
discrete aLgebraic closure of
153 Then ther-e exists a
I~.
PROOF. Let F1,f2"" be an enumeration of the nonconstant polynomials in k [X]. Let k 1 be a discrete splitting field for f lover k. Having constructed k 1 ~ k 2 ~ ••• ~ k j we let k j+1 be a discrete splitting field of f j +1 over k j • Finally let n = U k j , that is, the direct limit of the fields k j . We must show that rl is algebraically closed. Let F E rl [X] • By Theorem 3.3, there is a discrete field E containing rl, and a root a of
f
in E.
Then a is algebraic over k, by Corollary 1.5, so there is a
with f j (a) = O. As f j is a product of linear polynomial f j in k [X] factors in k j +1 it follows that a is in k j +1 !;; rl. 0 EXERCISES 1. Derive the world's simplest axiom of choice from the statement
that if h is a discrete field, and f is a nonconstant polynomial over k, then there is an extension field of k in which f has a root. Hint: For each two-element set S, let k S be the ring gene ra ted by S over ~ subject to the relations S2 = -1 for s in S and LsES s = O. For T a set of two element sets with at most one element, let k = USETkS and f (X) = X2 + 1. 2. Let a be a binary sequence with a most one 1, and let k = Unl!.l(ia n ).
Enumerate k[X) in such a way that i f a2n = 1, then the
polynomial X-i precedes X + i, while if a2n+1 = 1, then X + i precedes X-i. Use this enumeration to effect the construction of a splitting field E for X2 + lover k via (3.2) and (3.3). Show that E and I!.l(i) constitute a Brouwerian example of splitting fields for X2 + lover k that are not isomorphie over k. 3. Construct a Brouwerian example of a field k, that lies between ~ and C!l[iF3], such that the polynomial X3 - 2 does not have a unique splitting field over h, even though it is irreducible over k.
4. Nonisomorphie splitting fields. Let a be a binary sequence with at most one 1, and e a sequence in {-1,1}. Let Pe be the ideal in the polynomial ring ~(i)[X) generated by the elements (iX - nen)an .
(i) Show that Pe is a detachable prime ideal.
154
Chapter VI. Fields (ii) Show that P = Pe fl (Q[X]
is a detachable prime ideal of
(Q[X], and does not depend on e. (iii) Let k be the field of quotients of (Q[X]/P, and Ke the field of quotients of (Q(i)[X]/Pe. in Ke • (Q(x p
Let xe denote the image of X
Show that Ke is a splitting field for y2 + 1 over
=k.
)
(iv) Let e n = 1 and f n
(_1)11 for all n.
=
is an isomorphism.
Show that i f
for each even n, and if
Suppose
:
Ke
-.
,t iXf' then an
,t -ixf' then an
Kf
0
=
= 0 for each
odd n. 4. SEPARABILITY AND DIlIGCW'.LIZABILITY
If f = aO + alX + ••• + a n-lXn - 1 + 0 n XTl is a polynomial over a commutative ring R, then the formal derivative of f is defined to be
f' = 01 + 2a 2X + ••• + (n-l )on_lXn-2 + nanx n - l . A polynomial f E k [X]
is separable over k if i t is strongly relatively
prime to its formal derivative, that is, if the ideal in k[X] generated by fand f'
rf k
contains 1.
is a discrete field,
then the Euclidean
algorithm allows us to decide whether or not f is separable. 4.1 'IHEDREM.
Let I, be
(i) (f + g)' (ii) (Fg)'
=
= f'g
commutotive ring
0
and f ,g E k[X].
Then
f' + g'. + fg'.
(iii) fg is separabLe iF and onLy ir fand gare separ-abLe and str-ongly relativeLy pr-ime.
To show part (ii), let f
Part (i) is clear.
PROOF.
= aXm and g = bX n .
Then (fg)'
=
(m+n )abX m+n - l
and the general f'g
+ fg'.
(f,g).
=
rnabXm+n - l + nabXm-tn - l = f' g + fg' ,
result follows
from (i).
For
( iii ) compute
(f g
)'
=
Clearly (fg, f'g + fg') is contained in (f,f') and (g,g') and
Conversely,
(Fg,
F'g
+ fg') 2
(f,f'g)(g,fg') ~ (f,f' )(g,g' )(f,g)2.
(f,
f'g
+ fg' )(g, F'g + fg')
=
0
If a polynomial f with coefficients in a discrete field is separable, then we can find polynomials sand t, wi th coefficients in the field gene ra ted by the coefficients of f, such that sf + t f'
=
1; indeed the
4. Separability and diagonalizability
155
Euclidean algorithm does just that. Thus the notion of separability of polynomials over a field is absolute in the sense that it doesn't depend on the particular field in which we are considering the coefficients of the polynomial to He. We can characterize separability of polynomials over fie1ds in terms of the absence of multiple roots. 4.2
'lHEX)REM.
Let F be a polynomial over' a discr'ete Field, and let k be
the Field generated by the coeFFicients oF F. has no multipLe roots in any discrete
IF F is separable then F
field containing k.
IF
F is not
separabLe, then F has a multiple root in some extension field oF k.
PROOF. We may assume that F has degree at least 1. Suppose r is a reot of F in a discrete field containing k, and F (X) = (X - I')g (X) • Then F' (X) = g(X) + (X - r)g' (X), so f' (r) = g(r). If f is separable, then s(X)F(X) + t(X)F' (X) = 1 so t(r')F' (r') = 1 whence f' (r), and hence g(r), is nonzero. Conversely, let K be a splitting field for f over k, and write f(X) = aITi(X - r i ) where a t O. If f is not separable, then fand f' have a nontri vi al commen factor, hence a common root r j" But f' (r' j) = aITitj(l'j - r i
),
so
r'j
= I'i for some
i
t
j.
0
Let k be a discrete subfield of a commutative ring E. An element of E is separable over I. i f it satisfies a separable polynomial in k[X]; the ring E is a separable extension of k i f each element of E is separable over k, and k is separably closed in E if each element in E that is separable over k is in k. By the field of definition of a finite set of matrices over a field, we mean the (countable) field gene ra ted thei r entries . If matrices are linearly dependent over a discrete field h, then they are already linearly dependent over their field of definition: indeed if we treat the matrices as if they were row vectors, then (111.6.8) and (III.6.9) show that if the rows are independent over their field of definition, then they are independent over k; and rows are either Hnearly dependent or linearly independent ( II. 6.5) • In particular, the coefficients of the minimal polynomial of a matrix lie in the field of definition of that matrix, so the minimal polynomial of a matrix does not depend on the particular field from which the entries of the matrix are considered to be taken. Also, if a matrix A can be wri tten as a polynomial in a matrix B, then the polynomial may be chosen with coefficients in the field of definition of
156
Chapter VI. Fields
A and
The
B.
fundamental
relationship
between
separabili ty
and
diagonalizabilty is the following. 4.3 LEMMA.
A matrix ouer a discrete field is separable if and only if
it is diagonalizable ouer some discrete field.
If a matrix A is diagonalizable over a
PROOF.
field F,
minimal polynomial of A is separable by (V.4.3) and (4.2).
then the
Conversely, if
A is separable, let F be a splitting field for the minimal polynomial of A over the field of definition of A. (V.4.3) and (4.2). The
next
Then A is diagonalizable over F by
0
lemma
allows
us
to
translate
questions
about
algebraic
elements over discrete fields to questions about matrices. 4.4 LEMMA.
Let E be a commutatiue ring containing a discrete field k,
let fand 9 be polynomials in k[XI, and let a and ß be elements of E such
that F (a)
= O.
g(ß)
size ouer k
such
Then there are square matrices
that
F(A)
=
g(B)
= 0,
and a
A and B oF
r'ing map oF
the same
k[A,Bl
onto
k[a,ßl that is the identity on h and takes A to a and B to ß.
PROOF. k
and
The ring k[X,YI/(f(X),g(Y))
maps
onto k [a,ß I .
naturally
=
h[x,yl is finite-dimensional over
Let Tx
and T y
be
the
linear
transformations on k[x,yl given by multiplication by x and y, and let A and B be the matrices of T x and T y with respect to some basis of k [x,y I. Then the natural isomorphisms k[A,BI ;: IdTx,Tyl
= k[x,yl,
the map from k [x,y I to Ida,ß I, yield the desired results. 4.5
0
IF E is a commutatiue ring containing a discrete Field k,
'l'IIEX)Rm.
then the elements oF
PROOF.
together with
E that are separable ouer k form a ring.
It suffices to show that if a and ß are separable over k, then
every element of k [a,ß I is separable over k.
Let Fand 9 be separable
polynomials satisfied by a and ß respectively, and let A and B be as in (4.4).
It suffices to show that any element of k[A,BI
Given a matrix
e
field of definition of A, B and and B
factor
is separable.
in k[A,BI, we can construct a field F, containing the
into
linear
e,
over which the minimal polynomials of A
factors.
By
(V. 4.3)
simultaneously diagonalize A and B, and hence
by (V.4.3) and (4.2).
0
e,
and
(V. 4.4)
over F, so
e
we
can
is separable
157
4. Separability and diagonalizability We can characterize when an extension E of a discrete field k
is
separable in terms of the lack of nil potent elements when we (possibly) extend the field k
of coefficients of the commutative ring E.
This
characterization can serve as a definition of separability in the not necessarily algebraic case. for
countable k,
We will establish the characterization only
leaving the fornrulation of the general case as an
exercise. 4.6 'l'HOOREM.
Let k be a discret.e subfidd oF a corrunutaHue ring E, and
consider the FoUowing two condi Hons.
(i) E is separabte ouer k, (ii) For each discrete extension Fietd K oF k,
the ring
K 0 k E has no nilpotent elements.
Then (i) impties (ii); conuersell) , if k is countabte, and E is atgebraic ouer k, then (ii) impLies (i).
The ring K 0 k E is generated over K by elements that are
PROOF.
separable over k; hence each element 8 of K 0 k E is separable over K by (4.5).
If 8n
=
0, then 8 satisfies the polynomial Xn in addition to a
separable polynomial F(X) over K.
The GCD of F(X) and XIl divides XTl and
= 0.
is separable, hence is X, whereupon 8 Conversely, suppose 8
€ E
satisfies a nonzero polynomial F over k.
proceed by induction on the degree
of
n
F.
If
We
F is separable we are done.
Otherwise we can construct, by (4.2), a discrete extension field K of k such that F has a nrultiple root in K. g(r)
= 0,
so Flg 2 .
Then g(8)
means that the elements 1,8, ...
,8 n -
Write F(X) and g(8)2
= 0,
=
(X - r)g(X)
so g(8)
=
)\'here
That 1 of K ®k E are linearly dependent over
€ K ®/~ E
0.
K, hence over /{ (Exercise III.5.6), so we can find a polynomial in k[X],
satisfied by 8, of degree less than n.
0
EXERCISES
1. Let k k[X]. 2. Let k
E be commutative rings, and f a separable polynomial in Show that any square factor of f in E[X] is a unit in E. be a commutative ring, € hand fE /<[X]. Show that X - a ~
(l
is strongly relatively prime to f if and only if f(a) is a unit of k.
158
Chapter VI. Fie1ds 3. Let 01' ... ,an be elements of a conunutative ring 1<.
Show that
(X -(1)(X -a2)···(X -an) is separable i f and only i f a i -a j is a unit of k whenever i
je j .
4. Let a, band c be elements of a conunutati ve ring 1<. aX + b
is
separable
relatively prime. aX
2
if
and
Show that
only
if a
if b 2
-
Show that
and bare 4ae
is
a
strongly
uni t,
then
+ bX + c is separable.
5. Let E be a discrete conunutative ring that is algebraic over a subfield F.
Show that E is separable over F if and only if for
each finite subset S of h, and finite subset T of E, there is a countable subfield 1<'
of 1<,
containing S,
countable extension field K of h'
so that for
the ring K
~,
each
k' [T] has no
nilpotent elements.
5. PRIMITIVE ELEMENTS Let E be a conunutative ring containing a discrete field 1<. 9 of E is a primitive element of E over I< i f E if E
is
finitely
generated
and
separable,
An element
= h[9]. We shall show that then we
can
construct
a
primitive element provided either h is big enough, or E is a discrete field. We first look at the situation where h is big enough. show that h[A,B] 'l'HEXlREM 5.1.
The key is to
= I<[C] where A and Bare conunuting separable matrices. IF A arui Bare eommuting n-by-n matriees over a discr-ete
FieLd h oF ear-dinaU ty gr-eater-
than r1(r1-1 )/2, arui B is seporabte, then
there exists an etement c in h such
that
A und B ean be wri tten as
potynomiats in A + eB wi th eoeFfieient s in h.
PROOF.
We first treat the case where A is also separable.
By (4.5) we
may assume that A and B are diagonal with diagonal elements 01'·'· ,on and b 1 , ... ,bn .
Choose e distinct from (Cl j - 0i )/(b i - b j) for each pair i and
such that b i je b j. Then Cl i + eb j t- ° j + cb j whenever 0i t- ° j or b i tb j . It follows from (V.3.3) that A and B can be written as polynomials in j
A + cB.
For the general case, we may assume that B has the block-diagonal form
159
5. Primitive elements
"l
Bn
"i
where the block Bi is times an identity matrix, and the are distinct. As AB = BA, the matrix A has the same block structure as B, but the blocks At are arbitrary. Ne can put each block Ai in Jordan canonical form without affecting B. Choose CER, as in the special case, such that Band the diagonal of A can be written as polynomials in the diagonal of
A + cB. Note that A + cB is in Jordan canonical form, so its diagonal can be written as a polynomial in A + cB. As A + cB agrees with A off the diagonal, we are done. 0 5.2 COROLLARY.
Let E be a commutatiue ring containing a discrete fleld
Iz, and let a and f3 be elements of E such that a is algebraic and f3 is
Ir
separabte ouer k. k[o,13] =
PROOF.
Iz is big enough,
then
there exists
El such timt
1<[9]. Let fand 9 be polynomials over 1<, with 9 separable, such that
f(o) = g(fJ) =
O.
Let A and B be matrices as in (4.4).
By (5.1), i f I< is
big enough, then there exists c in I< such that A and B can be written as polynomials in A + cB. Set El = er + c~i. 0 If
the Held I, is tao small, then (5.2) need not. hold (Exercise 2)
However if E is a discrete field, then (5.2) holds for any k. Classically the proof divides into two cases depending on whet.her k is finite or infinit.e.
Ne have to be a Ettle more careful because we may not. be able
to determine which of these cases we are in. primitive element when E is a finite field. 5.3 I.EMM!\-.
Lei k be
Cl
discrete {[dd ond G
muLtipUcotive gr'oup of nonzE'J'O elemenis of k,
PROOF.
Let x and
y
The next lemma provides a
/"i ni LI' subgr'oup of the
0
Then Gis cycLic.
be elements of G of orders
Ne shall construct an element of C of order q
=
m
and
Tl
LCM(m,n).
respecti vely. Nrite q
= ab
where (a,b) = 1, and a In, and b Im. Ne shall show that xOyb has order Cl. Clearly (xoyb)q = 1; suppose (xo'jb)i 1. Then xCIi = lj-hi, so (xoi)o = 1 whence mlo 2 i, so bli.. Similarly, Clli, whence ql;. Thus if 9 is an element of G of maximal order N, then ~ = 1 for every x in G. As the polynomial XN - 1 has at most N roots (II.4.5), there are
160
Chapter VI. Fields
at most N elements of G, so each must be apower of <J.
0
The trick for the general case is to pass to a subfield that is either finite or contains a lot of elements. 5 . 4 'l'HEXJREM. positiue
k !Je n
Let
Th"n
[nteg"'".
finitel,!
r,
eUher"
gen'Tntcd fini (e,
is
discre(c
cw I<
emd
field
eontains morT
N n
tlenn
N
distinet elements.
PROOF.
Ei ther O,1,2, ••• ,N are distinct elements of k, or " has Uni te
characteristic at most N, so we may assume the latter. (finite)
prime
field
1<0 (nI"" ,a,,).
}, =
k,
~ /'0
[Xl"" ,X))] be the finite set of polynomials
Let S
and
let
n1'''''<\1
Let !co be the
of
generate
so
I"
such that the degree of each variable Xi at most N; and let S be the image of S in I, upcn taking Xi to a i #S
~
Ei ther #8
.
~
N or #8 > N, so we may assume
We will show that r;, is finite.
N.
Let p (
"0 [X l' ... ,Xn ].
We shall construct a po1ynomial q ( S such that
p(a1"" ,an) = q(al, .. ·,a n )· power of Xt in p by X t . elements
of
has
Ei
If n t f
N+l
multiplicative order of integer,
terms, G
then a~ = a{
xi,
occurence of X~ in p by
,n'i
0, then, as the sequence 1 ,n t ' ••• two of them must be equal, so
is at most N.
t
for
= 0, we can replace each positive
If a t
some unique J < N,
Therefore i f m > 0 is an so we
giving us an element q of 8.
can
replace each
As each element
of I, is a quotient of such polynomials q, the Held k is finite. 5.5 COROLIARY. n!gebnJ.ic
OUfT
Let k
~
E be disenne fields.
k, nn11 Pis sepaTnbLe aue'" k.
of the
0
Suppase a und ß E F are Then
the'T is 0 ( E such
that k[e] = k[n,f3].
PROOF.
We may assume that h is generated by the coefficients of the
separable pclynomials satisfied by a and (L enough so that k[cr,ß]
=
k[e]
by (5.2), or /,
finite, and h[cr,ß] = k[O] by (5.3).
By (5.4) ei ther k is big is finite,
so k[cr,ßl
is
0
I t suffices in (5.5) for 13 to be separable over k [a]: see (6.7).
EXERCISES
1. Show that the 3-by-3 diagonal matrices over a two-element Held I, form a finite separable extension of h primitive element.
that does not admi t
a
161
5. Primitive elements
2. Let E be a commutative ring containing a discrete field k., Let a and ß be elements of E such that a is separable over k, and ß satisfies a separable polynomial over k [a) • Show that ß is separable over k. 3. Construct a Brouwerian example of a field that is neither finite nor infinite. 6. SEPARABILITY
AN[)
CHARACTERISTIC p
extreme ca se of a polynomial that is not separable is a nonunit f such that f' = 0. This is possible even when k is a discrete field, i f k has finite characteristic p. An
6.1 THEOREM. and
q apower
homomorphism,
Let p be a prime, R a commutative ring such that pR of
whose
f E R[X), then f'
p.
Then
image
the
map {x q
Rq
taking : x E R}
0 if and onLy if FIX)
=
x
in is a
xq
is
subring
of
R
to
a R.
=
0,
ring If
fO(XP) for same fO E R[X).
(r]
PROOF. If 0 < < p, then the binomial coefficient is divisible by p, so (a + b)P = a P + bP . By induction, the map taking x to x g is a ring hornamorphism; the image of a ring homomorphism is a subring. If n is not divisible by p, then n is strongly relatively prime to p, so n is invertible in R. Thus if f' = 0, then the coefficient of Xn in F,
for n not divisible by p, is 0; whence F(X) = Fo(XP) for some Fa E R[X). Conversely, the derivative of Fa(XP) is clearly O. 0 The following is a criterion for the separability of an element over a discrete field of characteristic p. 6.2 THEOREM. positive
power
containing k.
Let k be a discrete fieLd oF characteristic p, Let g be a oF
p,
and
Let
a
be
an
eLement
of
a
commutative
ring
Then a is separable over k if and onLy iF a E klag).
PROOF. If a E klag), then writing a as a linear combination of powers of a g gives a polynomial f in k[X) such that F' = 1 and F(a) = 0, so a is separable over k. Conversely, suppose a satisfies a separable polynomial F in h[X). It suffices to show that the image ß of X in k[X)/(f(X)) can be written as a polynomial in ß g . Let Tß be the linear transformation of
induced by multiplication by ß, and let B be a matrix for Tß' I t suffices to show that B can be written as a polynomial in Bq. As B is
k [ß)
162
Chapter VI. Fields
separable, we may assume that B is diagonal. By (6.1) the map taking x in h to x q is one-to-one, so diagonal elements of Bq are equal if and only if the corresponding diagonal elements of Bare equal. be written as a polynomial in Bq.
Thus by (v.3.3) B can
0
The next theorem is used in situations where we would like to factor polynomials into irreducibles but can't. 6.3 THEX>REM.
Let /, be a diso"ete eonunutative r"ing with reeognizabte
units, and 8 a finite set oF monie potynomiats in heX].
Then either h has
a nonzer'o nonunit or we ean eonstnlet a Finite set T oF monie potynomiats in h [X] sueh that (i) Eaeh element oF T is oF the for"m F(Xq) where f q =
is separ'abte, and
1 or" q is apower of a pr"ime that is zer"o in h.
(ii) Distinet eLements oF T U/"e strongty r"eLativeL!J prime. (iii) Ever!J potynomial in 8 is a prodllet of pot!Jnomiats in T.
PROOF.
We will transform the set 8 into the desired set T, or find a
nonzero nonuit of
along the way, proceeding by induction on the sum of
h
the squares of the degrees of the polynomials in 8.
Given distinct
elements S"S2 E 8, the Euclidean algorithm constructs either a nonzero nonunit of h, or a monic polynomial h that generates the ideal (S"S2)' If h i' 1, then we can replace s, and S2 by h, S,/h, and s2;11, decreasing the sum of squares of degrees. So we may assume that the polynomials in 8 are pairwise strongly relatively prime. If 9 is monic of degree n > 0, and g' has a nonzero nonunit or some prime can write 9 = f(XP). we may assume that s
p
=
0, then n = 0 in 1<, so either /,
is zero in h.
In the latter case we
Repeating this argument we see that i f s E 8, then = F(xq) where q = 1 or q is a power of a prime that
is zero in h, and F' i' O. We may assume that the Euclidean algorithm constructs a monic polynomial h that gene rates the ideal (F ,F' ). If h i' 1,
then we can replace
the squares of degrees. 6.4 COROLIARY.
over' 1<.
s
by h (X q ) and S;11 (X q ), decreasing the sum of
0
Let I< !: E be diser"ete Fields, and a in E be integr"al
Then there exists q, ei ther' eqllat
to 1 or
Cl
power oF the Fini te
eharaeter"istie oF 1<, sueh that a q is separable over" h.
PROOF.
Let 9 be a nonzero polynomial in h[X] such that
g(a)
O.
Then
163
6. Separability and characteristic p
by (6.3) we can write 9 as a product of relatively prime separable polynomials of the form f (xq)n. rf f (a q ) -,!. 0 for each such polynomial, then each
f(a q )
6.5 LEMMA.
is a zero-divisor of E.
0
Let p be a prime und R a eommutative r-ing sueh that pR
=
O.
Let q be a positive power- of p, and let a E R. If xq - a = F(X)g(X), where f and gare strongly relativeiy prime manie polynomials over R, then either f or 9 is 1.
PROOF. As f'g + fg' = 0, and fand gare strongly relatively prime, it follows that there exist polynomials f, and g, such that FIX) = f,(XP) and g(X) = g,(XP). If q = p, then clearly either f , or g, is 1. Otherwise
= f , (X)g,(X)
xq/p- a
and we are done by induction on q, for if s(X)f,(XP) + t(X)g,(XP)
=
1,
then letting S,(XP) and t,(XP) be the sums of the monomials in sand t of degrees divisible by p, we have s,(XP)f,(XP) + t,(XP)g,(XP) so
S,
whence
F,
E
(X)F , (X) + t, (X)g, (Xl = 1,
and g, are strongly relatively prime.
6.6 THEOREM. a
=1 0
Let k be a diserete field oF finite eharaetertstte P, let IF xq - a ts redueible in
k, und let q be a positive power- of p.
k [X). then a E k q .
PROOF.
By Lenuna 6.5 the polynomial xq - a does not admi t strongly
relatively prime factors in k[X).
Thus, by (6.3) we can write xq - a as
for some monic h in k[X) and m > 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<.
and k[a,tlr)j = k[tl] far same Ei in F by (5.5).
over k, then 1<[8 P ] = h[oP.ßP]
= h[n,ßI']
= l<[o.ßl
Thus ß is separable over Iz by (4.5).
over k. Let I<
is separable
If u
so
= 1<[8],
e
is separable
0
E be discrete fields, and let u E: E be a1gebraic aver I,.
~
Then,
by Corollary 6.4, there exists q, ei ther equal to 1 or apower of the fini te characteristic of k, such that ar) is separable over /,. then
u
is purely inseparab1e over k.
If n q
k,
E:
extension E of /, is a purely
An
inseparab1e extension if each element of E is purely inseparable over I,. The next theorem shows that any finitely generated algebraic extension of a discrete field is the composi bon of a finitely generated separable extension and a plu:ely inseparable extension. 6.8 1'tIEOREM. vector-
space
cantaining h,
Let E be
over SliCh
a
discrete
0
subfietd
thot K
k.
field Then
thot t11ETe
iso
fini tdy gener-oted
isa
subfield
or
K
E
f(nltely genenlted nmi separable auer' k. nnd
(s
E is pur-ely ins,,!xu-able aver K.
that
For eaeh 0i there exists q(i )
Let 01'''' ,an generate E over h.
PROOF.
so
a i q (i )
is
separable
over
1, [u1 q(1) ' ... ,l'n q(n)].lS
separa'bl e
inseparable
follows
over K,
inseparable over K.
it
over at
By
/, . /, .
(4.5)
As
each
the 0
i
that k[a1'''' ,anl
anee
field
K
=
is
purely
is
purely
0 EXERCISES
1. Let I, be a diserete held and f ( h [X 1 a nonconstant polynomial.
g21r for 9 (Xl') If for
Show that f is not separable if and only if either nonconstant
gE:I< [X
l,
or
chor /,
=
p
and
some some
noneonstant 9 E h[X].
7. PERFECT FIELDS
If k is a discrete field of charaeteristie p, then /,11 = {a P : (] a subfield of h isomorphie to k. nonzero polynomial in h [X]
E /<)
is
A discrete held h is perfeet if each
is a product of separable polynornials.
The
165
7. Perfect fields
following adaptation of the classical characterization of perfeet fields does not require that the characteristic of the field be known. 7.1 THIOCIREI'l..
PROOF.
is perFect if and ontu if
A discrete FieLd k
prime p either Pt- 0 in h, or k P
For each
= k.
Assume that h is perfeet.
If p is a prime such that p = 0 in
h, and a E k, then xP - a is a product of separable polynomials, and so is
reducible. Therefore a E hP by Theorem 6.6. To prove the converse, let f E k [X) • By (6.3) we may assume that f (X) = 9 (xq) where 9 is separable, and ei ther q = 1, or k has finite characteristic p and q = pe. If q = 1, then F is separable. If q = pe, then the map x .... x q is an isomorphism from k [X) to (k [X))q = k q [xq) = Thus there is h in h[X) so that g(xq) = h(X)q. As n q n (mod p), the isomorphism preserves formal derivatives, so g' (xq) = h' (X)q. As 9 is separable, so is h, so f is a product of separable polynomials. 0
=
k[Xq).
If h is a discrete field of characteristic 0, then h is perfeet, as p t- 0 in k for each prime p.
If k is a finite field of characteristic p, then the map x .... x P is an i somorphi sm, so k is perfect; but k(X) is not perfect as X (k(XP). If k is a prime field, we don't need to know its characteristic to establish that it is perfeet.
Ir
7.2 COROLIARY. PROOF.
Let
p
k is a discrete prime FieLd,
be a prime.
elements so k = k P • 7.3 DEFINITICN.
If P
= 0
then k is perfect.
in k, then k is the field wi th P
0 Let k be a subFieLd of a discrete Fietd K.
Then
K is
a perfect closure of k iF (il K is perfect
(ii) IF a E
K,
then either a E
and there exists q
= pe
k, or
k has
finite characteristic p,
such that a q E k.
In classical treatments the perfeet closure is often constructed within an algebraic closure. Although we may be unable to construct an algebraic closure, we can always construct a perfect closure as a direct limit. If k, .... k 2 .... h 3 .... • • • is a sequence of rings and ring homomorphisms, then the direct limit koo is defined as follows. The elements of hoo are the elements of the disjoint union U "i; elements a E k i and b E k j are
166
Chapter VI. Fields
equal if there exists m such that a and b go to the same element in Rm • It is easily checked that Roo is a ring.
If the Ri are discrete fields,
then Rm is a discrete field and each Ri may be identified with a subfield of Roo ' If we
are
given
the
finite
characteristic
of a
construction in Exercise 1 gives a perfect closure. what
the
characteristic
of
the
field
field,
then we must
is,
then
the
If we do not know refine
that
construction. 7.4 THEDREM.
Euer!} disc!"ete Fidd has aperFeet dasure.
Let {Pn : 1
PROOF.
~
n} be the increasing enumeration of the primes.
For each n. let Rn be R, and define 'Pn : Rn 'Pn(x)
xP if p E {Pj : j ={ x
~
-+
Rn +1 by
n} and P
= 0 in R
otherwise.
Let K be the direct limit, and denote the image of x E Rn in K by x n ' may identify the image of R1 in K with IL
suppose a E K and P is a prime equal to 0 in K. x
R so that a = x n . Then 'Pn(x) = x p . so E
There exists nEIN and
We may assume that n large enough so that Pn 2 p.
a = xn
Thus K is perfect.
We
To show that K is perfect,
= 'Pn (x )n+l
=
(x p )n+1
= x~+l'
To verify condition (ii) of (7.3), let a E K and
choose nEIN and x E k so that a = x n ' rf Pl"" 'Pn are all nonzero in k. then a = x n = xl ER. Otherwise there exists j ~ n so that Pj = 0 in R. Then 'l'n_10"'0'l'l(x)
= x q , where q = pn-j.
oq = x~ = ('P n _1 0
"' 0
Thus
'P1(x1))n
=
xl E R.
0
The perfeet closure is essentially unique. 7.5 THEX>REM.
If K ond L ar"e pe!"fect dasures af the dise!"ete field I<,
then there is a unique isamarphism aF K and Laue!" k.
Define a function F f rom K to L as follows.
PROOF.
either a
E
k, in which ca se we set f
(0) =
0,
or there exists q
p is the finite characteristic of ", such that (lq
E
E K,
pe, where
As the map
is an isomorphism of L, the element ß is unique and does not depend
on the choice of q. L.
=
0
k.
In the latter case there is f3 in L such that f3q = oq. x -+ x q
Given
Setting
F(a) =
ß we obtain a homomorphism from
K
to
Similarly, interchanging the roles of K and L, we can construct a
7.
Perfect fields
167
homomorphism from L to K. and so are isomorphisms.
It is easy to see that these are inverse maps 0
Fieid k
A discrete algebraic
eiement
oF
an
arbitrary
is
perfect
discrete
if
fieid
and
oniy if each
extension
oF
k
is
separabie ouer k.
PROOF.
If k is perfect, then each polynomial in k[X] is a product of
separable polynomials, so each algebraic element of a discrete extension field of k is separable. Conversely let p be a prime such that p = 0 in k. Gi yen a € k, let K ;;! k be a perfect field and choose b in K with P b = a. As b is separable over k, i t follows, by Theorem 6.2, that b € k[b P ) = k. By Theorem 7.1 we conclude that k is perfect. 0 7.7
Let E be a discrete fteld containing the perfect fieid k.
THI!X)REM.
If a in E is aigebraic ouer k, then k[a) is perfeet . PROOF. Let P be a prime such that p = 0 in k. Then k[a]P = kP[a P ). As k is perfect, a is separable over k, and kP[a P ) = k[a P ) = k[a) by Theorem 6.2. Thus k[a) is perfect. 0
EXERCISES
1. Let k be a discrete field of finite characteristic p. Show that the direct limit of the sequence k .... k .... k ....... , where each map k .... k takes a to a P , is a perfect closure of k. 2. Construct a Brouwerian example of a discrete field k of finite
characteristic p such that neither k P = k nor k P f. k. 8. GALOIS 'lHEXlRY
Let k ~ K and k ~ E be commutative rings. A ring homomorphism a:K .... E is a k-homomorphism if a is the identity on k. We will often denote the image of an element x under the homomorphism a by x a , and the image of K under a by [(7. We extend a to a IdX]-homomorphism of K[X) into E[X) by setting Xa = X. The fundamental technique for extending a k-homomorphism of discrete fields is the following. 8.1 LEMMA. homomorphtsm.
Let
k
~
K
~
E
be
discrete
fieids,
and
a:K .... E a
k-
Let f € K[X) be irr'educibie, a E E a root of f, and ß E E a
168
Chapter VI. Fie1ds
,'oot of fO.
°
Then
exteru1.s to a homomorphism fr'om K[a) to E that
takes a
to ß.
PROOF.
Define a map from K[a] to E by taking g(a) to gO(ß) for eaeh
K[XJ.
To show that t.his map is weH defined, it suffices to show that.
gE
if g(a) = 0, then gO(f3) = O.
Suppose 9{a) = O.
Then f divides 9 beeause
f is irredueib1e and f (0) = 0; so FO di vides 90 , whence If S is a set that is bounded in number, and o:S
°
t.hen
is an isomorphism (see Exercise I.2.11).
gO (ß) ->
= O.
0
S is one-to-one,
There is an analogous
result in the eontext of fields. 8.2 LEMl'lA.
I<
Let
~
K be an aLgebraic extension of disCT'ete
fields.
Tilen each k-endomor'phi sm of K is an outomorph; sm.
Let
PROOF.
one.
If
E
CI
0
be a i<-endomorphism of K.
set of roots of f
rP ES.
As K is a held,
°
1S one-to-
is a root of a polynomial f in '~[X]. Let S be the in K. If ß E. S, then f (f3 o ) = fO (ß0) = f (ß)o = 0, so
K, then
CI
Therefore er induces a (one-to-one) map from S to S.
bounded by the degree of that ~ =
As
CI.
CI
E
r,
But S is
so a maps S onto S, so there is ß E S
Kwas arbitrary,
Let r, !: K be discrete
fields.
is onto.
Cl
If
~
K such
0
each element of K satisfies a
polynomial in r, [X] that is a produet of linear polynomials in K [X), then K is said to be normal over k. 8.3 'IHEOREM. vector
SlXlce
LeL k !: K be di.scn:te fields wUh K a Finitel!} generated
Tilen K is normal
Oller I<.
Oller k
if onIi oniy if K is
(l
spUtting Fleld of a poLynomiaL i.n k[X].
Suppose K is normal and generated by al' ... 'an over k.
PROOF.
monie
fi E
r~ [X]
such that f i. (ai.) = 0 and
polynomials in K[X].
r;
is a
CI
product of linear
Then K is a splitting field for f 1 f2"""f n "
Conversely, suppose K is a splitting fie1d for f lf!=l (X -
Choose
i ) with a; E K,
that x = p(al"" ,an) "
and given x Let S
be
E
the
E /, [X] •
Then f' (X) =
E r<[X 1 , ... ,Xnl so permutations of the
K, there is p group
of
indeterminates Xl" .. ,X n ' and set q(X,X1,···,X n )
= fIaES (X - p(Xl""'X~~))'
As q is symmetrie in Xl' ..• ,Xn , Theorem Ir. 8.1 says that the coefficient.s
of
q (X ,al' ... ,un )
are
polynomials
in
the
coeffieients
of
F,
so
169
8. Galois theory Clearly q (X ,01' ..• ,an)
= O. 0
polynomials in K[X], and q(x,ol, ... ,an ) Let k .:;; K be discrete fields.
is a product of linear
The set of k-automorphisms of K forms a
group 'tl (K/k) called the Galois group of K over k. separable over
then K is said to be a Galois extension of k.
The group
is a subset of the set of all functions ftom K to K, so has the
'!J (K/k)
natural 01(a)
I~,
I f K is normal and
#-
inequality given by
01
i f there
-F 02
exists a € K such that
When we say that 'tl(K/k) is finite, we mean finite with
02(a).
respect to that inequality. Let
8 • 4 'l'IIOOREM.
dimensional.
k
Gal.ois
~
K be
extension
discr-ete of
k,
fiel.ds. then
If
'tl(K/k)
K is a is
finite-
finite
and
#,fJ(K/k) = dimkK.
PROOF.
As K is sepatable ovet
I~,
thete is B such that K =
k[B].
The
minimal polynomial F of B aver k has degree n = dimkK and is separable. As K is normal, we can wtite f(X) = (X-9 1 )···(X-B n ) with Bi € K for each
i.
As
F is separable, the Bi are distinct.
fot each i K
°
= k[EI],
=
l, ••• ,n,
this
completely
'tl(Kfi{), then F(Elo)
€
0= 0i"
Thus 'tl(K/k)
are distinct.
thete is 0i € 'tl(K/k) determines
0i.
Lemmas 8.1 and 8.2 say that, such that 0i(B)
on
= fO(El o ) = F(EI)o = 0, so Elo
= {ol, ... ,on), and the
the =
other
= Bi. hand,
As
if
El i for some i, whence
0i ate distinct since the El i
0
We consttuct k-automorphisms of a normal extension of k by extending khomomorphisms of subfields. 8.5 LEMMA.
Let k .:;; K
~
E be discrete fieids with E normal. over k and
fini te dimensional. over K.
Then an!) k-homomorphism oF K into E can be
extended to a k-automorphism of E. PROOF.
Let o:K
~
E be a k-homomorphism.
As E is finite dimensional
over K it suffices, by induction, to show that if a € E, then E is finite dimensional over K[a], and
°
can be extended to a k-homomorphism of K[o]
into E. As E is finite dimensional ovet K, Theorem 1.13 says that K[a] fini te dimensional over K, (II.6.6). g(a)
=0
so E is
finite dimensional
over K[0]
is by
As E is notmal ovet k, thete is a polynomial 9 in k[X] so that and 9 is a ptoduct of linear polynomials in E[X].
As E is finite
170
Chapter VI. Fields
dimensional over K, Theorem 1.13 says that there is an irreducible polynomial f in K[X] with f(a) = 0. Clearly f divides 9 in K[X], 50 Fa divides gO ; g. As 9 is a product of linear polynomials in E [X], it follows that FO has a root ß in E. By (8.1) we can extend ° to a hhomomorphism from K[a] to E that takes a to ß. 0 Normality is related to invariance under automorphisms. 8.6 'l'HEXlREI'I.
Let I< !: K !: E be discr-ete fields.
tlIen ever-y k-eooomorplIism of E maps Konto K.
If K is nor-ma! over
I~,
If E is fini te dimensional
and normal over- 1<, and if ever-y k-
tlIen K
is normal over- 1<.
is normal over hand ° is a h-endomorphism of E. By Let a E K, and let f E I<[X] be a As product of linear pol ynomi al 5 in K[X] such that f (a) ; 0. f (ao) fO(ao) = f(a)o = 0, and f is a product of linear polynomials in K[X], it follows that aO E K. Now suppose E is finite dimensional and normal over k, and every hIf a E K, then a is a root of a automorphism of E maps K into K. polynomial f E h[X] that is a product of linear polynomials in E[X]. As E is finite dimensional over I~ we may assume that f is irreducible. It suffices to show that each root ß of f in E is in K. By (8.1) there is a PllOOF.
Suppose
K
(8.2) it suffices to show that ~ !: K.
k-homomorphism 0:1<[0] ~ Etaking 0 to ß. By (8.5) we can extend ° to a kautomorphism of E. But each k-automorphism of E maps K into K. Thus ß=aoEK.D
A key property of finite-dimensional Galois extensions h !: K is that each element of K~~ is moved by some element of ~(K/k). In fact, this is a characterizing property. 8.7 'l'HEXlREI'I. field k.
Let K be a
finite~timensional
extension of a
discrete
TlIen the fo!!owing are equivalent. (i) K is a GaLois exterlsion of k, (E) If a E K~,
(Ei) If
then there is
a E K, and
degree f(X)
=
n, (X -
distinet.
then
°
the minima! tlIere
are
0l(o»"'(X -
E 'fJ(K/k) such that o(a) ;f. o. po!ynomia! 01, .•• ,on
0n(a»,
und
f in
oF a ~(K/k)
over- h sueh
0l(o), ••• ,on(a)
110S
that are
171
8. Galois theory
Suppose that (i) holds, and let f be the minimal polynomial
PROOF.
over k of the element a in
~.
As K is separable and normal over k, and
the degree of f is greater than 1, there is a root ß F a of f in K. (8.1) a(a)
we
can
= ß, and
then
construct
a
k-homomorphism a:k[a]
~
such
K
By that
by (8.5) we can extend a to an automorphism of K.
Now suppose that (ii) ho1ds, and let a and f be as in (iii).
We show
by induction how to construct the desired elements a1, ... ,an of '!l(K/k). Suppose that we have constructed a1, ... ,as ' with s is
(X - a1(a))···(X - asta))
a
factor
f,
of
< n, such that
g(X)
and a1(a), ... ,as (a)
As F is irreducible, and 9 is a proper factor of f,
distinct.
coefficient of 9 is in gT F9 •
As Tai(a)
K~.
By (ii) there is
is a root of gT, for
Set a s +1 = Ta i That (iii) implies (i) is clear. 0
Tai(a) ({a1(a), ... ,as (a)}.
i
T
=
are some
in '!l(K/k) such that
=l, ... ,s, there is i with
•
If K is a discrete field, and 8 is a set of automorphisms of K, then {x E K : x a
= x for each
a E 8} is a field called the fixed field of 8.
Theorem 8.7 implies that if K is a finite-dimensional Ga10is extension of
k, then k is the fixed field of '!l(K/k).
Exercise 8 shows that (8.7.ii) is
stronger than the condi tion that k be the fixed field of '!l (K/k) • Let k
~
E be a finite-dimensional Galois extension.
The fundamental
theorem of Galois theory concerns the correspondence between subfields K of E containing k that are finite dimensional over k, and finite subgroups of '!l(E/k).
The subgroup of '!l(E/k) associated with K is '!l(E/K).
8.8 'l'HOOREM.
Let
discrete field k. dimensional ouer k.
E be
a
finit e-dimens ional
Galois extension of a
Let K be a subfield of E, containing k, that is finite
Then
(i) E is a finite-dimensional Galois extension of K. (ii) '!l(E/K) is a finite subgr-oup oF '9(E/k).
(iii) The fixed Field oF '!l(E/K) is K. (iv) If K is normal ouer k, then restrietion to K giues a map from '!l(E/k) onto
~(Kfi~)
with kernel
~(E/K).
80
~(E/K)
is a
normal subgroup of '{)(Efi<) (v) IF '!l(E/K) is anormal subgroup of
~(E/k),
then K is normal
ouer k.
PROOF.
As E is Galois over k,
it is Galois over K; so (i) holds.
172
Chapter VI. Fields
Condition (ii) is clear, while (iii) follows from (8.7.ii). To prove (iv) suppose that K is normal over k and a E 'fJ (E/k ) •
Lemma
8.6.i says that !
'!l(K;1
to '!l (Kjk) •
The kernel of this map is clearly 'i! (E/K) .
By (8.5), any
element of 'il(K;1<) can be extended to a k-automorphism of K.
Thus the
homomorphism is onto. To prove (v) note that i f TaT-i
E 'i!(EjKT);
a,T
thU5 y'!l(E/Kh-
then '!l(EjK) = 'g(EjKT).
'!l(Ejk),
i5 normal.
1
E '.§(E/ld, then
'.~(EjK)
a (
So i f 'i!(EjK)
= '!l(EjKT).
i f and only i f
is normal in
By (8.7.ii) this implie5 that KT ~ K, so K
0
It remain5 to show that every finite subgroup of '61 (E/I<) is of the form
In
'!l(EjK) •
fact
we
shaU
show
that
if
C
is
a
finite
group
of
automorphisms of a discrete field E, and K is the fixed field of C, then #Cl and E is a Galois extension of K with Galois group G.
dimKE :
Let E be a discrete field and M a monoid.
A monoid homomorphism from M
to. the multiplicative monoid E i5 ca11ed a character of M in E.
Any
automorphism of E i5 a character of the multiplicative monoid E in I. set EM of a11
functions
horn M to I
is a vector space over I
pointwise addition and scalar multiplication.
The under
The natural inequality on
EM is defined by setting f l' 9 i f there is m in M wi th F (m) t 9 (m) .
The
functions
f 1"" ,F"
in IM
a11 zew.
linearly independent i f alF1 +
are
a2f2 + ... + a r/ n t 0 whenever 0l, ...
n are elements of E that are not The fo11owing lemma gives a simple criterion for the linear ,<1
independence of characters. 8.9 LF2V'IA. clWf'acters
Let M be a monoLd wul E a disCf'ete FieLd.
of M in
E such
that
Gi t
Lei 01""
,an Loe
i f i t j.
0j
Lineady independer1t.
PROOF.
We proceed by induction on
elements of E that are not a11 zero.
Tl.
rf n
Suppose that al"" ,u n are 1, then 2ri1=10;oi (1) = al ~ 0,
so we may assume, after renumbering, that "2 t O. in
M
so that
subtract ( *)
from
CTl(x) ~ u2(x).
Let
m
As 01
~
02' there i5
be an arbitrary element of
M
x
and
173
8. Galois theory
(**)
ulal(xm) + •.. + unan(xm)
= ulal(x)al(m) + ... +
unan(x)an(m).
The terms involving ul cancel, leaving the term u2(a2(x) - al (x) )a2(m) plus a linear combination of the characters a3, ... ,an .
By induction there
is m so that this expression is nonzero, because u2(a2(x) - 01 (x»
~
o.
As this expression is the difference of (*) and (**), one of (*) or (**) must be nonzero.
Thus either m or xm shows that
~
uia i
~
O.
0
From (8.9) we see that distinct automorphisms of a discrete field E are linearly independent over E. 8.10 LEMMA.
Let
K
be
We refine this result a bit. u
discrete
Field
und
a1, ••• ,an
uutomorphisms oF K.
Then there exist w1' ••• ,wn in K such that matrix, whose ijth eLement is 0i(w j ), is inuertibLe.
PROOF. taking w1
We proceed by induction on n.
=
1.
The ca se n
distinct the nxn
= 1 is handled by
By induction there exist w1' ••• ,wn _1 in K such that the
(n-1)x(n-1) matrix
~n-1' whose ijth element is 0i(w j
),
is invertible.
So
there are u l' ... ,u n _1 in K so that u1a1 (w j) + ... + un-1an-1 (w j) for j
= 1, ••• ,n-1.
= an (w j)
By Lemma 8.9, there is wn # 0 in K so that
(*)
Let ~n be the nxn matrix whose ijth element is 0i(w j
).
To show ~n is
invertible, let the ith_row of 2 n be denoted by ui.
Replacing un by un (u1u1 + .•. + un_1un_1) we get a matrix with the same determinate as ~n'
whose last row is all zeros except for the last entry, which is nonzero by
(*).
The determinant of the modified matrix is det 2n _1 times the nonzero
entry in the lower right corner.
Thus de t 2n # O.
0
We can now complete the theory of the Galois correspondence between subgroups of 'B (E;1z), and fields between I< and E. 8.11 THEDREM.
Let
E be a discrete
FieLd,
automorphisms of E, and K the Fixed field oF G.
oF
E which ar"e
dimKE
= n,
PROOF.
Linear"ly
independent
over" K.
G u
set
of
n
distinct
Then ther"e ar"e n elements
IF
G is
a
gr"Ollp
Let G
=
{al'''' ,on l
. By (8.10) there exist
w1'''' ,wn in E such
that the matrix 2n , whose ijth element is ai(w j ), is invertible. be elements of K such that u1w1 + •.• + unw n = O.
a1"" ,an
thell
and E is aGalais extension of K with Galois gr"Ollp G.
Let Then
174
Chapter VI. Fields
= 0 for each
Because Ln is invertible, this implies that each a j is zero, so the wj's are linearly independent. Now assume that G is a group. We will show that the W /5 span E over
a1ai (w1) + ••• + arFi (wn )
K.
Let w
As Ln is invertible, there are a1' •.. ,an in K so that
E.
E
a1a i (w1) + .•• + ana i (wn )
(*)
for
i.
i
= 1, ... ,n. As
G is a group, aG
= a i (w)
= G for each
so applying a to
a E G,
(*) we obtain a(a1)a i (w1) + ... + a(an)ai(wn )
for
1, •.. ,n.
= a i (w)
Subtracting this from (*) gives
(a1- a (al»a i (wl) + ••• + (an -a(an»ai(wn )
=
O.
But Ln is invertible, so a j = a(a j ) for all a and j, whence al, •.• ,an are in K. Taking a i to be the identity in (*) we get Thus w1' ... ,wn span E over K. Theorem 8.7 shows that E is a Galois extension of Galois group follows from (8.4). 0
K.
That
G
is its
EXERCISES
1. Let K be an algebraic extension of a discrete field k.
Show that K is normal if and only if each polynomial in k[X] having a root in K has a factor in k[X] that is a product of linear polynomials in K[X].
2. Define what a splitting field of a set of polynomials iso Show that an extension K/k is normal if and only if there is a set of polynomials in k[X] whose splitting field over k is K. 3. Formulate and prove (8.9) and (8.10) for K a commutative local ring. 4. Let k be a discrete field and K = k (X) . Define automorphisms of K by a(X) = 1;X and T(X) = 1 - X. (i) Show that the group G generated by a and symmetrie group on three letters. (ii)
Let 1=
(X 2 -
X+
1)";X2(X -
1)2.
solution to the polynomial equation
two k-
T
is the
Show that X is a
175
8. Galois theory
(iii) Show that k(I) is the fixed field for G. 5. Let xl"" ,xn be distinct elements of a discrete field k, and uI"",un elements of k that are not all zero. Use (7.9) to show that there exists mE m so that u1xi + .•. + un~ ~ O. 6. Let k ~ K be discrete fields. Suppose K is factorial, and finite dimensional over k. Show that 'B(KjR) is finite and that II'B(KjR)
~
dimkK.
7. Let K be a splitting field of X3 - 2 over~. Show that if w is the quotient of any two distinct roots of X3 - 2 in K, then + w + 1 = O. By letting k be a field between ~ and ~[w], construct a Brouwerian example of discrete fields k ~ K such that dim''1K = 3, but 'B(KjR) is not finite. w2
8. Use Exercise 7 to construct a Brouwerian example, using Markov's principle, of discrete fields k ~ K such that K is finite dimensional and separable over k, and k is the fixed field of 'tJ (KjR ), but condi tion 8.7. i i does not hold.
A classically equivalent fonn of the weak Nullstellensatz is that if I is a proper ideal in k[X 1 , ••• ,Xn ], then there is an a1gebraic extension field K of k such that the polynomials in I all vanish on some point of Kn . This follows from our version by putting I in a maximal ideal M (Zorn's lemma) and setting K = k[X 1 , •.• ,Xn ]/M. with a suitable (constructive) restriction on k, we can prove this theorem directly. For a comparison of the constructive and the recursive approaches to splitting fields and algebraic closures see [Bridges-Richman 1987].
Chapter VII. Factoring Polynomials
1. FAC'IDRIAL
AN!)
SEPARABLY FAC'IDRIAL FIELDS
In this section we over a discrete field Brouwerian example of (IV.2.2). A discrete polynomial over k can
investigate the problem of factoring a polynomial into irreducible factors. We have already seen a a discrete field over which this is not possible field k having the property that each nonconstant be written as a product of irreducible polynomials
is called factorial. Equivalently, k is factorial i f every nonconstant polynomial over k is either irreducible or has a nontrivial factor. As k[X] is a principal ideal domain with recognizable units for any discrete field k, this is equivalent to saying that k is factorial in the sense of definition IV.2.1. We say that k is sepa.rably factorial i f each separabl.e polynomial can be written as a product of irreducible polynomials. Clearly any factorial field is separably factorial .. In some respects the notion of a separably factorial field is superior to that of a factorial field: Theorem 2.3 shows that a finite-dimensional extension of a separably factorial field is separably factorial, while Exercise 1.5 gives an example of a finitedimensional extension of a factorial field that is not factorial. Finite fields are easily seen to be factorial: gi yen a polynomial, divide it by each polynomial of lower degree and see if the remainder is zero. Algebraically closed discrete fields are also factorial for trivial reasons. The first nontrivial theorem about factorial fields is due to Kronecker who showed that the field III of rational numbers is factorial. This follows immediately from Kronecker 1 (Theorem IV.4.8), which implies that the ring Z of integers is factorial, and (IV.2.6), which implies that the field of quotients of a factorial domain is factorial. Many fields of interest are of the form k(al, ... ,a n ) where k is either a finite field or Ill. As the finite fields and III are factorial, we are led
176
177
1. Factorial and separably factorial fields
to investigate simple extensions k (a) of a factorial field k; the most important cases occur when a is algebraic or when a is transcendental. We first show that an extension of a countable factorial field by a separable element is factorial. Let k !;; E be discrete fields.
1.2~.
is algebraic over
k und
(j is separable over
Let a,{j E E be such that a
kral.
IF k is Factorial, then
a satisfies an ir-redueible polynomiaJ over- k[{j).
PROOF. There exists S in E such that k[a,{j) = k[S) by (6.7). If k is factorial, then k [(j) and k [e ) are both fini te dimensional over k by (VI. 1. 13) •
Hence k[S) is finite dimensional over k[ß) by (II.6.6), so by
(VI. 1. 13) we can find an irredueible polynomial satisfied by a.
Let
ß
that
is
Let E be a discrete held and /, a eountable faetm-ial
1.3 COROLLARY. subfi-eld.
over k[ß]
in E be separable ouer k.
Then k[{j) is Faetorial.
PROOF. Let g(X) be a noneonstant polynomial with eoeffieients in k[ß), and let E' ;;:! k[{j) be a eountable diserete Held eontaining a root a of 9 (Theorem VI.3.3). Then a satisfies an irredueible polynomial over k[{j), whieh must be a faetor of g(X) in k[ß)[X].
0
Corollary 1.3 shows that a finitely generated separable extension of a eountable factorial field is faetorial. Exereise 5 shows that the separability condition is neeessary. To remove the eountability eondition we will show that a diserete field is (separably) faetorial if and only if eaeh nonconstant (separable) polynomial either has a root in the field, or is nonzero when evaluated at any element of the field. construet,
from a
given polynomial
f,
To this end we
a polynomial q so that any
eeeffieient of a manie faetor of f is a root of q. To assure that the polynomial q is a produet of separable pelynomials in the ease that f is separable we prove the following. 1.4 LEMMA.
Let K be a separ-able extension fieid of a diser-ete field Iz.
If q E k[X] faetors into Unear- factor-s in K[X],
tI,en q is a
pr-oduef
of
separable polynomials.
PROOF. Let a E K be a reot of q. The element a satisfies a separable pelynemial f in k[X]. Let h = GCD(q,f). Then l1(a) = 0 and 11, being a faetor of f, is separable. The polynomial qfil is a produet of separable
178
Chapter VII. Factoring polynomials
polynomials by induction on degree, so q is is a product of separable polynomials .
0
Ne use a splitting field to construct q, although we could dispense with this device as the coefficients of of the coefficients of 1.5 THEDREM.
q
can be written düectly in terms
f.
Let I< be a diserete field, and let f E I<[X] be a monie
(separable) polynomial.
Then ther'e exists a polynomial q in k[X]
(which
is a produet of separable polynomials ) such that, for' any extension E of
k, the eoefflcients oF urlY monic Facto,' oF f in E[X] are ,'oots of q.
PROOF.
Let k o
coefficients of
F,
be
the
countable
subfield of h
gene ra ted by the
let Kf be a splitting field for f over k o , and write F(X)
=
(X - "l)(X - 1'2)··· (X - r n )
Let S ~ k[Y 1"" ,Yn ] be the set of elementary symmetrie polyin Kr' nomials in the finite subsets of the indeterminates {Y 1 , ... ,Yn ), and set q(X) = 1I0ES (X - o("l, •.• ,"n))'
As the eoeffieients of f are the elementary symmetrie polynomials in all the r t , it follows from (I1.8.1) that q E Ico[X], and that q is independent of the splitting field Kf . If f (1. 4)
.
is separable, then q is a product of separable polynomials by If 9 is a monie factor of F in E [X], then eonstruct a splitting
field Kr for f over the subfield of E generated by the coeffieients of 9 and f.
Clearly Kf is a splitting field for f over k o that eontains the coefficients of g. Each of coefficient of 9 is an elementary symmetric polynomial in some of the roots of f in Kf , henee is a root of q. 1.6 COROLIARY.
0
lf k is separably closed in K, and the monie separ'able
polynomial f in heX] has a monie faetor 9 in K[X], then gE heX].
lf k is
algebraieally closed in K and the monie pol ynomia 1 f in k [X] has a monic faetor 9 in K[X], then 9 E k[X].
PROOF.
The coefficients of any factor of f
(separable if f is separable) polynomial in h[X].
in k [X] satisfy a monie As k is algebraieally
(separably) closed in K, these coefficients must be in k.
0
Theorem 1.5 guarantees that the coefficients of a monie factor of a monie separable polynomial are separable.
Ne ean drop the separability
1. Factoria1 and separably factorial fields
179
requirement if we demand that the two factors be relatively prime. 1.7 CDROLIARY.
Let
k
~
E be discrete
g,h E E[X] ure monic polynomiuls such that f
fields.
= gh
If
f E k[X]
und GCD(g,h)
und
= 1, then
the coefficients of 9 und h are separable ouer k.
PROOF. By (VI.6.3) we can write f as a product of monic polynomials in k[X] of the form F(Xq), where F is separable and q is 1 or a power of the characteristic of k. Then F(Xq) = a(X)b(X) where a(X) = GCD(F(Xq),g(X)) and b(X) = GCD(F(Xq),h(X)). As 9 is the product of the a'S and h is the product of the b's, we may assume that f = F(Xq). We now induct on q. If q = 1 we are done by (1.5). If q > 1, then 0 = f' (X) = g'h + gh' so 9 divides g'h, whence 9 divides g' so g' = 0 = h'. Therefore 9 = G(XP) and h = H(XP), so F(Xq/P) = G(X)H(X). By induction the coefficients of G and H, which are the same as the coefficients of 9 and h, are separable over k.
0
We shall show that an arbitrary finite-dimensional extension of a separably factorial field is separably factorial. First we reduce the problem of showing that a field is (separably) factorial to that of deciding whether or not a (separable) polynomial has a root. 1.8 THEDREM (the root test). A discre te neid k is (separably) factorial if and only if for each (separable) f in k[X], either there exists a E k with f(a) = 0, or f(a) t 0 for all a E k. PROOF. If k is (separably) factorial, then we can construct all monic linear factors of f, so the condition is clearly necessary. To prove the converse we use Theorem 1. 5 to construct a polynomial q in k [X] so that the coefficients of any monic factor of f in k[X] are roots of q. Using induction on the degree of q, and the condi ti on of the theorem, we can construct the finite set of elements of k that are roots of q. Thus we can obtain a finite set of polynomials containing all the monic factors of f in k[X]. The elements of this set can now be tested to see which are factors of f. 0 The prototype of a root test is the rational root test that says that if s/t is a root in~, in lowest terms, of the polynomial aO + a1X + ••. + ~Xn in ~[X], then s must divide uO' and t must divide an' This narrows down the possible roots to a finite number, each of which can be tested.
Chapter VII. Factoring polynomials
180
The rational root test, together with Theorem 1.8, provides another proof of Kronecker's theorem that the rational number field is factoria1. In fact we get the fo110wing slight1y irnproved version of Kronecker 1. 1.9 THEXlREM.
If R is a unique factor'ization domain wUh finUeLy many
units, then so is R[X).
PROOF.
Let
K
Thus R is factoriaL.
be the field of quotients of R.
To show that each
element of R[XI can be factored into irreducibles in R[XI, it suffices, by Gauss's Lemma, to show that K is factorial. By Theorem 1.8, it is enough to show that K has a root test. The argument sketched above for the rational root test applies here because R is a unique factorization domain with a finite number of units. 0 If R is a unique factorization domain with a finite number of units, then (1.9) and induction show that R[X1, ... ,Xnl is a unique factorization domain. So i f F is a finite field, then F (Xl' ... ,Xn ) is factorial by (IV.2.6). Also ~(Xl"",Xn) is factorial because the ring of integers is a unique factorization domain with two units. This does not cover all the Kronecker 2 (Theorem IV.4.9) shows that i f k is a cases of interest. factorial field, then so is k(X). We present here a less elegant proof of this theorem, based on the root test, that includes information about separably factorial fields. 1.10 THEXlREM.
IF k is a
(sepambly) Factorial Field.
then so is k(X).
PROOF. Let f(X,y) E k[X,YI (be separable as a po1ynomia1 in Y over k[X I); it suffices, by the root test, to decide whether or not there exists a E k(X) such that f(X,a) = O. f(X,y) If we substitute Y
=
=
write
aO(X) + a1(X)Y + ••• + an(X)yn .
Zlan (X)
and multiply by an (X )n-l we get a monic
polynomial in Z, so we may assume that an(X)
=
1, and we need only look
for roots a in k[XI that divide aO(X). In the separable case, let f' (X,Y) denote the derivative of f(X,Y) as a polynomial in Y, and choose s(X,Y) and t (X,Y) in !?[X,Y] such that s(X,Ylf(X,Yl + t(X,Ylf' (X,Yl = g(X)
where 9 (X) is not zero. In the nonseparable case, let 9 (X) = 1. If k is finite, then k(X) is factorial (1.9) so we are done. So (VI.5.4) says
181
1. Factorial and separably factorial fields
that either we are done, or we can find distinct elements x1"",xm in k such that m exceeds the degree of aO(X), and 9 (xi) #- 0 for each i. The po1ynomia1 f(xi'Y) is separab1e, if necessary, so we can construct the finite set Ai = {y E k : f(xi,y) = O}. If f(X,a(X» = 0, then a(x i ) E Ai so we can construct, by unique interpolation, a finite set of candidates for a(X). 0 EXERCISES
1. Show that a discrete field k
is factorial i f and only i f each nonconstant polynomial over k is either irreducible or has a nontrivial factor.
2. Give a Brouwerian example of a simple extension
k(a)
with a
neither a1gebraic nor transcendental. 3. A prime field is a discrete field k such that every subfield of k
is k. Let p, ,P2 ,P3 ,P. , • •• be the sequence of prime numbers congruent to 1 mod( 4), and let a be a binary sequence wi th at most one 1. Define the ring R to be the ring of integers with the following equality relation: d = e i f there is n such that anPn Id-e. Show that R is a discrete integral domain, and that the field of quotients k of R is a Brouwerian example of prime field that is not factorial. 4. Show that if F, ~ F2 ~ F3 ~ ••• are factorial fields, and Fi is algebraically closed in F i +1 for each i, then F = U F i is factorial. Conclude that if k is factoria1, then so is the field k(X"X 2 , ••• ) in countably many indeterminates. 5. Let a be a nondecreasing binary sequence.
Let
Fi
= k of Theorem
VI.2.10, if a i = O. and F i = K of the same theorem i f a i = l. Use Exercise 4 to show that F = U Fi is factorial. Let e be an element of a discrete extension field of F such that e2 = a. By considering the polynomial X2 - b, show that F[e) is a Brouwerian example of a fini te-dimensional extension of a factorial field that is not factorial; and therefore a separably factorial field that is not factorial.
182
Chapter VII. Factoring polynomials
2. EX'J.'&IffiICH> OF (SEPARABLY) FAC'l'ORIAL FIELDS
Corollary 1.3 says that if k is a countable factorial field, and a is separable over k, then k [a] is factorial.
In this section we remove the
countability assumption from that theorem, show that i f k factorial
and a
is algebraic,
then k [a I
is separably
is separably factorial,
and
construct unique splitting fields for separable polynomials over separably factorial fields.
2.0 'l'lIEX>REX.
Let Iz t;; E be discrete fieLds.
Let a,{3 E E be such that a
is algebraic over k and {3 is separabte over klal.
If k
is separabty
factoriat, then {3 sat isfies an i rTedueibte polynomiaL over- k [a I.
PROOF. over k,
By Corollary VI.6.7, there is a primitive element S for k[a,{3] which is separable over Idal
by (VI.4.5).
To
show tha t
(3
satisfies an irreducible polynomial i t suffices, by Theorem VI. 1.13 , to show that Ida,(3] is a finite-dimensional vector space over k [a I.
Choose
q = pe, wi th e ~ 0, so that 1\ = sq, and hence a q , is separable over k. Then kll\l and kla q ) are finite dimensional over k, factorial.
Let l,I\,1\2, ... ,l\s be a basis for 1<[1\] over k[aql. S is separable over klal IdSI
=
lf=o
ail\i
as k
is separably
Therefore, by (II. 6.6), k [1\] is finite dimensional over k [a q I.
k[l\,al.
we have,
Thus 1,1\,1\2, ... ,I\S
= 0 with a i E kral.
As sq
= 1\ E k[l\,al, and
by Theorem VI.6.2, generate IdS]
S E kll\,a],
over k[al.
so
Suppose
Then
0= (2f=0 ail\i)q = 2f=0 (ai)ql\i q where (ai)q E kla q I. VI.6.2,
As 1\ is separable over Iz it follows, by Theorem that klaq,l\l = k[aq,l\ql. Thus 1,l\q,l\q2, ... ,A qs is a basis for
h[aq,A]
over Iz[aql.
Therefore (ai)q
=
0,
so a i
=
0 for all i.
Thus
1,1\, ... ,l\s is a basis for klSI over kla). so k[S] is finite dimensional overk[al.D 2.1 LEMMA.
Let k be a discr-ete subfield of a r-ing E.
satisfies a manie polynomial in k[XI of degr-ee n
> O.
Suppose S il1 E
If a
E
klS], then
either- SE kral, or a satisfies a polYl1omial in k[X] of degree less than 11.
PROOF.
Wrl'te a i
",n-1 = ~j=1
a i j sj f or i
we can put the matrix {aij)
l USlng ' = 0 , ... ,11-. row opera t'lons
into upper triangular form.
If there is a
zero on the diagonal, then 1,a, ... ,a"-1 are linearly dependent over k,
so
2. Extensions of (separably) faetorial fields a
183
satisfies a polynomial of degree less than n.
elements
oE
are
k[al.
nonzero,
then
the
If all the diagonal
determinant of
is
{a ij }
nonzero,
so
0
The following theorem, which has classieal content, will be used to remove the countability hypothesis from the classieally trivial Corollary 1.3. 2.2 THElJREM.
Let K
~
E be impotent rings and Radiscrete subfield of
K timt is separabty closed in K.
Let
klal is separably dosed in K[a].
If, moreave,' ,
0
in E be algebraic over k.
Then
is sepu'able and k is
0
algebraically closed in K, then klo] is algeb,'aicany closed in K[a).
Let ß
PROOF. is () in
k[a,ß]
€
Klo] be separable over klo].
so that
k[O] =
to show that 0 € klo I. whieh is in K[oq], eharaeteristie
separable over k. degree
By Theorem VI.6.7, there
To show that ß
i t is enough
E k[al
By Corollary VI.6.4, there exists q so that oq,
is separable over R_, where q
=
1,
0['
k
has finite klS q ] is
Then, by Theorem VI.4.5, kla q ]
~
Let o'l satisfy the separable polynomial f
E
and'l
p
kla,ßI.
=
pe.
k [X I,
of
We proeeed by induetion on n to prove that O'l E R [at] I; so
Tl.
EJ E k [0], by (vr.6. 2), as H is separable over k [0]. If n
=
1, then oq E k, so Bq
closed in K.
E
K and thus EJq
Now suppose that n
> 1.
E
k sinee R is separably
As Hq E K[a q
it
],
follows, by
Corollary VI.1.3, that EJq satisfies a polynomial of degree n over K. 8'1 satisfies a separable polynomial over k.
But
Taking GeD' s of these two
polynomials, and using Theorem VI.4.1.iii, we obtain a sepat-able polynomial of degree at most n in k[X] that eq satisfies. As a q E k[Oq], Lemma 2.1 says either eq E lz[u Q ], and we are done, or else u q satisfies a polynomial of degree less than n and we are done by induction.
eq
Thus
E k[u q ].
Now suppose that a is separable over h, and that k closed in K. () E k[al.
Let H
E
is aIgebraically
K[0] be algebraic over /du]; we want to show that
By Theorem VI.5.5, we may asstnne that h[a,8]
satisfy a manie polynomial of degree n in k [X] • on n to show that EJ
E h[a].
As
fl C Kla] i t
that 8 satisfies a monie polynomial in K[X] of degree
Tl.
to
over k.
Applying
Let u
follaws, by Corollary VI.1.3,
also
algebraic
= h[e].
We proceed by induetion
Corollary 1.6
The element 8 is the
r~
of
the
polynomials that H satisfies over I< and K we obtain a polynomial of degr'ee
184
Chapter VII. Factoring polynomials
at most n in Ic[X] satisfied by 8.
By
Lerrnna 2.1 either 8 E 1<[0'], and we
are done, or 0' satisfies a monie poIynomial of degree Iess than latter case induetion shows that 0
E "[0'
1.
In the
11.
0
Ne can now remove t.he countability restriction of Corollary 1 . .3 and prove the ana10gous result for separab1y factorial fields. 2.3 'I'HEOREM. (leId E, and
If
{actodal.
PROOF.
Let
k be a
separnb/lj {aetorlaI
0' E E be LltgebrLlir oue1- I"
I.el
of a
sub{ield
fhen K
= k[aJ
diso-ete
is se[xu"abty
is sepornbte and h is {aetar-iot, then K is {Cletm-ial.
0
Let. F
E
K [X 1 be separable.
Let. 1<0 be the separable closure
within I< of the countable field generated by the coefficients of a nonzero polynomial in k[X 1 satisfied by powers of
11,
together with the coefficients of the
that occur in the coeffieients of f.
11
faet.orial, 1<0 is a eountable field.
As I<
is separably
As we can determine whether separable
polynomials over le o have roots in k, and hence in /'0' it follows from Theorem 1.8 that h o is a separably factorial field.
To complete the proof
of the first statement it suffiees, by Theorem 1.8, to find the roots of f that He in 1<0 [11], as Theorem 2.2 says that k o [11
is separably closed in
J
h [11], and the coefficients of F He in 1<0 [11]. As h o [l1] is eountable, we ean eonstruct, by Theorem VI.3.4, a splitting held L of f
over ko[o].
Let rl, ... ,1's be the reots of f
in L.
By
Theorem 2.0, we ean find irreducible polynomials 9i in ko[a][X] that are satisfied by the r i . a root of f in Ra [a] •
If 9i is linear, then "i E Ro [a 1, and we have fOlmd If no gi is linear, then f has na raots in
I~a
[a
J•
The second statement is preved in exactly the same way as the first, except we take closure.
I~o
to be the algebraie closure instead of the separable
0
In Seetion 1 we showed how to polynomial over a countable field.
eonstruet a
spH tting field
for
a
For a separable polynomial over a
separably faetorial Held we can drop the eountabili ty assumption. 2.4 COROLlARY.
Let k be Cl seporC1bLtJ FCletodal
be Cl sepa.rabLe potynomia!. fleld K for
PROOF. K = k.
Then
[hen' i s
Cl
ffeld. und tet
f E k[X]
sepCl1'Clbllj faetar-lClI
spl. i t t i11g
of F.
= I, take
f ouer 1<.
Ne proeeed by induetion on the degree
Suppose
11
> 1,
11
If
11
As R is separably factorial, f has an irredueible
185
2. Extensions of (separably) factorial fields
factor p. Let F = k[Y]/(p(Y)). Then F is a field and the element a = Y in F is a root of F, so F(X) = (X - a)q(X) in F[X]. By Theorem 2.3, the field F is separably factorial. Thus, by induction, q has a separably factorial splitting Held over F. This splitting Held is the desired splitting field for F over k. Finally we
0
show that any two
splitting
fields
for
a
separab1e
polynomial over a separably factorial field k are isomorphie over k. 2.5 THOOREM. k2
Let k, and k 2 be separabty Factorial FIeIds, and
be an isamal"phism.
Let p, be a sepal"able palynamial
be the image af p, under
in
k,[X], and P2
Let K, and K2 be spli tt ing fie lds
k, and P2 auer k 2 respectiuely.
Far p, auer
Then
K ..... 1.
It is c1ear that the polynomial P2 is separab1e over k 2 • If deg p, = 0, there is nothing to prove, so let a, be a root of p, in K,. As k, is separably factorial, a, is a root of an irreducible factor q, of p, in k,[X]. Let q2 be the image of q2 under
EXERCISES
1. Let k be a countable separably factorial field. Show that there exists a countable separable extension K of k such that each separable polynomial over K has a root in K. The field K is called a separable closure of k. 2. Let k be a countable separably factoria1 field.
Show that if K
and L are separable c10sures of k. then there is an isomorphism of K and Lover k. 3. Use Theorem VI.2.10 to construct a classical example showing that the condition that a be separable is needed in Theorem 2.2. 4. Let k be a separab1y factorial field contained in a discrete conunutative ring R. Suppose that R is a finitely generated, separable, algebraic extension of k. Show that R is fini te-
Chapter VII. Factoring polynomials
186 dimensional over k.
(Look for nontrivial idempotents e E R; note
that ke is a subfield of the ring Re and is isomorphie to k)
3. SEIDmBERG FIELDS Let k be a discrete field of finite characteristic p.
Seidenberg' s
condition Parises from considering the problem of whether a given element of k has a P th reot in k, that is, whether the subfield k P of P th powers of elements of k
is detachab1e from k.
The condi tion introduced by
Seidenberg is the first condition in the following theorem. 3.1 'l'fIED.R&Il.
Let k
be a discr-et.e Field oF Finite character-istic p.
Then the FoUowing conditions are equivaLent.
(i) Given a ij E k for 1 ~ i ~ m and 1 ~ j <; n, ei ther there exist , wheneverx j E k P , not all zero, with L aijx j = 0 For all i i 2 aijx j = 0 Far- all i, with x j E k P , then x j = 0 for- all j. (ii) If K [s a Fini te-dimensiona I extension of k, then KP is or~
detachable fr-om K. (iii) If K is a Finite-dimensional purely inseparable extension of k, then KP is detachable from K.
(iv) Every finitely generated extension Field K oF k wUh KP ~ k is finite dimensional.
(v) Every finitely generated IzP-subspace oF k is Finite dimensional. PROQF.
To derive (ii) from (i) let w1, ... ,wn be a basis for K over k,
and let wf = 2r]=1 bijw j"
Then an arbitrary element 2 a jW j of K is in KP if and on1y if there exist xi in k P such that
r
a·w· j=1 J J
=
~' x.w~
Li =1
1
=
1
r' 2n
x.b. ·w· j=l i=l 1 1J J
that is, a j = 27=lxibij for all j or, since the wf are independent over k P , that the system of equations
xoa. J
2=1x.h .. 11
i
1
1J
o
has a nontrivial solution in k P . Obviously (iii)
follows from (ii).
To show that (iv)
follows from
( iii ), let K be a fini tely generated extension of k such that KP ~ Iz. w be one of the generators of K over k.
Let
Since k P is detachable from k,
either wP E k P whence w € k and we are done by induction on the number of generators, or
xP -
w is irreducible over k (Theorem VI.6.6) so k(w) is a
187
3. Condition P
p-dimensional extension of k. Clearly in the latter ca se k (w ) also satisfies (iii). But K is a finitely generated extension of k(w), so K is finite dimensional over k(w), whence over k, by induction on the number of generators. To derive (v) from (iv) let V be a finitely generated kP-subspace of k. Then V gene rates a subfield F of k which is finitely generated over k P • Let K = F-P d k. Then K is finitely generated, hence finite dimensional over k; so F = KP is finite dimensional over k P and therefore the finitely generated subspace V of F is finitely generated over k P • To prove (i) from (v) choose a basis for the kP-subspace of k generated by the aij' Then the question in (i) reduces to the existence of a nontrivial solution in k P to a set of homogeneous linear equations with coefficients in k P , which is decidable by (II.6.2). 0 We say that a discrete field k satisfies Seidenberg's condition P, or that k is a Seidenberg field, if, whenever P is a prime that is 0 in k, then the conditions in Theorem 3.1 hold. From (ii) of Theorem 3.1 we see that condition P is inherited by finite dimensional extension fields. transcendental extensions. 3.2~.
PROOF.
We show next that it is inherited by purely
IF k is a Seidenber-g Fietd. then so is k(X).
We shall show that (i) of Theorem 3.1 holds for k(X).
Lj
Suppose
aijx j = 0
is a system of equations over k(X) to which we are seeking a nontrivial solution in k(X)P = kP(XP). We may assume that the a ij lie in k[X], and that we are seeking a nontrivial solution in kP[XP]. separating the powers of X into residue classes modulo p, we get an equivalent system of equations in which the a ij lie in k[XP]. The coefficients of the a ij generate a kP-subspace of k which, by (v) of Theorem 3.1, is finite-dimensional.
Let
~1""'\n
be a basis for that
subspace, and write a ij = LkUijk~k with Uijk E kP[XP]. Then we can find a nontrivial solution in kP(XP) to the original system if and only if we can find a nontrivial solution in I,P [Xl'] to the system
Lj aijkxj
= 0,
but now the coefficients and the sought-for solution both live in the same
188
Chapter VII. Factoring polynomials
field kP(X P ), so the result follows from (II.6.2).
0
Exercise 1.5 shows that a finite-dimensional extension of a factorial field need not be factorial; the missing ingredient is condi ti on P.
Call
a discrete field I< fully factorial if any Hni te-dimensional extension k is factorial. 3.3 'lHEX>REM.
Let k be a discrete fLeld.
and onl!) if k is separabl!}
PROOF.
factor~lal
and saLisfies condltion P.
Suppose 1< is fully factoriaL
As k
extension of i tself, k is (separablyl factorial. eondition P we will verify (ii) dimensional extension of h. polynomial xP -
CL
E K [X
(f
Tflen k is falLy factorial
is a finite-dimensional To show that I< satisfies
of Theorem 3.1.
Let K be a finite-
Then K is factorial, so for each a
J is ei ther ineducible, in whieh ease
E K
a 'l
the
KP, or
it is reducible, in which case a E KP by (VI.6.6). Conversely, suppose h is separably faetorial and satisfies condition P. Let K be a finite-dimensional extension field of h.
Then K is separ:ably
factorial, by (2.3), and satisfies condition P by (3.1).
Let f be a monie
polynomial in K [X 1; we will show that either F has a root in K or i t By (VI.6.3) we ean write f as a
doesn't, so K is factorial by (1.8).
product of monie polynomials of the form g(XC!) where 9 is separable and '1 i5 ei ther 1 or apower of a prime that is zero in Iz, so by induction on degree we may assume that f is of this form.
As K is separably factorial,
we may assume that 9 1S irreducib1e.
Thus if g(xq) has a root in K, then
Xq - a.
We ean deeide whether X'i - a has a
9 is linear, whence 9 (xq)
=
root in K by (ii) of Theorem 3.1.
0
AS a corollary we have that every purely transcendental extension of a fully factorial field is fully factorial. 3.4 COROLLARY.
PROOF.
Tf I, is Cl
flll!~J
factor-ial
Field.
thell so is h(X).
Theorem 3.3 shows that I< is a Seidenberg field, so I«X) is a
Seidenberg field by (3.2). fully factorial by (3.2).
But I!(X) is factorial by (1.10), so I«X) is 0
EXERCISES 1. Show directly that the held F of Exercise 1. 5 does not satisfy
condi tion F.
189
3. Condition P 2. Show that any two algebraie elosures of a faetorial field k are isomorphie over k.
eountable
fully
3. Show that a eountable diserete Held is fully faetorial i f and only i f every Hnitely generated algebraie Held extension is finite dimensional. 4. Construet a Brouwerian example of a diserete eommutative ring extension of ~ that is finitely generated and algebraie, but not finite-dimensional, over ~. 5. Show that i f k satisHes eondition P, then so does the Held k(X " X2 , ••• ) in eountably many indeterminates. 4. TBE FUNIli\MENTAL THFDREM OF ALGEBRA
The eomplex numbers that are algebraie over ~ form a diserete field ~ (Theorem VI.1.9), ealled the field of algebraie numbers. In this seetion we will show that ~ is algebraieally elosed, using many of the results of Galois theory and, without proof, the existence of 2-Sylow subgroups of finite groups. A proof of the eorollary, that every monie polynomial of positive degree over ~ has a root in~, is outlined in the exereises. 4.1 LEMMA (Intermediate value theorem for ~[Xl). Let f E ~[Xl. and b are rational numbers with f(a) < 0 < f(b), then there exists with f(c) = O.
C
If a in rn
We may assume that a < b.
PROOF.
DeHne, by induetion, two sequenees Let aO = a and b O = b. To deHne an (an _1 + bn _1 )/2, and set.
{an} and {b n }, of rational numbers.
and bn , for
Tl
> 0, let c n =
(i) an = bn = c n (ii) an = cn and bn = bn _1 (iii) an = a n _1 and bn = cn
i f f (en)
0,
if f (c n ) if f (c n )
< 0, > O.
Clearly a n -1 ~ an ~ bn ~ bn_1 , and bn - an ~ (b - al/2n for eaeh n > O. Thus the sequenee {cn) is Cauchy, so represents areal number c. To show that f (c) = 0, we must show that f (en) eonverges to O. Applying the remainder theorem to f (X) as a polynomial over ~[Y 1, we get f (X) = (X - Y)g (X, Y) + f (Y) for some polynomial 9 E ~[X, Y] . It is easy to get abound M on {!g(x,y)! : a ~ x,y ~ b), so !f(x) - f(y)! ~ M!x - y! whenever a ~ x,y ~ b. As c n - an and bn - c n eonverge to 0, it
190
Chapter VII. Factoring polynomials
follows
f
that
f (on) \ 0
f
(en) -
f
and
(on)
Let f E lll[X] be of odd degr~ee.
4.200ROLLARY. As f
PR!XlF.
4.3 LEMMA. (e + d!)2 =
0
-
0
O.
As
Tllen f has
CL
If
+ bi
0
E
(["1,
0
in IR.
r~ool
and b
Cl
0
then there exist
+ di
C
in (La so that
+ bi .
= O.
We first prove the real case, b
>0
0 or a
=
X2
to
is of odd degree there are rational numbers
satisfying the conditions of (4.1).
o
converge
f (b n ) for each n, it follows that F(c n ) converges to O.
~
PROOF.
r (c n )
(b n ) -
or
0
< O.
If a
As ~~ is discrete, either
> 0, then by Lemma 4.1 the polynomial rf" < 0 then i k is a
has a root in IR", which we denote by /;;.
root of X2
For the general case we take c and d to be roots of
".
-
and Then c + di is the desired square root of Ta prove
the
final
theorem we will
Cl
+ bi.
need
0
two
resu1t.s about
fini te
groups. (i) If C is a finite group, and 2n divides #G, then G contains a subgroup of order 2n . (ii) If
G is a finite group, then G has a subgroup T, such that
#T is apower of 2 and the index of T in G is odd. Any subgroup T of G satisfying (ii) is called a 2-·Sylow subgraup.
are
standard
resul ts
in the
theory of
finite
groups,
These
and present no
constructive problems. 4.4 LEMMA. PR!XlF. in
a::"'.
TI,en
r
hHS
LI
r~oot
Let E be a splitting field of f over lll.
Let T be a 2-Sylow subgroup of G, and let K be the fixed field
is equal to the index of T in G.
a::"'.
It suffices to embed E
By (VI.5.5) there is an element a in K so that K
satisfies an it:reducible polynomial 9 of degree in
in a::"'.
As E is finite dimensional over lll, the Galois group G of E over III
is finite. of T.
Let fE lll[Xj.
dim~,
lll[aJ.
Then a
which is odd as it
By (4.2) the polynomial 9 has a root 8
By Lemma VI. 8.1 there is an embedding er:K
Suppose #T = 2".
=
->
a::'" that takes " to
By (i) we can find subgroups TOr'"
r T I1
= T
fJ.
such
that IIT i = 2i.. Let Ki be the fixed field of Ti' We shall inductively extend a, which is defined on ~" to an embedding of KO = E into (Ln.
191
4. The fundamental theorem of algebra
Suppose a is defined on Ki +1 . The dimension of Ki over Ki +1 is 2, so where f3 satisfies an irreducible quadratic polynomial h over
K i = L [f3)
Ki +1 . By Lemma 4.3, and the quadratic formula, ha has a root in~, so we can extend a to K i by (VI.8.1). 0 4.5 'lmDREM ('1he discrete fundamental theorem of algebra). ~
PROOF.
has a
Let F E ~[X) be a nonconstant polynomial; we will show that F
root in ~.
coefficients of F.
Let K be the subfield of ~ generated by the By Theorem VI. 3.3, construct a countable diserete
field containing K that contains a root 8 of f, and let 9 E a root. g. ~.
The field
is algebraically closed.
~[X)
have 8 as
Induetion on the degree of Fallows us to assume that f divides
By Theorem VI.3.4, construet a eountable splitting field E for 9 over As E is a finitely generated separable extension of
a E E so that E
= ~(a).
let f3 be a root of h in ~. ~(f3),
=E is normal.
as ~(f3)
~,
we ean find
Let h E IQ[X] be the minimal polynomial of a, and Then As
9
is a product of linear factors in
F divides
g, there is a root of f
in~.
o A proof of the fundamental theorem of algebra, in the form that every
monie polynomial of positive degree over
~
has a root
in~,
is outlined in
the exercises.
EXERCISES
1. Let F be a monie polynomial in
~[X)
of degree
for each x E ~ there exists a root r of f
n
> O.
Show that
in ~ such that
Ir - xl n ~ IF(x)l.
2. Let F be a monic polynomial in ~[X) of degree n > O. (i) Use Exercise 1 to show that for eaeh x E [;, and
Eo
> 0,
there exists rE 1Q(i) such that IF(r)1 < Eo and Ir - xl n <
(ii)
If (x) I + Eo. Use (i) to construct a Cauchy sequence r l l r 2 such that f(r i in [;.
)
converges to O.
, •••
in lQ(i)
Conclude that F has a root
192
Chapter VII. Factoring po1ynomia1s
Kronecker (1882) showed that algebraic number fields are faetorial. The result that finite separable extensions of faetoria1 fie1ds are faetoria1 is found in [van der Waerden 1953], but the proof is ineomplete (see the diseussion in [Mines-Riehman 1982]). An example showing that the separability eondition is neeessary (Exereise 1.5) [Seidenberg 1974). Condition P was introdueed in [Seidenberg 1970].
first appeared in In that paper it was
shown that finite-dimensional, and purely transeendental , extensions of (faetorial) fields satisfying eondition P also satisfy eondi tion P (and are faetorial). Theorem 3.1 is from [Riehman 1981). Brouwer and de Loor (1924) gave a eonstruetive proof that every noneonstant monie polynomial over ~ has a root in~. Bishop (1967) proved this assuming only that the polynomial had a nonzero coefficient for some positive power of
x.
Chapter VIII. Commutative Noetherian Rings
1. TI:1E HILBER'!' BASIS 'lmDREM
The Hilbert basis theorem states that R[X1 is Noetherian whenever R iso No one has given a constructive proof of this theorem for our present definition of Nae theriun,
but other definitions have
led
to proofs.
Standard classical proofs of the Hilbert basis theorem are constructive, if by Naetherian we mean that every ideal is finitely generated, but only trivial rings are Noetherian in this sense from the constructive point of view.
The first proof of a constructively interesting Hilbert basis
theorem was given by Jon Tennenbaum; we will present same of his ideas in Section 4.
In this section we prove the Hilbert basis theorem for
coherent Noetherian rings; in the classical context these are just the Noetherian rings. If R is a ring, then we denote the R-module {F E R[X1
: deg F
< n) by
R[Xl n .
ClearIy R[Xl n is a rank-n free R-moduie. If I is a left ideal of R[X), then I n R[X 1n = {F E I : deg F < n). If M is an R-submodule of R[Xl n , then XmM is an R-submodule of R[Xln+m' 1.1 LJ!2IIMA.
Le t R be u caherent Naetheriun ,'ing und Le t I be the LeF t
af R[Xl generated by F1 , .... Fs • IF Fi E R[Xl n Far euch i, then there is a FiniteLy generated R-maduLe M !;;; R[X1 n such that XM n R[X1 n !;;; M und I n R[X1 m = XiM Far euch m Z n. ideuL
tt:o
PROOF.
Corollary III.2.8 says that R[X1 m is a coherent Noetherian
R-module.
Construct a
submodules of I Nk +1 = Nk
n n
R[X1 n . As R[X1 n +1 is coherent, the modules Nk are finitely generated; as R[X 1n is Noetherian there is I~ such that Nk = Nk +l . Set M = Nk • C1ear1y XM n R[Xl n ~ M. As M !;;; I
+ XNk
chain NI !;;; N2 !;; ••• of finitely generated R [X 1n as fallows . Let N1 = Rf 1 + ••• + Rf s' and let
n R[Xl n
that I n R[Xl m ~ where gi
E
we have
tt:'O
XiM.
tt:ö
n R[Xl m for each m Z n. To show EIn R[X1 m. Write f = 21'=1 gJu
XiM !;;; I
suppose F
R [X ld for each i, and proceed by induction on d.
193
If d = 1.
194
Chapter VIII. Comrnutative Noetherian rings
€ M and we are done. If d > 1, define h i E R[X] by 9i = gt (0) + * s d Xh i an set f = 2 i =1 hifi € I. Note that h i € R[Xl d _1. Then F s 2i=1 9i(Olf i + Xf * , so Xf * € In R[Xl m whence f * E R[Xl m_ 1 . If m = n, then induction on d gives f * E M so Xf * E XM n R[X]n ~ M whereupon f E
then f
2T:6-n
NI + M = M. If m > n, then induction on d gives f* E NI + Xt!l-l-n XiM C 2~-fl XiM 0 1=0 1=0 • 1.2
integer-.
so f
is a finitely generated Left ideaL of R[XJ, und k is a
If I
positive integer,
n
particuLar, I
E
R be a coher'ent Noetller-ian r'ing and m a positive
Let
THEX)REM.
XiM,
then In R[Xl h is a finitely generated R-moduLe. R is a finiteLy generated Left ideaL of R.
In
PROOF. Let n L hand M I n R[X]n be as in Lemma 1.1. Then I n R[Xl k M n R[Xl h is finite1y generated because R[Xl n is coherent and M is finitely generated. 0 1.3 LEMMA. ring.
If R is a coherent Noetherian ring, then R[X] is a coher-ent
IF, in addit ion, R has de tachabLe LeF tideaLs, then so does R [X].
PROOF.
Let I be a finitely generated left ideal of the po1ynomia1 ring
and let f E R [X I .
R [X]
Choose n such that R [X In contains fand a finite
family of generators of I. generated.
By Lemma 1.1 we have I
n R[Xl n = M is
finitely
If R has detachable 1eft ideals, then M is detachable from
R[Xl n by Corollary 111.2.8, so we can decide whether f
€
I
n R[Xl r1 ,
whence
I is detachable from R[X]. Let g1' ... ,9h generate M n R[X]n_1' set gh+i let
= Xg i for i = 1, ... ,h, and M. Then g1, ... ,ge generates I as an Let e1, ... ,ee be the natural R[X]-module basis for R[XIl', and
g2h+1, ... ,ge
R[X]-ideal. let
generated
be the rnap from R[X]e onto I
that takes e i
construct a finite family of generators for her
to 9i.
We shall
thus showing that R[X]
is coherent. the R-module Re n her-
As R[Xl n is a coherent R-rnodu1e,
generated.
Let K be the R[X]-submodule of R[X]e
elements together with the elements Xe, - e"+i for i show that K = her Suppose that de9 r i
2f =1 r- i 9i ~
Clear1y K
=
~
her
1, ... ,h.
We proceed by induction on m.
each "i is in R, so we are done as Re n her Si + 0i Xm , where 0i € R and deg Si
< m.
~ K.
is finitely
generated by these =
2f =1 " i eiE K.
0 i we shall show that
m for each i.
If m
We shall
Choose m so If m = 0, then
> 0, write r i =
From}; si 9 i + Xm}; aig i = 0
we
195
1. The Hilbert basis theorem
e
'\ L i =l
is in K.
t
mI =1
+
s·e· t
X
Z
i.
b.e. I.
,
Because Xei - eicH is in K, for i = 1, ... ,kr it suffices to show
that
is
in K,
but this
pre5ented.
is true by induction on m.
Thus I
is
finitely
0
If R i5 a ring, and I i5 a left ideal of R[X], define L(I)
to be the set of formal leading coefficients of the polynomials in I. Note that a polynomial may have different fornal leading coefficients (one of them zero) depending on how it i5 written, and note that L(I) is a left ideal of R. 1.4 LEMMA.
Let R be a coherent Noetherian ring, and I a finitelv
generated left ideaL of
Let J
generated. let rn be a
~
R[XJ.
Then the
left ideaL L(I)
oF R
is finitely
I be a left ideaL of R[X] su.ch that L(I)
LU), and
lf I n R[Xl m generates I, then J n R[Xl m
posi.ti.ve integer'.
J.
gener'u t es
Let n and M = I n R[X ln
PR<XJF.
be as in Lemma 1.1, and let Ln (I) !;;
L(I) be the set of formal leading coefficients of polynomials in M.
n
XM
R [X ln
so is Ln (I).
=
Ln(I) f m-1
E
L(l).
Ln(I).
the
M,
set Ln (I)
As
the image of the map taking each polynomial in M to its coefficient of Xn - l . As M is finitely generated, ~
i
5
We complete the proof of the first claim by showing that
r = fm_1xm-1 + ••• + fIX + Fa E 1. If > n, then f E il':'ü xiM, so we can find
Suppose If
In
In
<; n, then
gE
M with
formal leading coefficient f rn-I' whence f m-.l E Ln (I). Now suppose J and m
In
are as in the second claim.
n, rechoosing n and M if necessary.
=
Let f
We may assume that
f d_1xd - 1 + '"' + fIX + f 0
be in }; we shall show that f is in the 1eft ideal generated by } n R[XJno If cl
s:
n, then F € }
we can find 9 f -
0-«g
€
=
n
R[X]no
9n_lxn-1 +
I f cl 0.0
> n, then, since Ln(I)
+ 91X + 90
E
~
L(I)
= L(J),
M with 9n -1 = f d - I "
Then
J n R[Xl d _ 1 , so is in the left ideal generated by J n R[Xl n by
196
Chapter VIII. Commutative Noetherian rings
induction on d. o
Therefore f is in the left ideal generated by J
1.5 'lHEDREM (Hilbert basis theorem). ring,
then so is R[X).
If,
n R[X)n'
R is a coherent Noether-ian
If
in addition, R has detachabLe
Left
ideaLs.
then so does R [X) •
PRODF. Lemma 1.3 says that R[X) is a coherent ring, and has detachable left ideals if R does. It remains to show that R[X) is Noetherian. Let ~ 1 2 ~ • • • be a chain of finitely generated left ideals of R[X). We construct positive integers v(l) < v(2) < ••• and nonnegative integers n (1) ,n (2), ••• as follows. Let v (I) = 1. If v (m) has been constructed,
11
choose v(m-1)
such that Iv(m) n R[X)n(m) generates the left ideal Iv(m)' If and n(m-l) have been constructed, use the fact that R[X)n(m_l) is
n(m)
Noetherian, and 1 i n R[X)n(m_l) is finitely generated for each 1.2) to choose v(m) greater than v(m-l) such that Iv(m)
The left ideals
n R[X)n(m_l)
I v (m)+1
n
(Theorem
R[X~l(m_l)'
L(I v (I)) ~ L(1 v (3)) ~ L(1 v (5)) ~ •••
ated by (1.4) so we can find m such that
i
L(I v (m_l))
are finitely gener= L(1 v (m+l))'
Then
also L(Iv(m_l)) = L(Iv(m))' But 1 v (m-l) n R[X)n(m_l) gene rates 1 v (m-l)' so Lemma 1.4 teIls us that 1v(m) n R[X)n(m_l) generates 1v(m)' Similarly 1 v (m)+1
n
generates
R[X)n(m_l)
Thus
1 v (m)+I'
1v(m) = 1 v (m)+I'
In particular, ~[Xl"",XnJ is a coherent Noetherian detachable ideals, as is k[X 1 , ••• ,Xn I for I< a discrete field.
0
ring
with
EXERCISES 1. Let G be the abelian group
~ ffi ~,
and let P be the submonoid of
G consisting of those pai rs (m ,n) such that m > 0, or m = 0 and n ~ O. Consider the monoid ring I< (P) where k is a discrete field.
As
P is additive we write elements of k(P) as formal sums
2: a yP where P € P and P
submonoid of
k
Cl
P
€
(P) of elements
h. 2:Cl p
Let S be the multiplicative YP wi th
Cl
(0,0)
"I
0, and let R be
s-lk (P).
(i) Show that R is a ßezout domain with recognizable units, in fact a valuation ring, hence a coherent ring wi th detachable ideals.
197
1. The Hilbert basis theorem (ii) Let I
be the ideal in R[X] generated by the elements Show that neither I n R nor
1 + y(O,l)X and y(l,O).
n
R[X
lz is a finitely generated R-module.
2. Show that every ideal in
z: [X 1
that is generated by a finite
number of polynomials of degree at most 1 is either principal or uniquely of the form
o~
b
< c.
a(X
+ b, c) with a and c positive, and
Develop a similar theorem for ideals generated by a
finite number of polynomials of degree at most 2.
2. IDETHER NORMALlZATIGI A homomorphism 'p
AN!)
THE ARTIN-REES LEMMA.
R .... S of commutative rings is said to re fleet
:
finitely generated ideals if ~-II is a finitely generated ideal of R for each finitely generated ideal I of S. coherent commutative Noetherian finitely R
-t
generated
R [Xl' ... 'X n 1.
ideals;
Corollary 1.2 says that if R is a
ring,
and
by
then
the map R .... R[X]
i nduc ti on ,
so
does
reflects the
map
Classically this is not a very exciting resul t because
eveqj ideal of R is finitely generated; but i t is a very useful tool for
constructing generators
for
ideals.
In
this
section we
give
a
few
important applications. First
we
note
that
the
remainder
theorem
admits
the
following
generalization to polynomials in several variables. 2.1 LEMMA.
Let R be a commutative ring, emd
~
a r-ing homomor-pllism from
RIXI, ... ,Xnl to R tM.!. Ls the identity on R (lnd takes Xi her' ~ =
1.0
0. 1
'
Then
(X I -{lI,X 2 -QZ' ••• ,Xn-{ln).
Consider the ring endomorphism 8 of R[X I , .. . ,X,,] that is the identity on Rand takes Xi to Xi - Gi. The map ~8 takes each Xi to 0, so PROOF.
its kernel consists of those polynomials with zero constant term, that is, the ideal (Xl'" .Xn )· Thus her-
But 8 is an automorphism, with fJ-1x.1 =
(XI')I'··· ,Xn-a n )·
Xl +
(1
i
~
0
We use (1.2) to compute relations in polynomial rings module finitely generated ideals; that is, given polynomials PI'."'Pn , and an ideal I, we construct generators for the set of polynomials f such that f(PI" .. 'P n ) E 1.
198
Chapter VIII. Commutative Noetherian rings Let R be a coher'ent
2.2 THEDREM.
'" : R[X1, ... ,Xml ident i ty on R. PROOF.
.... R[Y1, ... ,Ynl
commutatiue Noelher'ian r'ing.
be
Then '" ,-eflects fini
a
ring
horrwmor'phism
the
teLll generated ideats.
Extend '" to a map '" * horn R [X 1.' ... ,Xm , Y l' ... , Yn 1 by defining Then
= Yio
",*
the
is
identity on R[Yl,o .. 'yn l ,
(X l -
Let
is
that
so
]
n
R[X1"",XmJ
R[X1, ... ,Xml by (1.2).
",-Ir
=
(",*)-1 J
=
a
is
so ker / '
If I 15 a finitely
is a finitely generated
finitely
generated
ideal
of
0
Let k be a discrete field.
A commutative ring R containing k
1S said
to be finitely presented over k if R is isomorphie to k[X1, ... ,Xnl/I for some finitely generated ideal I.
If Rand S are two commutative rings
eontaining the discrete field k, then a k-algebra map is a ring map from R to S that is the identi ty on k 2.3 COROLLARY.
0
If R ls n finite!y pr'esented commutatiue "[ng ouer I"
then nny k--<J.tgebr·a map fr'om k[X1, ... ,Xnl into R Iws a finiteLy generated kerne! . PROOF.
«[Xll ... ,X"l
Let R to R.
=
h[Yl"",Yml/I,
and let
Then'P lifts to a map
re fleets fini tely generated ideals by (2.2). kernel of "'I is finitely generated.
'P
be a k-algebra map horn
t into
k[Yl,oo.,Yml,
whieh
Thus t - 1 I, which is the
0
Any finitely presented commutative ring over a diserete field k ean be viewed as an integral extension of a polynomial ring over h. 2.4 THEOREM (Noether normalization)
Let
0
R
fini lel.y p,'esented commutati.ue ring ave,' a disCl'ete
field I,.
exist zl"",znE R, und m <:11 , such that R=1<[zl" . .,z"l
Ther; the,'e is
integral
ouer' I;,[zl"o"zm], and zl, ... ,zm ar'e a!geb,'oically independent oue,",PROOF.
We proeeed by induetion n.
independent, which is decidable by (2.3)
If xl"" I
,x"
are algebraically
then we are done.
So we may
assume there is a nontrivial relation ):
where
j
is the n-tuple
ajx~l ••• x,~n
h, ... ,j"
and a j
Let d be an integer such that d >
j i
0,
0 for all but finitely many
for all ( and
j
wi th a j f. 00
j.
Let
199
2. Noether normalization and the Artin-Rees lemma
Yj
dn - i
= xi - xn
dn - i
< n,
and substitute Yj + xn Expanding the resu1t, we get
equation.
' for
i
2: ajx n
j*
+ f(U1"",Un-1'x n
for
xi
in the above
0,
) =
j1dn - 1 + j2dn-2 + ••• + .in' and the degree of x n in each term of f is less than j* for some j. Si.nee the j* are all distinet, this is
where j*
=
an integral equation for x n Over k[YI"" 'Yn-l]' So R = klu1"" 'Yn-1'xn ] is integral over IZ[Yl, ... ,Yn _ l l, which is a finitely presented ring by (2.3).
By induetion I<[Yl""'Yn-l1
k[zl"",zn_1 J is integral over
=
I<[Zl"",zm] with zl"",zm algebraieally independent over k.
A diserete
Set
Z
=
n
held K eontaining k is said to be a finitely presented
extension Held of k if K is the field of quotients of a ring that is finitely presented over k.
Noether normalization gives us a way to
eonstruct a particularly niee transcendence basis for a finitely presented field extension. 2.5 CORQLUlRY. pr'esenled over k.
Let k
K be discr'ete fields such that K is finitely
~
Then ther'e is a fini te tT'anscendence basis B of K over'
I< such that K is finite dimensionLll Quer k(B).
PROOF.
Let K be the field of quotients of the finitely presented
eornmutative
ring
R.
By
Noether
normalization
we
ean
write
R
as
h[X1""'X"'Y1""'YsJ;P = k[X,Y];P where I<[X] np = 0 and R is integral over k[X].
Then K = k(X)[YI!(k(X)P)
Iz(X)[Yl""'Y s ]
=
presented, and algebraie, over the field Iz (X) is hni te dimensional over Iz (X)
I
•
is
finitely
It suffiees to show that K
that is, to prove the eorollary when t.he
t.ranscendenee degree r is O. Suppose r'
=
0, so K
=
k[Y1, ••• ,Ys];P
k[Yl""'Y s ]'
The subfield
k [Y s 1 is a finite dimensional vector spaee over k by (2.3).
k[Y1'.'.'Ys-1,ysl;P', where P'
'l'hus K is finitely presented over the Held I< [Ys J, dimensional
over
dimensional over h.
k [!}s J
by
But K
=
is t.he image of P in /dY1, ... ,Ys-1'.'I s ]'
induetion
on
s.
and so is finite
Therefore
K is
0
We also get t.he following result. about fully factorial fields.
finit.e
200
Chapter VIII. Commutative Noetherian rings
2.6 'l'HEXlREM.
Let
K be a Finitdy pr"esented fidd extension oF the
IF k is Fully Factorial, then so is K.
discrete Field k.
Let B be a transcendence basis for K over k.
PROOF.
Then k(B) is
fully factorial by (VII.3.4), so K is finite dimensional over k(B), hence fully factorial.
0
As another application of (2.2) we prove the Artin-Rees lemma. 2.7 'l'HEXlREM (Artin-Rees).
r be a Finitdy generated ideal of a
Let
coherent commutative Noether"ian ring R.
Let N be a
submodule oF a fini tely presented R-module M. For all n
k we have
~
By passing to the ring R ffi M, where m, m2
PROOF.
may assume that M is a finitely generated ideal of R. and
Finttely generated
Then there is k such that
0 for m" m2 E M, we Let I = (a1, ... ,am )
: R[X 1 , ... ,Xm] ~ R[Y] be the identity on Rand take Xi to aiY' The kernel of ~ is finitely generated, by (2.2), so the image R[IY] of ~ is a ~
finitely presented R(X 1 , ... ,X m]-module, hence a coherent Noetherian ring. Now N[Y] is a finitely generated R[Y]-ideal, so R[IY] n N[Y] = ~~-l(N[Y]) is a finitely generated R[IY]-ideal by (2.2), whence M[IY] finitely
generated
R[IY]-ideal
as
R[IY]-ideal and R[IY] is coherent. Choose
so
k
n
2~=0 (IiM
that
M[IY)
N)y i •
n
N[Y)
is
M[IY)
is
But M(IY]
n
generated,
a
n
finitely
N[Y]
N[Y) = 200 i =0 (Ii M
as
is a
generated ) i n NY.
an R[IY)-ideal,
by
0
The ring R[IY] in the preceding proof is known as the Rees Ring.
2.8 'l'HEXlREM (Krull intersection theorem).
Let
M
be
a
Finitely
presented module over a coherent commutative Noetherian ring R, and let I
be a Finttely generated ideal of R. a E A, so IA
we have InM n N N = IN.
Then a E Ia for each
= A.
Let N
PROOF.
Let A = ~rnM.
=
= Ra.
By Artin-Rees there is k such that for all n
rn-k(IkM
n
N).
But N !;; rnM so, taking n
=
~
k
1<+1, we get
0
2.9 COBOLLARY.
Let M be a finitdy presented module over" a coher"erlt
commutative Noetherian ring R, and let I be a Finitely generated quasiregular ideal of R.
Then ~InM
=
O.
201
2. Noether normalization and the Artin-Rees lemma If a E r\inM, then a E Ia by (2.8). I, whence (1 - A)a = 0, so a = O. 0
PROOF. A E
Thus a
Aa for some
EXERCISES 1. Let R be a discrete field and R = R[X 1 ,X 2 ]/(X 1 X2 R as in the Noether normalization theorem. 2. Krull intersection
).
Find
2 1 ,2 2
in
a la Herstein. Let R be a commutative ring, I
a finitely generated ideal of R, and r E I. Let M be a finitely presented Noetherian R-module, and K and N fini tely generated submodules of M such that K n N = IN. (i) Show that L n (,-) = {x E M : ,-'lX E K} is an ascending sequence of finitely generated submodules. (ii) Show that i f Ln(r) = Ln +1 (r), then (rnM + K) n N = IN. (iii) Show that K is contained in a finitely generated submodule K' of M such that K' n N = IN and InM !;; K' for some n. (iv) Use (iii) to prove the Krull intersection theorem. 3. TßE NULLS'l'EI.I.mSA'I'Z
Let R be a discrete field and K an algebraic extension field of R. If a E Kn , then the natural map of k [X] = h [X l' ... ,Xn ] onto the field h[a1, ..• ,an ] has as its kernel M
(F
E h[X]
: F(a)
= O},
so M is a detachab1e maximal ideal. In this section we will show that, under suitable restrictions on h, every proper finite1y generated ideal of h[X] is contained in such an M, and M is finitely generated if K is finite dimensional. 3.1 THEDREM.
Let h !:;; K !:;; E be discrete fieids such that E is finiteiy
presented over K, and K is Finiteiy presented over R.
Then E is Finitety
presented over h. PROOF. Let X = (Xl, ... ,Xm) and Y = (Y 1 , ... ,Y n ) be sequences of indeterminates, and let x E Km and y E En be such that the maps h [X] .... h(x) = K and K[Y] .... K(y) = E have finitely generated kernels. Then each of the maps
202
Chapter VIII. Commutative Noetherian rings IdX,y] """ r,[x,YJ
->
k(x)[YJ
-> h(x)(y)
reflects finitely generated ideals; the first map is onto and its kernel is generated by the kernel
of "[X]
->
/< [x
J,
the
second map
reflects
finitely generated ideals by (2.2) because the localization k[x]
--,> /,(x)
does, and the third map is onto with a finitely generated kernel.
Thus
the composite map reflects finitely generated ideals; in particular, its kernel is finitely generated. We
have
seen
that
0
finitely
presented
extension
fields
aLe
purely
transcendental extensions followed by finite-dimensional extensions (2.5). The converse follows from (3.1). 3.2 COROLIARY.
Let
pr-esented ouer' k if
k r;;; K be
tllen'
is a
discr-ete
finite
fieLds.
Tlten
K is
finiteLy
tr'onscendence basis B oF K ouer- I,
such tllat K is (ini tp dimensional Olle,- h(B).
PRCXJF.
Suppose such a transcendence basis B exists.
finitely presented over /"
If K = h we are done i otherwise choose x E K\], and
dimensional over h. let f
E /,
[X] be an irreducible polynomial satisfied by x.
k [X J/( f) is fini tely presented over /" k (x)
by induction on dimension,
(3.1).
Clearly h (B) is
so by (3.1) we may assume that K is finite Then k(x)
~
and K 1S fini tely presented over
so K is finitely presented over h by
0
From (3.1) and (2.3) it follows that if K is finite-dimensional oveL /<, and 0: E Kn , then (f E "[X J : f (0:) = O} is a finitely generated ideal. Let I
be a proper finitely generated ideal of the polynomial
IdXl, ... ,XnJ.
ring
If we want to construct a finde1.u genero:ted maximal ideal
containing I, we need
I~
to be factorial, even if n = 1.
However, we can
get the following weaker result without being ahle to factor.
3.3 LEMMA. gene,-ated
idenl_
Let
I,
be
0:
di.scnete
k[X1, ••• ,X n ] =
oF
fjeld,
h[X].
am! Tilen
T a the,-e
p,-ope,'
hni telu
a
Finitel~j
is
gener-ated p,'oper- ideal] ;;J I sucl, Ihnt I'[XJ/.1 is i/llegr-ol Olle'- I,.
By Noether normalization there are U1' ••• ,U r-
E
k [x 1, algebraically independent over ", such that
h[x] is integral over k[Yl'.'.'YrJ.
We shall show that u1 is not a unit
in k[x], so we can r-eplace I by I + (Y 1) I where Y1 E h [X J maps onto Ul' and we are done by induction on r.
203
3. The Nullstellensatz Suppose zYI ;
I
for
some z
in k[xl.
As
k [x I
is
integral over
k[YI, ..• ,yrl, there exist a i € k[YI, ... ,yrl such that
L!rl + a m-I zm-I + ••• + 00 ; 0 hence I + am-IYI + ••• + aoY~ ; 0, but
this
says that YI
is a unit
algebraic independence of Yl' .•. 'Y r .
in k[YI'''.'Y r ),
contradicting the
0
A discrete Held k admits splitting fields i f for each polynomial f
k [X I
€
there
is a
discrete
splitting field
for
f
over k.
Fully
factorial fields and countable fields admit splitting fields. 3.4 LEMMA.
Let k be a discrete field admitting splitting fields, and
let I be a finitely generated pr'oper ideal of k[X I , ... ,Xnl ; ldXI.
Then
there 1S an al.gebraic extension fiel.d K of k such that tlle pol.ynomials in I haue a common zero in Kn • PROOF. k.
For
By Lemma 3.3 we may assume that k[xl ; k[XI/I is integral over each xi
there
is
a
polynomial
fi
€
k [Y I
such
that
Let K be a splitting field of f '" ITif i' and let J be the
f i (xi) ; O.
ideal generated by I in K[XI. (lII.3.3).
monic As I
nk
;
Let fl(Y) ; ITi(Y - ail in K[YI.
0, the ideal J is proper by As fl(X I ) € I,
J ; ] + IT i (XI-a i ) d IT i (] + (XI-aill,
so ] + (XI-ßI) t- K[XI for some root ßI of fl.
Replace] by J + (XI-ßI)'
and repeat the above procedure for f 2' .. . ,f m constructing a proper ideal
] + N, where N '" (X1-ßI, ••. ,Xn-ßn). Nd
J,
As N is a detachable maximal ideal,
and ß ; (ßI, ... ,ßn ) is a common root for the polynomials in
also for the polynomials in I.
J,
thus
0
3.5. THEOREM (NUllstellensatz).
Let k be a discrete fiel.d that admits
splitting fiel.ds, Ia finitely generated ideal of k[XI ; k[X 1 , ••. ,Xn ), and f € k[XI. Then either f E ../1, or there exists an o.lgebraic extension field K of k, and an el.ement a € Kn , such that f(a) t- 0 but g(a) ; 0 for each 9 € 1. PROOF.
Consider the ideal I' of ',-(X,YI generated by I and 1 - Yf.
If
I' is proper, then by Lemma 3.4, there is an algebraic extension field K of k, and elements a € Kn and ß € K such that 9 (a) ; 0 for each 9 € I, and 1 - ßf(al ; O.
If I' '" k[X,Y), then there are polynomials h i E k[X,Y) and
204
Chapter VIII. Commutative Noetherian rings
gi E I such that
1 - h O(l - YF) + 2 i >0 hig i . Substitute Y
Fr
Then
E
=
l/F and choose r- greater than the degree of Y in any ll i .
I, so F E
JI.
0
EXERCISES
1. In the proof of (3.3), use Exercise VI.I.I to show directly that
J - I + (Y 1 , •.• ,Y r ) is the desired proper ideal, where Yi maps to
'h· 2. Show that i f k is a discrete fie1d admitting splitting fields, and I is an ideal in k [X l' ... ,X r1 ] , then
JI
is the intersection of
the detachable maximal ideals containing I.
3. Let k
~
K be discrete fields with K algebraically closed.
k-variety in
Kn
is a set of the form {x
1, .. ,n} where F1, ••• ,F n are in k[X] a k-variety V is
{FEk[X]
E Kn
:
Fi (x) - 0 for
= 1{[X 1 , ... ,Xn ].
F(v)=O
for
A
The idealof
each vEV}.
A
k-variety V is irreducible if whenever V is the union of two k-varieties A and B, then either V
=A
or V - B.
(i) Show that the ideal of the k-variety corresponding to
F1, ••• ,F n
is the
radical of the ideal generated by
F1, ••• ,F n , and is detachable.
(ii) Show that a k-variety is irreducible if and only if its ideal is prime 4. TENNENBAlJM' S
~
TO '!'HE HILBERT BASIS THEDREM
Let R be a ring and M a discrete R-modu1e.
A Nöetherian basis function
for M is a function 'f' taking Mn to Rn-I, for n - Z, 3, . .. , such that if xl'xZ' ...
is an infinite sequence of elements of M,
arbitrarily
large
n
such
that
xn =
Irr:! rix i ,
then there exist
where
(r1, ...
,r- n _1)
=
'P(xI'··· ,xn )·
n-l r-ix as te h Ne often refer to the function p(x1' ... 'x n ) - x n - Li_I i basis function, suppressing reference to 'P which appears implicitly when
we wri te xn - I~:i
basis
Function
r- iX i
derives
as a consequence of p (xl' ... ,xn ) from
the
fact
that
P(x1)'
P(x1' ... ,x n ) generate the same submodule as xl' ... ,xn '
=
O.
The term
P(xI'xZ)' so P effects a
205
4. Tennenbaum's approach to the Hilbert basis theorem change of basis. We call a basis function P consistent if P (xl' ... ,xn )
=
0 whenever
P(xi(l), •.• ,xi(m)'xn ) = 0 for some sequence I S i(l) < i(2) < ••• < i(m) < n. Clearly any discrete module admitting a Noetherian basis function
admits a consistent one.
IF a discr"ete
4.1 'lmDREI'l.
R~oduie
M admits
a
Noetherian
basis
Function, then M is Noetherian.
PRODF. Let P be a consistent Noetherian basis function for M, and let I, ~ 1 2 ~ ••• be a chain of finitely generated submodules of M. Construct a sequence X"X 2 , ••• of elements of M, and a sequence all) < a(2) < ••• of
positive integers such that I j is generated by xa(j), .•• ,xa(j+l)_l. construct a sequence ß(1),ß(2), .•. of positive integers such that (i) a(j)
Now
ß(j) < a(j+l),
~
(ii) if P(xI",.,Xa(j)_l,xß(j» = 0, then P(xl, ... ,Xa(j)_I'x i ) = 0 for a(j) Si< a(j+l). There exists n such that P (xß (I) , .•. ,xß (n»
= 0, so
P(xl'··· ,Xa(n)-l,xß(n»
=
0
o
as P is consistent. a (n+l), so In = I n - l .
i f a(n)
i
~
<
0
The ring 7l oF integer"s, as a moduie ouer itsetf, admits a
4.2 'lmDREI'l.
Noetherian basis function.
PRODF.
Let
d
the
be
nonnegative
greatest
common
divisor
Xl,X2'·.· ,xn_l· If d = 0, set P(xI' ... ,xn ) = xn ; otherwise P(xl' ..• 'x n ) be the (nonnegative) remainder when xn is divided by d. 4.3 'lmDREI'l.
Let
submoduie of B. does B.
B
be
a
diser"ete
R~odute,
und
A a
of let 0
detachabie
IF A und B/A admU Noe therian basi s Func tions, then so
Moreouer the basis function For" B ean be chosen to extend the
basis Function for A.
PRODF.
Let
1f
denote the natural map !rom B to B/A, and let b l , .. · ,b n
be elements of B.
Let I = {j I' ... , jm} consist of those indices that PB/A(1fb 1 , ••• ,1fb j ) = O. For j E I, let
and set
j
such
206
Chapter VIII. Commutative Noetherian rings
a
=
so, letting (sl,s2, ... ,sm_1) bn
r"-
1
i =1
rr:'-b, 1
1
= ~A(aj1, ... ,ajm) we have
PA (a J, , ••• ,a J, ) 1 m
=
+
n
~-1
+ L
siaJ',
1=1
1
+
L,n-1 t=l
r~bi
•
Substituting for each a j the expression b j - l{:l"{b t we compute elements t i ERsuch that =
1
m
= (t1,···,t n _1) so
Set ~B(b1,···,bn)
~-1
+ L tib i • 1=1
PA (a J, , ••• ,a J, )
PB(b 1 ,···,b n )
= PA(aj1, ••• ,ajm)'
We may assume that ~B/A (xl"" ,xn _1'0) = (0, ... ,0), so i f all the b i are in A, then J = {l, ... ,n} and a j = b j for each j in J. Thus PB extends Given an infinite sequence b 1 ,b 2 , ... in B we get an infinite sequence a j1 ,a j2 , ... in A and PA(a j1 , ••• ,a jm ) = 0 for infinitely many m. 0 PA'
For M an R-module, coefficients in M.
let M[X ] be the set of polynomials in X wi th
Then M[X] is an R(X]-module in a natural way.
M = R the following is the Hilbert basis theorem for
For
rings that admit
Noetherian basis functions. 4.4 THEDREM.
IF
the
R--modu[e
M admi
ts a Noetherian basis Function,
then so does the R[X]--modu[e M[X]. PROOF.
In what follows we consider the degree of the zero polynomial
to be zero. than N.
Let M[X]N be the set of polynomials in M[X] of degree less
Let P be a consistent Noetherian basis function for M and let PN
be the basis function on the R-module M[X]N defined inductively by viewing M[X]N
91, ... ,9n as follows. i
< n.
For F1, ••• ,f n E M[X] define Let 91=F1 and suppose we have defined 9i for
as an extension of M[X]N_1 by M. Let N
=
max{l + deg gi
1
~
i < n},
and
define
gn
by
the
following iterative procedure. If deg Fn < N, then set gn = f n . If de9 Fn 2 N, let C j be the leading coefficient cf 9 j ' and let c be the leading coefficient of f n' If
P(c1, ... ,c n _1'c) ,t. 0, set 9n and
2'i":l T'ixe(i)9i'
= fn.
~(c1,
Otherwise let e(i)
=
Fn - deg gi Fn by Fn This lowers the degree of Fn , and we start again at the
(r1, ... ,rn _1)
... ,cn_1'c),
and
replace
deg
207
4. Tennenbaum's approach to the Hilbert basis theorem top of the paragraph.
Note that this construction guarantees that
P(C1""'C n ) # 0 if deg gn ~ N. Given f1, .•. ,f n E M[XJ, construct g1' ... ,gn as above, let N max{l + deg gi : 1 $ i < n}, and set p(f1,···,f n ) = PN(gl, ... ,gn)' If f1,f2"" is an infinite sequence in M[X], construct the sequence gl,92"" as above and let N(k) = max{l + deg gi : 1 $ i $ k}. For each k there exists n > k such that either N(k) < N(n) or p(F1, ••. ,f n ) = 0; indeed if lI"N(k) denotes the projection of exists n > k such that
M[X]
on
M[X]N(k)'
then there
PN(k)(lI"N(k)gl, ..• ,lI"N(k)gn) = O. Thus either
N(k)
< N(n) or
N(k)
= N(n) and
p(F1,···,F n )
= PN(n)(91, ... ,gn)
O. We can construct a sequence a(l) < a(2) < ••• of positive integers such that for each k (i) either N(a(k)) < N(a(k+1)) or p(f1, ••• ,F a (k+1)) (ii) N(a(k)) = N(a(k+1)) = ••• = N(a(k+1)-1).
0,
If c j is the leading coefficient of gj' then there exist arbitrarily large values of n such that P(ca(1)"" ,ca(n)) = O. N(a(n)) = N(a(n)-l), so p(f1, •.. ,f n ) = O. 0
But this can only happen if
EXERCISES 1. Show that any finite module admits a Noetherian basis function. 2. Show that any discrete module admitting a Noetherian basis function admits a consistent one. 3. Strike the phrase arbitrariLy large from the definition of a Noetherian basis function. Show that i f M admits a Noetherian basis function in this new sense, then it admits one in the old sense. Why require arbitrarily large n? It's used in the proofs of (4.3) and (4.4); can these be easily circumvented? 4. Let M be a Noetherian R-module with detachable submodules.
Must
M admit a Noetherian basis function? (probably not) 5. Call an R-module M a Tennenbaum module if there exists a discrete R'-module M' that admits a Noetherian basis function, a map ~ : R' ~ R of rings, and an epimorphism ~ : M' ~ M of abelian groups
Chapter VIII. Commutative Noetherian rings
208
such that
~(rx)
= ~(r)~(x)
for each r in R' and x in M'.
Show
that every Tennenbaum module is Noetherian. Show that the Hilbert basis theorem holds for Tennenbaum rings. 6. Let a be a binary sequence, and let I be the ideal in the ring of
integers 7L generated by the elements ann! • Show that 7L/l is a Brouwerian example of Tennenbaum ring that is not coherent. Is every Noetherian ring a Tennenbaum ring? 5. PRIMARY IDEALS A prime ideal in a commutative ring is analogous to a prime number-more precisely, to the set of all multiples of a fixed prime number. A primary ideal is like apower of a prime number, and the theorem that every number is the product of powers of distinct prime numbers has an analogy in more general rings, such as polynomial rings in several variables over the integers, in the Lasker-Noether decomposition theorem: every finitely generated ideal is a finite intersection of finitely gene ra ted primary ideals. In Sections 7 and 8 we study these Lasker--
rings.
Noe ther
In this section we present some basic properties of
primary ideals in a commutative ring. Let R be a commutative ring. An ideal Q of R is primary i f xy € Q implies x € Q or yn E Q for some n. Thus a detachable ideal Q is primary i f and only if RIO is a discrete ring in which every zero-divisor is nilpotent. 5.1 PBOPOSITICN. If Q is a primary ideal of a commutative r-ing, then
yfj"
is prime.
yfj", then (xy)n yfj" or y E yfj". 0
If xy
PROQF.
Thus either x
€
€
€ Q,
so x n
E Q
or ynm
If Q is a primary ideal of a commutative ring R, and
say that
Q
ideal
Let Pond Q be ideals of a commutative r-ing. belonging
to
P,
if
and
only
condi Hons hold.
(i)
(ii)
v6"
for some m.
=
P, then we
belongs to P, or P belongs to Q.
5.2 THEXlREM.
primary
E Q
Q
~
If r
P €
P, then r m E Q for some m
if
the
Then Q is a
following
three
209
5. Primary ideals (iii)
If rs E 0, then r E Q
01'
E P.
S
Clearly conditions (i) and (ii) are equivalent to Q ~ P ~ ~.
PROOF.
Suppose (iii) holds.
If rs E Q, then
l'
implies that r E 0 or sn E 0 for some n; ~ t;:; P,
let q E~.
Then qn E
qn-l E 0 or q E P;
Q
t;:; P for
some n.
To show that
By (iii) we have Thus ~
so by induction on n we obtain q E P.
Conversely, suppose
Q
sn E Q for some n;
but sn E 0 implies s E ~
holds.
As P ~~, this
E 0 or s E P.
thus Q is primary.
is primary and ~
=
P.
If rs E 0, =
P.
then
0
is a primary ideal beLonging to P.
5.4 R.
E Q or
I"
Hence condi tion (iii)
5.3 COROLIARY. Let 0, and Q2 be primary ideaLs beionging to P.
n Q2
= P.
~.
Then Q,
0
Let P be a detachabLe maximaL ideal of a commutative ring
Let 0 t;:; P be an ideaL such tllat pEP impiies pl1 E Q For some n.
Then
Q is a pI"imary ideaL belonging to P.
PROOF.
It suffices to verify condition (iii) of Theorem 5.2.
= 1.
maximal there exists x in Rand p in P such that p + xs that prt E Q. l'
=
Suppose
If s E P we are done, so we may assume that s E P.
rs E Q.
Then
rpn + yrs € Q.
(p+xs)n
=
pn + ys
= 1 for some
y
As P is
Choose n so
in R.
Hence
0
5.5 COROLIARY. Let P be a detachable maximal ideaL of a commutative ring R, and let 0 be an ideaL such tllat p n t;:; Q t;:; P for some n. a pr"imary ideal beLonging to P.
Then 0 is
0
The following proposition extends the characteristic property of a primary ideal from elements to finitely generated ideals. 5.6 LEMMA. Let Q be 0 primary ideal P =~.
of
0
commutotive
PROOF.
0
01'
Let I = (ol, ••• ,om) and I = (b1, •.• ,bn ). If b j E P for each j, then I t;:; P;
each i, so I t;:; Q. ~.
and Then
I t;:; P.
a i E Q or b j E P.
As 0ibj E Q, either otherwise ai E Q for
0
Let
Q
be
a
finitely
commutotiue ring R, and let P =~. of R.
R,
Let I ond I be finitely generoted ideols such that II t;:; Q.
either I t;:;
5.1
ring
generated
primary
ideal
of
0
Let I be 0 finitely generoted ideol
If Q:I is finitely generated, then either I t;:; Q
01'
Q:I is 0 primary
210
Chapter VIII. Commutative Noetherian rings
ideal belonging to P.
PROOF. As I(Q:I) ~ Q, Lemma 5.6 teIls us that either I ~ Q or Q:I ~ P. We shall show that in the latter ca se Q:I is a primary ideal belonging to P. We use the characterization of Theorem 5.2. We have just hypothesized condition (i) of that theorem, while condition (ii) holds because Q ~ Q:I. To check (iii) suppose rs E Q:I. Then rsI ~ Q, so either rI ~ Q or s E P; that is, either r E Q:I or s E P. 0 5.8 PROPOSITICW.
Let 'I'
P and Q be ideal.s of R'.
R .... R' be a map of commutative r-ings, and let If Q is detachable, then so is
If Q is
a primary ideal belonging to P, then
Suppose Q is detachable and xE R. Then xE
i f
E Q;
primary ideal belonging to
'1'-1 (P ) •
0
EXERCISES 1. Let k be a discrete field, and let R = kIX,Y]. Show that R is a coherent Noetherian ring wi th detachable ideals. Let P = (X, Y ) and Q - (X,y 2 ). Show that the inclusions p2 ~ Q ~ P are proper. Show that Q is a primary ideal belonging to P. Conclude that a primary ideal need not be apower of a prime ideal. 2. Let k be a discrete field, and let R = k IX , Y ,Z l!(XY - Z2). Let x,y,z ERbe the images of X,Y,Z in R. Show that P = (x,z) is a detachable prime ideal of R, but that p 2 is not primary. Compare with Corollary 5.5. 3. Let Q, and Q2 be primary ideals belonging to a detachable maximal ideal P. Show that Q, + Q2 and Q,Q2 are primary ideals belonging to P. Show that in the ring ~[X,Y,Z"Z2] the ideals Q,
= ((X,Y)3,
Z,X +
Z2Y)
and
Q2
= ((X,y)3, Z,Y
are primary belonging to (X,Y), but that Q, + because it contains Z,(X 2 - y 2 ) (Seidenberg).
Q2
+ Z2X)
is not primary
211
5. Primary ideals
4. Let 'fJ be a homomorphism fram a ring R onto a ring R', and Q an
ideal containing the kernel of belonging to P if and only if to 'fJ(P).
'fJ.
'fJ(Q)
Show that Q is a primary ideal is a primary ideal belonging
6. LOCALlZATIOO.
Let S a multiplicative submonoid of a cOIlUllUtative ring.
For each
R-module M we define the S-torsion submodule of M to be {x
E M :
sx
= 0
for some s in
S}.
It is readily checked that Ts(M) is indeed a submodule of M. If = TS(M), we say that M is S-torsion. The fo11owing theorem shows that the map R .... S-lR reflects finitely generated ideals if and only if TS(R/I) is finitely generated for each finitely generated ideal r of R.
M
6.1 THEOREM. Let S be a multiplicative submonoid oF a commutative ring R, Let r be an ideal oF R, and Let x E R. Then the FoUowing conditions are equivatent. (i) x/1 E S-1 r
,
(ii) sx E I For some s in S, (iii) x represents an eLement oF TS(R/I). PROOF.
Clearly (ii) and (iii) are equivalent. If (i) holds, then x/1 = y/S, for some y E rand s, E S. Therefore there exists S2 in S such that S2(S,X - y) = 0, so we can take s = S2S, in (ii). Conversely if (ii) holds, then x/1 = (sxJls E s-1r . 0 We say that a module M is S-bounded if there exists s in S such that sM = O. It is readily seen that any finitely generated S-torsion module is S-bounded. For S-torsion submodules of finitely presented modules over a coherent ring, the converse is true. 6.2 'lBOOREM.
Let
commutative ring R.
bounded, then
TS(M)
S be a muUiplicative
submonoid of a coherent
If M is a Finitely presented R-module, and TS(M) is S
is finitely generated.
PROOF. Let s in S be such that sTS(M) = O. Then TS(M) is the kernel of the endomorphism of M induced by multiplication by s, so TS(M) is
212
Chapter VIII. Commutative Noetherian rings
finitely generated by (111.2.2) and (111.2.6). 6.3 LEMMA. R.
0
Let S be a muUipiicative submonoid of a commutative ,-ing
Le t M' be a submoduLe of the R-moduLe M.
If TS (M' ) and TS (MjM' ) are S
bOlmded, then so is TS(M).
If x E TS(M),
0 and tTS(MjM')
=
O.
then x represents an element of TS(MjM'), so tx E M' whence
tx E TS(M').
Therefore stx
6.4 'lHOOREM. I of R,
S be
Let
commutatlve ring R. ideaL
=
Choose sand t in S such that sTS(M' )
PROOF.
=
= O.
0, so we have proved that stTS(M) a
submonoid
muLtipUcative
of
a
0
coherent
If TS(RjI) is S-bounded for each finitel.y generated
(in pa,-ticul.ar,
if R .... S-lR reFLects finitdy generated
ideaLs) then
(i) TS(M)
is finitdy generated
fo,-
each finitdy
p,-esented
R-modu Le M. (ii) IF R has detad1l1bl.e ideaLs.
then so does S-lR.
PROOF. To prove (i) let xl, ... ,x n generate M and let M' be the submodule of M generated by xl"" ,xn_l' By induction on n we have TS(M' )
is S-bounded. The ideal I = {r ER: rXn E M'} is finitely generated because R is coherent and M is finitely presented. By hypothesis TS(RjI) is S-bounded, so TS(MjM' ) '" TS(RjI) is S-bounded.
Thus TS(M) is S-bounded
by (6.3), hence finitely generated by (6.2). To prove J
= s-l1
tTS(R/I)
S.
let J be a finitely gene ra ted ideal of s-lR.
(ii)
for some finitely generated ideal I of R.
=
O.
Choose t in S so that
If x E Rand tx EI, then xjs = txj(ts) E ] for each s in
Conversely, if xjs E
(6.1), whereupon tx E I. see if tx E I.
Then
J, then x/I E ] so s,x E I for some s, in S, by Thus we can tell whether x/s E J by checking to
0
6.5 'lHOOREM.
Let P be a Finitdy generated detachabl.e p,-ope,- prime
ideal. of a coherent commutative ,-ing R, and let M be a fini tdy pr-esented R-modul.e such that pllM
=
0 Far- some positive integer- n.
Then TR\f(M) is
finitely generated.
PROOF. pn
=
Let S
=
R\f.
As M is a module over R/pn, we may assume that
0, and proceed by induction on n.
If
11
= 1,
then P
=
0 so R is a
discrete integral domain and S consists of the nonzero elements of R. I
If
is any finitely generated ideal of R, then TS(Rj1) = 0 i f 1=0, and
6. Localization
213
STS(R/I) = 0 fm: any nonzero element s in I;
we
may decide which of t.hese
alternatives holds because R is diserete and I
1S finitely generated.
Thus TS(R/I) is S-bounded for any finitely generated ideal I, so TS(M) is bounded by (6.4), and hence is finitely generated by (6.2).
> 1, then the modules
If n
annihilated by pn-l.
PM
and M,lPM are finitely presented and
Thus TS(f'M) and TS(M,lPM) are finitely generated by So TS(M) is S-bounded by (6.3) and
induction on n, and hence S-bounded. therefore finitely generated by (6.2). 6.6 THEOREM.
Let S be a muLtiplicative submonoid or
Naetherian dng R.
Then S-lR is Noethedan.
Then we ean eonstruct a chain I , ~ 1 2 ~
ideals of R such that J j = s-l1 j In
the commutative
Let J 1 !:;; L !:;; ••• be a chain of finitely generated ideals of
PR(X)F.
S-lR .
0
= I n +1 , so J n
= Jn + 1 .
6.7 COROLJARY.
Let
for each
j.
of finitely generated
•••
There exists
Tl
such that
0
R be a coherent Noetherüm r'ing wUh detaclw.ble
Let P be a FinUeLY gener'ated pr'ime ideal oF R such that p n = 0
ideals. for same
Then Rp i.s a coher'ent Naetherüm dng with detachable ideals.
Tl.
PROOF.
Theorem 6.6 shows that Rp is Noetherian.
from Exercise III.3.4.
As pn
Coherence follows
= 0, Theorem 6.5 says that
TR\f(M)
fini te1y generated for each fini te1y presented R-module M, detachable ideals by (6.4).
is
so Rp has
0
See Exercise 8.5 for a strengthening of Corollary 6.7.
We turn our
attention to the behavior of primary ideals under localization. 6.8 LEMMA. R, and
Let 8 be a multipl icatiue submolloid of a cammutati.ve r'ing
Let Q be a
phrru:u-y ideal
x/i E: 8- 10 if und only iJ xE: O. 8- 1 0 ls
(1
PR(X)F.
or
R stich
that
Q
nS
is empty.
'fhen
If Q belongs to the pr"ime ideal P, then
pr-imar"y ideal belanging 10 the pr-i.me ideal 8- 11'.
Obviously x/I E: S-lQ if x C Q.
sx E Q for some sES by (6.1).
Conversely if x/I E: 8- 1Q, then
Thus x E Q or sn E Q for some
1l,
but the
laUer is impossible as Q Il S is empty. C1early 8- 1Q ~ s-l p and same power of each element of 8- 1P lies in 8- 10. Suppose (X/SI )(Y/S2) E 5- 10. Then sX~1 E Q for some s in 8, so either x E P, whence x/s, C 8- 1P, or sy E Q, whence Y/S2 E S-IQ . 0
214
Chapter VIII. Commutative Noetherian rings Let 8 be a multiplicative submonoid of a commutative ring
6.9 THEOREM.
Let Q1' ••• ,Qn be detachable primary ideals of R such that Qi
R.
empty
I = Q1
= 1, ... ,m, n .•• n Qn' then
for"
PBOOF.
i
If Q n 8
n8
and Qi
8- 1I =
is nonempty
n'L l 8-lo i
for
i
=
n
m+1, ••• ,n.
8
is If
·
-1 then 8 -1~ '-I = 8 r. Clearly 8-1I ~ Conversely, suppose xis E n'f =1 8-lo i . By (5.8)
is nonempty,
rYf:=1 8-lo t = n'f=1 8-lo i · we have x E I, so xis E 8- 11.
0
From Lemma 6.8 and Proposition 6.9 it follows that if R is a commutative ring with detachable ideals such that each finitely generated ideal is a finite intersection of finitely generated primary ideals, and P is a finitely generated prime ideal of R, then Rp is a commutative ring with detachable ideals (but see Exercise 8). Let P be a detachable prime ideal of a commutative ring R, and let n be rf pn is a primary ideal, then pn belongs to P. a positive integer. Although pn need not be primary if P is not maximal (see Corollary 5.5 and Exercise 5.2), there is a closely related ideal that is primary. The symbolic power P (n) of P is the ideal
= {x ER:
p(n)
sx E pn for some s in R\P}.
Observe that p(n+1) ~ p(n) ~ p(l) = P, that p(n)/pn p(n) is the preimage of 8-1pn in R.
R,
and
let
n
and that
Let P be a detachable prime ideal oF a commutative ring
6.10 THEOREM.
be
belanging to P.
= T8(R/pn),
a
positive
integer.
Then
p(n)
is
a
primar"y
ideal
If pn is primary, then p(n) = pn.
PBOOF. rf xy E p(n), then sxy E pn for some s in R\p. rf x E P, then R\p so Y E p(n). rf pn is primary, and x E p(n), then sx E pn for some s E R\p, so x E pn or sm E pn for some m. Thus x E pn. 0
sx E
Let P be a prime ideal of a conunutative ring R.
Then P is a minimal
prime ideal over an ideal I if P 2 I, and for each prime ideal p' such that P 2 p' 2 I we have P = p'. The ideal P is a minimal prime ideal of R if p is a minimal prime over O. 6.11 THEOREM.
Let P be a
commutative ring R. prime ideal over O.
fini tety generated pr"oper prime ideal of a
If p(n) = p(n+l)
for
same n,
then P is a minimal
215
6. Localization Let Q be a prime ideal of R such that P ~ Q.
PROOF.
p(n+l) it follows that (pp)n
=
(pp)n+l
= pp(pp)n.
From P (n) =
The Rp-module (pp)n is
finitely generated and Pp is a quasi-regular ideal of Rp , hence by the Nakayama Lemma (III.l.4) we have (pp)n = Op ~ Qp. So pn ~ p(n) ~ Q, by
(6.8), and thus P
Q.
~
0
Under the addi tional assumptions of coherence and detachable ideals, p(n) is a finitely generated ideal. 6.12 'l'BEDRE2I.
Let P be a fini tdy generated proper prime ideaL oF a
coherent
commutative
positive
integer.
ring R with detachabLe Then
p(n)
is
a
finiteLy
ideaLs,
and
generated
Let
n
primary
be
a
ideaL
bdonging to P. PROOF. generated.
By Proposition 6.10 it suffices to show that p(n) is finitely But p(n)/pn
= TR\p(R/pn)
so p(n) is finitely generated.
is finitely generated by Theorem 6.5,
0
EXERCISES
1. Consider
the
rings 71.
~
7l[Xl/(2X-4).
Show that
the
ideal P
generated by 2, in either one of these rings, is prime. that 2 E p(2) in one ring and 2 ( p(2) in the other. Brouwerian example of a fini tely
generated
Show
Construct a
ring R with detachable ideals, and a
prime
ideal
P
such
that
P (2)
is
not
detachable. 2. Let S
be a
finitely generated multiplicative submonoid of a
coherent commutative Noetherian ring R.
Show that i f M is a
finitely presented R-module, then TS(M)
is finitely generated.
(Hint: Let s be the product of the generators of Sand consider
Mn = {x E M : smx = O}) 3. Let S be a
finitely generated multiplicative submonoid of a
coherent commutative Noetherian ring R with detachable ideals. Show that S-lR is a coherent Noetherian ring with detachable ideals. 4. Let
R
be
the
indeterminates s elements sx i '
polynomial
,x, ,X2, • ••
ring
over
module the
the ideal
integers
in
the
generated by the
and let S be the multiplicative submonoid of R
216
Chapter VIII. Commutative Noetherian rings generated by s. Show that T8(R) is 8-bounded but not finite1y generated. Why doesn't (6.2) app1y. 5. Let a be a binary sequence and let 8 be the mul tiplicati ve submonoid of the ring of integers 7L gene ra ted by {lian : n = 1,2, ... }. show that 8- 171 is a Brouwerian example of a ring that does not have detachable ideals.
Why doesn't (6.4) apply?
6. For p a prime, let Ap denote the ring of pairs (x,y) where x E 71 and y E 71 p (the ring of integers module p), and multiplication is defined by (x, ,y, ) (x 2 'Y2) = (x,y, ,X'Y2+X2Y' +Y'Y2). Let p(n) be the n th odd prime, and let a be a binary sequence. If a i = 0 for each i ~ n, set ~l = 71i otherwise let ~ be the ring Ap(i) where is the first index not exceeding n such that a i ~ O. Let R be the union (direct limit) of the rings ~. Show that R is a coherent Noetherian ring with detachable ideals. Let P be the
i
ideal of R gene ra ted by 2. Show that TR\f (R) is a Brouwerian Why example of an R-module that is not finitely generated. doesn't (6.5) apply? 7. Let 8 be a multiplicative submonoid of a commutative ring R, and let", be the natural map from R to S-lR. Show that TS(R/I) is finitely generated for each finitely generated ideal I of R i f and only if ",-l(J) is finitely generated for each finitely generated ideal J of S-lR. 8. Use the rings 71 ~ 71(Xl/(2X) to construct a Brouwerian example of a ring R with a prime ideal P generated by the element 2 such that R has detachable ideals but Rp is not discrete. 9. Let K be a discrete field and R = K[S,X,Y,Z 1/(SXY-Z 2 ). Use the two prime ideals I = (x,z) and J = (x,!J,z) of R to construct a Brouwerian example of a prime ideal P of R such that xy E P (2) but it is not the case that x E P or Y E p(2). Why doesn't (6.10) apply? 7. PRIMARY DECCJIIPOSITlOOS
An ideal I of a commutative ring has a primary decomposition if there
are fini tely generated primary ideals 01' ••. ,On' belonging to fini tely generated prime ideals, such that I = nio i . Classically every ideal in a
7. Primary decomposition
217
Noetherian ring has a primary decomposition (see Exercise 4). A primary decomposition is irredundant i f no primary ideal of the decomposition contains the intersection of the other primary ideals, and no two primary ideals belong to the same prime ideal. In a coherent ring with detachable ideals we can replace primary ideals belonging to the same prime ideal by thei r intersection (5.3), and delete primary ideals that contain the intersection of the other primary ideals, so that any ideal that has a primary decomposition has an irredundant one. Let I be an ideal of a commutative ring R, and P a proper finitely generated prime ideal of R. We say that P is an associated prime ideal of I i f P = ..;r:;. for some a in R. 7.1 THOOREM. ideaLs.
Let I
ideaL I of R.
Let R be a
= niQi
coherent
commutative ring mUh detachabLe
be an ir-redundant pr-imary decomposUion of a proper
Then the set of associated prime ideaLs of I consists of
the ideals v6'i'
PROOF.
To see that v6'i is an associated prime ideal of I, choose
such that a E Qi' Then I:a = nj(Qj:a), and Qj:a = R if j ~ i. Theorem 5.7 teIls us that I:a = Qi:a is a primary ideal belonging to v6't' Conversely..;r:;. = ~ which, by (5.7), is equal to the intersection of those ideals v6'i such that a E Qii so if ..;r:;. is a prime ideal, then ..;r:;. = v6'i for some i by (II.2.4). 0 a € nj~iQj
7.2 THOOREM.
Le t R be a
coherent commutat i ve ring mUh de tachabLe
ideaLs, and Let I be an ideaL of R having a primary decomposition.
Then
each minimal prime over I is an associated prime of I.
PROOF. Let P l , ..• 'Pn be the associated prime ideals of I, and let P be a minimal prime over I. Then Pd JI = niP i , hence P d Pi for some i. By minimality P = Pi' 0 In the situation of (7.2) the associated prime ideals of I that are not minimalover I are called embedded prime ideals. The primary ideals belonging to embedded prime ideals need not be unique (see Exercise l), but the primary ideals belonging to minimal prime ideals are unique. 7.3 THOOREM. ideaLs.
Let I
Let
= niQi
R be a
coher'ent
commutatil1e
r'ing mUh detachable
be an irredundant primary decomposUion of an ideal
I mi th associated prime ideals Pi'
For' each i deFine Qi
=
{x € R : sx € I
218
Chapter VIII. Commutative Noetherian rings same s ( R\1'j)'
fal'
Qi
= Qi
Then Qi
cantained in Qi' and
is a deladwble ideal
if Pi is a mirlimal pr'ime ideal oper' I.
We easily verify that
PROOF.
Gi
is an ideal of R.
As R is coherent,
the ideal I:x is fini tely genera ted, so we can decide whether I:x Thus
Qi
is detachable.
If x
0i
If Pi
Thus
t;; Qi .
E
Gi, then
n itt P j \l't .
Qi t;;
Gi.
7.4 COROLLARY.
are
distinct
so
\1\
using
because
the
coherence
we
primary
can
find
for some m,
R be a
Let
Thus
p1'ime ideaL over' I.
0
phmar'y decompositlon, (md Let P
Then th." pr'imaq} Ideal
pr'i.mar'y decomposUioT! of I,
in an iTTeclundant
some s E R\j').
comntutative rirlg will! detadwble
colleT'ent
Let I be an ideal of R hauin9
be a minimal
fIT'e
Gi for some s Il F\, so
I:;:
0
ideals.
P,
a m E nj;tt Gj
Then
I
E
is a minimal prime over I, then Pi does not
contain flJi'i Pi (the ideals p,J decomposition is irredundant), o E
sx
I:;; Pi'
Hence
is
Le t I be an ideo 1 of
Fr
hos
primar'y
(l
FOT'
pr'imes
0
camntu tu ti ve r'i ng R.
0
(i) If I has a pr-imar-y decompositiofl,
(ii)
I
sx E
the pT'imorv idECll s be/onging to the minimal
[he sm"e 1'01' all irTedtmdan! IwLmw'[J decompasitions.
7.5 J:.ElIlMI\..
that belangs to
{x ER:
then so daes
decompasUion
if
and
Fr.
anly
,)1
if
the
is
inter'sectiofl ar finitely many hnitely gener-ated pdme ideals.
(iii) If R is primar[J
Cl
cohencnl
ring wilh delachabl.e
decomposi t ton,
pneci.sel.!)
[he
then
associated
the
primes
ideal.s, p,'i.mes
minima!
Fr,
af
and
Cllld
Fr ha.s
over'
,)1
is
I
0
Q.f'C
thei,'
intel-secl fon ..
PROOF.
If I, or
Fr, is equal to n Qi' then -ii
=
n vtlL •
This, and the
observation that prime ideals are primary and belong to themselves, proves (i) and (ii).
To prove (iii) suppose that JI has a primary decomposition.
From (ii) we can write generated
prime
representation of
Fr as an intersection of finitely many finitely
ideals,
hence
we
can
get
an
Fr from among these prime ideals.
irredundant
By (7.2) these are
associated primes of Jl as they are clearly all minimal. primes of (Ei).
primary
As the minimal
Fr are the same as the minimal primes of I, we have established
0
Let R be a coherent commutative ring with detachable ideals, and let I and P be finitely generated proper ideals of R such that P i5 a minimal
219
7. primary decomposition prime over I.
Consider Q
verify
Q
that
is
Oll = TR\P(Rl1).
=
a
Ix ER: sx E I for some s E R\P}.
detachable
ideal
containing
If 0 is finitely generated, then 0
so 1:0 is not contained in P.
We easily and
I,
= l:s
that
for some s E P,
If I has a primary decomposition, then
Theorem 7.3 tells us that 0 is the primary ideal of I helonging to P. Under certain other circumstances it is also true that the ideal 0 is a finitely generated primary ideal. 7.6 LEMMA. commuta ti ve p rt Q
Le t r'ing
!;;; J !;;; P for
=
{x eR:
I
and P be
R wUh
generated ideals of a
Uieals.
Suppose
P
is
coherent
prime
and
Then P is the unique mini.mal, prime over I, and
some n.
sx E I
flnUeLy
detachable
For
some
s E R\P l
isa
finUe ly
generated
primary
ideal belonging to P.
PROOF.
From p n ~ I it follows that P is contained in each prime ideal
containing I.
It remains to show that Q is a finitely generated primary
ideal belonging to P. Then Q
P (n)
=
As
I is finitely generated, we may assume I
Proposition 6.12. 7.7 THEOREM.
0
tet
R be a
coher'ent
comlllutative
dng wUh detachable
Let I be a finUely generated ideal such tM.!
fdea,ls.
= O.
is a hni tely generated primary ideal belonging to P by
,ff has a prima"!,} 0* = {x ER:
If P is a lIlinimal pr'im" ideal over I, then
decolllposition.
sx E I for some s E R\Pl is a fini teLy gener'ated primary ideal beLonging to
P. Let K he
PRlX>F.
the product of the minimal prime ideals over I There is n such that (PK)n !;;; I. Let J = I :Kn . Then
different tram P. p n !;;; J !;;;
P.
By Lemma 7.6 the ideal
Q
=
{x ER: sx E J for some s E R\p}
is a fini tely generated primary ideal helonging to P. in ], the ideal 0* is contained in 0; If r
E Q,
whence r The
E
then sr-Kfl !;;; I Q*.
in R\P.
1ft
E Kn\p,
then s t r
E I
0
ideal Q*
helanging to P.
for some s
As I is contained
we shall show that they are equal.
in
(7.7)
is called
the
isolated pdmary ideal of I
220
Chapter VIII. Commutative Noetherian rings EXERCISES 1. Let R = ~[X].
Show that (2) is a prime ideal in R, and that are primary ideals in R belonging to the prime (2,X). Conclude that (4,X) n (2) and (4,X-2) n (2) are irredundant primary decomposi tions of (2X, 4); thus the primary ideals belonging to ernbedded prime ideals need not be unique. (4,X)
and
(4,X-2)
2. Let k be the ring of integers modulo 2 and consider the pair of rings k ~ k[X]/(X 2 ) . Construct a Brouwerian example of a coherent, Noetherian ring R with detachable ideals, such that every finitely generated ideal of R is primary, R has a detachable proper prime ideal, hut R has no finitely generated proper prime ideals; thus 0 is a primary ideal with no primary decomposition. 3. Let R be a coherent Noetherian ring, I a finitely generated ideal of R, and a,b E R with ab EI. Show that there exists n such that I = (I + (a)) n (I + (b n )). Hint: look at the ascending chain of ideals I: (b n ) • 4. Call an ideal I irreducible if whenever I is written as the intersection of two ideals, then one of the ideals is equal to I. Use Exercise 3 to show that, if R is a coherent Noetherian ring, and I is a finitely generated irreducible ideal, then I is primary. Give a classical proof that every ideal in a Noetherian ring is an interseetion of primary ideals using the principle that every ideal is either primary or it isn't, and that every set of ideals in a Noetherian ring contains a maximal element. 8. IASKER-NJE'l'HER RINGS
A Lasker-Noether ring is a coherent Noetherian ring with detachable ideals such that the radical of each finitely generated ideal is the intersection of a finite nurnber of finitely generated prime ideals. Classically every Noetherian ring is a Lasker-Noether ring (see Exercise 7.4) . Discrete fields are Lasker-Noether rings, as is the ring of integers. The name Lasker--Noether- refers to the Lasker-Noether decornposition of Theorem 8.5. If k is a discrete field, then k[X] is a coherent Noetherian ring with
221
8. Lasker-Noether rings detachable
Each finitely
ideals by the Hilbert basis theorem (1.5).
generated ideal of I< [X 1 is principal, so if the radical of the principal ideal (f) is the intersection of a finite set of finitely generated prime ideals, then every prime factor of f is an associate of a generator of one of those (principal) prime ideals.
Thus if h [X J i8 Lasker-Noether, then I<
is a factorial field. Theorem 7.5.iii guarantees that in a Lasker·-Noether ring we can find lots of minimal prime ideals over a finitely generated ideal. of
Lasker-Noether
rings
is
closed
under
localizing
with
The class respect
to
fini tely gene ra ted prime ideals, and under passing to quotients module finitely generated ideals. Let S be a muLtiplicativc submonoid of a Lasker-Noet.her
8.1 THEOREM. r-i.ng
R such
.r n S
that
generated ideal I of R.
PROOF.
is
eith.e," empty or" nonempty
For"
each
finitely
Then S-IR is a Las/<er'-Noether' ,··lng.
The ring S·-IR is Noetherian by (6.6) and coherent by Exercise
III.3.4.
Let] be a finitely generated ideal of s-lR.
We must show that
.Jj is the interseetion of a finite number of fini tely generated prime ideals of s-IR. Then .Jj
=
write ]
for some finitely generated ideal I of R.
s-IJI = S-I(P I n ••• rl Pn ) where Pi
prime ideal of R. i
= s-11
is a
finitely gener'ated
We may assume that Pi has empty intersection with S for
<; m, and nonempty interseetion for i. > In. Then (6.9) says that we can . j ,.,m -1 -1 = Ift=l S Pi' But S Pi is a (finitely generate(1) prime ideal by
wnte vJ (6.8).
0
8.2 THEOREM.
Le t R be a Lasher"-Noether" r"t ng, and Le t I be a fi.nit e 1.y
generated i.deal of R.
PROOF.
Trivial.
Then [{lI
es
a Lasker"-NoeLher" rin.g.
0
A composition series for a coherent finitely generated module M' with detachable submodules is a maximal finite chain in the lattice of finitely generated submodules of M.
It is easy to see that a finite-dimensional
vector space over a field has a composition series.
As the lattice of
finitely generated submodules of M i5 modular, the Jordan-lliilder-Dedekind theorem applies , so a module wi th a cornposi ti on series i5 Noetherian and also
satisfies
submodules .
the
descending
chain
condition
on
finitely
generated
222
Chapter VIII. Commutative Noetherian rings 8 • 3 'ffiEOREM..
ideaL
Let R be a Lasker'-Noether' dng and P a minimaL
of R such
that
evenj
elE'ment
R\f is a
of
unit.
pr'ime
the
Then
left
R-moduLe R Iw.s a composi t,ion series.
PRCX:lF.
The ring F = RjP is a discrete field.
minimal prime ideal over 0, theLe exists
11
As F is the unique
such that p ll
O.
=
The modules
pijPi+l are vector spaces over F, and are finite dimensional because R is
coherent.
Thus R admits a
detachable ideals.
8.4 LEl'lMA.
se ries of
finitely
generated
0
R be a
Let
,°
composition
commulativE'
r-tng,
I
ideaL
an
R,
of
OIld
PI"" '/'n 1 "" ,On detochable (deaLs or R such that I ~ niQi' and Qi 1.s a pl'imary ideeü belanging to the pr'ime ideal Pi for each i. Let
f E (I :niQi )\(LJ j Pi) •
TIlen
(i 1
I :f 1:1'
(ii)
If
ni
=
1:('.
i .
I=(I:flr1(l,{l.
(iii)
PRCX:lF.
°
=
then xf ( I,
x ( f\Qi'
xf E I, then xf E Gi' so x (Qi since
t
so r1 jQ i
Conversely,
I:f.
~
Thus I:f S nio i .
'l Pi'
AS f2 E (I:r1iOil\(UjPi)' part (i) says I:f
=
Obviously I <.; (I:f) n (I,tl.
n (I,fl, then there is a
and r E I< such tha t x
= a
+
l'
If x (
f ( I: F.
By (ii 1 this implies d' EI, so x
=
(l:f)
r1 i Oi = I:[2.
Thus a F +
a + I'f CI.
8.5 'ffiEOREM. (Primary decOlllpOsition theorem). Tilell
each
fini tel!)
gener'ated
if
tderd
I'
f 2 EI, hence
I'
E
I
f 2 EI.
0
Lei R /Je uf
R
0
Lasher'-Noether
has
a
pr-i,mar'lJ
decomposi t ion.
PRCX:lF.
Let I
be a finitely gene!:ated proper ideal of R.
We shall
construct fini tely generated ideals ] and K such that J has a primary decomposition, I
=J
n K, and K properly contains I.
By Theorem 7.7 the isolated primary ideals 01"" ,01< of I belonging to the minimal prime ideals PI'''' 'P h over I are fini tely genera ted. Let J = 01 n ••• n 01,' As 1:0, is not contained in Pi' the ideal I:J is not contained K
=
(I, f
in
any
Pi'
so
there
is
E
(J
:J)\(U;f'i)
by
II.2.3.
Tf
), then I," J n K by Lemma 8. 4 .
We construct an ascending chain of fini tely generated ideals Hn of R as Let Hl = I. Let Hn +l R if Hn = R; otherwise choose Hn + l so that Hll = ] n Hn + l , where J has a primary decomposition and Hn + l properly follows.
8. Lasker-Noether rings contains Hn .
223
Note that if Hn has a primary decompo-sition, then so does
Hn - 1 , and so on down to I.
As R is Noetherian there exists n such that
but that can only happen i f I, has a primary decomposition. 0
Hn = 1In +1 ;
8.6 'HIEXJREM. Noether
ring R,
Hn = R
in which ca se 1In , and hence
Let P be a detachabLe pr-oper prtme Iden.l and
I.et
'" : R .... Rp
be
the nnt.tlral
mnp.
of n Lasker'Then Rp is a
Lasker-Noether rIng, nn.d '" r-eFLects finitely generated i.deals.
PROOF.
That Rp is a Lasker-Noether ring follows from (8.1).
If I is a
finite1y generated ideal of Rp , then there is a finitely generated ideal J
~
R
such that I
°
=
Jp •
By Theorem 8.5 the ideal J has a primary
°
decomposi bon J = 1 r1 ••• n Qn' If 1 "" ,Os are the primary ideals of this decomposition that are contained in P, then by (6.8) and (6.9) the ideal ",-1(1) = ",-1Up ) = 01 (l ••• n Qs is finitely generated. 0 EXERCISES 1. Let R be a Lasker-Noether ring that is a principal ideal domain.
Show that R is a unique factorization domain. 2. Let F be the Brouwerian example in Exercise VII.1. 5 of a field
that is factorial but not fully factorial. Show that F[X,Y] is a unique factorization domain and a coherent Noetherian ring wi th detachable ideals. Show that each principal ideal of F[X,Y] has a primary decomposition. Show that F[X,Y] is not a LaskerNoether ring. 3. Let h be a discrete field.
Show that if h[X) is a Lasker-Noether ring.
4. Let I
k
is factorial if and only
be a fini tely generated ideal of a Lasker-Noether ring.
Show that of vI.
vI
is finitely generated, and that
I
contains apower
5. Let R be a coherent Noetherian ring with detachable ideals. P be a finitely generated prime ideal of R such that p"
Let = O.
Show that Rp is a Lasker-Noet.her ri.ng 1Ilith a composition series. 6. Let I be a fini tely generated ideal of a Lasker-Noether ring R. Let P ~ I be a detachable proper prime ideal of R. Show that P is a minimal prime ideal of I if and only if there is 1l such that
224
Chapter VIII. Commutative Noetherian rings (P p )f1 C;;; I p in Rp .
7. Let R be a Lasker-Noether ring and I an ideal of R eonsisting of zero-di visors.
Show that ,-r
use (II. 2.3) .
=
0 for some nonzero
t-
EI.
(Hint:
One should be able to weaken this hypothesis to
coherent Noetherian, or just Noetherian.) 9. FULLY LASKER-WETHER RINGS
The
property of
polynomial rings:
being
a
Lasker-Noether
ring
is
not
inherited
by
any discrete field h is a Lasker-Noether ring, but «[X 1
is a Lasker-Noether ring only if k is factorial. Let ), be a ring such that R[XI, ••• ,Xnl is a Lasker-Noether ring for every n.
Let P be a finitely generated proper prime ideal of R, and K the
field of quotients of R/P. extension field of K. R/p for eaeh i.
Let E be a
We ean \vrite E
finite-dimensional algebraic
= /{[01""'0,,]
Then E is isomorphie to the
with
a,
integralover
held of quotients of
R[X 1 , ... ,X n l/I where the prime ideal r is generated byP and, for- eaeh i, the
preimage
K[al"'. ,ai---11.
in
R[X 1 '.'.'X i 1 of
the
minimum
polynomial
of
0;
over
So eaeh finite-dimensional algebraie extension field of K
is factorial, hence K is fully faetorial.
This suggests the following
definition. Call R a fully Lasker-Noether ring if it is a Lasker-Noether ring and if for eaeh finitely generated prime ideal P of R, the field of quotients of R/P is fully factorial.
Note that the ring of integers 7L is a fully
Lasker-Noether ring, as is any fully faetorial field.
9.1 THEOREM. NoetlH'T ,-ing R.
PROOF.
Let
be (1 fini tdu genEToted
By Theorem 8.2 the ring R/I is a Lasker-Noether ring.
finitely
generated
(RjI)/p
R/P' is fully tactorial.
9.2 'I'HIDREM.
If
prime
f'
is
ideal
Let r be
The preimage P' of P in R 1S a
a finitely generated prime ideal of R/I. ~
of Cl {ullU Lnsl<er--
ideat
Then R/I lS Cl full\! LaskET-Noether ,-ing.
of N,
so
the
field
of
quotients
of
0
G detoclwhle
pt'ime
"l('u)
oF a
fully
LClSI,,,,--
Noether- ,-ing R, then Rp is a fullu Lasker'-Noetltet- '-;'l1g.
PROOF.
By Theorem 8.1 the ring Rp is a Lasker-Noether ring.
a finitely generated prime ideal of Rp • finitely generated prime ideal by (8.6),
Let Q be
The preimage Q' of Q in R is a so the held of quotients of
9. Fully Lasker-Noether rings Rp/Q,
225
whieh is isomorphie to the field of quotients of R/Q',
faetorial.
is fully
0
If R[X I' ... ,X n J is a Lasker-Noether ring for eaeh n, then R is a fully Lasker-Noether ring.
We shall show that the eonverse holds.
Let S be a multiplicatilJe subset of
9.3 LEMMA.
commutatiue ring R.
Ir
(!
cohe1'ent, Noether"ian
R .... s-lR r"eflects (inUelv gellETaled idea.ls, then
so does R[XJ .... S-1R[Xl. As R .... S-lR reflects finitely generated ideals, the kernel of
PROOF.
R .... s-IR is finitely generated, so we may assume that R ~ S-1R •
S-1R is coherent and Noetherian.
The ring
Let I be a finitely generated ideal of
S-lR[XJ, and let R[Xl m = {f E R[X1 : deg r < m}. By (1.1) there exists m such that M = 1 (1 s-11~ [X 1m is a fini tely generated s-1R-module, and
r n S-1R [X ln
= 27:0 MXi.
for each
n
2 m.
Let M' be the R-module generated
by a finite set of generators of the S-1R-module M. Then TS(R[XJmjM' ) = (R[X Jm n IljM' is finitely generated by 'rheorem 6.4, so the R-module R[XJ m n I = M n R[X] is finitely generated. Let] be the ideal of R[XJ generated by M n R[XJ. ThenJ is a finitely generated ideal of R[XJ contained in 1 genera ted )
ideal
elements of M.
Ag
Let A be the (finitely
(1 R[X].
in S-1R consisting of
the
coefficients
of Xm-I
of
R ..., s-IR reflects finitely generated ideals, A n R is
fini tely generated, so there exist f l' ... ,fh in M whose Xm-l coefficients generate An R.
Choose sES so that sr i E R[X1 for each i..
coherent and Noetherian, show that I Suppose hEl
n
R[XJ = J:sP, so I
Sl11 E ]
n R [X J .
there is
for
such that ] :sp
p
n R[X]
Suppose So sl1
n
> m.
h E R[XJ.
Then
Conversely suppose hEl Then h
sPb EI,
n R [X J •
2i"isfiXn--;n + sg, whence sg EIn R[XJ n _ l . have sg E ]:sp. Thus hE J:sP+l = J:sp. 0 Let
R
I
n
R,
let
K
generated by I
the
/Je
fi.eLd
i.n KIX).
hence
hEl,
so
Then hER [X ln for some
If n Sm,
=
9.4 LEMMA.
We shall
then h
E ]
I:;; ]
:sp.
2i."JiXn-m + g, where 9 E S-1R[XJ n _ 1 and "l E R.
=
finite1y generated i.deal
As R[Xl is
:sp+l.
is finitely generated.
and we proceed by induction on n.
n,
= ]
(1
(,,1tU
of R[X).
l.osl,ef'-Noethef' Let
By induction on
ring
M be n mini.m(1l
oF quot[ents of RjM, and let
(md
prime
J be
If ] is pr'oper', tflen the Fweimage P o(
I
a
we
P"O[JE'f"
i.denl the
Fr
n
Dve,'
ideal
i.n R[X)
is a finite inter'sectlon oF finiteLy gener'ated pr'ime ideals, and I:P ;t I.
226
Chapter VIII. Commutative Noetherian rings
PROOF. As K is a factorial field, v7 is the intersection of finitely many principal prime ideals in K[X). From Lemma 9.3 we see that the intersection of each of those prime ideals wi th (RjM) [X ) is fini tely gene ra ted , 50 P is the intersection of finitely many finitely generated prime ideals. To prove that I:P ~ I, consider first the special case when M ~ ~ and K = RjM. As v7 is a principal ideal of K[X), there exist 9 in R[X) such that the image of 9 generates v7. As M ~ .../R n I, there exists n ~ 1 such that Mn ~ I and Mn-l\I is nonempty. Choose t € Mn-l\I and note that tM ~ I. As 9 € .,11 +M [X), there exists m € IN such that tg m f/. I and tgm+l € I. Then tg m € (I :P)\I. Notice that we did not need (9.3) for
this case as P is an intersection of prime ideals of the form (p) + M[X) which are clearly finitely generated. For the general case set S = R\M. Then S-lR is a Lasker-Noether ring, S-lM is the radical of S-lI n S-lR, and s-lR;S-lM = K. The preimage of v7 in S-lR[X) I:P
~
I.
is S-lp.
so
0
If
9.5 LEMMA.
cohel'ent
The special case gives us S-lI :s-lp ~ s-lI,
R
Noethel'ian
is ring
a
fuUy wUh
finUety generated pr-ime ideaL
Lasl~el'-Noether-
detachabLe
P of
R[X],
ideaLs
.-ing,
then
such
that
R[X)
is
the fieLd of quotients of
a
each
for-
R[X]/P
is fuLLy factoriaL.
PROOF. From the Hilbert basis theorem (1.5) it follows that R[X] is a coherent Noetherian ring with detachable ideals. If P' = P n R, then P' is a finitely generated prime ideal of R by (1.2),
so the field of
=
quotients of R/P' is fully factorial. The field of quotients of R[X)/P (R/P' ) [X l/(P /P' [X]) is a finitely presented field extension of a fully factorial field , so by Theorem 2.5 is fully factorial. 9.6 'lHEDREM,
PROOF.
0
If R is a fuUy Lasker--Noether- ring, then so is R[X].
By (9.5) it suffices to show that if I is a proper finitely
generated ideal of R[X], then there exist a finite number of finitely generated prime ideals containing I 50 that some power of the intersection of those prime ideals is contained in I. As some product of minimal prime ideals over R n I is contained in R n I, there exists a minimal prime I f K is the field of ideal M over Rn I such that I + M[X] ~ R[X). quotients of
RjM,
then the ideal] generated by the image of I in
K[X]
is
227
9. Fully Lasker-Noether rings
proper. By (9.4) the preimage P of v7 in R[XJ is a finite intersection of finitely generated prime ideals, and I:P t I. Let L be the intersection of all those ideals P as M ranges over the minimal prime ideals of R such that J is proper. Replace I by 1 2 = I:L d I:P t I and start again. This provides us with an ascending chain of ideals I = I, ~ 12 ~ ••• and finite intersections of finitely generated prime ideals L = L"L 2 , • • • such that I n +1
= In :Ln , and In
In
I n +1 so R[XJ
=
=
In
In = R[X J. There is n such that 1:L 1L2···Ln _1 whence L1L2···Ln _1 ~ I. 0
f. I n +1 unless ~
EXERCISES 1. Show that a discrete field is fully factorial if and only if it is a fully Lasker-Noether ring. 2. Show that the ring of integers is fully Lasker-Noether. that the field of rational numbers is fully factorial.
Conclude
3. By considering the rings ~ ~ ~[Xl, construct a Brouwerian examp1e of a fully Lasker-Noether ring that does not have a fini tely generated maximal ideal. 4. A bad Noetherian ring (Nagata).
Let k be a discrete field and
X"X 2 , ••• a countable set of indeterminates. Let m" m2,'" be a sequence of positive integers such that 0 < mi - mi -1 < mi +1 - mi for each i. Let Pi be the (prime) ideal in K = k [X "X 2 , ••• 1 generated by {Xj : mi ~ j < mi +1}' let S be those polynomials in K that are not in any Pi' and let R = s-lK. Show that (i) K, and hence R, is coherent. (ii) If I is a proper, nonzero, finitely generated ideal of R, then {i : I ~ S-lP i } is nonempty and finite. (iii) The localization Ri of K at Pi is Noetherian. (iv) R is a (fully) Lasker-Noether ring i f k is (fully) factorial. (v) R has arbitrarily long chains of finitely generated prime ideals. 5. Let S be a finitely generated multiplicative subset of a fully Lasker-Noether ring R. Show that s-lR is fu1ly Lasker-Noether. 6. Let K be a fully factorial field. Let R = K[X,Y,l J. (X 2 - Yl, y 2 - Xl). Determine the minimal primes of I.
Let I
228
Chapter VIII. Cornmutative Noetherian rings
10. 'lliE PRINCIPAL IDEAL TflEORm By the
height of a finitely generated detachable prime ideal of a
Noetherian ring R, we mean i ts height in the set of fini tely generated detachable prime ideals of R ordered by inclusion (see the definition of depth and height on pages 24 and 25). 10.1 'l'HIDRm (Principal ideal theorem).
Le t a b e
Grl
e tement
of
(1
LasP.er-Noether' dng R, aM P a (ini te1.y gener'ated IJr'oper pr'ime ideaL that Ls minimalover (a).
Trten P Iws Iteight
ul
most 1.
Let Q be a finitely generated prime ideal of R such that P d Q.
PROOF.
We shall show that Q = P or Q is minimalover O.
By Lemma 6.8 we may
localize at P and so assume that every element of R\f is a unit. u
E Q, then
P by the minimali ty of f', so we may assume
Q =
If
Q.
a If
We
shall show that 0 is minimalover O. Consider the decreasing sequence of symbolic powers 0(1) d 0(2) d ... of Q. The ideals O(i) + (a) form a decreasing sequence of ideals containing (u).
Since P/«(1) is a minimal
prime of R/(a), the ring R/(a) is a Lasker-Noether ring wi th a composi ti on series by
(8.3).
generated,
so there is
By Pwposi ti on 6.12
Cl
E
CI
If Q, while xa = Cl -
0 (n), then there exist r I'
E
the
E
Q(n),
ideals Q(i)
= Q(n+1) + Q (n+l) and x ERsuch that Cl =
such that Q(n) +
Tl
so
xE
Q{n).
(a)
Thus O(rr)
=
are
finitely So if
(0).
r'
+
Now
XCI.
0(n+1) + O(n)o.
By the, Nakayama lemma Q (n) = Q (n+1), so () is a minimal prime over 0 by Theorem 6.11.
0
Let R be a Lasker'-Noether dng.
10.2 LE21MA. be
Ci
so'tet l.y
decr'eastng
sequenee
E PO'
Then then:
ideaLs, and let
* •.• 'Pn*- 1 such PI' and x E Pn*- 1 .
thot Fa ;;)
or W'f'
Po
d ••• d P n - 1 d P"
genera ted
F,nit,,!y gellenlled prIme
x I[
Pl
,
and that P 2 = O.
be a minimal prime ideal over (x) that is contained in Po' is nonzero.
theorem (10.1) has height at most 1. 10.3 COROLLARY. be
0
pr'ime ideals
Let P~
Obviously P ~
The prime Po has height at least 2, while the prime P~ is a
minimal prime over the principal ideal
I'n
proper
P~ d .•• d Pr~-l d Pn i s str'ict:ly decr'eosing,
We may assume that n = 2, that
PROOF.
Let
f;nttel.y
(x),
Thus Po
so by the principal ideal properly contains P:.
Let R be 0 Loskcl'-Noether' 1'ing.
str'ictl.y decr-eosing sequence of
Let
0
Po d •.. :2 Pn - 1 d
{ini t"l.U genenl.ted pr'oper' prime
10. The principal ideal theorem ideals, and Let Q1"" ,Om
Po
be nnitely generated proper pr'ime ideals not
""
*
Ihen ther'e are finitely generated prulte ,deals PI""
containing PO' such timt
229
'Pn*- 1
* :2 P n is strictly decreasing, and such thnt :2 PI* :2 ••• :.2 Pn-1
no Qi contains Pn* - 1 ·
By proposition II.2.3 the ideal Po is not contained in the
PROOF.
union Q1 U
U Gm'
Take x E P O\(Q1 U ••• U Qm) and apply Lemma 10.2.
10.4 'l'HOOREJI!. (Generalized principal ideal theorem). Noe ther" r"ing.
Le t I = (al""
,an)'
0
Let R be a Lasher-
Ihen euery minimal. pl'ime ideal ouer" I
hns height at most fl.
PROOF.
on n.
If
Let P be a minimal prime ideal over I, and proceed by induction n =
0, then I = 0, and P is a minimal prime ideal of R, so P has
height O. Let ] = (al"" ,an _1) .
If P is a minimal prime ideal over ], then P
has height at most n - 1 by induction.
Thus we may assume that P is not
contained in any of the minimal prime ideals over J. We must show that there cannot exist a strictly decreasing sequence P
=
Po
2 Pi 2 ••• :2 Pfl 2 Pn +1 of finitely generated prime ideals. Given such a sequence, by Corollary 10.3 we could construct a strictly decreasing
sequence of Hni tely generated prime ideals P = P o * ] F *l ] ... 2 Pn* :J- Fn+1 wi th F~ not contained in any of the minimal primes over J. The prime ideal FI] of RI] is a minimal prime ideal over the principal ideal II], so has height 1 by the principal ideal theorem (10.1).
The ideal (F~ + J)IJ
is not contained in any minimal prime ideal of RI], but F I] d (Pl~ + J )IJ . Thus PI] is a minimal prime over (Pr~ + .T)j], so F is a minimal prime over . a mJ.nlma ". lpn.me ' * In R/Pn* t h e Pn* + ] , so F /Pn* lS over (P n" +) ] /l'n*.Hl R/Pn' ideal (P~ + .Tl/Pr~ is generated by n - 1 elements. By induction P/P~ has Thus 1'*/l0* 1'*i+1 /p*fl ror ~ i 11 = some t. < Tl, so F*i = p*i+1'
. h t at most n - 1. h elg
a contradiction.
0
The converse of (10.4) is not true; a prime ideal of height at most 1 need not be principal (Exercise 3). 10.5 THEX>REM.
Let
Lasker-Noelher" ring R.
F be a Then
However the following is true.
finitdy (]enen1ted pr·oper' pr'ime (/l(T('
is m such that P
IlOS
ideal
o( (1
height m, and F is
a minlm111 prime ouer' same ideal gener"(1ted by m elements.
PROOF.
As P is minimalover F, by (10.4) there is
Tl
such that P has
230
Chapter VIII. Commutative Noetherian rings
height at most n.
We proceed by induction on n.
There are finitely many
minimal prime ideals Ql" •• ,Qk of R contained in P.
=
If P
Qi
for some i.,
then P has height 0 and is minimalover the ideal 0, which is generated by
o
elements.
Otherwise for each i we can choose Xi
E
P\Oi'
There are
finitely many prime ideals that are minimalover 0i + (xi)' and if
Q
is
one of them, then 1';0 has height at most n - 1, so by induction P;O has a height.
Let m - 1 be the maximum of the heights of
minimal prime ideal over some Qi + (xi) for some
i.
1';0
where
6~
P is a
We shall show that P
has height m. Clearly P has height at least m because each Q has height at least 1. Let P = pO :;;l ••• :;;l Pn be a strictly decreasing sequence of fini tely generated prime ideals. Then Pn :;;l Qi for same i. By Lemma 10.2 there is a strictly decreasing sequence P = x'1 E p* n- l'
Po
* :;;l Pn such that :::l PI* :::l ••• :::l P,,-l
Thus p* n- 1 contains one of the primes ,
Q minimalover
Q.1 + (x,, ),
so n <; m + 1 and I' has height at most m. Let P have height m.
We sha11 show that F is a minimal prime ideal
over an ideal I generated by m elements. contained in P. is x
E
We proceed by induction on m.
> O.
Let li 1 , ... ,On be the minimal primes of R As P properly contains each 0i' by Theorem 11.2.3 there
We may assume that m
1'\(01 U •.• U On) •
Then I' /(x) has height at most m - 1 in R/(x).
By induction there is an ideal .I <:: R such that ] is generated by m - 1 elements and P/(x)
is minimal prime over (J + (x) )j(x) •
minimal prime over 1 = ] + (x). 10.6 COROLIARY.
Hence P is a
0
Each fini tdy-generated pr-oper pr-ime Idea!
Noether- r-ing hns a height.
in n Lnsker-
0
EXERCISES 1. Let R be a coherent Noetherian ring with detachable ideals.
Show
that each finitely-generated prime ideal of R of height 1 is a minimal prime ideal over same principal ideal. 2. Let R be a Lasker-Noether domain.
Show that R
is a unique
factorization domain if and only if each prime ideal of height 1 is principal. 3. Let R be E[~-51, and P the ideal generated by 2 and 1+~-5. that R is a Lasker-Noether domain.
Show
Show that R/p is a discrete
231
10. The principal ideal theorem
field with 2 elements, so P is a finitely-generated prime ideal. Show that P has height 1 but is not principal. mTES
A constructive version of the Hilbert basis theorem was proved by Jonathan Tennenbaum in his 1973 dissertation directed by Errett Bishop at the University of California, San Diego. Tennenbaum employed a basis operation rather than a basis function. This seems to be one of the few instances where Bishop's notion of an operation from A to B cannot be interpreted as a function from A to the set of nonempty subsets of B. The original scope and spirit of Tennenbaum's result can be retrieved by ignoring the function ~ in Section 3 and considering the function p to be an operation. Exercises 4.5 and 4.6 on Tennenbaum rings are an attempt to restore the scope of Tennenbaum's result without relying on the notion of an operation. The Hilbert basis theorem for coherent Noetherian rings with detachab1e ideals was proved by Richman (1974), who relied on Tennenbaum's result. Seidenberg (1974a) showed how to remove the dependence on Tennenbaum' s theorem, and at the same time showed that the Hilbert basis theorem also holds for coherent Noetherian rings (with no reference to detachability of ideals). Note that this latter Hilbert basis theorem is neither more nor less general than the one requiring detachable ideals: both the hypothesis and the conc1usion are weaker. The main theorem on primary decomposition of ideals in Noetherian rings -- that R[X] is a fully Lasker-Noether ring if R is -- was first proved by Seidenberg (1984) who cannot, however, be b1amed for the termino1ogy. The bad Noetherian ring in Exercise 9.4 is from [Nagata 1962]. It provides an example of a fully Lasker-Noether ring that is not built up from a discrete field by adjoining finitely many indeterminates, taking quotients, and loca1izing.
Chapter IX. Finite Dimensional Algebras
1. REPRESENTATICWS.
Let k be a discrete Held.
A
k-algebra is a ring A that is also a
vector space over k, satisfying A(ab) a_,b in A.
=
(Aa)b
a(Ab) for each A in k and
=
rf A and B ace k-algebras, then a hcmomorphism fLom A to B is a
ring homornocphisrn that is also a k-linear transfocmation.
The term finite
dimensional, when applied to a structure S that is a vector space over k, like a k-algebra, signifies that S is a finite-dimensional vector space over k. If
1>1
is an n---dimensional vector
space over k
I
then
the k.-linear
transformations horn M into M focm a finite-dimensional k-algebra E (M) whieh may be identified wi th the algebca of n x finite
dimensional Iz-algebra
H
matriees over k.
A is isomorphie to a
My
finite-dimensional
subalgebra of E (A) by associating wi th each element a of A the linear transformation Ta from A to A given by Ta(x) representation of a
A
dimensional vector space ~
: A~
1>1
=
ax.
Hni te-dimensional h-algebra A is a over /"
rf we drop explicit referenee to
E(M).
~,
and write am for
'P
is zero, then we say that
'P
(or
~(a)m,
If the kernel K
then we have the notion of a finite-dimensional A-module. of the cepcesentation
finite-
together with a k-algebra homomorphism
M)
is faithful.
If M is a faithful A-module, then A may be identified with a subalgebra of E(M).
The following theorem gives a number of irnportant constructions
that yield finite-dimensional vector spaces. 1.1 THEOREM.
Let A be a finite-dimensional 'c-algebnl. and M a fin!
dimensionaL A-moduLe. a
fini te-dimensionaL
dimensionaL
(i) The
subspace
of
M.
Then
the
Followi.ng
an,
uector- spaces ouer" ". subspace
[e-
Le t I be a f Ini te-d.Imens lana 1 subspace of A, and N
IN oF M generated by {ax
: a E I and x E N}.
(ii) The subaLgebra l + 1 2 + [" + ••• of A generated by (iii) The center {a E A : ax = xa for- a11 x in Al of A.
232
r.
f'i.ni t e-
233
1. Representations (iv) The f'epresent.atton henlPt K = {o E A : uM = (v) The centraHzer {b E E(M)
PROOF.
For each
the subspace Sn
Tl
dimensional, so we can find fx : A
oF N.
Al
of
AlK.
The subspace IN is c1early fini tely generated, hence finite
dimensional. m ? n.
Ol
; ab = bel FOT' aLL a E
Tl
I +
=
[2
+ ••• + In is finite
such that Sn = Sn+l' and so Sm
= Sn
for each
The center of A is the intersection of the kernels of the maps ->
given by Fx
A
(0)
= ax -
where x ranges aver a basis for A.
xa
The representation kernel K of M is the intersection of the kernels of the maps gm : A
-> M
given by gm (a)
= um
where m ranges aver a basis for M.
The centralizer of AlK is the intersection of the kernels of the maps h a E(M) ~
E(M) given by ha(b)
= ab
- ba where a ranges over a basis for A.
0
Sy (1.1.iv) we can decide whether or not M is faithful; in either ca se we may view M as a faithful module over AlK.
The endomarphism ring of M
is the centralizer of the algebra AlK viewed as a subalgebra of
E(M).
lf a and bare nonzera elements of a finite-dimensional I<-algebra A, and ab
then we say that a is a left zero-divisor, and that b is a
0,
=
right zero-divisor.
1.2 CORDLLARY.
Let A be a Fint te-;:HmensionoJ Iz-o.l geb1'a.
nonzero eLement of A is eLther
PROOF.
0
For a E A, let Pa:A
[mit,
01'
a
left (md
T'ight zero-divisOT'.
A be defined by Pa(x)
->
either Pa has a nonzero kernel, in which case is b
Pa(ab)
=
such that ba
= 1,
so Po
Pa (1), so ab = 1, whence
a right zero-divisor. divisor.
(1
Similarly,
is onto,
is a unit. 0
=
xa.
By (II.6.2)
is a fight zero-divisor,
0
or Pa is anto, in which ca se a has a left inverse. there
Then each
In the latter case
hence one-to-one.
Sut
Thus a is either a unit or
is either a unit or a left zero-
0
AS Act = A if and only i f a 1S a unit, a finite,,,dimensional algebra A
has a nontrivial finite-dimensional left ideal if and only if A has a zero di visor.
We cannot always decide whether eller!) nonzero element of A has
an inverse, that is, whether A is a division algebra. 1.3 EXAMPLE.
Let a be a binary sequence and let /,
= UnQ(io n ).
Let X
be an indeterminate and A t.he two-dimensional algebra k [X I/(X 2 +1) over 1<. I t is easy to verify that. A is a division algebra precisely when an = 0
for all n.
0
234
Chapter IX. Finite dimensional algebras
A finite-dimensional h-algebra is a division algebra if and only if it
has no zero-divisors. It is a classical theorem that if A is a division algebra, then any A-module is free. Thus, classically, ei ther A has a zero-divisor, or every finite-dimensional left A-module is free.
Although
we may not be able to determine which of these alternatives holds, we can, given an A-module M, either construct a basis for M as a free A-module, or construct a zero-divisor of A. 1. 4 THOOREM.
Let
A be
a
Fini te-dimensional
h-algebra.
dimensional A-module. and u a nonzer'o element oF M. zer'o-divisor',
01'
M a
Fini te-
Then ei ther A !las a
M is a Free A-module wUh a basis containing u.
PROOF. The set (a E A : au = O) is a finite-dimensional left ideal of A. If it is nonzero we are done by (1.2), so we may assume that Au is a rank-1 free A-module. Then N = M/Au is a finite-dimensional A-rnodule of dimension less than the dimension of M. If N = 0, then {u} is basis for M. Otherwise, by induction on the dimension of M, either A has a zerodivisor, or N is a free A-modulei in the latter case M is a free A-module with a basis containing u.
0
Let K be a finite-dimensional k-algebra.
Many constructions that could
be effected i f K were a division ring can be attempted whether K is a
division ring or not, the result being either the desired construction or the construction of a zero-divisor of K. Theorem 1.4 is an exarnple of this. A related exarnple is the construction of minimal polynomials. Let A be a finite dimensional h-algebra, K a subalgebra of the center of A, and a E A.
A monic polynomial F E K[X] is called the minimal poly-
namial of a over K if for each 9 E K[X] we have g(a) = 0 if and only if F divides g. It is easily seen that the minimal polynomial is unique, if it exists, and that K[a] ~ K[X]/(F). 1. 5 THOOREM.
Let
A be a
finite-dimensional
k-algebra,
dimensional subalgebra of the center oF A, and a E A.
struct either a zero-divisor in K,
PROOF.
01'
K a
finUe-
Then we can con-
the minimal polynomial of a over K.
For each nonnegative integer n let Mn
=
K + Ka + ••• + Ka n .
Consider the map 'i'n K...., Mnj1rfn-1 that takes 1 to an, and choose the least n such that her 'l'n ~ 0; we will find such an n ~ 1 + dirn A, because we must have Mn = Mn - 1 at some point. If her
235
1. Representations
Mn_lover K, and i f a n = Co + cla + ••• + cn_Ia n-1 then p(X) = Xn - clX - Co is the minimal polynomial for a over K. 0 cn_lxn - l Mn
=
EXERCISES Show that Hni te-dimensional k-algebra. coherent Noetherian ring with detachable ideals.
1. Let A be a
A
is a
2. Show that the algebra A of Example 1. 3 is a division algebra
precisely when an
=
0 for all n.
3. Construet a Brouwerian counterexample to the theorem that if A is a finite-dimensional k-algebra, then either A has a zero-divisor, or every left A-module is free. 4. The quaternion algebra H over I, is the four-dimensional k-algebra
with basis (l,i,j,ij) satisfying .2
.J i j
-1
= -ji.
For what h is H a division algebra?
For what k does H have a
zero-divisor? 5. Show that i f k
is algebraically closed, and A is a finitedimensional k-algebra, then either A is a division algebra or A has a zero-divisor.
2.
'!'HE
DENSITY THIDREM.
A module M is reducible if it has a nontrivial submodule--otherwise it
is irreducible (or simple).
The elassical density theorem asserts that if
isa fai thful i rreducible A-module, then Ais a dense ring of endomorphisms of M, when M is viewed as a vector spaee over the centralizer B of A (which is a division ring). The terminology refers to the situation where M is infinite dimensional over B, and we conclude that A is dense in
M
the sense that any finite B-independent set can be mapped anywhere by multiplying by some element of A.
In the finite-dimensional case, which
is what we are dealing with here, we conclude that A is the full ring of B-module endomorphisms of M, that is, A is the centralizer of B. From a constructive point of view, the density theorem is much more useful if we can appIy it to a possibIy reducible M.
236
Chapter IX. Finite dimensional algebras 2.1 mEX:lREM (density).
LI" t
I inear transformations on a nonzer-o eLement of M.
f ini. t e-d i"mens i onaL
a
A !Je
/,-aL gebr-a
of
(inite-dimensi onaL h-space M, and Lel u be a
Let B be file centr-alizer of A.
Then either
(i) M has a non t rot viaL A-sllbmodu I e.
(ii) M ls a fr-eI" B-module wUh a basis containLng u, and A is the centr-aLiZE'r' of B.
PROOF.
By Theorem 1.4, either B has a zero-divisor or M is a free
B-module with a basis containing
If bEB is a zero-divisor, then bM
tl.
is a nontrivial A-submodule of M.
Thus we may assume that M is a free
B-module wi th a bas i s "1' ... ,u n containing Lj
{n
E A :
aU i
1I.
For
0 for each i
=
Then L j is a fini te-dimensional left ideal of A.
0, ... ,n set
j
=
~
j}.
We may assume that the
finite-dimensional A-submodule L j l l j of M is ei ther 0 or M for each j and Let
i.
Mj
Then Mj
is a B-submodule of M containing 2{ =1 Eu;; we shall show,
induction on j, j
that Mj
ll=1
=
BU t .
because u j +1 E: Mj' Let xE Mj + 1 . As L j + 1x jJ E L j defines bEB, with Lj(x - bU j +1)
,]
Li=l
BH
by
As L O = A, both sides are 0 for Suppose the equality holds for some j < Tl. Then L /'}+1 = M
O.
=
{xCM:Lr,=OJ
0, setting b(i'u j +l) = fJx for
Thus x -
O.
E ~j+l B i , sox" Li=l u f '
bU j
+l
E
Mj
=
To show that A is the centralizer of 13 we construct, given vl""'v" in M, an element n in A such that n" l
As Li _1"[
1'[.
= M,
we can construct
n i E Lt _ 1 inductively such that
We use the density theorem to reduce the question of whether an algebra
1S simple to the comrnutative case. 2.2 THEOREM. Then
Let
e;ther- A hns n
between ideuls 1
A be
u
finlte-climensi()nul
nontrilliu?
J
I PROOF.
ideCll,
01' A und ideals J
ur (
or-
fhe/"!'
/'-iC11gebr-u is CI
witb cente," C.
1-1 rorTespondence
givE'll iJu
nc AJ. I
Let R be the (finite-dimensi.onal) subalgebra of F,(A) generated
237
2. The density theorem
The centralizer of R
by left and right multiplication by elements of A.
is the center C of A, and the R-subrnodules of Aare precisely the ideals of A.
By the density theorem, either A has a nontrivial ideal, or A 1S a
free C-module wi th basis containing 1, and R is the centralizer of C. Assuming the latter alternative holds, let T be an ideal of A and let] be an ideal of C. is a
n
That] = (AJ) n C follows immediately from the fact that A We must show that I
free C-module wi th basis containing 1.
Let 1 = uD"" ,u n be a basis for A over C, and let r i ERbe such that riu i = 1 and T'iUj = 0 i f j t i. If x E T, then x =;;: etUi so (I
C)A.
r-ix = ci whence x E (I
n C)A.
0
We shall improve on this theorem in the next seetion,
replacing the
condition "A has a nontrivial ideal" by "T'ad A t 0". EXERCISES 1. Schur's lemma.
Let M be a finite-dimensional faithful irreduc-
ible module over the finite-dimensional h-algebra A.
Show that
the centralizer of A is a division ring. 2. Bumside's theorem.
Let M be a finite-dimensional vector space
over an algebraically closed field h, and let A be a subalgebra of E (M) •
Show that either M has a nontri vial A-submodule, or
A = E(M).
3. Let A be a finite-dimensional h-algebra.
Show that ei ther the
center of A has dimension greater than one, or A has a nontrivial ideal, or every ideal of A is trivial. 4. Refer to the first two sentences of the proof of Theorem 2.2. Show that the centralizer of R is the center C of A. 3. TIlE RADICAL AND Sl.lMMl\NDS.
Let A be a
finite dimensional algebra.
nilpotent i f Ln = 0 for some ntc! A
Tl.
A left ideal L
of A is
The radical of A is defined by
{x " A
Ax is nilpotent} .
It is readily verified that nId A is an ideal of A.
Moreover rad A is
detachable from A, for i f L 1S a left ideal of dimension
Tl,
finite dimensional, and In = 0 i f and only if L is nilpotent.
then Ln i5
238
Chapter IX. Finite dimensional algebras The
classical
Wedderburn
algebras A with md A We
want
the
structure
theorem
for
finite-dimensional
= 0 hinges on the construction of idempotents in A.
additional
information
provided
by
an
constructs either an idempotent or a nonzero element of 3.1 LEMMA.
L
!Je
Th1'n 1'i
tllET
Let
h-olgeb,-a A.
n
n01lZ1'ro
ideal
Left
of
algorithm
a
that
A.
T'ad
finite-dimensümal
L eontains a nonzeT'O nt tpotent left ideal or' a.
r1Onze,'O idempotent.
Passing to Ax r:; L we may assume that L is finite-dimensional
PROOF.
and proceed by induction on the dimension of L. our nonzero nilpotent left ideal; if 0 Thus we may assume that L2 As L2
=
l'
L"
l'
If L2
=
0 we have found
L, we are done by induction.
L.
0, we can find x in L such that Lx # O. Define f : L ... L by füll = yx. By (II.6.2) eUher t,eT' f is a nonzero 1eft ideal properly contained in L, or f is an isomorphism. In the former case we are done by = L l'
induction, so we may assume the latter. yx =
tJ2 y
1..
whereupon
x
(!:j2 -
lJ)x =
Hence
0 so 11
= lJ2
yx =
x for some y in L, so
is a nonzero idempotent in
0
The following is the standard result relating idempotents to summands. 3.2 LEMMA. dimensionaL
only if A
Let
L
I~-algebr'a
be
a
A.
finite-dimensional
Therl L = Al'
L EIl K for' some [eft ideal.
for'
some
left
ideal.
of
n
finite
e E A if
idempotent
and
K.
K, If x E L, then x = xe + xf so weite 1 = e + f where e E L and f c K. Conversely if x = xe, then x E L. Thus L = Al' and xf and x = xe. PROOF.
If e
is an idempotent, then A
=
Al'
Tf A = L EIl
(j) A(l-e) .
°
3.3 THFDREM.
tet
A be a
finite-(limensiono[
dimensiona.L (left) ideol 01' A. same
h~lgebn!
and Ln
The" ei ther L r1 f'(!d A # 0 01' A = L
finite
EIl N 1'01'
(left) ideal N.
PROOF.
We may assume L # O.
By
Lemma 3.1 either L
(1 "'lei A ,t
0 or L
contains a nonzero idempotent and hence, by Lemma 3.2, a nonzero 1eft ideal summand K of dim
A.
So
11.
= I< EIl M and L = K Gl (M n L) •
By
L we may assume that M n L is a summand of A, hence of Gl N = L f1) N. If L i 5 an ideal, and R = {,- E A
(M
n L)
N ~
R.
As R
nL
induction on
M, LI'
so A = K Gl O}, then
=
is a finite-dimensional nilpotent right ideal contained
239
3. summands and the radieal
in the ideal L, eUher L contains the nonzew nilpotent ideal A(R Cl L), or R Cl L
=
0 so N
=R
is an ideal.
0
If A, and A2 are k-algebras, then A, x A2 is a k-algebra in a natural way, ealled the product of A, and A2
The algebras A,
•
and A2
may be
identified with the ideals in A, x A2 generated by the idempotents (1,0) and (0,1) respectively.
Conversely, if an algebra A ean be written as the
di reet sum of two-sided ideals Land N, as in (3.3), then L and N are algebras in their own right, and A is isomorphie to their product. rf L in Theorem 3.3 is an ideal, then we either get a nonzero element of L n rru:l A, or A decomposes into a product of k-algebras.
This latter
deeomposi ti on will be used frequently for proofs by induction on the dimension of A.
An
example is the following stronger version of Theorem
2.2. Let A be a
3.4 mEJOR.EM.
Then ei. thfT rod A A ami i dea l s 1
;t
finUe-dtmensioru:d. k-i:d.gebr'a wUh center C.
0 or there is (t t-l c01Tesporu:ience between idealS I of
f C gi ven by
0
1
I
n
C
Al PRIXlF.
By Theorem 2.2 we may assume that A has a nontrivial finite-
dimensional ideal L.
By Theorem 3.3 either rru:l A t 0 or A is a product of
algebras, in whieh case we are done by induetion on the dimension of A. 3.5 LEMMA. rod K
Let K be (t finUe-dimensional commutaUve h-«lgebra mUh
If F [s a separable [Jolynomi(tl in K[XJ, then rod K[Xl!(f)
O.
=
0
=
O.
If K has a zero-divisor, then K has nontrivial ideal, so K is a
PRIXlF.
produet by Theorem 3.3, and we are done by induction on the dimension of K.
We shall show that if 9 E K[X] and f divides g2, then either f divides
9 or K has a zero-divisor.
The Euclidean algorithm constructs ei ther a Let f = da and
zero-divisor of K, or a generator d of the ideal (f,g). 9
= db.
to d.
As f
divisor.
is separable,
= da
As f
But
CI
CI
is separable and strongly relati vely prime
divides g" ~ d 2 b 2
,
either
whenee either K has a zero ·divisor or degree of f.
CI
divides
or K has a zero-
d!)"
and cl are strongly relatively prime, so CI
CI
divides
!)2,
divides b by induction on the
Thus f divides 9 or K has a zero-divisor.
0
Chapter IX. Finite dimensional algebras
240 3.6 LEMMA.
Let p be a prime and " an element of
O.
such timt pR
f
01'
commulal i ue ,'ing R
Let '1 be apower of p, ami fand 9 be monic polynomicrls If f emd gare st,'ongly ,'elatiuely pr';me, thell
in R[Xj wUh Fg = x'1 - ". ei tller
(1
9 llas degf'ee O.
If '1 = 1, then the theorem is clearly truer so we may assume
PROOF. t.hat q
> 1.
(g'
o.
Let
sr
+ t 9 = 1.
As sf + tg
o.
Similarly 9
Differentiating f 9 = X'1 - " we get f' 9 +
1 we conclude that f
so f' = 0.
divides f'
Thus we can wri te f (X l = F0 (Xp land 9 (X) =
s(X)
=
1f;:5 Xisj(XP)
f 090
=
X'1 / p - ", so we are done by induction on '1.
and I(X)
=
1f:5 Xit; (XIl).
Then sofa +
90
(xl') •
t 0 90
Let
~ 1 and
0
The next lemma generalizes (VI.6.6) from fields to finite-dimensional commutative
algebras
wUh
zero
radieal.
Clas5ically
products of fields and the generalization is routine.
the
latter
are
As we may not know
how to write the algebra as a produet of fields, the constructive proof is more delicate. 3.7 LEMMA.
Let /, be
xP
If
t hnn
-
0
0
(Hso'ete field of clu::wocterisHc p, let K be
commultItlue k-algebnt, ond let a be on element
finite-dimensional
0
of K.
= (g whenc fond 9 ewe monic in K[X J (md IWlle deg,'ees g,'ealer
ze "0,
tllefl ei t he,' (i)
(ii)
PROOF.
a E KP
mdK", 0.
If we encounter a nontrivial ideal L. of Kr then by Theorem 3.3
either md K '" 0 or K = L fIl N for nontdvial ideals Land N.
In the
latter ease the theorem holds for Land N by induction on dimension. Hence ei ther r'ad L or rod N is l1on2ero, or the components of a in Land N are in LP and NP respeetively, so a
E
KP.
Thus we may assume that every
nonzero element af K that we construet is invertible. We want to wd te X'1 - n as a nontd vial power. with fand 9 manie polynomials of positive degree. of fand g. deg
9
=
d > O. g, d •
If cl -
Let d be the monie GCD
1, then Lemma 3.6 is violated, so we may assume that
=
Let. f
Suppose X'1 - n = f 9 i
be ffionic polynomials such that r = f ,d and f,d(g,d)i = (f,9\ ld i +1 . If (,91 = 1, then we have
and
1
0
=
91
writt.en X'1 - a in the desired form; otherwise we eontinue with new f, and d as the new g. xq -
[,g;
as t.he
After at most '1 steps we will have written
n = h(X)m for same monie h(X) and m > O.
Clearly m divides q, so m
241
3. Summands and the radieal is a power of p.
3.8 LEI'Ii'IA.
Let K be a
K[XI/(XP - a).
=
E
KP.
0
finite~Hmensional
commutative
k~lgebra.
(ii)
rad L f. 0
01-
I-ad L f. 0 implies rad K f. O.
Let e l' .•• ,e n be a basis for K over k.
eondition P,
either
a E K.
Then either
(i )
PROOF.
= (-h(O)m/p)p
Let k be a discrete fieLd of characteristic P satisFying
condition P. and L
Therefore a
e~, ... ,eR
are
independent
k
As
kP
over
or
satisfies they
are
If 2 ajej = 0, then 2 aje j E rad K, so we may assume that they are independent, in whieh ease the natural map from K to KP is
dependent over k P •
an isomorphism of rings. As k satisfies eondi ti on P, we ean decide whether elements of Kare dependent over k P , so the KP[al is a finite-dimensional
linearly
kP-algebra.
By Corollary 1.5 we ean either eonstruet a zero-divisor of
KP, henee of K, or the minimal polynomial of a over KP. ease
( 3.1)
ei ther gi ves us an idempotent of K,
induetion on dim K,
or
an element
of rad K,
In the former
and we are done by
and we
have
trivially
established (ii). KP.
Otherwise , let 9 be the minimal polynomial of a over We may assume that 9 is monie. Either 9 = xP - a P or xP - a P faetors
over KP.
In the latter ease either ,-ad K f. 0 or a E KP, by Lemma 3.7, so rad L f. 0 and (i) holds. If xP - a P is the minimal polynomial of a over KP, then KP[al is isomorphie to KP[X]/(XP - a P ) whieh is isomorphie, as a ring, to L.
Sinee KP[al ~ K we have (ii).
0
EXERCISES 1. Show that i f L is an n-dimensional
dimensional algebra,
left ideal of a
then Ln is finite dimensional,
nilpotent if and only if Ln
finite-
and L is
= O.
2. Show that the radieal of a finite-dimensional algebra A is the Jaeobson radieal of A.
Show that rad(A, x Az )
= rad A,
x rad Az .
3. Let A be a finite-dimensional algebra with a finite-dimensional radieal
J.
The left soele of A is (x E A : Jx
=
Ol.
Show that
the left soele is a two-sided ideal that has nonzero interseetion with every nonzero left idea1.
Identify the radieal, and the
left and right socles, of the algebra of lower triangular n x n
Chapter IX. Finite dimensional algebras
242
matrices over a discrete field. 4. Show that the following four condi tions on a finite-dimensional algebra Aare equivalent. (i)r-adA=O (ii) each finite-dimensional left ideal of A is a summand (iii) each
finite-dimensional
submodule
of
a
finite-
dimensional A-module is a summand. (iv) each finite-dimensional A-module is projective. 5. Maschke's theorem. Let G be a finite group, Iz a discrete field such that n = #G is not zero in 1" and A the group algebra of G over 1,. Let M be a finite-dimensional A-module and
=
1 ( -1 ) n2gEG g
.p2 = .p, and
'I' (M)
Show that .p is a map of A-modules,
Conclude that r-ad A = 0 (see Exercise 4).
= M.
4. WEDDERBURN'S THEOREM, PART a.E.
In
order
to
prove
Wedderburn's
algebras with zero radical, 4.1 LEMMA.
we
theorem
about
finite-dimensional
must put some restriction on the field
Let A be a finUe-dimensiona!
dimensional subatgebr'a oF the center' of A.
k~lgebr'a,
k.
and K -;t A a finite
Then either' K has a nontrivia!
idempotent or a nontrivial nitpotent, or' there is an element in A\j{ whose minima! po!ynomia! 9 over' K i seither separab!e, or 9 (X) =
xP -
r where
P = char k. or 9 (X) = Xm •
PROOF. We may assume that A = Kla) and construct, by Corollary 1.5, either a zero-divisor in K or the minimal polynomial g(X) of a over K. If K has a zero-divisor, then K has a nontrivial idempotent or a nontrivial nilpotent by Lemma 3.1. By (VI.6.3) either K has a nontrivial ideal or we can factor 9 into pairwise strongly relatively prime polynomials of the form fm(Xq) where m is a positive integer, q is either 1 or apower of the finite characteristic p of k, and f separable. This induces a decomposition of A into a product of algebras of the form KIX)/fm(xq), so we mayassume that g(X) = fm(Xq). If a q E K, then g(X) = xq - r'. If P P q = p we are done; otherwise a ([ K while (aP)q/p E K, so Kla ) is a nontrivial subalgebra of A and we are done by induction on dim A. If is the minimal polynomial of a q over K and is
a q ([ K and m = 1, then f
4. Wedderburn's theorem, part one
243
separable. If a g ( K and m > 1, then the minimal polynomial of f(a g ) over K is Xm. 0 4.2 THEOREM.
A diserete fieLd k is separabLy faetoriaL if and onLy if
eaeh finite-dimensionaL k-aLgebra A with rad A = 0 is either simpLe or has
a nontrivial ideaL. PROOF. By Theorem 2.2 we can restrict our attention to commutati ve k-algebras. Suppose A is a finite-dimensional commutative k-algebra. If A = k we are done; otherwise we can find a E A~ with minimal polynomial
g.
First suppose k is separably factorial and rad A = O. Then 9 cannot be of the form Xm. If g(X) = xP - r, then 9 is irreducible by (VI.6.6) lest r E kP in which ca se rad A t O. If 9 is irreducible, then we can replace k by k[X]/(g) and we are through by induction on dim A. So, by Lemma 4.1, we may assume that 9 is separable. Factor 9 into irreducible polynomials. If a satisfies one of these factors, then k(a) is a field and, as A is commutative, A is a finite-dimensional k(a)-algebra. Thus, as k(a) is also separably factorial , we are done by induction on dim A. If a does not satisfy any of the factors f of g, then each f (a) gene rates a nontrivial ideal of A. Conversely let K be a finite-dimensional extension field of n. If f E K[X] is a separable polynomial, construct the finite-dimensional commutative k-algebra A = K[X]/(f). By Lemma 3.5 we have rad A = O. The finitely generated ideals of A are in 1-1 correspondence with the monic factors of f, so f is either irreducible or has a nontrivial factorization. 0 We now characterize separably factorial fields in terms of decomposing algebras into products of simple algebras. This is the first part of Wedderburn's theorem. A diserete fietd k is sepambLy faetoriaL if and onLy iF every finite-dimensionaL k-aLgebra with zero radieaL is a produet oF simpLe aLgebras. 4.3 THEOREM.
PROOF. Let A be a finite-dimensional k-algebra. By Theorem 4.2, A is either simple or has a nontrivial ideal L. In the former case we are done. In the latter we may apply Theorem 3.3 and write A as a nontrivial product of algebras, and we are done by induction on the dimension of A.
244
Chapter IX. Finite dimensional algebras
=
Conversely, if each A with rud A
0 is a product of simple algebras,
then A is either simple or has a nontrivial ideal, factorial by Theorem 4.2. 4.4 'l'HroREl'!.
50
h
is separably
0
A disc,-ete fieLd
satisfies condition P if and onLy if
I~
each fini te-dimensionaL h-aLgebra has a fini te-dimensionaL rudicaL.
PROOF.
Suppose I< satisfies condi tion P.
element a
of rud A,
induction on dim A.
If we construct a nonzero
then we can pass to A/(AaA) and we are done by Let C be the center of A.
rud A -F 0, or rud A = A'rod C.
By Theorem 3.4 either
Thus we may assume that A is commutative.
If we construct a nontrivial idempotent of A, then we can factor A into two algebras of smaller dimension and we are done by induction on dim A. Let K be a fini te-dimensional h-subalgebra of A wi th ,-ud K
= O.
Ini tially
we take K = h, and we proceed by induction on dim A - dim K. Either K = A, in which case we are done, or we can apply Lemma 4.1 to construct an element a in A\}\. whose minimal polynomial 9 is either separable, or g(X) = xP - r, where p = char 1<, or g(X) = Xm. If 9 (X) = Xm , then a is a nonzero element of ,-ud A.
If 9 (X) = xP - '-, then,
as I< satisfies condition P, Lemma 3.8 says that either rod K[a) -F 0 or rad K[a]
=
In the latter case we replace K by K[a] and we are done by
O.
induction on dim A - dim K.
If 9 is separable, then rud K[a] = 0 by Lemma
3.5, and we replace K by K[a] as before. Conversely suppose K is a finite-dimensional extension field of 1<, and Consider the I<-algebra L
a E K.
=
a polynomial
f
E K[X) such that xP -
0
4.5 COlIOLLARY. euery
0
f2
but does not divide
f.
and f is a proper factor of
0
A disc,-ete fieLd I< i s fully focto,-ioL
fini te-dimensional
then
If rod L -F 0, then there is
divides
Then the greatest common divisor of Xl' xP - 0, 500 E KP by (VI.6.6).
If rad L = 0,
K [X lI(XP - a) •
a rt KP for if a = r P , then 0 -F X - ,- E rud L.
algebra
A
ouer
I.
has
0
if ond onLy i f
fini te-dimensional
ni/potent ideaL I such that AlL is a l'roduct of simple k-atgebras.
PROOF.
As k is fully factorial if and only if I, is separably factorial
and satisfies condition P, this follows immediately from Theorems 4.3 and 4.4.
0
245
4. Wedderburn's theorem, part one EXERCISES
1. Let K be a finite-dimensiona1 eommutative k-algebra and 9 E K[Xj. Suppose 9 ~ F1F2···Fn where the Fi are pairwise strongly relatively prime. Show that K[X1/(g) is isomorphie to the produet of the k-a1gebras K[Xj/(F i ). 2. Let k be a faetorial field and f E k [X j.
of A ~ k [X 1/( F ) , k-algebras.
and describe
All
as
Identify the radical I a
product
of
simple
3. Let G be the symmetrie group on {I, 2,3} and A the group algebra of G over IQ. Deeompose A into a produet of simple IQ-algebras (see Exercise 3.5). 5. MATRIX RI~
AN!)
DIVISICfi ALGEBRAS.
The seeond part of Wedderburn's structure theorem for
semi-simple
algebras says that a finite-dimensional simple algebra is isomorphie to a full ring of matriees over a division algebra. being able to eonstruet nontrivial left ideals.
This theorem hinges on Onee we have a nontrivial
left ideal we ean use it to deeompose the original algebra. 5 .1 '1mX>RE7'I.
Let
A be
nontrivial LeFt ideaL oF A.
a
Fini te-dimensionaL
k-üLgebra,
and L a
Then either'
(i) A has a nonzero radieaL (ii) A is a produet oF Finite dimensionaL k-üLgebras (iii) A is isomorphie to a FuL! matrix ring ouer some k-üLgebra oF dimension less than A. We may assume that L is finite dimensional by passing to the prineipal left ideal generated by a nonzero element of L. If L is not a faithful A-module, then the representation kernel of L is a nonzero ideal PROOF.
of A; whenee by Theorem 3.3, either rad A t 0 or A is a produet.
Thus we
may assume that L is faithful. Let B be the centralizer of A on L (that is, the A-endomorphism ring of
L). By the density theorem (2.1), either L is redueible (and we are done by induetion on the dimension of L) or A is a full matrix ring over the opposite ring of B. It remains to show that the dimension of B is less than that of A. By Theorem 3.3 either rad A t 0 or L is a summand of A. In the latter ease the algebra B is a proper subalgebra of the eentralizer
246
Chapter IX. Finite dimensional algebras
of A on A, which is the opposite ring of A.
0
The fundamental problem is to be able to recognize whether a given finite-dimensional algebra is a division algebra or not, in the sense of being able either to assert that it is a division algebra or to construct a nontrivial left ideal.
If we could do that,
then Theorem 5.1 would
imply that every fini te-dimensional h-algebra has a
finite dimensional
radical, and modulo its radical it is a product of full matrix rings over division
algebras.
recognize whether
This
condition
an arbitrary
is
equivalent
finite-dimensiona.1
to
being
able
to
representation of a
fini te-dimensional k-algebra is reducible. 5 • 2 'lHOOREM.
The
foUowing
condi Lions
on
a
disCl-ete
field
kare
equivalent. (i) fach finite-dimensionnl I<-algebra is eithel' a division aLgebr'a or
hns a IlOntr'ivial left ideal.
(ii) Each finite-dimensional
lefl module M ovel' a finite dimensionnl
h-algebr-a A is ei ther' r'educible
(iii) Each
finite-dimensionnl
r'adica!,
and
01'
ilTeducible.
h-algebnJ
A/nw A is
A hns
pI'oduct
Cl
of
a
full
finite-dimensi,onal matrix
f'i,ngs
over
elf vi sion al gebras.
PROOF.
Clearly both (ii) and (iii) imply (i).
we wish to establish (ii).
Suppose (i) holds and
We may assume that M is a faithful A-module.
Let B be the eentralizer of A on M. a nontrivial A-submodule of M.
If R has a zero-divisor b, then bM is
If B is a division algebra, then by the
densi ty theorem ei ther M is reducible or A is the centralizer of B so M is i rreducible.
Now suppose
(i) holds and we wish to establish (iii).
division algebra we are done.
If A is a
If A has a nontrivial 1eft ideal, then by
Theorem 5.1 either A has a nonzero finite-dimensional nilpotent ideal I, in whieh ease we pass to A/l and are done by induetion on dim A, or A is a produet of fini te-dimensional k-algebras and we are done by induction on dirn A,
or A is isomorphie to a full matrix
induction on di.rn A.
ring and we are done by
0
For what fields k do the conditions of Theorem 5.2 hold?
Finite fields
and algebraieally closed fields provide trivial examples.
The field of
algebraie
real
numbers
admits
only
three
finite-dimensional
division
247
5. Matrix rings and division algebras
algebras, and a constructive proof of this statement shows that this field satisfies the conditions of Theorem 5.2. 5.3 THEOREM.
Let k be a diserete subFieLd oF ffi that is aLgebraieaLLy
dosed in ffi, und H = k(i,j)
IF A is a
the quaternion aLgebra ouer k.
Finite-dimensionaL aLgebra ouer k, then either" A has a zero-divisor, or A is isomorphie to k, to k(i), or to H.
If A
PIlOOF.
= k we are done; otherwise let a
As A is finite
E A\lz.
dimensional we can construct a nontrivial polynomial satisfied by a. field k(i)
!;;
ce is discrete and algebraically closed, so
and no irreducible polynomial over
I~
k
The
is factorial,
has degree greater than 2.
Thus
either A has a zero-divisor or a is of degree 2, so we may assume that a E k(i) where i E A and i
2
-1.
=
The centralizer of i in A is a finite-
dimensional algebra over the algebraically closed field k either has a zero-divisor or is k{i) itself. If A
= k{i) we are done.
we may assume that iß + ßi set j = ß + rij2.
We may assume the latter.
Otherwise we can construct 13
13 2 = -1 just like we constructed i E A. E kU)
Then ij + ji
As iß + ßi
n k{ß) = = 0 and
k. j2
j
to get
commutes with i, so Does the field 5.2?
~
jj' E
j2 =
-1.
If j'
k{i) whereupon
j'
E A\lz{i)
commutes with
Let iß + ßi
=
r
i
with
and 13,
E k,
and
= S2
for some
Otherwise j2
< 0 so we
E k.
s E k, then j - s is a nonzero zero-divisor of A. can normalize
hence
(i ) ,
If j2
is another such E k{i,j).
j,
then
j j'
0
of rational numbers satisfy the conditions of Theorem
Certainly we are not going to produce a Brouwerian counterexample
when k
=~.
algebras
probably a close analysis of the classical theory of division
over~,
in analogy with Theorem 5.3, will yield a proof.
EXERCISES 1. Show that every finite-dimensional commutative algebra over a
fully factorial field is either a division algebra or has a nontrivial left ideal.
Show that over a factorial field every
algebra of prime dimension is either a division algebra or has a nontrivial left ideal.
248
Chapter IX. Finite dimensional algebras OOI'ES
Most of the material in this chapter appeared in [Richman 1982]. might be interesting to see how the theory develops for composition series (Artinian rings).
It
rings wi th
Frobenius algebras, quasi-Frobenius
algebras, and Lie algebras are also natural topics for investigation.
Chapter X. Free Groups
1. EXISTEK:E AND UNICUFNESS.
A group F is a free group on a subset 8 of F if for any group Hand any function f from 8 to H, there is a unique homomorphism from F to H extending F. If F is a free group on 8, then 8 is called a (free) basis for F. Before showing how to construct free groups, we show that there is, up to i somorphi sm, only one free group on a set. 1.1 THEOREM.
Let F, and F 2 be free 9roups on 8, and 8 2 respectiuety.
IF F is an isomorphism from 8, to 8 2
,
then there is a unique isomorphism
from F, to F 2 that extends f.
PROOF. Let 9 be the inverse of F. is a unique homomorphism f* from F, to a unique homomorphism 9* from F2 to F, from F 2 to F 2 extends the identity map 82
,
F, is a free group on 8" there F2 extending F. Similarly there is extending 9. The homomorphism F*9* on 8 2 , Since F 2 is a free group on As
and the identity map on F 2 also extends the identity map on 8 2
identity map on F2 must equal F*9*.
,
the
Similarly 9*f* is the identity map on
F,. Thus f* is the required i somo rphi sm; F* is unique because F, is free on 8,. [] To construct a free group on an arbitrary set 8, define the set 8 U 8-1 to be 8 x {l,-l}, with 8 identified with 8 x {I}, and (s,-t) denoted by (s,~)-'. Let F(8) be the free monoid on 8 U 8-1 • We define two equalities on F(8). The first, denoted by the symbol '==', is the usual equali ty on a free monoid. To define the second equali ty, we take the notation x -, seriously:
Call two words in F (8) adjacent i f one can be
written as uw and the other as uxx-'w with u and w in 8 U 8-1 .
F(8),
and x in
words u and ware equal in F(8), written u = w, if there exists a sequence of words u == wl,w2, ... ,wn == w, such that wi is adjacent to wi +l for each i < n. We must show that the natural map of 8 into F(8) embeds 8 as a subset Two
249
250
Chapte~
X. Free groups
as elements of F(S),
of F(S): that is, if sand s
are in S, and s
then s = s' as elements of S.
To this end, and for i ts general utili ty,
= s
we establish the Church-Rosser property for F (S) .
If xi
(C
S U S-l for i =
1, ... ,Tl, then the length of z:= x1x2"'xn is deHned to be P(z)
The
= n.
1ength is a function on F (8) wi th respect to ':=' but not wi th respect to
'=', as the
wo~ds
x
and
Xy-l y
are equal but their lengths are land 3
respectively. 1.2 THEOREM (Church-Rosser property). TheT! we can find an integer- n, and
< I'(w i
wor-ds such that if P(wi_l)
Lel
=
U
lJ
chain u ::: w1""
0
< Tl,
) for i
be
words
in
F(S).
,wn := v of adjacent
tllen I'(w;)
< 1'(w i +1)'
PROOF.
As U = l) ther·e is a sequence of adjacent words u := w1"" ,wn We shall prove the result by induction on the total length N = I I'(w i ) of the sequenee. We may assume that n > 1. If for same i. we have
u.
< 1'(10;) and fl(lV i
i'(wi_1)
>
)
deleting a part xx - 1 , and
then
i'(w i +1)'
IV i
IVi_1
If these two parts eoineide, then!V i and
y,,-1
is obtained from lV i
+1 is obtained fr om
by
!Vi
by deleting a par-t
!Vi+1
may be omitted from
the sequenee and we are done by induetion.
If these two parts overlap
without eoineiding, then wi has apart
and lVi-l and wi.+l are both
ZZ-,Z
obtained by replacing this part by z, so and
!Vi
+1 may be ornitted.
IVi---l := !Vi+l
onee again, and "'i
In the remaining ease, where the two parts are
disjoint, we can delete both parts from!V i
to obtain a new sequence with
total length N-4 and we are done by induetion.
0
It is an easy eonsequence of (1.2) that S is a subset of F(S).
The (assoeiati ve) multiplieation on F (S) that is, lJ
- 1
if
LJ
= LJ'
... xz'xi';
_
=
and w
= w',
then vv-'
=
then V-lU
=
Ulv
respects the equality
- ,
'_I
~
= v'w'
1, where 1 is the empty word.
Thus
F(S) is a group. Ta show that F (8) 1.S a free gmup on 5, let f be a funebon from S to a group H.
If
f
is a
hornomorphism from F(S) ),
setting
-
f
(lv)
-
= f
f(x1)"'f(x n )
F(x1"'xn)
(w')
if w '"
w'.
defines
a
then
Moreover,
hornomorphism
because
Thus F (5) is a free group on S, whieh we will
refer to as the free group on S. 1.3 THEDREM.
to Hextending f,
so f i s unique.
We summarize the above in the following.
If 8 is a set, then F(8) is a free group on S_
0
251
1. Existence and uniqueness A
ward
x i x i +1
=x1x2"
w
= 1 for same
'xn
in F(S), with eaeh
xi
in S U S-l, is reducible i f
= 1, ... ,n-1.
If w is not reducible we say that w is Note that if u and v are in F(S), and if uv is reduced, then u
reduced.
i
and v are reduced.
If S is a diserete set, and
u
and
u
are reduced words
in F(S), then lW is either reduced or u == ax and v == x- 1 b with ab reduced. If w = w', and w' is reduced, then w' is called the reduced form of Wi the Church-Rosser property of F(S) implies that i f w = w', and wand w' are both redueed, then
IV
= w',
so the reduced form i s unique.
If S is
discrete, then every element of F(S) has a unique reduced form.
From this
we immediately get 1.4 'HIEOREM.
If S is a discrete set,
1.5 THEOREM.
If w E F(S) and wrt = v wUh l!(u) ~ I'(w) for
= 1 or w is reducibIe. Thus u if w" = 1 for same n > 0, then IV = 1. then w
PROOF. l!(w , ).
Write w ==
U-'W , ll
then F(S) is a diserete group.
(for example:
II
==
1 and w , ==
If wfl = v with 1'(0) ~ l'(w) for some n
f!(U-1W~1l)
\
l!(u),
or
w
same
f,'ee grollp is torsion f,'ee;
is reducible, or
='
!Vi
X-1W2X
> 1,
tllllt is,
anel induct on
w)
> 1, then
11
0
by (1.2) either
for some x in S U S-l.
In the first ease w, == 1 so w = 1, in the second we are done, and in the third we are done by induetion on 1.6 THEOREM.
respectivety.
PROOF.
Let
Fand F'
l!(w , ).
be
(,'ee
0
gnJUps on fLnite
Then Fand F' ar'e [somm'pllie if und only if
sets S und S'
*''
=
#S'.
If #S = #S', then Fand F' are isomorphie by Theorem 1.1.
prove the eonverse we show how to recover the number *,S from F. the subgroup of F generated by the elements v 2 with v in F.
To
Let N be
Clearly N is
anormal subgrouPi the quotient group FjN is abelian because the square of every element is 1, so xyx - ' y -' = X!JXY = 1. We will show that FjN is a finite set with 2#S elements. If w == XIX2"' *x n i
such that
vs(w')
Xi
modulo 2.
E F, and s E 8, let V s (w) denote the number of indices = s or Xi = S-1 rf w =w', then vs(w) is congruent to
Let D = {w E F :
vs(w)
is even for eaeh sES}.
Clearly
D and D is a detachab1e subset of F. Conversely, as FjN is abelian, and the square of anY element of FjN is 1, we have D ~ N, and each element of FjN ean be written uniquely as a product of distinet elements of 8. 0
N
~
If F(8) is the free group on a finite set Sr then #8 is an invariant of
Chapter X. Free groups
252
F(S), ealled the rank of F(S).
rf S is a eountably infinite set then F(S)
is said to be of countable rank. 1.7 THEXlREM. morphism
IF G is a group, then there is a Free group F ami a epi-
f : F ..... G.
IF G is discrete,
then F can be
tul~en
to be discr'ete.
Let F = F(G) be the free group on the set G. Using the identity function from G as a set to G as a group we obtain, by the PRCKJF.
definition of a free group, a unique homomorphism F from is the identity on G. 0 Let U be u subset oF a group G such tlwt U
1.8 LEMMA.
F(G)
n U- l
to G whieh
= <1>.
Then
U is a Fr'ee basis for a subgroup of G iF and only iF whenever' u1u 2·· ·u n 1 wUh n ? 1 und each u i E U U U-1 , then u i u i +1 = 1 Far' some i.
PRCKJF.
rf U is a basis for a free subgroup of G, then U satisfies the
eonditions of the lemma by (1.2). Conversely suppose that U satisfies the eondi tions of the lemma, and let F (U) be the free group on U. The inelusion map from U to G extends uniquely to a group homomorphism f from F(U) to G whose image is the subgroup generated by U. Suppose
=
1
where eaeh
H
i
E U U U- 1 •
F(u1···un) As f
=
f(u1)···F(u n ),
is the inclusion map on U I we have F(ui.)
U U U- 1
1.
E
in G, so there is i with f (u i )F (ui.+1) = 1, and therefore u i lli +1 = Thus the kernel of f is trivial, so F(U) is isomorphie to the subgroup
generated by U.
0
EXERcrSES 1. Show that if F(S) is abelian, and a,b E S, then a
2. Show that if
w
E F(S) and sES, and sw =
ws,
= b. then
w
sn
for
some integer n. 3. Show that every word in F(S) has a redueed form if and only if S is diserete. 4. Show that F(S)
=F(T)
implies S is isomorphie to T if: (i) S is finite, (ii) S is IN, (iii) S is a detachable initial (no gaps) subset of IN.
253
1. Existence and uniqueness 5
Let 8 be a finite set of cardinality m.
Show that i f T is a
fini te subset of F (8) of cardinali ty greater than m, then there exists a nonempty product of distinct elements of T that is equal to a product of squares. 6. Let
8
{sl"",sm}
and T {tl, ... ,t n } (not Use Exercise 5 to show that if F(8)
discrete). m
< n, then
= t j for some
t{
i
necessarily F (T),
~
and
< j.
2. NIELSFN SETS.
Let 8 be a discrete set.
We study condi tions on a subset U of F (8)
that ensure that U is a free basis for the subgroup
generated by U. The reduced Iength of an element w in F(S) is the length of its reduced form, and is denoted by Iwl.
A subset U of the free group F(S) is called a Nielsen set if (NO)
U
n
U- l =
and for all x,y,z E U U
u- l
we have
{I xl,
(NI)
I f xy #- 1, then Ixy 1 ~ rnax
(N2)
If xy #- 1 and yz # 1, then Ixyzl
Note that NO ensures that I ( U.
1y 1}
>
lxi - Iyl + Izl.
We will show that i f U satisfies NO
and N2, then U is a free basis for . Conditions
2.1 EXAMPLES.
Then
{s,t,u,v}.
(st)-1(S2)(tS)-1
=
the t- 2
set
NI U
and
N2
are
independent.
{S2,st,tS}
satisfies
it does not satisfy N2.
,
satisfies N2 but not NI.
The set V
Let 8
NI
=
but,
as
{tuv, suv}
0
If u and v are reduced words, then there exist unique reduced reduced words a, band e such that u == ab- 1 and v == be, and ae is reduced. call b the part of v that cancels in the product uv. called the part of u that cancels in the product
[IV.
We
Similarly b - 1 is The following lemma
helps to explain the meaning of NI. 2.2 LEMMA.
the
f,'ee
Le tUbe
grollp F (8) .
If
fini te se t of ,'educed wonls, so ti sfyi ng NI, i
0
II
and " aJ'e wo,'ds
i
U, emd bis
11
that eaneets in the pr'oduct uv, then 21bl ~ min(lul,lvl). PROOF.
This is an easy consequence of NI.
0
t he
pa,' t
of
11 V
254
Chapter X. Free groups
The following lemma explains the significance of N2, and sets the stage for invoking Lemma 1.8. 2.3 LEMMA.
Let S be a discrete set, and let U be a set of l'educed
words in F(S) satisfying NO and N2. U U U- 1 , and u i u i +1 t 1 Fm' each i
Let w '= u1u2···un wUh each u i in < n. Then u i '= aibLc i . wUh b i t 1,
und b 1b 2 •• ·bn is the reduced fOl'm of w.
PROOF.
Set aO '= e n '= 1. Let ci be the part of u i ' and a i +1 '= ci' the part of Ui +1' that cancels in the product uiui+1 for i < n. By N2 we have
> lu i _ 1 1 - IU i I + IU i +11 so u i '= aibic i , with b i t 1, i f 1 < i < n. From (1.5) we know that u1ul t 1 and Unlln t 1, so N2 says that Ill 111 1U2 1 > 111 2 1 and Illn_111nUnl > IUi-1uiui+11
lun _ 1 1; therefore b 1 t 1 and bn t 1.
As ci is the part of u i that cancels
in the product u i u i +l' we have aibibi+lci+1 is the reduced form of uiui+l' so b 1b 2 •• 'bn is reduced. 0
IF U is a set of words satisFying NO and N2 in a free
2.4 COROLIARY.
group on a discl'ete set, then
PROOF.
U is
a free basis For
.
This follows from Lemma 2.3 and Lemma 1.8.
0
As the set V of Example 2.1 satisfies NO and N2, it follows that V is a
free basis for . Unlike in the abelian case, a free group of small rank can have a subgroup that is free of large rank.
In fact there are infinite-rank free
subgroups of a finite-rank free group. 2.5 THEDREM.
A Fl'ee gl'Ollp on a Fini te set oF cQ1'dinal Uy two contains
a countable-l'ank fl'ee subgroup.
PROOF.
Let {x,y} be the basis for the free group, and consider the set
It is readily verified that U is a Nielsen set, hence a free basis for the subgroup it generates.
0
EXERCISES 1. Let F be the free group on the two-element set {x,y}. {~ynx-my-n
: m,n E ;r\{0}}
is
free
basis
for
the
Show that commutator
255
2. Nielsen sets subgroup of F.
2. Show that i f U is a Nielsen set of words in a discrete free
group, and if w : ul" 'un wi th lI t in U U U- l and each UtUt+l f. l. then Iwl ~ max { Iull ..... IUn I) .
3. Show that the subgroup constructed in (2.5) is detachable. 3. FINITELY GEm:RATED SUBGBOOPS OF FREE GROOPS.
In this section we show how to transform a finite set of generators for a subgroup of a free group on a finite set Sinto a Nielsen set. Let U and V be Finite sets oF wor'ds.
3.lDEFINITIOO.
We say tllat V is
obtained From U by a Niel.sen transformation iF either
(TO)
v
(Tl)
V
(U \
{u)) U {U-I) where U € U
(T2)
V
(U \
{u)) U {v) where v is either UU· or u·u. For' some
u' A
{I)
= U \
in U U U- 1 diFferent (rom u and u- I
•
transformation of type T2 is referred to as replacing
u
by v in U.
Note that i f V is obtained from U by a Nielsen transformation,
the
=
and #U
2 #V.
cardinali ty of
then
(Transformations of types Tl and T2 may decrease
the
set U.
For
example
if
U
=
{a ,ab,b) ,
the
transformation of type T2 replacing b by ab decreases the cardinali ty. ) Also, if V is obtained from U by a transformation of type Tl or T2, and
#U = #V, then V is obtained from U by a transformation of the same type. If S is a finite set, we can linearIy order the words of F(S), with Fix a linear ordering of S U s-1 and extend
respect to ':', as follows.
lexicographically to an ordering on the words of F (S) .
If u and v are
words of F(S), define u < v if either (i) e(u) (ii)
<
e(v)
or
e (u) = e (v)
and
1I
comes before
v
in the lexicographic
ordering. Note that each word has a
finite number of predecessors under
this
ordering. 3.2 'l'HEDREM.
Let S be a Fini te set and U a Fini te subset oF F(S).
Then there ts a sequence oF Nietsen U'ansFor'mations tllat u'ansForms U to a Nietsen set.
256
Chapter X. Free groups
First we set up a measure of the size of U on which to base our induction. If W is a word, then we can write the reduced representative of W uniquely as wL wR where IWL I is the greatest integer not exceeding (lwl+1)/2. We define the function ~ from F(S) to~, somewhat cryptically, PROOF.
by setting ~(w) equal to the number of words v such that v < wLwR'.
Note
that if lu, I < IU21, then ~(u,) < ~(112)' Let ~U = 2:uEU~(u). We proceed by induction on ~U. By aseries of transformations of type TO and Tl we may assume that U n U- 1 =~. If Ix!) I < lxi for X,!) E U U U- 1 and xy f. 1, then x f. y by (1. 5) and we can replace x by xy (er x -, by and decrease ~U. Thus we may assume that U satisfies N1. Let x, y and z be reduced words in U U U- 1 wi th xy f. 1 and
y-1 X -')
Suppose Ixyzl ~ lxi - Iyl + Izl.
yz f. 1.
Then we can write -
ap
- 1
Y -
pg
- 1
X
z - gc.
Ipl > Igl, then Ixyl < lxi; i f Ipl < Iql, then Iyzl < Izl; i f Ipl > lai, then Ixyl < Iyl; i f IClI > !cl, then Iyzl < Iyl; so none of these cases occur as U satisfies N1. Thus Ipl == Igl ~ mint la I,!c I), and Ixy I == Ix I and Iyz I == Iz I . Note that p f. q since y f. 1. If p < g (in the lexicographic ordering) , then ~(yz) < ~(z), while if q < p, then
If
~(xy)
< 'I'(x).
In either case we can apply a Nielsen transformation that
reduces 'I' (u) for one element alone, thus decreasing ~U. 0 3.3 COROLLARY.
11
in U while leaving the other elements
Ever'y Fini teLy gener-ated subgraup aF a Fini te-r-ank Free
graup 1.s Free, and has a Nielsen set as a Fr-ee basis. 3.4 'l'HOOREM.
0
IF F is a Fr-ee gr-aup aF fini te r-anh n, and U 1.s a set oF
generatar-s Far F, then U cantains at least n elements.
Moreaver, iF U has
exactly n elements, then U is a Free basi s Far- F.
PROOF. As each element of a free basis for F is a product of finitely many words in U, we may assume that U i s finite. By ( 3 .2) we can transform U into a Nielsen set V by a sequence of Nielsen transformations. Thus = and #V ~ #U. As V is a Nielsen set, and U is a generating set, V is a free basis for F, so V has n elements. If U has n elements, then no transformations of type TO were used in transforming U to V.
Therefore V can be transformed to U by transformations of types Tl and T2.
257
3. Finitely generated subgroups of free groups
But these transformations can be used to define a function from the free basis V onto U, that when extended to a homomorphism of F is an i somorphi sm.
0
3.5 'l'HIDREM.
Let
F
generated subgroup oF F.
be
a
Finite-r'anl,
Free
gr'oup and G a
FiniteLy
Then G is detachabLe.
PROOF. Since G is finitely generated, (3.3) says that G has a Nielsen set U as a finite free basis. Let w E F. From (2.3) it follows that if wEG, then w can be written as a product of not more than Iwl elements of U U U- l . As F is discrete, we can check to see if w can be so written. 0
EXERCISES 1. Show that the commutator subgroup of a free group on a twoelement set is not finitely generated (see Exercise 2.1). 2. Gi ve a Brouwerian example of a countable subgroup of a fini terank free group that is not free. 3. Show that a finite-rank free group F is Hopfian in the sense that any map from F onto F is one-to-one. 4. Show that finitely generated subgroups of discrete free groups are free and detachable. 4. DETACHABLE SUBGRalPS OF FINI'l'Fr-RANK FREE GRCUPS.
In this section we prove that detachable sugroups of finite-rank free groups are free.
Also, a subgroup of finite index n in a free group of
finite rank r is free of rank n(r-1) + 1. Let F be a group and G a subgroup of F. A function T from F to F is a (right) transversal for G if T(x) € Gx for each x in F, and if T(x) = T(y) whenever Gx = Gy. In other words, T is a choice function for the set of right cosets of G in F. Note that T(T(x)y) = T(xy) for all x,y in F. Let F be the free group on a finite set S. If w == uv is a reduced word in F, then u is called an initial segment of w, and v is called a final segment of w. A transversal T for a subgroup G of F is a Schreier transversal i f T(w) is reduced for each w in F, and if T(F) = (T(w) : w E F} is closed under taking initial segments. If T(F) is also closed under taking final segments, then T is a two-sided Schreier transversal. Note that a Schreier transversal is a function trom F with the equality
258
Chapter X. Free groups
'='
to F with the equality 4.1 THEOREM.
'=='.
F is the Fr-ee grollp on a finite set S, and G is a
IF
If G is also
detachabte sllbgroup of F, then G 'ws a Sdweier tr'ansver'sal.
anormal sllbgroup, then G /las a two-sided Schr'eier tr-ansver'sal. PROOF.
Each
Equip F with the total ordering defined just prior to (3.2).
element
in
F
has
a
finite
number
of
predecessors,
and G
is
detachable, so for each w in F we can define T(w) to be the first word in
Gw (note that Gw is the set of all words that are equal to a word of the form gw).
As 1 is the first ward in Go 1, we have 1 E T(F).
Also each
word in T(F) is reduced. If w
==
uv is the smallest element of Gw, then u is the smallest element
of Gu, for if there is 9 in G with gu = c Thus T is a Schreier transversal. the
smallest
transversal.
of Gv
element
gw =
If G is normal, then Gw
vG,
=
< u, then
whence T
is
a
=
Cl'
< uv
we,
twa-sided
w.
so v is Schreier
0
4.2 LEMMA.
Lel F be the Fr'ee group on the Finite set S. and let T be a
Sdu'eier- transversal For- a subgroup G oF F.
sand s' be elements of
Let
S U S-l, and t and t' be elements of T(F) such that neither- ts nor t's' is equal to an element oF T(F).
Let
u be the reduced form oF T(tS)-l t ,.
Then
(i) (ii) (iii) PROOF.
and t's'T(['s' )-1 are ,-educed,
tsT(tS)-1
IF tsT(tS)-1 sus
=
t's'T(t's' )-1, then t == t' und s _ s',
is reduced unless
1I
=
=
1 und s'
S-I.
If ts is not reduced, then t == t"s-1 with t n E T(F), as T is a
Schreier transversal.
Thus t s = t" E
T (F),
contrary to the hypothesis .
If sT(tS)-1 is not reduced, then T(ts) == t"s with t" E T(F), as T is a Schreier transversal.
But then
t" =T(I") =T(T(ts)S-I) =T(tss- 1
=T(t) = t,
)
so ts = t"s E T(F), contrary to the hypothesis. both
reduced,
it
follows
that
t sT (t s ) - 1
is
As t sand sT ( t s ) - 1 are reduced.
Similarly
t ' s' T( t ' s' ) - 1 i s reduced. If tsT(tS)-l = t's'T(t's' )-1, then, as they are both reduced, and as ts is not an initial segment of t', and t's' is not an initial segment of t', it follows that t
t' and s
= s'
.
To show (iii) it suffices to show that su and us' are reduced.
We have
259
4. Detachable subgroups of finite-rank free groups
tL
= T(ts)-'t',
and
and sT(tS)-1 are reduced.
tL
If su is not reduced, then
must cancel in the product T(tS)-l t , leaving an
T(tS)-1
so T(tS)S-1 is an initial segment of t ' . since T is a Schreier transversal. whereupon ts
T(ts)
~
T(F),
E
on the left,
S-l
Therefore T(tS)S-l 1S in T(F)
Hence
=
t
T(T(ts)s"')
T(tS)S-l,
contrary to the hypothesis.
So
is
Stl
reduced. Similarly,
if us
is not
T(ts)-','s'
=
reduced,
then,
a5 ['s'
i5
reduced, t's' i5 an initial segment of 1'(ts), hence is in T(F), contrary to the hypothesis. 4.3 LEMMA. t,t'
0
Let l' oe
E T(F) and s E F,
(i)
Moreover',
Let
= T(ts)
t'
(iil
for a subgroup of a gr"otLp F.
U'ansver"saL
(1
Then the foUowing condiLions are equivalent:
t =T(t's-l).
[f these conditions hold, then
(iii)
PROOF.
f(ts)f(t's-1) = 1, wher'e f(lV) = WT(W)-l.
If t'
then
= T(ts),
T(t's-1) =1'(T(ts)s-') =T(tss-
1 )
=T(t) = t.
so (i) is implies to (ii) whence (ii) implies (i).
If (i) and (ii) hold,
then f(tS)-l =T(tS)S-l t T(t')5-
4.4 'IHElORElIJ. S.
1
t- 1
Let
=
Let T be a Sc/weie,"
=
=
(t's-1)[-1 = f(t's-l).
G be a subgr"oup of the (,-ee group F on
to G by f(w) = w1'(w)"',
Y
=T(T(t's-')s)s-'t-' =
-l
t's-1 t -1
tr"ansver"sal For" G.
0 0
discr-ete set
[JeFine the functi.on f from F
Then tI-,e set
(f(ts) : sES,
tE 1'(F), and f((s);< I}
is a ba.si.s Far" G.
PROJF.
If
w
E G,
then 1'(w) = 1,
so
it is enough to show that f (w)
G
and sES.
E
f((Tw)s) E Y
f (w)
E
w in F.
Let w
F
E
Then
f(w)f(T(w)s) = wT(w)"T(w)sT(T(w)s)-t Ag
Thus to show that
F(w) = w.
for all
U {I}
it
follows
that
= wsT(WSJ-1
f{ws) «Y>
= f(ws).
if
and
only
if
.
Now let w be a reduced ward in F. for same s in S, or Iwsl
<
Either- w
Jwl for some s in S.
=
1 €
or Iws -1 I < I
Iw
I
In the latter two cases,
260
Chapter X. Free groups
.
Hence
We will show that Y is a basis by appealing to Lenuna 1.8.
First
by induction on length,
either
f{ws-
1 )
€
or
f{ws) €
.
w €
observe that y- l = (F(ts- 1
)
follows from Lenuna 4.3.
T(F), and F(ts- 1
So if lJ
Y U y- 1 , then
€
Next we note that Y n y-1
S U s-1, and t in T{F).
4.2.ii.
s € S, t €
=
Now suppose that Y1Y2"' Yn
}"# 1}
y =
F{ts} with s in
= ~ follows from Lenuna
I with each Yi in Y U y- 1
({tisi) == t i sJ{t i s i }-l, with si in S U s-1, and t i in T{F). be the reduced form of T(tisi}-lti+1' Then
lJi ==
Write Let
ui
(*)
Were tis i equal to an element of T(F}, then lJi would equal 1; therefore we can apply Lenuna 4.2.iii and conclude that either siuisi+1 is reduced for each
i
< Tl, or for some
i
< n we have
In the former case
"i = si si+1 = 1.
the right hand side of {*} is reduced, so the left hand side cannot equal
1.
In the 1atter case
ti
= T(tis i }, so Yi Y i+1 = 1 by lenuna 4.3.
+1
0
Theorems 4.1 and 4.4 imply that detachable subgroups of finite-rank free groups are free.
If the subgroup has fini te index,
then we can
compute its rank as follows. 4.5 THEDREM.
oF f ini te index
PROOF.
Let F = F{S) be a free gl'oup of ,'ank " and G a subgrollp Tl.
Then Gis a fn~e g"oup
Tl
(,'-1) + 1.
Let T be a Schreier transversal for G, and Y be the basis We only need to show that Y has n{I' - 1} + 1
defined in Theorem 4.4. elements.
of ,'ank
Define maps A and M A :
T{F}\{1} M :
Y
~
T(F)
x S
T(F} x S
~
by A{t}
{
( I " S } i f t - t' s
with s
€
if
with s
€ S
{t,s)
t
-
t's-l
S
and M{Y)
=
(t
,s) such that F{ts}
Lenuna 4.2 guarantees that the map one.
~
=
y.
is weIl defined; it is clearly one-to-
The map A maps into T{F} x S because T is a Schreier transversal; is
easily seen that A is one-to-one.
The range of A is the set of pairs
261
4. Detachable subgroups of finite-rank free groups (t,s)
in T(F) x S for which ts E T(F), which is the same as the set of
pairs (t,s) such that f (ts) = 1. Thus the ranges of " and 11 partition x Sinto a union of two disjoint sets, the first containing n-1 elements and the second containing the same number of elements as Y. As T(F) x S has nr elements, we see that #Y = nr - (n-1) = n(r-1) + 1. 0
T (F)
EXERCISES
1. Show that if G is a subgroup of a discrete free group, then G is detachable if and only if G has a Shreier transversal. 2. Construct a Brouwerian example of a free subgroup of a finiterank free group that is not detachable. 3. Let G be a fini tely generated subgroup of a finite-rank free group. Show how to determine whether or not G has finite index. 4. Use ( 2.5) to show that a detachable subgroup of a countable discrete free group is free. 5.
Let S be a set and u,v E F(S). Then u and v are conjugates in F(S) i f there exists c in F with u = c-Ivc. A word w == x1···xn' with Xi E S U S-1 is cyclically reduced if it is reduced and x n x1 t 1. If S is discrete, then each reduced word w of F(S) may be written uniquely as w == v-Iw'v where w' is cyclically reduced. 5.1 'lHEXlREM.
If F is a free gr-oup on a discr-ete set S, and u,v E F,
then either u and v are conjugates or thell are not.
PROOF.
We may with u' i f and only i f u' = c-Iv' c, then v == b-Iv'b
reduced.
assume that u and v are reduced. Write u == a-Iu'a and and v' cyclically reduced. Then u and v are conjugates u' and v' are conjugates. We will show that i f u' is a cyclic permutation of v' We may assume c is
As u' is cyclically reduced, either
c-Iv'
or
v'c
is not reduced.
If the first ca se holds, then c == sd and v' == sw, with s in S U s-l. Then u' = d- 1 s- 1 swsd = d-Iwsd and ws is a cyclic permutation of v'. As
Id I < Ic I, we are done by induction on Ic I. The second case is treated similarly. So u and v are conjugates if and only if u' and v' are cyclic permutations of each other. 0
262
Chapter X. Free groups
E
If U is a subgroup, then
U}.
W""'
{w-tuw
For w E Flet w-tUw =
Let F be a group and U a subset of F. II
UW is a subgroup.
The subgroups G
and H are conjugate if there exists w in F with H = w- 1 Gw. 5.2 'l'HEXJREM.
finitel!)
subgnJup oF H
PROOF. for H.
Let
gener'oted
(Ln; te-r'onl< fr"ee
0
of F.
Tllcn
9r"0I1P, und ret G artd H be
c; ther'
Gis
conjuga t e
t0
a
i.s not; und eitller G is conjllgate to H or' it is not.
U
01-
F be
sllbgr'oups
Let U be a fini te free basis for G, and V a finite free basis
We may assume that V is a Nielsen set of generators for Hand, by
replacing G wUh a conjugate subgroup, we may assume that U contains an element u which is cyclically reduced (the case U = m = max {lw I
w E U U V}.
(*)
As w' w'
E
C;; H,
and
If w-
with
Iw' I < Iwl.
GwH we have w'-'Gw ~ H.
as follows.
Let
We will show that, for each w in F I
1 Gw
in GwH so that Iw'
is trivial).
I S m.
Iwl >
m,
then we can find w'
in GwH
It follows by induction that we can find
Once we have shown (*), the theorem is proved
Since F has finite rank there are only a finite number of
Iwl
words w with
Sm.
For each such]V we test t.O see if w- 1 Uw ~ Hi we can
do this because U is finite and H is detachable.
If there is no such w,
then G is not conjugate to a subgroup of HI while i f such a w exists, then If lJ)-'UW <:; H, then we can decide
w- 1 Cw is a subgroup of H conjugate to G.
if <w-'Uw> = H because 11 is finitely generated and <w-'Uw> is detachable. If so, then
C;
and H are conjugates; if not, then they are not conjugates.
Iwl > m,
'1'0 prove (* ) suppose that w is reduced ,
Then one of u
in U be cyclically reduced.
assume, replacing " by u -
lu -, w I ,
1
i f necessary,
then we may take w' =
U
.. 1
- 1
and w-'Cw ~ H.
w or wv is Leduced.
that uw is reduced.
w, so we may assume tha.t
Therefore no more than half of u cancels in the product
Iw- I
reduced, and
1
=
IlJJl > m
2
lul
I
Iw
UJ - ' "
!l
We may If
- 1
Let
111 2 • As
Iwl > Iw- I. 1
UUJ
is
we see that the reduced fom of the
product w- 1 {lW begins wi th more than half of the factor w" 1, and ends wi th the factorw.
Iw-'uwl > Iwl >mo
Inparticular,
write w- uw in terms of the basis V of 11; that is, w- l uw 1
with ''i E V u v-I and Uj"i+I must have
11
assume that
> 1.
Iwl
~
je
1 for all
i
< 11.
Iwull < Iw I, then we Iwv11. As V is a Nielsen If
As
=
"l{l2'" "['"
Iw- uwl > miluli, l
we
may take w'., wUl' so we may set, (2.2) and (2.3) say that
the reduced form cf w-'uw = u1···on begins with at least half of LJI' and
5. Conjugate subgroups
263
by the above it also begins with at least half of w- 1 0 As Iwi > m ~ lVII this
implies
that w- 1
begins
with
at
least
half
of
Iwl ~ Iwvll, so w- 1 begins with no more than half of vI'
However
vI'
Thus the reduced
form of w-'uw begins with exactly half of VI' so half of VI cancels in Vl1l2'
Similarly, i t ends with exactly half of un0
Suppose that Iull S
Iv n I ° As w- , begins wi th half of vI and w ends with half of v n ' i t follows that half of VI cancels in the product v n vl' and all of it in the As Dlv2 -t I, and V is a Nielsen set, it follows that VI =
product vnDID2' vr~',
Similarly, i f Ivnl S IDll, then VI
in half of v n =
product IV'
=
vi
1 I
wUI' so IIV'
by induction on n.
= vr~'.
Thus VI = D~'.
As w ends
i t follows that at least half of VI cancels in the
I
S Iwl.
But
w' -'uw·
= v2'
"vn_l and we are done
0
EXERCISES. 1. Show that (5.2) is true i f F is a discrete free group.
2. Construct Brouwerian counterexamples to generalizations of (5.2), one
with
G
countably
generated
and
one
with
H
countably
generated.
The word problem for a group G is to decide whether or not two elements of Gare equal; that is, to solve the word problem for G is to show that G is discrete.
The terminology comes from considering quotients F/N where F
is a free gwup and N anormal subgroup of F i in this setting the problem becomes how to decide whether or not a word in F is in N or not.
It would
seem plausible that we should be able to solve the word problem when F is of finite rank, and N is finitely generated as anormal subgroup, that is, there is a fini te subset A of N so that each element of N ean be wri tten as a product of conjugates of elements of A.
However a famous result of
Novikov and Boane constructs such Fand A in such a way that the word problem cannot be solved by a Turing machine, and therefare no algori thm for deciding whether a ward in F is in N can be written in any standard programming language. The generalized word problem for a group G relative to a subgroup H is to decide whether or not an element of G is in H or not; we solve the generalized word problem by showing that H is detachable.
The generalized
264
Chapter X. Free groups
word problem for a finite-rank free group relative to a finitely generated subgroup is solved by the Nielsen construction (3.5). The Schreier construction shows, classically, that any subgroup of any free group is free. To construct a Schreier transversal, weIl-order the elements of 8 U 8-1 and proceed as in the finite case.
Chapter XI. Abelian Groups
1. FINITE-RANK 'lORSICfi-FREE GRalPS.
abelian group is a module over the ring 7L of integers, so when studying abelian groups we may appeal to the general facts about modules developed in Chapter III, and about modules over a PID developed in Chapter V. The structure theorem for finitely presented abelian groups is a special ca se of the structure theorem (V.2.3) for finitely presented An
modules over a PID. In this seetion we are concerned with the simplest kinds of torsion-free abelian groups that are not finitely presented. If G is a module over a commutative ring R, and r E R, then rG = {rx x E G} is a submodule of G. Given x E G, it is of interest to know for what r E R we have x E rG. If R = 7L, thi s question reduces to the question of when is x E qG for q a prime power. 1.1 LEMMA.
If G is a moduLe ouer a commutatiue ring R, and a and bare
strongLy reLatiueLy prime eLements of R, then abG
PROOF. Clearly abG C;;; aG n bG, so suppose x We can write 1 = sa + tb, so x sax + tbx ty} E abG. 0
= aG n
bG.
= ay = bz =
is in aG n bG. = ab(sz +
sabz + tbay
abelian group G is torsion-free if, for each nonzero n E m and each = 0, then x = O. If G is a torsion-free abelian group, then the natural map G ~ ~ ® G, that sends x to 1 ® x, is a monomorphism. Note that ~ ® G is discrete i f and only i f G is discrete, and that, in any An
x € G, if nx
case, ~ ® G is a vector space over~. A torsion-free group G is said to be of rank n if ~ ® G is an n-dimensional discrete vector space over~. A group is a torsion-free group of rank one if and only if it is isomorphie to a nonzero subgroup of the additive group of 0). Classically the rank-one torsion-free groups are classified by equivalence classes of functions from the set of primes to mu {ro}. If x is an element of a torsion-free group G, then we define the 265
266
Chapter XI. Abelian groups
p-height of x to be hpx
=
Slip
pnG) , where the supremum is taken in
{rt : x (
Of course there is no reason to believe that we can compute hpx
IN U ('»).
in general; if we can, for each prime P, and G has heights.
and only i f A and B have heights 1.2 LEMMA. prime,
+ tm
(1
G, we say that the group
Ili;x
in which ca se
I
[hen
IN. then h l;Px
hpmx ~ hpx
equaU
with
tu
1f
p
is
holding
(1
if
Suppose (p,m)
1 and mx = p"y.
Write
1. Then x = SIPX + tmx = pn (sx + u); thus hpx ? hpmx. = ptl+l y if and only if x = /PU as G is torsion-free. px 0
Finally r
type is a function from the set of primes to IN LJ
A
(hj;x ,ll~X) •
+ 1.
= h / l
Clearly hpmx ? hpx.
= mi n
gr'oup wi th Iteights.
tot'sio"-f,'ee
m E IN, F
hpx
be
G
(111[j
IF
PRClOF. spn
Let
x E GI
(p,m) = l.
x (
If C = A ffi fl, then i t is easily seen that C has heights i f
{~));
two types are
equal if they are equal except at a finite number of places where each is finite. ~ T2
Tl
The set of types admits a natural partial order by setting if
finite.
T ,
(p) ~
T2
(p)
except at a finite number of places where each is
It is easily seen that
the set of types forms a distributive
lattice with a greatest element and a least element. The type of an element x in a torsion-free group G with heights is the function hpx of p, viewed as a type. and
m
From (1.1) it follows that if x
E C,
is a nonzero integer, then the type of x is equal to the type of mx.
As any two nonzero elements of a rank-one torsion-iree group have a common nonzero multiple, we can define the type of a rank-one torsion-free group with heights to be the type of any nonzero element.
The next theorem
shows that two such groups with the same type are isomorphie. 1.3 THEOREM.
Let G artd C' be ,'anlz-one tOf'si on-Fr-ee abet tan grOllfJS Wt th
heights, of types
G
10
G'
nnd
T
i Fand only i f
PROOF.
Suppose
all x in G, so
T
T'
r-espectiuelu.
Then ther'c is a nonzero mDfl f,'om
TI'
the" G and C' are i somOf'phic.
~ T'.
T
=
T',
Clearly Itp
~
and x and x' are f1onzero e1ements of
C;
G -') C'
:
T
is a nonzero map.
Now suppose the
T
~ T
,
From (1.1) there exist a f1onzero integers m and m' such that hpmx for all primes P, with equality if
T
= T·.
= {w E C;
6)
G'
: mv
E
and C' . ~ hpm' x'
Let H be the subgroup of
C ffi C' generated by (mx,m'x' ), and let K
hpx ior
~I
H for some nonzero
nEIN)
267
1. Finite-rank torsion-free groups
If Y E G, then ny = I!mx for some Clearly K is a subgroup of G al G' . Therefore em'x' EnG' by (1.1) and nonzero E and n as G is rank-one. (1.2). Thus there exists y' E G' such that ny = Ern' x', so (y ,y' ) E K. If T T ' , then, by symmetry, for each y' E G', there exists y E G such that
(y,y') E
K.
If
and
(~J,Y')
are in K, then
(~j,z')
z' as G and G' are torsion-free. map which is an isomorphism if T = T'.
y
Setting
~(y)
=
(O,y'-z')
E K, so
gives a nonzero
y'
0
The rank-one torsion-free abelian groups with heights form a semirigid class in the following sense. 1.4 COROLLARY.
with heights.
Let
A and B be r"anl,-one
tor"sion-free obel ian gr'oups
If ther"e exist nonzer"o mops f,"om A to B
then A arld B ar'e isomor'phic.
arm
fr"om B t.o A,
0
The hypothesis that A and B have heights cannot be dropped in (1.4); an example is sketched in Exercise 7. Classically every fini tely gene ra ted torsion-free abelian group is a direct sum of cyclics.
This is not the case constructively (see Exercise
2) but the following theorem holds. 1.5 THIDRm.
FiniteLy generated subgr"oups of finUe-r"ank tor"sion-Free
obeUan gr"oups are direct sums of cycUc groups.
PROOF. It suffices to consider finitely generated subgroups G of ~n. Multiplying G by a common denominator of the coordinates of the generators of G we may assume that G
~
il',n.
Then G is finitely presented, being a
submodule of a finitely presented module over a coherent Noetherian ring, hence G is a direct sum of cyclic groups by the structure theorem.
0
The simpiest kind of finite-rank torsion-free groups are finite direct sums of rank-one groups with heights. These groups may be specified, up to isomorphism, by a finite family of types. I t is a pdori possible that distinct families of types might give rise to isomorphie groups, but this turns out not to be the case.
In the ciassical context we can compute, in
an invariant manner, the number of rank-one summands of type considering the subgroup
T,
by
G(T) = {x E G : type(x) ~ Tl
and the subgroup G(T*) generated by (G(a) [J > T}. quotient G(T)jG(T*) is the number of summands of type
T.
The rank of the But since the
268
Chapter XI. Abelian groups
set of types is not discrete, constructi ve context.
we cannot determine this
rank in the
OUr approach hinges on the following lemma about
bases of finite-dimensional vector spaces. 1.6 LEMMA.
1"1"" ,e n
Let
and
e n +1"" ,eZn
dimensional vector space over a discrete Field.
bases
be
oF
Let 1f i (xl be
a
Finite
the
scalar-
in the expression Far- x retative to the appropriate basis.
mut tiple oF e i
Let R be the transitive dosw'e oF the relation 1fi(ej' t- 0 on {l, ... ,2n}.
Then
each
class
equivalence
elements
oF
(I, ... ,2n).
oF
under
the
'= j i f RU,j) and RU,i), has exactLy haLF oF its elements
equivatence
in {I, ... ,n}.
PROOF.
suppose C
Let C be an equivalence class of elements of {I, ••• ,2n} and
=A U B
where A = C
equal to A or to B let 1fS {ei:
i
n
{l, ••. ,n} and B = C
= 2 iES1f i
{n+1, .•• ,Zn}.
For S
and let Vs be the subspace generated by
ES}, that is, the image of 1fS'
same dimension as VB'
n
We shall show that VA has the
By syrnrnetry it suffices to show that 1fAVB
= VA'
We
shall show that e i = 1fA1fBei for each i E A. Now 1fA1fBei = 2kEA2jEB1fk1fjei' But 1fk1fjei = 0 i f i,k E A and j E {n+1, ... ,2n}\B by the definition of the equivalence classes.
So 1fA1fBei
= 2kEA2f;n+11fk1fjei = 1fAei = e i
•
0
1. 7 '1'HEDREM. each Hi
und Kj is a
torsion-F,-ee group with heights.
I'anl~-one
and there is apermutation a oF {I, ... ,nI such that Hi
='
Then m
= n,
Ka(i)'
Let 1"1"" ,e n be nonzero elements of H 1 , ••• ,Hn respectively, and e n +1"" ,en+m be nonzero elements of K1""'~ respectively. Then PROOF.
= n.
For notational
Let the 1f i
be as in Lemma
1"1"" ,e n and e n +1"" ,en+m are bases of III ® G, so m
convenience, set Hn +i
1.6, and note that from Hj to Hi
.
= Ki for
1f i G
= Hi
,
i
=
1, ... ,n. .
so if 1f i e j
~
0 then there is a nonzero map
The result now follows from (1. 6) and (1. 4).
0
EXERCISES 1. Show that any finite abe1ian group is finitely presented. 2. Let a be a binary sequence with at most one 1. subgroup of 71. EIl 71. gene ra ted by {( 1 ,nan ) (71. EIl 71.ljS
is
a
Brouwerian
example
of
Let S be the
nEIN} . a
finitely
Show that generated
discrete torsion-free abelian group that is not a direct surn of cyclics.
269
1. Finite-rank torsion-free groups
3. For eaeh nonnegative integer i, let Ai be the subgroup of ([l generated by the set (I/pi : p is a prime). Show that Aj has heights, and that Ai and Aj are not isomorphie i f whieh values of
i
i
t-
j.
For
is Ai finitely generated?
4. Let a be a binary sequence, and A be the subgroup of © generated by 1 and the set (1/2n : an = 1). Show that A is a Brouwerian example of a rank-one torsion-free group that does not have heights. Show that A is detaehable from but not in general.
m if
5. Given a funetion f from the set of primes to rank-one hpx
=
torsion-free
F (p)
group,
for all primes
and
an
a 1S deereasing,
mU
element
(a,), construet a
x,
sueh
that
p.
6. A subgroup A of an abelian group B is full i f B/A is torsion. Show that a torsion-free abelian group is of rank n if and only if it eontains a full subgroup isomorphie to ?Ln. 7. Let a be a binary sequence, and let A be the subgroup of !ll generated by land (a n /2 : n € IN). Construct nonzero maps from lL to A and from A to lL, hut show that A 1S a Brouwerian example of a group that 1S not isomorphie to lL.
Why doesn't (1.4) apply?
Construet an example of this kind where A and B have the property that mA and mB are detaehable subgroups for every m. 8. Construet a Brouwerian example showing that the hypothesis that the groups in (1.7) have heights is necessary. 9. An example, of an indeeomposable rank-two torsion-free group with heights.
Let G be the subgroup of © 9 !ll generated by elements of
the form (Z-m,O), p
and (5-,5-111).
(O,3-m),
> 5, compute h p ( 3omx) in
ignore
(2--i1l ,0),
and
To eompute hp(x) for
In computing 11" (x), we may for ft 3 and 1>5' That G is
7L ill?l.
similarly
indecomposable follows from the faet that i t has elements of three pairwise ineomparable types. 2. DIVISIBLE GRCXJPS. A
group is p-divisible if pG
=
Gr divisible if it is p-divisible for
each prime p. From (1.1) we see that G is divisible if and only i f nG = G for eaeh nonzero integer '1. The additive group of rational numbers ~ is a
270
Chapter XI. Abelian groups
divisible
torsion-free
group.
The
simplest example
of
a
nontrivial
divisible torsion group is the p-primary subgroup of the torsion group IlU'"', which is called 7l(p"'); the cyclic group of order pro, so to speak.
In
a
torsion-free
divisible
group,
multiplication by a nonzero integer an i somorphi sm.
r<
the
endomorphism
induced
by
is both one-to-one and onto, hence
'I'herefore a torsion-free divisible group admits a unique
structure as a vector space over the held
Conversely, it is clear
(J}.
that the additive group of a vector space over the held lIi is torsion-free and divisible. A coherent abelian group is one that is coherent as all-module. easily seen that a coherent group has detachable
It is
(finitely generated)
subgroups, and that any discrete torsion group is coherent. Let D be a di.visibLe subgroup of a group G such tlwt G;D
2.1 'l'HEDREM.
is
Tllen we can cons/cuct a countnble subgr-oup K
and coher-enf.
coUtltabl~e
o F G such tJw t G = K al D.
PROOF.
Let X,
,X2 , • ••
be elements of G that enumerate G;D.
As GID i s
eoherent, we ean arrange so that either <Xj+l> n «xl'''''x[> + D) = 0, or We
shall
pXl+l E <xl""'x[> + D for some prime p. construct
a
sequence
of
hni tely
emunerable
of G such that Ki + D = <x1""'x i > + D,
K, !;; K2 !;; •• ,
subgroups
and Ki nD = 0.
Then K = U K i is as desired. Set Ko = O.
Given K[, construct Ki+l as folIows,
decidable question), then set K(+l = Kr' <x(+l> n (K i + D) = 0, K i +1
= <x i +1> +
and cl E D. !J
=
xi
pXi+l
E Kj
Then P!1
= hi ,
<xl"" ,Xi +1> + lJi
and we set K(+l we must
w E K[+l
n D, write!Li = ny + z, where z
But PU
Kr' so if (p,n)
E
to our assurnption. Kj
n
D
=
0.
In
+ IJ.
the
=
K i ; in the latter case write pX i +1
former
=
1, then
Therefore
p
=
show that E
K, .
then either case,
set
/'i + cl, where /\ E K i
As [) is divisible, we can find d' E D such that d
+1 - d'.
K(+1 + D
or
If xi+l E Kr + D (a
If xi+l ([ K[ + D,
pd'.
Let
+ Kj .
Certainly
K( +1 n D
0.
So TlY
=
For
w - z E D + Ki .
Ki + D, so xi+l E K( + D, eontrary divides n, so tllJ <: Kr whereupon !Li E
~J E~
0
We ean use (2.1)
to get a strueture theorem for eountab1e coherent
2. Divisible groups
271
divisible groups. 2.2~.
a
Let G be a eountable eoherent divisible group.
eountable direet
swn of subgroups isomorphie
to CQ und
Then G is
to ~(poo)
for
various primes p. l'BOOF. Let T be the torsion subgroup of G. Then Gif is coherent because, as G is coherent, finitely generated subgroups of Gare direct sums of finite and infinite cyclic groups. Thus (2.1) says that we can write G = T m F, where F is torsion-free. Thus it suffices to prove the theorem under the assumption that G is torsion, or that G is torsion-free. If G is torsion-free divisible, then G admits a unique structure as a vector space over the field CQ, so each nonzero element of G is contained
in a unique subgroup of G that is isomorphie to the additive group CQ. Let xO,xl' ... be an enumeration of G. Define a detachable subset S of IN by putting i E S if Xi is not in the vector space generated by xO, ... ,x i _1; this is decidable because G is coherent. It is readily seen that G is a direct suro of the subspaces CQx i CQ for i E S.
=
If G is so we may element of isomorphie
torsion, then G is a direct suro of its primary components Gp ' assume that G is a p-group. It suffices to show that every a discrete divisible p-group is contained in a subgroup that is to ~(poo); we then apply (2.1) repeatedly. But given such an x
we can construct a sequence x = YO' Yl' ... such that PYi+l = Yi for each i. The subgroup generated by the y's is the desired subgroup. 0 A
subgroup A of a discrete abelian group B is essential if for each
nonzero
bEB
there is n
E ~
such that nb is a nonzero element of A; in
particular, B/A is torsion (but that is not enough).
A divisible hull of
a discrete abelian group A is a discrete divisible abelian group B containing A as an essential subgroup. 2.3~.
Any eountable diser-ete abelian group has a eountable
di sere te di vi sible huU. l'BOOF.
We may assuroe that the given group is F/K, where F is a
countable-rank free abelian group, and K is a detachable subgroup of F. Let QF = Q®F and construct a countable subgroup N of QF as follows. First note that i f A is a finitely generated subgroup of QF, then A n F is finitely generated, as A and the relevant basis elements of F can be put inside a finite-rank free.subgroup of QF. Let ao,a" ••• be an enumeration
272
Chapter XI. Abelian groups
of QF and set
NO = 0 Ni +l
=
Ni +
!in.(
if
+ !in[) r1 F t;; K
(Ni
otherwise.
= Ni
The decision as to whether to put
in Ni +1 can be made because Ni +
Ct i
is finitely generated. Set N = U Ni and let D
QFjN.
The subgroup N is detachable from QF
~ai) n F ~ K; therefore D is discrete. = K, so we may view F;K as a subgroup of QFjN = D. Finally,
because a i E N if and only if (Ni + Clearly N
n
F
if a i E QF\N is a nonzero element of D, then there exist x E Ni and n E ~ such that x + na i E F\}\.. Thus rlll i is equal to a nonzero element of F;K.
o The divisible hull of a cohe,·C'ttt countable discrete abelian group is coherent.
In fact the following more general theorem holds.
2.4 THEOREM..
Lct Ac:: B be discr'ete I.lbelil.ln groups.
If BIA is tor'sion,
and Ais coher'cnt, then Bis coher'cn t •
PROOF.
We
first show that i f B/A
presented, then R is discrete.
is
torsion,
Given bEB there is
Tl
and A is
If nb 'l nA (decidable as A finitely presented), then b 'l A.
for some (]
E A,
fini tely
t 0 so that nb E A.
then i t suffices to decide whether or not
If nb = no, b--o E A.
But
there are only finitely many torsion elements of A. For the general case, let 'p map a finite-rank free abelian group F into R.
Then there is n t 0 such that
discrete, by the first paragraph, and thus
the
A/
~
hypotheses B/
of
the
theorem are
The group
is coherent by
inherited
by
the
R/
is
(III.2.5)i
situation
If B/
we may assume that
But F/nF is finite, and /3 i5 discrete, so
the induced map from F/nF to B has finite kernel, whence the kernel of is finitely generated.
0
EXERCISES
1. Show that any torsion group i s equal to the direct
sum i ts
p-primary subgroups. 2. Show that every finitely generated subgroup of ~(pro) is cyclic,
273
2. Divisible groups that the finitely generated subgroups of 7l(pO» inclusion, and that for each of 71 (pro)
of order p".
Tl
form a chain under
there is a finite cyclic subgroup
Show that any group G wi th
properties is isomorphie to
these
7l(pw).
3. Let a be a binary sequenee with at most one 1, and let H be the
=
1,2, ... }..
Show that H
is a Brouwerian example of a countable divisible subgroup of in faet a countable direct sum of copies of summand.
Why doesn't (2.1)
apply?
(Q,
([l,
that is not a
Show that G
(1);11
=
is a
Brouwerian example of a eountable discrete torsion-free divisible group that is not a eountable direct sum of copies of
Why
(1).
doesn't (2.2) apply? 4. Find two places where the axiom of dependent ehoiees is used in the proof of (2.2), other than (at one remove) in the appeal to (2.1) •
5. Let G be a diserete divisible p-group, and let X be a basis for {x
E
G : px
Ol, viewed as a veetor space over the p-element
=
For each x E X, eonstruet a sequence x = !JO' !J1""
held iZ/(p).
so that PYi+l = PUL' and let Ax be the subgroup of G generated by the y's. Show that each Ax is isomorphie to 71(1'00), and that G is the direct sum of the subgroups Ax ' 6. Construct a Brouwerian example of a countable abelian group between 71 and
(Q
that is not detachable from any divisible hull.
3. HEIGHT FUNCTlooS 00 p-GROJPS.
Let G be an abelian group and p a prime.
We say that G is a p-group i f
for eaeh x in G there is a positive integer n sueh that rl'x
=
O.
A
high
point in the theory of abelian groups is the classification of eountable p groups by the dimensions of certain veetor spaees defined in terms of the notion of /wight.
If
11
is a nonnegative integer, and x
that the height of x is n if x
E
pnC and x
r,n+1 c .
E
C, then we say
We have already seen
that there are problems eomputing heights in torsion-free groups. Moreover, to get the classification theorem for eountable p-groups I we must extend the notion of height to include transfinite values. Let G be an abelian group, p a prime
I
A an ordinal and Aro
=
A U 1m }.
A
274
Chapter XI. Abelian groups
p-height function on G is a function h horn C onto \XI such that (i) rf In < 00, then
> ilx
hpx
(ii ) If (m ,1') = 1, then hmx
(iii) If
<
0
hx
hx
=
(1
=
then the re exi ts
al,
The ordinal A is called the p-length of G. 110
=
00;
if hx
=
implies x
00
such tha t
y
and hU ;:: o.
== x
PU
or
hx,
=
Note that (i) implies that
0, then we say that G is p-reduced,
If G is
a p-group, then (i) and (ii) imply that if q is a prime other than p, then the q-height of each element of
is m; in this ca se we drop the prefix
C
'p' from 'p-height' and 'l'-length'.
Sueh a height funebon is unique; in
fact, both 1\ and Il are isomorphism invariants of G in the following sense. 3.1 THEOREM,
emd
h'
ond
Let G emd G'
l'-!engths
i somoqJh i sm .
Tllen
/1" obeliclIl gn)lJps with p-l"cight fllnctions 11
A "nd
then?
1\'
exists
u'sjJPctilJel!J.
Le t
on
/J:
isomor'phism
: G
-t
G'
be
on
h<{i = ph' •
We say that
PRQOF.
then
hy = a,
p
is defined at a if whenever x,U E G such that hx rf p is defined at a, we set po
h'
any x such that hx
Let [0 ,0 J
= a.
S = {o E 1\
: p
=
{b E A ; b
~
= 11''I'X
for
0), and let
is defined at each element of [0,0 J ,
and p is an injection on [O,oJ). We want to show that S
Suppose
= !\.
0
S for all
C
< b.
0
If
b we
11X
shall show ( i) If () (ii) If c
< b, then po < h'
< b such that
0
This will show that b ES; so S is hereditary, whence S To show (i) suppose a < b. that (Ja
a ~ hz
< band
= 1\.
As h is a height function, there is z such So P'l'z
<{Jx
and h' '1'7
=
phz ~
Thus
pa,
< h' <{JX. To show (ii),
suppose c < h' <{JX •
exists z such that b
pz = x.
c.
pa
there is a As S
h'z ;:
'I'herefore hz
= hx.
=
~
c and
= h'fl'fl-
1Z
d < b such that pa
1'7 =
As h' cpx.
is a height function,
Then
l',p-1 Z
= x,
=
into A.
=
d <
Hence
c.
A, the function p is an injection of A into 1\'.
get an injection from 1\'
so h~)-17
= pd and p is an injeetion on [0 ,d 1 .
there
Similarly we
Their composition is an injection of A
275
3. Height functions on p-groups (or A')
into i tself, hence the identi ty by ( I. 6.6) •
poo =
0
00.
Finally we set
To illustrate the kind of structures we shall be dealing with, we consider the simplest example of an abelian p-group with elements of transfinite height other than the generators xO,xI'... n
This group is constructed by subjecting
00.
to the
relations pxO
=
0,
and pnxn
= Xo for
> O. Let F be the free abelian group on the discrete set
3.2 EXAMPLE. {x n : nEIN}.
Let P be the quotient of F by the subgroup R of F generated It is a routine exercise
to establish that R is detachable from F, so P is discrete, and that each element of P has a cananical representative in F of the form 2 nix i with o ~ no < p and 0 ~ n i < pi for i > O. Clearly P is a p-group. Let A be the well-ordered set {0,1,2, ... ,w}. P
to Aoo as follows.
Define a function h from
Let 2 n i xi be the canonical representati ve of the
= 00 i f all the n i are 0; set h(y) = w i f nO is the unique nonzero n i ; set h(y) = min{upn i : i t- O} otherwise, where vpm is the p-adic value of m, that is, the exponent of p in the prime
element Y in P.
Set h(y)
decomposition of m. P.
It is readily verified that h is a height function on
0
The next theorem shows that homomorphisms (weakly) increase heights, provided one can make sense of that statement. Le t G and H be groups with p-height functions, whose
3.3 'lHEXlREK.
lengths
are
initial
segments
of
a
common
ordinal
A.
If
is
a
homomorphism from G to H, then h
= a, then h gy, we have h h hy, a contradiction. Thus S = A. Now let S = {a E A : h a i f hx = ro}, and suppose a E S for each a < b. If hx = then there exists y such that hy = ro and py = x. If h b, so we have shown b ES. Otherwise h a as a E S. 0 PROOF.
Let S
{a
E A : i f hx
00,
So far the only interaction between a p-height function on a group and
276
Chapter XI. Abelian groups
the additive structure of the group that we have considered concerns relations between hmx and hx.
We turn now to the fundamental relation
between height and addition. Let h be a p-height function on a group G.
3.4 THIDREM.
h(x + y) For
b
By symmetry we may assume that hx
and let S =
min(hx,11Y),
~
any pair x,y in C, with equatity holding i.F hx t hy. PROOF.
C,
Then
hx
~
be
hy.
the
set of b
E A
for
~
hy.
Let A be the length of
whieh h (x + y)
We shall show that if a E S for all a
~
hx whenever
< b,
then b E S.
Suppose, on the eontrary, that h (x + y)
~
hy.
= pz, where 11W ~ h(x + y) and hz ~ h(x + v), so > h(w + z) ~ min(hw,hz) ~ h(x + v), a eontradietion. Thus S = A, so the claim is true if hx t 00. If hx = 00, then h(x + y) = 00 lest hy = h((x+y) + (-x)) t 00. Finally, if hx < hy, then hx = h((x+y) + Then
=
< hx = b
x
h(x + y)
pw
and
= h(pw
y
+ pz)
min(h(x+y),hx), so hx
(-x)) ;:>
~
h(x + v).
0
I t follows from (3.4) that {x E C : hx ~ a) is a subgroup of C.
EXERCISES
1. Let p be a prime and C a finite abelian group. p-height function.
Show that C has a
What is the p-length of C (in terms of the
invariants of V.2). 2. Let C be a group with a p-height funetion h. a
p-divisible
subgroup of G that
Show that h- 1 (00) is
eontains every p-divisible
subgroup of G. 3. Let G = A ID B be an abelian group with a p-height function hG . Show that A has a p-height
funetion whieh is equal
to the
restrietion of hc to A, and that the p-length of A is an initial segment of the p-length of G. 4. Give
an
funetion,
example
of
an
abelian
p-group G,
with
a
p-height
and a subgroup A of G sueh that A has a p-height
funetion, but the p-height function on G does not restriet to a p-height function on A.
277
3. Height functions on p-groups 5. verify the claims made about the p-group P in Example 3.2. 6. Let u be a binary sequence.
Let C be the subgroup of the group P
in Example 3.2, generated by the elements anx n . Show that C is a countable discrete p-group with a height flIDction. What is the length of C? 4. ULM r S THEXJ.REM
If C is a countable p-group with a height function, then the subgroup D
= h-'(ro)
is a countable divisible detachable subgroup of C, and C/D is a
p-group, hence coherent.
Therefore by (2.1) we can write C
=
D
~
R, where
D is a countable coherent divisible group, hence of known structure (2.2), and R is a countable reduced p-group with a height function.
Thus we are
led to the study of such groups R, which we caU ulm groups. p-groups are clearly Ulm groups.
Finite
The following theorem shows how to
construct lots of big Ulm groups. 4.1 T.HEDREM.
[f A is an ordinal.
Ir
o Ileight funclion, of !ength A. i
then the,-e is A is
0
0
,-educed p-g'-oup, wUh
cOlilltable ordinol,
then liiere
s an 111m g'-oup of lengtll A. PROOF.
Let F be the
free abelian group on
finite
sequences
(ul' ... ,u n ) of elements of A such that al < u2 < ••• < u n and n ::: 1. K be the subgroup of F generated by the elements of the form
G
=
Let
P(ol)' and p(al,u2, ... ,a,,) - (u2, ... ,Cin ), for n and let G be the quotient group F/K.
We say that an element of F is in
standard form if i t can be written as 2: free generators of F, and 0
~
nf
< p.
> 1,
"lG i ,
where the Gi are distinct
Each element of G comes from a
unique element of F in standaLd form: that such an element can be found is clear fLom the nature of the generators of K; that such an element is unique follows from the fact that each nonzero element :>: J Ä,,,, by set.ting hx equal to the least element of ;\ occurring in a sequence that has a nonzero coefficient in the standard form of
x,
and
/1.< =
w
if
x =
O.
It is readily seen that h is a reduced height function on Gunder which G has length A.
Clearly G is an ulm group if A is a countable ordinal.
0
278
Chapter XI. Abelian groups
A complete set of invariants for Ulm groups is provided by certain countable discrete vector spaees ove!: the p-element Held :a:/(p) , caIIed Ulm invariants.
It will be convenient to define these invariants in the
more general setting of valuated groups. 4.2 DEFINITlOO.
A valuated p-group is a p-group H together with an
ordinal 1\ and a funetion v : H
<
(i) rf tJx
00,
(ü) If (p,m)
->
\x,
satistying
< vpx.
then vx
1, then vmx
vx.
=
(iii) u(x + y) ;: min(ux,uy).
We say that H (or
is redueed i f
v)
V-i
(00) =
o.
The model for a valuated p-group is a subgroup H of a p-group with a height funetion h: the funetion v is h restricted to H.
Every p-group
= h.
with a height funetion becomes a valuated group upon setting v is an ordinal, and a < b are in there exists c such that a 4.3 DEFINITlOO.
"'n'
< c < h.
then we write a
<(
b if b
> a, then
Note that if ux
=
00,
If 1\ or if
llPX» a.
Let H be a valuated p-group with values in 1\.
For
eaeh " E 1\ we define the a th Ulm invariant of H to be the group
Ix
EH: vx ;:
{x EH:
a and upx lIX
» Ct l
> Ct}
The funetion f H is called the Ulm function of H.
Note that fH{a) is a
discrete vector space over the fie1d :a:/(p). A countable discrete vector space over a finite field has a countable basis, so an ulm invariant of an Ulrn group is determined by its dimension, which is the cardinality of some detachable subset of
m.
In the classical
ease the Ulm invariants may be thought of as elements of
[N
U
Im 1.
One
virtue of defining them as vector spaces is that they are then additive functors. If H is a finite cyclic group of order rP, with function,
then the length of H is
while f H (a)
=
0 for
Ct
< n-I.
lJ
{O,1, •.. ,n-1J,
equal to the height and dirn fH(n-l)
f G(0) is the direct sum of the vector spaces f H (i ) (a) • is
a
finite
direct
1,
The Ulm invaLiants are clearly addi ti ve in
the sense that if G is a direct sum of a family of subgroups p-group
=
sum of
finite
cyclic
H(i ),
then
As each finite
p-groups,
the
Ulm
invariants provide a complete set of invariants for finite p-groups; more gene rally, they provide a complete set of invariants for direct sums of
279
4. Ulm's theorem
finite cyclic p-groups. We now turn to the proof of Ulm's theorem: an Ulm group is determdned, up to i somorphi sm, by its ulm invariants. If H is a subgroup of an Ulm group G, and x E G, we say that x is H-proper if x is of maximum height among the elements of x + H. 4.4 'lHEDREM (U!m's theorem). isomorphie
with
ULm
Let
G
und
invariants,
G' be
and let
~
ULm groups oF
be
a
Length "
preserving
height
isomorphism from a finite subgroup H oF G to a Finite subgroup H' of G' • Then
~
extends to an isomor'phism from G to G' •
PROOF. Let xI,x2' •.• be an enumeration of G such that pX i subgroup generated by xl, ..• ,xi_l for each i, and let xi,xi, ... enumeration of G'. We shall construct sequences of finite H1 ~ H2 ~ ••• of G, and Hi ~ Hi ~ ••• of G', and height isomorphisms
E H 2n _1 and x~ E Hin, and so that ~n+l extends opn. By symmetry, it suffices to show that if ~ is a height preserving isomorphism between a finite subgroup H of G and a finite subgroup H' of G', and if x E G and px EH, then ~ can be extended to a height preserving isomorphism of the subgroup generated by Hand x. If x E H there is nothing to do. Otherwise, by choosing an element of maximum height in the finite set x + H, we may assume that x is H-proper. Among such x, pick one that maximizes hpx. Note that h(x + z) OPn : Hn ... H~,
so that
is in the be such an subgroups preserving
xn
for all zEH because x is H-proper.
min(hx,hz)
Let
hx =
a.
We must
define ~x. Since h~px = hpx > a, we can find x' E G' such that hx' ~ a and px' = oppx. If hx' = a, and x' is H'-proper (note that these questions are decidable), then we can extend op by setting opx = x'. If hx' = a but x' is not H'-proper, then there exists zEH such that h(x'
so h ( px ' + hz = hopz = hpx ~ h(px hpx
= hpx'
~pz) ~
+
~z)
> a,
a, and so h (px + pz)
~
a.
But
h(x +
z)
~
a, since
a, so x + z is H-proper. Hence, by our choice of x, we have + pz), so hpx ~ a. Finally, if hx' > a, then we also have ~
a.
So we turn our attention to the case when hpx ~ a, and hence x € FG(a) but, since x is H-proper, x f. fH(a) !;; FG(a). We need to find an element x;" € FG' (a) that is not in FH' (a). Let a be an isomorphism from FG(a) to
280
Chapter XI. Abelian groups
Note that the finite set FH' (a) is a detachable subspace of the discrete set Fc,(a). rfax E fH'(a), then we are done. Otherwise, by repeated application of ~-1 and a we can find a finite subspace V of FH(a) such that ax E
E aV, then x E V which contradicts the fact that x get our x;'.
ax
Il FH' (a) •
Thus we
We can choose x;' so that px;' = O. Moreover, x;' is H'-proper, for if h(x;' + z') > a, and hence hz' = a, then hpz' ~ min(hp(x;'+z' ) ,hpx;') > a, so z' E FH' (a), and x~ == -Zl as elements of Fc ' (a), a contradiction. Set , where px' = ~px and hx' > a, noting that x~ + x ~x = is +
x* x
H'-proper because x;' iso
0
We can use Ulm's theorem to prove Prüfer's theorem characterizing the countable direct sums of finite cyclic p-groups. 4.5 oF
'J.'HEX)REM
Finite
C\{O}
=
(Prüfer).
A
countabLe discr'ete
cycLics iF and onLy
iF
pllC
p-group C is a
is detachabLe
For each
dir'ect sum nEIN, and
UnE IN pnC\pn+lC.
PROOF. AB finite cyclic p-groups C have the specified properties, so do direct sums of them. Conversely, if the specified properties hold, then we can define a height function on C by setting hx = n i f x € pnC\pn+~, and hO = 00. We can then construct a countable direct sum of finite cyclic p-groups with the same Ulm invariants as C. By Ulm's theorem, C is isomorphie to this group, hence is a direct sum of cyclics. o An
immediate corollary to (4.5) is that finite p-groups are direct sums
of cyclic groups. EXERCISES
1. A torsion forest is a discrete set X with a detachable subset R and a function rr : X\R ~ X such that for each x E X there exists n such that ~x E R. Define what a height function on a torsion forest iso Let S(X) be the free abelian group on X modulo the relations px = 0, if x E R, and px = rrx if x E R. Show that S(X) is a discrete p-group, and that if X has a height function then so does S(F). What is X in the proof (4.1)? Groups of the form Show that every S(F) are called sÜIIIly presented p-groups.
281
4. Ulm' s theorem finite p-group is simply presented. 2. Let e be the generator of a finite cyc1ic group H of order p3
I
and let v be the valuation on H such that ve = 0, vpe = 2, and vp2e = 3. what are the Ulm in'Zariants of H? Embed H in a group G in such a way that the valuation on H is the restriction of the height function on
G,
and the Ulm invariants of Hand Gare
isomorphie. 3. What are the Ulm invariants of the group in Example 3.27 4. Let A
=
{O,1,2,3}.
What are the Ulm invariants of the Ulm group
of length A construeted in the proof of (4.l)? 5. The c1assical formulation of (4.5) has ~'EIN pn G = 0 instead of G\{O) = UnEIN pnc\pn+lc. Show that this moclified version of (4.5) is equivalent to Markov's prineiple. 6. Construet A
~
a Brouwerian example of an ulm group with length
(O,l) which is not a direct sum of cyclies.
group of length A 7. A
~
A
Show that an Ulm
is a direct sum of cyc1ics if and only i f
w.
subgroup
nEIN.
H
Use
of a p-group (4.4)
G
is pure if r/'u
= H
n pne for each
to show that a pure subgroup of a finite
p-group is a direct summand. 8. Let G be an Ulm group, and x,y E Gelements of order' p.
Show
that hx = hy if and on1y if there is an automorphism of G taking x to y. 5. cn.STRUCTlOO OF ULI'l GRa.JPS
It simp1ifies things a great deal to put a mild condition on the length of G. Let A be a we1l-ordered set. If G < b are elements of A, we say that b is the successor of G, and write b = 0 + 1, if there is na c E A For n a nonnegati ve integer we dehne b = u + n sueh that 0 < c < b. inductively by setting a + 0 = G, and Cl + (n + 1) = (u + n) + 1, i f the latter exist.s.
A well-ordered
set A is said to have successors i f
whenever there exists b > a E A, then 0 + 1 exists. Let f be a function that assigns to each element of a countable ordinal with successors, a countab1e discrete vector space over u/(p). When can we construct an ulm group G such that f G = f 7 A necessary condi ti on is
282
Chapter XI. Abelian groups
given in the following theorem. 5.1 'l'HOOREl'l.
-or'dered se t wUh Sl1ccessors I and
Le tAbe a we U
gr'oup of !ength A.
e
an U!m
If a E A, then we can find a nonnegative integer' n
such that a + n exists and fe(a + n)
je
O.
= a,and let n be the least nonne ga ti ve or h p n+1x > hrPx + 1. Then pnx represents a
Choose x such that hx
PROOF.
integer such that hpn+1x =
00
nonzero element of fe(a + n).
0
Let A be a countable well-ordered set with successors , and f a function that assigns to each element in A a countable discrete vector space over the p-e1ement field a:/( p) . a
E A,
there is
nEIN
We say that f is a U-function if
such that
a
+ n exists and f
(a
+ n)
I
for each
je O.
Theorem
5.1 says that an ulm function of an Ulm group whose length is a wellordered set with successors ,
is a U-function.
We
shall
show that ,
converselYI if f is a U-function on A, then we can construct an Ulm group
e
of length A, such that fe(a) is isomorphie to f(a) for each a E A.
This
result is known as Zippin's theorem. Let H be a reduced valuated p-group wi th values in a countable wellordered set A with successors.
Let f be a U-function on A.
We say that H
is f-admissible i f we can ernbed fH(a) as a subspace of f(a) a E A.
for each
Thi s condi tion is certainly necessary for H to be ernbedded as a
subgroup of an Ulm group with Ulm function f that extends the valuation on It turns out that it is also sufficient if H is finite; the ca se H
H.
is Zippin's theorem.
=0
The key construction is to enlarge an f-adrnissible
finite valuated group to an f-adrnissible finite valuated group in which a given instance of Property (iii) of a height function is satisfied. 5.2 LEMMA.
Let
H be a finite reduced valuated p-group wUh values in a
weU-ordered set A wUh successors.
a
fH(C) ... f(c)
< vx
=
be
one-to-one
maps
Let
f be a U-function on A, and let
for
each
c E A.
IF x E H, and
b, then we can construct a fini te va!uated p-group K containing
H, so that
(i) the valuation on K extends that on H, (E)
the maps
(iii) py
=x
PROOF.
By increasing a,
For same y E K such that vy 2 a.
b = a + 1, or that F(a)
je
if necessary,
we
rnay assurne either that
0 and there is no element zEH such that
5. Construction of Ulm groups vz = a.
283
We mayaiso assume that x has maximal value in the set {x + pz :
zEH and hz ?:
a},
since if py
x +
=
then p(y - z)
pz,
'=
x.
Adjoin an
element y to H subject only to the relation py = x, and set v(z +mlj) =min(uz,a)
< m < p.
for zEH and 0
is easily checked that this defines a
It
valuated p-group K containing H, whose valuation extends that on H.
We
must now extend the maps ~e. The embedding H!;; Kinduces one-to-one maps f H(c) .... f K(e ) e E A.
for
each
Suppose z + my represents an element of f K(c), where zEH and
o < m < p.
Then e = min (uz,a)
As we only care about the subspace
::; a.
generated by z + my, we may assume that m in f K(e ), hence z + y
= 1.
rf e
< a, then z + y = z
represents an element of f H(c ) .
If e = a,
and
> b, so v(pz + x) > vx, contrary to the choice of x. So i f e < 0., or b = 0. + 1, then FH(e) = f'K(e) and there is no problem extending ~e. There remains the case when e = a, there is no element zEH such that vz = c, and F(c) t' O. Then FH(e) = 0 and FK(e) is
b = a + 1, then
vp(z + !})
one-dimensional, so we simply embed the latter in f(e) any way we like. 5.3 'l'HEXJREM.
Let. A be 0. countabie weiL-moder'ed set with successors,
and f a [J-function defined on A. vaLues i.n A.
0
If H
Let H be a fi.nite vaiuated p-group wUh
Ls f-admi.ssibLe,
then H can be embeckied in an ULm group
G of length A so tlw.t the height function on G induees the vatuation on H, amI so that fG(a) '" (a) for each a E A.
I f I.L
PROOF.
is a finite subset of A,
one-to-one map for each 0.
E
and
<pa
: fl/(a) ....
f (0.)
is a
A, then by repeated applications of (5.2), we
can embed H in a finite valuated p-group E(1f,p) such that
(i) the valuation on
extends the valuation on H,
E(l/,F)
( ii) the maps <pa extend to one-to-one maps from f K(0.) to f (a ) , (iii) i f
a E I.L
and
x
E l/
with
satisfying py = x and ViJ
~
a
<
vx,
then
there
is
y E E(H,I.!)
a.
We shall construct a chain of finite valuated groups H = Ho !: 1f t !;; with a common valuation v, and one-to-one maps 'P~.: fH.(a) .... f(a) l
such that
E
the disjoint union of
IN and a
E
A.
{f (u) JaEA ,
Let x lIX2 let a i
, • ••
,U2' •••
be an be an
284
Chapter XI. Abelian groups
enumeration of A, and let JJ.n = {al"" ,an} .
= E(H 2n _ l ,JJ.n )· H2n _ l = H2n _ 2 if x n
If x n € f (b ), set
H2n
H 2n -
l
= H2n _ 2 6l
b € '1'211-2 (
fR
211-2
(b)),
where py = 0 and vy = b otherwise.
In the last case we extend the map 'I'~-1 by setting
does not have successors. 2. Show that a well-ordered set has successors i f and only i f whenever a < b, then either a < b or b = a + 1. 3. Let A be a countable well-ordered set with successors, and f a U-function on A. Using arguments similar to the proofs of (5.2) and (5.3), show that there is a simply presented Ulm group with Ulm function isomorphie to f. 4. Let A be a countable well-ordered set with successors, and f a U-function on A. If f(b) # 0, show that there is a U-function 9 on [O,b), and a U-function f' on A such that (i) f
(a) =
(U) f(a) =
9 (a)
6l
f' (a) for a
~
b,
f' (a) for a > b,
(iii) dim 9(b)
=
1.
Conclude that each element of order p and height a of an Ulm group G of length A is contained in summand H of G of length [O,a)
such that dim
fH(a)
= 1.
5. Construction of Ulm groups
285
5. Use Exercise 4 to show that i f A is a well-ordered set wi th successors, and G is an Ulm group of length A, then G is a direct sum of Ulm groups of lengths [O,a] for various a E A. 6. Let A be a well-ordered set with successors such that UJ ~ A. Show that any Ulm group of length A contains a summand that is an unbounded direct sum of cyclics. Can you find such a summand in Examp1e 3. 2?
OUr construction of a divisible hull for a countable discrete abelian group i s due to Ri ck Smi th ( 1981 ) who was working in the context of recursive algebra. In that same paper he shows that the divisible hull of a p-group is unique if and on1y if pG is detachab1e from G. The proof of the "only if" part of this theorem provides a dramatic illustration of the difference between recursive and constructive mathematics. To establish the existence of an algorithm that decides for any element of Gwhether it is in pG, he constructs an algorithm that doesn't quite do the right thing for each element, but cannot be wrong infinite1y often. From a recursive point of view such an algorithm is easily modified to yield the desired one: simply change a finite number of answers. From a constructive point of view this algorithm is essentially worthless: not only do we have the wrong answer in a finite number of cases, but we have no idea at what point the answers can be counted upon as being correct. Smi th' s treatment of uniqueness of divisible hulls is mode1ed on the treatment of the uniqueness of algebraic closures in [Metakides & Nerode 1979] where it is shown, for countab1e discrete fields of characteristic zero, that the algebraic closures is unique if and only if the field is factorial. For more on this issue see the treatment of uniqueness of splitting fields in [Bridges-Richman 1987]. A recursive algebra counterexarnple to Ulm's theorem is allegedly constructed in [Lin 1981a], but it relies on a faulty proof of zippin's theorem.
Mal'cev (1971) has abrief discussion of types of rank-one torsion-free abelian groups in the context of recursive algebra. The material on Ulm's theorem and Zippin's theorem is a simplified version of IRichman 1973].
Chapter XI. Abe1ian groups
286
Rogers (1980) studied p-groups that are not necessarily either discrete She investigated those p-groups G such that pnG is a or countable. detachable subgroup of G for each
11,
and proved, for groups with countable
dense subsets, that this property is equivalent to having a basic subgroup. Classically every p-group has a basic subgroup; constructively i t is already a problem to construct a desnse discrete subset. Untl.l we have some examples in mind, it is idle to specu1ate about extensions of this resul t.
Chapter XII. Valuation Theory
1.
VALUAT!
If k is a Heyting field, then a (rank-one) valuation on k is a function assigning to each element x in k a nonnegative real number lxi such that lxi # 0 if and only if x is a invertible
Ixyl = Ixllyl Ix + yl ~ lxi + Iyl The set {lxi: x € k is invertible} forms a group called the value group of the valuation. A field with a valuation is called a valued field. A valued field is a metric space with metric Ix - yl. A general (rank-one) valuation on a Heyting field k is a function assigning to each element x in k a nonnegative real number vx such that (i) vx # 0 if and only if x is invertible (ii) v(xy)
=
(vx)(vy)
(iii) There is a constant B such that v(x + y) and y.
~
B·sup(vx,vy) for all x
The constant B in (iii) is called a bounding constant for v. Note that (i) and (ii) imply that vI = 1, so taking x = 1 and y = 0 in (iii) shows that B ~ 1. Note also that a valuation is a general va1uation with bounding constant 2. 1.1 THEOREM. A general valuation is a valuation if and only if 2 as a bowuiing constant.
it
Iws
PROOF. Suppose v is a general valuation with bounding constant 2. An easy induction show that v(al + ••• + a) ~ m.sup~=lvai if m is apower of 2. Ag there is a1ways a power of 2 between m and 2m, we have v(a1 + ••• + am) ~ 2m.suP'i=lvui for any m; in particular v(m) ~ 2m. Therefore, for each n > 1,
287
288
Chapter XII. Valuation theory
~ 2(n+llf'
t=O
2(7lvxiv~n-i
Taking n th roots and letting n ~ A general
valuation
nonarchimedean if v (x archimedean if vn
is
= 4(n+l)(vx +
gives v(x + ~l ~ vx + v~.
w
nontrivial
+ yl
if
sup (vx I vy l
~
v~ln.
> 1 for
vx
for
> 1 for some integer n.
0
some
all x and y
x
in
in 1<,
I< I
and
We say that two general
valuations are equivalent if they induce the same uniform structure, that t > 0 there is 6 > 0 such that < 6 implies V,X < c. We say that v, and are inequiva1ent if there exists x such that either v,x < 1 and ~ 1, or VzX < 1 and v,x ~ 1.
is, v , v ,x Vz VzX
is equivalent to
< 6 implies
VzX
1. 2 'llIEOREM.
if for each
Vz
< 10, and
VzX
Let v , and V z be generaL vaLuations and consider the
foL Lowing three condi. t ions.
(il There is a positive l'eat number
l'
such that v,x
invertibte x, (ii) v, is equivalent to
(iii) V,X
< 1 implies
V
z
I
<1
VzX
For all x,
Then (il impUes (ii) impUes (iii).
Moreover iF v, is nontrivial, then
(iii) implies (i). PROOF.
If (il holds and
> 0, choose 6 = Er.
t
< E or v 2 x > then VzX < Eo. A
Either v 2 x
0, and in the latter case x is invertible so if v,x < 101', similar argument works with the roles of v , and
< 1.
Suppose (ii) h01ds and v,x
< 1. < 1.
vzy VzX
Then for some n
>0
interchanged.
Vz
Choose 6 such that if
we have (v,x)n
>1
v,~
<1
so vzy
and hence
to show,
V2~
for x t
(log vzx)j( log vzy). with V
n
positive,
> 1. > 1.
0,
Set
Letting
~
have
z (ym x "') ~ 1 implies m/h ~ 12'
1.3 'llIEOREM. and
V
Note that if
v,~
> 1,
As v, is nontrivia1 we can choose y such that l'
=
(Log VdJ)/(log
(Log v,x)/(log
that I,
we
< 6, then
< 6 so (vzx)n < 1 whereupon
Now suppose (iii) holds and v, is nontrivial. then v,y-l
V,~
V2~)
v,~)
> O.
It suffices
is equa1
to 12
=
stand for < or >,and m and n be integers m/n~l,
implies
implies
v,(~mx"'l'v1
0
Let v, and Vz be nontrivial gener'al valuations.
z are inequivalent then there exists x such that v,x
< 1 and
VzX
If u,
>L
289
1. Va1uations If v, and
V2
PROOF.
are not inequivalent, then they ar'e equivalent.
v, and
Suppose
V2
are inequiva1ent.
As rep1acing x by
x-' in
the conclusion shows that the first claim is symmetrie, we may suppose that there exists y such that v,y < 1 and vzz
have
> 1 and set x V,x < 1.
Suppose v, and
=
V2
ynz •
Then
V2X
Choose z so that
1.
;:>
> 1, and, for sufficiently large
are not inequivalent.
that if v,x < 1, then V2X < 1.
V2Y
Let
y
Then there is n € IN such that v, (xny)
By (1.2) it suffices to show
be an element such that V2Y > 1.
< 1.
If
Therefore
valuations would be inequivalent,
n, we
V2(X n y) V2
> 1, then the two
(xny) ~ 1
(see Exercise
The nontriviality requirement in Theorem 1.2 is essential even if condition (iii) reads "v,x < 1 if and only if V2X < 1."
In fact without
the nontriviality condition we can prove neither (iii) implies (ii), nor (ii) implies (i). t
~
To see that we cannot prove (iii) implies (ii) let
0 be areal number and define va1uations v, and
V2
on the rational
= Ixl t and V2X = (v,x)t i f x t O. Here lxi is the absolute value of x. Then v, x < 1 if and only if v 2 x < 1, while if we could find ö > 0 such that v,x < ö implies V2X < 1/2, then either t > 0 or ö t > 3/4, and in the 1atter case t = 0, for i f t > 0 we cou1d find x such numbers by setting v,x
that v,x < ö is arbitrarily close to ö so to ö t > 3/4. t
would be arbitrarily close
V2X
Thus we would have a procedure for deciding whether t > 0 or
= 0, which is LPO. To see that we cannot prove (ii) implies (i) let t be areal number and
define v, and
V2
> 0 and
v,x 2 i f t
continuity.
It is readily verified that
that v,
V2
v,x
=
and
(V2X)r,
LLPO, as r
= Ix II t I for < 0 and extend to all t by
on the rational numbers by defining v,x
x t 0 and V2X to be v,x if t
are equivalent.
V2
is a general valuation and
But if we could find r
then we could determine whether
< 1 implies
Note that if v,
t
= v~,
bounding constant for v,.
S 0 while
r'
t ;:>
> 1/2 implies
t
0 or ;:>
t
such that S 0, which is
O.
and B is abounding constant for v 2
,
then Br is a
Hence any general valuation is equivalent to a
valuation. 1.4 '1'HI!XlREM.
Let v be a general valuation on k.
and m and n integer's greater than 1.
rhen
Let x and y be in k,
290
Chapter XII. Valuation theory (i) um<; sup(l,un)[og m/log n
(iil sup(l,v2) ts" bounding const"nt for u.
PRCX>F.
Let log = [ogz and note that BI +
slwi.' 1 [" 1::::
(ag j
VX, • l
To prove part (i), wri te mS = 0.0 + aln + ••• + ar.n T' where 0 <; a l < n and r log n <;
log m.
S
Then ums ~ B 1 +[og(n-l)(rtll sup(l,vn)/'.
Raising both sides to the l/s power, and letting s result.
To prove part (ül, let
u(~) <;
u(x + y)
<;
gives the desired
Bl+log(s+l)sup~=O qSv(~).
<;
sup(1,u2)log(~).
But
(~) <; ([s/2]) so
B(1+10g(s+1))/5'1'SUp(1,v2)(109 ([s/2]))/s
and (1 + log(s+l))/s .... 0 while ([s/2])/s
=
u(x + y) <: q'sup(1,lJ2)
As
00
sup(vx,uy) and consider
I:=O(~)xiyS-1
u By part (i) we have
q =
~
corollaries
we
-->
1 (use Stirling's formula), so
sup(1,u2)·sup(ux,vy).
have
the
0
following
characterizations
of
nonarchimedean and ar-chimedean valuations. 1.5 COROLLARY.
Let v be " gener·".l ualuat i. on on ".
Thert the fol !owing
equivalent.
QT'e
( i)
II
i s n OIw.T-ch imedean
+
(ii) v(l
(iiil um <: 1
(iv)
ll2
) < 1 For f(H"
PROOF.
(lU
x such th.at ux <: 1,
all integers m,
<; 1,
(v) un <; 1 for
(v) .
,
Clearly (i)
same
integer n
implies (U)
> 1.
implies (iii)
If (v) holds, then u 2 <; 1 by ( 1. 4 . i ),
(1.4.ii).
implies (iv)
implies
so we can take B = 1 by
0
1.6 COROLLARY.
Let v be a geTter'(d
llCllu"tion on Cl
(ield h.
Then the
foLLowing ur'e equivulent.
(i) v(x + y)
>
(ii) Ther'e is q B
>1
+ q,
sUp(UX,llY)
>
fot· same
0, such that
x und [/,
if 13 is abounding constant Fot' v, then
1. Va1uations
291
(iii) u2 > 1, (iv) um > 1 For some integer m (u is archimedean), (v) un > 1 for all integers n > 1, (vi) For some x we haue ux ~ 1 and u(l+x) > 1. PRDOF.
C1early (i) imp1ies (ii).
That (ii) implies (iii) fo1lows from
the fact that u2 is abounding constant for u (1.4.ii). C1early (iii) implies (iv). That (iv) implies (v) follows from (1.4.i). Clearly (v) implies (vi) upon taking x = 1. Finally (vi) implies (i) upon taking y
= 1.
0
It is a consequence of (l.S.iv) and (1.6.iii) that to assert that k is nonarchimedean is the same as to deny that k is archimedean, for to assert u2
~
1 is the same as to deny u2 > 1.
The following is a Brouwerian example of a nontrivial valuation that is neither archimedean nor nonarchimedean. 1.7 EXAKPLE. Let a be a binary sequence with at most one 1. Let R be l[Xl/I, where I is the (detachable prime) ideal generated by {an(X - 2n ) : n E ~}. Let k be the field of quotients of R. We define a valuation u on R by associating wi th each polynomial f (X ) the limit of the Cauchy sequence If(zn) 11;!n
i f an
0 for all n
IF(zn) 11/ n
if an
1 for some n
Here I I is absolute value. nontrivial because uX = 2.
m
~ ~
m.
The valuation u extends uniquely to k. It is However to setUe the archimedean question
would be to settle the question of whether an = 1 for some n or not. 1.8 'lHI!lOREM.
Let u be a map Fr'om a Heyting field k to the nonnegatiue
reals satisFying (i) and (ii) of the deFini tion oF a generaL ualuation. Suppose
ther'e
is a constant B such tImt iF vx
~
1, then v(l + x)
~
B.
Then v is a generaL ualuation with bounding constant B.
to show that we cannot have u(x + y) > this is the case. Suppose ux > O. If
PRDOF. It suffices B·sup(ux,uy), so suppose v(y/x) < 1, then u(x + y)
= vx·u(l +
a contradiction, so u(y/x)
~
y/x)
~
B·ux
~
B·sup(ux,uy),
1, whereupon uy > 0 and u(x/y)
~
1.
But then
Chapter XII. Valuation theory
292 v(x
+
y)
vY'v(l + x/y)
a eontradietion; so vx = 0 whenee x then v(x + y) = vy ~ B·sup(vx,vy).
~
B'vy
B'sup(vx,vy),
~
o beeause
k
is a Heyting field.
But
0
EXERCISES
1. Let k be a eOllUnutative ring in the definition of a valuation.
Show that if k admits a (general) valuation, then k is a loeal ring. Observe that (1.1) through (1.6) still hold in this more general situation. Show that (1.8) ho1ds if k is a loea1 ring, but eonstruet a (elassieal) eounterexample to (1.8) for general k.
2. 1he trivial valuation. Show that every diserete field admits a valuation v such that v(x) = 1 for all nonzero x. 3. '1t1e p-adic valuations on tQ. Let p be a prime. Every nonzero rational number can be wri tten uniquely as r = ±pnajb, where a and bare relatively prime positive integers not divisible by p; set Vpf" = P-n. Show that vp is a nonarehimedean valuation on IQ. Show that every nontrivial nonarehimedean valuation on IQ is equivalent to some v p ' 4. Let v be a nonarehimedean valuation. v(x
Show that if vx < vy, then
+ y) = vy.
5. Let k be a diserete field. field k (X) by setting v (F ) v(f/g)
=
v(f)/v(g).
Define v on the rational function 2-deg f for F E k [X 1, and setting
Show that v is a nonarchimedean valuation on
k(X) •
6. Let S be the set of nontrivial valuations on a Heyting field k, with equality in S being equivalence of valuations. Show that inequivalence is a tight apartness on S. 2. LOCALLY PRE{DIIPAC'!' VALUATlOOS
A subset B of a metric space X is bounded if there exists N such that d (x,y) ~ N for all x,y E B. Ametrie space is locally precompaet i f we can approximate bounded subsets wi th finite sets, that is, i f for each bounded subset B, and t > 0, there is a finite subset Y such that if x E B then d(x,y) < t for some y in Y. A loca11y precompact spaee is locally
293
2. Locally precompact valuations
compact if it is complete. It is readily seen that a valued field k is locally precompact if for each positive integer N and positive number ~ there is a finite subset Y of k such that if lxi S N then Ix - yl < ~ for some y in Y. Such a subset Y is called an E.-approximation to the N-ball. Absolute value and the p-adic valuations on ~ have this property. 2.1 THEOREM.
A locatty precompact uatuation is either archimedean or
nonar"chimedean •
Choose a finite subset Y of k such that i f Ix I s 2, then Ix-yl < 1/3 for some y in Y. Consider the integers O,l, ... ,cardY. Viewed in k, either one of them has value greater than 1, so k is archimedean, or they all have value less than 2. In the latter case, two must be within 1/3 of the same y € Y, hence within 2/3 of each other, so their difference has value 1ess than 2/3; so k is nonarchimedean by (l.5.v) as 111 = 1. 0 PROOF.
Let k be a nonarchimedean valued field.
The residue class field of k
is the set k = {x € k : Ix I s 1} where x = y i f Ix - y I < 1. Note that x € k is invertib1e if and only if lxi = 1, and that k is a denial field. The residue class field k need not be a local ring as we shall see in the example following Theorem 6.2. However if k has a discrete value group, then k is a discrete field. A nontrivial nonarchimedean valuation on k is discrete i f the value group of 1~ is cyclic. Note that a discrete valuation has a discrete value group and a discrete residue c1ass fie1d. 2.2 THEOREM.
Let
I~
haue a nontriviat, nonarchimedean ualuation.
Then
k is tocal ty precompact if und only if the valuation is discr"ete an.d the
residue class fietd is finite. PROOF. Suppose k is 10cally precompact. As the valuation is nontrivial we can find z in k such that 0 < Izl < 1. Choose a Izl-approximation x1, ... ,xn to the N-ball for N ~ 1/lzl. Ne will show that the value group is a discrete subset of the positive real numbers. For x in I~ either Ix n I -I 1, in which case Ix I -I 1, or Iz I < Ix n I < 1/lz I,
in which ca se Izl < Ixtl < l/lzl for t = 0,1, ... ,n. In the 1atter case there must be s, t and i, with s -I t, such that Ix s - xi I ~ Iz land Ix t - xi I s Iz I. As Ix s I > Iz land Ix t I > Iz land the va1uation is nonarchimedean, Ixsi
= lXi I =
Ix t I, so lxi
=
1.
294
Chapter XII. Valuation theory Thus t.he value group is a discret.e subset. of the positive real nurnbers. Izl
As
<
1
< N,
there
than 1. S 1.
such
1S
sup ( Iz-x i I , Iz I) = Iz I < 1.
that
Iz - xi
I
~
Izl,
lXi
so
I ~
Choose" among x I' ... ,x n of biggest value less
I" 1 <
For any invertible y there 1S an integer m such that
Choose j so that IXj - "-tll,JI ~ Izi ~ 1"1.
= IX j
Then IXjl
I"-my I
- v-tll y +
I"-tll y !, so Irr-myl cannol be less lhan 1 by the ehoiee OfIT. Thus Irr-tllyl = 1, so Iyl = Irr1 m. This shows that the vaIuat.ion is discrete. To
rr-myl show
=
thal
the
residue
class
field
i-approximation Y to the I-ball.
is
finite,
ehoose
a
finite
The elements of Y form a system of
representatives for the residue class field, which is therefore finite sinee it is discrete. Conversely, if the valuation is diserete and the residue class field is finite, let
1"1
< 1 generate the value graup and let A be a finite system
of representatives for the residue elass field. must find a finite set Y such that if in Y.
lxi;
Given positive N and
N then
Ix - yl <
we
E
far some y
E
Choose m so that Irr Im < , and Irr l-ffi i N and let Y
{""
L'i =-m
aO"i: a i ( l
Al.
If Ix I ; N then l,In x I ~ 1 so
~x
< 1"1 2m .
where Ibl
a-tll + 0-tll+lrr + ••• + 0m"2m + b Thus x
27=-mai"1 + b/~t, and Ib/~I < 1"lm
=
< E.
0
Although in the locally precompaet ca se you can distinguish archimedean from nonarchimedean,
you
cannot
necessarily distinguish
trivial
from
nontri vi al. 2.3 EXAMPLE.
Let r< be a finite held and X an indeterminate.
a binary sequenee wi th at most one l, and ao = a , = O.
Let
be
Let F be the
subfield of the rational funetion field "(X) generated by hand nEIN).
°
(anX
:
We define a nonarchimedean valuation on F by setting Ix I = 11 if Thus Ir I = nd f'9 f if f is a nonzero polynomial in F. Note that
an
= 1.
if
on = 0, then for any
ease Ix of F is I<.
t
y
in Feither
I < 1;11 for same
t
(
h.
As
lxi> 11 0
=
n or
(1,
= 0
lxi<;
In the latter
1.
the residue class Held
To get an f-approximation to the N-ball choose
n
>
sup(N,l/L).
If an = 1, then the valuation is discrete and Theorem 2..2 shows that F is
locally precornpact.
If on
by the previous argument.
0, then k is an tc-approximation to the N-ball 0
295
2. Locally precompact valuations
EXERCISES 1. Show that absolute value and the p-adic valuations on CI! are locally precornpact.
Show that the trivial valuation on any
discrete field is locally compact. 2~eg F is not
2. Show that the valuation on CI!(X) given by v(f) locally precornpact (see Exercise 1.5). 3. Show that residue class fields are denial fields.
4. Construct a Brouwerian example of a nonarchimedean valuation whose value group is cyclic but not discrete. 3. PSEUDOFAC'IORIAL FIELDS
A field k wi th a valuation is pseudofactorial i f inf{ IF(a) exists for each polynomial F in k[X].
I :
a E k}
The reason for this terminology is
that the root test (VII.1.B) says that a discrete field k is factorial if and only if for each polynomial F in k[X], either f has a root in k, or f (a) I- 0 for all a
when
Thus a discrete field k is factorial precisely
E k.
is pseudofactorial under the trivial valuation.
I~
An example of a
factorial field that is not pseudofactorial is provided by taking k to be U kn where kn is either CI! or CI!(lTi) polynomial X2 + 1 cannot be computed.
~
IR.
Then the infinimum of the
If k is a valued field we may form the caupletion
k
of k in the same
way we formed IR from CI! using the absolute value valuation on CI!. valuation on k extends naturally to The
p-adic numbers
are
valuation on the rationals. between
k
and
k,
then
k
k,
constructed and
and
k
in
this
The
is a complete Heyting field. manner
from
the p-adic
If k is nonarchimedean, and E is a field E
have the same the residue class fields and
value groups. 3.1 THIDREM.
Let k be a discrete Fidd wUh a vaLuation.
separabLe poLynomiaL in h[X], and Let there
is ö
construct
PROOF. deg F
> O.
>0
such that, given
0 0
k denote in k
Let F be a
the compLetion oF k.
for which
IF(ao)1
< ö.
Then
we can
~
0
raot of f in k.
If F is constant, then the conclusion is clear, so assume Choose N so that IF(a)1
whenever If(a)1 < 1.
Write
L1
whenever
101
2 N, and thus
101
~ N
296
Chapter XII. Valuation theory f(Y + Z)
f(Y) + Zf' (Y) + Z2 g (y,Z)
=
and choose M so that Ig(a,b)1 ~ M whenever lai ~ N and Ibl ~ 1. Find sand t
o < r' I F (a ) I
in k[Xj
such that s(X)F(X) + t (X)F' (X)
= 1.
Choose
1 so that I t (a) I ~ r' - 1 and Is (a) I ~ r' -1 whenever la I ~ N.
~
< 1, then
la
If
I ~ N, so
r- 1 If'(a)1 i It(a)F'(a)1 i 1-ls(a)IIF(a)1 i 1-r- 1 IF(a)1. whereupon
2 r-lf(a)l.
If'(a)1
Choose 0
~
2
r/2 such that if u
1/0, then
~ ~.
2M[ru : 1)2
< 0, then we can use Newton's method to construct a root of f as fo11ows. Define an inductive1y by ~+1 = an + h n , where h n = If IF(ao)1
-f (an l/f' (an) , n
= 0,
We shall show that
If IF(an)1 ~ 2-flo
If (an) I ~ 2-fl O•
< 1, then lanl
This is true for
~ N and IF' (an)1 2 r - 2-fl o
so
Ihnl ~ (rZn/o - 1)-1 ~ 2-fl, But If(an +1)1 = If(an + hn)1 = Ih~llg(an,hn)1 ~ Ihn 12M ~ r(n+1)0 by taking u = 2n /0, Thus {arJ is a Cauchy sequence converging to a root of F in 3.2 COROLLARY.
k,
PROOF.
k
be
0
a discr'ete
field wUh a
IF f is a separ'abte polynomial in k[X],
valuation.
root in
Let
k.
or If(a)1 is bounded away from 0 far alt
Choose 0 as in Theorem 3.1.
0
pseudofactorial
then either f has a in k.
Either inf{IF(a)1 : a E k} 2 0/2,
in which case IF (a) I is bounded away from 0, or inF{ If (a) I : a in which case we can find a root of f in
k
by (3.1).
E k}
<0
0
Using the notion of winding number, we can get another proof of the discrete fundamental theorem of algebra. 3.3 COROLLARY. dosure of k in [;. PROOF.
Let k be a discr'ete subfield of [;, and K the algebr'oic
Then K is algebraically dosed.
Let f be a monic polynomial of degree n wi th coefficients in K,
which is a discrete field by (VI. 1. 9) ,
Choose r
complex number of modulus r, then Izn - f (z) I
> 0 so that
< r,n.
if z is a
Then the winding
number with respect to 0 of the path given by restricting F to the circle of radius r is n. would be O.
If
inF{IF(x)1
Hence inF{lf(x)1
:
:
lxi ~ r}
> 0, then this winding number
lxi ~ r} = O.
Ag
f i s a product of
297
3. Pseudofactorial fields
separable polynomials (VI. 6.3), we can find a root of F in the complex numbers by Theorem 3.1. is proved.
By (VI.1.S) such a root is in K, so the corollary
0
Pseudofactorial fields arise as follows. 3.4 'J.'HEDREfII.
IF k is a discl'ete fieLd with either a 10caHy p,'ecompact
vaLuation, or a discrete valuation with a factorial
residue class Field,
then k is pseudoFactorial.
Let F be a polynomial in k IX 1•
PRCXlF. choose IF(a)1
N
such
that
IF(a) I ;. If(O) I
< IF(O)I, then lai
is locally precompact
If k
whenever
lai
;.
N,
so
if
Let Yn be a l/n-approximation to the N-ball. Then the sequence r n = inf{lf(y)1 : y E Ynl is Cauchy and converges to inf ( IF(a) I : a E k l, so k is pseudofactorial. ~
N.
Suppose the valuation is discrete with a factorial residue class field. Replacing f (X) by dc deg f f (X/c) we may assume that f is monic and the values of the coefficients of F do not exceed 1, generated by r
< I, and choose 1/" in
,~
Let the value group be
such that
I1/" I
= r.
construct a sequence of (possible empty) finite families Bm Ibl ~
1 and
If(b)1
and positive integers n(m) IF(a)1 ~ I,n(m), then la - bl ~ r m for some bin Bm • ~
rml,
We shall !;;
(b
E k
such that
:
if
Let BI consist of one representative of each of the roots of F read as a polynomial over the (factorial) residue cl ass field, and set n(l) If IF(a)1 ~ r, then lai
s
= 1.
1 because f is monic with small coefficients, so
a represents a root of F in the residue class field, whereupon la - bl < r for some b in BI'
To construct Bm+1 from Bm we proceed as follows.
For
each b in Bm write
f(b + ~X)
= 1/"e(b)gb(X)
where the maximum value of the coefficients of gb E h [X 1 is 1.
Aß
F(b + ~X) - F(b) is divisible by ~X, and IF(b)1 ~ I,m, we have e(b) 2 m.
Let cf,., "c~ be representatives of the roots of gb read in the residue class field, and let Bm+l consist of all elements of the form b +
for b in Bm ,
Let n (m+1) exceed n (m) and e (b) for each b in Bm • Suppose Then la - bl S r m for some b in Bm , so a = b + ~c for in k such that Ic I s 1. Since If (a) I s r n (m+l) and f (a) =
IF(a) I S r n (m+1).
some c
~c'?1
298 Tr
Chapter XII. Valuation theory
e (b ) gb (e ), we have 19 b (e ) I .:; r' so e represents a root of gb read in the
Hence le - e~1 .:; r for some i, so la -
residue class field. !rrme - rrme~! .:; r m+1 .
If B 1 is empty, then 1 compute inf { I f (a II : a E dn
=
If
rrme~)!
=
If(all for each a in k. Otherwise we by taking the limit of the Cauchy sequence
= If(Oll .:; ")
inf {I f (b II : b E Um':;nBm).
3.5 COROLIARY.
(b +
0
a
is
I~
diser'ete
fieid
wUh
a
pseudofaetoriaL
vaLuatton, then the aLgebr'aie (separabLe) eLosure of 1< in Us comptetton
h
is separabLy faetoriaL. ~
The algebraic (separable) closure K of h in k is discrete PROOF. (VI.1.9). Moreover K is pseudofactorial since k is dense in K. Hence, if f is a separab1e polynomia1 in K[X], then either f has a root in = h, or If(a)! is bounded away from 0 for all a in K, by Corollary 3.2. But any root of F in k is in K, so K is separably factorial (VII.1.7). 0
K
EXERCISES
1. Let k be a discrete field, and equip the rational function field K = k(X) with the valuation of Exercise 1.5. The completion of
K
is the field of formal power series over identify with formal sums of the form
K
K
I~.
Show that we may
F
where m E 71 and
ai
E k,
and that if
am
-F 0, then v(F) = m.
2. The stipulation that f be separable cannot be eliminated from the hypothesis of Theorem 3.1.
Let {an} be a binary sequence of with
at most one 1. Let K be the field of formal power series in Y over the two element field 71 2 • Let ( be an element of K that is transcendental over 71 2 (Y). J)
=
Show that
(1 +
L0r/ 2n +l )(2
is transcendental over 71 2 (Y) • Let k = 71 2 (Y , J)) and let f (X) = f). Show that f has a root in K = k i f and only i f an = 0 _ for all n. Show that k is pseudofactorial by Theorem 3.4, so Corollary 3.2 would allow us to decide whether f had a root in K
X2
or not.
3. Pseudofactorial fields
299
3. Describe the valuation on the completion
k
of a valued field k. Show that the residue class fields and value groups of k and k are the same.
4. Show that the sequences r n
and d n in the proof of (3.4) are
Cauchy. 4. IDRMED VECTOR SPACES.
Let k be a valued field.
A
normed vector space over k is a vector
space V over R, with a tight inequality, and a map, called a noen, taking each x E V to a nonnegative real number lxi satisfying (i) (ii) (iii)
lxi ~ 0 i f and only laxl = lailxi Ix+yl ~ lxi + Iyl
if
x ~ o.
If V has no inequality, and we replace (i) by lxi = 0 only if x = 0, then can use (i) to define a tight inequality on V. A normed vector space is a metric space with the metric Ix - ul. The strong inequality on k n is given by setting x ~ y if Xi ~ Yt for some i. Ne turn Rn into a normed vector space by setting lxi = sup{lxil : i = I, •.• ,n}. If we put the absolute value valuation on ~ and rn, then any subfield of rn is a normed vector space over ~. Let V be a normed vector space over a valued field k. Ne say that vI' ... ,vn in V are metrica1ly independent i f for every t > 0 there is we
> 0 so that if I~ atv t I < ö, then latl < t for each t. If, in addition, every element of V can be written as a linear combination of the v t ' then the Vi are said to be ametrie basis of V. Thus V admits a metric basis if and only if there is a continuous isomorphism from V to kn , the latter equipped with the supremum norm. The subfield ~(J2) of rn is a finite-dimensional normed vector space over a discrete field that does not admit a metric basis. On the other hand, as linear transformations of finite-dimensional normed vector spaces ö
are continuous, if a normed vector space over a discrete valued field admits a metric basis, then any basis is ametrie basis. Two norms I I, and I 12 are equivalent if there exists t > 0 such that tlxl, ~ Ixl2 ~ t-'lxl 2 for each x in V. I t is an easy exercise to show that the norms I I, and I 12 are equivalent i f and only if the identity map from (V,I I,) to (v,1 12) is bicontinuous.
Chapter XII. Valuation theory
300
Let V be a normed vector space over k. A set of elements vI"" ,vn of V is said to be a basis for V if the linear transformation ~ from kn to V defined by ~(ei) = Vi is an isomorphism that preserves inequality. We shall see that for locally compact fields k, any basis for V is a metric basis; this is the same as saying that any two norms on kn are equivalent. warning: A discrete field is locally compact under the trivial valuation, but not necessarily under others--for example,
Ql
wi th absolute value is
not complete. LEMMA 4.1.
Let k be a valued fieid, and el"" ,e n the natur'al basis A norm on k n is equivalent to the supremum norm if and onty if e i is bouruled away from k i-I = kel + ... + ke i _l for i = 2, ... ,n.
for k n •
PROOF.
The 'only if' is clear from the definition of the supremum
For the 'if', it suffices to show that the coordinate projections
norm.
are continuous. Suppose le n - 01 ~ 0 > 0 for all 0 in ,~n-l. Then olxnl ~ IXlel + ••• + xnenl so projection of ,p onto its last coordinate, and hence also onto kn - 1 , is continuous. We are done by induction on n. 0
lTi(xlel + ••• + xne n )
=
Xi
A subset A of a metric space X is located if we can find, for each x in > 0, an element ao in A such that d (x ,ao) < d (x ,a) +" for all a € A; thus d(x,A) = inf{d(x,o) 0 € A} exists. If A is a nonempty X and "
10cally precompact subset, then A is 10cated: Choose N so that the bounded set B = {o E A : d(x,o) < N} is nonempty. Approximate B to within ,,/2 bya finite set F ~ A. Choose 0 0 E F so that d(x,oo) is within ,,/2 of inf{d(x,a) : a E F}.
'HIEXlREM 4.2.
Let I< be a 10caHy compuct valued fieid.
Then
,,11
is a
locally compuct l1or'med spuce aveI''' under the supr'emum norm, and any nor'm on k n is equivatent to this one.
PROOF.
Clearly kn is 10cally compact under the supremum norm.
kl1
Suppose
that is equipped with another norm. By induction "n-1 is locally compact and so is 10cated. It suffices by Theorem 4.1 to show that d (e n "p-l) > O. using the fact that "n-l is 1ocated, define a sequence of elements S.1 E kn - l so that for each i > 0 either ISi - enl
or
d(e n ,l
< I/i
> l/i+l and
Si
= Si-I'
301
4. Normed vector spaces
!t is easily seen that the 0i form a Cauchy sequence in I 0 whence le - enl > I/i for some 1. As le - enl <; Is[ - enl, we cannot have lei - enl < I/i.,
> 1/( 1+1).
whereupon d (e n ,hn - 1 )
0
The condition in Theorem 4.2 that the field I< be locally compact cannot be weakened to read that h is complete, as in the classical context; the
to the locatedness of h n - 1
appeal
in the proof is essential as the
following Brouwerian counterexample shows. EXAMPLE 4.3.
Let
i 0 be areal number and let k be the completion of
T'
the subfield lR(ri) of IC.
Then k is a closed subfield of IC.
Ne shall
construct a norm on k 2 such that you cannot bound e 2 away from /,'. If e"
is the natural basis for
e2
fj,:2,
then every element of 1C 2 can be
written uniquely in the form z = a(e, - ie z ) + ß(e, + fe2) with a and ß in IC.
Equip 1C 2 wi th the seminorm Ilzll = I(51 +
a norm on k
IIm
zl = la -
sup( la I,
r' la
I.
To see that this defines
let Im z denote the irnaginary part of z
2 ,
pI.
Iß I )/2,
If z t 0, then either a t 0 or
f3
f- O.
and note that
pI <
Either la -
in which case l{ll t 0 so Ilzll f: 0, or Im z t 0 so
I'
>0
whence Ilzll t O.
> 0, then r = Ile, - fezll i d(ez,h'). By < d(e2,h'), and in the latter case I' = 0 > 0 is impossible. Thus either /' > 0 or T' = 0, which is
Suppose d(ez,k')
> O.
cotransitivity either because
T'
equivalent to LPO.
T'
If
r'
> 0 or
T'
0
EXERCISES 1. The trivial metrie on a discrete set X is defined by setting d(x,y}
=
0 if x
=
H, and d(x,y) = 1 otherwise.
Show that a
subset of X is detachable if and only if it is located in the tdvial metde. 2. Show that the space ~(V2) is discrete. 3. Define a positi ve notion of inequi valence of norms on vector spaces, and show that the supremum norm
on~"
is inequivalent to
the norm on ~2 induced by an isomorphism with ~(v2).
302
Chapter XII. Valuation theory
5. REAL
AN!)
COMPLEX FIELDS.
All archimedean fields contain a copy of the rational numbers there is essentially only one archimedean valuation on
m.
m,
and
The following
theorem characterizes all nontrivial valuations on ID. THEOREM 5.1.
ArlY nontn:vioJ
absolute votae or to the
PROOF.
If
In/pi>
on m is
t)(ll!lotion
equivolent
1 for positive integers
Tl
and
p,
Inl > 1 or Ipl < 1. > 1 by (1.6). Furthermore
cotransitivity implies that either
Im I > 1
for every integer m
by (1.4) so
Im I Iml
<;
ci ther-
10
voluation for- some prime p.
p~dic
In II.Og m/log Inl log rn/tag
n
and
If
Inl > Ipl so Inl > 1, then
Im Ilog n/log m
<;
mr- for fixed r; thus the valuation is
Tl
equivalent to absolute va1ue.
In I
then
If, on the other hand,
Ipl
< 1, then the
valuation is nonarchimedean (1.5) and we may assume that p is a prime.
If
sm + tp = 1, then
Itpl = It Ilpl <
hut
Ism + so Isml
I
1 =
tp
1
= 1
Ism I , It pi) , whence Iml = 1.
~ sup (
'l'hus the value of an
arbitrary positive integer mit is 11'11" if (m,p) = 1; so the valuation is equivalent to the p-adic valuation.
0
Note that every nontrivial valuation on ID is locally precompact, and that absolute value is the only archimedean valuation on ID with 121 = 2. There are two important kinds of archimedean fields. DEFINITlOO. that Ö
10
2
Let f, be an archimedean Held.
+ 11 ,> {; for all a in f"
> 0 there exists
0
then k is said to be reaL
in I, such that
la?
+ 11
<
> 0 such
If there is {; Ö,
If for every
then k is said to be
complex. C1assically every archimedean Held is either real or complex.
However
the Held k of Example 4.3 cannot be said to be real or to be complex. 'I'he construction of the valued Held U of complex numbers from the valued field IR of real numbers can be used to construct a unique valued splitting field for X2 + 1 over any real field. THEOREM 5.2.
IF
h is a n,ol. Ficld, then I, can be embedded in.
fieLd K containing an element i
ir "
such thot i
2
= -land K
= k(i ).
is embedded in a uoLued field E containing W1 element
j
0
lIoLued
~loreover·
such tlwl
303
5. Real and complex fields j2
= -1,
is bounded away from I< and h(j) is isomorphie to K as a
then j
voLued fieid.
Let K be the ring of all formal sums
PROOF. i
-1.
2
If we define inequality on K bya + bi ~ 0 if a ~ 0 or b ~ 0,
then K is a Heyting field.
Jla 2 +
b2
wi th a,b E k and
o. + bi
1.
la + bi
We can va1ue K by setting
I
To show that this is a general valuation it suffices to find a
1(1+0)2 + bZI
s
s
l(lto.)2 + b 2 1
constant B so that
B whenever
+ b2 1
10 2
s
But
1.
1 + 10 2 + b 2 1 + 21al, so we need only bound 101.
lai> 0, then 101-
Z 11 + b /o 1 is bounded away from O.
2
2
As 121
2
If
s
2,
it's a valuation. To prove the second claim we first show that j is bounded away from k. I f a E k,
then either
> 1;
101 - lil
if
lai> 2 or
lai< 3.
lai< 3, then la - jl
=
la - jl 2
lai> 2, then
If
la 2 + li/la +
jl >
102 + 11/4,
and the latter is bounded away from 0 because h is real. If'P
: K
-+
k(j)
is defined by
homomorphism of Konto k(j). If a + bj
~
0, then a
'P(o
+ bl)
We !!lUst show that
0 + bj, then'P is a
=
'P
preserves inequality.
0 or bj # 0 as E is a Heyting field, so a
~
~
0 or
b # 0, whence 0 + bi. # O. Conversely if 0 + bi # 0, then a ~ 0 or b l' 0, so 0 ~ a 2 + b 2 = (a + bj) (0 - bj) whence a + bj # O. The valuation on
induces, via 'P,
k(j)
a valuation on K.
Theo rem 4.1 says tha t
thi s
valuation is equiva1ent to the given valuation on K, so one is a positive power of the other (1.2). COROLLARY 5.3.
AB
they agree on k, they are equal.
There is onLy one vaLuaUon on the fietd
a::
0
timt exlends
absol.ute vaLue on IR. PROOF.
Immediate from Theorem 5.2.
'lBEOREM 5.4 (Gelfand-Tornheim). k =
0
If k is a normed (Leid over-
a::,
then
a::. PROOF.
As
a:: is eomplete, it suffiees to show that a:: is dense in h.
Suppose we want an element of IC that is elose to a given a E k.
By
cotransitivity, either 0 is elose to a, or lai> 0, so we may assume the latter.
Construct a square of width
ISa I,
centered at 0, eut it into 4n 2
li tUe squares of width 140 I/n, and let S be the polygonal path formed by the edges of the li tUe squares that lie on the perimeter of the big square.
Let 0 be the infimum of
Iz -
01 as z ranges over the midpoints of
the sides, and the centers, of the little squares.
We will show that
304
Chapter XII. Valuation theory
0 3 ~ 314a1 3 ;h,
SO
we can get complex numbers that are arbitrarily close to By cotransitivity we may assume 0 > O.
a by choosing n large.
Given a function f vertices
zO, .. "zn
IC .... h,
in
and a polygonal path P with successive
we
IC,
define
2,1~=1(zi - zi_1)F((zi + zi_1)/2). F(z)
the
Clearly I p
'integral' is
linear
= cz + d, then Ip(f) = (zn - zO)((zn + zO)c/2 + d).
if P
is a
closed path,
that is,
if
Zo
= zn'
Ip(f)
to
be
in F,
and
if
In particular,
then I p
vanishes on
po1ynomials of degree at most one. Consider the function 9 (z) = 1/(z -
from IC to
a)
I~.
First we get an
upper bound on I Q(g) where Q is a square wi th sides of length
o >0
Let
Co.
be at most the infimum of the distances from a to the midpoints of
the sides, and to the center, of Q.
Our bound will depend only on
0, we may assume that 0 is the center of Q. _1_ z - a
g(z)
Co
and
As _!L.±...L
Z2
a 2 (z _ a)
a2
and 1 Q vanishes on po1ynomials of degree at most one, we have IIQ(g) 1 ~ 4t(Co/2)2/0 3 = t 3 /0 3
•
As 1S(g) is equal to the sum of the integrals 1Q(g), where Q ranges over the 4n 2 little squares of width Co = 14aI/n, we have I1S(g)1 ~ 414a 13 /0 3 n.
We use this to get an upper bound on 1S (1/z) .
1
_1_
z
z - a
so 3 414a1 3
+
on
_
Now
a
z(z - a)
la 141aa 1 14a I ( 14a 1 -
la 1)
We now compute a lower bound for I S (1/z) directly from the definition. Each of the four sides of the square contributes
.Lf [ n k=1 1 +
1 i (k - 1/2 )/n
+
1
1 - i (k - 1/2 )/n
which, in absolute value, is equal to
>\n~=l. Lk =l 2n 2
-
Combining this with the upper bound on II s (1/z) 1 we get 4
~ ~ 03 n
+
~ 3
)
305
5. Real and complex fields COROLIARY 5.5.
Let I, be an Q/'chimedeo.11 Field mUh
real iF on!!) iF h == IR os
PROOF.
°
Clearly any subheld of IR is real.
to assume that IR ~ k.
Aß IR (i )
==
c:
Then h is
Conversely if h is real, Theorem 5.1 allows us
by Theorem 5.2, we have k(i) = IRU) by
But i is bounded away from
THEOREM 5.6.
2.
vaLued neId.
then k i s real and we can form k (i) by Theorem 5.2. Theorem 5.4.
121
k
as
k
is real, so
Le t h oe an ar eh lmedean fi eid mUh
121
k= =
IR.
2.
0
Then
the
fol1.oming are equivalent. (i) (ü)
(iU)
(iv)
PROOF. I f 10
2
h is campl.ex,
la 2
11
+
< 3/4
Thene is
for' same
in h such that
.2 1
= -1
Clearly (i) implies (ii), and (iv) implies (i).
+ 11< 3/4, then
1,,1
> 1/4 so
2
11
if b
=
(a
Suppose (ii).
-1/a)l2, then I/)2 + 11 =
Thus i f we dehne a sequence by
a,,+l = (an - 1/an )l2, we have ar~ .... lar~ +
in k,
I< - C:.
1(0 2 +1)2/40 21 ~ 1(t2 + 11 2 •
<
a
-1,
= 0,
00
'while la,,+l - anl = la~ +
and
11/2 1a n l
shows that t.he sequence is Cauchy.
Now suppose (iU).
By (5.1) we may assume IR (;: R,
( 5 • 2), so I< == IR (i) by (5. 4 ).
and lR(i) == ([ by
0
Using condition (5.6.ii), and cot.ransitivity, we see immediately that an archimedean field is real if and only if it cannot be complex. COROLIARY 5.7 (Ostrowski).
121
mith
PROOF.
la
2
= 2,
then
k ==
IR ar
precampact vaLued fi.eld
As h 15 locally precompact we can calculate the infinimum of
+ 11 as a ranges over k.
is complex by Theorem 5.6. real.
IF I< 1s a [ocalt!)
k == c:. If this infinimum is less t.han 3/4, then I<
If the infinimum is greater than 0, then k is
0
As t.he completion
k
of a discrete held is not discrete in general, it
is often convenient to work inst.ead wi th t.he separable closure Iz.
Toi
of il in
Because in the nonarchimedean case the field h i5 intimately tied up
with Hensel's lemma, we call k the Henselization of k.
The following
characterizes when a discrete archimedean held /, is real or complex in terms of its Henselization.
306
Chapter XII. Valuation theory 'IHEOREfI1 5. 8 .
ir
factor'ial
h
is )~(i)
Ir
and. ortly if k is T'eal or" /, is complex.
is aLgebraicaUy c!osed.
If k is n!ol,
is
k is cOlllplex, then
then the Hensetization of k(i)
h
which is aLgebralcaUli c!osed, amt euer'y poilinomial. ouer
I
h
riten
i.s a
pr"oduct of ir'T'educible 1 inear and quadT'at ic factoT's.
h. is factorial, then either X 2 + 1 has a root in r{, or X + 1 is irreducible over h.. In the first case k is comp1ex by PROOF.
If
2
(5.6.iii); in the second case h is real because (5.6.ii) cannot hold, so
la
+ 11
2
> 1/2 for all
is comp1ex,
If /,
a E /,.
a1gebraieally closed, henee faetoLial. may assume /, = IR.
h ~ 0:: so I:; then h ~ IR, so
then
If k is real,
Let f be a polynomial of positive degree in 1{[Xj.
f has a complex root a + bi, so a - bi is also a root of f.
Thus
is we
Then and b
Cl
are a1gebraie over hand so are in h (so the Henselization of I< (i) is
h (l ) ) • a
2
+
1;2
If
I; =
0, then X -
Cl
is a factor of f.
i5 an i rreducible factor of
r
in r;rX J.
If
I; ;i
0, then X2
2nX +
-
0
EXERCISES 1. Construet a Brouwerian example of a nonarchimedean valuation on
~
that is neither trivial nor equivalent to a p-adie valuation. 2. Show that I S ( l/z) in the proof of (5.4) approaehes 27ri as n
~
00,
using
f~(l
+ X 2 )-ldx = rr/4.
3. Show that if every arehimedean field ean be embedded in 0::, then
the world's simplest axiom of ehoiee holds (use the construetion in Exereise VI.3.l).
6. HENSEL'S LEMMA Let k be a nonarehimedean valued field, and I< i ts residue class field. Let f be a monie pclynomial in Ie. (X 1 a11 of whose eoeffieients have value at most 1.
Then f determines a polynomial
Hensel' s lemma is that if then f
has a
relatively
F has
LOot in h.
prime
faetorization of f.
factors,
r
in /, [X 1.
A standard form of
a simple root in 1, (X 1, and I< is eomplete,
More gene rally, then
this
if f
faetors
faetodzation
is
into strongly indueed
by
a
Dur version of Hensel's lemma eoncerns how to irnprove
approximate factorizations of f
into approximately strongly relatively
307
6. Hensel's lemma prime factors. First we show how to extend a nonarchimedean valuation on k valuation on k(Xl, with any specified positive value for X. lemma
to a
The following
is needed because we cannot determine which coefficients of a
polynomial have maximum value. Let sl""'sp be ,'eal nwnbers with supno:mwn a, und tet m be
6.1 LEMMA.
Then for aU but at most (m+l) (!'-1) positive integeT's
a positive integer.
n we can constr'uct a finite subset ..I. of {l,. .• ,!') such timt
if i.
E ..I.
and i
For each positive integer
PROOF.
Tl
F
i. r{. ..I.
canstruct a finite subset An af
(l, •.. ,!') such that s.
>a
_ 2-(n-l)
si
- 2-n
a
descending
t
and
Then
the
form
An
if i
r{.
An'
chain of nonempty
finite
subsets
of
Since #..1. 1 S l!, there are at most l!-l values of n for which
(l, ..• ,l') .
An + l '" An' and hence at most (m+l)(l!-l) values of n such that Arn +fl+1 '" An' Far the remaining values of n we let ..I. = An = Am+f1+l' 0 6.2 THEOREM.
A be
posi tive
a
each c
nwnber.
I t suffices to show that we have defined a valuation on k [X].
PROOF.
Let g(X)
reat
Then thi.s defines a lJol.uation on k(Xl.
sup lajlAi.
Clearly
Let k be a fietd with a nonarchimedean lJa l.ua ti on , and let
=
IFgl > O.
Ir 11 gl > 0,
biX i .
L
~
The only problem is in showing that
Irllgl If
whence
so it 5uffices to show that
1f 1I gl
Irgl
+ c
Ifgl
=
Irllgl.
> Irllgl for
< (, then we are done, so we may assurne that and 19 1 > O. Choose rn and n so that
If I > 0
2-msup (1,lr!+l91)
< inf(lfl,lgll
and
and use (6.1) to construct finite StIDsets A and B of the indices of the coefficients of fand g respectively, so that
308
Chapter XII. Valuation theory lai I lai I Ibjl Ib j I
Let r
If j ~ B,
then
f2921
Z-(n+m) 2--11
E A
if
~
A
if j E B if j
~
B.
If i E A and j E B, then
(Ifl - Z-(n+m»(191 - T(n+m»
>
Ifl191 - 2-(n+m)(lfl + 191)
Let f
=
< Ifl(191 f, + f 2 and
2--11)
=
>
Ifl191
9 = 9, + 92'
-
r. 2-;11fl
where f,
~ r.
= 2iEAaiXi
j
Then f9 = f,9, + f,92 + f 29, + f 292 and If'92 + (29, + But considering the monomial of highest degree in f,9, shows
= 2 j €B b
< r.
l
>0
if
>
laixibljl
Similarly for i ~ A.
>0
2-n
If 119 I - 2--11 i nf ( I f I, 19 I ) • laiXibljl
and 9,
2-(n-tm)
> If I < I FI > 19 I < 19 I
that I f ,9, I > '".
Hence I f 9 I = I f ,9, I > '" > I f 119 I -
Ec.
0
We can now give an example of a residue class field that is not a local ring. Let k be a field with a nonarchimedean valuation, and let 0 and ß be positive real numbers such that sup (0 ,ß) = 1. Applying Theorem 6.2 twice we get a valuation on the rational function field
k(X,Y)
such that
and IYI = ß and Ix + YI = sup(o,ß) = 1. In the residue class field k we have X + Y is invertible but we cannot assert that either X or Y is invertible (see Exercise II.3.5). The following theorem bounds the value of the remainder polynomial in the division algorithm. lxi =
0
6.3 'lHOOREM.
Let k be a field with a nonarchimedean valuation and Let
and .p (X) = aO
+ alX + ... + a,/n.
be eLements of k[X) such that a,l is a unit. M = max(m-;l+l,O).
PROOF.
< deg
1.p(X)I/lanxnl. and
Then the,"e exist q,'" E k[X) such that
9(X) de9 r(X)
Let s =
= q(X).p(X)
+ r(X)
.p(X) and IdX) I <:; sMlg(X) I.
We induct on m, which we may assume is at least n.
Write
9,(X) = 9(X) - .p(X)Xm--l1bm/a n •
Then 1.p(X)Xm-;lbm/anl = slbmXml ~ slg(x)I, so 19,(x)1 ~ SI9(X)I. By induction, g, (X) = q, (X).p(X) + dX) where deg r(X) < de9 .p(X) and ,,"(X) I ~
309
6. Hensel's lemma
The setting for Hensel's lemma is an approxirnate factorization of a polynomial into approximate1y strong1y re1ative1y prime factors. Let k be a field wi th a nonarchimedean va1uation,
6.4 DEFINITlW.
extended to k[XI via (6.2).
An Henselian context is
F,.,t,h,A,B,C E k[XI,
L,M E rn,
dER,
c E
m
such that (i)
F
= .t
+ h,
( ii) Aop + B>jJ
= d + C,
(iii) 1>jJ1 ~ 1 and IBI ~ 1, (iv) op = !~=O aiX i and lanl ~ 0,
(v) deg f, deg h, n (vi) deg A
~
~
L,
M -L -1 anddeg B
~
M - 2(L-n) -1,
(vii) 0 < Idl ~ 1 and t < I, (viii) sM IC/d I ~ t and s2MIh/d 2 .I ~ c where s = I.I/Ianxnl. condition (i) says that op>jJ approximates f to within h, and (viii) says that h is small compared to op.
Condition (viii) also says that le I is
sma11 compared to d, so (ii) says that op and t are approximately strongly relatively prime. 6.5 LEfiIIMA.
Let f ,op,>jJ,h,A,B,C,d,L,M,t be an Hensetian cohtext. Then
(i) IAop I ~ I, (ii) deg >jJ ~ L - n, (iii) deg C
~
M,
(iv) deg 01, deg Bh
PROOF.
~
M+
n -
1.
From (6.4.ii) we get IAop I ~ sup!l B>jJ I, Id I, ICl} ~ 1,
using the bounds on IBI,I>jJI,ldl, and Ici in (6.4).
=
As deg op>jJ
deg f-h ~
L, and the leading coefficient an of op is a unit, we have deg >jJ S; L - n. Therefore deg C
so deg 01
~
~
sup (deg A +
M + n - 1.
Tl,
Finally
deg B + L -
Tl)
~
M - L - 1 +
Tl
310
Chapter XII. Valuation theory deg Bh S M - (L-n) + n - 1 S M + n - 1.
0
The heart of Hensel's lemma is the refinement of Henselian contexts. 6.6 'I'HOOREM.
There is a Function From HenseUan contexts to Henselian
contexts, taking F,~,+,h,A,B,C,d,L,M,~
to
F ,~*,+*,h*,A,B,C*,d,L,M,~ such that (i) Ih*1 ~ ~Ihl, (ii) ~* has the same degree und leading coeFFicient as ~, (Ei)
I~* -- ~I
(iv) 1+* -- +1
s sMlh/dl s s-Mcld~l, s sMlh/d~1 s s--M~ldl.
so I~*I
I~I,
=
We shal1 construct ~* = ~ + ß and +* = + + a satisfying (ii), (iii) and (iv), and define PROOF.
h*
= F --
~ *+ *
and C*
=
A~ * + B+ * -- d.
We must then show (i), that deg h* S L, and sM IC*/d 1 S~. That 1+*1 S 1 follows from (iv). In view of (6.5.iv) we can apply (6.3) with 9 = Bh/d and with 9 = Ol/d obtaining Bh/d = Ol/d
+ ß
q~
+ r
= 11
where deg ß < n and deg r < n and 1ß I S sM 1Bh/d 1 ~ sM 1h/d I 1,,1
Setting ~* = ~ + ß
we
s sMlol/dl s ~Ihl·
have (ii) and (iii).
Let
a = Alt/d + q+ -- p.
Then multiplying (6.4. ii) by h/d we get a~ +
so deg L - n.
a~
ß+
=
h + "
S L, because deg ßt ::; n + (L-n) by (6.5.ii), whence deg
Also la~1
s sup{Ißtl,lhl,lrll s sup(IßI,lhj,I"ll s sup {sM Ih/d I , Ih I , ~ 1h I)
=
sM Ih/d I
CI
S
311
6. Hensel's lemma so
lai
s sMlh/d
Thus (iv) holds for /' = ,p + a. Let h* = f -
Then
h*
=h
(h + r) - af3
-
=-
r - af3
so deg h* S sup(deg r, deg a(3) S L. Also
Ih*1
s supe Ir!, laf3l) s sup{sM ICh/d I ,(sMlhl/ld
which is (i).
Define e* by A
=d
+ e*
Then the only thing left to check is the bound (6.4.viii) on le*l.
We
have C*
=C +
Af3 + Ba
so
sM Ic* I S sMsupfic I, lAß I.IBa I l = sMsup!lc I, IA
Let k be a field with a nonarchimedean valuation,
extended to k[X] via (6.2).
We say that k
is Henselian i f whenever
f,
context f,~,~,O,A,B,C,d,L,M,E such that ~ has the same degree and leading coefficient as <po Note that this definition depends on the value A = lxi chosen for ~he extension of the valuation on k to k [X] via Theorem 6.2.
In fact this
dependence is only apparent, at least for discrete fields, as we shall show in Theorem 6.11. 6.8 HENSEL'S LEMMA.
If k is a complete nonarchimedean field, then k is
Hensel ian.
PllOOF. we
can
Let f ,,,,,"',It,A,B,e,d ,L,M,E be an Henselian context. construct
sequences
{"'i}'
{"'i}'
{h i
l
and
{Ci
From (6.6)
l
so
f ''''i ,,pi ,h i ,A,B,C i ,d,L,M,E is a Henselian context, "'0 = "', ,pO = ,p, h O Co
= C and
that
= h,
Chapter XII. Valuation theory
312 (i) Ih i +1 1 ~ tlhil, (ii)
has the same degree and leading coefficient as
~i
~,
(Ei) I~i+l - ~il 5. sMlh/dl ~ s-Mtld~l, (iv) l,pi+l -,pi I ~ sMlh/d~1 ~ s-Mtldl. From (i), (iii) and (iv) we see h i Cauchy.
,pi ->~, and Ci are
-> 0,
and that the other sequences are
As all degrees are bounded, and k is complete, we have '!'i
equations
->
C,
with ;,~,C
and weak
E
k[X 1.
inequalities
conclusion easily follows.
->;,
As all the conditions on ~,~ and satisfied by
~i,,pi
and Ci'
C
the
0
Note that i f k is a discrete field, then the algebraic closure of k in A
k is Henselian by (VII.lo6), and discrete by (VI.lo9).
ample supply of discrete Henselian fields:
Thus we have an
for example,
the algebraic
closure of the rational numbers in the p-adic numbers. 6.9 CDROLLARY.
Let f(X)
on Hensel ion field k.
= ar,X r' + ... + alX + aO
be 0 polynomiol ovel'
Suppose there i sn< r such tlw t an
-je
0 ond
I (olj/onXn)2(2(r'-n)+1)omxml < sup {loiXil : 0 ~ i ~ n} whenever j
~
n < m Sr.
Then f hos 0 foc tor of degr'ee n in k [X 1•
We set up an Henselian context. Let '!' = 0nXn + ••• + 00 and n +1 = f - .~/. = 0 Xl' + ••• + 0 lx ~~ r n + · Let A = C = 0 and B = d = 1 rand M = 2(r~l) + 1. The hypothesis says that there is t < 1 so
PROOF.
,p = 1, so h and L
=
that s 2M lhl S cl'!'l, so F,,!,,,p,h,A,B,C,d,L,M,t is an Henselian context. k is Henselian, we get the desired factor. 6.10 CDROLLARY.
Hense U on fi eld k.
Let F
=
AS
0
0rXr' + ••• + 0lX + 00 be 0 polynomiot over' on
lf
o < < r}, then f is ,'educible over k. PROOF.
By (6.1) we can construct a finite subset A of {O, ... ,r} so
that z-(m+t)
if
E A
and i f i Il A.
Furthermore we can choose m and t so that N Then
= 4r + 2 < zm and 2- t < IFI.
313
6. Hensel's lemma ( If I - T(mH))N > (I f I _ Tl IN)N > IriN - 2- t IrI N- 1 so If I (mH) ] NI' I -t I I [ IF 1-2 (f - 2 ) < f . We can also require
lar_xr-I
biggest integer in A.
+ T(m+t)
As Irl =
< I f I, so ['
Let
q A.
be the
n
iloiX' I : 0 < l < r-) we have
SlIp
Tl
> O.
The hypotheses of Corollary 6.9 are met, so f has a factor of degree n in I<[X].
0
6.11 THEOREM.
Let k be a discr-ete lIensd ian fieLd wUh a
nontr',vial VOluoUon, and f 0 sepcwable pO!.YflOmüll in h[X].
o >0
If
if
such tim t
-
I
<
Ö,
ond de 9 f ~ de 9
t
tri.vi.aL or
Then ther'e is
hen f has a f oe t 01'
in h[X] of the same degree os <po
The trivial case is trivial so we may assume that there exists
PROOF.
such that 0 < le I < 1.
e E "
For any nonzero polynomial
be the value of the leading monomial in p.
L (p)
in k [X 1, let
p
As the map taking the
polynomial F(X) to F{Xle m ) is an automorphism of h[X] that is uniformly bicontinuous on polynomials of bounded degree, we may assume that Ir I = Using the Euclidean algorithm, and multiplying by an appropriate
L(f).
power of
er
we can find
and b in k[X] so that lai ~ 1 and Ibl S 1 and
0
of
aF + bF' = cI E h.
with Idl ~ 1, and lori< lel 2 and Ibfl < 1e121xl. Ö =
<
Now suppose IF -
I
IdI2inf(IXI,lfl)
0, and deg f = deg
= I
Set h = f -
If I
Ib'l'l
I].; I <
< Ibfl/l,,1
2
<
l
< lxi.
'I",p -
Eldl
2
so setting
'1'''''
I
s
2 loh +
E
,
If II Ie 1
As 0
< Irl
that i s, s
=
we
= L{f)
1 in (6.4).
is
-
t <1
In'l' I < 10 F 1I Ie 12 < 1 so
that
111/
~
and
reo
<;
Note that i f F (HX], then Ir' I ~ Irl/lxl, so
Ici 12 and Ih'l" I < 1.
b(f'
we may assume that le 12 < 1>].;1 < 1, Then
2 •
There
Eld2Iinf(lxl.rll ~ tld 2
= L(
By multiplying >].; by an appropriate power of e, and
~.
multiplying
Let
'1">].; - ,p'p'
Thus
11 = Id -
[(0'1'
+ b'l")>].; + b>].;'
314
Chapter XII. Valuation theory A
= b-p'
= a",
B
+
, b",' ,
C = A", + B-p - d, L =max(deg F, deg h, n),
and taking M as large as necessary (s = 1), we get an Henselian context. As k is Henselian, the desired factor can be constructed.
0
Is this theorem true without the hypothesis that the valuation on k is trivial or nontrivial? Recall
that i f k
is a discrete valued field,
Henselization lZ of k terminology
is
then we define the
to be the separable closure of k
justified by the
fact
that a
discrete
in
k.
This
field with a
nonarchimedean va1uation is Hense1ian exactly when it is separably closed in its completion. 6.12 LEMMA.
Let
"'(X)
= anX n + •••
+ aO oe a
poLynomiaL
ove'-
a
Suppose lan I ;I! 0 and set s = I", I/Ianxn I. Let C E k[X] and d E k be such that sMlcl < Idl and deg C ~ M. Then '" and
nonarchimedean vaLued FieLd k. d + C are reLati veLy prime. PROOF.
Let S = So + SlX + ••• + SrX'- be a common factor of '" and
Letd+C=Sawherea=00+a1X+··· +atX t •
d+C.
AB lei<
Idl we
have Isoool = Idl = Id + cl = Isllol, so Isl = Isol and we may assume that So = 1. Suppose, by way of contradiction, that ,- > 0 and sr f. O. Because S is a factor of "', and I",;sl = 1",1, we have Isrx'-I l Thus 10txtl ~ slcl. sl+llel.
Thus
induction on
I,
I
vaLuation.
+ 1 > M l deg e whence a l
getting a contradiction when
6.13 THEDREM.
lanxnl/I",I = l/s.
Similarly IOt_1xt-11 ~ s21cl and so on until 1001 ~
Let
k
be
a
discf-ete
t
= 0 and we are done by = O.
FieLd
0
wUh
a
nona,-chimedean
Then k is Hensel ian i Fand only iF k is separably closed in
fts compLetion. PROOF.
Suppose k is separab1y closed in its completion k.
Since I, is
Henselian we can construct the required ~ and ~ with coefficients in /,. The leading coefficient of ~, and hence the leading coefficient of -P, is in k.
;j; are algebraic over IL ;j; are relatively prime, hence strongly
By (VI. 1. 6) all the coefficients of ~ and
Lemma 6.12
says
that '"
and
relatively prime as their coefficients lie in a discrete subfield.
As k
315
6. Hensel's lemma
1S separably closed in
k,
the coefficients of
'P
and ~ in fact lie in k by
(VII. 1. 7). Now suppose k is Henselian and e in h is separable over h. satisfies a separable polynomial f in I
If deg f = 1, then e E k and
Otherwise we can find a in k so that 1{(a)1 is as small as So ei ther f
we please.
has a root in 1<, and we get a polynomial of
smaller degree satisfied by e, or the valuation on k is nontrivial. f (a)
=
f -
But
(X - a)>J; for same >J; in Iz[X], so by Theorem 6.11 we can find a
linear factor of f satisfied bye. As
e
Then
remarked
in k [X], and so get a polynomial of smaller degree
0
earlier,
this
theorem
shows
that
the
definition
Henselian for discrete fields is independent of the choice of
of
lxi.
EXERCISES 1. Let fE&: [X J and let f be the image of f in &:p IX 1.
Show that i f
f has a simple root in J'p' then there exists x in the p-adic completion of
msuch
2. Hensel' s le1lllla.
that f(x)
Let k
=
O.
be an I-Jenselian field and f
a monic
polynomial in h[X 1 all of whose coefficients have value at most If the
1.
image f
of f
in k IX]
is a product of strongly
relatively prime monie polynomials A and of monie polynomials
'P
and t such that
~ =
~,
then f is a product
A and
~ = ~.
7. EXTFNSIOOS OF VALUATlOOS
We turn to the question of extending valuations from k to 1«8) where 8 is algebraic over k. when 1«8)
In general we must restriet ourselves to the case
is finite dimensional over k,
irreducible polynomial (VI.l.13).
that is, when 8 satisfies an
The following example illustrates this
for both the archimedean and the nonarehimedean eases. EXAMPLE.
Equip the field of rational nurnbers
value or the 7-adic valuation, and let eompletion
of~.
/2
be a fixed root of 2 in the
Let a be a binary sequence with at most one 1.
The
Chapter XII. Valuation theory
316
is diserete, eountab1e, and loeally preeompaet.
DeHne a Held E
h(8)
with 8 algebraie over k by
E
=2 am = 1
where 8 2 or if
(s + tS : s,t E k)
and equality is defined by setting s + t8 and s
=
(-l)m t h.
=
0 if s
Note that E is a diserete Held.
=
t
=
0,
But i f
we eould extend the valuation to E, then in E either 18 - ,/21 "I- 0 or 18 + hl "I- 0, so either a m giving us LLPO. 0 7.1 LEMl'IA.
=
0 for all odd m, or a m
=
0 for all even m,
Let K be a nonarchimedean vatued Fidd and k a discrete
lIensetian subFieLd oF K on which the uatuation is tI-iviaL or- nontriuiaL. Let 8 E K satisFy aseparabte ir-reducibte poLynomiat oF degree n over- k. Then ther-e is t iess than n.
> 0 such that Ig(8)1
~ tlgl wheneuer 9 E k[X) is oF degr-ee
Thus the dements 1,S, ... ,Sn-1 Form a metr-ic basis Far- 1«8)
ouer- k. t1 ?: ••• ?: t n _1 > 0 so that go + glX + ••• + gm xm . As Igl ~
We eonstruet a sequenee 1 = "0
PROOF.
Ig(8)1 ?: tmlgl i f deg 9 sup(l,lxln) sup{lgil
=
m.
Let
g(X) =
~
: 0 ~ i ~ m}, we mayassume that lxi = 1.
We may
also assume that 9 is monie, so Igl ?: 1. Let F be the minimal polynomial of 8 and suppose we have eonstrueted tm-I'
Set Jl
=
sup(1,18I n ) and set tm =
where m.
Ö ~
ö(E m_1/2Jl
)n-m+2IF 1-1 Write F
1 is gotten from Theorem 6.11.
qg + r where deg r- <
By (6.3) we have Ir-I ~ Igl n -m+1IFI so
IF 11 gln-m
Iq I = IF - ,-1/ I9 I ~ whenee
Either
Ig I
> Jl/t m_1
or
Ig I
< 2Jl/"m-l'
If
Igl
> Jl/tm_l ?: 1, then, by
induetion,
As the valuation is nonarehimedean we have
Ig(8)1 Otherwise Igl
=
Ig(8) - 8m l ?: tm_1lg1 ~ tmlgl.
< 2Jl/tm_1 and Ig(8)1
=
jr(8)1/lq(S)1 ?: tm_1Ir-I/lfllgln-mJl
7. Extensions of valuations since deg r
< m.
317
As f is irreducible, Theorem 6.11 says Irl Z 0 so
Em_11r1/Ifllgln-m/1 > Em_1olgl/lfl(4L/Em_1)n-m+1/-l
= 2Emlgl > Emlgl.
0
The hypothesis that the valuation is trivial or nontrivial in this lemma was needed in order to apply (6.11).
Is the lemma true without that
hypothesis? 7.2 LEMMA. dimensionaL
Let
be a discr-ete HenseUan
k
fieLd
Then any
separabLe extension of k.
and K a
finite-
two uaLuations
on K
extending the vaLuation on k. are equaL. Let K
PRODF.
= k(O) where
0 is separable (VI.4.3) of degree n and let
f be the minimal polynomial of O.
Let I I, and I 12 be valuations on K
extending the one on k, and suppose Ib I, > Ib 12 for some b in K. f(O)
= 0,
if we take lxi
be written uniquely as a
= 1,
then loli ~ If(X)I.
= ga(O) where
9 E '~[Xl
As
Each element a in K can has degree less than n.
If the valuation on h is trivial or nontrivial, then by Lemma 7.1, for some E > 0 we have If(X)l n lga (x)12 lali Z Elga(x)l. But there is m such that Ibml, So if Ibl, nontrivialj
> Ibl 2 then it is impossible that the valuation be trivial or but
a
nontrivial, so Ibl, in K.
> (lf(X)ln/E )lbmI2 , which is impossible.
va1uation
is
trivial
if
and only
> Ibl 2 1 is impossible, whence Ibl,
if
it
is
~ Ibl 2 for each b
0
Let E be a finite-dimensional extension of a discrete field k.. field
not
IlOrm
NE,Ih fraa E to k
The
is defined on elements x of E by setting
NEjk(x) equal to the determinant of the linear transformation of E induced by multiplication by x. 7.3 THEDREM.
Let
k
be a discr-ete Henselian fieLd and E a finite-
dimensional extension of k.
Then the valuation on hextends uniquely to
E, and Vi!;; Vk for some positiue integer- P.
Thus if the value grouIJ of I~
is discrete, so is the ualue grollp of E. PRODF.
Let F be the separable c10sure of I< in E.
Since Ern !;; F for
some positive integer rn (VI.4.3), any valuation on F extends uniquely to E so we may assume that E is separable.
Lemma 7.2 takes care of uniqueness.
To construct a valuation on E, for 0 in E set
318
Chapter XII. Valuation theory 181
=
IN(8) l1!n
where N is the field norm from E to k and n is the dimension of E over k. The on1y problem in showing that this defines a (general) valuation, is verifying that IN(l + 8) I ~ 1 if IN(8) I ~ 1. f(X)
= Xm +
be the minimal po1ynomia1 of 8.
have laol ~ INk.(8)/k(8) i, as sup lai f (X-I)
is
I >
I
~ 1.
Let
a m_1xm - 1 + ••• + aO Since N(8) is apower of INk.(8)/k(8)1, we
By Coro11ary 6.10 we have lail ~ 1 for all
1 implies f
is reducib1e, which is impossible.
the minimal polynomial of 1 + 8
and
its constant
±(1 + 2 ±ai ). As the valuation is nonarchimedean 11 + 2 ±ai I ~ 1. N(l + 8) is apower of this term, we have IN(l + 8)1 ~ 1. Let
e
=
mn.
But
term is Since
0
7.4 THEDREM.
Let I< be a discrete HenseUan field wUh a
tr-i.viat or"
nontrivial valuatton, and E a finite dimensional separ"abte extension of IL Then Eis Hense t ian .
PROOF.
By Theorem 6.13 it suffices to show that E is separably c10sed
in i ts completion.
Let E
=
h (8) .
As k is separably c10sed in
fol10ws by (VIII.2.2) that E is separab1y c10sed in k(8).
h,
it
Lemma 7.1 shows
that the norm on k(8) given by the va1uation is eguivalent to the supremum
k (8) = E.
norm on "n, so By
Coro1lary
3.5,
0
the
Henselization
of
a
discrete
pseudofactorial valuation is separably factorial.
field
with
a
Thus there are plenty
of situations in which the following theorem app1ies. 7.5 THEDREM.
Let
k be a nontt"ivial discr"ete vaLued
either reat, complex, or" nor1llrchimedean.
u denote
Let
of k, and let E = k(8) be a finite-dimensional
If
the
minimal
irTeducible
poLynomial
factors
in
of
ulX],
8
over"
then
"
tllere
11 1 ,.",1 Is on E, extending the valuation the HenseUzation of E under" I li' then
=
PROOF.
Let a be a reot of gi
is
the HenseUzation
p,"oduc t
isa
exist
g1g 2'" gs
distinct
of
valuations
I Ion k, such that if t i is ~
te i
•
Euery ualuat ion on E extending I I is equal to some
(iii) n i
tl1l1t
sepm"able extension of k.
(i) gi is the minimal polynomittl of 8 ouer' ), (ii)
fieLd
Iti:kl is the degree of gi' and 2 n i
I
li'
= n.
in a discrete extension field of
u
7. Extensions of valuations
319
(possible since gt is irreducible over R). By (7.3), in the nonarchimedean case, and by (5.2) and (5.8) in the archimedean case, we can extend I I uniquely to k (a) • As a satisfies the minimal polynomial of 8 over k, there is a monomorphism ~ : E ~ kral, that is the identity on k and takes 8 to a. For x E E, set lxii = I~(x) I. As kral is the Henselization of kral by (7.4) and (5.8), we can identify ~i with kral, so (i) and (iii) are clear. To prove (ii), suppose we have a valuation on E extending the valuation on k, and let ~ be the Henselization of E with respect to that valuation. Then there is i such that 9t(8) = 0, so the valuation on ~ must be I li by (7.2) and (5.2). We want to show that I li -;f. I Ij i f i -;f. j. As gigj is separable, we can find polynomials s and t such that s9i + tg j = 1. Then gi(8) = 0 in ~i and gi (8) -;f. 0 in ~ j" By choosing a sufficiently close approximation
g;
E
k[X)
to
9i'
we can make
Ig~(9) Ij bounded away from 0, so
Ig;(8) li I li
-;f.
as small as we wish and keep
I Ir
0
EXERCISES
1. Show that if 8 is in a field K with a nonarchimedean valuation, and fE K[X) has 9 as a root, then 181 ~ Ifl, where we take
lxi
=
1.
2. Find all archimedean valuations on the following fields (i) ~[X)/(X2 - 2) (ii) ~[X)/(X2 + 1) (iii)
~[X)/(X3
- 2)
Find all extensions of the 5-adic valuation on Cl to the same fields. 3. Let K be a finite-dimensional extension field of G:l. element x of K is integral over 7l if and only i f every nonarchimedean valuation on K.
Show that an
Ix I
5 1 for
8. e AND f
Let I< be a discrete field wi th a discrete valuation, and E a finite dimensional separable extension of I< with a valuation extending that of 1<. If the quotient VE/Vk of the value groups of E and k is finite, define the ramification index e = e (Ejk) to be the number of elements in VE/Vk .
320
Chapter XII. Valuation theory
Similarly i f the residue class field E is finite dimensional over k,
E over
define the residue class degree to be the dimension f = F(E/k) of
t:
As passing to the Henselizations h' !:
k.
leaves the value groups and
residue class fields unchanged, we may assume that Iz is Henselian. The classical method for constructing e is to choose Tr in k such that
ITrI < 1 gene rates the value group
Vh of k.
Then (Theorem 7.3) VE is a subgroup of the cyclic group generated by ITr 11/[1, hence is cyclic and contains Vk as a subgroup of finite index e.
As we cannot establish that
a nontrivial subgroup of a cyclic group is cyclic, our construction of e The problem with computing F is that finitely
must be more elaborate.
generated field extensions need not be finite dimensional.
We need to
impose Seidenberg's condition P on the residue class field. 8.1
Let
suppose
residue
the
Iz be a
Let
'l'HEX)RElI'l.
uaLuation.
E
be
an
discr'ete
HenseUan
cLass
(feLd h
field
separ'abte
n-aimensiorwl
WUll a
extension
satisfies Seidenberg's
diso'ete
of
h,
and
condUion P.
Then the quotient gr'oup VE/Vk Iws fini te cardinat i ty e, the Field E Iws finite dimension f ouer k, and n
= eF.
In par'ticutar, the ualua t ion on E
is discrete, and E satisFies condUion P.
By Theorem 7.3 the group VE is discrete so E is discrete. Also VE/V k is discrete because Vk is cyclic and !: k • We shall construct a finite subset S of E that maps one-to-one onto
PROOF.
VE/Vk'
vi v
and a finite subset W of {w
onto a basis for for E over
I~.
E
E : Iwl
E over k, so that sw =
=
I} that maps one-to-one
{sw : sES and w E W} is a basis
We build these sets up inductively starting with S
= W=
{I} .
At each stage of our construction we will have a finite set S of nonzero elements of E containing 1 and mapping one-to-one into VE/Vk' and a
finite
subset W of
independent family
{w E E
Win E.
that maps
Iwl = I}
to a
k-linearly
Moreover we will require that kW, the k-vector
subspace of E spanned by W, be a field. If b w E k
Ibw.1 = max among W.
fm: Ibwl,
w E W,
then
12wEW bww I
then, dividing by bw"
=
max Ibw 1 for
if
12: bww 1 <
we get a dependence relation
So if a sw E k, for sES and w E W, then l2:sxwas1Os101
=
maxW l2:s aswsl
=
maxSxW las10sl
as the values of the nonzero asws are distinct for fixed w.
So SW is
321
8. e and f linearly independent over k. spanned by SW.
x of
the image
Let I<SW be the k-vector subspace of E
la5w51 = 1,
If
E is
x in
then
5 = 1,
so if
xE
kSW and
= kSW, then we are done as S maps onto
If E
lxi
~ 1, then
in hW.
Otherwise there is a in E'\j<SW.
VE/Vh
and E
= hW.
As h is Henselian, E has ametrie basis
over h by Theorem 7.1, so {al U SW is metrically independent, whence a is bounded away from kSW by some positive distance.
Let
~
in h be such that
I~I
< 1 generates Vk . We may assume that if s E S\{l}, then
1.
Let r be the maximum of the values of elements of S\{l}, with r = I~I
if S = {1}. (ii)
la - b' I ~ r-Ia - bl. lol/Isl is not in Vh for any 5 in
such that that
(iii) A field extension K !;;; E of
I~
S.
of dimension m > 1 such that K is
k
(so VK = Vk ). (iv) A proper finite-dimensional field extension of hW in m-dimensional over
As VE/Vh is discrete, either ß k and s in S such that e
shall show that either see what to do i f
e = uso le - III le - III ~ lei = 1.
< Isl <
Given b in kSW, we shall construct one of the following:
b' E I<SW ß E E so
(i)
I~I
=
eE
e E kW.
=a
E.
- b satisfies (ii) or we can find
(a - b)/st has value 1.
hW or (iii) holds or (iv) holds.
If e E kW,
in
t
In the latter case we First let's
then there is u in kW such that
< 1. Either le - ul ~ r- or 0 = e - u satisfies (ii). If b' = b + 5tU, and (i) holds since a - b = 5te and
r we set
We return to the problem of whether
e
E kW.
As Vk is discrete, and
is algebraic over k, we can divide the minimal polynomial of largest coefficient to show that
e is algebraic over k.
e by
e
its
By (VI.6.3) there
is a manie polynomial 9 with coefficients in k, each of value at most 1, so that
9
is separable and
characteristic of k. such that g(w)
=
g(eq )
=
0 where q is 1 or apower of the finite
Since E is Henselian (Theorem 7.4), there is
0 and
w=
eq.
w
in E
The minimal polynomial of w, which exists
as E is finite dimensional over k, divides 9 and, by Gauss's lemma, all its coefficients have value at most 1.
Thus we may assume that 9 is
As E is Henselian and 9 is separable, 9 is irreducible. k(w) is finite dimensional over k. If deg 9 > 1, set K = I, «,l) and we get (iii). If deg 9 = 1, then e is purely inseparable over k, and hence
irreducible.
Thus
over
',W.
over kW.
By condition P we then have that kW(e) is finite dimensional rf this dimension is 1, then e E kW.
Otherwise I~W(e) is the
322
Chapter XII. Valuation theory
field needed in (iv). If (ii) occurs we can increase Si if (iv) occurs we can increase W.
As
a is bounded away from 18W, situation (i) can occur only finitely many times.
Finally, suppose (iii) occurs.
(Theorem 7.4).
Then VK
=
VI< and K is Henselian
Moreover K satisfies condi ti on P since
dimensional (VII.3.1.ii).
We are done by induction on n.
is finite
K;1~
0
The requi rement in Theorem 8.1 that k satisfy condi ti on P cannot be removed, at least not entirely.
To show this we first prove the following
theorem. 8.2 'l'IIOOREM.
Le t
K be a
di scr-e te
fi eid of
chDTacter-i s t ic
p.
Then
ther-e exists a discrete Hensetian fieid k wUh a discr-ete valuation and residue class fieid k
=K
such tlwt
dimensi01wl separable extension E
or
for each y k wi th y €
in
I~
there is a
fini te-
EP.
If h is a field of characteristic P, then by h l / p we mean a
PROOF.
field containing I~ such that (I> ljp)p = 1<.
That such a field exists is
obvious upon contemplating the isomorphism k ~ I~P .;;; I>. Henselization valuation.
of
the
If y
=
0,
rational let E
function = IL
field K(T)
If y t- 0,
Let h be the
under
the T-adic
consider the
separable
polynomial f (X) = xP + TX + y in IdX 1. That f (X) is irreducible (over h l / p ) is easily seen by substituting X = Z - A, where A is the pth root of y in h l / p , and applying Eisenstein's criterion.
E = h(e) where f(e) = O.
Clearly lei
Thus we can construct
= Iyll/p = 1 and y = 9P.
0
A discrete field h of characteristic p satisfies condi tion P i f and only if KP is detachable from K for any finite-dimensional extension field K of
k
(VII.3.l.ii), that is, for each y in K either y
Thus the following corollary shows that the
€ KP
or y ~ KP.
k in Theorem 8.1 must at least
satisfy a weak form of condition P. 8.300ROLLI\RY.
Let h be a discr-ete Fieid.
Suppose tImt whenever-
I~
is
a discrete Henselian Field wUh residue dass field h, and E is a finitedimensi01ml separable extension field of h, that the r-esidlle dass field E is finite dimensiollQl ouer- h. PROOF.
y
E
EP.
over
kP.
Let y
€
Then h"P is detachable fr-am I<.
hand apply Theorem 8.2 to construct h and E with
Then E is finite dimensional over h so EP is finite dimensional Thus y E ~ is either in kP or it is not. 0
323
8. e and (
The requirement in Theorem 8.1 that the extension be separable cannot be removed. Referring to Example 3.6 let k = ~2(y2,ry2) and E = ~2(y2,ry) with the valuation inherited from the field of formal power series in Y. Then the value group of E equals the value group of k i f and only if = 0 for all n. Henselizing doesn't change the situation. The p-adic valuations on (ij satisfy the hypotheses of the following theorem.
an
8.4 'l'HEDRE2'f.
Let
I~
be a discr-ete {ieLd wUh a valuation such timt
(i) The valuation is discrete, (ii) The Henselization
is separably factorial
(for- example,
if k
is
pseudofactorial),
(iii) The residue class field satisfies condition P. If
E
is
a
finite
separ-able
extension
of
k,
then
any
valuation
on
E
extending the one on k satisfies (i), (ii) and (iii).
Let! be the Henselization of E, and R the Henselization of k. We may assume that R ~!. Choose a in E such that E = k(a). As R is separably factorial, R(a)~ is finite dimensional so R(a) is Henselian by PROOF.
Theorem 7.4. Hence R(a) =!. Thus! is separably factorial by (VII.2.3), so (ii) holds. As the value groups and residue class fields of E and ! are the same, (i) and (iii) hold by Theorem 8.1. 8 . 5 'l'HEDRE2'f.
Let
hypotheses of Theorem of k.
be
I~
8.4.
a
field
wuh
a
0
valuation
the
satisfying
Le t E be an n-dimensional separ-able extension
Then there are a fini te number s of valuat ions on E extending the
valuation on k. valuations.
Let El, •.. ,E s denote
the
field E equipped wUh
these
Tllen e(E Ul1z) and f(EUlk) are defined and
n
2 e (Ei/lldf (Ee'k).
PROOF. Theorem 7.5 constructs the finite number of valuations on E. We have e(!i~)f(!i~) = n i = [!i :Rl by Theorem 8.1, and n = 2 n i by Theorem 7.5. But Henselization does not change the value group or the residue class field. 0
EXERCISES 1. Calculate e and f for each extension of the 5-adic valuation on to (ij[X]/(X 3 - 2). Verify Theorem 8.5 for this case.
(ij
Chapter XII. Valuation theory
324 NOTES
For a more complete treatement of the constructi ve theory of real numbers and metric spaces the reader is referred to [Bishop 1967) and [Bishop-Bridges 1985). The theory of valuations plays a central role in the development of the theory of fields from a constructive point of view. The study of fields with valuations provides an interesting mix of discrete, purely algebraic constructions combined with analytic constructions such as the completion. Although it is commonly feIt that algebraic number theory is essentially constructive in its classical form, even those authors who pay particular attention to the constructive aspects of the theory employ highly nonconstructive techniques which nullify their efforts. In [Borevich-Schafarevich 1966), for example, it is assumed that every polynomial can be factored into a product of irreducible polynomials (every field is factorial) and that given a nonempty subset of the positive integers you can find its least element. The definition of a general rank-one valuation on a field was adopted by Staples (1971) in his constructive treatment of valuations, and the proof of Theorem 1.6 is in that paper. The problem here is that we cannot decide whether ux S vy or vy S ux, or whether ei ther is zero. Our proof of Theorem 1.1 is taken from [Weiss 1963, Proposition 1-1-8). The last two sentences of our proof of Theorem 1.2 follow [Weiss 1963, Theorem 1-1-4) The notion of a pseudofactorial field is a purely constructive one having no classical counterpart (all fields wi th valuations are pseudofactorial from a classical point of view). However it provides just the information we need either to construct a root of a polynomial, or bound the polynomial away from zero, in the completion of the field (Corollary 3.2). The proof of Corollary 3.3 using winding number is from [Brouwer-de Loor 1924). Theorem 3.4 shows that there are lots of pseudofactorial fields. Corollary 3.5 shows how to factor separable polynomials over the algebraic or separable closure of a pseudofactorial fie1d in its completion. So, for example, we can factor polynomials over the algebraic p-adic numbers. Our proof of the Gelfand-Tornheim theorem fo110ws [Artin 1967, page 24] The somewhat elaborate Hensel's lemma of Section 4 is a modification of
Notes
325
[Artin 1967]. The hypotheses of the leIlUl1a are relaxed slightly by not requiring the leading term of ~ to have the same value as ~ (not requiring s = 1), to cover situations where we cannot determine which coefficients of a polynomial have maximal value, and we must worry about some constructive peculiarities. The definition of HenseUan, when s = 1, is the conclusion of Hensel's leIlUl1a as given in [Artin 1967, Theorem 5]. The main theorem is Theorem 6.13 that a discrete fie1d with a nonarchimedean valuation satisfies the conclusion of Hensel's leIlUl1a precisely when it is separably closed in its completion. The construction of the valuation in (7.3) Theorem 14:1].
follows [O'Meara 1963,
Chapter XIII. Dedekind Domains
1. DEDEKIND SETS OF VALUATICfiS.
If S is a set of valuations on a Heyting field k, then we will denote a member of S by I I with a subseript. Instead of writing I Ip E S, we will often write pES. For eaeh prime number p we get a diserete valuation on the field ~ of rational numbers by defining Iplp=l/p, and Iqlp=l i f q is a prime different from p, and extending multiplieatively. This family of valuations forms a Dedekind set in the sense of the following definition. 1.1 DEFINITlOO. A nonempty diserete set S of nontrivial diserete valuations on a Heyting field k is a Dedekind set if (i) For eaeh x
E
I<
there is a finite subset T of S so that
Ixlp ~ 1 for eaeh p E SV. (ii) If q and q' are distinet valuations of S, and c > 0, then there exists x E k wi th Ix Ip ~ 1 for eaeh pES, such that Ix - llq < c and Ixlq. < c. inequivalent.
Henee distinet valuations are
Note that a nonempty detaehable subset of a Dedekind set is a Dedekind set. Let S be a Dedekind set of valuations on a Heyting field k. If pES, then, beeause p is nonarehimedean, the set R(p) = {x E k : Ixlp ~ l} is a ring, whieh is loeal as p is diserete.
We call R(p) the loeal ring at p.
The elements of the ring npESR (p) are called the integers at S.
A ring a
Dedekind daDain if it is the ring of integers at a Dedekind set of valuations on a Heyting field. 1.2 'lHEX>REl'I. a Heyting fieLd
The r'ing of integer-s nt " Dedeldnd set S of ll"luations on I~,
is a detachable subset of h.
PROOF. Let x E k. There is a finite subset T of S so that Ix Ip ~ 1 for eaeh pE S\T. Then x is an integer at S if and only if Ixlp ~ 1 for 326
1. Dedekind sets of valuations
327
each p in the finite set T of discrete va1uations. Given
inequiva1ent
va1uations
I
0
... , I In
11 ,
on
and
k,
elements
x1, ... ,xn in k, we will be interested in constructing an element of k that simultaneous1y approximates each xi with respect to the valuation I li.
As a first step, 1. 3 LEMMA.
we
~
In be inequiuaLent nontriuiat uatuations on
rhen there is x € k such that' lxiI> 1, and
If n
lxiI> 1.
= 1, then, as I 11 is nontrivial, there is x € k such that If n = 2, then since I 11 and I 12 are nontrivial
inequivalent valuations, there exists x
with lxiI< 1 and Ixl2 > 1.
€ k
For n > 2 we proceed by induction, so we may assume we have y that Iy 11 > 1, and Iy Ii < 1 for 2 $ i < n. n = 2,
Iz 11 > 1
with
Iyml i Izli < 1
for
2 $
and i
Iz In < 1,
< n.
Iyml n > 1 or Iyml n < Izl~'. the xm
lxii< 1
1.
PROOF.
of
11 , ... ,1
I
Le t
a Heyting fieLd k. for i
have the following lemma.
desired
element.
As
Choose z
and
choose
Izl~' > 1,
it
m
such
€ k
using the ca se
€ k,
so
large
fo11ows
that
that either
In the latter case, Iymzln < 1 so x = ymz is If,
on
the
other
hand,
Iyml n > 1,
let
= zym/(l+ym). Then the sequence Ixmli converges to Izli if i = 1 or n,
and converges to 0 otherwise.
Set x
=
Ixml1 > 1 and Ixmli < 1 for 2 ~ i ~ n.
xm where m is large enough so that 0
If we let k p denote the field k with the metric given by the valuation Ip ' then the next theorem says that the diagonal is dense in ITpkp • 1.4 'lmDREM (weak approximation).
Let
I 11 , ... , I
In be
inequiuatent
nontri uiat uatuat ions on a Heyting fie td k, te t xl' ••• ,xn be e tements of
> O.
k, and tet t
Then there exists x
such timt
€ k
Ix -
Xi
li <
t
for
each i.
PROOF.
For each
use Lemma 1.3 to construct Yi
IYili > 1, and IYilj < 1 if j ~ i.
For each m
t
€ k
such that
~ define
zim = y~xi/(l + Y~).
Then
li~lzim - Xi li
zm - L~=lzim· The
strong
approximations.
=
0,
and
approximation
=
li~~lzimlj
rf m is large enough, IZm theorem
Xi
li <
allows
E
0
if
for all to
j ~ i. i.
choose
Let
0
integer
Chapter XIII. Dedekind Domains
328
1.5 'lHEDREM (strong approximation).
Let T be a
Dedekind set S of uaLutions on a Heyting field Iz. pET Let x p E k.
Let
finite E
>
subset
of a
0, and for' each
Then ther'e exists y E k such that
(i) Iy - xplp <
E
for each pET,
(ii) IYl q ~ 1 for each q E S\T.
PROOF. We may assume E ~ 1. For eaeh pET, there are on1y finite1y many q E S with IXplq > 1. For eaeh p in T and q in S\T such that IXplq
> 1, add
q
to T and define x q
O.
=
Thus we may assume that
IXplq ~ 1 for eaeh pET and q E S\T.
Fix pET.
For eaeh q E T\{p}, eonstruet, by Definition l.1.ii, an element of k that is an integer at S, that is elose to 1 at p, and that is elose to 0 at q. Let Y p be the produet of these elements (the empty produet is 1). The elements Yp are e10se to 1 at p and elose to 0 at all valuations q in T\{p}. Finally let Y = 2pETY px p ' Then IY I q ~ 1 fot eaeh q E S\T, and Y is elose to xp at p. 0 The strong approximation theorem allows us to write elements of
k
as
quotients of integers. 1.6 'HIEDREI'I.
Let S be a Dedekind set of uaLuations on a Heyting held
k, and let R be the set of integer's at S.
Then each element of k is a
quotient of elements of R, and R is integral ly closed in k.
PROOF.
Let x E k; we shall write x as a quotient of elements of R.
By
Definition 1.1.i there is a finite subset T of S so that Ixlq ~ 1 for eaeh q E S\T. If Ix Ip ~ 1 for each p in the fini te set T, then x ER, so we may assume that x ~ O. Ag eaeh valuation in the finite set T is diserete, we may assume that Ixl p > 1 for eaeh pET. By Theorem 1. 5 there is y Iy - x-'I p
for eaeh p in T, and I y Iq
~
E k
with
< min{lx-'l p
: pET}
1 for eaeh q E S\T.
<1 As Iy
- x -, Ip
<
Ix -, Ip
for eaeh pET, and I Ip is nonarchimedean, it follows that IYlp = Ix-'l p < 1 for each p in T. In particular, y E R. If pET, then IxYlp = IxlplYlp = Ixlplx-'Ip = 1, and if q E S\T, then IxYl q <; 1. Therefore xy E R, so x = {xy)y-' is a quotient of elements of R. As each valuation in S is nonarchlmedean, it fo1lows that any element of k, satisfying a monic polynomial over R, is in R. 0
329
1. Dedekind sets of valuations EXERCISES
1. Let S the set of all p-adic valuations on~. Show that the ring of integers at S is l. What is the ring of integers if S {3}?
2. verify that if I I is a discrete valuation, then {x : lxi
~
I} is
a local ring. 3. Complete
~
in the p-adic valuation to show that a Dedekind domain
need not be discrete. 4. Chinese remainder theorem. Use the strong approximation theorem to prove that if a, and a2 in ~ are relatively prime, and b, and b2 are in l, then there exists x in l congruent to b i module a i
for i = 1,2. 2. IDFAL THEX>RY.
Throughout this section k will denote a Heyting field, S a Dedekind set of valuations on k, and R the associated Dedekind domain. A fractional ideal is a nonzero R-submodule A of h such that IA Ip = IIl1lx{ Ix Ip : x E A} exists for each pES, and is 1 for all but finitely many pES. That is, for each pES, there exists x(p) E A such that Iy Ip ~ Ix(p) Ip for each y E A, and Ix(p)1 = 1 for p outside of a finite subset of S. 2.1
'1'HFDRE2{.
Any nonzero
FinUdy generated R-submodule
of k
is a
fractional ideal.
PROOF. Let A be a nonzero R-submodule of k generated by a1' •.• ,an' As is a Heyting field, one of the a i is nonzero, and if a i ~ 0, then either a j ~ 0 or a i + a j ~ 0 i thus we may assume that the a i are all nonzero . h
Suppose a = '27=lriai is an element of A, and pES. As I Ip is nonarchimedean, lal p = 1'27=lriai Ip ~ IIl1lX Iria i Ip ~ max lai Ip . As each I Ip is discrete, we have IA Ip = maxI lai Ip : i = 1, ... ,n) exists. We can choose a finite subset T of S so that lai Ip Then IAI p = 1 if pE S\T. 0
=
1 for p E S\T and i
=
1, ... ,n.
We shall show later that, conversely, every nonzero fractional ideal is generated by two elements as an R-module. Let A and B be fractional ideals. Define the sum of A and B by A + B = {a + b a E A and bEB), and the product of A and B to be set AB of finite sums of elements of the form ab, with a E A and bEB.
330
Chapter XIII. Dedekind Domains 2.2 THIOClREM.
Let A and B be fr'ac ti anaL ideaLs.
Then A + B and AB and
A n B ar'e Fr'actionaL ideaLs such timt (i) IA + Bl p ~ max (IAlp,IBl p ) (ii) IABl p = IAlplBl p (iii) IA n Bl p = min( IAlp,lBlpl for each pES. PROOF. Choose a E A and bEB so that IAl p = lal p and IBl p = Iblp' Then Ixlp ~ la + bl p ~ max (IAlp,IBlpl for each xE A + B, so IA + Bl p = la + bl p exists and (i) holds. If IAl p = IBl p = 1, then clearly
= 1, so A + B is a fractional ideal. For x E AB we have Ixlp ~ labl = IAlplBl p ' because I Ip is multipli-
IA + Bl p
cative and nonarchimedean, so (iil holds and AB is a fractional ideal. If xE An B,
then clearly Ixlp ~ min(IAlp,IBlpl,
construct an element x of A n B such that Ixlp
so it suffices to
= min(IAlp,IBl p )'
As I Ip is discrete, the values IAl p and IBl p are comparable, so we may assume that IAl p ~ IBl p ' Now T
=
< 1}
{pI U {q ES: la/bl q
is finite, so, by Theorem 1.5, there exists y E k with IYlp and IYl q ~ mirl(l,la/blq) for all q E S.
=
la/bl p ~ 1,
The 1atter implies that yb/a E R,
so yb E A, and also that y E R, so yb E B.
But min(IAlp,IBl p )
lal p = IYblp' so x = yb is the desired element.
= IAl p =
0
A fractional ideal A is completely determined by its values IAl p ' 2.3 THIOClREM.
Let A ruw B be fractionaL ideaLs.
Then
(i) A = {x E k : Ixlp ~ IAl p for each pES}. (ii) A ~ B if und onLy if IAl p ~ IBl p for euch pES. PROOF.
To prove (il let xE /< be such that Ixlp " IAl p for each pES.
Choose a nonzero element z E A and consider xz- 1 a
finite
(xz-
1
)z
set,
E A, or
and xz-
1
•
As Ixz- 1 Ip " 1 outside xz- 1 E R, whence x
I Ip is discrete, either t- 0, whence x t- O. Thus we may assume that x t- O.
Replacing A by x-lA, it suffices to show that 1 E A whenever A is a fractional ideal such that IAl p 2 1 for each pES. Replacing A by Rn A, and using (2.2l, this is the same as showing that 1 E A whenever A is a fractional ideal contained in R such IAl p Let z be a nonzero element of A.
=
1 for each pES.
By Definition 1.1.i there is a finite
2. Ideal theory
331
subset T of S so that Izlq
=
Iz - 1 Iq ~ 1 for
1 for each q E S\T.
If
1,
=
E S\T.
q
As z E A ~ R we have
1 for each pET, then
Z-1
E R, so
1 E A.
Therefore, as T is finite and each I Ip is discrete, we may assume that l' is nonernptyand that IzlI' < 1 for each I' E T.
For each pET choose x p E A with Ix p Ip = 1. As Iz Iq = 1 outside of a finite set, Theorem 1. 5 says that there exists Yp E k such that I,)p - x~'lp ~ Izlp
< 1, and
IYpl q ::; Izlq ~ 1
for
each q E S\{p}.
IX~'lp ~ 1, and I Ip is nonarchimedean, it follows that ')p E R.
As
If q E T,
then 11 - 2pETxpYplq ~
1(1 - xqYql - 2pET\{r!lx p Yp l q max(11 - xqYqlqlmax!IYplq : pE T\{q}))
~
max(IXq(X~' - Yqllq,lzlql
Izlq.
and i f q E 8\1', then
11 -
2pETxpYplq
Therefore (1 - 2pcrXpYp)z'" so 1
E
S 1
=
Izlq.
E.R, so 1 - 2pETXp!ip C zR
~
A.
But xpY p E A,
A.
statement (ii) follows immediately from (i).
If A !: B
2.4 COROLLIIRY. subset of B.
PROOF.
Cln'
0
fr'aeti.mwl ideaLs, then A !s a detaehabl.e
Henee discrete Dede1dnd domains have detaclwble ideals.
If x E B, then (2.3) says that x E A i f and only i f Ixl p ~ IAl p
There is a finite subset T of S such that IA Ip = IB Ip = 1 Then x E A if and only if Ixl p ~ IAl p for each pET, which
for each pES. for I' E T\8. is decidable.
Take B
=R
to establish the second claim.
0
We can now show that fractional ideals are finitely generated. 2.5
I Ipl
'lHEX)EEM.
For- euch pES let d p be an element oF (he value group of
such thnt d p = 1 outside af u fin! te slLbset of S.
Let A
{x C k :
Ixlp ~ d p for each pes}. Then A -je 0, and if Y is any nonzer'o element of A, t./ler'e is nonzer-o x C A such that A is generated by x [md y as an R-module. PROOF.
Morever,
IAl p
=
dp
for' eacll P (
S.
By Theorem 1.5, there is a nonzero element
fini te subset of S so that IY Ip there exists x E k such that
= el p
u
(A.
1 for each p E 8\1'.
Let T be a
By Theorem 1.5
332
Chapter XIII. Dedekind Domains (i) Ixlp = dp for each pET (ii) Ixlq ~ 1 for each q E S\T.
Let B be the submodu1e of A generated by x and y. IBl p
=
so IAl p
= max(
Ixlp' IYlp)
=
Then
d p for each pES
IBl p for each pES, whence A = B by Theorem 2.3.
Let I be the set of functions a : S
~
7L such that a(p)
0
= 0 outside a
finite set (depending on a). Then I is the free abe1ian group on the set S. Let 9 p be the generator of the va1ue group of I Ip that is less than 1.
Let
map the monoid of fractional ideals under rnultiplication to T, Theorem 2.2.ii says that ~ is a homomorphism, Theorem 2.3 says that i t is one-to-one, and Theorem 2.4 says that it is onto. Moreover, if ~(A) t ~(B), then A t B as ~
by taking A to the function a such that IA Ip = 9~(P).
subsets of 1<; that is, either there exists an element x E A such that x t y for all y E B, or there exists an element y E B such that y t x for all x E A. The natural basis of the free group I on S consists of the functions 0p such that 0p(p) = 1, and 0p(q) = 0 for q t p. The corresponding fractional ideals Mp are a maximal ideals in R, for if x E R, and Ixlp < 1, then x E Mp ' while if Ixlp = 1, then the ideal A of R generated by x and Mp has IAl q = 1 for each q E S, so is equal to R. EXERCISES
1. Show that any fractional ideal is detachable from k. Construct a Brouwerian example of a finitely generated submodule of the p-adic completion of
~
that is not detachable.
2. Show that a discrete Dedekind domain is an integrally closed Lasker-Noether ring in which every nonzero proper prime ideal is maximal. 3. FINITE EXTFNSIOOS
We shall be interested in discrete valuations p on a discrete field /, that satisfy the c1assical1y trivial condition: (3.1)
The Henselization l
and the residue
3. Finite extensions Let
p be
333
a discrete valuation on k satisfying (3.1), and let E be a finite
separable extension of 1<.
Then XII. B. 4 and XII. B. 5 say that the each
valuation on E that extends
is discrete,
p
and that the set of such
valuations is finite. The p-adic valuations on
satisfy (3.1).
(Q
The fields of algebraic
nurnber theory are finite-dimensional extensions of
(Q,
and they acquire
Dedekind sets of valuations as folIows. 3 • 2 'J.'HOOREM.
Let
u discr'ete
be
k
dimensional
separabLe
extension
oF
1<,
und S u Dedel
field
on 1<, each of which satisfies
vatuations
(3.1).
und
S'
let
valuations on L that extend a valuation on S.
E be
Let be
the
Finite
a
of alt
set
Then S' is u Dedekind set
of vuluations. PRODF.
Let P and Q be in S'.
We first show that S' is discrete.
If P
and Q are extensions of distinct valuations in S, then they are distinct. So suppose that P and Q both extend pES.
By XII. 7 . 5 the set of
valuations of S' that extend p is finite, so ei ther P
= Q or P
f. Q.
That each valuation in S' is discrete follows from XII.8.4. Given x
Ixlp'
E E,
we must construct a finite subset T'
~ 1 for each P E S'\T'.
k,
then
f(X)
= Xn + uIxn-1 + ••• + an
by
VI.l.13
of S'
such that
As E is a finite dimensional extension of
there
exists
an
such that
E k[Xj
irreducible f(x)
polynomial
= O.
Let
T
finite subset of S such that lailp ~ 1 for each i and each p E S\T. be the finite subset of S'
T'
extend elements of T.
be a Let
consisting of those valuations on E that
If P E S'\T', then lailp ~ 1, so
Ixlp
~ 1 as P is
nonarchimedean. Finally we show that i f P and P' c > 0, then there exists
xE E
c, and
Ix IQ
and p'
in S respectively.
are distinct valuations in S', and
such that 11 -
~ 1 for each 0 E S' \{P ,P' ).
xlp
and
Ixlp'
are 1ess than
Let P and P' be extensions of p
If p f. p', then, since S is a Dedekind set,
there is x E k satisfying the required propertiesi so we may assume that p
=
p'.
We mayaIso assume that
valuations on E that extend p. that
lyl Q < c
for
lyl Q ~ 1 for each Q E
each
SV.
E
~
1.
Let S;J be the finite set of
By Theorem 1.4 there exists y E E such
and Iy - 11 p < Ei consequently, By the preceding paragraph, there is a finite
Q E S;J\{PJ,
set T' of S', disjoint from Sp' such that Iy 10 ~ 1 for each Q E S' \T' . Let T be the finite subset of S that have extensions in T'. Then p f. T,
334
Chapter XIII. Dedekind Domains
and,
by Theorem 1.5,
there is xE k
extending q E T, and 11 - x Ip Ixyl Q
=
< c,
and
so that
Ix Iq
Ixlq ~ lyI6 1
IxlQlylQ ~ 1 for each Q E S', and 11 - xylp
I
c, and Ixy p '
< E.
for each Q
~ 1 for each q E S\T.
=
Then
11 - Y + (l-x)yl p
<
0
Suppose R is a discrete Dedekind domain with fie1d of quotients k, and that E
is
a
finite
dimensional
separab1e
extension of k.
If
the
va1uations that determine R satisfy (3.1), then Theorem 3.2 shows how to eonstruct a Dedekind domain wi th fie1d of quotients E.
We now gi ve a
pure1y a1gebraic characterization of that Dedekind domain. 3.3 'l'HEDREM.
Let E be a
diserete fieLd k.
pES satisfies (3.1). those valuatians
Fini te dimensional
separable extension oF a
Let S be a Dedelünd set of valuntions on k so that eaeh Let S'
be
the Dedekind set
that extend a ualuation in S.
on E eansisting of
Then the Fottowing ar'e
equivalent For an element x E E. (i) x is an integer' at S' ,
(ii) x satisFies a manie irTeducible palynamial wUh eaeFFieients in
R,
(iii) x is integral aver R.
PROOF.
C1early (ii) implies (iii), while (iii) implies (i) because
eaeh valuation in S' is nonarehimedean and extends a valuation on S. prove that (i) implies (ii), suppose x E E is an integer at S'. finite
To
As E is
dimensional, (VI.1.13) says that x satisfies an irredueible = Xn + a1xn-1 + ••• + an E k[Xl. Thus kral is a finite
polynomial F(X)
dimensional extension of 1,-, so we may assume that Ida 1 to show that lai Ip ~ 1 for each p ( S and each be the Henselization of k
j.
= E.
It suffiees
Let pES, and let hp
As E is separab1e, so is F.
at p.
As p
satisfies (3.1), the field h p is separably factorial, so (VII.2.4) says that there is a splitting field Kp for f over )'p' Let r'l"" ,r'n be the roots of f in Kp ' and let P be the unique extension of the valuation p to Kp guaranteed by XI!.?3. Define monomorphisms Gi : E .... Kp ' for i = 1, ... ,n, by setting GiX = r'i' By (XII.? .5), each valuation I I on E which extends p is of the form lvi = laiYlp for some
As IxlQ ~ 1 for each
j.
Q E S', it follows that Ir'i Ip <:; 1 for i = 1, ... ,n. The coefficients of f (X) are symmetrie polynomials in the roots r j , and sinee P is nonarchimedean,
Ir; Ip
~ 1 for eaeh i.
But
rj
E k,
so
Ir i Ip
=
Ir'; Ip
<:;
1.
0
Bibliography
Actin, E.
1967 Atgebraie numbers and atgebrate funetions, Gordon & Breach.
Baumslaq, G., F. B. cannonito and C. F. Miller, 111 1981 Computab1e algebra and group embeddings, ]. ALgebra 69, 186-212. Bishop, E.
1967 Foundations of eonstruetive wvatysis, McGraw-Hill. 1973 Sehizophrenia in eontemporary mathematies, AMS Lectures, Missoula, Montana.
Bishop, E. and D. S. Bridges 1985 Construetive ruvatysis,
Grundlehren Wissenschaften 279, Springer-Verlag
der
Colloquium
mathematischen
Borevich, Z. 1. and 1. R. Schaferevich 1966 Number' theory, Academic Press, New York.
Bourbaki, N. 1961 Atgebre eommutative: I. ModuLes ptats, Hermann, Paris. Bridges, D.
s.
1979 Construetive funetionaL ruvaLysis, Research notes in mathematics 28, Pitman, London.
Bridges, D. S. and F. Richman
1987 Varieties of eonstruetive mathematies, London Math. Soc. Lecture Notes 97, Cambridge Univ. Press.
Brouwer, L. E. J.
1981 Cambridge teetures on intuitionism, D. van Dalen editor, Cambridge Univ. Press.
Brouwer, L. E. J. and B. de Loor
1924
Intuitionistischer Beweis des Fundamentalsatzes der Algebra, Proe. Aead. Amsterdam 27, 186-188.
Diaconescu, R
1975 Axiom of choice and complementation, Pr oe . Amer'. Math. Soe. 51, 176-178.
Feferman,
1975
s.
1mpredicativity of the existence of the largest divisible subgroup of an abelian p-group, ModeL theory and aLgebra, Springer Lecture Notes 498, 117-130. 335
Bibliography
336
Fourman, M. P. and A. Scedrov 1982 The "wor1d' s simp1est axiom of choice" fails, Manuser-ipta Mo th. 38, 325-332. Greenleaf, N. 1981 Liberal constructive set theory, Springer Lecture Notes 873, 213-240.
Constr-uetiue
Heyting, A. 1941 untersuchungen über intuitionistische algebra, Akad. Amsterdam, 1. sectie 18. 1971 Intuitionism, an introduction, North-Holland.
mathematies,
Ver-handel ingen
Julian, W., R. Mines and F. Richman 1978 Algebraic numbers, a constructive development, Pae. 1. Math. 14, 91-102. 1983 Alexander duality, Pae. 1. Math. 106, 115-127. Kap1ansky, I. 1949 Elementary divisors and modules, Trans. Amer. Math. Soe. 478-479. 1969 InFinite abelian groups, Univ. of Mich. Press.
66,
Kronecker, L. 1882 Grundzüge einer arithmetischen Theorie der algebraischen Grossen (section 4), lournal für die reine und angewandte Mathematik 92, 1-122. Lin, c. 1981 Recursively presented abelian groups: effective p-group theory l, ISL 46, 617-624. 1981a The effective content of Ulm's theorem, Aspeets oF eFFeetiue algebra, J. Crossley (ed), Upside down A book company, Yarra GIen, Victoria, Australia Magnus, W., A. Karrass and D. Solitar 1966 Combinatorial group theory, lnterscience, New York Ma1'cev, I. A. 1971 On recursive abelian groups, The metamathematies oF algebraie systems, North-Ho1land, Amsterdam. Metakides, G., and A. Nerode 1979 Effective content of field theory, Annal s oF Math. Logie 17, 289-320. Mines, R. and F. Richman 1981 Dedekind domains, ConstrLLetive mathematics, Springer Lecture Notes 873, 16-30. 1982 Separability and factoring polynomials, Roeky Mtn. 1. Math. 12, 91-102. 1984 Va1uation theory: a constructive view, 1. Number Theory 19, 40-62. 1986 Archimedean valuations, 1. London Math. SOC. 34, 403-410.
337
Bibliography Myhill, J. and N. D. Goodman 1978 Choice imp1ies excluded midd1e, Zei t. Ma th. Log. 23, 461 Nagata, M. 1962 LocaL rings, Interscience, New York
DIson, P. L. 1977 Difference reLettlons etnd etLgebr-a: et consU"uctive study, Univ. of Texas at Austin dissertation. O'Meara, O. T. 1963 Intr-oduc t i on to qUlldm ti c forms, Springer-Verlag, Ber1in. Richman, F. 1973 The constructive theory of countable abelian p-groups, Pac. ]. Moth. 45, 621-637. 1974 Constructive aspects of Noetherian rings, Pr oc . Ame,.. Ma th. Soc. 44, 436-441. 1975 The constructive theory of KT-modules, Poc. ]. Math. 61, 263-274 1977 A guide to va1uated groups, AbI' 1 ian gr-oup theoqj, Springer Lecture Notes 616, 73·-86. 1977a Computing heights in Tor, llouston J. MO.th. 3, 267-270. 1981 Seidenberg's condition P, ConslnlctllJe matitemaUcs, Springer Lecture Notes 873, 1-11. 1982 Finite dimensional algebras over discrete fieIds, L.E. J. Brouwercentenary symposium, A.S. Troelstra and D. van Dalen (editors), North-Ho11and Pub. Co., 397-411. 1988 Nontrivial uses of trivial rings, Proc. Amer. Metth. Soc.
Rogers, L. 1980 Basic subgroups from a constructive viewpoint, Communicat ions in algebra 8, 1903-1925. Rootse1aar, B. van 1960 On intui tionistic difference relations, Irwog. motl<. 22, 316-322. Corrections: lndetg. math. 25, 132-133. Rudin, W. 1985 Unique right 489-490.
inverses
are
two-sided,
Amer-.
Math.
Monthly
92,
Scott, D. 1979
Identity and existence in intuitionistic logic, Springer Lecture Notes 753, 660-696.
Scedrov, Andre 1986
Diagonalization of continuous matrices as a representation of intuitionistic rea1s, Ann, Pw-e Appl. Logi.c 30, 201-206.
Seidenberg, A. 1970 Construction of the integral closure of a finite integral domain, Renn. Sem. Mat. Fis. MUano 40,100-120. 1971 On the length of a Hilbert ascending chain, Pr'oc. Amer", Math. Soc. 29, 443-450.
Bibliography
338 1972 1973 1974 1974a 1975 1978 1984 1985
Constructive proof of Hilbert's theorem on ascending chains,
Trans. Amer. Math. Soc. 174, 305-312.
On the impossibility of some construcUons in polynomial rings, Proc. Int. Cong. Geom, Milano 1971, 77-85 Constructions in algebra, Trans. Amer. Math. Soc. 197, 273-313. What is Noetherian?, Rend. Sem. Mat. e Fis. MiLano, 44, 55-61. Construction of the integral closure of a finite integral domain. II, Pr'oc. Amer. Math. Soc. 52, 368-372. ConstrucUons in a polynomial ring over the ring of integers, Amer. J. Math. 100, 685-703. On the Lasker-Noether decomposition theorem, Amer'. J. Math. 106, 611-638. Survey of constructions in Noetherian rings, Pr'oc. Symp. Pure Math. 42, 377-385.
Smith, H. J. S. 1861 On systems of linear indeterminate eguaUons and congruences, Phil. Trans. 151, 293-326, in the collected mathematical papers of H. J. S. Smith, J. W. L. Glaisher (ed) Chelsea, NY 1965. Smith, Rick L. 1981 Two theorems on autostability in p-groups, Logic year 1979-1980, Springer Lecture Notes 859, 302-311. SOUblin, J-P. 1970 Anneaux et modules coherents, J. ALgebra 15, 455-472. Staples, J. 1971 On construcUve fields, Proc. London Math. Soc. (3) 23, 753-768. Stoltzenberg, G. 1968 Constructive normalization of an algebraic variety, Bull. Amer. Math. Soc. 74, 595-599. Uspenskii and 5emenov
1981
Springer Lecture Notes in Computer Science 122
Waerden, B. L. van der 1930 Eine Bemerkung über die unzerlegbarkeit von Polynomen, Math. Annalen 102, 738-739. 1953 Modern Algebra, Ungar, New York. Wang, H. 1974 From mathematics to philosophy, Routledge Weiss, E.
&
Kegan Paul, London.
1963 Algebr'aic number theory, McGraw-Hill, New York.
Index
in number, 11 metric space, 292 module, 211 set, 11 Bounding constant, 287 Brouwerian example, 5 Burnside's theorem, 237
Acyclic relation, 28 Additive monoid, 35 Adjacent words, 249 Adjoint matrix, 71 Admits splitting fie1ds, 203 Algebra, 232 Algebraic, 139 numbers, 189 Algebraically closed fie1d, 152 (in)dependent, 145 Algorithm, 2, 30 Apartness, 8, 30 domain, 64 Approximation to the N-bal1, 293 Archimedean valuation, 288, 290, 291 Artin-Rees lemma, 200 Ascending chain condition, 23 Associated prime ideal, 217 Associates, 108 Automorphism, 18 Axiom of choice, 14 countable, 15 dependent, 15 unique (nonchoice), 10, 31 world's simp1est, 15, 33 Azumaya theorems, 97, 98
Cancellation monoid, 108 Cardinality, 10 Cartesian product, 9 category, 16 categorical coproduct, 20 product, 19 Cauchy sequence, 48 Cayley-Hamilton theorem, 72 Center, 232 Centralizer, 233 Chain, 22 Character of a monoid, 172 Characteristic function, 13 of a ring, 45 polynomial, 71 Chinese remainder theorem, 329 Church-Rosser property, 54, 250 Cofactor, 70 Coherent, 82 Column operations, 66 Commutative diagram, 93 monoid, 35 ring, 41 Companion matrix, 136 Complement, 9 Complementary summand, 57 Complete metric space, 51 Completion of a valued field, 295 of ametrie space, 51 Complex field, 302 numbers, 50 Composition series, 221
Basis of a free module, 56 of a free group, 249 of a normed vector space, 300 Belongs to, 208 Eezout domain, 115 Bifunctor, 90 Bijection, 10 Bilinear map, 89 Bimodule, 53 Binary sequence, 4 Bounded cancellation monoid 109 depth, 24 discrete domain, 109 element, 109 339
340 Computational interpretation, 2 Condition P, 187 Conjugate, 261, 262 Consistent basis function, 205 inequality, 8 Content, 123 Converges, 51 to zero, 48 Coproduct, 20 Coset, 39 Cotransitive inequality, 8 Cotransi ti vi ty of a < b, 49 Countable axiom of choice, 15 set, 11, 32 Cycle of apermutation, 37 Cyclic group, 37 module, 56 Cyclically reduced word, 261 DH-ring, 121 Decidable relation, 7 Dedekind-Hasse map, 118 Dedekind-infinite set, 14 Dedekind domain, 326 set of valuations, 326 Degree of a polynomial, 60, 61 Denial field, 45 inequality, 8 Dense subspace, 51 Density theorem, 235, 236 Dependent vectors, 68 Depth, 23, 24 Derivative (formal), 154 Descending chain condition, 23, 28 Detachable ideals and submodules, 83 subset, 9 Determinant, 69, 70 Diagonal matrix, 128 Diagonalizable, 137 Difference relation, 12, 30 Dimension of a vector space, 67 Direet limit, 20, 165 sum, 53, 54 summand, 57 product, 53 Discrete set, 9 valuation, 293 Disjoint union, 18
Index Distributive 1attice, 21, 24 Divides, 63, 108, 113 Divisible abelian group, 269 hull, 271 Division algebra, 233 algorithm, 61 ring, 41 Divisor chain condition, 110 Domain, 42 Dual category, 20 Eigenvalue and eigenvector, 136 Eisenstein criterion, 117 Elementary divisors, 134 matrix, 66 row and colUJlU1 operations, 66 symmetrie polynomials, 73 Embedded prime ideal, 217 ~1domorphism, 18 ring, 41 Epimorphism, 18 Equality relation, 7 Equivalent matrices, 128 norms, 299 valuations, 288 Essential subgroup, 271 Euclidean algorithm, 63 domain, 118 map, 118 Even permutation, 37 Exaet sequence, 90 Exchange property, 145 Existence, 1, 2 Extensional, 7, 9 strongly, 10 Factor, 63 Factorial domain, 114 held, 176 Faithful module or representation, 52, 232 Family, 11, 18 Field, 42 generated by, 43 norm, 317 of definition, 155 of quotients, 42 Finitary tree, 6, 7 Finite cardinality, 10, 11
341
Index depth, 24 dimensional, 67, 232 family, 11 presentation, 130 rank free module, 56 rank torsion-free group, 265 routine, 2 set, 11, 32 support, 26, 53 Finitely generated, 37, 43, 56 enumerable, 11 presented extension field, 199 presented module, BI presented ring, 19B Fixed fie1d, 171 Flat module, 92 Fraetional ideal, 329 Formal power series, 64, 298 Free group, 249 module, 56 monoid, 36 Full subgroup, 269 Fully faetorial field, 1BB Lasker-Noether ring, 224 FUnetion, 9 FUnetor, lB FUndamental theorem of algebra, 191 GCD (domain and monoid), 108 Galois extension and group, 169 Gauss's lemma, 123 Gelfand-Tornheim theorem, 303 General valuation, 287 Greatest eommon divisor (see GeD) Group, 36 ring, 60
Height (see depth) of a prime ideal, 228 in an abelian p-group, 273 in a torsion-free group, 266 Hensel's lemma, 311, 315 Henselian eontext, 309 field, 311 Henselization, 305, 314 Hereditary subset, 24 Heyting field, 42, 77, 96 Hilbert basis theorem, 196 Homomorphism, 18 of algebras, 232 of modules, 52
of monoids, 35 of rings, 43 Hopfian, BO, 85, 257 Ideal, 43, 44, 53 Idempotent, 57 Identity fune ti on , 9 matrix, 65 Image, 10 Impotent ring, 104, 141 Index of a subgroup, 39 Independent over a loeal ring, 99 submodules, 53 veetors (diserete ease), 68 Inequivalent valuations, 288 Indeterminate, 60 Inequality, 8, 31 on a loeal ring, 98 Infimum, 21 Infinite set, 11 Inhabited, 30 Injeetion of ordinals, 28 Integral closure, l39 domain, 42 element, l39 extension, 139 Integrally elosed, 139, 140 Intermediate value theorem, 189 Interseetion, 8, 11 Invariant faetors, 138 Inverse, 36 funetions, 10 Irreducible element, 109 ideal, 220 module, 235 Irredundant primary decomposition, 217 Isolated primary ideal, 219 I somorphi sm, 18 Jaeobson radical, 78 Jordan eanonieal form, 106, 136 Jordan-HOlder-Dedekind theorem, 22 k-homomorphism, 167 Kernei, 38, 52 König' s lemma, 7 Kripke's schema, 32 Kroneeker 1, 125 Kroneeker 2, 126 Krull interseetion theorem, 200, 201
342
Index
LLPO, 5, 33, 51 LPO, 4 Lagrange interpolation formula, 64 Lasker-Noether ring, 220 theorem, 227 Last-difference relation, 26 tattice, 21 Law of excluded middle, 4 Leading coefficient, 61 Least common multiple, 113 Left ideal, 43, 53 inverse, 36 Length of an abelian group, 274 of a word, 250 Limit, 51 Linearly independent, 99 independent functions, 172 ordered, 22 Local ring, 85, 96 Localization, 85 Locally (pre)compact, 292, 293 Located subset, 300 Uiroths's theorem, 148
Nakayama Lemma, 78 Newton interpolation formula, 62 Newton's method, 296 Nielsen set, 253 transformation, 255 Nilpotent element, 78, 141 left ideal, 237 Noether norma1ization, 198 Noetherian basis function, 204 module, 80 ring, 80 Nonarehimedean valuation, 288, 290 Nonempty, 7, 8, 30 Nongenerator, 79 Nontrivial homomorphism, 35 valuation, 288 Nonzero left ideal, 43 Norm, 118, 299 Normal extension field, 168 subgroup, 38 Normed vector spaee, 299 Nullstellensatz, 203
Map of partia11y ordered sets, 20 Maschke's theorem, 242 Markov's princip1e, 33 Matrix product, 65 Maximal ehain, 22 ideal, 45, 77 left ideal, 79 Metrie basis, 299 space, 50 Metrically independent, 299 Minimal po1ynomial, 135, 144, 234 prime ideal, 214 Modular lattice, 21 Module, 52, 232 of fraetions, 85 Monie polynomial, 61 Monoid, 35 Ling, 60 Monomial, 60 Monomorphism, 18 Multiplieative funebon, 118 monoid, 35 norm, 118 submonoid, 85
Odd permutation, 37 ('l-bounded set, 14 Omniscience principle, 4 One-to-one funetion, 10 Onto function, 10 Operation, 31 Opposite ring, 58 Order of a group element, 36 Ordinals, 25 equali ty of, 28 Grayson, 29 Ostrowski's theorem, 305 p-adie met.rie on (Q, 51 valuation, 292 p-group (abelian), 273 partially ordered set, 20 Perfeet closure, 165 field, 164 Permut.ation, 37 matrix, 67 Piecewise isomorphie, 22 polynomial, 60 positive real number, 49 Power set, 7 Preimage, 10
343
Index Primary decomposition, 216 decomposition theorem, 222 ideal, 208 module, 134 Prime field, 43 ideal, 45 in a cancellation monoid, 109 Primitive element, 158 polynomial, 123 Principal ideal, 44, 110 domain, 115 theorem, 228, 229 Product of algebras, 239 Projection, 58 Projective module, 57 set, 16 Proper element in a p-group, 279 ideal, 43 subset, 9 prüfer's theorem, 280 Pseudofactorial field, 295 Pseudonorm, 118 Pure subgroup, 281 submodule, 84, 92 Purely inseparable, 164 transcendental, 146 Quasi-
110 factorization theorem, 111 regular, 78 Quaternion algebra, 235 Quotient of an ideal, 44 group, 38 UFO,
Radical Jacobson, 78 of an ideal, 44 of an algebra, 237 Ramification index, 319 Rank
function, 34 of a free group, 252 of a free module, 56 of a torsion-free group, 265 relation, 29 Rational canonical form, 136
function, 102 quaternions, 46 reot test, 179 Real field, 302 numbers, 48 Recognizable units, 41, 108 Reduced abelian group, 274 form of a word, 251 length, 253 valuated p-group, 278 word, 251 Reducible word, 251 module, 235 Rees ring, 200 Reflect fini tely generated ideals, 197 Relation, 7 Relatively prime, 63, 108 Remainder theorem, 62 Representation, 52, 232 Representable functor, 20 Residue class degree, 320 field, 293 Right ideal, 43, 53 inverse, 36 Ring, 41 of fractions, 85 with detachable ideals, 83 Root of multiplicity n, 65 test, 179 Row operations, 66 space, 66 Saturated multiplicative submonoid, 115 Scalar multiplication, 52 Schanuel's trick, 57 Schreier transversal, 257 Schur's lemma, 237 Seidenberg field, 187 Semirigid class, 267 Separable closure, 163, 185 element, 155 extension, 155 polynomial, 154 Separably closed in E, 155, 163 factorial field, 176
344 Set, 7 Sequenee, 11 Simple module, 235 root, 65 Simply existential statement, 6 presented p-group, 280 Smith normal form, 129 Soele of an algebra, 241 Span operation, 150 Splitting field, 152 Standard inequality, 8 Strietly bounded domain, 121 Strong approximation theorem, 328 inequality, 99 Strongly extensional funetion, 10 relatively prime, 63 Subfinite, 32 Subgroup, 37 Submodule, 53 Submonoid, 35 Subring, 42 Subset, 7, 31 Sueeessor, 281 Summand, 57 Support of a cyele, 37 Supremum, 21 Surjeetive funetion, 32 Sylow subgroup, 190 Symbolie power, 214 Symmetrie group, 37 inequality, 7 po1ynomial, 73 Tennenbaum, 204, 231 module, 207 Tensor produet, 89 Tight inequality, 8 Torsion, 133, 211 forest, 280 Torsion-free abelian group, 265 Transeendenee basis, 146 degree, 146 Transitive set (relation), 25 Transversal, 257 Trivial metrie, 301 ring, 41 valuation, 292 Type, 266
Index UFD, 114 U-funetion, 282 Ulm funetion, 278 group, 277 invariant, 278 Ulm's theorem, 279 Union, 8, 11 Unique ehoiee, 10,31 faetorization domain (see UFO) interpolation, 62 Unit, 36 Valuated p-group, 278 Valuation, 287 Value group, 287 Valued field, 287 Valuation ring, 130 Vandermonde determinant, 73 Variety, 204 Veetor spaee, 67 Von Neumann regular ring, 69, 96 Weak approximation theorem, 327 Nullstellensatz, 142, 143 Wedderburn's theorem, 243, 246 We11 founded, 25 ordered, 25 WLPO (weak LPO), 6 Word, 36 problem, 263 World's simplest axiom of ehoiee, 15, 33 Zero-divisor, 233 Zippin's theorem, 282
