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Ellis Horwood Series in
Mathematics and its Applications
Series Editor: G. M. BELL. Professor of Mathematics,
King's College (KaC), University of London
Numerical Analysis, Statistics and Operational R8earch
Editor: B. W. CONOLL V, Professor of Operational Research,
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COMMUTATIVE NOETHERIAN AND KRULL
RINGS
STANIStAW BALCERZVK and TADEUSZ JOZEFIAK, Insti-
tute of Mathematics, Polish Academy of Sciences, Torun,
Poland
Translation Editor: D. KIRBY, F-eculty of Mathematical
Studies, University of Southampton -
The fundamental concepts of commutative ring theory are
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integral extensions are emphasized. The book contains a
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rates two important results which unite an
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any Abelian group is isomorphic to an ideal.
Dodekind domain; and a Nagata-Mori TI
normalization of a Noetherian domain is a 
Much of the material closely connects moder
methods with classical algebraic number tl
braic geometry. Highlights include a systema'
geometric and number theoretical sources 4
ring theory, the proof of finiteness of ideal (
rings of algeb .3ic integers, as well as a detail
of e)t,;,"ple of rings with some special pro
", ,J: Mathematics, algebra, commutativi
ge!Jt, (, algebraic number theory.

COMMUTATIVE NOETHERIAN AND KRULL RINGS
J/ ..
.-  1 1-
:11,1

MATHEMATICS AND ITS APPLICATIONS
Series Editor: G. M. BELL, Professor of Mathematics,
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Commutative Noetherian
and Krull Rings
STANISLAW BALCERZYK
Professor, Institute of Mathematics,
Polish Academy of Sciences
TADEUSZ JOZEFIAK
Professor, Institute of Mathematics,
Polish Academy of Sciences
Translation Editor
DAVID KIRBY
Faculty of Mathematical Studies
Unit'ersit)' of Southampton
, . t .
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,
,J 4"..L.J.i\'_
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Balcerzyk, S. (Stanislaw)
Commutative Noetherian and Krull rings
1. Commutative rings
I. Title II. J6zefiak, T. (Tadeusz)
III. Kirby , David IV. Series
512.4
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Table of Contents
Preface . . . . . . . . . . . . . . . . .
Chapter I-PRELIMINARY CONCEPTS . . . . . . .
1.1 Rings and Ideals . . . . . . . . . .. ........
1.2 Algebraic Sets and the Prime Spectrum of a Ring
1.3 Modules . . . . . . . . . . . . . . . . .
1.4 Localization of Rings and Modules. . . . . . . . . . . . .
1.5 Graded Rings and Modules . . . . . . . . . . . . . .
Chapter ll-NOETHERIAN RINGS AND MODULES
2.1 Basic Concepts and Properties . . .
2.2 The Basis Theorem. . . . . . . .
2.3 Primary Decomposition of an Ideal. .
2.4 Primary Decomposition of a Module .
2.5 The Artin-Rees Lemma. . . . . . .
2.6 Completions of Rings and Modules. .
2.7 Artin Rings and Modules . . . . . .
7
11
11
19
28
36
47
53
54
58
61
71
80
83
96
Chapter DI-INTEGRAL EXTENSIONS AND DEDEKIND
DOMAINS ....... .... ... 104
3.1 Integral Extensions . . . . . . . . . . 103
3.2 Normal Domains ... ... .......... 114
3.3 Dedekind Domains. . . . . . . . . . . . 1 18
3.4 Ideal Class Group and Picard Group . . . 126
3.5 Modules over Dedekind Domains.. ... ... 134
3. 6 Valuations ................ . .. 138
Chapter IV-DIVISORS AND KRULL DOMAINS . . . . . . 153
4.1 Divisors and the Semigroup of Divisor Classes . ... 154
4.2 Krull Domains. . . . . . . . . . . . . . .. .... 159
4.3 Krull Domains as Intersections of Discrete Valuation Rings 165
4.4 Divisors in Krull Domains ......... . . . . . . . 174
4.5 Induced Homomorphisms of Class Groups. . . . . . . . .. 181
4.6 A Representation of an Abelian Group as a Divisor Class Group 187
4.7 Normalization of Noetherian Domains . . . . . ... . . . . 192
References . . . . . . . . . .. . . . 205
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

Preface
The theory of commutative rings is connected not only with many branches
of algebra but also with other fields of mathematics, especially with number
theory, algebraic topology and functional analysis. Its main sources are alge-
braic geometry and algebraic number theory.
The starting point of algebraic geometry was the study of the set of solutions
of a system of equations
fi(X1, ..., x n ) = 0
J",(Xl' ..., XII) = 0,
where 11' ..., fm are elements of the ring K[X 1 , ..., Xn] of polynontials with
coefficients in an algebraically closed field K (e.g. in the field of complex
numbers). This set is completely determined by the ideal Iof the ring K[X 1 , ...
. . . , X n ], generated by the polynomials 11' ..., fm, and the geometrical properties
of the set of solutions are faithfully reflected in the properties of the factor
ring K[X 1 , ..., XlI]!I (e.g. to the points of the set of solutions correspond
maximal ideals and to the subsets given by additional equations-ideals).
It turned out that a large part of the geometrical theory of rings of the form
K[X 1 , ..., X n ]! I carries over to a much wider class of rings, called Noetherian
rings. They are characterized by the most essential property of polynomial
rings over fields: any ideal is finitely generated. Thus the theory of Noetherian
rings, which constitutes the predominant part of the theory of commutative
rings, takes up the problems and methods inspired by algebraic geometry.
Continuous enrichment of the methods of algebraic geometry has considerable
influence upon the theory of commutative rings. We present elements of the
theory of Noetherian rings in Chapter II, showing in detail its connections
with algebraic geometry.
Algebraic number theory was created in the second half of the 19th century
in the works of Kummer, Dedekind and Kronecker. Its main object of study
is the class of rings of algebraic integers, i.e., rings R K defined for every field K
which is a finite extension of the field of rational numbers. The ring Rx consists
of those numbers (X of K which satisfy an equation of the form (Xm + Qm-1 «",-1 +
+ ... + ao = 0, where 00, ..., a m -l are integers. The theory of ideals of rings
of algebraic integers was created to provide new methods for solving classical

8
Preface
problems of number theory. The development of methods of algebraic number
theory still continues to the present day. The abstraction of the most essential
properties of rings of algebraic integers has led to axioms of a new class of
rings called Dedekind domains. We present in Chapter III basic properties
of Dedekind domains which form the introductory material for modern
algebraic number theory. The class of Dedekind domains is much wider than
the original class of rings of algebraic integers. Indeed, the basic invariant
of a Dedekind domain, its ideal class group, can be an infinite Abelian group
in general whereas it is always finite for rings of algebraic integers.
The last Chapter IV of the book is devoted to still another class of rings,
called Krull domains. Dedekind domains are Krull domains, not every Krull
domain is Noetherian. The notion of a Krull domain has its roots in certain
arithmetic aspects of the theory of Dedekind domains related to the notion
of a divisor and a valuation of a field. The class of Krull domains is closed
with respect to some natural algebraic operations .and its properties are used
in the proof of a rather surprising result of Claborn that every Abelian group
can be realized as an ideal lass group of a suitable Dedekind domain.
Chapter I contains introductory material as well as background material
for later chapters. A few additional facts which are needed can be found in
S. Lang's Algebra [L]. Exercises are included at the end of almost every section,
the more difficult ones being furnished with hints in brackets [ ]. With few
exceptions the material contained in the exercises is not used in the text.
The reader of the English edition is entitled to a few explanatory remarks.
This book constitutes, with minor changes, a translation of the first four
chapters of our book Commutative Rings (Pierscienie Przemienne) published
in Polish in 1985. The remaining four chapters will form another book Commu-
tative Rings: Dimension, Multiplicity and Homological Methods, to be published
in this series. Our aim has been to write the first textbook in Polish presenting
the fundamentals of the theory of commutative rings. This has influenced the
character of the book.
The partition of the Polish original into two separate books in the English
translation has caused certain problems. We have made some effort to handle
them. However, there is at least one sign of this partition left. The content
of the introductory Chapter I has not been altered, and the reader will notice
that some material is never used in this book. This material can serve as
a reference for the second book and we have decided to keep it unchanged
to make both parts, treated as a whole, as self-contained as possible.
In English there are numerous books devoted to commutative rings differing
considerably in the choice and presentation of material. Some of them are
listed at the end of the book. At this point, however, we should mention two
books: M. Nagata, Local Rings [N], N. Bourbaki, Algebre commutative [D],
that were very useful for us in writing Chapter IV.
We wish to express our thanks to Professor Andrzej Bialynicki-Birula for
encouraging us to write the book, for a careful inspection of the first version

Preface
9
of the manuscript, and for many valuable remarks and discussions concerning
the choice and presentation of the material. We are indebted to those of our
colleagues who contributed to the improvement of the text and helped us to
prepare the exercises.
Stanislaw Balcerzyk
Tadeusz J6zefiak

Chapter I
Preliminary Concepts
This chapter is an introductory one. Sections 1.1 and 1.3 contain the basic
definitions concerning rings, ideals and modules. Most of the facts presented
there are either known from a course of algebra or can easily be transferred
from the case of linear spaces to the case of modules, or follow directly from
the definitions. Less obvious facts are singled out as theorems and supplied
with proofs. In Section 1.2, we introduce elementary concepts concerning
algebraic sets and their connections with the theory of commutative rings.
We also study in it the set Spec(R) of all the prime ideals of a ring R. The whole
of Section 1.4 is devoted to the localization of rings and modules, which
is a generalization of the construction of the field of fractions of a domain.
In Section 1.5, we define graded rings and modules.
1.1 RINGS AND IDEALS
A triplet (R, +, .) is called a ring if: R is a nonempty set, + and · are binary
operations, + , .: R x R -» R, and the following conditions are satisfied
for all elements r, s, t e R:
(a) R is an Abelian group with respect to the operation +,
(b) r(s+t) = rs+rt,
(c) (9+t),. = sr+tr,
(d) r(9t) = (rs) t,
where, as usual, we write rs for r · s. For simplicity we usually say that R is
a ring. We shall only consider rings with a unity element, i.e., rings R which
contain an element 1 #- 0 such that rl = lr = r for all r E R where 0 is the
neutral element for +. A ring R is called commutative if the condition r9 = sr
is satisfied for all r, s E R. In the sequel, only commutative rings with a unity
element will be discussed; thus, for brevity, we shall write "a ring" instead
of "a commutative ring with a unity".
We denote by Z, Q, R, C the ring of integers, the field of rational numbers,
the field of real numbers and the field of complex numbers, respectively.
A mapping of rings,!: R -+ T, is called a homomorphism if, for all elements
r, 9 E R, the conditions f(r+ s) = f(r) +/(9) and l(r9) == f(r) · 1(9) are satisfied

12
Pre limin ary Concepts
[Chi
and moreover, /(1) = 1. Rings, together with homomorphisms of rings, form
the category of (commutative) rings; the class of its objects consists of all
rings, the class of its mappings consists of all homomorphisms of rings, and
the composition of mappings is the usual composition of functions. A homo-
morphism/: R -+ T is called an isomorphism if there exists a homomorphism
g: T -+ R such that gf = l R and fg = IT' where IR: R -+ R denotes the
identity homomorphism. A homomorphism f is an isomorphism if and only
if it is a one-to-one mapping, and every element of T belongs to the image
1m(/) of the homomorphism f; we then write f: R  T or R  T.
A ring R is called a subring of a ring T if R c T, the operations in Rare
identical with the operations in T restricted to R, and the unity element of the
ring Tbelongs to R. Given a subring R of the ring Tand elements t 1 , ..., t k E T,
we denote by R[t 1 , ..., tt] the smallest subring of T containing R and the
elements t 1, ..., tt. The elements of this ring have the form* L "1... 1 " t' 1 ... t,,,,
where 'it' ..., '1" E R, and i 1 , ..., ilc range over a finite set of non negative
integers. In general, the representation of an element in this form is not unique.
Let R be a ring. We define an R-a1gebra as a pair (A, '1/) where A is a ring
and '1/: R  A is a ring homomorphism. For simplicity, we usually say that A
is an R-algebra provided the homomorphism 'YJ is specified. We often say also
that the homomorphism 'YJ: R --+ A establishes the structure of an R-algebra
on A. The canonical homomorphism Z -+ R, which sends any integer n to
the elemet nl, allows us to regard any ring R as a Z-algebra. The characteristic
of the ring R, char(R), is 0 if nl ¥: 0 for all positive integers n; it is the least
positive integer m such that ml = 0 otherwise.
A homomorphism of R-algebras (A, 'YJ) -4 (A', 'YJ') is a homomorphism of
rings f: A -+ A' such that f'YJ = 'YJ'. We often consider R-algebras where R
is a field; we can then identify the ring R with a subring of the ring A, since
the homomorphism 'YJ: R -+ A is an injection. We say that an R-aIgebra
(A, 'YJ) is finitely generated if A = 17(R)[al' ..., ak] for some a1, ..., ak E A.
An element a of a ring R is called:
(a) invertible, or a unit, if there exists an element b E R such that ab = I,
(b) a zero-divisor if there exists a non-zero element b e R, such that ab = 0,
(c) nilpotent if there exists a positive integer n such that an = o.
If a is an invertible element of R, there exists a unique elelnent b E R
such that ab = 1; it is called the inverse of a and denoted by a-I or 1/ a. We
write R* for the set of all invertible elements of R. Every nilpotent element
is a zero-divisor. If an = 0, we have (1 +a)(1-a+a 2 - ... + (_I)n- 1 a rr - 1 )
= I, and consequently the element 1 +a is invertible. A zero-divisor, or a
nilpotent element, is called proper if it is different from zero. A ring which
does not contain proper zero-divisors is called a domain (the term integral'
domain is also in use); a ring without proper nilpotent elements is called
1\' We write only Land n if there is no doubt as to the range of summation and multi-
pIication.

I]
Rings and Ideals
13
a reduced ring. All non-zero elements of a ring R are invertible if and only if R
is a field.
A non-zero element of a domain is said to be irreducible if it is not a unit
and cannot be represented as the product of two factors neither of which
is a unit.
A domain R is called a unique factorization domain if, for any non-zero
element ,. E R, the following conditions are satisfied:
(i) r is the product of irreducible elements and a unit,
( .. ) . f ' , , h ' - t d
11 1 r = UPI ... p& = U PI ... Ps" were u, U are unl s an PI'...' Ps,
P, ..., p;, are irreducible elements of R, then s = s', and there exist a permu-
tation (J of the indices 1, ..., 8 and units Ul, ..., Us E R* such that p<,) = U,Pi,
i = t, ..., s.
We define an ideal I of a ring R to be a non..empty subset of R which
satisfies the following conditions:
(a) ifr,sE/thenr+8e/,
(b) if rEI and t e R then rt E /,
and we call it proper, if I ¥: R.
To a proper ideal I of a ring R we can assign a factor ring R/I whose ele-
ments are all residue classes r = r+l = {r+x; x E I}, where r ranges over R,
and the operations of RII satisfy the conditions (r+.1) + (8+.1) = (r+s)+1
and (r+1) (s+1) = rs+I. We define the natural homomorphism v: R -+ R/I
by the formula ,,(r) = r+l. It is easy to observe that, associating with each
ideal I:=J I the ideal v(J) = III = {r+l; r E J}, we obtain a one-to-one
correspondence between the set of all ideals of the ring R which contain
the ideal I and the set of all ideals of the ring RI I. Sometimes we use the tradi-
tional notation, r == s (mod]), and say that r is congruent to 8 modulo I, if
r+I = s+1 in the factor ring RII, i.e., if r-s E I.
The kernel Ker(f) = {r E R; f(r) = O} of a homomorphism of rings
f: R -+ T is an ideal of R. The image Im(f) = If(r); r E R} of the homo-
morphism f is a subring of T, isomorphic to the factor ring RIKer(f). If
I c Tis an ideal, thenf-l(J) is also an ideal. If I c: R is an ideal, thenf(I)
is an ideal of the ring Im(f), but not in general of the ring T. We use the term
injection for a homomorphism f: R -+ T which is one-to-one, i.e., for which
Ker(f) = 0; similarly, we call it a surjection if Im(f) = T.
Each element a e R determines a principal ideal, {ra; r e R}, written Ra
or (a). The ideal which consists of the zero element only is denoted by (0)
or O. A domain in which all the ideals are principal is called a principal ideal
domaiD.
If U is a non-empty subset of a ring R, the smallest ideal of R which contains
the set U consists of all elements which can be written in the form r1 Ul + ... +
+r"u", for Ul'''', u" e U, r1' ..., r" E R, k = 0, 1, ..., or, in other words,
it is the intersection of all ideals which contain the set U. We say that this ideal
is generated by the set U. To denote the ideal generated by a finite set of ele-

14
PreHminary CODcepts
[Chi
ments, U1, ..., Uk, we write RU1 + ... +RUk or (U1, ..., Uk). An ideal I is said
to be finitely generated if I = (U 1, ..., Uk) for some u 1, ..., Uk E I.
If {l(X }exeA is a family of ideals of a ring R, then the ideal which is generated
by this family (more exactly, by the union of these ideals) consists of elements
of the form 00:1 + ... + aex", where a(X, E l(/.l' i = 1, ..., n. We denote it by
L: I(/., or, in the case of a finite family{I 1 , ..., In}, also by 1 1 + ... +1". Obvi-
exeA
ously, the intersection n 1(1, of any family of ideals is also an ideal. Given two
exeA
ideals, I, I, of a ring R, we write IJ to denote their product, i.e., the idea
which consists of all elements of the form a1 hi + ... +a"h n , where a1, ..., an
E I, b 1 , ..., h" e I, n = 1, 2, ... We write rl instead of (r)l, and I" to denote
the product of n identical factors I, ..., I, assuming 1° = R. Obviously, we
have I :::> [2 :::> 1 3 ::> ... ::> In ::> In + 1 ::> ... We define the quotient of the ideals
I and J as the ideal I:J = {r e R; rJ c I}, and the radical of the ideal [
as the ideal rad(!) = {r e R; r" E I for some n > O}. The latter is indeed an
ideal since, if r" E I, sm E I, then
(r+s)"+m-l L ("+'f-l)r l s J el (i+j = n+m-l).
Now, from the condition i+j = n+m-l it follows that either i  n orj  m,
whence ,iSJ E I.
The set of all nilpotent elements of a ring R forms an ideal, rad(O), of R,
called the nilradical of R.
We sum up the basic properties of the operations on ideals introduced
so far in the theorem below, leaving the simple proofs to the reader. Other
properties will be presented in the exercises at the end of the section.
Theorem 1.1.1
Let /, 1(/., I, IfJ be ideals of a ring R. Then the following conditions are satisfied:
(i) 1 nl ::> II,
(ii) (L: lot) (L: JfJ) = L: I(1.JfJ'
(X fJ ex, fJ
(iii) (n lex): I = n (I(/.: I),
(/. ex
(iv) I: (L: Ip) = n (I:/fJ),
fJ fJ
(v) rad(l1 ... I,,) = rad(I I n ... nl,,) = rad(I 1 )n ... nrad(I,,),
(vi) if rad(/) is a finitely generated ideal, then there exists a number n such
that (rad(I))" c I.
A proper ideal I of a ring R is called
(1) prime if the ring R/I is a domain,
(2) maximal if the ring R/ I is a field.
We usually denote prime ideals by the letters P, Q, and maximal ideals
by the letter m with suitable indices. A proper ideal P is prime if and only

I]
Rings and Ideals
15
if the condition r9 E P implies r e P or 8 E P. An ideal m is maximal if and only
if the condition m c I * R, where I is an ideal, implies I = m.
A set S c R is said to be multiplicative if 1 e S, 0 , Sand 81' 92 e S implies
S182 e S. Thus, an ideal P is prime if and only if the set R'P is multiplicative.
Theorem 1.1.2
Let S c R be a multiplicative set. Write J to denote the family of all ideals I
of the ring R such that InS = 0. Then, for every ideal 10 E J, there exists
an ideal P, maximal in the family J, such that P ::> 10. The ideal P is a prime
one.
Proof
Since the family J satisfies the hypotheses of the Kuratowski-Zorn Lemma,
there exists an ideal which satisfies the first part of the assertion. We claim
that this ideal, let us call it P, is prime. Let a, b e R'P. Then the ideals P + Ra
and P + Rb do not belong to the family J, and consequently there exist elements
81 E (P+Ra)r\S and 92 e (P+Rb)nS, whence we have 8182 e (P+Ra)(P+
+Rb) c P+Rab and 8192 e S. Hence we get P+Rab =1= P, or ab  P, which
means that the ideal P is indeed prime. 0
orollary 1.1.3
Every proper ideal of a ring R is contained in a maximal ideal of R.
It is sufficient to set S = {I} since then J is the family of all proper ideals
of R, and the corollary follows immediately from Theorem 1.1.2.
Corollary 1.1.4
The nilradical rad(O) of a ring R is the intersection of the family of all prime
ideals of R.
Proof
If r e Rand r" = 0, then r E P for every prime ideal PeR. Suppose, con-
versely, that 8 e R is not a nilpotent element. Then the set S = {I, s, 9 2 , S3, ...}
is multiplicative, and, by Theorem 1.1.2, there exists a prime ideal P such that
PnS = 0, whence 8  P. 0
Corollary 1.1.5
The radical rad(I) of an ideal I is the intersection of all prime ideals containing
I.
In fact, rad(I)/I is the nilradical of the ring R/I, and so it is the intersection
of all ideals of the form P /1, where P ranges over all prime ideals containing I.
Hence the corollary follows.
We define the Jacobson radical (or the radical) of a ringR as the intersection
J(R) of all maximal ideals of R.

16
PreUminary Concepts
[Ch.
Theorem 1.1.6
The Jacobson radical J(R) of a ring R is the ideal which consists of all elements
r e R such that, for every x E R, the element 1 + rx is invertible.
Proof
Indeed, r e J(R) if and only if the elelnent rx belongs to all maximal ideals
of R for every x e R, i.e., if and only if the element 1 + rx does not belong
to any maximal ideal for every x E R. The latter condition is equivalent to
the invertibility of 1 + rx. 0
Theorem 1.1.7
Every ring R satisfies the following conditions:
(i) if 1 1 , ..., Is c R are ideals, PeR is a prime ideal, and II ... Is C P
(or, in particular, II (j ... nI, c P), then 1" c P for some k, 1  Ie  s,
(ii) if s  2, and I, P1, ..., P II c::: R are ideals, where P 3 , ..., Ps are prime
and 1 c Pt U oo. uP" then I c::: PIc for some k, 1  k  s,
(iii) if J  I c R are ideals, P 1, ..., P s c::: R are prime ideals, and I". J
c P1 U ... uPs, then I c P" for some k, 1  k  s.
Proof
(i) Suppose the ideal P does not contain any of the ideals 1 1 , ..., Is. Then
there exist elements ale E I",""P, k = 1, ..., s. The product al ... as belongs
to the ideal 1 1 oo. I.. Now, P being prime, one of the factors ak belongs to P,
contrary to the choice.
(ii) Suppose that the ideal I is not contained in any of the ideals P 1, ..., Ps.
Moreover, without loss of generality, we can assume that the union of less
than s ideals P l' ..., P II does not contain I. Thus there exist elements aI, ...
..., all eI such that ale ePIc' ale (:P I u oo. uP"-t UP"+l U ... uPs, k = 1, ...
... , 9.
For 9 = 2, the conditions at E PI' a2 , P 1 imply al +a2 ,PI and, similarly,
the conditions at ,P 2 , a2 E P 2 imply at +a2 ,P 2 , which contradicts the con-
dition at + a2 e I.
For s > 2, we have a = a1, ..., a s - 1 + a.t e I; moreover, the conditions
a1, oo.,a s -1 ePt n .oo nP s - 1 , as (:P 1 u... UPS-l yield a(:P I u... uPs-to
The ideal Ps is prime and at, ..., a s -1  Ps, whence at, ..., a s -1 (: Ps, and
therefore, by the condition as e Ps, we have a, Ps. Thus a E I and a, Pt U
u ... uPs, contrary to the assumption.
(iii) We have J c PI U ... uPs, whence I c JuP 1 U ... uPs, and the
statement follows from (ii). 0
Usually the following weaker version of (ii) is required.
Corollary 1.1.8
If Ie R is an ideal, P1' ..., Ps are prime ideals, and 1 c PI U ... uPs, then
I c PIc for some k, 1  k  9.

I]
Rings and Ideals
17
Let {ROC}OCEA be a family of rings. The product, n Ra.t of this family is
fleA
a ring whose elements are all functions (! defined on the set A such that e(lX) e Ra.
for all tX eA, i.e., n R« is the Cartesian product of the sets. The addition
ex
and the multiplication of functions are defined by the formulae
(e+e')(rt) = e(rt)+e'(rt), (ee')(rt) = e(rt)(!'(a), rt eA.
The homomorphism 1'lfJ: n Rf'J.  RfJ is defined by the formula
oc
n{J(e) = e{P).
Products have the following properties: for every family of homomorphisms
of rings, {fa.: R -+ R(J. }oceA, there exists a unique homomorphismsf: R  n 
ex
such that nJ = fa. This property characterizes (up to isomorphism) the ring
n R« together with the family of the homomorphisms na..
«
If the index set A is finite, A = {tXl, ..., (X,,} say, then we often write
Roc! x . . . X Ra.1I instead of n Rf'J..
oceA
The following result is known as the Chinese Remainder Theorem:
Theorem 1.1.9
If 11' ..., Is are ideals of a ring Rand IJ+lk = R for j #- k, then 11 n ... nI.,
= /1 ... Is, and the rings RI(ll n ... f"'\l s ) and RIll x ... x RIIlJ are iso-
morphic.
Proof
Suppose 9 = 2. There exist elements al ell, a2 e 1 2 such that al +a2 = 1.
The inclusion 11 f"'\1 2 ::) 1 1 1 2 is obvious. In fact, II r'\I 2 = 1 1 / 2 , since if b e / 1 n1 2 ,
then bal, ba2 E 1 1 / 2 and b = bal +ba2 e 1 1 1 2 . We define a homomorphism
of rings/: R -+ RIll X RI12 by the formulaf(x) = (x+/I' x+/ 2 ); obviously,
Ker(f) = 11r'\1 2 =1 1 1 2 . Given rl,r2eR, we have 'la2+r2al=rla2
= 1"1 a2 +1"1 al =='1 (mod 11). Hence/(rl a 2+ r 2 a l) = ('1+ 1 1' '2+ 1 2), andf
proves to be a surjection.
Suppose 9 > 2, and assume that the assertion is true for numbers smaller
than 9. Clearly,
(11 + Is) (/ 2 + Is) ... (IS-l + Is) C II ...ls-1 +Is.
Hence, since 1,,+ls = R for k = 1, 2, ..., 9-1, we have 11 ...1&-1 +Is = R,
and consequently, by the first part of the proof and the induction hypothesis
we deduce that
/ 1 ... Is-1 Is = 11 ... Is-tnls = I 1 n ... n/s_lnIs
and
RI(/ l n ... n/s)  RI(I 1 n ... n/ s -l) x RIIs  RIll x ... x RII.. 0

18
Preliminary Concepts
[Ch,
Example 1.1.10
Forevery ring R and every positive integer n > 0 (as well as for every cardfnal
number) there exists a ring of polynomials in n indeterminates Xl' ..., X n
with coefficients in R, denoted by R [Xl' ..., X n ]. These rings may be defined
by an inductive formula, R[X l ,..., Xn] = R[X l , ..., X n - l ] [XII]' where the
ring R[XIJ consists of all sequences r = (ro, r1, ...) of elements of R such that
r" = 0 for almost all k. Let s = (so, Sl, ...) be another sequence of this type;
we define the sequences r+s and rs by the formulae
(r+s)" = I'"+s,,, (rs)" = L r,sj, k = 0, 1, ...,
i+j=k
and put Xl = (0, 1,0, 0, ...). Identifying elements a E R with the sequences
(a, 0, 0, ...), we obtain the formula
r = (Yo, 1'1' ...) = rO+rl X + r 2 X2 + ... +rmX"',
where r" = 0 for k > m. The number m is called the degree of the polynomial
r, deg(r), provided rna #= O.
The ring R[X 1 ,..., Xn] is characterized by the following property: if
f: R --. T is a homomorphism of rings and t l' ..., t n E T, then there exists
a uniquely determined homomorphism f: R[X 1 , ..., Xn] --+ T such that
fi R = fand f(X ,) = t" i = 1, ..., n. Indeed, every element h of R[X l , ..., Xn]
has a unique representation in the form L ri1,...,inX11... Xn, where the sum
ranges over systems of non-negative integers i 1, ..., i". Obviously, the homo-
morphism / satisfies the condition
f(h ) = L.f(r'l.... .,,.} tl 1 ... tn,
and one can easily verify that the mapping thus defined is the homomor-
phism with the required properties.
Every sequence a 1, ..., an of elements of the ring R determines a homo-
morphism g: R[X 1 , ..., Xn] --. R which satisfies the conditions g(X,) = a"
i = 1, ..., n, glR = l R . The element g(h) of R is called the value of the poly-
nomial h at (a1, ..., an) and denoted by h(al' ..., QlI)'
Example 1.1.11
Proceeding as in the case of rings of polynomials, we define rings of (formal)
power series in n indeterminates, Xl' ..., XII' with coefficients in a ring R,
which are denoted by R[[X l , ..., XII]]' The ring R[[X 1 ]] consists of all sequences
r = (ro, r1, ...) of elements of R, the operations being defined as before. The
00
element (ro, '1' ...) is written L 1"nX.
11=0
Exercises
1. Prove that if 1, la, J are ideals of a ring R then
(i) (I+J)n+m-1 c ]n+Jm, where m, n are positive integers,
(ii) 13 ::> 12, implies (1 1 +12,)nI 3 = 1 1 nI 3 +1 2 ,

I]
Algebraic Sets and the Prime Spectrum of a Ring
19
(Hi) (II +1 2 )(1 1 +1 3 )(1 2 +1 3 ) = (1 1 +12+13)(1112+1113+1213),
(iv) 11 :1 2 1 3 == (11: 1 2 ):1 3 .
2. Show that the radical has the following properties:
(i) if I c: J, then rad(I) c rad(J),
(ii) rad(rad(I) = rad(I),
(iii) rad(I+J) == rad(rad(I)+rad(J),
(iv) rad(I) == R  I = R
(v) if rad(I)+rad(J) = R, then I+J = R.
3. Let R be a ring such that the image of the number n! by the canonical homomorphism
Z -+ R is an invertible element of R. Prove that, if I, 11, ..., I, are ideals of R, s =E; n + 1 ,
and 1 c: 1 1 V ... vI., then leI" for some k, 1  k  s.
4. Let R be a ring, let R [X] be the ring of polynomials in one indeterminate with coeffi-
cients in R, and let h = rO+rlX+ ... +rnX" e R[X]. Prove that
(i) h is nilpotent if and only if '0, ..., r n are nilpotent,
(ii) h is a zero-divisor in R [X] if and only if there exists a non-zero element s of R such
that sh = o.
s. Let R be a ring and N be its nilradical. Prove the equivalence of the following state-
ments:
(i) R contains exactly one prime ideal,
(H) any element of R is either invertible or nilpotent,
(Hi) tbe ring R/N is a field.
6. Prove that if R is a domain, a, b, c, d are non-zero elements of R, and ab = cd, then
(a):(c) == (d):(b).
7. We define the law of multiplication in the Abelian group ZfI)Q/Z by the formula
(a, q)(a', q') = (aa', aq'+a'q) for a, a' e Z, q, q' e Q/Z. Describe all ideals in the ring
thus obtained.
8. Let Rl' ..., Rn be rings. Describe all prime ideals of the ring R 1 X ... x Rn .
9. Prove that if a ring R has exactly one maximal ideal, then the characteristic of R i-
either zero or a power of a prime number. [Observe that the decomposition of the characteriss
tic into a product of two relatively prime numbers induces a decomposition of the ring R
into a product of two rings.]
1.2 ALGEBRAIC SETS AND THE PRIME SPECTRUM OF A RING
Our aim in this section is to exhibit the basic relations between the theory
of commutative rings and geometry by establishing a correspondence between
(affine) algebraic varieties and a certain class of rings.
It is this very connection with geometry and the possibility of applying
algebraic concepts and theorems to the study and classification of algebraic
varieties which contributed, as an important incentive, to the development
of the abstract theory of commutative rings.
In the second part of this section we present the essential properties of the
set of all prime ideals of a ring, called the prime spectrum of a ring. It is the
first step towards the definition of so called schemes, which are studied in
contemporary algebraic geometry.
Classical algebraic geometry dealt with the sets of solutions of systems
of polynomial equations. In order to give the basic definition of an algebraic
set, denote by K an arbitrary field and by Kn the standard affine space of
dimension n over K.

20
Preliminary Concepts
[Ch.
Definition 1.2.1
Let 11, ..., I, be elements of the polynomial ring K[X 1 , ..., X n ]. The algebraic
set determined by It., ..., I, is the set of all solutions in the field K of the system
of equations
It (t 1, ..., f n) = 0, .. · , 1 1I ( t 1, ..., tn) = 0,
(1)
i.e., the set of those (t 1, ..., I,,) E Kn for which Ji(t 1, ..., In) = 0, i = 1, ..., p
If I is the ideal generated in K[X 1 , ..., Y,,] by f1, ...,1;" then the point
I = (t 1, ..., t,.) e K" is a solution of system (1) if and only if every polynomial
f of [ vanishes at t, i.e., I(t) = 0 for all f E I. Hence the set of solutions of
system (1) depends only on the ideal I = (/1' ... ,1,); thus we shall denote
it by V(I). There is no need t take into consideration infinite sets of polynomial
equations since, as stated in Corollary 2.2.3, every ideal of the ring K[X 1 , ..., Xn]
is finitely generated.
Theorem 1.2.2
The mapping I 1-+ V(I) of the set of all ideals of K[X 1 , ..., X,,] into the set
of algebraic sets in K" has the following properties:
(i) V(O) = K", V(K[X 1 , ..., X,.]) = 0,
(ii) if [ c J, then V(l) :::> V(J),
(iii) for any family {l(l.} of ideals in K[X 1 , ..., Xn]
v (,2 IcJ = n V(la),
« ex
(iv) V(Ir.J) = V(IJ) = V(I)uV(J).
Proof
The proofs of (i), (ii) and (iii) are trivial. We shall prove (iv). Since IJ c: [(1J,
it follows in view of (ii) that V(IJ) ::) V(If1J) :::> V(1)u V(J). Hence it is sufficient
to show that V(IJ) c V(I)uV(J), Le., that t  V(I)uV(J) forces I, V(IJ).
The assumption I , V(l)u V(J) implies the existence of polynomials f, g such
that 1 e I, gEl, I(t) -:F 0, g(t) -:F O. Thus we have fg e II and (fg) (t) -:F 0,
and consequently t , V(IJ). 0
Properties (i), (iii), (iv) make it possible to establish a topology on K",
with algebraic sets as the only closed sets, which is known as the Zariski
topology.
Examples 1.2.3
(1) Ifm is a maximal ideal of K[X 1 , ..., X,.] of the form (X t -11' ..., X" -
-t,,), t = (tt, ...,1,.), then V(m) = {t}. Thus finite sets of points are closed
in the Zariski topology. They are the only closed sets in Kl except the entire
space Kl ; if the field K is infinite, Kl endowed with the Zafiski topology is not
a Hausdorff space.

I]
Algebraic Sets and the Prime Spectrum of a Ring
21
(2) Algebraic sets V(J) can be empty, even for proper ideals I; this depends
on the field K. If, for example, K is the field of real numbers, we have V«X 2 +
+ 1» = 0. If, however, K is an algebraically closed field, any maximal ideal
of K[X 1 , ...,X n ] is of the form (X l -t 1 , ...,Xn-t n ) for some teKn (see
Theorem 1.2.6). Since any ideal I is contained in some maximal ideal (see
Corollary 1.1.3), we conclude by Theorem 1.2.2 (ii), that V(I) i: 0.
(3) An algebraic set determined by one equation, f(X l , ..., X n ) = 0, is
called a hypersurface. A line (Fig. 1), or a circle and a line (Fig. 2), provide
examples of hypersurfaces in K2.
There exists also a mapping which assigns
Fig. 1
Fig. 2
to any subset in K" some ideal of the ring K[X 1 , ..., X n ]. Given E c: KR, we
write I(E) to denote the set of all polynomials equal to zero at all the points
of E. Clearly, I(E) is an ideal, and
(i) E c E' implies I(E) ::> I(E'),
(ii) if E = E 1 uE 2 , then I(E) = I(E 1 )(")/(E,).
An important connection between the operators I and V is presented
in the following:
Theorem 1.2.4
(i) If W c: KR is an algebraic set, then V(I(W» = W. Thus the mapping
E I I(E) is injective on the set of algebraic sets in Kn.
(ii) If K is an algebraically closed field and J is an ideal of the ring K[X 1 , ...
... , X n ], then I (V(J) ) = rad(J), i.e., if a polynomial f vanishes at all common
zeros of polynomials from J, then some power of f belongs to J.
Proof
(i) It follows directly from the definition that W c: V{l(W»). Suppose
that t E V(I(W)). Clearly, I(t) = 0 for any feI(W); on the other hand,
W = V(J) for some ideal J, whence I(W) :::> J because every polynomial from
J takes value zero at the points of W, while I(W) consists of all polynomials
with this property. Thus, if f(/) = 0 for any f E I(W) then f(t) = 0 for any
f E J, Le., t E W.
(ii) It is obvious that rad(J) c I (V(J) ). To prove the opposite inclusion
let us consider a non-zero elementf E 1 (V(J) ) and the ideal j of the polynomial
ring B = K[X 1 , ..., X n , X n + 1 ] generated by the ideal J of K[X 1 , ..., Xli] and.

22
Preliminary Concepts
[Ch.
the element I-X n + 1 f We claim that j = B. Assume, on the contrary, that
j -:F B so that the ideal j is contained in some maximal ideal m of B by Corol-
lary 1.1.3. By Hilbert's Nullstellensatz 1.2.6 m is of the form (Xl - aI' ..., X;,+ 1-
- a n + 1) for some ai' ..., a,. + 1 E K. This implies that all polynomials from j
vanish at the point (al'...' a n + 1), in particular 1- a,. + 1f(a1, ..., an) = o.
Since (aI' ..., an) E V(J) we get l(a1, ..., an) = 0 which leads to a contradic-
tion.
From the equality j = B it follows that there exist polynomials h 1 , ..., "m,
h e B, g 1, ..., g m E J such that
h 1 g 1 + ... +h m g m +h(1-X n + 1 f) = 1.
Denote by rp a homomorphism B --. K(X 1 , ..., X..) of K-algebras such that
rp(X i ) = Xi for i = 1, ..., n, and tp(X n + 1) = 11f, where K(X 1 , ..., Xn) is a
field of fractions of K[X 1 , ..., Xn]. Applying tp to the above equality we get
hi (Xl' ..., XII' 1 t.n g 1 + ... + h m (Xl' ..., XII' 1 t.n g m = 1
and consequently, multiplying by sufficiently high power of f, we obtain that
f E rad(J). 0
orollary 1.2.5
If Kis an algebraically closed field then there exists a one-to-one correspondence
between algebraic sets contained in K n and radical ideals of the ring K[X., ...
... , Xn] (Le., ideals J such that rad(J) = J) determined by the mappings
WH I(W) and JH V(J).
The proof follows immediately from Theorem 1.2.4 because, for any
subset E c Kn, l(E) is a radical ideal.
Theorem 1.2.6 (Hilbert's Nullstellensatz)
Let K be an algebraically closed field. An ideal m of the polynomial ring
K[X 1 , ..., Xn] is maximal if and only if there exist a 1, ..., an e K such that
m = (Xl -ai' ..., Xn-an).
Proof
An ideal of the form (Xl - at, ..., X,. - an) is the kernel of the K-algebra
homomorphism of K[X l , ..., Xn] onto the field K sending the indeterminate
X, to a" i = 1, ..., n, hence it is maximal.
Assume now that m is a maximal ideal of K[X 1 , ..., X n ]. Then the finitely
generated K-algebra K[X 1 , ..., XnJ/m = K[X1' ..., xn] (where Xl' ..., x,. are
residue classes of Xl' ..., X n modulo m, respectively) is a field containing K.
It is enough to prove that all the Xi are algebraic over K. Indeed, this implies
that Xl' ..., X n E K since K is algebraically closed, and consequently m = (Xl -
-Xl' ..., X" -X n ).

I]
Algebraic Sets and the Prime Spectrum of a Ring
23
We are going to prove the following more general result.
(*) If L is a field and a finitely generated L-algebra A = L [y 1, ..., YII]
is a field, then the elements Y l' ..., Yn are algebraic over L.
We proceed by induction on n. For n = lour claim follows from the fact
that if Yl is not algebraic, it is not invertible in L[Yl].
Suppose that (*) holds for arbitrary fields L and all numbers less than n,
and assume that at least one among elements Yl, ..., Y.., say Y1' is not algebraic
over L. Then A contains a subfield of rational functions L(Y1). Since A = L(Yl)
[Y2, ..., Yn] our induction hypothesis implies that the elements Y2, ..., YII are
algebraic over L(Yl). Hence there exists a non-zero polynomial W E L[Yt]
such that each of the elements WY2, ..., }VYn is integral over L[Yl], i.e., satisfies
an equality of the form
t m + / 111-1 + + - 0
gl1l-1 ... go-
for some grn-l, ..., go E L[Yl].
We shall prove in Chapter III, Theorem 3.1.7, that the set of all elements
tEA which are integral over L[Yl] is a subring of A. It follows from this that
for any a E A there exists a natural number q such that Wi a is integral over
L [Yl]. Indeed, any a E A is a sum of monomials in Y2, ..., Yn with coefficients
in L[Yl] and for a monomial b = uy2 ... y", U E L[Y1] an element
w'b = U(WY2)i 2 ... (wYn)in, where p = ;2 + ... +;.., is integral over L[y.].
Now let h E L[Yl] be a non-constant polynomial relatively prime to IV.
Applying what was said above to the element I/h E A we conclude that there
exists q such that t = }yq/h satisfies an equation of the form
(}yll/h)"'+gm_t(w"/h)11I-1 + ... +go = O.
Multiplying this equality by h m we easily deduce that h divides w. contrary
to the assumption that hand ware relatively prime. This contradiction com-
pletes the proof of (*). 0
Geometrical intuition allows us to discern a substantial difference between
the hypersurfaces represented in Figs. 1 and 2. The second one can be repre-
ted as the union of two proper algebraic sets (a circle and a line). On the other
senhand, it seems impossible to do the same with the algebraic set in
Fig. 1. These simple observations result in the following:
Definition 1.2.7
An algebraic set W is said to be reducible if it can be represented as a union
of two algebraic sets, both different from W. Otherwise we call Wan irreducible
algebraic et, or an algebraic variety.
Theorem 1.2.8
An algebraic set W is irreducible if and only if l( W) is a prime ideal.

24
Preliminary Concepts
[Ch.
Proof
Let the algebraic set W be irreducible, and suppose fg e /( W). Since (V(f)r.
f'\ W)u (V(g)r. W) is a representation of W in the form of the union of two
algebraic sets, one of them, say the first one, is equal to W. Thus W c V(f),
i.e.,fis equal to zero at every point of the set W, whencefe/(W), and con-
sequently I( W) is prime.
To prove the opposite implication, assume that I(W) is a prime ideal,
and let W = W 1 U W 2 be the representation of W in the form of the sum
of two algebraic sets, W 1 and W 2 . It follows (see property (ii) preceding
Theorem 1.2.4) that I(W) = I(W 1 )f'\I(W 2 ). Now, since the ideall(W) is prime,
we infer by Theorem 1.1.7 (i) that for instance, I(W 1 ) c I(W), whence W
c W 1 . Thus W = WI' and the set W is irreducible. 0
Any polynomial g E K[X 1 , ..., XII] determines the polynomial function
g/: Kn --+ K given by the formula g'(t) = g(/ 1 , ..., t ll ) with t = (t 1 , ..., I,,)
E K" (different polynomials may determine the same polynomial function, for
instance in the case of a finite field K). Its restriction to an algebraic set W is by
definition a polynomial function on W. All polynomial functions defined on W
form a ring K[ W], called the ring of polynomial functions of an algebraic set
W (or the doe coordinate ring of an algebraic set W). Since I( W) is the ideal
of those polynomials in K[X 1 , ..., XII] which vanish on W, we have K[W]
 K[X I , ..., X,.]/I(W). If W = V(l) for some ideall, then, by Theorem 1.2.4
(ii), we have K[W]  K[X 1 , ..., XII]/rad(I). In particular, if P is a prime ideal,
then K[V(P)]  K[X 1 , ..., Xn]f P is a domain. In the sequel, we shall identify
the rings K[W] and K[X 1 , ..., Xn]fI(W).
As we already know (see Example 1.2.3), if a maximal ideal m has the form
(Xl - t l' ..., X n - tn) for some t 1, ..., t n e K, the set V(m) consists of one
point t = (t 1 , ..., tn) e Kn. In the case where K is an algebraically closed field,
every maximal ideal of K[X 1 , ..., ,] has this form (see Theorem 1.2.6); thus
the mapping m i-+ V(m) establishes a one-to-one correspondence between
all maximal ideals of the ring K[X 1 , ..., XII] and all points of the space Kn.
For every algebraic set W c Kn, the mapping m H- V(m) induces a one-to-one
correspondence between points of the set Wand maximal ideals of the
ring of polynomial functions K[W] since V(m) c V(I) if and only if
Iem.
Thus we have replaced a' geometrical object, namely the algebraic set W,
with an object which is fully describable in terms of the ring K[W], Le., with
the set of maximal ideals of K[W]. Yet, the knowledge of the points of an
algebraic set W is not the only source of information about W. In various
geometrical problems an important part is played by the algebraic varieties
contained in W, and their mutual relations (for example, the way they intersect
one another). Let W = V(/); to any algebraic variety W' contained in Wthere
corresponds a prime ideal P' = [(W'); obviously P' :::> rad(/). Thus such
a variety, Le., W', determines a prime ideal P' frad(l) of the ring of polynomial

I]
Algebraic Sets and the Prime Speetrum of a Ring
2S
functions K[W] = K[X 1 , ..., XIIJ/rad(I). The converse is also true: any prime
ideal P'/rad(I) in K[W] determines a variety W' = V(P') contained in W.
To sum up, the set of all prime ideals of the ring K[W] and the set of all
algebraic subvarieties of the algebraic set W, are in one-to-one correspondence
with one another and, consequently, the latter acquires a purely algebraic
interpretation. Moreover, it can be endowed with a topology which induces
the Zariski topology in the set of maximal ideals of K[W] (identified with W
by the mapping m i-+ V(m»). In defining this topology we make no use of any
particular properties of the ring K[W]; thus, from the algebraic point of view,
it would be quite natural to replace K[W] by an arbitrary ring. In order to
make effective use of the notions in question, we adopt the following notation
and definitions.
DefiDition t .2.9
The prime speetrum (or, briefly, the spectrum) of a ring R is the set Spec(R)
of all prime ideals of R. The subset Max(R) c Spec(R) consisting of all
maximal ideals of R is called the maximal spectrum of R.
A homomorphism of rings!: R --. T induces the mapping
Spec(f): Spec(T) --. Spec(R)
given by the formula Spec(f) (Q) = f-l(Q) for Q E Spec(T).
The ideal f-l(Q) is indeed prime since the ring R/f-l(Q) is isomorphic
to a subring of T/Q and therefore is a domain.
If g: T -+ T' is also a ring homomorphism, then Spec(gf) = Spec(f)
Spec (g), and Spec(lR) = lspec(R). Hence Spec is a contravariant functor
defined on the category of rings with values in the category of sets.
We now proceed to the establishing of a topology on the set Spec(R).
For any set B c R, we define
V(B) = {P E Spec(R); P ::> B}.
If I is the ideal generated by B, then, clearly, V(B) = V(I) = V(rad(I)). It
follows from Theorem 1.1.7 (i) that, for ideals II' ..., I k , we have
V(/ 1 )u ... u V(I,,) = Vel! n ... nIt).
Furthermore,
VCR) = 0, V(O) = Spec(R), n V(B«) = V (U B«),
«eA «eA
where {Bcx }«EA is any family of subsets of R. Thus the family of subsets of
Spec(R) which are of the form V(B) is closed with respect to any intersections
and finite unions, contains the empty set and the entire Spec(R). There exists,
therefore, a unique topology of the set Spec(R) in which closed sets are of the
forln V(B) for B c R.

26
Preliminary Concepts
[Ch.
Definition 1.2.10
The set Spec(R) of all prime ideals of a ring R, together with the family
{V(B)}Bc:R as the class of closed sets, is a topological space. We denote it also
by Spec(R), and call it the (prime) spectrum of the ring R. The topology of
Spec(R) is called the Zariski topology.
The mapping Spec(f): Spec(T) --. Spec(R), which is induced by a homo-
morphism of rings f: R -+ T, is a continuous function. Indeed, given a set
B c R, we have
(Spec(f) )-1 (V(B») = {Q E Spec(T); Spec (n (Q) E. V(B) }
= {Q e Spec(T); f-l(Q) e V(B)}
= {Q E Spec(T); j-l(Q) :::> B}
= {Q E Spec(T); Q :::> 1(.8)} = V (f(B) ).
We may thus consider Spec as a contravariant functor from the category
of rings to the category of topological spaces.
The next theorem gives an important property of the topological space
Spec(R).
1lheoremm 1.2.11
The space Spec(R) is quasi-compact, i.e., for any family of open sets {U«}OCEtt
such that U Ury. = Spec(R), there exist elements CXl'. . . , (Xn E A such that
«sA
n
Spec(R) = U Ua.,. The space Spec(R) is a Hausdorff space if and only if
i=1
Spec(R) = Max(R).
Proof
Given any element r E R, we write D(r) to denote the set Spec(R)"V(r)
= {P E Spec(R); r, P}. Clearly, the sets D(r) are open and, for any set
B c: R, we have Spec(R),,",V(B) = U D(r). Thus the family of open sets
reS
{D(r) },eR is a base of open sets in the space Spec(R).
Without loss of generality, we may assume that U« = D(rry.), with some
ra. e R. By the assumption, for every ideal P e Spec(R), there exists an element
(X e A such that P E D(r«), Le., such that ra.  P. Thus the ideal generated by all
the elements 1"« is equal to R. Hence there exist elements CXl, ..., CX" E A and
Sl, ..., Sft e R such that slr OCl + ... +snra.n = 1, and consequently R = Rr«l +
+ ... + Rr«II' whence it follows that V(r OC1 )n ... f1 V(r ocn ) = V(R) = 0. This,
however, means that Spec(R) = U«l U ... U U«,., i.e., the space Spec(R) is
quasi-compact.
To prove the latter part of the theorem, observe that prime ideals P, Q
e Spec(R) and an element r E R satisfy the conditions P e D(r), Q, D(r)
if and only if r rf: P, r E Q. Hence it is easy to verify that the hypothesis of
Spec(R) being a Hausdorff space is equivalent to the following condition:
Q c P forces Q = P for any two elements P, Q E Spec(R), which implies
Spec(R) = Max (R). 0

I]
Algebraic Sets and the Prime Spectrum of a Ring
27
The set Spec(R) is endowed in a natural way with an order relation defined
by inclusion. For a chain of distinct prime ideals Po c: Pt c... c: P d the
number d is called the length of the chain.
We define now an important invariant of a prin1e ideal in terms of the
inclusion relation.
Definition 1.2.12
The height ht(P) of a prime ideal P is the upper bound of the lengths of
chains of distinct prime ideals
Po c Pt c: ... c P
descending from P.
Ideals of height zero in R are exactly minimal prime ideals of R.
The fundamental property of height is contained in the Krull theorem which
implies that in a Noetherian ring (this class of rings is studied in Chapter II)
a prime ideal generated by n elements has height not greater than n (for a proof
see [B], Chapter I). In particular a principal prime ideal is of height  1. Since
any ideal of a Noetherian ring is finitely generated, Krull theorem implies
that height of any prime ideal in such a ring is finite. .
Exercises
1. Show that W = {(a, a 2 , a 3 ); a E C} c C3 is an irreducible algebraic set. Find a set
of generators of the ideall(W), and prove that the ring C[W] is isomorphic to the ring of
polynomials in one indeterminate over the field C.
2. Let W be an algebraic set in C., determined by the polynomials XI- X 2 YJ ,
Xl X 3 - Xl. Prove that W is a union of three algebraic varieties and find the prime ideals
which correspond to them.
3. Let K be a field. Prove that a K-algebra A is isomorphic to an algebra of polynomial
functions on a certain algebraic set in K", for some n, if and only if A is a finitely gener-
ated K-algebra without nilpotent elements.
4. Let K be a field. Show that the closure If of a set E c Kft in the Zariski topology
is equal to V{I(E»).
s. Let R be a ring, P, Q e Spec(R), and P $ Q. Prove that there exist Po, Pie Spec(R)
which satisfy the following two conditions: (i) P C Po $ P 1 C Q, (ii) there is no P' E
Spec(R) such that Po * pi  Pl.
6. Prove that an ideal 1 of a ring R is an intersection of a family of prime ideals if
and only if rad(I) = I.
7. Prove that a space Spec(R) is connected if and only if the elements 0 and 1 are
the only idempotents in R (i.e., elements e e R such that e Z = e).
8. Let X be a compact topological space. Denote by C(X) the ring of all real-valued
continuous functions on X. Prove that every maximal ideal of the ring C(X) has the fornl
mxo = {Ie C(X); f(xo) = O}.
for some point Xo eX. Show that the correspondence x t-+ m determines a homeoo10r..
phism between the spaces X and Max(C(X»).

28
Prelimioary Concepts
[Ch.
1.3 MODULES
Let R be a ring. An R-module is by definition a triplet (M, +, .) where M
is a non-empty set, +: M x M -. M, · : R x M -. M, the following condi-
tions. being satisfied for all elements r, s e Rand m1' m2 EM:
(i) M is an Abelian group with respect to the operation +,
(ii) r(m1 +m2) = rml +rm2'
(iii) (r+s)m1 = rm1 +sml,
(iv) r(sm1) = (rs)m1'
(v) Im1 = m1.
The ring R is called the coefficient ring of the module M. For simplicity,
we say that M is an R-module, or, if there is no doubt about the coefficient
ring, that M is a module. In the case where R is a field, the notion of R-module
coincides with the well-known notion of linear space over R.
A mapping of R-modules, f: M  N, is called a homomorphism (or an
R-homomorphism) provided it satisfies the condition f(rml +sm2) = rf(m1)+
+sf{m2) for all r, s e R, m1, m2 e M. The set of all R-homomorphisms
f: M -+ N is denoted by HomR(M, N).
For a fixed ring R, all R-modules, tgether with the homomorphisms and
their usual composition, form the category of R-modules, R-Mod. A homo-
morphism f: M -. N is an isomorphism if there exists a homomorphism
g: N -+ M such thatgf = IM,fg = IN; in this case we write M  N. A homo-
morphismfis an isomorphism precisely if it is both an injective and a surjecti-
ve mapping onto N.
A subset Mo of an R-module M is said to be a submodule if Mo is a sub-
group of the group M and the conditions r E R, m E M 0 imply rm E Mo.
To a submodule M 0 of a module M we can assign a factor module M / M 0
whose elements are all residue classes m = nz+Mo = {m+x; x E Mo} of Mo,
where m ranges over M, and the operations satisfy the conditions
(m1 +M o )+( ln 2+ M O) = (m! +m2)+M o , r(ml +Mo) = rm1 +Mo.
The mapping v: M -+ M/Mo given by the formula v(m) = m+Mo is a homo-
morphism, called the natural homomorphism. It is easy to observe that associ-
atingwith a submodule M' ,:::> Mo of Mthe sub module v (M') = M' /Mo = {m+
+Mo; m EM'} of M/Mo we obtain a one-to-one correspondence between
the set of those sub modules of M which contain Mo and the set of all sub-
modules of M/Mo.
For any homomorphism of tnodules f: M -. N, its kernel Ker(f) = {m
EM; f(m) = O} is a submodule of M and its image Im(f) = If(m) EN; m
EM} is a submodule of N, isomorphic to the factor module M/Ker(f). If
M' c: M is a submodule, f(M') is a submodule of N; if N' is a submodule
of N, j-1(N') is a sub module of M. The factor module N/Im(f) is called the
cokernel of the homomorphism f, written Coker(j). A homomorphism
f: M -+ N is called a monomorphism or an injection, Le., if Ker(f) = o.

I]
Modules
29
A homomorphism f is called an epimorphism (or a surjection) if Im(f) = N.
The embedding mapping of a submodule M' c M into the module M (Le.,
the mapping i: M' -+ M given by the formula i(m) = m for m EM') is denoted
by M' ( M.
A sequence of modules and homomorphisms
1 ,,-1 I " I n+1
. .. --.. M" -1 -=----+ Mn _0) M"+ 1 -_. ...
is by definition an exact sequence if and only if Ker(f") = Im(f'.-1) for all n.
In particular, a sequence 0 -+ M'  M !.... M" -+ 0 is an exact sequence
if and only if C( is a monomorphism, fJ is an epimorphism, and Ker({J) = Im( lX).
Such a sequence is called a short exact sequence.
Every element m eM determines a cyclic submodule, {rm; r E R}, written
Rm. Similarly, for every ideal I in R and every element m e M, we denote
by 1m the set {rm; reI}, which is a submodule of the module M. For any
subset A c M, the smallest submodule of M containing A consists of all ele-
ments of the form.rlat + ... +rna", where a1, ..., an E A and rt, ..., r.. E R;
it is the intersection of all the submodules of M which contain the set A.
The submodule generated by a finite set {a l' ..., ak} is denoted by Ra 1 + ... +
+ Rak and called a finitely generated submodule.
A ring R is an R-module with respect to ordinary multiplication; its sub-
modules are ideals.
If M is an R-module, then, for every element m eM, there exists a unique
homomorphism of R-modules, 1m: R -+ M, such that fm(l) = m; clearly,
Im(r) = rm for r e R. The kernel of this homomorphism is called the annihilator
of an element nz and denoted by Ann (m). We thus have an isomorphism
Rm  RIAnn(m). In the general case, if B is a subset of a module M, then, by
definition, Ann(B) = {r E R; rb = 0 for all b E B}; it is an ideal of the ring R,
which is called the annihilator of the set B.
A module M is said to be simple if 0 and M are the only submodules of M,
The module M is then generated by any of its non-zero elements; the annihilator
of every non-zero element is identical with the annihilator of the entire module,
and it is a maximal ideal. Thus non-zero simple R-moduJes have the form
RIm, where m is a maximal ideal.
If {MexlxEA is a family of sub modules of a module M, then the module
generated by this family (more exactly, by the union of the submodules Mex)
consists of elements of the form m ex1 + ... +mex", where mex, E Mex" i = 1, ..., IJ.
We denote this submodule by L Mex, and in the case of a finite family {M 1 , ...
ex
... , M,,} also by M 1 + ... + M n . The intersection n Mex of any family of
ex
sub modules is obviously a submodule.
Let I: T -+ R be a ring homomorphism and M be an R-module. Defining
the multiplication by elements t E T by the formula tm = f(t)m for m E M,
we turn M into a T-module. This procedure is called a change of coefficients
by means of the homomorphism f.

30
PreUminary Concepts
[Ch.
Given an ideal I c: R, the sublnodule 1M of an R-module M is by definition
the submodule generated by all the elements of the form am, where a E I,
m E M, Le., the submodule which consists of sums of elements of the form am.
If I c Ann(M), then 1M = 0, and M becomes an R/I-module with respect
to the multiplication (r+J)m = rm. The change of coefficients by means
of the natural homomorphism R  R/l leads to the original R-module M.
In particular, for any ideal I c R and any R-module M, the factor R-module
M/IM satisfies the condition I(M/IM) = 0 and therefore may be regarded
as an Rjl-module.
Let M be an R-module, M 0, M t submodules of M, and I c: R an ideal.
The following quotients can be constructed: the ideal M 0: M 1 = {r e R; r M 1
C Mo} and the submodule Mo:l = {m eM; 1m c: M o }. We have Ann(M o )
= 0: M 0, where 0 denotes the zero submodule.
Theorem 1.3.1
Let Mo, Ml' M 2 be sub modules of a fixed R-module M. Then the following
are satisfied:
(i) if Mo c M 1 C M 2 then (M2/Mo)/(Ml/Mo)  M 2 /M 1 ,
(H) (Mo+Ml)/Mt  Mo/(Mo t1M t),
(Hi) if M 1 c M2' then (Mo+Ml)t1M2 = (MonM2)+Mt.
We leave the proof to the reader as a simple exercise.
An element a of a ring R is said to be a zero-divisor on an R-mod-
ole M if there exists a non-zero element m e M such that am = 0, i.e., if
a E U Ann(m). For such an element a, the homomorphism M  M
O:l:me¥
which carries an element m E M to the element am is not a monomorphism.
The set of all zero-divisors on M is denoted by a(M).
A module M. is by definition the direct sum (or the coproduct) of a family
of its submodules {MfA }cxeA provided every non-zero element m E M has a unique
representation in the form m = mcx J + ... +m cxll , where eXt,..., an E A are
pairwise different and m cx , are non-zero elements of Mcx" i = 1, ..., n. In this
case we write M = E9 Ma.. It is easily seen that, for every R-module Nand
cxe..4
every family of homomorphisms fa.: Mcx -. N, there exists a unique homo-
morphism f: E9 Mcx -+ N such thatflMfA = lex.
cxeA
If {Mcx}cxeA is a family of R-modules which are not supposed to be sub-
modules of any module, then there exists a module M which is the direct
sum (the coproduct) of this family, i.e., the direct sum of the family of its sub-
modules {M} such that Mcx  M. We can build the module M as the set of
functions p, defined on A and such that p,(a) E Mcx for a E A, #(a) = 0 for
almost all a EA. We define the operations +, · as usual. The following condi-
tions characterize (up to isomorphism) the module M: there exist homo-
morphisms Wa.: Ma. -. M, a E A, such that for every family of homomorphisms

I]
Modules
31
fa.: M(J. -+ N there exists exactly one homomorphism f: M -+ N such that
fw« = fr,. for all (X E A. Clearly, we have M = EB Im(w lX ).
lXeA
The product of a family {M«}«eA of R-modules is defined dually as the
module n MrJ. which consists of all the functions p defined on A such that
cceA
p,(a.) e M« for a. EA. We define the operations +, · as usual. The homo-
morphisms nfJ: n MtX -+ MfJ, called projections are defined by the formulae
tXeA
n{J(p) = #(P). The module M = n MIX can be characterized (up to iso-
«eA
morphism) by the following conditions: there exist homomorphisms n tX : M
-4 M«, eX E A, such that, for every family of homomorphisms f: N -+. MtX,
there exists exactly one homomorphism I': N -+ M such that nrt.f' = f
for every ex E A.
The direct sum of a finite family of modules M 1, ..., M" is isomorphic
to the direct product of this family and is denoted also by M I EB ... E9 M k ,
its elements being written as sequences (ml' ..., m,,), with m, e M" i = 1, ...
. . . , k. A submodule M 0 of a module M is said to be a direct summand of M
provided there exists a submodule M 1 of M such that M = M 0 EB MI. This
condition is equivalent to the following one: there exists a homomorphism
e: M -+ M such that e 2 = e and Im(e) = Mo. We say that an exact sequence
o -+ M '  M !.+- Mil -+ 0 splits if Im(ex) is a direct summand of M. This is
equivalent to each of the following conditions:
(a) there exists a homomorphism y: M -. M' such that yex = 1 M "
(b) there exists a homomorphism : M" -+ M such that P = I M ".
A subset B of an R-module M is called a basis of M if and only if B gener-
ates M, and the condition rl b 1 + ... + r,.b n = 0, where b l , ..., b n are pairwise
different elements of Band r I, ..., r n E R, implies r I = ... = r" = o. Under
the above condition, every non-zero element of the module M has a unique
representation of the form rlb l + ... +rnb". A module M is said to be free
if there is a basis .of M. The reader can easily prove that the following conditions
are equivalent: .
(i) a set B is a basis of M,
(ii) for every R-module N and every function/: B -+ N, there exists exactly
one homomorphism f: M -. N such that fiB = f,
(Hi) M =  Rb and Ann(b) = 0 for all b E B.
beB
For every set X, there exists a free R-module with X as a basis. Hence,
for every R-module N, there exists a free R-module F and an epimorphism
F -. N.
If R is a field, there exists a basis of every R-moduIe; thus, every R-module
is free. A basis of a free module is not uniquely determined; however, every
two bases of a given module have the same cardinality (this is not true in the
case of noncom mutative rings). Indeed, suppose that B is a basis of a free

32
PreUminary Concepts
[Ch.
module F and meR is a maximal ideal; it is easily seen that the residue classes
b +mF E FlmF, b e B, form a basis of FlmF regarded as an RIm-module.
Now, the ring RIm is a field, whence the cardinality of the set B is equal to the
dimension of the linear space FlmF over the field RIm and consequently does
not depend on the choice of a basis.
The rank of a free module F, written rk(F), is by definition the number
of elements of any basis of F.
An R-module M is called projective if and only if M is a direct summand
of a free R-module. It is easy to see that M is projective if any diagram with
an exact row
"M
JI/// !
N - ..- N" . 0
can be completed to a commutative diagram by a homomorphism - - -+.
There is a notion dual to projective: a module M is called injective if and
only if any diagram with an exact row
o . N' .. N
! ;(////
M
can be completed to a commutative diagram by a homomorphism - - - .
Lemma 1.3.2
If an R-module M is generated by n elements, 1 c R is an ideal, r e R, and the
condition r M c 1M is satisfied, then there exist elements ao, ..., a n - 1 E 1
such that
,... + a n -l ,...-1 + ... + ao e Ann( M) .
In particular, if 1M = M, then there exists an element a e I such that
1 +a E Ann(M).
Proof
Let m 1, ..., m n be generators of M. By the assumption, each element r1ni
n h
can be written in the form rm, = L a'l m} where al) e 1, whence L (,)r-
J=1 J=l
- au) m J = 0 for i = 1, ..., n. Let us denote the cofactor of the ij-th element
of the nxn matrix (,}r-a,}) by A,) and its determinant by A. Then, for
k = 1, ..., n, we have

I]
Modules
33
o = LA 1t L (dljr-alj)mj = L (L A,k(dljl'-a,j))m j
; j ) ;
= L dtJAmj = Amt,
j
whence A E Ann(M). Using basic properties of determinants, we deduce that
A = r n +a"-1 r ll - 1 + ... +ao for some 00, ..., 0..-1 E I.
If M = 1M, we set r = 1, a = On-1 + ... +ao, and obtain 1 +a E Ann(M),
the desired result. D
Lemma 1.3.3. (Nakayama)
If M is a finitely generated R-module, I c R is an ideal contained in the
Jacobson radical J(R) and 1M = M, then M = O.
Proof
It follows from Lemma 1.3.2 that there exists an element a E I c J(R) such
that (1 +a)M = O. By Theorem 1.1.6 the Iement 1 +a is invertible, i.e.,
b(l-f-a) = 1 for some b ERe Thus M = 1M = b(l+a)M = O. 0
A sequence
Mo c: M 1 C ... C M k
(2)
of distinct submodules of a module M is called an ascending chain of sDbmod-
ules; the nunlber k is called the length of chain (2). A chain which is obtained
by adjoining extra submodules is called a refinement of the original cbain.
A chain which has no proper refinement is called a composition chain (or
a composition series). The chain (2) is a composition chain (or series) if and
only if Mo = 0, M k = M, and the modules M'+l/Mi are simple, i = 0, 1, ...
...,k-l.
Not all modules have composition chains; for instance, among Z-modules,
only finite ones have them. The conditions for the existence of composition
chains will be given in Section 2.8.
We shall prove now the fundamental theorem on chains of submodules:
Theorem 1.3.4 (Schreier)
Let
Mo C M 1 C ... C M k ,
No C N 1 C ... c: N,..
(3)
(4)
be chains of submodules of a module M, and Mo = No = 0, M" = N m = M.
Then there exist refinements
M c: M c ... c M,
N c: N c ... c N;
(3')
(4')
of chains (3), (4) and a permutation f1 of the set {I, 2, ..., s} such that
M;/M;_l  N;(p)/N;(p)_l, p = 1, 2, ..., s.

34
Pr eJjotln ary Concepts
[Chi
Proof
Write
At) = M,+M'+lnN J , B'J = N j +M,nN J + 1
for i = 0, 1, . I . , k - 1, j = 0, 1, ..., 111- 1; thus we have
M, = Ai,o c: A t ,l c: ... c: A'.m-l c: A'm = M'+l,
N J = B o . J C B l . J c: ... c: B k - 1 ,J c: B kJ = N J + 1 .
Using Theorem 1.3.1, we obtain
A',J+l/ A ij = (M,+M,+lf1Nj+l)/(M,+M,+lnNJ)
= (M, + M'+l nN J + M'+l nN J + l )/(M, + Mi+l nN J )
 (M'+lf1N)+l)/(M,+M,+lnNj)nM,+lnNj+l
= (M'+l nN J+l)/(M,nM,+l nN J+l +M'+lnN J )
= (M'+lnNj+l)/(M,f1+l +M'+lnN J ).
SimiJar1y, we prove that
B , + l . J /B'J  (M'+l nN J+l)/(M,+l n Nj + M,nN J + l ).
It follows that A',J+l/A'J  B'+l,J/B'j, and consequently we see that the
refinements obtained by adjoining the modules A iJ, i = 0, ..., k - 1, j = 1, ...
..., m-l, and B'J' i = 1, ..., k-l, j = 0, ..., m-l, respectively, have the
desired properties. 0
Corollary 1.3.5 (Jordan-Holder)
If, in a module M, there exists a composition chain, then every chain of sub.
modules of M can be refined so as to become a composition chain. The
lengths of all composition chains of the module M are equal.
Definition 1.3.6
If, in a module M, there exists a composition chain, its length is called the
length of the module M, and denoted by 1(M). If there are no composition
chains in M, we write 1(M) = 00.
Thus the number /(M) is the upper bound of the lengths of chains of sub..
modules of M. In the case where R is a field and M is finitely generated, I(M)
is equal to the dimension of M.
Theorem 1.3.7
If a sequence of R-modules 0 -+ M' -+ M -+ Mil -+ 0 is exact, then /(M)
= I(M') + l(M").
Proof
Write «: M' -. M, p: M -+ Mil for the mappings in the exact sequence under
consideration. Let
O M il M il M il
= 0 c ... C k = ,
o = M C ... c: MI = M'
(5)

I]
Modules
35
be chains of submodules of M" and M', respectively. Then
o = cx(Me» c: ... c cx(M) c P-1(M') c: ... c fJ-1(lvJ') = M (6)
is a chain of sub modules of M, its length being equal to k+m. Thus it follows
that 1(M) = 00 in the case where either I(M') = 00 or I(M") = 00. If I(M')
< 00 and I(M") < 00 and (5) are composition chains, then so is chain (6),
and, consequently, 1(M) = I(M') + I(M"). 0
Example 1.3.8 .
Let L be a linear space over a field K, and rp: L -+ L be a linear mapping.
The space L becomes a K[X]-module, where K[X] is the ring of polynomials
in the indeterminate X if, for a polynomial f = ao + al X + ... + QkX" and an
element Y E L, we set
fy = °01+°1 ,(y) + ... +atrp"(y).
Every K[X]-module can thus be obtained from the linear mapping given by the
formula 9'(y) = Xy. Now, suppose that the linear mappings rpl, ..., ffJ,. of the
space L satisfy the condition l/li rpJ = rpJ 9'" j, j = 1, ..., n; proceeding as before,
we may turn the space L into a module with K[X 1 , ..., .] as its coefficient ring.
This is the way in which all K[X 1 , ..., Xn]-modules can be obtained. In the
case where the mappings do not satisfy the condition of commutativity a poly-
nomial ring in noncommuting indeterminates should be employed.
Similarly, any endomorphism rp: M -+ M of an R-module endows the
module M with the structure of an R[X]-module.
Exercises
1. Let I, lex be ideals of a ring R and let N, Np be submodules of an R-module M. Prove
the following formulae:
(Ll ex ) (LNIJ) =: L1exNp,
ex fJ ex, fJ
(n Np ):1 = n (Np :1),
fJ {J
(nNp):N= n (N{J:N),
fJ p
N: (L loe) = n (N:I«),
ex ex
N: (L Np) = n (N:NfJ).
{J {J
2. Let 0 -+ M' -+ M -+ Mil  0, be an exact sequence of modules. Prove that, if the
module M" is projective, then the sequence splits.
3. Let 0 -+ M' -+ M -+ Mil -+ 0 be an exact sequence of modules. Prove that, if the
modules M', Mil are finitely generated, then so is M. Does the opposite implication hold?
4. Prove that, if a module M is projective, then there exists a free module F such that
M63F F.
5. Prove that, if an ideall c R is finitely generated and 1 2 = I, then I = Re, where e
satisfies the condition e 2 = e (i.e., is an idempotent element).

36
" Preliminary Concepts
[Ch.
6. Let I Co R be a finitely generated ideal of a ring R. Prove that, if the R-module Ill'}.
is generated by m elements, then 1 is generated by nl1-1 elements. [Show that, if the residue
classes of 1 2 determined by the elements r1, ..., "m generate the module 111 2 , then the ideal
J = 1/(r1, ..., r m ) satisfies the condition J2 = J.]
7. Prove that, for any elements r, s of a ring R, there exist isomorphisms of R-moduIes:
«r):(s»)/Ann(s)  «s):(r»)/Ann(r),
«r):(s»/(r)+Ann(s)  «s):(r»)/«s)+Ann(r»),
R/«r):(s»)  (r, s)/(,.).
8. Prove that, if a ring R has a unique maximal ideal m (i.e., R is a quasi-local ring,
cr. Definition 1.4.4), then any finitely generated projective R-module M is free. [Construct
a homomorphism f:F M such that the induced homomorphism FlmF MlmM is an
isomorphism, then apply the Nakayama Lemma 1.3.3 to Coker(f) and Ker{f)].
9. A module which is a direct SUln of simple modules is called a semi-simple module.
Prove that:
(i) a module is semi-simple if and only if each of its submodules is a direct sumn1and,
(ii) every factor module of a semi-simple module is also a semi-simple module.
Describe those rings R which, regarded as R-modules, are sen1i-simple. [Given a sub-
module N of the direct sum EB Mrx. of simple modules Ma., consider a maximal subset Ao
oceA
such that Nn Ea Ma. = 0.]
oceA o
10. In the ring of all real-valued continuous functions defined on the interval [0, 1],
we choose a subring R which consists of all functions f such thatf(O) = 1(1), and an R-module
M which consists of all functions f such that 1(0) = -/(1). Prove that MEaM  REa R,
and M is not a free R-module. [Hint: find a two-element set of generators of M.]
1.4 LOCALIZATION OF RINGS AND MODULES
The reader familiar with elementary algebra should remember that any domain
R can be embedded in its field of fractions Ro. Each element of the field Ro
has the form I' Is, where 1', S E R, s #= 0, and is the only solution of the equation
sx = r. The operations + and · are given by the formulae
r/9+r1/ s 1 = ( rs l +1'1 S)/ 8S 1, (r/s) (rl/Sl) = "'l/SS1- (1)
The elements 'l/S1, '2/92 are equal if and only if rl S2 = r2s1. An element
I'/s of the field Ro is defined as the equivalence class of the pair (r, s) with
respect to the equivalence relation given by the formula
(r, s) = (1'1' SI) <=> 1' 9 1 = "1 J' .
From formulae (7) it can easily be inferred that a set S c R, closed with
respect to multiplication and such that 0  S, determines a subring Rs of the
field Ro which consists of all elements rls such that S e S. Rs is the smallest
ring containing R in which all the elements S E S are invertible. Our aim is to
show that, proceeding similarly, we can assign to a set S and a ring R the" ring
Rs without the requirement that R should be a domain. This construction
is of fundamental importance for the theory of rings.

I]
Localization of Rings and Modules
37
Theorem 1.4.1
Let S be a multiplicative subset of a ring R. There exist a ring Rs and a homo-
morphism w: R -. Rs which satisfy the following conditions:
(i) for all 8 E S, the elements W(8) are invertible in Rs,
(ii) given a homomorphism of rings f: R -4 T such that, for every S E S,
the elements 1(8) are invertible in T, there exists a unique homomorphism
f: Rs  T such that fw = f.
The ring Rs is determined by the above conditions up to isomorphism.
Proof and Con8truction
Let us consider the set R x S of all pairs (r, s) where r E R, 8 E S, and define
the operations +, · in it by the formulae
(r,8)+(r 1 ,81) = (r81+r18,881)' (r,8). (r1,81) = (rrl,88 1 ).
The pairs (r, 8), (rl' 81) E R x S are called equivalent (we write (r, s) = (rl, 81»)
if and only if there exists an element 82 e S such that 82 ( r8 1 - rl s) = O. We
shall prove that the relation thus defined is a relation of equivalence. Indeed,
it is obviously both reflexive and symmetric. Suppose now that (r, 8) == (r1, 8t)
and (rl, 81) = (r2, 82); thus there exist elements 83, 84 E S such that s3(rs 1 -
-r1s) = 0, 84(r182-r281) = O. It follows that
84 S 3 8 1 (rs 2 ) = 84 8 2 8 3( r8 1) = 84 8 2 8 3(rl s) = s3884(r182)
= 83 s8 4(r2 s 1) = S4 8 3 s 1(r 2 8),
whence (r, s) = (r2, 82) which is the desired result. It is equally easy to verify
that, if (r, s) == (r',8'), (rl, 81) = (r, 8), then
(r, 8)+ (rl, 81) == (r', 8')+(r, 8),
(r, s). (rl, 81) == (r', s'). (r, 8).
The equivalence class of a pair (r, 8) e R x S with respect to the
relation == is denoted r8- 1 or r/8.and called a fraction. In view of the above
formulae, the set of all fractions of the form r/s, r E R, S E S, can be endowed
with the operations +, .; we define
r/8+rl/8 1 = ( r8 1 +rI 8 )/ssl,
(r/8). (rl/81) = (rrl)/(881).
The set of all fractions r /8 with r e R, S E S, forms a ring denoted by Rs.
Fractions r/s, r' /8' e Rs are equal if and only if there exists an element s" e S
such that 8" (rs' - r's) = O.
We define the ring homomorphism w: R -+ Rs by the formula w(r) = ,./1
for r E R.
We shall show that the ring Rs constructed above and the homomorphism
}v: R -. Rs satisfy conditions (i), (ii) in Theorem 1.4.1. Indeed, given 8 e S,
the element 1/8 is the inverse of the element W(8) = s/l. Suppose that a ring
homomorphism/: R  T satisfies the hypotheses in (ii). We define the function

38
Preliminary Concepts
[Ch.
f': RxS T by the formula/'(r,s) =/(r). ([(S»-1. It is easily verified
that (r, s) = (r', s') yields f'(r, s) = f'(r', s'). Hence the function f' induces
the mapping f: Rs  T which satisfies_the conditio.n f( rls) =f(r) ([(S»-I.
The mapping fis a homomorphism, and fw = f. We may argue for the unique-
ness of the homomorphism f as follows: suppose that a homomorphism
g: Rs  T satisfies the condition gw = f; it follows that, for every rls e Rs,
we have fer) = gw(r) = g(r/l) = g(rls. s11) = g(rls)g(s/l) = g(rls)gw(s)
= g(rls)/(s), whenceg(rfs) =/(r) (/(S»)-1. 0
Definition 1.4.2
Let S c: R be a multiplicative subset of a ring R. The ring Rs constructed
in the proof of Theorem 1.4.1 is called the ring of fractions of the ring R with
respect to the set S, or the localization of the ring R with respect to the set S.
Note that in the symbol of a fraction, rls, the dependence on the set S
or the ring R is not expressed; this ambiguity however, should not lead to
a misunderstanding.
If PeR is a prime ideal, the set R"P is multiplicative; the ring of frac-
tions RR'-..P is denoted also by R p and called the ring of fractions or the localiz-
ation of R with respect to the prime ideal P. Another example of a multiplicative
set is a set of the form R"U Pa., where Pa. are prime ideals, as well as the set
a.
R'3(R) of all elements of R which are 110t zero-divisors in R (let us recall
that in Section 1.3 we denoted by 3(R) the set of all zero..divisors of R). The
ring of fractions RR'-..a(R) is called the total ring of fractions of R.
If R is a domain, the ring R(o) = RR'-..(O), denoted also by Ro, is the field
of fractions of R.
In the case where R has zero-divisors, the ring Rs does not necessarily
contain a ring isomorphic to R, and, consequently, the homomorphism
w: R  Rs is not necessarily an injection (see Exercise 1, Section 1.4).
The kernel of the homomorphism w: R -+ Rs consists of all elements
r e R for which there exists an element S E S such that rs = O. Thus the homo-
morphism w is an injection if the set S does not contain zero-divisors.
The homomorphism w: R  Rs allows us to regard the ring Rs as an
R-module (even as an R..algebra) with respect to the multiplication r(rl/rl)
= rrl/s.
Example 1.4.3
Consider the algebraic variety V c: K" determined by a prime ideal P of the
polynomial ring K[X t ,..., X n ]. The ring of polynomial functions K[V]
= K[Xt, ..., Xh]1 P is a domain. The field of fractions of the ring K[V] is called
the field of rational functions of the variety V, and denoted by K(V). Elements
of K(V) are called rational functions on V. Let a = (at, ..., an) e Kn be a point
of V. Every rational function on V has the form fIg , where f, g e K[V] are

I]
Localization of Rings and Modules
39
residue classes determined bypolynomialsf, g. If fli = fl /g1, then n 1 = ft C
and f(a )gt(a) = fl (a)g(a). If, moreover, g(a) :/:. 0, 8'l(a) :f:: 0, we get f( a)lg(a)
= 11 (a)/il (a) and this element of the field K is called the value of the rational
function fIg E K(V) at the point a. The value of a rational function p e K(V)
at the point a is ,veIl defined if and only if there exists a representation of the
- -
function rp in the form of a fraction p = fIg, wheref, g e K[V] and g(a) :/:. O.
In this case we write rp(a) = f( a)lg(a). A rational function may be not defined
at some points of the variety.
Write ma to denote the maximal ideal of the ring K[V], generated by the
residue classes of elements Xl - a1, ..., X II - a,.. Then g(a) = 0 if and only
if g E ma, and consequently the set of those rational functions from K(V)
which are defined at the point a is the localization of the ring K[V] with respect
to the maximal ideal ma. This ring, called the local ring of the point a on the
variety V, is denoted by mea, V). Thus we have
mea, V) = (K[V)m.. = if{g eK(V); gma}.
The set of all rational functions from (!)(a, V) which are equal to 0 at the point
a is an ideal denoted by mea, V). It is easily seen that mea, V)"m(a, V) is the
set of invertible elements of the ring mea, V). Hence mea, V) is the only maximal
ideal of mea, V).
Our consideration can easily be generalized by replacing the point a E V
with the subvariety V' c V, determined by a prime ideal pi :;:) P of the ring
K[X 1 , ...,X n ]. To the prime ideal pi corresponds the prime ideal Q c K[V];
the ring
(!)(V', V) = (K[V])Q = {IlK e K(V); g, Q},
consisting of those rational functions which are defined at some point of the
subvariety V', is called the local ring of the subvariety V' on the variety V.
The ring (!)(V', V) also has only one maximal ideal.
I>eiUUtion 1.4.4
A ring R is called quasi-local if it has exactly one maximal ideal.
We also say, that (R, m) is a quasi-local ring, where m denotes the only
maximal ideal of R.
Corollary 1.4.5
A ring R is quasi-local and m is its only maximal ideal if and only if every
element of R which does not belong to the ideal m is invertible.
Proof
<= If r e R"m, then the ideal Rr is not contained in any maximal ideal; it
follows that Rr = R, whence the element r is invertible.
<= It follows immediately from the assumption that every proper ideal is
contained in m, and thus m is the only maximal ideal. D

40
Preliminary Concepts
[Ch.
Corollary 1.4.6
If P is a prime ideal of a ring R, then the ring R p is quasi-local.
Proof
The set of elements of the form rls E R p , where r E P, s rt P, is an ideal in R p .
An element which does not belong to this ideal has a representation of the form
r/s, where r, s  P. Hence it is invertible since s/r E R p and (r/s)(slr) = 1.
Tbe assertion follows from Corollary (1.4.5). D
Of fundamental importance for various applications of ring localizations
are the following relations between ideals of a ring R and ideals of the localized
ri ng.
Let R be a ring and S be a Dlultiplicative subset. Every ideal I c R deter-
mines the ideal IRs of Rs:
IRs = {r/sERs; reI,seS}.
The ideal IRs is called the extension of the ideal I. Observe that IRs is generated
by the set w(I), where w: R --. Rs is the canonical mapping given by the formula
w(r) = r/l, r e R.
Conversely, with any ideal J c Rs we can associate. the ideal w- 1 (J)
of R:
w- 1 (J) = {r E R; r/l E I}.
The ideal w- 1 (J) is called the restriction of the ideal J.
We sum up the basic properties of both operations in the following
Theorem 1.4.7
Let S be a multiplicative subset of a ring R, and write I, 1 1 , 1 2 for ideals of R,
and I, J 1 , 1 2 for ideals of Rs. Then
(i) w-1(IRs) ::> 1, W-1(J)Rs = J,
(ii) (I 1 nI 2 )R s = 1 1 RsnI2 Rs,
w- 1 (J 1 nJ 2 ) = w- 1 (/ 1 )nw- 1 (/ 2 ),
(Hi) (11 +I2)Rs = 1 1 Rs+12 R s,
W- 1 (J 1 +/ 2 ) = w- 1 (/ 1 )+W- 1 (J 2 ),
(iv) rad(I)R s = rad(IRs), w- 1 (rad(J)) = rad(w- 1 (J»),
(v) if P E Spec(R) and PnS = 0,
then PRs e Spec (Rs) and W-1(PRs) = P.
Proof
(i) If reI, then every element of the form r/s, s e S, belongs to IRs, whence
r e w-1(IRs). To prove the second formula, we write the following sequence
of equivalences:
rls e J <:> rll e 1 <=> r e w- 1 (J) <:> rls e W-1(J)RS.

I]
Localization of Rings and Modules
41
(ii) Clearly, we have (/ 1 n/ 2 )R s c: I1Rsn/2Rs. Now, if a1/s1 = a2/82 e
1 1 Rsn/2 Rs, a1 e 1 1 , a2 e 1 2 , 81' S2 e S, then SS2 a1 = SS102 E 1 1 n/2 for some
S E S. Hence al/sl = 882a1/ss182 e (/lnI2)Rs.
We leave to the reader the pr90fs of the remaining formulae in (ii), (Hi),
(iv), which are equally easy and direct.
(v) Let (rl/sl)(r2!s2)ePRs, where S1,S2eS. Then r1r2EP, and,
since P is prime, one of the factors '1' r2 belongs to P. It follows that also one
of the factors rl/s1, r2!s2 belongs to PRs, and consequently the ideal PRs
is prime.
In view of (i), it is sufficient to prove the inclusion W-1(PRs) c: P. If
r E W-1(PRs), then r/l = w(r) e PRs, i.e., r/l = a/s for a E P, s E S. Hence,
for some s' e S, we have s'r E P and, as P is prime, PnS = 0 implies,. E P. 0
In subsequent corollaries, S denotes a multiplicative subset of R.
Corollary 1.4.8
The correspondence P --.. P Rs between the set of those prime ideals of a ring R
which satisfy the condition PraS = ((J and the set of all prime ideals of the
ring Rs is one-to-one and preserves the relation of inclusion. The formula
p't-+ w- 1 (P') defines the inverse mapping.
Corollary 1.4.9
If R is a domain, so is the ring Rs. There is a natural isomorphism between
the ring Rs and a subring of the field of fractions Ro.
Indeed, since 0 is a prime ideal of R, its extension is by Theorem 1.4.7 (v),
also a prime ideal, i.e., the ring Rs is a domain. The mapping r /8 H> r /s defines
an injection Rs --. Ro which allows us to identify Rs with a subring of the
field Ro.
Theorem 1.4.10
If R is a domain, the localizations R p with respect to prime ideals P c: Rare
subrings of the field of fractions Ro, and
R = n R p = n Rm,
P m
where P ranges over Spec(R) and m ranges over Max(R).
Proof
Clearly, we have R c nR p c nRm. If x e nR ln , then for every maximal
ideal m there exist elements r m e R, Sm e R"m, such that x = rm/s m . The ideal
generated by all the elements 8m is not contained in any maximal ideal, and
thus it is equal to R. Consequently, there exist elements t 1 , ..., tt e R such
that t 1 sm! + ... + tlc8ma: = 1 for some maximal ideals m1, ..., m,p This yields
x = t 1 s m1 X+ ... +t"SmtX = t 1 r m ! + ... +ttrmt' i.e., x e R, which proves
the theorem. 0

42
Pr elimin ary Concepts
[Ch.
Theorem 1.4.11
Let R be a domain. For every prime ideal PeR,
R p = {x e Ro; x :F 0, RnRx-1q:P}u {O},
where Ro denotes the field of fractions of R.
Proof
Let 0 :/:. x E Ro and Rr.Rx-l q: P. Then there exists an element S E RfJRx- 1
such that s  P. Thus, for some r e R, we have s = rx- 1 , whence x = rls e R p ,
which proves the inclusion :;:).
Now, let 0 :/:. rls e R p , where r e R, S e R,P. For x = rls, we have
S = rx- 1 e RnRx-1, s rJ P, i.e., the inclusion c also holds. 0
We now pass to the problem of defining the localization Ms of an R..
module M with respect to a multiplicative set S c R.
Theorem 1.4.12
Let S be a multiplicative subset of a ring R. For every R-module M, there
exists an Rs-module M s and a homomorphism of R-modules w(M): M  M s
(we regard Ms as an R-module with respect to the multiplication rm = }v(I") m)
which satisfies the following condition:
(i) for every Rs-module N and every homomorphism of R-modules f: M
--. N, there exists exactly one homomorphism of Rs-modules f: Ms  N
such thatfw(M) = f.
This condition uniquely determines the module Ms up to isomorphism.
Proof and Construction
We shall proceed as in the proof of Theorem 1.4.1. In the set M x S, we define
the operation + and the multiplication by elements of R x S by the formulae
(m,s)+(m1,s1) = (s1 m + sm 1,sSI),
(r, s) (m1, S1) = (rml, SS1),
where m, m1 eM, r e R, s, S1 E S. Pairs (m, s), (m1, S1) e Mx S are said
to be equivalent (we write (m, s) = (m1, S1») if there exists an element S2 e S
such that S2(S1 m-sm1) = O. The relation = is a relation of equivalence and
satisfies the following condition: if (m1, S1) == (m, s), (m2, S2) == (m;, s),
and (r, s) == (r', s'), then
(m1, S1)+(m2, 82) == (m, s;)+ (m, s),
(r, s) (m1, S1) == (r', s') (m, s).
The equivalence class of a pair (m, s) e Mx S with respect to the relation =
is denoted by !!!. or mls. In view of the above formulae, we see that the set
s

I]
Localization of Rings and Modules
43
of all the classes mls can be endowed with the operation + and the operation
of multiplication by elements of Rs given by the formulae
mI1 s 1 +m21 s 2 = (S2 m I +S1 m 2)/S1 s 2,
(rls)(1/s1) == ( rfJn 1)/(ssl).
Thus we have obtained a module, written Ms, which consists of all elements
of the form mis, its coefficient ring being Rs. The fractions mls and mIls!
are equal if and only if there exists an element S2 E S such that S2(s1m-Smt)
= o.
We define a homomorphism of R-modules w(M): M  Ms by the formula
w(M)m = mil for m e M.
We shall show that the Rs-module Ms and the homomorphism w(M): M
 Ms satisfy condition (i). Given an Rs-module N and a homomorphism
of R-modules I: M --. N, we define the function!': M x S  N by the formula
f'(m, s) = (lls)f(m), where lis e Rs. It is readily verified that (m, s) = (m', s')
implies I'(m, s) = I'(m', s'). Thus the function I' induces the mapping
f: Ms --. N satisfying the condition f(m ls) = (lls)f(m). Hence the mappingl
is a homomorphism of Rs-modules and fw (M) = f. We argue for the unique-
ness of the homomorphism f as follows: if a homomorphism g: M s -+- N of
Rs-modules satisfies the condition gw(M) = f, then, for every mls eMs, we
have
j(ln) = gw(M)(m) = g(m/l) = g(s(mls») = sg(m/s),
whence g(mls) = (l/s)f(m) = f( mls).
o
Definition 1.4.13
Let S be a multiplicative subset of a ring R,' and M be an R-module. The
Rs-module Ms constructed in the proof of Theorem 1.4.12 is called the module
of fractions of M with respect to the set S, or the localization of M with respect
to S.
.A homomorphism of R-modules g: M.... L induces the homomorphism
of Rs-modules gs: Ms -+ Ls given by the identity gs(mls) = g(m)ls for
m e M, s e S.
The assignments M H> M s, g H> gs determine the covariant localization
functor with respect to the multiplicative set S( - )s :R-Mod -+ Rs-Mod.
Corollary 1.4.14
The kernel of the homomorphism of R-modules w(M): M -+- Ms given by the
formula w(M) (m) = mil consists of all elements m e M such that sm = 0
for some s E S.
In the next theorem we shall need the notion of tensor product ,
which can be found in [0, p. 16] and [L, p. 408].

44
PreUminary Concepts
[Ch.
Theorem 1.4.15 .
If M is an R-module, the homomorphism of Rs-modules 1p(M): M(8)RR s
-+- Ms given by the formula 1p(M)(m@rls) = rm/s is an isomorphism.
Proof
According to Theorem 1.4.12, the homomorphism 1p'(M): M --. M(8)RR s
defined by the formula 1p'(M)(m) = m@ 1 induces the homomorphism
ip(M): Ms --. M(8)RR s such that 1jJ(M)(mls) = m(8) lIs. The homomorphisms
1p(M) and ip(M) are mutually inverse. D
Corollary 1.4.16
If a sequence of R-modules,
ex {l
o  M' -+ M... M" --. 0,
is exact, then the sequence of Rs-modules
O M ' OCs M (ls M il 0
--. S --. S --. s--'
is also exact.
Proof
Since the tensor product functor is right exact [0, p. 22], we conclude by
Theorem 1.4.15, that the sequence
M ' «5 II Ps M " 0
S --. JY..i S --. s-+-
is exact. If m'ls E Ker(<<s), then «(m')ls = O. Thus there exists an element
SI E S such that S1 «(m') = 0, i.e., «(SI m') = O. This implies SI m' = 0, whence
m'/l = 0, and therefore also m'ls = O. It follows that «s is a monomorphism.
D
The result of Corollary 1.4.16 can be obtained without using Theorem
1.4.15, through a technique similar to that used in proving that Ker(<<s) = O.
If M' is a submodule of M, then, on the basis of Corollary 1.4.16, we can
identify M with a submodule of Ms.
Theorem 1.4.15 and Corollary 1.4. I 6 imply
Corollary 1.4.17
If S is a multiplicative subset of a ring R, then Rs is a flat R-module (i.e., the
functor - (8) R Rs is exact).
Corollary 1.4.18
If M l' ..., M 1 are submodules of a module M, then
(M 1 ,", .0. nMk)s = (M 1 )sn ... n(M 1 )s.

I]
Localization of Rings and Modules
45
Proof
We may restrict ourselves to the case k = 2. The inclusion c is obvious.
Given an element x = ml/s1 = m2/ s 2' where ml E M 1 , m2 E M 2 , SI, 82, e S,
there exists an element S E S such that 892m1 = SS1m2. This element belongs
to M 1 r.M 2 , whence x = SS2ml/8s2s1 e (M J f1M 2 )s, which proves the corol-
 0
Theorem 1.4.19
Let M, N be Rs-modules Regarding M, N also as R-modules with respect
to the homomorphism R --. Rs, we have
MRN  Mf8; Rs N, HomR(M, N) = HomRs(M, N),
and, in particular, M @ R Rs  M.
Proof
The homomorphism OG: M @ RN --. M (8) RsN given by the formula tX(I: nlj @.R
0Rn,) = L m,0R s n" is well defined and is an epimorphism. It is easily seen
that Ker(o:) is generated by elements of the form x = (m/s)@n-m@(n/s)
where m e M, n e N, S e S. However, n = s(n/s), whence
x = (m/s)@n-m(8)(n/s) = m/s@s(nls)-m(8)(n/s)
= s(m/s)f8;n/s-m@n/s = 0,
and therefore (X is an isomorphism.
Evidently, every Rs.homomorphism is an R-homomorphism. Let/: M  N
be an R-holnomorphism, then, for r E R, s E S, m e M, we have
sf«r/s)m) = t(s(r/s)m) = f(rm) = rf(m),
whence f«r/s)m) = (r/s)f(m) , and consequently f is 'also an Rs.homomor-
phism. 0
Corollary 1.4.20
Ifm c R is a nlaximal ideal, then there exists an isomorphism of Rm-modutes,
(mR m )n/(mR m )n+l  mn/mn+l, n = 0, I, ...
Proof
Every Rim-module N is endowed with a natural structure of an Rill-module
with respect to the multiplication defined by the formula (r/s)n = ;;(i)-I"
for r e R""- m, s E R, n E N, where r, S E Rim denote the residue classes of r, s.
The exact sequence
o --. mn+l --. m n --. mn/m"+l --. 0
induces the exact sequence
o --. m n + I @ R R.n -+ mil <8> R Rm --. mil /m n + 1<8> R Rm -+ 0
and consequently the sequence
o --. m n + 1 Rm --. m'l Rm --. mil /mfl+ 1 --. 0

46
Preliminary Concepts
[Ch.
is exact. Indeed, using the formula 1Rm  I@R m for I = m,,+l and I = m n
(see Theorem 1.4.19) we get first two terms of the last sequence. We obtain
the third term by applying Theorem 1.4.19 to the module mn/mn+l regarded
as an Rm-module. 0
Theorem 1.4.21
If M, N are R-modules, then
(MQ!)RN)s  Ms@RsN s  MS@RN s .
If, moreover, there exists an exact sequence 0  L -+ F --. M -+ 0, where
L, F are finitely generated R-modules and F is free, then
(HomR(M, N))s  HomRs(M s , N s ) = HomR(M s , N s ).
Proof
The tensor product being associative and commutative, (1.4.15) and (1.4.19)
yield
(M@RN)s  M@RN(i!;)RR s  M@RN(8)RR s Q9 R R s
 (M@RRs)@R(N@RR s )  A((i)RNs  Ms0 Rs N s .
Fix the module N. For every R-module X, the functor of localization with
respect to S induces an R-homomorphism HomR(X, N) -+ HomRs(X S , N s ),
which, by Theorem 1.4.12, has the form
A(X)
HomR(X, N) --. (HomRX, N)s  HOJDRs(X S , N s ).
The homomorphism A(X) is defined as follows: if E: X --. N, S,81 E S, then
(J..(X)(E/s»)(X/Sl) = (SSl)-l(X) for x eX. It is easily seen that A(F) is an
isomorphism provided F is a finitely generated free module.
The exact sequence 0 --.. L -+ F --. M --.. 0 induces the diagram
o
 [HomR(M N)]s
 [Hom R (F, N)]s
 [Hom R (L, N)] s.
A(M)
A (F) ::::
Ji(L)
o
:. HomRs(M s , N s )
:. Hon1R s (F s , N s )
.. HomRs(L s , Ns)
the rows of which are exact, by the left exactness of the functor Hom (see
[0, p. 27]). It is readily verified that the diagram commutes, whence it follows
that A(M) is a monomorphism, the proof being based merely on the fact that M
is finitely generated. Thus, A(L) is also a monomorphism. It is easily deduced
from these two facts that A(M) is also an epimorhism. 0

I]
Graded Rings and Modules
47
Theorem 1.4.22
If M is an R-module, and M m = 0 for every maximal ideal meR, then
M= O.
Proof
Let Y E M be an arbitrary elelnent. Since yll = 0 in M m , it follows that, for
every maximal ideal m, there exists Sm e R".m such that 3mY = 0, i.e., 8m
E Ann(y). The ideal generated by all the elements Sin is not contained in any
maximal ideal, whence Ann(y) = R, and therefore y = 0 and M = o. D
kerdses
1. Let R be the factor ring of the polynomial ring K[X, YJ, with the coefficients in the
field K by the ideal (XY). Write P for the prime ideal of R, generated by the residue class
of X. Determine the ring R p and show that the homomorphism R  R p is not an injection.
2. Prove that, if R is a domain, then the field of fractions Ro is a finitely generated
R-module if and only if R == Ro.
3. Prove that, if S is a multiplicativ set in a ring R, P c: R is a prime ideal, and Pr.S
= 0, then Rs/PRs  (RIP).,(s), where v:R -+ RIP is the natural homomorphism. Deduce
that if Q => P is also a prime ideal then RolPRo  (RIP)o/p.
4. If t is a non-nilpotent element of a ring R, write Rt for the ring of fractions of R with
respect to the multiplicative set {1,t,t 2 , ...}, and denote by S the set {reR:rKer(R
 R,)  Ann(r)}.
(i) Prove that S is the inverse image of the set of non-zero-divisors by the homomorphism
R .... R,.
(ii) Prove that the total ring of fractions of R, is isomorphic to Rs. Determine the units
of the ring Rs.
5. Prove that if an R-moduJe F is finitely generated and projective then, for any R-module
N and any multiplicative set S, we have (HomR(F, N»)s  HomRs(Fs, Ns). Generalize this
result to the case where F is the cokemel of a hOlnomorphism of finitely generated free
modules.
6. Let f: M -+ N be a homomorphism of R-modules. Prove that I is an epimorphism
(monomorphism, isomorphism) if and only if, for every maximal ideal m c: R, the induced
homomorphism 1m: M m -. N m is an epimorphism (monomorphism, isomorphism).
7. Let f: R -. T be a ring homomorphism. Every ideal I c: R determines an ideal
I- c T generated by the setf(I), and every ideal J c: T determines an ideal JC == 1- 1 (J) C R.
Prove the formulae I c: I ec , J  J ce , 1= Iec., JC =:: J cec .
In Theorem 1.4.7 (i)-(iv), replace ideals of the form IRs, w- 1 (J) with the ideals Ie, J C ,
and the sign of equality with inclusions c:, =>; verify which of the inclusions thus obtained
are valid.
8. Prove that, if K is a field, the localization of the polynomial ring K[X 1 , ..., X n , Yl, ...
..., Y.,J with respect to the ideal generated by Xl' ..., X n is equal to the localization of the
polynomial ring K(Yl' ..., Y m ) [Xl, ..., Xn] with respect to the ideal generated by Xl' ..., XII.
1.5 GRADED RINGS AND MODULES
The prototype of graded rings is a polynomial ring. Every element f of a poly-
nomial ring T = R[X I , ..., X,.] in indeterminates Xl' ..., X n , with the coeffi-

48
Preliminary Concepts
[Ch.
cients in a ring R, can be uniquely represented in the form f = 10 + /1 + ... +
+ fm, where fp, p = 0, 1, ..., is a homogeneous polynomial of degree p,
i.e., the sum of monomials of degree p which have the form rXfl ... Xtl, i 1 +
+ ... + in = p, r E R (we regard the zero polynomial as a homogeneous
polynomial of every degree). Thus the ring T is the direct sum of R-modules
CX)
T = E9 Tp, where T, is an R-module which consists of homogeneous polyno..
p=O
mials of degree p (including zero). Clearly, we have To = R. If gET", we
write deg(g) = p. Hence, for homogeneous polynomials g, h, we have deg(gh)
= deg(g)+deg(h), and" consequently Tp Tq c T p + q .
Definition 1.5.1
A graded ring is a ring R together with the decomposition of R, regarded
CIO
as a Z-module, into a direct sum R = E9 Rp which satisfies the condition
p==o
R"Rq c R"+fl for all p, q  0, Le., such that r E Rp, S E Rq implies rs E R,,+q.
A homomorphism of graded rings (or a homogeneous homomorphism)
00 00
is by definition a homomorphism of rings f: E9 R"  €a Tp which satisfies
p=O p=o
the condition f(R,) c: T", P = 0, 1, ... Graded rings \vith homogeneous
homomorphisms form the category of graded rings.
.
It follows directly from the definition that 1 E Ro, Ro is a subring of R,
00
and R, are Ro.submodules of R. We shall write, briefly, "a graded ring Ee R,".
p=o
Although in Section 1.4 we used the symbol R(o) or Ro to denote the
field of fractions of a domain, this ambiguity should not cause any misun-
derstanding.
fIJ
Given an element r e R =  Rp, we write r = I"0+r1 + ... +r m ; this
p=o
means that ro E Ro, '1 E R 1 , ..., r m E Rm. We call 'p the homogeneous com-
ponent of degree p of the element r. The elements of Rp are called homogeneous
of degree p and we write deg(a) = p for a e R".
Graded rings occur in a natural way in studies on projective algebraic
sets and also in algebraic topology.
Example 1.5.2
For any ring R, the polynomial ring T = R[X 1 , ..., Xn] becomes a graded
ring if we take as Tp the R-module consisting of all homogeneous polynomials
of degree p including O. In the same ring T, another grading can be defined,
by first assuming deg(X 1 ) = k 1 , ..., deg(X n ) = k", where k 1 , ..., k,. are any
integers, and then extending the function deg onto the monomials according
to the formula deg(rXfl ... X") = k 1 i 1 + ... +kni n . We define T; as the direct
sum of free R-modules RX1 1 ... X" such that k 1 i 1 + ... +kni n = p.

I]
Graded Rings and Modules
49
Example 1.5.3
We may regard any ring R as a graded ring, setting Ro = R, R" = 0 for p > o.
There are situations leading in a natural way to graded rings of the form
R = E9 Ryt where r is a group, e.g., Z or Z/2Z, or a semi-group with zero.
yer .
00
The kernel I == Ker(f) of a homomorphism of graded rings f:  R,
1'=0
00
-+ E9 T, is clearly an ideal and has an additional property: if an element
1'=0
r = ro+ ... +rm belongs to I, then ro, r 1 , ..., r", also belong to I. Indeed,
we havef(r,) E T, and 0 = .(ro) + ... + f(r m ), whencef(r p ) = Oforp = 0, 1, ...
. . . , m.
Definition 1.5.4
00
An ideal I of a graded ring R =  R, is called homogeneous if the homo-
p=o
00
geneous components of an element from I belong to 1. We then have 1 = Eel",
p:.O
where I, = If1R,.
An ideal is homogeneous if and only if it is generated by homogeneous
elements.
co co
If I =  1, c R = E9 Rp is a homogeneous ideal, the factor ring RII
p=o p=o
is endowed with the structure of a graded ring such that the natural homo-
co 00 co
morphism R -+- R/I is homogeneous. Identifying E9 R,/  1" with E9 R,II"
peO p:.O p=O
\ve regard the module R,,/I, as the component of degree p of R/f; we then have
the following identity:
(r,,+I,,)(r+Iq) = r"r+Ip+q for r" E R", r E Ril.
The concept of grading can be generalized to modules.
Definition 1.5.5
co
Let R = E9 R, be a graded ring. A graded R-module is an R-module M together
,=0
co
with the decomposition M = E9 M p of M regarded as a Z-module which
p=O
satisfies the condition R,Mq c: M,+q for allp, q  0, Le., is such that r e R"
me Mq implies rm eM p + q .
A homomorphism of graded R-modules (or a homogeneous homomorphism,
GO
or a homomorphism of degree 0) is by definition a homomorphism I: E9 M,
I' D O
GO
-+  N, of these modules which satisfies the condition f(M,) eN"
p=o
p = 0, 1, ...

50
Pr elimin ary Concepts
[Ch.
Graded modules over a fixed graded ring R form the category of graded
R-modules.
It follos directly from the definition that the components Mp of a graded
00
R-module are Ro-submodules of M = E9 M p .
p=o
We define a homomorphism of degree s of graded R-modules as an R-
eo ()()
homomorphism f: E9 M, -+ e N p sa tisfying the condition f(M p ) c N p + tl
1'=0 p=o
for p  O.
In the sequel, the following construction will play an important role.
Definition 1.5.6
Let I be an ideal of a ring R. The descending sequence of ideals
R = 1 0 :;:) 1 1 :::::> 1 2 ::> ... :::::> 11' :;:) /1'+ 1 :;:) ...
determines a graded ring
eo
Grl(R) = E91P/Il'+1,
p=o
with multiplication defined by the formula
(r+l"+I) (s+ Ifl+ 1) = rs+ll'+q+l for rEI", s E [q.
We call this ring the graded ring associated with the ideal I.
In any R-module M, the descending sequence of submodules
M = [OM;::) [1-M ::> [2M:;:) ... :;:) [I'M => [1'+ 1 M ::> ...
determines a graded module over the graded ring Grl(R)
GO
Grl(M) = E91"MJIP+1M,
p=o
in which multiplication by homogeneous elements of Grl(R) is defined by the
formula
(r+ll'+l) (m+[t+ 1 M) = rm+ll'+fl+lM for reII', mE [flM.
Every homomorphism of R-modules g: M -+ N induces a homomorphism
of graded modules Grl(g): OrI(M) -+ Grl(N) which is given by the formula
Grl(g) (m+[P+1M) = g(m)+/P+IN for m E ["M.
The assignment Orl is a covariant functor from the category of R-modules
and their homomorphisms into the category of graded Grl(R)-modules and
their homomorphisms.
Example 1.5.7
Let T = R [Xl' ..., Xn] be the ring of polynomials in indeterminates Xl' ..., XII'
with coefficients in the ring R, and write I = (Xl' ..., X n ). Then the R-module

I]
Graded Rings and Modules
51
/P/IP+l is isomorphic to a free R-module whose basis consists of monomials
X1 ... X" of degree p. It is easily seen that the ring Orl(T) is isomorphic to the
ring T regarded as a graded ring, described in Example 1.5.2.
Example 1.5.8
Let U = R[ [Xl' ..., Xn]] be the ring of formal power series in indeterminates
Xl' ..., X"' with coefficients in the ring R, and write 1 = (Xl' ..., XII). The
reader will find it easy to verify that GrJ(U)  Grl(T) (the isomorphism being
that of graded rings), where T denotes the polynomial ring from the preceding
example.
We sum up the basic facts concerning graded rings and homogeneous ideals
in the following
Theorem 1.5.9
co
Let R = E9 R, be a graded ring. Then
p=o
(i) a proper homogeneous ideal Iof R is prime if and only if any two homoge-
neous elements a, E R, b q E R satisfy the following condition: a,b q E I, a, ,p I
imply b q E I,
(H) a minimal prime ideal of a homogeneous ideal is also homogeneous,
(Hi) if ideals I, 1 are homogeneous, so are the ideals 1+1, IJ, 1(11, rad(I),
(iv) a homogeneous ideal I is maximal if and only if it has the form I
go
= m E9 E9 R", where m is a maximal ideal of the ring Ro.
p=1
Proof
(i) The implication => is evident; we shan prove the converse one. Let a, b e R,
ab e I, a,p I, and a = ao + ... +a", b = b o + ... +b q . Without loss of gener-
ality, we can additionally assume that a,  I (if a, e I, we can replace a with
a-a, without disturbing the hypotheses, etc.). The component of degree p+q
of the element ab is equal to a,b q e I, whence, in view of the assumption,
.
b q e I. The element b ' = b-b q satisfies the condition ab ' E I; thus, proceeding
as before, we show that b q - l e I, ..., b o e I, whence it follows that bel.
(ii) Let P be a minimal prime ideal of the homogeneous ideal I, i.e., P is
minimal among all prime ideals containing I. Write pi for the ideal generated
by all homogeneous elements of P. Thus we have I c pi C P. Let a" b q
..
be homogeneous elements, and suppose that a,b q epl, a" P'. Thenapb q eP
and a, ,p P, and consequently b q e P. This implies that b q e pi since b q is homo-
geneous. Now, the ideal P being minimal, the condition P :;:) pi :;:) I implies
p = P'. Hence the ideal P is homogeneous.
(Hi) The homogeneity of the first three ideals follo\vs directly from the
definition; the homogeneity of rad(J) follows from Corollary 1.1.5 and from
(H).

52
Preliminary Concepts
(iv) Suppose that a homogeneous ideal I is maximal. If a homogeneous
element ,." E R" of degree p  1 did not belong to I, then, by the maximality
of I, we would have l-r,9 E I for some s e R. By the homogeneity of I, this
would imply I e I, contrary to the assumption of I being maximal. Thus we
00
have I:::>  R" whence the assertion follows immediately, in view of the
p=1
C)
isomorphism R/ E9 R,  Ro.
p=1
D
Exercises
CIO
1. Let R = ED Rn be a graded domain and assume that Rn:P 0 for some n > O. The
,,=-o
set S = U R.." {O} is then multiplicative. Define, in the ring of fractions Rs, a grading with
n>O
indices ranging over Z, R, = ED (R s )... Prove that Rs, as a graded ring, is isomorphic with
IIeZ
the ring K[X, X-I], where K  (Rs)o, and tho degree of the indetenninate X is suitably
chosen.
2. Let R be a graded ring, and I be a homogeneous ideal of R. Let us denote by "(I)
the set of all homogeneous elements belonging to 1. Prove that if
h(I) c Pt u ... up"
where P t , ..., P" are homogeneous prhne ideals, then, for some k, 1 C Pt. This is an analogue
of Theorem 1.1.7 (ii) for graded rings.
3. Prove that an analogue of Theorem 1.1.7 (iii) does not hold for graded rings, i.e.,
give an example of a graded ring R, a homogeneous ideal I, and pritne homogeneous ideals
Pt, P 2 such that
h(/"-.1') c PtUP2,
whereas I cj: PI, 1 cj: P2' where h(l"'-IZ) denotes all homogeneous elements belonging
to 1"'-.12.

Chapter II
Noetherian Rings and Modules
We continue the study of commutative rings by introducing the fundamental
concept of a Noetherian ring, so-called to honour Emmy Noether, who in
the 19208 formulated the basic definitions and properties of this class of rings.
She found inspiration for general considerations in specific examples, her point
of departure being the wish to extend the theorem on the factorization of an
integer into a product of powers of prime numbers to ideals of a possibly large
class of rings.
It had been known by then that the elements of rings of integers of algebraic
number fields admit such factorizations. On the other hand, Hilbert's theorem
stated that, in a ring of polynomials in a finite number of indeterminates
over a field, every ideal is finitely generated (the basis theorem). Using these
facts, E. Lasker proved that in a polynomial ring of this kind, every ideal has
a primary decomposition, i.e., that it can be expressed as an intersection
of a finite number of so-called primary ideals, that is, ideals corresponding,
in a sense, to the powers of prime numbers in the ring of integers.
E. Noether's great contribution was to observe the connection between the
existence of primary decomposition and the validity of the basis theorem:
and to develop an abstract theory of rings generally called Noetherian rings.
In Section 2.1 we present the basic concepts of this theory, extending
them simultaneously to modules.
In Section 2.2 we formulate the basis theorem in a modern form, Le., we
prove that if a ring R is Noetherian then the polynomial ring R[X] ring is also
Noetherian.
Before starting a thorough analysis of primary decompositions in Noetherian
rings (Section 2.3) and modules (Section 2.4) we continue the geometrical
considerations initiated in Section 1.2. Throughout the book the basic notions.
of algebraic geometry will serve us as an illustration and motivation for purely
algebraic theorems and concepts. Such a geometrical prototype of primary
decomposition is the representation of an arbitrary algebraic set in the form
of a union of a finite number of irreducible algebraic sets.
In Section 2.3 we present the classical theory of primary decompositions
of ideals in Noetherian rings, pointing out irredundant decompositions and
analysing the problem of their uniqueness. The concept of an associated

54
Noetherian Rings and Modules
[Ch.
prime ideal which appears here is the starting point for the considerations
in Section 2.4, where, entirely independently of the contents of Section 2.3,
we present a general theory of primary decompositions of submodules of
arbitrary modules over a Noetherian ring.
The concise Section 2.5 contains the very useful Artin-Rees Lemma and
its consequences, including Krull Intersection Theorem. We use these facts
in Section 2.6, devoted to completions. Considering topologies of a special
type of rings and modules, we study the operation of completion, defined
in terms of Cauchy sequences and known from elementary analysis and
topology. Complete rings and modules have many remarkable properties.
In the second part of Section 2.6 we concentrate upon the so-called adic topo-
logies determined by the powers of a fixed ideal.
Section 2.7, the closing section of Chapter II, deals with Artin rings,
which may be defined dually to Noetherian rings by means of the descending
chain condition for ideals. Surprisingly, it turns out that every Artin ring is
Noetherian, and it is possible to give a relatively full description of those
rings. We also state some connections between modules of finite length and
Artin modules.
To simplify the exposition, we assume in the whole chapter that a ring R
is always a ring with a unity and M always denotes an R-moduIe.
2.1 BASIC CONCEPTS AND PROPERTIES
Definition 2.1.1
A module M satisfies the ascending chain condition for submodules if any
sequence of submodules M 1 C M 2 c: ... of M becomes stable, i.e., if there
exists a positive integer n such that M n = M n + 1 = ...
Definition 2.1.2
A module M satisfies the maximum condition if every nonempty falnity of
sub modules of M, ordered by the inclusion relation, has a maximal element.
Theorem 2.1.3
Let R be a ring and let M be an R-module. The following properties are
equivalent:
(i) M satisfies the ascending chain condition for submodules,
(ii) M satisfies the maximum condition,
(Hi) every submodule of M (including M itself) is finitely generated.
Proof
(i) => (ii). Let  be an arbitrary nonempty family of sub modules of M.
If the family t'{j has no maximal element, then for any M 1 E fd there exists
M 2 E  such that M 1 $ M 2 . Iterating this process, ,ve find an infinite
sequence M 1 $ M 2 * M 3 $ ..., contrary to (i).

II]
Basic Concepts and Properties
55
(ii) => (iii). Let N be any submodule of M and denote by l'd the family
of all finitely generated submodules of N. 1he family is nonempty since it
contains the zero submodule. By (ii), in (9 there exists a maximal element N'.
If N' =F N there is x E N, x , N', and the submodule N' + Rx belongs to r.9
and contains N', which contradicts the maximality of N'., This shows that
N' = N, i.e., the submodule N is finitely generated.
(Hi) => (i). Consider an arbitrary ascending chain of submodules M 1
00
c M 2 C ... of M and denote the submodule U M p by N. If N is generated
p=1
by elements Xl' ..., Xk, there exists n such that Xl' ..., Xk E M n . Thus M"
= M n + 1 = ..., and the sequence becomes stable. 0
Definition 2.1.4
A module satisfying the equivalent conditions of Theorem 2.1.3 is called
a Noetherian module. If a ring R regarded as an R-module is Noetherian,
it is called a Noetherian ring.
Since every R-submodule of a ring R is an ideal, and vice versa, Theorem
2.1.3 implies the following
Coronary 2.1.5
A ring R is Noetherian if and only if one of the following equivalent conditions
is satisfied:
(i) any ascending sequence of ideals of R becomes stable,
(ii) any nonempty family of ideals of R has a maximal element,
(iii) any ideal of R is finitely generated.
Examples
1. The ring of integers Z and the polynomial ring K[X] over a field are Noe.
therian because they have property (Hi) of Corollary 2.1.5. More generally,
every principal ideal ring (in particular every Euclidean ring) is Noetherian.
2. The ring K[X 1 , X 2 . ...] of polynomials in a countable number of indeter-
minates is not Noetherian since the sequence (Xl)  (Xl' X 2 ) * (Xl' X 2 , X 3 )
$ ... is strictly increasing. The ring K[X 1 , ..., Xn] on the other hand, is Noe-
therian if K is a field (see Section 2.2):
Theorem 2.1.6
If 0  M'  M !. M"  0 is an exact sequence of R-modules, then the
Inodule M is Noetherian if and only if the modules M' and M" are Noetherian.
Proof
Assume that M is Noetherian. Since each submodule of M' is isomorphic
(via a) to a submodule of M, M' is Noetherian by condition (Hi) of Theorem
2.1.3. Similarly, every submodule of M" is a homomorphic image of a sub-
module of M, \vhence it is finitely generated.

56
Noetherian Rings and Modules
[Ch.
Conversely, suppose that M' and Mil are Noetherian and let N be an arbi-
trary submodule of M. By the hypothesis, the modules «(M')nN and P(N) are
finitely generated. Let Yl, ..., Y" be generators for «(M')r.N, and let Xl' ..., xfJ.
be elements in N such that P(x 1), ..., P(Xq) generate peN). For every X E N,
II
there exist at, ..., a q e R such that P(x) = L a,fJ(x,), whence x- L aiXi
1=1
e Ker(fJ)r\N = «(M')r.N. Thus x- L a,x. = L bJYi for some b 1 , ..., b p
e R, whence N is generated by a finite number of elements Y 1, ..., y p, Xl' ..., Xq.
o
Coronary 2.1.7
p
If R-modules M 1 , ..., M" are Noetherian, then their direct sum E9 M, is also
'ct
Noetherian.
Proof
We prove by induction onp. If p = 2 then the corollary follows from Theorem
2.16 applied to the sequence 0 -+ M 1 -+ M 1 E9M2 -+ M 2 -+ O. Suppose that
p-l
p > 2 and that E9 M, is a Noetherian module. Applying Theorem 2.1.6 once
1=1
p p-I
more, this time to the sequence 0 -+ M, -+ E9 M i -+  M, -+ 0, we complete
1= 1 1= t
the proof.
o
oroll 2.1.8
A finitely generated module over a Noetherian ring is a Noetherian module
Proof
If M is an R-module generated by p elements and if F is a flee module
of rank p over R, then there exists an epimorphism F  M. Since F
 R e ... Ea R, our hypothesis and Corollary 2.1.7 show that F is Noetherian.
p
In view of Theorem 2.1.6, M is Noetherian as a homomorphic image of a
Noetherian module. D
Coronary 2.1.9
A homomorphic image of a Noetherian ring is also a Noetherian ring.
Proof
A homomorphic image of a ring R is isomorphic to R/ f for some ideal I.
By Corollary 2.1.8, R/f is a Noetherian R-module. But every R-submodule
of R/f is also an RII-submoduIe, and so Rff is a Noetherian ring. D
Coronary 2.1.10
Let R be a subring of a ring T. If R is Noetherian and if T is a finitely generated
R-module, then the ring T is also Noetherian.

II]
Basis Concepts and Properties
57
Proof
By Corollary 2.1.8, Tis a Noetherian R-module, i.e., every ideal of Tis a finitely
generated R-module. Clearly, such an ideal is also finitely generated over T
(because R c 71, whence T is Noetherian. 0
1nIeoremm 2.1.11
If R is a Noetherian ring and S is an arbitrary multiplicative subset of R,
then the ring of fractions Rs is also Noetherian.
Proof
By Theorem 1.4.7 (i), any ideal of Rs has the form IRs for some ideal I of R.
H I is generated over R by elements Xl' ..., x"' IRs is generated over Rs
by their images under the natural homomorphism R -. Rs; accordingly IRs
is finitely generated, and hence Rs is Noetherian. 0
Coronary 2.1.12
If R is a Noetherian ring and P is a prime ideal of R, then the localization R p
is also Noetherian.
Definition 2.1.13
A Noetherian quasi-local ring (i.e., a ring having only one maximal ideal)
is calle a local ring. A Noetherian ring with a finite number of maximal
ideals is called a semi-local ring.
As a consequence of Corollaries 1.4.6 and 2.1.12 we have
Corollary 2.1.14
The localization R p of a Noetherian ring R with respect to a prime ideal P
is a local ring.
To conclude this introductory section, we provide a simple but rather
surprising characterization of Noetherian rings in terms of prime ideals.
Theorem 2.1.15 (Cohen)
R is a Noetherian ring if and only if each prime ideal of R is finitely generated.
Proof
It is enough to show that if every prime ideal of R is finitely generated then all
the ideals of R are finitely generated. Suppose that this is not true, i.e., that the
family .!II consisting C?f all ideals which are not finitely generated is nonempty.
The family d satisfies the hypothesis of Kuratowski-Zorn Lemma, whence
there exists a maximal element in .91, sayan ideal I. We shall show that I is
a prime ideal and thus arrive at a contradiction.
Assume that the ideal I is not prime and let xy e I, x, y rp I. Then I * (I, x)
and, in view of the maximaJity of I, the ideal (I, x) is finitely generated, say

58
Noetherian Rings and Modules
[Ch.
(I, x) = (a1 +b 1 x, ..., an + b,. x), ai E I, b , E R, 1  i  n. Setting J = I:(x),
we have J  (I, y) ;;J2 I and therefore J is also finitely generated, J = (Cl, ...
..., cl'). We shall prove that I = (aI' ..., an, CI X , ..., cpx). Indeed, if U E I,
then U E (I, x), and
n
U = Lr,(o,+b,x) = L r,a,+(Lr,b,)x.
1=1
Consequently L ri hi E J and u is a linear combination of the elements at, ...
..., an, Cl x, ..., c"x. The fact that the ideal I is finitely generated implies
a contradiction; hence I is prime, which in turn contradicts the initial assump-
tion that all prime ideals are finitely generated. 0
Exercises
1. Let R be a ring and let 11' ..., II' be ideals of R such that Itn ... nIp = 0 and such
that each of the rings RIIt is Noetherian. Prove that R is a Noetherian ring.,
2. Let M be a Noetherian R-module and I the annihilator of M in R. Prove that R/I
is a Noetherian ring.
3. Let M be a Noetherian R-moduJe and tx: M -+ M an epimorphism. Prove that IX
is an isomorphism.
00
4. Let R be a Noetherian ring and let f = L OnX n e R[[X]]. Prove that the element f
n=O
is nilpotent if and only if each on(n  0) is nilpotent.
5. Prove that, if a ring R satisfies the ascending chain condition for finitely generated
ideals, then R is a Noetherian ring.
6. Prove that the ring K[X 1 , X 2 , ...], where K denotes a field, satisfies the ascending
chain condition for principal ideals.
7. Let R = Z[2X, 2X 2 , 2X 3 , ...], be the subring of the ring of polynomials Z[X)
generated by 2X, 2X 2 , 2X3, ... Show that R is not a Noetherian ring.
8. Let the notation be as in the preceding exercise. Denote by I the ideal of R generated
by 2X, and by J the ideal of R generated by 2X 2 . Prove that I ()J is not finitely generated.
2.2 THE BASIS THEOREM
Theorem 2.2.1 (Hilbert Basis Theorem)
If R is a Noetherian ring, then the polynomial ring R[X] is also a Noetherian
ring.
Proof
Let I be an ideal of R[X] and suppose that I is not finitely generated. We
construct a sequence of polynomials 11,/2' ... E R[X] in the following way:
we take /1 to be a non-zero polynomial of minimal degree in I; when 11' ..., fk
have already been defined, we take for fk+ 1 a polynomal of I such that fk+ 1
 ifl' ... ,fie) (this is always possible because 1 is not finitely generated) and
h+ 1 is of the lowest degree among the polynomials of  ifl, ..., fie). Let ni
= deg(ti), and let a, denote the coefficient of x n , in Ii. From the construction

II]
The Basis Theorem
59
of the sequence {Ii} it follows that nj  ni+ 1 for any i. 'We shall show that,
for every k, we have
(at, ..., ak) * (ai, ..., ak, ak+l).
Ie
Indeed, if we had ak+l = L C,Oi for some Cj E R, then the polynomial g
;=1
k
= L C j X"Ic+l- n 'li would belong to I and would have the form g = ak+ 1 X"t+l
1=1
+ ... Thus deg(h+ 1 - g) < nk+ 1 = deg(h+ 1), and obviously h+ 1 - g  if!, ...
· · · , It). This is in contradiction to the choice of the polynomial fk+ 1. Accord..
ingly, the sequence (at) * (a1, a2) $ ... is strictly ascending, which in turn
contradicts the hypothesis that R is Noetherian (Theorem 2. 1.3 (i». The ideal I
must therefore be finitely generated. 0
Corollary 2.2.2
If R is a Noetherian ring, then the polynomial ring R[X 1 , ..., Xn] is also
Noetherian.
Proof
We argue by induction on n, using the equality R[X 1 , ..., Xn] = R[X t , ...,
.. . , X n - t ] [X n l and Theorem 2.2.1. 0
Since a field is a Noetherian ring, we obtain
Corollary 2.2.3
The ring of polynomials K[X 1 , ..., X,,] over a field K is a Noetherian ring.
Corollary 2.2.4
Let T be a finitely generated R-algebra. If R is a Noetherian ring, then T is
also a Noetherian ring. In particular, every finitely generated algebra over
a field is a Noetherian ring.
Proof
The ring T is a homomorphic image of the polynomial ring R[X 1 , ..., Xn] for
some n. According to Corollaries 2.2.2 and 2.1.9, Tis a Noetherian ring. 0
Corollaries 2.2.4 and 2.1.14 yield
Corollary 2.2.5
Let Vbe an algebraic variety contained in K n . For any variety V' c V, the ring
m(V', V) (see Example 1.4.3) is local. In particular (when V' consists of one
point a), the ring m(a, V) is local.
Using the idea of the proof of the basis theorem, one can prove an analogous
property of the formal power series rings.

60
Noetherian Rings and Modules
[Ch..
Theorem 2.2.6
If R is a Noetherian ring, then the formal power series ring R[[X]] is also
Noetherian.
Proof
00
For any f e R[[X]], f ¥= 0, f = L Ci X' , we denote by d(j) the number min
;=0
{;  0; C i ¥= O} and call it the degree of the series f, we define d(O) = 00. Let I
be an ideal of R[[X]]. We construct a sequence of elements fl e R[[X]]; let
d(li) = n" and let at be the coefficient of x n , in the seriesji. We let!1 be a series
of minimal degree in [; if 11, ..., Ji have already been defined, we choose
Ji+l to be a series satisfying the conditions: 1 0 fi+1 eI, 2° ai+l' (a1, ..., aa,
3° h+ 1 is of the lowest degree among the series satisfying 1 ° and 2°. Clear]y
n,  ni+1 by definition. Since the sequence of ideals a1 c (a1, a2) c... is
strictly increasing and R is Noetherian, this sequence is finite. This means that
there is k  1 such that, for any h e 1, if h = aX n + ..., n = d(h), then a
E (a1, ..., ale)' We shall show that [= (f!, ... ,fie). Let h = aX n +... e [,
1J = d(h); we shall consider two cases:
k
1. n  n". Then we know that a = 2: aib"O for some b"o e R. Let It o
;= 1
k
- 2: bi, oX"- n 1i; then
;=1
ho = aX n + ..., d(h-h o ) > 1l.
We define inductively a sequence 11 0 , h 1 , ..., hi' ... E (fl, ... ,fk) as follows.
Assume that ho, ..., It, e (/1' ... ,h) have already been defined and have the
property
p
d(h- L: hi) > n+p,
1=0
p
11- .l: hi = CX"+l>+l + ...
;=0
k
Then C E (a1, ..., Ok), and let C = 2: aibi J p+ 1 for bi,p+ 1 E R.
1=1
k
If h p + 1 = L b " p + 1 x n + .p + 1- n'Ji, then
;=1
p+l
hp+l = CX"+P+l+ ..., 0(11- Lh , ) > n+p+1.
;=0
00
It follo\vs from the last property tbat h = '2: h j . Furthermore,
j=O
00 00 k k 00
)  h j = L L bl,JXn+J-VI = Lft (L: b',Jxn+J-n,),
j=O j=O i = 1 ; = I jr:=O
i. e., It E (It , ..., h).

II]
Primary Decomposition of an Ideal
61
2. n < nip Then there exists a minimal r, r  k, such. that a E (a J.' ..., a r ).
r
The choice of Ii shows that n  n,. Let 0 = L OJ b"o, b i . O e R, and let ho
1-1
,
= L b"oxn-ntfi. Then d(h-h o ) > n.. Repeating this procedure at most
1-=1
nk-n times, we shall get an element of degree  n", and thus the problem is
reduced to case 1. 0
Corollary 2.2.7
If R is a Noetherian ring, then the power series ring R[[X 1 , ... J X,,]] is also
Noetherian.
Proof
We argue by induction on n, using the equality R[[X 1 , ..., XII]] = R[[X 1 , ...
..., X n - 1 ]][[X II ]], and Theorem 2.2.5. 0
Corollary 2.2.8
The power series ring K[[X 1 , ..., Xn]] over a field K is a Noetherian ring.
Exercises
1. Must R be Noetherian if the ring R[X] is Noetherian?
2. Let P be a prime ideal in R[[X]] and p. the image of P under the natural homo-
morphism R([X]]... R which maps X onto o. Prove that P is finitely generated if and oo1y
if p. is finitely generated. Show that, if p. is generated by s elements, P can be generated
by s + 1 elements, and even by s elements when X rF P.
3. Using Exercise 2 and Theorem 2.1.15, give another proof of Theorem 2.2.6.
4. Prove that, if M is a finitely generated R-modute and N c M a submodule, then every
epimorphism f: N... M is an isomorphism. [Examine first the case where M is Noetherian;
consider the sequence {f- n (Ker(f))} of submodules. Reduce the general case to this special
case by selecting a suitable Noetherian subring of the ring R.]
2.3 PRIMARY DECOMPOSITION OF AN IDEAL
The prototype and motivation of general decompositions of ideals in Noeth..
erian rings occurs in a decomposition of an arbitrary algebraic set into a union
of irreducible algebraic sets. With a description of the latter decomposition
we begin the present section.
Theorem 2.3.1
For any algebraic set W, there exists a representation in the form of a union
W = Wtu ... uW" (1)
such that
(i) WI' ..., W, are irreducible algebraic sets,
(ii) Wi q: U W J for any i.
Jrl:l

62
Noetherian Riags and Modules
[Ch.
The sets W 1 , ..., W p satisfying conditions (i) and (ii) are uniquely deter
mined by Wand are called the irreducible components of the algebraic set W.
Proof
Suppose that W does not admit a representation of the form (1). Then W is
reducible, that is, there exist algebraic sets WI' W 2 , WI :F W, W 2 :f::. W,
such that W = W 1 U W 2 . One of the sets WI' W 2 must be reducible; if it is
WI' there are algebraic sets W I1 ' W 12 , W 11 :f::. WI' W 12 :/= WI having WI
as their union, i.e., WI = W I1 UW 12 . Proceeding in this way, we obtain
a strictly decreasing sequence of algebraic sets W  W 1 ;j2 ..., which, by
Theorem 1.2.4, induces a strictly increasing sequence J(W) $ J(W 1 ) $ ...
of ideals of the ring K[X 1 , ..., X n ]. Since K[X 1 , ..., Xn] is Noetherian, this
contradiction establishes the first part of our theorem.
If a given decomposition of the form (1) does not meet condition (ii),
then, by deleting successively those components which are contained in a union
of the remaining ones, we finally arrive at a decomposition satisfying condition
(ii); such a decomposition is said to be irredundant.
Assume that we are given two irredundant representations of the set W,
say W = W 1 u ... uW p = V 1 u ... uV q . Then Wi = Wif"'\W = W i f"'\(V I U
V ... uV q ) = (W i f1V t )u ... u(W,f"'\V q ). Since Wi is an irreducible algebraic
set, we have W, = Wif"'\V J for some j, and WI c: Vj. Applying the same
argument to V J , we deduce that there exists i' such that V} c: Wi'. Thus W, c:
c: V J c: W,,, and from the fact that the decomposition is irredundant it follows
that i' = i and WI = VJ.
The decomposition (1) implies the equality I(W) = I(W t )f1 ... f"'\1(W,)
of the corresponding ideals in the polynomial ring. Since each ideal [(Wi)
is prime by Theorem 1.2.8, every ideal of the form J(W) (and hence every
radical ideal of the ring of polynomials in a finite number ofi indeterminates
over a field) is an intersection of a finite number of prie ideals. 0
More general decompositions, also in arbitrary Noetherian rings, will be
discussed further on in this section.
In .the introduction to the present chapter, we mentioned the factorization
of any integer into a product of powers of prime numbers as the prototype
of more general factorizations (decompositions) in arbitrary Noetherian rings.
Our purpose now is to describe and analyse precisely such decompositions.
To do this, let us state the following property of a power pS of a prime p:
If x, ye Z, and pSlxy and p',(x, then pSly (2)
for some positive integer n. The reader will easily verify that the above property
characterizes powers of prime numbers among all positive integers.
The general concept of a prime ideal corresponds to the concept of a prime
number in ordinary arithmetics. Now we shall define ideals which, in this sense,
correspond to powers of prime numbers, and as a point of departure we shall
take property (2).

II]
Primary Decomposition of an Ideal
63
Definition 2.3.2
We call a proper ideal Q of a ring R a primary ideal if it has the following
property:
for all x, y e R, xy E Q and x rp Q imply y" e Q
for some n > 0 (in general depending on x and y).
Clearly, every prime ideal is primary.
Passing to the factor ring, we can say that Q is primary if and only if
R/Q :f:. 0 and any zero-divisor in R/Q is nilpotent. This formulation makes the
following simple but useful property obvious.
Lemma 2.3.3
Let Q and lbe ideals of a ring R, I c:: Q. Then Q is primary in R if and only if
Q/I is primary in RII.
The lemma follows from the existence of an isomorphism RIQ  (R/I)
I(QII).
From the remraks preceding Definition 2.3.2 it follows that, in the ring
of integers, a non-zero ideal is primary if and only if it is generated by a power
of a prime number. Thus it is natural to ask about the connection between
powers of prime ideals and primary ideals in the general case. The examples
below show that there are no general relations.
1. An example of a primary ideal which is not a power of a prime ideal.
Let R = K[X, Y], where K is a field, and let Q = (X, y2); then R/Q
 K[YJ/(y2). In the ring K[YJI(y2), every zero-divisor is a multiple of the
element Y + (y2); hence it is nilpotent, which means that Q is primary. If
P = (X, Y), then p2 c: Q c: P, both inclusions being proper, and so Q is not
a power of P. Also, the ideal Q is not a power of any other prime ideal P'.
If we had Q = P''', then p2 c: p'n c: P, and, by Theorem 1.1.1 (v), we would
have P c: P' c: P, or P' = P, but this, as we have already seen, is impossible!
2. An example of a prime ideal some power of which is not a primary.
ideal.
Let R = K[X, Y, z]/(XY-Z2), and let x, y, z denote the residue classes
in R of X, Y, Z, respectively. The ideal P = (x, z) is prime since RIP  K[Y]
is a domain. We shall show that p2 is not a primary ideal. Indeed, we have
xy = Z2 E p2, but x ; p2 and no power of y belongs to P2. In fact, if yS E p2 c:
c: P for some s, then YEP, which is impossible in view of the isomorphism
RIP  K[Y].
However, certain important properties of powers of prime numbers hold
also for any primary ideal. It turns out that, in the general case, every primary
ideal uniquely determines some prime ideal. More precisely, we have the
following

64
Noetherian Rings and Modules
[Ch.
Lemma 2.3.4
If Q is a primary ideal of a ring R, then rad(Q) is a prime ideal; this is the only
prime ideal minimal among all prime ideals of R containing Q.
Proof
Let xy e rad(Q); then (xy)rJ E Q for some n > O. The ideal Q is primary,
hence either x" e Q or Y" e Q for some s > O. This means that either x e rad(Q)
or Y E rad(Q), i.e., rad(Q) is prime.
To prove the second part of the theorem, note that if P is a prime ideal and
Q c: P then rad(Q) c rad(P) = P. 0
If P is the radical of Q, we can reformulate the definition of primary ideals
in the following way: Q is primary if and only if xy e Q, x tI Q imply yeP.
If Q is primary and P = rad(Q), we also say that the ideal 0Q is P-primary.
The next lemma provides us with a large class of primary ideals, and shows
that, under certain assumptions, there exists nevertheless a connection between
powers of prime ideals and primary ideals.
Lemma 2.3.5
Let R be a Noetherian ring, m a maximal ideal of R, and Q any ideal of R.
The following properties are equivalent:
(i) Q is m-primary,
(ii) rad(Q) = m,
(iii) m k c Q c m for some k > o.
Proof
The implication (i) => (ii) is obvious.
(ii)  (i). If rad(Q) = 11t, the image of m in R/Q is equal to the nilradical.
Hence R/Q is a local ring whose only maximal ideal is the nilradical m/Q.
Thus R/Q contains only invertible or nilpotent elements, and therefore every
zero-divisor in R/Q is nilpotent.
(ii) => (Hi). This follows from Theorem 1.1.1 (vi).
(iii) => (ii). Passing to radicals in (iii), we get
m c rad(m k ) c: rad(Q) c rad(m) = m, hence rad(Q) = m. 0
Observe that the equivalence of properties (i) and (ii) does not require the
assumption that R is a Noetherian ring.
In connection with Lemma 2.3.5, it is worth while to note that there exist
non-primary ideals whose radicals are prime ideals (Exercise 3).
In the discussion connected with the representation of an ideal as an
intersection of primary ideals, we shall need two more lemmas stating the
properties of primary ideals.
Lemma 2.3.6
If Q 1, ..., Q are P-primary ideals of a ring R, then the intersection Q 1 tl
f"'\ ... f"'\Q, is also P-primary.

II]
Primary Decomposition of an Ideal
65
Proof
Put Q = Qln ... nQs. By Theorem 1.1.1, we have rad(Q) = rad(Ql)tl ... n
r.rad(Qs) = P. If xy E Q and x  Q, then x rp Q, for some i, 1  i  s. Because
Qi is P-primary we have YEP, i.e., Q is P-primary. 0
Lemma 2.3.7
Let Q be a P-primary ideal of a ring R, x E R. Then:
(i) If x  Q, then Q:(x) is also P-primary,
(ii) If x E Q, then Q:(x) = R.
Proof
(i) If a E Q:(x), then ax E Q; since x  Q and Q is P-primary we see that a E P.
Consequently Q c Q:(x) c P. Passing to radicals, we obtain P = rad(Q) c
c rad(Q:(x») c P, i.e., rad(Q:(x») = P. The assumption ab E Q:(x) and
a  Q:(x) mean precisely abx E Q and ax  Q. Since Q is P-primary, we get
b E P, and therefore Q:(x) is P-primary.
Implication (ii) is obvious and does not require that the ideal Q should
be primary. 0
Equipped with the required properties of primary ideals, we now proceed
to a description of arbitrary ideals in a Noetherian ring in terms of primary
ideals. To this end it is useful to have a concept of an irreducible ideal.
Definition 2.3.8
A proper ideal I of a ring R is called an irreducible ideal if it is not an intersec-
tion of two ideals properly containing it.
Lemma 2.3.9
Any ideal of a Noetherian ring is a intersection of a finite number of irreducible
ideals.
Proof
Assume that the lemma is not true, and consider a nonempty family d
consisting of all ideals of R which are not intersections of a finite number
of irreducible ideals. R is Noetherian, and so, by Theorem 2.1.3, there exists
an element maximal in .91, say J. Of course, the ideal J is not irreducible
since it belongs to d. Hence there exist ideals 11, 1 2 properly containing J
and having J as their intersection, J = II r\12. Neither 1 1 nor 1 2 belongs
to .91 in view of the maximality if J in .91, and therefore they are intersections
of a finite number of irreducible ideals. Thus J is also a finite intersection
of this kind, contrary to the choice of the ideal J. This contradiction completes
the proof of the lemma. 0
Theorem I. I .7 (i) implies that in any ring a prime ideal is irreducible. For
Noetherian rings, the following result is true:
Lemma 2.3.10
Every irreducible ideal of a Noetherian ring is primary.

66
Noetherian Rings and Modules
[Ch.
Proof
Let I be an irreducible ideal of a Noetherian ring R. By Lemma 2.3.3, we may
assume that I = O. Suppose that xy = 0, x, y e R, x :f: O. We shall show that y
is nilpotent. In order to do this, let us consider the chain of annihilators of
powers of the element y, i.e., O:(y) c: 0:(y2) c: 0:(y3) c: ... As the ring is
Noetherian, this sequence becomes stable, that is, there exists a number n
such that O:(y") = 0:(1'+1) = ... We claim that (x)n(y") = O. If z e (x)r\(1')
then z = ax = by" for some a, b E R. From the assumption that xy = 0
it follows that byn+1 = 0, and hence, the annihilators of yn and 1'+1 being
equal, we get by" = 0, i.e., z = o. Since the zero ideal is irreducible and x :F 0,
we have y" = O. D
As an immediate consequence of the last two lemmas, we have
.Theorem 2.3.11 (Lasker, Noether)
Any ideal of a Noetherian ring can be represented as an intersection of a finite
number of primary ideals.
To facilitate further discussion, we adopt the following definition:
Definition 2.3.12
By a primary decomposition of an ideal I we mean a finite set of ideals Ql, ...
... , Qs such that:
(i) Ql, ..., Qs are primary,
(ii) I = Qlt1 ... r\Q,.
If the above conditions are satisfied, we shall refer briefly to a primary decom-
position I = Q 1 n ... nQs of the ideal I.
In connection with Theorem 2.3.11, we are faced with the question of the
uniqueness of a primary decomposition, or, more generally, of the classification
of all primary decompositions of a given ideal. Formulated in this way, the
question cannot be reasonably answered. Indeed, given a primary decomposi-
tion I = Qtr\ ... nQ of an ideal I and any primary ideal Q => I, we obtain
another primary decomposition 1= Q1t1 ... t1Q s r\Q. Another reason is
provided by Lemma 2.3.6. If some of the primary ideals Ql, ...,Qs, say
Ql, Q2, are P-primary, then I = (Qlr\Q2)r\Q3t1 ... r\Qs is also a primary
decomposition of I. The classification problem for primary decompositions
should therefore be restricted to "reasonable" decompositions. Accordingly,
we introduce the following definition:
Definition 2.3.13
A primary decomposition I = Qlr\ ... t1Q, is said to be irredundant if:
(i) n QJ q: Q" for each k, 1  k  S,
j;/:k
(ii) rad(Q,,) :f:. rad(QJ) for k =F j.

II]
Primary Decomposition of an Ideal
67
As a consequence of Theorem 2.3. I 1 and Lemma 2.3.6 we get
Corollary 2.3.14
Any ideal of a Noetherian ring admits an irredundant primary decomposition.
Proof
Starting from any primary decomposition of a fixed ideal, we first group
together all the components accoring to their radicals, which ensures that
condition (ii) in Definition 2.3.13 is satisfied. In this new decomposition, we
remove superfluous ideals (i.e., those which contain the intersection of all the
remaining ones) so that condition (i) is also satisfied. 0
But even the restriction to irredundant primary decompositions does not
lead to the theorem on their uniqueness. Consider the ideal (X2, XY) in the
ring K[X, Y] of polynomials in two indeterminates with coefficients in a field K.
We have two primary decompositions,
(X 2 ,XY) = (X)n(X, Y)2 = (X)n(X 2 , Y),
and both are irredundant. Note, however, that rad(X, Y)2 = rad(X 2 , Y)
= (X, Y), that is, the radicals of the respective primary components coincide.
This fact is true for any irredundant primary decompositions of a given ideal.
More precisely, we have
Theorem 2.3.15
Let 1= Q 1 r. ... r.Q, be an irredundant primary decomposition of an ideal I
in a Noetherian ring R, and let Pi = rad(Q,), 1  i  s. A prime ideal P
of R is equal to some Pi if and only if there exists an element x E R such that
I:(x) = P.
Proof
We shall first show that every ideal Pi is of the form I:(x). To simplify the
notation take i = 1 and consider the following family d of ideals of R: d
= {I:(y); y rp Ql, y E Q2n ... r.Q,,}. d is nonempty (property (i) in Definition
2.3.13) since the decomposition in question is irredundant. Observe moreover
that if I:(y) E d then 1:(y) = Q1 : (Y). Indeed, if ay e Q1, then ay E Qlr. ... r.
r.Qs = I, i.e., I:(y) ::) Ql :(y); the opposite inclusion is trivial. Consequently
Qt c: I:(y) c: PI since, by Lemma 2.3.7, the ideal I:(y) = Qt :(y) is P 1 -
primary.
Summing up, d is a nonempty family of certain PI-primary ideals lying
between Qt and Pt. Since R is Noetherian, d has a maximal element, say
1:(x). We shall prove that it is a prime ideal. Let ab e I: (x), a tI 1: (x). In view
of I:(x) = Ql : (x) we have abx e I, ax tI Ql and clearly ax e Q2r. ... nQs.
Accordingly I:(ax) e d and b E I: (ax). From I:(x) c: 1:(ax) we deduce, in
view of the maximality of 1: (x) in d, that I:(x) = 1:(ax), and finally b E I:(x).
Since the only prime ideal between Qt and PI is P1 (Lemma 2.3.4), we have
PI = I: (x).

68
Noetherian Rings and Modules
[Ch.
Now assume the ideal P to be equal to l:(x) for some x E R. Applying
Theorem 1.1.1 (Hi) to the decomposition I = Ql" ... "Qs, we obtain
I:(x) = (Ql : (x) )n ... n (Qs:(x»). (3)
By Lemma 2.3.7 the ideal Q,:(x) is either Pi-primary or equals R. Taking the
radicals on both sides of formula (3), we get the equality P = Pit" ... "Pi"
where {it, ..., it} is a subset of the set {I, ..., s}. Using Theorem 1.1.7, we
finally conclude that P = Pin for some in, 1  in  s. 0
Coronary (and definition) 2.3.16
If 1= Qt("\ ... nQ& is an irredundant primary decomposition of an ideal I
of a Noetherian ring R, then the set of prime ideals Pi = rad(Q'), 1  i  s,
does not depend on the decomposition but only on the ideal I. The set is
denoted by AssR.(RII), and its elements are called prime ideals associated
with I.
Theorem 2.3.15 enables us to describe the elements of the set AssR(R/I)
as those prime ideals P for which there exists a monomorphism of R-modules
R/P  RII. Indeed, if P = l:(x) then the multiplication by x determines such
a monomorphism. On the other hand, given a homorphism qJ: R/ P -+ R/ I,
we have qJ(r+P) = rqJ(1 +P). If x is a representative of 9'(1 +P) in R, then
the homomorphism qJ is just the multiplication by x and if it is also a mono-
morphism then P = I:(x).
In the next section, by means of this very interpretation, we shall define
the set AssR(M) for any R-module M and present the theory of primary
decompositions in an arbitrary module over a Noetherian ring.
Corollary 2.3.17
If I = Qlf1 ... nQs and I = Qn ... nQ; are two irredundant primary decont-
positions of an ideal I of a Noetherian ring, then s = I, and there exists a permu-
tation (J such that rad(Q,) = rad(Q(l» for 1  i  s.
Corollary 2.3.18
An ideal I of a Noetherian ring is primary if and only if the set AssR.(R/l)
consists of one element.
In spite of the irredundancy of a primary decomposition there may occur
proper inclusions between associated prime ideals. As a simple example we
may take the ideal (X 2 , XY) of the polynomial ring K[X, Y], which we have
already considered. Here (X 2 , XY) = (X)t'1(X, Y)2, for example, is a primary
decomposition, and so the associated prime ideals are (X) and (X, Y).
Since the structure of the ordered set AssR.(R/l) plays an essential role in
solving the uniqueness problem for an irredundant primary decomposition
of an ideal I, we introduce the following definition:
Definition 2.3.19
An associated prime ideal of an ideal I is called

II]
Primary Decomposition of an Ideal
69
(a) isolated if it does not contain any other prime ideal associated with I,
(b) embedded if it contains another prime ideal associated with 1.
This terminology is traditional and arises from the correspondence between
ideals of the polynomial ring and algebraic sets; we wrote about that corre-
spondence in Section 1.2. If I is an ideal in the polynomial ring K[X 1 , ..., X n ],
then a primary decomposition I = Qln ... nQs gives rise to a decomposition
of the algebraic set V(I) into the union V(Ql)U ... uV(Qs). If Pi = rad(Q,),
then obviously V(Qi) = V(P,), and so
V(I) = V(P 1 )u ... uV(Ps). (4)
From Section 1.2 we know that the sets V(P i ) are irreducible; the decomposi-
tion itself, however, need not be irredundant since P, c: Pj implies V(Pj)
c V(P,). To the irreducible components of V(l) correspond the isolated prime
ideals, whereas the algebraic sets corresponding to the embedded ideals are
just those superfluous sets (embedded in irreducible components) which should
be deleted in order to make decomposition (4) irredundant.
Isolated ideals admit the following characterization:
Lemma 2.3.20
In a Noetherian ring, the set of isolated prime ideals of an ideal I is identical
with the set of all ideals minimal among the prime ideals containing I.
Proof
It is sufficient to show that every prime ideal P => I contains some isolated
prime ideal of the ideal I, for then every prime ideal minimal over I is isolated.
Conversely, if P is an isolated prime ideal of I and I c P 1 C P, then P 1
contains the isolated prime ideal P 2' that is, P 2 C P 1 C P. By the definition
of isolated prime ideals, P = P2 and thus P = PI' which shows that P is a'
prime ideal minimal over I.
Let I "- Ql n ... nQs be a primary decomposition of the ideal I. Since
I
P :::) I, we have P = rad(P) => rad(I) = n rad(Q,). By Theorem 1.1.7, we
;==1
have P :::) rad(Qi) for some i, i.e., P indeed contains one of the isolated prime
ideals of I. 0
In connection with Lemma 2.3.20, the isolated prime ideals of I are also
called the minimal prime ideals of the ideal I.
Corollary 2.3.21
For any ideal I of a Noetherian ring R, the set of all minimal prime ideals
containing I is finite. In particular, the set of all minimal prime ideals of the
ring R is finite.
In terms of associated prime ideals one may formulate one more result.
which is very useful and will often be applied in the sequel.

70
Noetherian Rings and Modules
[Ch.
Theorem 2.3.22
Let I be an arbitrary ideal of a Noetherian ring R and let P 1, ..., Ps denote
all the prime ideals associated with I. Then the set of all zero-divisors in the
R-module R/lis equal to the union Pi U ... uPs, i.e., 3(R/l) = P 1 u ... UPse
Proof
We recall that 3(R/I) = U I:(x). Since, by Theorem 2.3.15, we have P, = 1:(XI)
X;I
for some x" X, f/:l, it follows that P 1 u... uPs C 3 (R/I). We shall show tha,
any zero-divisor on R/I belongs to some P,.
Let b e 3(R/I), and consider a family of ideals db = {l:(y); y  I and
b e I:(y)}. Since b is a zero-divisor on R/l, the family is nonempty. Let l:(x)
be an element maximal in db (such an element exists since R is Noetherian).
We shall prove that l:(x) is a prime ideal. Let ac E 1: (x), a f/: l:(x), that ist
acx E I, ax f/: I. Since bax E I, it follows that I:(ax) E db. The maximality
- of the ideal l:(x) in .!lib' in conjunction with the obvious inclusion l:(x)
c I:(ax), yields l:(x) = l:(ax), i.e., C E l:(x). Since l:(x) is prime, it is asso-
ciated with I in view of Theorem 2.3.15. Accordingly b E PI U ... uPs. 0
Lemma 2.3.20 and Theorem 2.3.22 yield immediately
Corollary 2.3.23
In a Noetherian ring, each element of a minimal prime ideal is a zero-divisor
We now proceed to settle definitively the question of the uniqueness of an
irredundant primary decomposition. To this end we shall prove the following
Lemma 2.3.24
Let 1 = Qln ... nQs be an irredundant primary decomposition of an ideall
in a Noetherian ring R. If Pi = rad(Q,) is an isolated prime ideal of I, then
Q, = {x ER; l:(x) ct: P,}.
(5)
Proof
To simplify the notation, assume i = 1. If X E Ql then Ql : (x) = R, and
s s
l:(x) = n (QJ:(x»)::> n QJ. If we had l:(x) C PI' then PI ::> QJ for some
)=1 J=2
j #: 1, and therefore PI ::> rad(QJ) = PJ, which contradicts the assumption
that P 1 is isolated.
If I:(x) cf: PI' then there is y f/: PI = rad(Qt) such that yx E I c: Qt.
From the fact that the ideal Ql is primary it follows that x E Ql. D
Since the right-hand side of equality (5) depends only on I and P" we
obtain as a corollary

II]
Primary Decomposition of a Module
71
Theorem 2.3.25
Let R be a Noetherian ring and I an ideal of R. If P is an isolated prime ideal
associated \vith I, then a P-primary ideal occurring in an arbitrary irredundant
primary decomposition of 1 does not depend on the decomposition (but only
on I and P).
Corollary 2.3.26
If all prime ideals associated with an ideal I of a Noetherian ring are
isolated, then there exists exactly one irredundant primary decomposition of I.
Unlike isolated ideals, primary ideals corresponding to embedded prime
ideals may vary in irredundant primary decompositions in an infinite number
of ways (see Exercise 4).
Exercises
1. Prove that in the ring R = Z[X] the ideal m = (2, X) is maximal and the ideal Q
= (4, X) is m-primary but not equal to any power of m.
2. Find a primary decomposition of the ideal p2 from Example 2, page 63, and the
prime ideals associated with that ideal.
3. Let R = K[X 1 , ..., X n ], where K is a field, let P, = (X t , ..., X,), 1 E; IE; n and put
I = Pt()p() ... nP:. (6)
Prove that
(a) the ideal pI is P ,-primary, i == 1, ..., n,
(b) the representation (6) is an irredundant primary decomposition .of 1,
(c) AssR(RII) = {P t , ..., Pn},
(d) rad(I) is a prime ideal.
4. Let R == K[X, Y], m == (X, Y), where K is a field. Show that m 2 is not an irreducible
ideal although it is primary (i.e., that the converse of Lemma 2.3.10 is not valid). Show that,
for each a e K, (X)m = (X)n(X2, aX - Y) is an irredundant primary decomposition and
ideals (X2, aX - Y) are distinct for distinct a's.
5. In the polynomial ring K[X, Y, Z] over a field K, let us consider the prime ideals
Pt == (X, Y), P 2 = (X, Z), m == (X, Y, Z). Put 1= P t P 2 . Show that 1== P 1 ()P 2 ()m z
is an irredundant primary decomposition of I.
6. Let R be a Noetherian ring. For any ideal I of R, denote by IR[X] the ideal of the
polynomial ring R[X] which consists of all polynomials with coefficients in I. Prove that:
(a) if P is a prime ideal of R, then PR[X] is a prime ideal of R[X],
(b) if Q is P-primary ideal of R, then QR[X] is PR[X]-primary ideal of R[X],
(c) if I == Qt n ... nQ, is an irredundant primary decomposition of I in R, then IR[X]
== QtR[X]n ... nQ,R[X] is an irredundant primary decomposition of IR[X] in R[X],
(d) if P is a minimal prime ideal of I, then PR[X] is a minimal prime ideal of IR[X].
7. Show that, in a Noetherian domain every element is a product of a unit and a finite
number of irreducible elements (the proof is similar to that of Lemma 2.3.9).
2.4 PRIMARY DECOMPOSITION OF A MODULE
All the definitions and results presented in the preceding section can be extended
to finitely generated modules over a Noetherian ring. This can be done by

72
Noetherian Rings and Modules
[Chi
defining primary modules in a suitable way and using the methods of Section
2.3 (see, for example, the Appendix to Chapter IV in [S]). Here, we shall adopt
another point of view and apply another technique; namely, we shall employ
the whole category of modules and not only ideals.
Previously, we were mainly interested in primary decolnposition, and only
incidentally-in the set of associated ideals. In this section, on the other hand,
it is the associated prime ideals of a module that we choose as our starting
point, their definition being inspired by Theorem 2.3.15.
Formally, this section is entirely independent of Section 2.3, and so it
provides, among other things, a different approach to the theory of primary
decompositions of ideals in a Noetherian ring.
Definition 2.4.1
Let R be a ring and M be an R-module. A prime ideal P of R is said to be
an associated prime ideal of M if there exists a monomorphism RI P  M.
The set of all associated prime ideals of a module M is denoted by AssR(M),
or simply by Ass(M), when it is clear which ring is involved.
This notation agrees with that considered in Section 2.3, Ass R (RI1), and
defined for an ideal I of a ring R. The only difference is that in Section 2.3
the elements of AssR(RII) were called the prime ideals associated with the ideal
I, whereas, according to the present terminology, they will be referred to as
the prime ideals associated with the module RII. We believe that this change
in terminology will not cause confusion.
Alternatively, one may say that AssR(M) = {P e Spec(R); there is x EM
such that Ann(x) = P} since every monomorphism cp: RIP  M determines
an- element x = cp(l +P), Ann(x) = P, and conversely.
Immediately fron1 the above statement and the definition of a prime ideal
we get
orollary 2.4.2
(i) If N is a submodule of a module M, then Ass(N) c Ass (M).
(ii) If P is a prime ideal of a ring R, then Ass(R/P) = {P}.
Lemma 2.4.3
Let R be a Noetherian ring and M be an R-module. Then Ass(M) = 0 if
and only if M = o.
Proof
Clearly Ass(M) = 0 if M = O. Suppose M #: 0, and let I = Ann(x), x E M,
be an ideal in R maximal among all ideals of the form Ann(y) for y E M,
Y =F O. We shall show that 1 is a prime ideal. Let a, b E R, ab e I, a f: 1; then
ax :F 0 and b E Ann(ax). Since 1 c: Ann(ax), it follows that in view of the
maximality of 1 the latter inclusion is an equality, i.e., bEl. Hence I E Ass(M),
and therefore Ass(M) :F 0. 0

II]
Primary Decomposition of a Module
73
Corollary 2.4.4
The set of zero-divisors on any module over a Noetherian ring is a union
of all associated prime ideals of the module.
Proof
Let M be a module over a Noetherian ring R. Since a(M) = U Ann (x),
xeM, xo
the definition shows immediately that
U P c: 3(M).
PeAss(M)
Conversely, suppose that ax = 0 for some non-zero x E M, a E R. Since
Rx i= 0, there exists P E Ass(Rx) by Lemma 2.4.3. Thus P = Ann(bx) for
some b E R, and P E Ass(M). Since abx = 0, we have a E P. 0
Lemma 2.4.5
Let M be an R-module and Nits submodule. Then
Ass(M) c Ass(N)uAss(MIN).
Proof
Let P E Ass(M), and let E be the image of the monomorphism RIP -+ M.
Further, denote F = EnN. If F = 0, then E is isomorphic to a submodule
of MIN, i.e., P E Ass(MIN). If F =F 0 and x E F is non-zero, then Ann(x) = P.
Indeed, x e E  RIP, and the annihilator of any non-zero element of RIP
is P. Since x E N, we have P E Ass(N). 0
Corollary 2.4.6
If {MrJ,}cxeA is an arbitrary family of R-modules, then
Ass( MJ = U Ass (Mcx) .
oceA oceA
Proof
In the case where the set A is :finite, the required equality follows from the
obvious inclusion Ass(Moc) c Ass( E9 Moc) and Lemma 2.4.5.
OteA
In the general case, if P E Ass( E9 MrJ,)' there exists a monomorphism
«eA
RIP  ffi M, the image of which is contained in the sum of a finite number
rJ,eA
of summands Mrx,. Hence, by the above arguments, P E U Ass(M(J,). D
cxeA
Lemma 2.4.7
Let M be a module over a ring Rand n any subset of the set Ass(M). Then
there exists a submodule N of M such that Ass(N) = Ass(M)"Q and Ass(M IN)
= D.
Proof
If Q = Ass(M), then we take the module N to be the zero module. Suppose
that Q =F Ass(M), and consider the family .s;I consisting of all submodules N'

74
Noetherian Rings and Modules
[Ch.
of M which satisfy the condition Ass(N') c Ass (M)"D. If P E Ass(M)"n,
then the image of the monomorphism RIP  M belongs to .91, and so the
family d is nonempty. Observe, moreover, that every inductive subfamily
{N(/. }oceA of the familye!1l has its least upper bound in .91. Indeed, if N' = U Nfl.
(leA
and P E Ass(N'), then there exists a monomorphism RIP  N' whose image
is contained in some N(I.. Accordingly, we can apply Kuratowski-Zorn Lemma
to d and pick a maximal element N of e!1I. By Lemma 2.4.5, to conclude the
proof of the lemma it is sufficient to show that Ass(M/N) c: Q; indeed then
Ass(M) c Ass(N)uAss(MIN) c (Ass(M)"!J)uQ = Ass(M), which yields
the required equalities.
If P E Ass(MIN), then MIN contains a submodule isomorphic to RIP.
It is of the form N'IN, where N' to a submodule of M. From Lemma 2.4.5
and Corollary 2.4.2 we see that Ass(N') c Ass(N)u {P}. If we had P f: D,
then N' would belong to the family .91, contrary to the maximality of N.
Therefore we must have P ED, and Ass(MIN) c {J. 0
Motivated by the characterization in Corollary 2.3.18, we introduce the
following definition:
Definition 2.4.8
A submodule E of a module M is called primary with respect to M if Ass(M I E)
consists of one element.
If Ass(MIE) = {P}, we also say that E is P-primary with respect to M.
There exists another description of primary modules, similar to that one
we used to define a primary ideal in Definition 2.3.2 (see Exercise 1).
Definition 2.4.9
Let M be a module and Nits submodule. By a primary decomposition of N
n M we mean a set {Ecx}cxeA of submodules of M such that:
(i) E«, « E A, are submodules primary with respect to M,
(ii) N = n Ecx.
«eA
If the above conditions are satisfied, we also say shortly that N = n E"
«eA
is a primary decomposition of the submodule N in M.
A primary decomposition {Ecx}cxeA is said to be irredundant if:
(iii) U Ecx ct: EfJ for every pEA,
«fJ
(iv) Ass(M/E(/.) =F Ass(MIEfJ) whenever OG -:F p.
Thus, in contrast to Section 2.3, we consider also," infinite primary decompo-
sitions. But in the same way as in Section 2.3, and for similar reasons, we
shall restrict ourselves to irredundant primary decompositions. Here is their
simple characterization in the case of modules over a Noetherian ring.

II]
Primary Decomposition of a Module
75
Lemma 2.4.10
Let M be a module over a Noetherian ring, and N a submodule of M. Let
N = n Ea. be a primary decomposition of N in M. Furthermore, denote
aceA
by Pac the only element of the set Ass (M/Eac), ex EA. Then:
(i) Ass(MIN) c {Pa.}cxeA,
(ii) the decomposition N = n Ea. is an irredundant primary decomposition
a.eA
of N in M if and only if:
(a) Ass(M/N) = {Pcx}oceA,
(b) Pcx :P PfJ whenever ex #: fJ.
Proof
SinceN = nEcx, the mapping M/N  $M/Er/., x+Nt-+ (x+Ea.), is a mono-
a.eA oceA
morphism. Thus (i) follows from Corollaries 2.4.2 and 2.4.6.
To prove (ii) suppose first that the decomposition N = n Eoc is irredundant
aeA
and set Fp = n Erx. for each pEA. Then FpnEfJ = Nand Fp/N = FfJ/(FfJn
OC:F{J
(',EfJ)  (Fp+EfJ)/Ep c M/EfJ' i.e., Ass(FfJ/N) c {P p ) by Corollary 2.4.2.
Hence Ass(Fp/N) = {Pp}, because Fp/N :F 0 in view of the irredundancyof the
decomposition and Lemma 2.4.3. On the other hand, Fp/N c M/N so that
Ass(FfJ/N) c Ass(M/N). Thus we have proved that PfJ E Ass(M/N) for any
peA. Conversely, assume that conditions (a) and (b) of the lemma are satisfied.
If the decomposition were not irredundant then N = n Ea. for some pEA.
a.:FP
But then, by (i), we would have Ass(M/N) c {Pa.}rx.:FP, contrary to hypothesis
(a). Therefore the decomposition N = n Ea. is irredundant. 0
a.sA
1nmeoremm 2.4.11
Let R be a Noetherian ring, Many R-module, and N a submodule of M.
Then there exists a primary decomposition of N in M such that N = n E(P),
where P runs over all the elements of Ass(M/N) and E(P) is P-primary with
respect to M. Consequently, every submodule of M has an irredundant primary
decomposition.
Proof
Let P E Ass(M/N); by Lemma 2.4.7 applied to the module M/N, there exists
a submodule E(P) of M containing N and such that
Ass(M/E(P») = Ass«M/N)/E(P)/N) = {P},
Ass(E(P)/N) = Ass(M/N)" {P}.
Let E = nE(p), where P ranges over all the elements of the set Ass{M/N).
Since" Ass(E/N) c Ass(E{P)/N) for every P and nAss(E(P)/N) = 0,
we have Ass(E/N) = 0, and hence, by Lemma 2.4.3, E = N. Therefore
N = n E(P) is a primary decomposition of N in M, which is irredundant by
Lemma 2.4.10. 0

76
Noetherian Rings and Modules
[Ch.
Finitely generated modules are the most interesting modules and will play
an essential role in the sequel. It turns out that in this case irredundant primary
decompositions are finite. In order to show this, we shall first prove a lemma.
Lemma 2.4.12
Let R be a Noetherian ring and M a non-zero finitely generated module over R.
Then
(i) there exists a chain of submodules 0 = M 0 c: M 1 c: ... c M n = M
such that M J IM J - 1  RIPj, where the Pj are prime ideals in R, 1  j  n,
(ii) for any such a chain we have Ass(M) c: {P 1, ..., Pn}; hence the set
Ass(M) is finite.
Pl"oof
(i) Since M =F 0, by Lemma 2.4.3. we have Ass(M) =F 0. Let P l E Ass(M),
PI = Ann(x1)"' Xl EM. We take M 1 = RXI  RIP!. If M #: M 1 , then
MIMI  0 and Ass(MIM 1 )  0. Suppose that P 2 E Ass(MIM 1 ) and P 2
= Ann(E), E e MIMt. If X2 is any representative of E in M, then we take
M 2 = M 1 +Rx2. We have M 2 1M 1 == RE  R/P 2 . Proceeding in this way,
we shall obtain a strictly increasing sequence of submodules 0 = Mo C M 1
C M 2 , ..., which must stop at M after a finite number of steps according to
Corollary 2.1.8, because M is a Noetherian module.
(ii) Using Lemma 2.4.5 and (i), we see that Ass(M J ) c: Ass(M j _ 1 )u {P j }
for 1  j  n, which proves assertion (ii). 0
Corollary 2.4.13
Let R be a Noetherian ring, M an R-module, and N a submodule of M such
that MIN is finitely generated. Then any irredundant primary decomposition
of N in M is finite.
In particular, if M is finitely generated, then every irredundant primary
decomposition of any of its submodules is finite.
Proof
By Lemma 2.4.10, the primary components in an irredundant decomposition
of N in M can be indexed by the set Ass(MIN). If MIN is finitely generated,
the set is finite in view of Lemma 2.4.12. 0
To settle the question of the uniqueness of a primary decomposition one
should investigate the behaviour of such decompositions with respect to
localization. First of all, we shall prove
Lemma 2.4.14
Let R be a Noetherian ring, S a multiplicative subset of R, and Q a subset
of Spec(R) consisting of all prime ideals disjoint from S. Then, for any R-
module M, the mapping P.... PRs is a bijection between AssR(M)".Q and
AssRs(M s ).

II]
Primary Decomposition of a Module
77
Proof
We know from Definition 1.4.8 that the mapping PH- PRs is a bijection
between Q and Spec(R s ). Moreover, the monomorphism R/P -+ M induces
a monomorphism (R/P)s  Rs/PRs -+ Mssince Rs is a flat R-module (The-
orem 1.4.17). Consequently, if P E AssR(M), then PRs E AssRs(M s ).
Now we shall show that, conversely, PRs E AssRs(M s ) implies P E AssR(M)
(clearly P E Q). When PRs E AssRs(M s ), there exist x E M, S E S, such that
PRs = Ann(x/s). Since P is finitely generated (R is Noetherian), there is an
element t E S such that P c: Ann(tx). Indeed, if at, ..., a" are generators of
P, the equality ai(x/s) = 0 implies the existence of ti E S such that tia,X = 0,
1  i  k. For the element t = t 1 ,..., tt we have Pc Ann(tx). Let aEAnn(tx),
i.e., at x = 0; then (afl)(x/s) = (atx)/(ts) = 0 and, by hypothesis, a/I E PRs,
i.e., a E P. Thus P = Ann(tx), and therefore P E AssR(M). 0
Corollary 2.4.15
Let E be a sub module of an R-module M and R a Noetherian ring. If S is
a multiplicative subset of Rand E is P-primary with respect to M, then:
(i) Es is PRs-primary with respect to Ms if PnS = 0,
(ii) Es = Ms if PnS #: 0.
Proof
Applying Lemma 2.4.14 to the module M/E, we deduce that
{ PR s if PnS = 0,
AssRs(MsIE s ) = 121 if PnS #: 121.
These equalities are equivalent to (i) and (ii), respectively.
o
For a module M and a multiplicative set S, let us consider the canonical
mapping }v = w(M): M  Ms. If N is a submodule of M then the submodule
w- 1 (N s ) of M is denoted by S(N) and called the saturation of N with respect
to S. Clearly N c S(N), and S(N) consists of all elements x E M for which
there exists s E S such that sx E N.
Lemma 2.4.16
Let E be a submodule P-primary with respect to a module M, and let S be
a multiplicative subset of a ring R. If PnS = 0, then SeE) = E.
Proof
If S E S, then s  3(M/E), since 3(M/E) = P by Corollary 2.4.4. Accordingly,
sx E E implies x E E, and S(E) = E by the description of the saturation given
ow. 0
Theorem 2.4.17
Let R be a Noetherian ring, M an R-module, and N a submodule of M such
that M/N is finitely generated, e.g., M is finitely generated. Denote by A the

78
Noetherian Rings and Modules
[Ch.
set AssR(M IN). Let S be any multiplicative subset of R, and r the subset of A
consisting of all prime ideals disjoint from S.
If N = n Eg. is an irredundant primary decomposition of N in M, then
(leA
N s = n (Ea.)s is an irredundant primary decomposition of N s in Ms and
cxel"
S(N) = n E« is an irredundant primary decomposition in M of the saturation
ocel"
S(N) of N with respect to s.
Proof
By Corollary 2.4.13 the set A is finite, and so N s = (n (Ecx»)s = n (Ecx)s
«eA cxeA
by Corollary 1.4.18. Using Corollary 2.4.15, we see that N s = n (Ecx)s,
a.er
and that the (Ea.)s are primary with respect to Ms for (i E r. From Lemma
2.4.14 it follows that the set r has the same cardinality as AssRs(MsIN s ) and
hence that the decomposition N s = n (Ea)s is an irredundant primary decom-
ocer
position by Lemma 2.4.10.
In order to establish the part of the theorem concerning the saturation
S(N), observe that, making use of the decomposition N s = n (Ea)s, we find
aer
S(N) = n S(EJ, because the inverse image operation commutes with the
cxel"
operation of intersection (that is to say, the inverse image of an intersection
coincides with the intersection of the inverse images). By Lemma 2.4.16 we
have S(E«) = EO'o for each a E r, and consequently S(N) = n E(J.. This decom-
g.er
position is irredundant in view of Lemma 2.4.10, since the isomorphism
MIS(N)  MsIN s implies that the set AssR(MIS(N)) has the same cardinality
 0
As in Section 2.3 with regard to ideals, we say that a prime ideal P E Ass(M)
.
IS
(a) isolated if P does not contain any other ideal from Ass (M),
(b) embedded if P does contain some other element of Ass(M).
Corollary 2.4.18
Let R be a Noetherian ring, M an R-module, and N a submodule of M such
that MIN is finitely generated (e.g., M is finitely generated). If Po is an isolated
prie ideal of MIN, N = n E(J. is an irredundant primary decomposition
aEA
of N in M, and E(J.Q is Po-primary, then ErJ.o is uniquely determined by N (that is,
it does not depend on the decomposition N = n Ea.).
OCEA
Proof
If S = R"P o , then Sf1P ¥= 0 for any P E Ass(MIN) different from po,.
since Po is isolated. Hence, applying Theorem 2.4.17 and the notation

II]
Primary Decomposition of a Module
79
used in its formlation, we deduce that r = {Po}, and so the saturation S(N)
is equal to ECl.o. Thus Ea.o does not depend on the decomposition but only
oo D
Corollary 2.4.19
Let the assumptions be as in the preceding corollary. If the set Ass(MIN)
consists only of isolated prime ideals, then there exists exactly one irredundant
primary decomposition of N in M.
Concluding this section, we shall introduce the notion of the support of
a module, and show its connection with the set of associated ideals.
Definition 2.4.20
By the support Supp(M) of an R-module M we mean the subset of the set
Spec(R) which consists of all prime ideals P such that M p ::f= O.
It follows from Theorem 1.4.22 that Supp(M) = 0 if and only if M = o.
1beorem 2.4.21
(i) For any ideal I c: R, ,ve have
Supp(R/l) = V(I),
where V(I) = {P e Spec(R):P ::J I}(see Section 1.2).
(ii) If a sequence of R-moduIes 0 -+ M' -+ M -+- M"  0 is exact, then
Supp(M) = Supp(M')uSupp(M").
(Hi) If M is a finitely generated R-module, then Supp(M) = V(Ann(M)),
and so Supp(M) is a closed set in Spec(R). .
(iv) If M, N are finitely generated R-modules, then
Supp(M@ RN) = Supp(M)nSupp(N).
Proof
(i) The exactness of the localization functor and Lemma 1.4.6 yield (R/1)p
 Rp/l p . Accordingly (R/l)p ::f= 0 if and only if P ::J I.
(ii) This is a consequence of the exactness of the localizati9n functor.
(iii) Let elements Xt, ..., x" generate the module M; then M p = (Rxt)P+
+ ... + (Rx,,)p for every prime ideal P, whence
n n
Supp(M) = U Supp(Rx,) = U Supp(R/Ann(x,))
;=1 1=1
11
= U V{Ann(x,») = V(Ann(M).
1=1
(iv) Observe first that, if (R, m) is a quasi-local ring and M ::f= 0, N :/:. 0,
then, by the Nakayama Lemma, M/mM::f= 0, N/mN::f= 0, and therefore
M/mM@R/mN/mN =F O. We have an epimorphism M@RN -+ M/mM(8)
(i)RN/mN  M/mM@R/mN/mN ::f= 0, and so M(8)RN ::f= O.
The inclusion c: in (iv) is obvious. To prove the opposite inclusion, assume

80
Noetherian Rings and Modules
[Ch.
that P e Supp(M)nSupp(N). Then M p #- 0, Np::F 0, and hence, by the
preceding remark, we have M p (29 Rp N p #= O. From Theorem 1.4.21 it follows
that (M@RN)p  Mp@RpN p #- 0, i.e., P E SUPP(M<8>RN). 0
Theorem 2.4.21
Let R be a Noetherian ring and M an R-module. Then
(i) Supp(M) consists of all prime ideals of R which contain some prime
ideal belonging to Ass(M).
In particular
(ii) Ass(M) c: Supp(M), and any minimal element of the set Supp(M)
is in Ass(M).
Proof
(i) By Lemma 2.4.3, P e Supp(M) if and only if AssRp(M p ) #= 0. On the
other hand, Lemma 2.4.14 shows that AssRp(M p ) :f:. 0 precisely if there is P'
e AssR(M) such that P' n(R"-P) = 0, i.e., if and only if P :::> p' for some
P' E AssR(M).
(ii) This follows immediately from (i). 0
Exercises
1. Let R be a Noetherian ring, and Man R-module. Prove that the following properties
are equivalent:
(a) Ass(M) has only one element;
(b) M ¥= 0 and, if c e R is a zero-divisor on M, then c is locally nilpotent on M, i.e.,
for any x e M, there exists n(x) > 0 such that c"(X)x = o. [Prove the implication (b) => (a)
by showing that every element locally nilpotent on M belongs to every prime ideal of Ass(M).)
2. Prove that, if M is a finitely generated R-moduIe and Ass(M) = {P} has only one
element, then the annihilator Ann(M) is a P-primary ideal in R.
3. Let R be a Noetherian ring and M a finitely generated R-module. Prove that if I
is an ideal such that each element of Supp(M) contains I, then there exists n> 0 such that
In M = O.
4. Let R be a Noetherian ring, M a finitely generated R-moduIe, and N any R-module.
Prove that Ass (HomR(M, N» = Ass(N)nSupp(M).
2.5 THE ARTIN-REES LEMMA
The result mentioned in the title of this section is one of the fundamental
facts concerning modules over Noetherian rings. As a corollary to the Artin-
Rees Lemma, we present the classical Krull Intersection Theorem.
Lemma 2.5.1 (The Artin-Rees Lemma)
Let R be a Noetherian ring, I an ideal of R, M a finitely generated R-module,
and B, C submodules of M. Then there exists a nonnegative integer Ie such
that
l"BnC = [n-"([k Br'\C)
for any n  k.

II]
The Artin-Rees Lemma
81
The proof requires some preliminary observations. Let
T = R+IX+1 2 X2+ ... +1"P+ ...
be the subring of the polynomial ring R[X] consisting of all polynomials for
which the coefficient of X" is in I" for each n. Given an R-module M, let us
write the R-module M@RR[X] symbolically in the form
M+MX+MX 2 + ...,
by analogy with the polynomial ring R[X], we denote it by M[X]. Just as we
determined the subring T of the ring R[X] by means of the ideal 1, we now
define a submodule
E = M+IMX+1 2 MX"+ ...
of the module M[X] formed by all elements of M[X] for which the coefficient
of X" belongs to 1" M. E can be endowed with a structure of a T-module by
setting
(aXD). (uX") = (au)X"+II, a el D , u e PM,
and then extending by linearity to arbitrary elements of T and E.
Lemma 2.5.1
(i) If R is a Noetherian ring, then the ring T is also Noetherian.
(ii) If M is a finitely generated R-module, then E is a finitely generated
T-module.
Proof
(i) The ideal I is finitely generated; let I = (a1' ... J as). Consider the homo..
morphism R[X 1 ,..., Xs] -+ T, X, I.... a,X, i = 1, ..., s, defined on the ring
of polynomials in s indeterminates. It is surjective since any element of In
is a sum of monomials of degree n in a 1, ..., as with coefficients in R. Therefore
T is Noetherian as a homomorphic image of a Noetherian ring (see Corollaries
2.1.9 and 2.2.2).
(ii) If Xl J ..., XII generate Mover R, and e = l: m,X', m, E I'M, is a typical
a
element of E, then there exist a" e 1', i = 1, ..., s, such that m, = l: a,p Xi.
lal
. .
Then e = l: (l: a"x,)X' = l: (l: a"Xp)x" where l: Q"X' e T, and so
p 1=1 1=1 p p
Xl' ..., Xs generate E over T.
o
Proof of the Art;n-Rees Lemma
Obviously PBnC ::> I"-"(I"BnC) for any n, k, n  k. To prove the opposite
inclusion, we define a submodule
D = (BnC)+(IBnC)X+ ... +(PBnC)X"+ ...
of E. It is a T-submodule because I" (IftBn C) c /11+' BnC. Since Tis Noetherian,
it follows from Lemma 2.5.2, Theorem 2.1.6 and Corollary 2.1.8 that D is a

82
Noetherian Rings and Modules
[Ch.
finitely generated T-module. Let d 1 , ..., d s be a set of generators of Dover T,
and let Xk be the highest power of X which occurs with a non-zero coefficient
in any of the generators d,. Furthermore, let n  k, m E In B()C. Then mX n
ED, and consequently
mX tl = 1 1 d 1 + ... +tsd s , I, E T.
By the choice of k we deduce that mX" E L [n-J(IjB() C) X", whence
J<k
me L In-J(IJBnC) c In-"([kBnC), since Ik-j(]JBnC) c [kBnC. 0
J<k
orollary 2.5.3
Let R be a Noetherian ring, I and ideal of R, and M a finitely generated R-
module. If
C = n PM, then IC = C.
nO
Proof
Applying the Artin-Rees Lemma to B = M and C = n In M, we get, for
n>O
n = k+ 1, Ik+l MnC = I(I"MnC). Since I"M()C = Ik+l M()C = C, we have
C = IC. 0
Corollary 2.5.4 (Krull Intersection Theorem)
Let the assumptions be as in the preceding corollary. Then we have n l'IM = 0
n>O
if and only if no element of the form I-x, x E I, is a zero-divisor on M.
Proof .
If there were x e I such that (l-x)m = 0 for some m =F 0, then m = xm = x 2 m
= ... and mEn InM; hence n ]nM ::F O. For the converse implication,
nO n>O
observe that for C == n In M we have IC = C by Corollary 2.5.3. Since C
n>O
is finitely generated (as a submodule of a finitely generated module over
a Noetherian ring) the Nakayama lemma (Lemma 1.3.3) implies the existence
of an element x E I such that (1 +x) C = O. By the hypothesis, 1 +x is not a
zero-divisor on M, which shows that C = o. 0
Corollary 2.5.5
Under the assumption of Corollary 2.5.3, if, additionally, [is contained in the
Jacobson radical J(R) of the ring R, then n In M = o. In particular, if R
n>O
is a local ring, then for any proper ideal I of R we have n r M = O.
nO
Proof
Since by Theorem 1.1.6 every element 1 + x, x e J(R), is invertible, the first
part of the corollary follows from Corollary 2.5.4.

II]
Completions of Rings and Modules
83
The second part of the corollary follows from the first one and from the
observation that, for a local ring R, the Jacobson radical J(R) is the only
maximal ideal, and hence it contains any other proper ideal of R. 0
Corollary 2.5.6
If R is a Noetherian domain and I a proper ideal of R, then n In = o.
n>O
Exercises
1. Let R be a Noetherian ring and I, J ideals of R. Show that n (J + I") _ == J if and only
n>O
if 1+ P #: R for every prime ideal P associated with J.
2. Let R be a Noetherian ring, 1 an ideal of R, and M a finitely generated R-module.
Prove that the submodule n 1 ft M consists of all elements m E M such that (1- x)m == 0
n>O
for some x e I.
2.6 COMPLETIONS OF RINGS AND MODULES
In this section we give the basic concepts and facts concerning topological
rings and modules with a linear topology. Of greatest importance is the case
of the topology determined by the powers of a fixed ideal of a ring, the so-
called adic topology. The basic construction of the completion of a module
in such a topology is obtained by a method analogous to that used in the
well-known description of the field of real numbers as the set of equivalence
classes of Cauchy sequences of rational numbers. Our ideas however, are
more directly derived from another example-from the way of describing
formal power series by means of polynomials. A series is regarded as "small"
if it begins with monomials of "high" degree. Every series can be approximated
by a sequence of polynomials which are partial sums of that series. Since the
successive differences in such a sequence "decrease", it is a Cauchy sequence
converging to the given series. The power series ring is therefore the completion
of the appropriate polynomial ring in the adic topology determined by the
powers of the ideal generated by the indeterminates.
Another example is provided by number theory-it is the construction
of p-adic numbers. Two integers are said to be p-close if their difference is
divisible by a large power of p. As every positive integer can be expressed
in the form
ao+atP+ a2p 2+ ... +a"p", 0  a, < p, (7)
two integers are p-close if the initial coefficients in their expression in form (7)
are identical. Therefore a Cauchy sequence in the topology determined by the
00
powers of a number p gives rise to a series L: aiP', 0  a, < p, called a p-adic
ic:O
integer.

84
Noetherian Rings and Modules
[Ch.
First (in Subsection 2.6.A) we shall present some general facts relating
to linear topologies, and then (in Subsection 2.6.B) we shall confine our atten-
tion to adic topologies on Noetherian rings and finitely generated modules.
2.6.A. Linear Topologies
We recall that a family fJ4 = {U}, of open sets of a certain topology is a base
for this topology if the sets in f!I form a covering of the whole space, and if,
for any U, V e f!4 and any Z E U r. V, there exists W e fJI such that Z E W,
We UnV. Conversely, if the above condition is satisfied for a family of
subsets f!)J of a set E, then on E there exists a unique topology with dI as a
base.
All the topologies on a module M which we are going to consider wilt
be determined by a (decreasing) filtration {M n } consisting of submodules of M,
i.e., by a descending sequence of submodules in the way described in the
following lemma:
Lemma 2.6.1
Let M be a module and {Mil}' n = 0, 1, ..., a filtration on M consisting
of submodules. Then the family f!I = {x+ M n }, X EM, n = 0, 1, ..., is a base
or some topology on M.
Proof
Let U = x+M II , V = y+M" and let Z E U("\. If p  n then M,. ::> Mp,
and so W = z+M, e tI, and Z E W c: UnV. 0
Definition 2.6.2
By the linear topology on a module M determined by a filtration {M,.}, n
= 0, 1, ..., of submodules of M, we mean the topology which has a = {x +
+M n }, x eM, n = 0, 1, ..., as a base.
The elementary properties of algebraic operations imply the following
corollaries:
C:orollary 2.6.3
A module M endowed with the linear topology determined by {M n } is a
topological module, that is, the operations of addition, taking the negative, and
mapping determined by the R-module structure on M are continuous map-
pings (on Mx M we consider the -product topology).
Proof
(1) If rp: Mx M -+ M, rp(x, y) = x+ y, then obviously rp(x+M", y+M II )
c: x+y+M n .
(2) If a: M -+ M, tX(x) = -x, then oc(x+M,,) = -x+M n .
(3) If, for r E R, r: M -+ M is a homotethy determined by r, then r(x+
+M n ) c rx+M II . 0

II]
Completions of Rings and Modules
8S
Corollary 2.6.4
A ring R with the linear topology determined by a descending sequence of
ideals {In} is a topological ring, i.e., the operations of addition, taking the
negative, and multiplication are continuous.
Proof
In view of Corollary 2.6.3, it is sufficient to sho,v that the multiplication
'I: RxR -+ R, 1p(x,y} = xy is continuous. We have 'lJ1(x+In, y+In} c xy+I"
since I" are ideals in R. 0
We note one more simple fact, whose proof is analogous to the proofs
of the preceding corollaries.
Corollary 2.6.5
If R is a ring with the linear topology determined by a descending sequence
of ideals {In}, and if M is an R- module with the topology determined by the
submodules {I,.M}, then M is a topological module over the topological
ring R, that is to say, the mapping R x M -+ M, in which. (a, x) 1-+ ax, is
continuous.
Corollary 2.6.6
The linear topology on a module M derived from a filtration consisting of
00
submodules {M,,} is a Hausdorff topology if and only if n M,. = o.
,,=-0
Proof
00
It is clear that no two distinct points of n M n can be separated by a neighbour-
11=0
00
hood from the base {x+M,,}. On the other hand, if n M n = 0 and x =F y,
11=0
then there exists apsuch that x-y tJM,; then (x+M,)n(y+M,) = 0. 0
Definition 2.6.7
A sequence {X,.} of elements of a module M is called a Cauchy sequence
in the linear toplogy determined by {M n } if, for any nonnegative integer n,
there exists a nonnegative integer s(n) such that x,-x q eM" for allp, q  s(n).
.
The sequence {xn} is convergent in such a topology if there exists x e M
such that for any n there is s(n} such that x,-x E M" provided that p ;?; s(n).
The element x is then called a limit of the sequence {XII}; it is determined
uniquely by {x.} if the topology under consideration is Hausdorff.
Let M be a topological module endowed with the linear topology determined
by a filtration {M n }. We denote by C the set of all Cauchy sequences in M.

86
Noetherian Rings and Modules
[Ch.
By defining addition of sequences and multiplication by scalars of R in a
natural way:
{x lI }+ {YII} = {xn+Yn}, a {y,,} = {aYn},
we endow C with R-module structure. Every convergent sequence is clearly
a Cauchy sequence. Denote by Co the subset of the module C consisting of all
sequences which converge to zero. Then Co is a submodule of C. We let
0: M  CICo denote the mapping defined by the formula O(x) = {x} + Co,
00
where {x} is a constant sequence determined by x. Of course, Ker(O) = n M,,;
n::sO
moreover O(x) = {x,,} + Co precisely if x is a limit of the sequence {XII}.
Definition 2.6.8
By the completion of a module M in the linear topology determined by a filtration
{M lI } we mean the module CIC o ; we denote the completion of M by M.
The module M is said to be complete if the above mapping (J: M -+- AI
is an isomorphism. This means in particular that the topology in question
is Hausdorff and that every Cauchy sequence is convergent.
If we are given another topological module N with the topology determined
by {N,,}, then a homomorphismf: M -+- N preserving the appropriate filtra-
tions, i.e., such thatf(M,,) c: N p for eachp, induces a homomorphism between
"- '" ",,-
the completions f: M -+- N, f( {x,,} + Co) = {f(x p )} + Co. Thus, the mapping
M 1-+ AI, f t-+) defines a covariant functor. Moreover, the diagram
f
M
. N
o
o
A
M
A
f
A
. N
is commutative.
If M = R is a ring and M n = In are ideals, then we can define multiplication
on C by setting
{xn}. {y,,} = {xnY,,}.
This defines a structure of a ring on C. The subset Co is then an ideal of C,
and consequently CIC o has a natural structure of a ring.
Definition 2.6.9
By the completion of a ring R in the linear topology derived from a descending
sequence of ideals {I,,} we mean the ring CICo; we denote it by R. A ring R is

II]
Completions of Riogs and Modules
87
called complete in a given topology if the appropriate homomorphism 0:
00 .
R  R is an isomorphism. It means that n In = 0, and every Cauchy sequence
11=1
in R is convergent.
'"
The completion M of a module M with respect to the filtration {I"M}
can be endowed with the structure of an R-module by defining for Cauchy
sequences {r ll }, {x n } in Rand M, respectively,
{rn} {x,,} = {r"x,,}.
From the definition we see at once that the homomorphism! AI -+- N induced
by f: M -+ N is a homomorphism of R':.modules. If Rand T are topological
rings and their topologies are determined by filtrations {In} and {In}, respect-
ively, then a homomorphism of rings f: R -+ T preserving these filtrations
(i.e., f(/,,) c: J II for all n) induces the homomorphism of rings J: R -+- T.
Example 2.6.10
We shall show that the completion of the polynomial ring K[X I , ..., X,,]
in the topology determined by the powers {In}, where I = (Xl' ..., X,,), is the
formal power series ring K[[X I , ...,X,,]]. We define a mapping rp: C
-+- K[ [Xl' ..., X,,]] as follows: if if,,} is a Cauchy sequence in K[X I , ..., X k ],
then f" - It E I" if p, q  s(n) for some s(n). This shows that the terms in I"
which are sums of monomials of degree  n-l are identical in alII, provided
p  s(n); let us denote this common polynomial by gn-l. Since gn-l is a
summand of K" for every nand K" - Kn-l is a sum of monomials of degree n,
the sequence {gn} determines uniquely a series g e K[[X I , ..., X,,]] such that
the sum of the monomials of degrees  n - 1 of the series g is equal to g,,- 1 .
We define cp( {J;.}) = g. Clearly, rp is an epimorphism, and its kernel is precisely
th e set of th ose Cauchy sequences which converge to zero, i.e., Co. Hence
K[ Xl' ..., iJ = C I Co  K[ [Xl' ..., X,,]].
We recommend the reader to prove directly that the power series ring
K[[X l , ..., X t ]] is already complete in the topology determined by the powers
of the ideal (Xl' ..., X,,). This fact will also be derived later as a corollary
to the general theory.
In order to establish the basic properties of the completion operation, we
shall introduce another, more convenient, description in terms of inverse
limits. We recall that an inverse system of modules and homomorphisms,
indexed by the natural numbers, {L i , hi}}' consists of modules L, and homo-
morphisms h,}: L} -+ L j for i  j such that hu = 1, h,Jh JIc = h,,, for i  j  k.
Observe that the homomorphisms hlj are uniquely determined by the homo-
morphisms h n , n+ 1. We shall denote these briefly by hll+ 1, and the whole system
simply by {Ln, h n }.
The basic example is provided by a module M with a decreasing sequence
of submodules {M n }, n = 0, 1, ... If we denote by hll+l = h",n+l: M/M"+l

88
Noetherian RIngs and Modules
[Ch.
-. MfM" the homomorphism h ll + 1 (x+M n + 1 ) = x+M", then the system
{MIM n , h n } is an inverse system of modules.
A map of inverse systems rp: {LI h,}  {Ln, h n } consists of a sequence
of homomorphisms fJ,,: L -+ L" such that h" f!J,. = qJn- 1 h' " .
A limit of an inverse system {L", hit} is the submodule of the product
ilLn, which consists of those elements (XII) for which h,,(x,,) = X n -l,
n = 1, 2, ... This limit is denoted by lim {L", h n }, or shortly lim {Ln} if it is
- t------
clear which homomorphisms are meant.
Any map of inverse systems qJ: {L, h}  {Ln, h,,} induces a homo-
morphism rp*: lim{L, h} -+ Iim{L", h ll } given by a formula 9'*«x»
- -
= <qJn(x». It is easy to see that a covariant functor results.
For a topological module M with the linear topology derived from a filtra-
tion of submodules {M,,}, there is a mapping
oc: lim {MIM n } -+ if
-
defined as follows. If (E,,) elim{MIM n } and En = x"+M,,, then the sequence
..---
{XII} is a Cauchy sequence since X"+l-Xn eM" for every n. We put «(En)
= {xn}+C o . Note that 0: is well defined for if we have at the same time
En = Yn+ M " for all n then Xn-Yn EM", whence lim{xII-Yn} = 0 and {xn}+
+C o = {Yn}+C o .
Theorem 2.6.11
The mapping ex: lim{MIM n } -. M is an isomorphism of R-modules.
4---
Proof
If {xn} is a Cauchy sequence, then for each n there exists s(n) such that x p -
-x t e M n whenever p, q ;:J= s(n). It follows that the element xp+M" e M/Mn
does not depend on p for p  s(n); let us denote it by En. Obviously (En>
E Um {M/M n }. Thus we have a mapping
--
P': C-.lim{M/M,,}, P'({x n }) = (En),
4--
which is a homomorphism of R-modules.
Observe that if {xn} converges to zero then x" E M" for k sufficiently large,
"-
that is, En = 0 for every n. Accordingly P' induces a homomorphism {J: M
 lim{M/M,,}. We claim that {J = «-1.
+--
If (E,,) eIim{M/M 1 ,} and En = xn+M,., then (Jex(E,,) = (J({x..}+C o )
--
= (E), where E = x,+M" for p large enough. Sincexn+l-Xn E M n , we have
xp-x" eM" for all p  n, i.e., En = E, and finally fJa. is the identity map.
Now let {x,,} be any Cauchy sequence and let (E,,) = fJ( {xn} + Co), i.e.,
suppose that for n there is s(n) such that En = xp+M,. for p  s(n) (in par-
ticular En = xs(n) + M n ). Then a,{J( {XII} + Co) = «(En) = {Xs(n)} + Co. Of course,

II]
Completions of Rings and Modules
89
the sequence s(n) may be chosen so as to be increasing, and so that s(n)  n
for any n. Then X,,-XS(If> eM" for p  s(n). For such p we have s(p)  s(n),
whence also x 6 (,,)-X S (II) eM". Consequently x p -X S (", EM" for p  s(n),
which means that the sequence x,,-xsCp) converges to zero. Finally «P is the
identity, 0
Observe that, if '1: M -+ lim{M/M,.} is given by 'YJ(x) = <x+M,,), then
--
the diagram
Iim{M/M n }
---
oc
A
... M
M
is commutative.
If M = R is a ring and M" = In are ideals, then Jim {RII,,} has the natural
....-
structure of a commutative ring and IX is an isomorphism of rings.
"
Further on we shall identify M with the inverse limit Iim {M/M n } by means
+---
of the isomorphism (X.
The inverse limits under consideration have an important exactness prop-
erty. We say that a sequence of inverse systems
o -+ {L, (t}.!. {LII' «n}  {L/, (X'}  0 (8)
is exact if, for every n, the diagram
lpn
"PII
o
,
 L"
)II- Ln
:.- L'
 o
,
OC n + 1
an 4- 1
"
ex" + 1
rp" + 1
1p,,+1
o
.. L + 1
 L" + 1
"
 L"+1
 o
has exact rows (and is commutative).
Theorem 2.6.12
If every homomorphism « of an inverse system {L, oc} is an epimorphism,
then the exact sequence of inverse systems (8) induces an exact sequence of
their respective limits
o  lim {L:,}!; lim {LIS}  Jim {L'}  o.
+-- -- _.-

90
Noetherian Rings and Modules
[Ch.
Proof
It is only necessary to show that 1p* is an epimprphism. The remaining equalities
follow directly from the definitions and the assumptions.
We recall that cp*(a) = (cp,.(a» and a similar equality holds for 'fl..
If <a') e lim {L'}, then, by hypothesis, there exists, for every n, an element
4--
b n e Ln such that "P,.(b n ) = a'. To show that 1p*(c,.> = (a'> for some (c n >
E lim {LII} we seek it in the form C II = CPII(x n ) + b n , XII e L. Then the condition
.-.--
(c n ) e Um {Ln} can be expressed as the equality
+--
(Xn+ 1 (CPn+ 1 (X n + I) + b,.+ I) = CPn(Xn) + bit
for any n. The commutativity of diagram (9) shows that this equality is equiv-
alent to the equality
CPn«.(+I(X"+I)-Xn) = bn-rtll+l(b n + I ).
Since cP" is a monomorphism, finding such X n reduces to solving the system
of equations
(X+I(Xn+l) = X n +d:', n = 0, 1, ..., with d E L.
We find such a solution by induction on n, using the assumption that all
mappings oc are epimorphisms. D
Theorem 2.6.13
Let M be a topological module with a linear topology determined by a
filtration {Mil} of submodules. Let 0  M' .!,. M .!. M"  0 be an exact
sequence, and regard M' and M" as endowed with the induced topologies,
i.e., the topologies determined by the filtrations {cp-l(M n )} and {1J'(M II }}
respectively.
A A
'" tp '" VJ J'\
Then the sequence 0  M' -+ M -+ M" -+ 0 is exact.
Proof
We use Theorem 2.6.11 and apply Theorem 2.6.12 to the exact sequence
of inverse systems which are determined by the exact sequences
0-+ M'/cp-I(M n ) --. M/Mn --. M"I"P(M,.) -+ o. 0
If M' c: M then M' c M. This fact can be used in particular for M' = Mil.
On the other hand, the induced topology on M" = M / Mil is discrete since
the image of Mil  M" is equal to zero. Thus M" = M", and from Theorem
2.6.13 we get
Corollary 2.6.14
If M is a topological module with the linear topology determined by a sequence
'" '"
of sub modules {M n }, then MfM n  MfM n .

II]
Completions of Rings and Modules
91
The filtration {M,,} on a module M gives rise to a filtration {M,,} on the
module M. If we regard AI as endowed with the linear topology derived from
{M,,}, and take into account the fact that the isomorphism in Corollary 2.6.14
is induced by the mapping 0: M -. M, then, from Theorem 2.6.11, we conclude
 '"
that the mapping (J: AI -+ M is an isomorphism, Thus we have
Corollary 2.6.15
The module M is complete in the topology determined by the filtration {M,,}.
It follows in view of Definition 2.6.6 that this topology is Hausdorff.
2.6.B Adie Topologies
Definition 2.6.16
Let M be an R-module and I and ideal of R. By the I-adie topology on M we
mean the linear topology determined by the filtration {I" M}.
00
The topology is Hausdorff if n In M = O. This condition is fulfilled, in
":::110
particular, when R is a Noetherian ring, I c: J(R), and M is finitely generated
(Corollary 2.5.5). In this case we can identify M \vith a submodule of M via
the monomorphism (J.

Farther on M will stand for the completion of M in the I-adic topology.
Lemma 2.6.17
Let R be a Noetherian ring, and let M be a finitely generated R-module. If
M' c: M, Mil = M I M', then the topology induced on M' and Mil by the
I-adic topology on M is also 1-adic.
Proof
Since the image of In M under the mapping M  M" is equal to In M", the
lemma is proved for Mil.
To prove the lemma for M' we shall make use of the Artin-Rees Lemma
2.5.1. By taking (in the notation of Lemma 2.5.1) B = M, C = M', we conclude
that there exists k such that InMr.M' = In-"(I"MnM') for all n  k. Thus
InM' c: [nMr.M' c: In-kM', which completes the proof of the lemma for M'.
D
In view of what has just been said, the results of Subsection 2.6.A, in relation
to I-adic topologies lead to the following corollary:
Corollary 2.6.18
Let R be a Noetherian ring.
(i) An exact sequence of finitely generated R-modules 0 -+ M'  M
-+ M" -. 0 induces an exact sequence 0 -+ M' -+ M -+ M" -. 0 of the comple-
tions in the 1-adic topology.

92
Noetherian Riags and Modules
[Ch.
------- '" jI\
(ii) (MEaN) = MeN for arbitrary R-modules M, N.
----
(Hi) MllnM  M/(PM) for every n.
(iv) If M' is a submodule of M and the /-adic topology on M', M, MIM'
is Hausdorff, then M'nM = M'.
Proof
By Lemma 2.6.17 (i) follows from Theorem 2.6.13, (ii) is obvious, and (iii)
is a consequence of Corollary 2.6.14. In proving (iv), one should make
use of the commutative diagram with exact rows
o
l
 M'
!
o
o
!
 M
1
o

 MIM'
!
. 0
A 1\ 
o . M', :.. M  (M/M') . 0 0
J\" At
The homomorphism 8: M  M and the R-module structure on M enable
us to define a homomorphism
'" '"
1J(M): R<8>RM -+ M, with <rll>x<rnx).
Theorem 2.6.19
If M is a finitely generated module over a Noetherian ring R, then the homo-
morphism 1](M): R 0 R M  M is an isomorphism of R-modules. The family
of isomorphisms fJ(M) defines an isomorphism of the appropriate functors
on the category of finitely generated R-modules.
Proof
We note that 1J(.M) is an isomorphism if M is a free module of finite rank
(by Corollary 2.6.18 (ii»). In the general case, we shall first show that '1J(M)
is an epimorphism. To this end, let us consider an exact sequence of finitely
generated R-modules 0 -. N -+ E -+ M  0, where E is free, and the commuta-
tive diagram induced by it:
A
R@RN
'"
,. R@RE
A
... R(g)RM
. 0
'YJ (N)
o.
'"
.. N
'YJ (E)
1](M)
1\
:. E
A
:. M
.. 0

II]
Completions of Rings and Modules
93
The upper row of the diagram is exact by the right exactness of the tensor
product, and the lower row-by Corollary 2.6.18. As 'leE) is an isomorphism,
1J(M) is an epimorphism. Applying this assertion to the module N, we deduce
that fJ(N) is an epimorphism. Using these observations, we easily check that,
if Y E Ker (1J(M»), then y is the image of some element z e R 0 E, and this in
turn is the image of x e RN because fJ(N) is an epimorphism. Finally, y is
the image of x under the composition of the two upper homomorphisms, and
so y = 0, which shows that fJ(M) is an isomorphism. 0
As a consequence of the above theorem and Corollary 2.6.18 we obtain
Corollary 2.6.20
If R is a Noetherian ring, then the completion R of R in the /-adic topology
is a flat R-algebra.
Coronary 2.6.21
Let M be a finitely generated module over a Noetherian ring R with the l-adic
topology; let I be any ideal of R. Then:
(i)IM IM= 1M,
_____ A A
(ii) (M/IM)  M/IM.
In particular, we have
A "
(iii) M/lnM  M/lnM for each n > O.
Proof
A "
(i) By Corollary 2.6.20, the homomorphism R  1M -. R  M is a mono..
A
morphism, and its image is equal to IR (i) M. This image is at the same time
------ A - A
isomorphic to the image of 1M -. M via fJ(M), whence 1M  1M.
(ii) follows from (i) and from the exactness of the completion functor. 0
--- A
According to Corollary 2.6.21 we have [nM = [11M, which shows that the
A
linear topology determined on M by the [-adic topology on M is also l-adic.
Therefore Corollary 2.6.15 yields
Corollary 2.6.22
A
A A A
The completion M is a module complete in the [-adic topology, i.e., M = M.
Consequently, the formal power series ring K[[X 1 , ..., X,,]] is complete
in the adic topology determined by the powers of the ideal (Xl' ..., X t ).
Moreover, Theorem 2.6.19 implies immediately
Corollary 2.6.23
A finitely generated module over a complete Noetherian ring is also complete.
A very important property of completions is stated in

94
Noetherian Rings and Modules
[Ch.
Theorem 2.6.24
If R is a Noetherian ring, then the completion of R in the l-adic topology is
also a Noetherian ring.
Proof
The ideal I is finitely generated, say 1 = (a1, ..., all). Set A = R [Xl' ..., X,],
A
m = (Xl' ..., X,). We know that the completion A of the ring A in the m-adic
topology is isomorphic to the formal power series ring R[[X I , ..., X,,]], and
A
hence it is Noetherian by Theorem 2.2.6. We shall show that R is a homo-
morphic image of R[[X 1 , ..., X,]].
For any n, the homomorphism A -+ R/I", fr-+ f(al' ..., a,,) + In, induces
a homomorphism AIm" -+ R/I". Consequently, we obtain a mapping n AIm"
-+ n R/ln which induces a homomorphism of rings
A A
cp: A -+ R, cp(f;. + m") = (f,. (a 1, ..., a p) + P) .
We shall prove that rp is an epimorphism. Let t = (t,,+]") e R. Then
1,.+ 1 - I,. E ]n for all n, and therefore there is a form I" E A of degree n such
that t,,+ 1 - t,. = !n(al' ..., a,,), n = 1, 2, ...; further, take 10 = t 1. Thus t,,+ 1
= 10 + 11 (ai, ..., a,,) + ... + In(al' ..., ap), and by setting h,. = 10 + 11 + ... +
A
+ /,,-1 we obtain an element h = (h,.+m n ) E A such that qJ(h) = t. As a
A A
homomorphic image of the Noetherian ring A, the ring R is also Noetherian.
o
A A
Any ideal J of a ring R extends to IR in R; on the other hand, if the l-adic
A A
topology on R is Hausdorff, R c R and the contraction of an ideal in R is an
ideal in R. The properties of these operations relating to the basic operations
on ideals are given in
Theorem 2.6.25
If R is a Noetherian ring, and the I-adic topology on R is Hausdorff, then the
following equalities hold for any ideals I, 11' 1 2 of R:
A
(i) IRnR = I,
A A A
(ii) I1R+/2R = (11 +/ 2 )R,
1\ A A
(Hi) (/1n/2)R = I1Rn/2R,
A A A
(iv) (/ 1 :/ 2 )R = I 1 R:/ 2 R.
Proof
Equality (i) follows from Corollaries 2.6.18 (iv) and 2.6.21, equality (ii) is
evident. To verify equality (iii), let us consider the exact sequence 0 -+ 1 1 r'.1 2
-. 11 Ee / 2 -. 11 +/ 2 -. O. In view of the exactness of completion, of the equal-
A A
ity I = IR, of the fact that the completion operation commutes with the
operation of taking a direct sum, and of property (ii) we obtain the assertion.

II]
Completions of Rings and Modules
95
p
(iv) By using the formula 1 1 :(a1, ..., a,,) = n II :(0,) and property (iii)
1=1
it is enough to prove (iv) in the case where J 2 = (a) is a principal ideal. Then
the sequence 0 -+ 1 1 : (a) -+ R!.. (aR+J t )/1 1 -+ 0 is exact. Now, the exactness
of the completion operation and Corollary 2.6.21 (ii) imply (iv). D
Theorem 2.6.26
A A
If R is the completion of R in the /-adic topology, then the ideal IR is contained
A
in the Jacobson radical I(R).
Proof
A A
By Corollary 2.6.22, the ring R is complete in the IR-adic topology. We shall
A A
show that every element of the form 1 - x, x e IR, is invertible in R. Indeed,
A
the sequence {Ylt = 1 +x+ ... +x"} is a Cauchy sequence in the IR-adic
topology; hence it is convergent in R. If Y is its limit then (I-x)y = 1. From
the characterization of the Jacobson radical in Theorem 1.1.6 we deduce that
A A
IR c: I(R). 0
Corollary 2.6.27
A
If R is a local ring with maximal ideal m, then the completion R of R in the
A
m-adic topology is also a local ring with maximal ideal mR.
Proof
A A A
By Corollary 2.6.21 we have RlmR  RIm, and so mR is a maximal ideal
A A A A
of R. By the preceding theorem, mR c I(R) and I(R) is the intersection of all
A A A A
maximal ideals in R, whence mR = I(R), and mR is the only maximal
A
ideal of R. 0
If (R, m) is a local ring, then the m-adic topology on R or on an R-module
is called the natural topology.
Conuluding this section, we state the following
Theorem 2.6.28
Ifmt, ...,mJ' are all the maximal ideals of a ring R, ifm = mln ... nmp,
A A A
and if R is the completion of R in the m-adic topology, then R = R 1 X
X ... x i.", where R, == R m " and R, is the completion of the quasi-local ring R,
in the m,R,-adic topology.
Proof
Since m7+m1 = R for i :F j, it follows from Theorem 1.1.9 that mil == m'l...
... m; == mn ... nm; and Rlm ft = (Rlm) x ... x (RIm;) for every n. Further-
more, RIm? is a quasi-local ring and Rlm7  R,/(m,R,)II. By making use of the
interpretation of completion in terms of inverse limits (Theorem 2.6.11) we
obtain the theorem. 0

96
Noetheriaa Rings and Modules
[Clt.
Exercises
00
t. Let (R, m) be a local ring and I an ideal of R. Prove that n (J+nt") = J. Deduce
11.0
from this that I is closed in the m-adic topology.
2. Let R be a Noetherian ring, I an ideal of R, and M a finitely generated R-module.
Prove that the completion of M in the l-adic topology is a zero module precisely when
Supp(M)nV(I) = 0.
A
3. Prove that every p-adic integer (i.e., every element of the ring Z, which is the comple-
tion of the ring of integers Z in the p-adic topology) can be represented uniquely in the form
00 k
L QnP" == lim L anp., where the integers an meet the requirements 0  an < p.
n=O k-+eo n=O
00
4. Prove that a p-adic integer L anpft(O  a" < p) is invertible precisely when OO:F o.
n=O
'"
Deduce from this that the ring Z, contains a ring isomorphic to the localization Z(P) of the
ring Z. Express the rational numbers 1/(1- pic), 1/(1 + p-), k = 1, 2, ..., and the number
00
-1 in the form L anpn.
n=O
5. Prove that a non-zero p-adic integer
00
pIc L a"pn (0  all < P, 00 fI: 0)
n-O
. is a square of some p-adic number precisely when k is an even number and there exists an
integer b such that b 2 == 00 (modp ) and, in the case p == 2, 01 = a2 =a O.
2.7 ARTIN RINGS AND MODULES
In this section we shall analyse concepts dual to those given in Definitions 2.1.1
and 2.1.2 for rings and modules.
Definition 2.7.1
A module M satisfies the descending chain condition for submodules if for an
arbitrary descending sequence of submodules M 1 ::> M 2 ::> ... of M there
exists n such that M" = M.+ 1 = ... (we then say that the sequence becomes
stable) .
Definition 2.7.2
A module M satisfies the minimum condition for submodales if any nonempty
family of submodules of M ordered by inclusion has a minimal element.
Theorem 2.7.3
Let R be a commutative ring and M an R-module. The following conditions
are equivalent:
(i) M satisifes the descending chain condition for submodules,
(ii) M satisfies the minimum condition for submodules.

II]
ArOO Rings and Modules
97
Proof
(i) => (ii). If M does not satisfy the minimum condition, then there is a family
of submodules of M which has no minimal element. This enables us to con-
struct in M an infinite descending sequence of distinct modules, contradicting
hypothesis (i).
(ii) => (i). If M 1 ::> M 2 ::> ... is a decreasing sequence of submodules, then
the family consisting of the submodules M, has a minimal element, say M n .
Then M,. = Mn+l = ... 0
Definition 2.7.4
A module satisfying the equivalent conditions in Theorem 2.7.3 is called
an Artin module. If a ring R regarded as an R-module is an Artin module,
it is called an Artin ring.
Example 2.7.5
1. A domain is an Artin ring if and only if it is a field. In particular, the ring
of integers Z and the polynomial ring K[X 1 , ..., XII] over a field are not Artin
rings. In order to prove this statement, let x be any non-zero element of the
Artin ring under consideration. A descending sequence of ideals (x) ::> (x 2 )
::J ... becomes stable; hence (x") = (x"+1) for some n. We thus obtain the
equality XII = x"+1r, and, since the ring is a domain, we have xr = 1, and x
is invertible.
2. Every ring which has only a finite number of eletnents is an Artin ring;
for example, rings of the form Z/nZ are Artin rings for any positive integer n.
It follows from the fact that, under this assumption, the ring in question
has only a finite number of ideals, so that condition (ii) in Theorem 2.7.3 is
fulfilled.
3. Let p be a prime number, and let Z(pOO) be a subgroup of the group
Q/ Z consisting of the residue classes of elements of the form nlpk, neZ,
k = 0, 1, ... Let G" denote the subgroup of Z(pCfJ) generated by the residue
class of the element l/pk; we have of course a chain G 1 c: G 2 C ... and it is
easily verified that every subgroup of Z(pOO) is identical with one of the sub..
groups a". Consequently, every decreasing sequence of subgroups becomes
stable, and therefore Z(pCfJ) is an Artin Z-module. Observe that (unlike Noe-
therian modules) Z(pOO) is not a finitely generated Z-module.
Some of the properties of Artin modules are analogous to those of Noe...
therian modules, which were proved in Section 2.1.
Theorem 2.7.6
If 0  M'  M  M" -+ 0 is an exact sequence of R-modules, then M is an
Artin module if and only if M' and M" are Artin modules.
Proof
With any descending sequence of submodules of M' or M" we can associate
an appropriate descending sequence in M; hence, if M is an Artin module.

98
Noetherian RiDgs and Modules
[Ch.
such a sequence becomes stable which shows that both M' and M" are Artin
modules.
Suppose now that M' and M" are Artin modules, and let M 1 :::> M 2 ::::> ... be
a descending sequence of submodules of M. By the hypothesis, the descending
sequences a- 1 (M 1 ) ::> a- 1 (M 2 ) ::::> ... and P{M 1 ) ::::> P{M 2 ) ::::> ... in M' and
in M", respectively, become stable, and thus it is possible to choose n such that
«-l(M,.) = oc- 1 (Mn+l) = ... and P{M,.) = (J(M n + 1 ) = ... In this situation,
we have a commutative diagram
o
.. a- 1 (M,,)
 M n
 (J(M n )
:lP' Q
Q .
 ex -l(M"+l)
:r- M"+l
:. f3(M n + 1 ) -..... 0
in which the rows are exact and the corresponding mappings are induced by (X
and p, respectively. Thus M n = Mn+l = ..., i.e., M is an Artin module. 0
The proofs of the following three corollaries are analogous to those of
Corollaries 2.1.7-2.1.9 for Noetherian rings and modules.
Corollary 2.7.7
p
H M 1, ..., M" are Artin modules, then their direct sum E9 M i is also an Artin
1=1
module.
Corollary 2.7.8
A finitely generated module over an Artin ring is an Artin module.
Corollary 2.7.9
A homomorphic image of an Artin ring is also an Artin ring.
In Chapter I we introduced modules of finite length. At present, we can
give their description with the aid of the notions already known.
Theorem 2.7.10
A module M is of finite length if and only if it is both a Noetherian module
and an Artin module.
Proof
If M is of finite length, then, by Schreier Theorem (Theorem 1.3.4), every
sequence of submoduIes of M is of finite length, and therefore M satisfies
the descending and ascending chain condition for submodules.

II]
Artin Rings and Modules
99
Assume now that M is a Noetherian and Artin module. We shall construct
a finite sequence M 1 ::::> M 2 ::> ... in the following way: We take M 1 = M,
and then define M n + 1 as a maximal proper submodule of M n for n = 1, 2, ...
(which exists because M is Noetherian). This sequence must stop at a zero
module after a finite number of steps since M is an Artin module. The con-
struction shows that we have thus obtained a composition series of M, whence
M is a module of finite length. 0
Theorem 2.7.11
Let R be a Noetherian ring and M a finitely generated R-module. Then M
is of finite length if and only if the set Ass(M) consists only of maximal ideals.
Moreover, if M is of finite length, then Ass(M) = Supp(M).
Proof
From Section 1.3 we know that any non-zero simple module is of the form
RIm where m is a maximal ideal in R. If M is of finite length, then there exists
a composition series, of length say p, whose factors are simple modules, and
hence of the form RIm, for some maximal ideals m" i = 1, ..., p. Lemma 2.4.12
shows that Ass(M) c {ml'...' m p }. Furthermore, by Theorem 2.4.22, we
have Ass(M) = Supp(M).
For the proof of the converse implication, suppose that Ass(M) consists
of maximal ideals, and consider a chain of the form Mo c M 1 c... c M n
= M, where M}IM J - 1  RIP} and PJ are prime ideals whose existence is
guaranteed by Lemma 2.4.12. By the same lemma, Ass(M) c: {PI' ..., Pn}.
In addition, PJ e Supp(MJ) since (RIPJ)pJ ¥= 0 and the localization functor
is exact. It follows that M pJ :f:. 0, hence Pj E Supp(M). These inclusions, to-
gether with Theorem 2.4.22 and the assumption that Ass(M) consists of maxi-
mal ideals, show that P 1, ..., P n are maximal ideals, and, in consequence,
Mo c M 1 c: ... c M n is a composition series of M. This proves that M is
of finite length. 0
From item 3 of Example 2.7.5 we know that there are Artin modules
\vhich are not Noetherian. For rings the situation is quite different. Here is
a characterization of Artin rings:
Theorem 2.7.12
A. ring R is an Artin ring if and only if the following two conditions are sat-
isfied :
(i) R is Noetherian,
(ii) every prime ideal of R is maximal.
If R is an Artin ring, then it has only a finite number of prime ideals, and
rad(R) is nilpotent.
Proof
If R has the properties (i) and (ii), then Ass(R) consists only of maximal
ideals. Therefore it follows from Theorem 2.7.11 that the ring R is of :finite

100
Noetherian Rings and Modules
[Ch.
length, i.e., R is an Artin R-module in view of Theorem 2.7.10. This lueans
that R is an Artin ring.
Assume now that R is an Artin ring. Let d denote the family of finite
products of maximal ideals in .R. d is nonempty and so has a minimal element
IDl. We shall show that rol = O. Assume that, on the contrary, rol ¥= 0, and
observe that under this assumption 9Jl = 9)12 since 9)12 c: rol, rol 2 E d, and
m is minimal in .91. Hence, the family fJl consisting of all ideals I in R such that
.run -:F 0 is nonempty since it contains 9)1. If I is an element minimal in f!l
(R is an Artin ring), then J9R2 = 19J1 =F 0, i.e., IIDl E dl; the minimality of the
ideal I yields the equality 1m = I. We note, moreover, that I is a principal
ideal; indeed, in view of 1m =#: 0 there is x E I such that xIDl =#: O. It follows,
again by the minimality of I in 14, that I = (x). Since the definition of 9Jl
implies that rot is contained in the Jac.obson radical of R, we conclude from
the equality roll = I and Nakayama Lemma that I = 0, which contradicts
the assumption J9R =F O. Hence the family fJ is empty and 9Jl = o.
Let 9Jl = ml ... mp where m, are maximal ideals in R (not necessarily
distinct). If P is a prime ideal of R, then 0 = ml ... mp c: P, and from Theorem
1.1.7 and the maxima1ity ofm, we see that P = mJ for somej. Accordingly,
every prime ideal in R is maximal, and ml, ..., mp are the only prime (maximal)
ideals in R.
Put It = ml ... mk for 1  k  p. Then R has a finite filtration 10 = R
::::> 1 1 ::> 1 2 ::::> ... ::::> 1p = 0, and by the assumption and Theorem 2.7.6 each
of the ideals 1" is an Artin R-module. Hence the factors It-lIlt are also Artin
R-modules and are of finite dimension as linear spaces over Rlmt. Conse-
quently, the chain 10 :::> / 1 ::> ... ::::> Ip = 0 can be refined to a composition
series, which shows that R is of finite length, and it is Noetherian by Theorem
2.7.10.
From the above considerations it follows that all maximal ideals of R
are among the ideals ml, ..., m,. Since ml ... mp = 0 and rad(R) is the
product of all distinct maximal ideals then some power of rad(R) is equal
to zero. 0
Corollary 2.7.13
If R is an Artin ring and M is a finitely generated R-module, then M is of finite
length, and Ass(M) = Supp(M).
Proof
By Corollary 2.7.8, M is an Actin D10dule, and by Theorem 2.7.12 and Corollary
2.1.8 M is a Noetherian module. Therefore it follows, in view of Theorem
2.7.10, that M is of finite length. Since aU prime ideals of R are maximal, the
equality Ass(M) = Supp(M) is a consequence of Theorem 2.4.22. 0
Corollary 2.7.14
If R is a Noetherian ring, then a finitely generated R-module M is of finite
length if and only if RIAnn(M) is an Artin ring.

II]
Artin Rings and Modules
101
Proof
By Theorem 2.7.11, it is sufficient to show that Ass(M) consists of maximal
ideals if and only if R/ Ann(M) is an Artin ring. In view of Theorem 2.7.12, this
is the case if and only if the only prime ideals containing Ann(M) are finitely
many maximal ideals. The above equivalence follows from Theorem 2.4.22
and the description of Supp(M) as the set of all prime ideals containing Ann(M),
given in Theorem 2.4.21. 0
Concluding this section, we state a structural theorem for Artin rings.
Theorem 2.7.15
(i) Any Artin ring is isomorphic to a product of a finite number of local
Artin rings.
(ii) A local ring is an Artin ring if and only if its maximal ideal is nilpotent.
Proof
(i) Let ml,..., mp be all distinct maximal ideals of a ring R. The proof
of Theorem 2.7.12 shows that there exists a number n > 0 such that m ... m
O S . n n R C · -L · h n n n n b
= . IDee m, +mJ = J.or I r J, we ave mI... mp = mjn ... nmp y
Theorem 1.1.9, and the mapping R  (Rlm) x ... x (RIm:) is an isomorphism
by Theorem 1.1.9. Each of the rings R/m7 is local and Artinian, which completes
the proof of part (i) of the theorem.
(ii) The radical of an Artin ring is nilpotent by Theorem 2.7.12. In a local
r-'"
ring the radical is equal to the only maximal idea], whence follows part (ii)
of the theorem.
If the maximal ideal of a local ring R is nilpotent, then it is the only prime
ideal of R by Corollary 1.1.4. Hence, by Theorem 2.7.12, R is an Artin ring.
o
We shall use Theorem 2.7.15 to describe the total ring of fractions.
Theorem 2.7.16
Let R be a Noetherian ring without nilpotent elements, and let PI' ..., P:J
be all the minimal prime ideals of R. Then the homomorphism
a
R  II (R/P,)o, fH(r+P,) 
1=1 d6
is the localization of the ring R with respect to the set of all non-zero-divis s'i r
to . I
Proof
Let S denote the set of all non-zero-divisors of R. Theorem 2.3.22 and the
assumption show, that R,",S = P 1 u ... uPs and that 0 = Ptn ... nPs is an
irredundant primary decomposition. From Corollary 1.1.8 we deduce that
PIRS' ..., PaRs are all the prime ideals of the ring Rs, and hence they are
maximal ideals. By Theorem 2.7.12, Rs is an Artin ring. Rs has no nilpotent
.

102
Noetherian Rings and Modules
[Ch.
m
elements, and so from Theorem 2.7.15 we see that Rs . II k" where k i
1=1
are fields. Denote by n, the homomorphism R -+ Rs  llk. -+ k" i = 1, ...
... , m. Then P; = Ker(n,) is a prime ideal, and 0 = n P;. It is easy to see
that the ideal P;R s is the kernel of the homomorphism Rs  llk, -+ k i ,
and to conclude that the decomposition 0 = n P; is irredundant. Moreover,
,
we have (see Ex. 3., Section 1.4) k,  Rs/PtRs  (R/P;)o. Since P 1 n ... nPs
= P n ... nP = 0 are both irredundant the set of Pi and the set of Pi coincide
and the theorem follows. 0
Exercises
1. Let R be an Artin ring and m a maxin1al ideal in R. Prove that, for n large enough,
we have an isomorphism Rm  R/mn.
2. Prove that the decomposition of an Artin ring in Theorem 2.7.15 is unique up to
a permutation and an isomorphism of the components. [Use the existence of a primary
decomposition of the zero ideal].
3. Prove that if R is a Noetherian ring and the irredundant primary decomposition
of the zero ideal 0 = Q 1 ('\ ... ('\Qs has no embedded components, then the homomorphism
s
R -+ II (R/Q,)p 10 , r I (r+Q,),
1= 1 ' ,
where P, = rad(Q ,), is the localization of R with respect to the set of all non-zero-divisors.
Show that the hypothesis on the embedded components cannot be omitted. [Consider
the factor ring of the polynomial ring K[X, Y] by the ideal (X, XY).]
4. Compute the length of the ring K[X 1 , ..., X,J/(X 1 , ..., Xn)P where K is a field, and
n, p are natural numbers.
5. Let T be a subring of the polynomial ring R = K[X] generated over a field K by
X n ,X"+ 1 , ..., xn+1' for fixed natural numbers n, p. Moreover, let 1= T('\R(X n + p + 1 ). Prove
that the factor ring T/ I is an Artin ring and calculate its length.
NOTES AND REFERENCES
The class of rings in which increasing chains of ideals become stable was first distinguished
by Emmy Noether in [22]. The paper contains also the basic theorem on the representation
of an ideal as an intersection of primary ideals and the uniqueness of this representation.
This comprehensive result was based on previous achievements, first of all on Hilbert basis
theorem, given in [9], for polynomial rings over the field of complex numbers. It had been
suggested by the theory of invariants and was used in that theory. Making use of it, Lasker
proved in [15] that there exists a primary decomposition of any ideal in a polynomial ring
over the field of complex numbers. It was also Lasker who introduced the notion of a primary
ideal, although some examples of such ideals which are not powers of prime ideals are implicit
in papers by Dedekind and Kronecker.
A distinction between primary ideals which correspond to isolated prime ideals and those
which correspond to embedded ones was first introduced by Macaulay in [17], where
the uniquness of primary decomposition was proved in the case where the prime ideals
associated with all primary ideals are isolated (Corollary 2.3.26). It was certainly also the
theory of ideals in rings of algebraic integers, created by Dedekind [6], which suggested the
axiomatic approach of E. Noether.

II]
Artin Rings and Modules
103
Cohen's elegant characterization of Noetherian rings (Theorem 2.1.15) comes from [5].
The converse of Corollary 2.1.10 says that for a ring T and its subring R, if Tis finitely
generated as an R-module and T is Noetherian then so is R. This is called Eakin-Nagata's
Theorem, see [7], [20] and [11] for a simple proof using injective modules.
In [8] Gilmer describes an influence of E. Noether's ideas on the development of commu-
tative ring theory.
According to Nagata [N], the Artin-Rees Lemma (Lemma 2.5.1) was formulated by
E. Artin in his lectures in Kyoto, 1955. A particular case is to be found in a paper by Rees
[25]. In proving the Artin-Rees Lemma, we have used notes of I. Kaplansky [K]. The con-
struction of the ring
T== R+IX+12X2+... +lnx"+ ...,
which is involved in it, is due to Rees. The ring T is usually called the Rees ring associated
with the ideal!.
The Krull Intersection Theorem (orollary 2.5.4) is a generalization, due to Chevalley
[3], of the first version (Corollary 2.5.5) given by Krull in [14].
The concepts of adic topologies and complete rings derive from the general ideas con-
tained in papers by Hensel. In his book [H], one can find the presentation of algebraic number
theory from the point of view based. on the investigation of rings of P-adic numbers, where P
denotes a prime ideal of a ring of algebraic integers of a number field. Further development
of the general theory of valuation (e.g. in [12]) has paved the way for structural results
concerning complete fields and local rings.
It was Artin who first studied systematically, in [2], rings (not only commutative) satisfy-
ing the condition of stabilization for decreasing chains of ideals. The characterization of
commutative Artin rings (Theorem 2.7.12) is due to Akizuki [1].

Chapter III
Integral Extensions and Dedekind
Domains
In number theory one studies equations, or systelns of equations, of the form
f(x I, ..., x n ) = g(x 1, ..., x,,), where f, g are polynomials with integer coeffi-
cients. Such equations are called Diophantine. The first question is whether
there exists a solution of the equation in the ring of integers. If the answer
to the first question is positive, there arises a second one: what is the set of all
such solutions? The most famous Diophantine equation is Fermat's equation
x +xS = x for p  3 (in the case of p = 2 we know all solutions). We may
restrict ourselves to exponents p which are prime numbers. All the results
obtained so far show that for some prime numbers p Fermat's equation has
no non-trivial integer solutions.
The methods devised to determine whether a Diophantine equation has
solutions have influenced essentially not only the development of number
theory itself but also that of algebra (the theory of rings and modules) and
other branches of mathematics, e.g., analytic functions. These methods are,
of course, very diverse; here we shall only show, using Fermat's equation as
an example, in what way they lead to the theory of Dedekind domains.
The first step in the investigation of Fermat's equation consists, in almost
all known cases, in factorizing the forms occurring in it. Suppose that integers
Xl' X2, X3 satisfy the equality X+X = x, where p is a prime number,
p  3. Denote by C the complex number C = e 27U / P ; then - 1, - C, ..., - C p - 1
are all the roots of the equation X'+ 1 = 0, whence
p-l p-l
x P + 1 = IT (x+C k ), and x +x = II (Xl +.C kX 2).
k=O k=O
We can write consequently
(Xl +X2) (Xl + Cx 2 ) (Xl +C2X2) ... (Xl +C,,-IX2) = x. (1)
The factors on the left-hand side of equality (1) are no longer integers, but
belong to the subring R of the field Q(C), generated by the ring of integers Z
and the number C, R = Z+ZC+ ... +ZCp-l. If the ring R were a unique
factorization domain, it could be demonstrated relatively easily that there
exist no solutions of Fermat's equation x +x = x not divisible by p (see
[C]). This amounts, roughly speaking, to half of the proof of non-existence
of solutions.

III]
Integral Extensions
105
The study of Fermat's equation, and many other Diophantine equations,
proceeds in most cases as follows: we first form a subfield K of the field of
complex numbers (Q(C) in the above reasoning) by adjoining to the field
of rational numbers Q the comple numbers needed to factorize polynomials
occurring in the equation into less complicated factors. Then we choose in K
a subring R, a unique factorization domain if possible, which contains the
factors obtained, our aim being to compare the resulting factorizations of
elements of the ring R (equality (1) above). Therefore one should bear in
mind that factorizations in R have to be non-trivial; thus the choice R = K
is useless. The choice of the ring R is discussed in greater detail in Section 3.1.
The basic difficulty lies in the fact that it is only exceptionally that the ring R
may be chosen as a unique factorization domain and at the same time not
too large. For example, for prime numbers p and the roots C" = e 21C1 / P of the
p-th degree of 1, the ring R = Z[,",] is a unique factorization domain only
for p == 3,5,7,11,13,17,19. In spite of this, a detailed study of the non-
uniqueness of factorizations enables us to derive the final result.
It turns out that the most convenient choice of the ring R in the field K,
which is a finite extension of the field of rational numbers, is the integral
closure Z of the ring Z in K, that is, the set of all numbers IX E K satisfying
an equation of the form «"'+rm-l «.-1 + ... +ro = 0, where "0' ..., r m - 1 E Z.
In those rings the theorem on the uniqueness of factorization into a product
of irreducible elements is not valid, in general. However, it is possible to develop
good methods for studying Diophantine equations, based on the theorem
asserting that in the ring Z every ideal can be factorized uniquely into a product
of prime ideals (in the classical version known as "ideal numbers"). This new
version of the uniqueness theorem proves to be a useful and efficient tool
for the study of Diophantine equations.
Investigation of the rings Z c K described abov started in the second
half of the 19th century (Kummer, Dedekind). It was found later that their
essential properties hold for a much larger class of rings, now called the
Dedekind domains.
In Chapter III we discuss the concepts of integral extension, integral
closure, normal domain, and Dedekind domain, which are all a development
of the classical concepts of algebraic number theory.
3.1 INTEGRAL EXTENSIONS
In algebra and number theory a fundamental role is played by algebraic
field extensions, i.e., field extensions K c: L such that, for each element IX E L,
there exists a polynomial f(X) = a"X n + ... + ao E K[X] subject to the condi-
tions an =F 0 andf(a) = o. We may additionally assume that an = 1, and this
form of the definition of an algebraic field extension is most suitable for
generalization to the case of rings, leading to the concept of an integral ring

106
Integral Extensions and Dedekind Domains
[Ch.
extension. At the beginning of this section we prove the basic properties
of integral extensions. The most important of them is the theorem asserting
that the set R of all elements of a ring T ::::> R integral over R is a ring, called
the integral closure of R in T. A ring R is said to be normal if it is a domain
and its integral closure in the field of fractions Ro is equal to R. We give the
proof of a simple but important theorem stating that a unique factorization
domain is normal. We also study the connections between the ideals of rings R
and T which form an integral extension R c T.
Assume that a ring R is a subring of a ring T. We recall that then the unity
element of T belongs to R and is necessarily the unity element of the ring R.
We call the ring T an extension of the ring R. We also refer to the inclusion
homomorphism R eTas an extension. The ring T may then be regarded as an
R-module.
Example 3.1.1
For any ring R and any set A, the ring of polynomials in indeterminates
{Xcx}cxeA with coefficients in R is an extension of R. If S is a multiplicative subset
of a domain R, then the ring Rs of fractions is an extension of R.
Definition 3.1.2
Let R c T be a ring extension. An element 1 E T is called integral over R
provided that there exist elements '0' ..., , m-l E R such that the following
equality
t m +rm_ 1 t m - 1 + ... +r11+ro = 0 (2)
holds. Equality (2) is said to be a relation of integral dependence for t over R.
If every element of T is integral over R, then we call R c T an integral
extension.
Example 3.1.3
The reader will easily verify that, in the extension Z c Q, the only elements
integral over Z are the elements of the ring Z.
Theorem 3.1.4
Assume that R is a domain with the field of fractions Ro. If a field K is an
extension of Ro, and if an element 0; E K is algebraic over Ro, then there exists
a non-zero polynomial fER [X] such that f( tX) = 0, and there exists an element
o #= r e R such that the element ro; is integral over R.
Proof
If 0 =F g E Ro [X] and g(tX) = 0, then by multiplying the polynomial g by the
product of the denominators of the coefficients of g we obtain a polynomial
f(X) = rX m +'m_l xm - 1 +ro e R[X] such thatf(tX) = 0 and, =F O. The element
ro; is integral over R since
(roc)m+ rm _ 1 (roc)m-1+'m_2r(ra)m-2+ ... +'l,.m-2(ra)+r o rn- 1 = o. 0

III]
Integral Extensions
107
As an immediate consequence of the definition of an integral extension
we get
Corollary 3.1.5
If R c T is an integral ring extension and if S c R is a multiplicative subset,
then Rs c Ts is an integral extension.
OUf first goal is to show that in a given ring extension  c: T, the set of all
elements integral over R forms a ring. In proving this, we shall find useful
the criterion established below. Recall that for any elements t l' ..., tk e T,
by R [t l' ..., tk] we denote the subring of T generated by R and the elements
t 1, ..., tk. The elements of this ring are sums of elements of the form rt1 ...
... t:", where r E Rand n 1  0, ..., nk  0 are non-negative integers.
Theorem 3.1.6
Let R c T be a ring. extension. For every element t e T, the following condi-
tions are equivalent:
(i) the element t is integral over R,
(ii) the ring R [t] is finitely generated as an R-module,
(iii) the ring R [t] is contained in a subring of the ring T which is finitely
generated as an R-module,
(iv) there exists a finitely generated R-module MeT satisfying the follow-
ing conditions:
(a) tM c: M,
(b) ifu e R[t] and uM = 0 then u = o.
Proof
(i) => (ii). Let an element t satisfy the integral dependence (2) and denote by N
the finitely generated submodule N = R+Rt+ ... +Rt rn - 1 c T. Equation (2)
shows that t rn eN. If, for a fixed number /e  m, we have t" eN, then tic = r +
+ r t + ... + r-1 t rn - 1 for some r, r, ..., r-1 e R, and therefore tlc+ 1
= rt+rt2+ ... +r_1trn eN. Thus all the powers oftlieinNandN = R[t].
(ii) => (iii). Obvious.
(iii) => (iv). We take M to be a subring described in condition (Hi). Then
tM c M, since t EM. If u e R[t] and uM = 0, then u = u. 1 = 0 because
1 eM.
(iv) => (i). This follows from Lemma 1.3.2. 0
If we strengthen condition (a) by assuming that, for a certain ideal 1 c: T,
the inclusion tM c 1M holds, then there is an integral dependence t n +
+rn-1 t n - 1 + ... +ro = 0 with the coefficients ro, ..., r"-1 in I.
Theorem 3.1.6 enables us to prove the following theorem, essential for our
further considerations.

108
Integral Extensions and Dedekind Domains
[Ch.
Theorem 3.1.7
Let R c T be a ring extension. The set of all elements of T, integral over R,
forms a subring of T. If elements t 1, ..., t" e T are integral over R, then the
ring R[t 1 , ..., I,,] is a finitely generated R-module.
Proof
Let elements t 1 , 1 2 E T be integral over R. It follows from Theorem 3.1.6
(ii) that the ring R[t 1 , 1 2 ] = R[1 1 ] [1 2 ] is a finitely generated R[t 1 ]-module,
and hence a finitely generated R-module because R[t 1 ] is a finitely generated
R-module. The rings R[t 1 +1 2 ], R[1 1 -t 2 ], R[t 1 .t 2 ] are all contained in the
ring R [t 1 , t 2], and so Theorem 3.1.6 (iii) shows that the elements t 1 + t 2 ,
11 - t 2, 1 1 1 2 are integral over R. Thus we have established the first part of the
theorem, and the second part for k = 2. We finish the proof by obvious
induction. 0
Corollary 3.1.8
H R c R 1 , R 1 c: T are ring extensions and the extension R c: R 1 is integral,
then each element of T which is integral over R 1 is also integral over R.
Proof
If 1 e T is integral over R 1 , then there are elements Yo, ..., r"'-1 E R such that
t m +rm_1 t m - 1 + ... +ro = O. The elements '0' ..., r na -l are integral over R,
whence the ring R 2 = R ["0' ..., l- na -1] is a finitely generated R-module. The
ring R 2 [t] is a finitely generated R 2 -module since the element t is also integral
over R 2 . This shows that R [ro, ..., r"._ 1, t] is a finitely generated R-module,
and t turns out to be integral over R. 0
Corollary 3.1.9
If R c: R 1 , R 1 c: T are integral ring extensions, then R c T is also an integral
extension.
Given an extension R c: T we denote by R the subring of T consisting of all
elements integral over R. We clearly have R c: R c T, and, by Corollary 3.1.8,
every element of T integral over R belongs to R, i.e., R = R . The ring R is
called the normaUzation (or the integral closure) of R in T. If R = R, then we
say that R is normal (or integrally closed) in T.
Of particular importance is the case where R is a domain and T is the field
Ro of fractions of R. A domain is said to be normal (or integrally closed)
if it is normal in its field of fractions. By the normalization of a domain we
mean the normalization in its field of fractions.
In Chapter IV we shall make use of the following generalization of the
notion of normality. Observe first that, if a ring R is Noetherian, then condition
(ii) in Theorem 3.1.6 is equivalent to the following condition:
(ii') the ring R[t] is contained in a finitely generated R-submodule of T.

III]
Integral Extensions
109
Indeed, by the assumption that R is Noetherian, a submodule of a finitely
generated R-module is finitely generated.
An element t satisfying condition (ii / ) is called almost integral over R.
Proceeding as in the proof of Theorem 3.1.7, we can show that the almost
integral elements form a subring of the ring T. A domain is said to be completely
normal if each element t E Ro almost integral over R belongs to R.
A connection between two important classes of rings is formulated in the
theorem which follows.
Theorem 3.1.10
A unique factorization don1ain is normal.
Proof
Let R be a unique factorization domain, and let 0 :/: x E Ro be an element
integral over R. Then there exist relatively prime elements r, S E R such that
x = r Is. Furthermore, there exist elements "0' ..., r m- 1 E R such that x'" +
+rm_lX".-l+ ... +r1x+rO = O. By substituting x = r/s in this equality
and multiplying by s'" we obtain rm+rn._trm-1s+ ... +rtrs'"-l+ros''' = o.
If the element 8 were not invertible in R, there would exist an irreducible
element p dividing 8, and from the above equality we would deduce that p
divides r m , and hence that p divides 1", which contradicts the assumption that
the elements r, s are relatively prime. The element 8 is therefore invertible
in R, and x E .R. 0
1nIeoremn 3.1.11
Every ring of fractions of a normal domain is normal.
Proof
Let S be a multiplicative subset of a normal domain R. Then (Rs)o = Ro
and, if an element x E Ro satisfies an integral dependence
x'"+tm_tx".-1+ ... +t 1 x+t o = 0, (3)
where to,..., t rn - 1 e Rs, then there exist elements ro,..., r m -l E R, s E S
such that to = To/S, t 1 = rl/s, ..., tm-1=rm-1/S. Multiplying relation (3)
by 8 m , we get
(sx)m +rm_1S(Sx)rn-l + ... +"1 s".-1(sx)+ros"l = 0,
whence, by the normality of R, sx e R and finally x = sx/s E Rs. D
Let us now return to the problem of choosing, in a finite field extension
K:::> Q of the field Q of rational numbers, a ring useful for the study of a given
Diophantine equation, for which the field K has been selected. Suppose that
a ring R c K is a unique factorization domain, and Ro = K. Since Z c R
and R is normal (by Theorem 3.1.10) we have R :::J Z, where Z denotes the
integral closure of the ring Z in the field K. The ring Z is not in general a unique
factorization domain, which is the main (though not only) difficulty in studying

110
Integral Extensions and Dedekind Domains
[Ch.
Diophantine equations. Nevertheless, as we shall show in Section 3.5, the
localizations Zp of the ring Z with respect to prime ideals P are principal ideal
domains, and hence unique factorization domains, and of course Z = n Zp .
- p
The ring Z can therefore be relatively well approximated by unique factoriza-
tion domains. It might be thought that the choice of one of the rings Zp
would be appropriate; it turns out, however, that the rings of this form contain
too many invertible elements, which oversimplifies factorizations.
Summing up, the best and generally accepted choice of a ring serving for
the investigation of a Diophantine equation in a field K => Q chosen for this
purpose is the integral closure Z of the ring Z in K. The ring Z is called the
ring of integers of K (or the ring of algebraic integers of K) with its elements
called integers of K.
We have pointed out the r61e of ideals in the study of Diophantine equations.
In the classical case, one studied the relations between the ideals of the ring Z
of integers and those of the ring of integers of a field K, which is a finite exten-
sion of Q. However, it turned out that many of those connections carryover
to integral ring extensions. In what follows, we concentrate on the connections
in question.
eorem 3.1.12
If a ring extension R c T is integral, then a prime ideal Q of T is maximal
!f and only if the ideal QnR of R is maximal.
Proof
Suppose first that Q = 0; then T, and hence R, have no zero-divisors. We
must show that T is a field if and only if R is a field.
Let T be a field and consider an element 0 :/= r E R. There is t E T such
that tr = 1. The element t is integral over R, and so there exist elements
ro, ..., r m -1 e R such that t'"+r m _1,,,,-1+ ... +ro :s: O. By multiplying this
equality by r m we obtain 1+rm-1r+ ... +r1rm-1+rorm = 0, which shows
that 1 = -r(r m - 1 +rm-2r+ ... +ro,'1I-1). Accordingly t = -(r m -1 +rm-2r+
+ ... +ro,....-l) e R, and R is a field.
Let R be a field and consider an element 0 =1= t E T. The element t is algebraic
over the field R, whence the homomorphism ({J,: R[X] -+ T, given by f!Jtif)
= I(t) for 1 e R[X], has a non-zero kernel I. The ideal I is prime, and hence
maximal. It follows that Im( f!Jt) is a field. Since t e Im( ((Jt) c T, t is invertible
in T and T is a field.
Now we pass to the general case. Identifying the ring Rf(QnR) with the
subring of the ring T/Q consisting of residue classes of the form r+Q for r E R,
we apply the special case proved above to the integral extension R/(QnR)
c TfQ of domains, and the theorem follows. 0
The assumption that Q is a prime ideal is essential (see Exercise 7).

III]
Integral Extensions
III
Theorem 3.1.13
If a ring extension R c: T is integral, and if Q c Q1 are prime ideals of T
and QnR = QlnR, then Q = Q1.
Proof
Assume first that the prime ideal P = QnR = Q1nR is maximal. Then it
follows by Theorem 3.1.12, that Q is maximal, whence Q = Q1.
We pass to the general case. Denote by S the multiplicative set S = R"-.P.
Then S c: R c T and SnQ = SnQ1 = 0, and thus, by Corollary 1.4.8,
it is sufficient to show that the ideals QTs, Q1 Ts of the ring Ts coincide. The
extension Rs c Ts is integral and RS:F Q1 TsnRs ::> PRs. The ideal PRs
c Rs is maximal, whence Q1 TsnRs == P Rs, and analogously QTsnRs = P 1"\ .
Applying to the ideals QTs, Ql Ts the established particular case, we deduce
that they are equal, whence also Q = Qt. 0
Corollary 3.1.14
If an extension R c T of domains is integral, and if Q c T is a non-zero
prime ideal of T, then QnR :/= O.
Proof
There exists an element 0 :F t eQ, and if t m +r"'_lt m - 1 + ... +ro = 0, where
ro, ..., r m -l e R, is an integral dependence of t of minimal degree, then it
follows from the minimality of the number m that '0 =1= 0, and moreover
ro e Qr.R. 0
The assumption that T is a domain is essential (see Exercise 7).
1beorem 3.1.15
If an extension R c: T of domains is integral, and if every non-zero prime ideal
of R is maximal, then every non-zero prime ideal of T is also maximal.
Proof
Let 0 :F Q c Q be prime ideals of T. Corollary 3.1.14 shows that 0 :/= Qn
nR c Q1nR, whence, by assumption, QnR = Q1nR since both ideals are
prime. From Theorem 3.1.13 we can therefore conclude that Q = Ql, and
thus the ideal Q is maximal. D
Theorem 3.1.16
If a ring extension R c T is integral, and if P is a prime ideal of R, then there
exists a prime ideal Q of T such that QnR == P (i.e., the extension j: R c: T
induces a surjection Spec(j): Spec(T)... Spec(R»).
Proof
We first assume that P is a unique maximal ideal of R. Let Q be any maximal
ideal of T. Then, by Theorem 3.1.12, the ideal QnR is maximal, which yields
QnR = P.

112
Integral Extensions and Dedekind Domains
[Cn.
We now pass to the general case. Let S denote the multiplicative set S
= R","P. The extension Rs c: Ts is integral, and we have a commutative
diagram.
R
c
T
w
)Vt
Rs
c
Ts
Let Q' be an arbitrary maximal ideal of the ring Ts and let Q = Wi"I(Q').
Then Q'nRs = PRs by the above special case, whence QnR = wl l (Q')nR
= W-1(Q'nRS) = W-1(PRs) = P, and the theorem is proved (see notation
in Section 1.4). D
Theorem 3.1.17 ("Going up")
If a ring extension R c: T is integral, and if
PI c: ... c: Pm c: ... c: P II
is a chain of prime ideals of R, and if further
Ql c: ... c: Qm
is a chain of prime ideals of T, 0  In < n, such that
Q1 fiR == PI' ..., Qnr nR = Pm'
then there exists an extension of the chain (4) to a chain of prime ideals of T
of the form
Ql c: ... c: Qm c: ... c: Q"
such that
(4)
Q,nR=P t , i=1,2,...,n.
Proof
If m  1, then, applying Theorent 3.1.16 to the extension RIP," = R/(Qmr.R)
c: T/Qm and the ideal P m + 1 / P,n, we conclude that there is a prime ideal Qrn+ l/Q".
of the ring T/Q,n such that (Qm+l/Qm)n(R/P lII ) = Pm+t/P".. This equality
implies Qm+lnR = P lII + 1 . Repeating this construction we successively find
the ideals Qm+2' ..., QII.
If m = 0, an application of Theorem 3.1.16 reduces the proof to the case
nl = I. D
The following theorem will be needed in Chapter IV.

III]
Integral Extensions
113
Theorem 3.1.18
Let R c T be a ring extension. Further, assume that (R, m) is a local ring and
that the ring T is finitely generated as an R-module. Then:
(i) T is semi-local,
(ii) if m' c T is the intersection of all maximal ideals of T, then the m' -adic
topology of the ring T is identical with the m-adic topology of T as an R-
module.
Proof
The ring T is Noetherian by Corollary 2.2.4. Moreover, for any t e T, the
subring R[t] is a finitely generated R-module according to Corollary 2.1.8.
This means that the extension R c T is integral. If Q c T is a prime ideal
and Q ::> m then QnR= m. Consequently Q is a maximal ideal by Theorem
3.1.12. Using Theorem 3.1.12 once more, we deduce that the maximal ideals
of T are the same as the minimal prime ideals of the ideal mT, and thus T
is a semi-local ring.
It follows from Corollary 1.1.5 that m' = rad(mT). The m'-adic topology
is therefore identical \vith the mT-adic topology, which is the m-adic topology
of T as an R-module. 0
Exercises
1. Let R c T be an extension of domains and R the integral closure of R in T. If S is
a multiplicative subset of R, then by (if;) we denote the integral closure of the ring Rs in Ts.
Prove that ( R) s = ( Rs) . Deduce from this that a localization of a normal domain is a
normal domain.
2. Let d be an integer not divisible by a square of any prime number. Prove that the
integral closure of the ring Z in the field Q(yd) is "the ring Z[s] = Z+Zs, where e = y d
if d :t: 1 (mod 4), and 8 = (1 + vd)/2 if d == 1 (mod 4). [Show that a number a+by d
e Q( y'd) is integral over Z if and only if 2a, a 2 - b 2 d e Z. Consider separately the case
where 2a is even and the case where it is odd.]
3. Prove that a domain R is normal if and only if for every finitely generated ideal I C R,
I:F 0, we have the equality {x E Ro; xl C I} = R.
4. Let K be a field of characteristic :F 2, and let f e X[X] be a polynomial of positive
degree which is not divisible by any square of a polynomial of positive degree. Prove that the
normalization of X[X] in the field X(X, VI) is the ring K[X, y j].
5. Let K[X 1 , ..., Xn] be the ring of polynomials with coefficients in a field K of charac..
teristic :F 2. Prove that the factor ring of this ring by the principal ideal (X: - Xl ... X.- 1 )
is normal.
6. Let R be a normal domain which is not a field. Prove that the subring T = R + RoX +
+ ROX2 + ... of the field Ro(X) is normal but not completely normal. Show that if R is, in
addition, completely normal, then T is completely integrally closed in Ro [X] and Ro [X]
is the complete integral closure of T.
7. Using as an example the extension Z C Z[X]/(X 2 ,2X), show that the assumptions:
(a) of Theorem 3.1.12 that the ideal Q is prime, d (b) of Corollary 3.1.14 that the ring T
is a domain are essential.

114
Integral Extensions and Dedekind Domains
[Ch.
3.2 NORMAL DOMAINS
In the first part of this section we continue to investigate the connections
between ideals of the rings of an integral extension R c T under the assump-
tion of the normality of the ring R, by proving Theorem 3.2.4 on going down
froin a sequence of prime ideals.
Generalizing the previously discussed situation Z c: Zo = Q c: K, where
K :::> Q was a finite field extension, we study the integral closure R of a domain R
in a field K, hich is a finite extension of the field Ro of fractions. We shall
first prove that R is contained in a finitely generated R-module if the domain R
is normal and the extension Ro c K is finite and separable. This theorem implies
that under those assumptions if R is a normal Noetherian domain, then so is R.
We state the basic properties of rings of algebraic integers, which will be used
to define the class of Dedekind domains. Many of the classical theorems on
rings of algebraic integers carryover to Dedekind domains, which shows that
this important class of rings is natural.
In the proof of the "Going down" Theorem 3.2.4 for chains of prime ideals
we shall use the following lemmas:
Lemma 3.2.1
If a ring extension R c: T is integral, and if I is an ideal of R, then the radical
of the ideal IT consists of all elements t e T for which there exist elements
ro, ..., r m -1 E I satisfying the condition
t m +r t lll-1 + + / " - 0
m-1 · · · 0 - ·
(5)
Proof
If an element t satisfies equation (5), then t m E IT, and hence t E rad(IT).
If t E rad(IT), then t n E IT for some positive integer n and there exist elements
S1, ..., Sk E I, 1 1 , ..., tk E T such that t ll = L s,t,. The subring T 1 = R[t 1 , ...
I
... , t k ] of T is a finitely generated R-module and t n T 1 c: L Si ti T 1 c: L S, T 1
. I i
C IT 1 . The remark following the proof of Theorem 3.1.6 implies that the
element t satisfies equality (5) for suitably chosen elements '0' ..., r m-1 E I.
o
Lemma 3.2.2
If a domain R is normal and if the product of two monic polynomials f, g
E Ro[X] belongs to the ring R[X], thenf, g E R[X].
Proof
It is known that there is a field extension K :::J Ro in which the polynomials f,
g can be factorized into a product of linear factors I(X) = IT (X - ai), g(X)
I
= IT (X -hJ). Since ai, h J are the zeros of the monic polynomial fg E R[X],
j

III]
Normal Domains
115
they are integral over R. The coefficients of / and g, as symmetric functions
of a1, ... and h 1 , ..., respectively, are integral over R and lie in Ro, and thus
in R because the domain R is normal. 0
Lemma 3.2.3
Let R be a normal domain and let R c T be an integral ring extension such
that no element of R is a zero-divisor in T. Then, for each t E T, the kernel
of the homomorphism ({Jt: R[X] --+ T defined by rpt(f) = I(t) is a principal
ideal generated by a monic polynomial.
Proof
We denote Ker«({Jt) by I; then IRo[X] is a non-zero ideal in the principal
ideal domain Ro [X]. Let g E Ro [X] be a monic polynomial generating the ideal
IRo [X]. The element t is integral over R, whence there is a monic polynomial
h E R[X] such that h(t) = O. This yields h = gh 1 for some monic polynomial
h 1 e Ro[X]. Lemma 3.2.2 implies that g E R[X]. We shall show that g is a
generator of the ideal I.
Since g E IRo [X], there exists an element 0 :/: r E R such that rg E I, whence
rg(t) = O. The element r is not a zero-divisor in T, and so g(t) = 0, i.e.,
gEl. Let/e I; thusfe IRo[X] and there exist 0 :/= r1 E R,/1 E R[X] such that
f = (1/r1)!lg, i.e., r1f = Ilg. From this equality we obtain in the polynomial
- -
ring (RfRrl) [X] the equality 11 g = 0, where /1' g are the residue classes of the
polynomials 11 , g. As the polynomial g is monic, it is not a zero-divisor and
therefore 11 = 0, which means that all the coefficients of /1 are divisible by 1.1
in R. It follows that (lfrl)/1 e R[X] and, in consequence, I is divisible by g
in R [X]. Thus g is a generator of the ideal Ker( lpt). 0
The assumption that the elements of Rare non-zero-divisors in T is essen-
tial (see Exercise 1).
1beorem 3.2.4 ("Going down")
Let R c: l' be an integral ring extension. If R is a normal domain no ele-
ment of which is a zero-divisor in T, if
P 1 c ... C Pm C ... c: P n
is a chain of prime ideals of R, and if, further,
Qm c: ... c Qn (6)
is a chain of prime ideals of T, 1 < m  n, such that
Qm nR = Pm' ..., QnnR = Pn,
then there exists an extension of chain (6) to a chain of prime ideals of T of
the form
Q1 C ... c Qm c ... c: Qn
such that
QtnR=P" i= 1,...,n.

116
Integral Extensions and Dedekind Domains
[Ch.
Proof
It is sufficient to prove the theorem in the case n = m = 2. Denote by S the
multiplicative subset of T consisting of elements of the form rl, where r E R","P l'
t e Q2. From the assumption that no element of R is a zero-divisor in T
it follows that 0  s. Moreover, we have R"P1 c: S, T"Q2 C S.
We shall show how to find, under the assumption Sr. P 1 T = 0, an ideal Q1
satisfying our assertion. This assumption implies that the ideal P 1 Ts of the
ring Ts is proper, and thus it is contained in some maximal ideal m of Ts.
We let Ql denote the contraction of m to T, i.e., t e Q1  tl1 em. We then
find QlnS = 0, whence Ql fl (R"P1) = 0 and Qln(T'Q2) = 0. It follows
that QlnR c P 1 and Ql c Q2. Since m :::J PITS (the contraction of the
ideal PITS to Tis equal to P 1 T), we have QlnR :::J P 1 TnR :::J P 1 , and finally
Q1 nR = Pl.
Thus it remains to prove that SnP 1 T = 0. Assume that there exists
an element rt E SnPl T, where r E R"P 1 , t E T'Q2. Lemma 3.2.1 shows
that there are elements ''0' ..., "".-1 E P 1 such that' the polynomial f(X) =
= X"'+rm_1X...-1 + ... +ro E R[X] satisfies f(rt) = O. Let f1(X) = rmx m +
+r m -1 r m - 1 X...-1 + ... +ro e R[X]; then /1 (t) = 0 and, by Lemma 3.2.3, there
exists a monic polynomial g E R[X] generating the kernel of the homomor-
phism f/Jt: R[X] --+ R. Consequently, /1 = hg for some h e R[X]. Denote
by /1, g, h the residue classes of the polynomials /1' g, h in the factor ring
- -
(RIP 1 )[X]; we then have hg = 11 = imX m . The polynomial g is monic, and
so the above equality implies that g = X" for some k, 1  k  m, and
hence that g(X) = Xk+r_lXi-1+ ... +,., where r, ..:, r-l EP 1 . Since
g(t) = 0, it follows from Lemma 3.2.1 that t e rad(Pl T). Accordingly, for
some q > 0, we have t q E PtT c: P 2 T c Q2, which results in t e Q2' contrary
to the assumption. . D
Theorem 3.2.5
Let R be a normal domain. If Ro c K is a finite, separable field extension
and if R denotes the integral closure of R in K, then there exists a basis Xl' ...
. .. , X n of Kover Ro such that R is contained in the free R-module Rx 1 + ... +
+ Rx,. .
Before we proceed to prove the theorem, let us recall the definition and
properties of the trace (see [L, p. 210]). Let /, c K be a finite, separable field
extension. Denote by (J l' ..., G q all the homomorphisms from K into the
algebraic closure k of k. Then the trace of an element a e K is defined by the
formula
q
TrK/1c(a) = L rJi a ).
}=.1
The mapping TrKI": K -.. k is k-linear. The k-linear mapping T:
K --+ Homk(K, k) defined by the formula (r(a))(b) = TrKI,,(ab) for a, b e K

III]
Normal DomaiDs
117
is an isomorphism of linear spaces (here we make use of the separability of the
extension k c K).
If k = Ro then, for every_ a E R and j, the element C1J(a) is also integral
over R (in the extension R c k), whence TrK/k(a) is integral over R and belongs
to Ro; it belongs to R if the domain R is assumed to be norma].
Proof
From Theorem 3.1.4 it is easily seen that there exist elements V1, ...  VII E R
constituting a basis of Kover k = Ro. The mapping -r: K -+ Homk(K, k)
described above is an isomorphism, and so there exist elements Xl' ..., X,. E K
such that -r(x 1), ..., -r(X,.) form a basis dual to the basis V I' ..., '0,., that is,
'J = (r(xi))('OJ) = TrK/k(x, 'OJ) for i,j = 1, ..., n. Ob viously Xl' ..., x,. form
a basis of Kover k, whence for any element X E R there are aI, ..., a,. e k
such that x = La,Xt. TheelementsxvJ = L.aixlvJbelongto R forj = 1, ...
. . . , n, and thus the elements
TrKI"(X'OJ) = L 01 TrKI,,(xl'lJJ) = L 01 lJ 'j = OJ
are in R, and so R c RX1 + ... +Rx,.. 0
From Theorem 3.2.5 we readily obtain
Theorem 3.2.6
If R is a normal Noetherian domain and if Ro c K is a finite separable field
extension, then the integral closure of R in K is also a normal Noetherian
domain.
Proof
By Theorem 3.2.5, there exist elements Xl' ..., X n E K such that the integral
closure R of R in K is contained in the R-module Rx I + ... + Rx,.. . Every ideal I
of R is an R-submodule of RXI + ... +Rx,., and hence it is finitely generated
as an R-module. The ideal I is therefore also finitely generated as an R-module,
whence the domain R is Noetherian. 0
Corollary 3.2.7
Rings of algebraic integers (i.e., rings which are integral closures of Z in a field
which is a finite extension of the field of rational numbers) are Noetherian.
The theorems proved previously allow us to formulate the most important
properties of rings of algebraic integers. Let K => Q be a finite extension of the
field of rational numbers and denote by Z the integral closure of Z in K;
then:
(i) Z is a Noetherian domain,
(ii) Z is a normal domain,
(Hi) non-zero prime ideals of Z are maximal.

118
Integral Extensions and Dedekind Domains
[Ch.
Property (Hi) follows from Theorem 3.1.15. The above properties of the
ring Z will be taken as the defining properties of a Dedekind domain (see
Section 3.3).
Those algebraic varieties V whose rings of polynomial functions are normal
play an important role in algebraic geometry. Such a variety has a compara-
tively small set of singular points [R].
Exercises
1. Using the example of the extension Z C Z[X]/(X 2 ,2X) show that the assumption
in Lemma 3.2.3 that the elements of the ring Rare non-zero-divisors in T is essential.
2. Let p be a prime number :F 2. The number  = e 2 rr,'J" is a p-th root of 1. Prove that
the ring Z[C] is the integral closure of Z in the field Q(). [First prove that Z( l - )f1Z = (p).
Deduce from this that if x = 001-a1,"+ ... 1-a"_2"-2 e Z, then Tr«'-l)x) e(p) and
Tr(C-x) = -pao.]
3. Let R c T be an integral extension of domains. Prove that if R is a normal domain,
s e R, and Ts is a prime ideal, then Ts n R = Rs. [Establish first that Rs is a prime ideal,
and then use Theorem 3.2.4.]
4. Prove that a domain R is normal if and only if the ring of polynomials in any number
of indeterminates with coefficients in R is normal. [Reduce to the case of one indeterminate.
Observe that if an element t e Ro[X] satisfies the relation t"+O"-1 t"-1 + ... +00 = 0, where
00, ..., a"-1 e R[X], and if the non-negative integer m is larger than the degrees of the poly-
nomials t, 0,,-1, ..., 00, then the element u = t-X'" satisfies the relation U"+h_lU"-l+
+ ... + fo = 0, where fo, ... ,h-l E R[X] and fo is a monic polynomial. Apply Lemma
3.2.2.]
5. Prove that an element t e Ro is almost integral over the domain R if and only if there
exists a non-zero element a e R such that all the elements at, at 2 , at 3 , ... belong to R.
6. Prove that a domain R is completely normal if and only if the ring of polynomials
in any number of indeterminates with coefficients in R is completely normal. [Reduce to the
case of one indeterminate. Let elements t == To+ ... +rmX'" e Ro[X], 0 #= a e R[X) satisfy
the conditions at n e R[X] for n  O. Prove successively that: 1 0 To E R, 2 0 a(t- ro)ft e R[X)
for n  0, 3 0 Yl e R, ...].
3.3 DEDEKIND DOMAINS
At the end of the previous section we formulated three properties of rings
of algebraic integers. We now adopt them as a characteristic of a class of rings
called Dedekind domains. All rings of algebraic integers are thus Dedekind
domains. We shall prove the principal theorem on Dedekind domains: every
non-zero ideal is a product of prime ideals, and the representation of an ideal
in this form is unique up to the order of the factors. As has already been men-
tioned, this theorem plays a very important role in the study of Diophantine
equations. In the sequel we provide two characterizations of Dedekind do..
mains, in Theorems 3.3.10 and 3.3.11. Referring to the subject matter of Section
3.2, we shall establish that the integral closure R of a Dedekind domain R
in a finite extension of the field Ro offractions (without the separability assump..
tion) is also a Dedekind domain. We also investigate connections between the
ideals of R and those of R .

III]
Dedekind Domains
119
In studying ideals of a domain R, we shall make use of non-zero R-sub-
modules of the field of fractions Ro regarded as an R-module.
If M, N c: Ro are non-zero R-submodules, then by M. N (or MN) we
denote the R-submodule of Ro consisting of elements of the form L mint,
where mi E M, ni EN. If M = Rnl is a cyclic module, we write mN instead
of (Rm)N. The non-zero R-submodules of the field Ro form an associative,
comlnutative semi group with identity R. If, for a module M c: Ro, there
exists a submodule M ' c: Ro such that MM' = R, we call the module M
invertible. Observe that an invertible module is' finitely generated. Indeed, if
k
L mim = 1, In, E M, n1 E M ' , then for each element m E M, we have n1
1=0
Ie
= L mi(m m) and m fn E R, whence ml, ..., mk form a set of generators
1=0
for M.
Submodules of the field Ro of fractions will be used extensively in Chapter
IV.
Definition 3.3.1
A Dedekind domain is a ring R satisfying the following conditions:
(i) R is a Noetherian domain,
(ii) R is normal (i.e., integrally closed in its field of fractions),
(iii) non-zero prime ideals of R are maximal.
From Theorem (3.1.10) we conclude that every principal ideal domain is a
Dedekind domain.
Theorem 3.3.2
If P is a non-zero prime ideal of a Dedekind domain R and if pi = {x E Ro; xP
c R}, then PP ' = R.
Proof
It is immediately seen from the definition of P' that P' is a submodule of Ro,
pi :;:) Rand PP' c R. We pick an element 0 :F peP; then P'  pip c: R,
and consequently pi is a finitely generated R-module since the domain R is
Noetherian.
We shall show that pi i= R: the ideal Rp contains the product P1 ... p..
of prime ideals because Rp is an intersection of primary ideals, and every
primary ideal Q contains some power of the prime ideal rad(Q). Assume the
number s to be minimal. Since P :::> Rp ::> P 1 ... Ps the ideal P contains one
of the ideals P 1, ..., Ps, say P :::> P 1. The maximality of P 1 implies that
P = Pl.' whence Rp :::> PP 2 ... Ps and, by the minimality of s, one has Rp
:p P 2 ... PS. Thus there is an element r E P 2 .0. Ps such that r  Rp, which
yields rip E Ro"R. Since
(rlp)P = (1Ip)rP c: (1Ip)PP 2 ... Ps c: (1Ip)Rp = R,
we get rIp E pi, and eventually P' :F R.

120
Integral Extensions and Dedekind Domains
[Ch.
The definition immediately implies that P c: P' PeR, and thus, in view
of the maximality of the ideal P, to complete the proof it is sufficient to show
that P' P =1= P. Suppose that P' P = P; then for every element x e P', we have
xp c: P. It therefore follows from Theorem 3.1.6 that the element x is integral
over R, whence x E R, which contradicts the previously proved condition
P' * R. Accordingly we have P' P = R and the theorem is proved. 0
Theorem 3.3.3
Every non-zero ideal of a Dedekind domain has a unique (up to the ordering)
representation in the form of a product of prime ideals.
Proof
Let R be a Oedekind domain. Denote by J the family of those non-zero
ideals in R which have 110 representation as a product of prime ideals and
suppose that the family f is nonempty. Then there exists an ideal I maximal
in the family J. The ideal I is not maximal, whence it is properly contained
in some maximal ideal P  I. Using the previous notation P' = {x E Ro; xP
c: R}, we have P' P = R, and from the condition P :::> I we obtain IP' c: P P'
= R, while from the condition P' ::> R we get IP' ::> I. The ideals IP ' , I are
distinct. Indeed if we had IP ' = I, then for each element x E pi we would
have xl c: I, and it would follow from Theorem 3.1.6 that the element x is
integral over R, and consequently x E R, which is impossible by the condition
pi * R. Thus IP' * I, and the ideal IP' is proper since the equality IP' = R
implies I = IPP' = RP = P, contradicting the assumption P;;2 I. Con-
sequently, the ideal IP ' does not belong to the family J, and so [P' = PI... Ps
for some prime ideals PI' ..., P;o Hence we obtain [ = [R = IP'P = PI' ...
. . . , Ps pi, which contradicts the assumption that I E J. Therefore the family J
is empty, and every non-zero ideal of the ring R is a product of prime ideals.
It remains to prove that the representation is unique. Assume that
PI ... Ps = QI ... Qr (7)
and that all the ideals PI' ..., PH QI, ..., Qr are prime and non-zero. Since
we have PI :::> PI... Ps = QI ... Q" it follows by Theorem 1.1.7 that there
exists j, 1  j  r, such that PI:;:) QJ. By a similar argument, the ideal Qj
contains one of the ideals Pi. Therefore PI::> Qj ::) P" and consequently,
by the maximality of Pi, we have PI = Pi = Qj. Multiplying (7) by P,
we obtain P 2 ... Pf; = Ql ... QJ-IQJ+I ... Qr. Continuing this reduction,
we arrive at the conclusion that r = s, and the sequence Q 1 , ..., Qs turns out
to be a permutation of the sequence PI' ..., P s. D
Theorems 3.3.2 and 3.3.3 yield
Corollary 3.3.4
A non-zero ideal of a Dedekind domain is invertible.

III]
Dedekind Domains
121
Theorem 3.3.3 shows that every non-zero ideal of a Dedekind domain R
can be written uniquely in the form
I = Il pnp(l),
p
where P runs through the non-zero prime ideals of R, the np(/) are non-negative
integers and np(/) = 0 for almost all P. Thus the ideal I determines a function
n(I) defined on the set of non-zero prime ideals of R. From Theorem 1.1.9
we readily deduce that also I = n pnp(l). Furthermore, it follows from Lemma
p
2.3.5 that if n > 0 then the ideal P" is P-primary. Theorem 3.3.3 is thus a
strengthening of Theorem 2.3.11 on the primary decompositions of an ideal,
because it describes the structure of primary ideals (see Example 2, Section
2.3).
Theorem 3.3.5
Let I, J be non-zero ideals of a Oedektnd domain R. The functions n(I), n(J)
satisfy the following conditions:
(i) n(IJ) = n(I)+n(J),
(ii) I c: J <=> n(I)  n(J),
(iii) n(I+J) = min{n(I), n(J)),
(iv) n(I(")J) = max{n(I), 11 (J)).
Proof
(i) is obvious.
(ii) Let I c: J; by Theorems .3.3 and 3.3.2, there exists a submodule
J' c: Ro such that JJ' = R. Then we have IJ' c: JJ' = R, which shows that
IJ' is a non-zero ideal of R and, in addition, I = (IJ')J; thus, from (i) we obtain
n(I) = n(IJ')+n(J)  n(/). The converse imp1ication follows from the inclu-
sion pm c P" for m  n.
(iii) Assume first that I+J = R; if, for some P, we had np(l) > 0 and
np(/) > 0, we would have I c: P and 1 c: P, and hence 1+1 c: P, which
contradicts the assumption. Thus min (n(T), n(J)) = O.
Now let I, 1 be any ideals and set M = 1+1. There exists a submodule
M' c: Ro such that MM' = R, whence R = (/+J)M' = 1M' + 1M'. The
proof of formula (ii) sho\vs that n(IM') = n(I)-n(M), n(JM') = n(/)-n(M),
and the case discussed previously yields
o = min(n(IM'), n(JM'» = min{n(I)-ll(M), n(/)-n(M»
= min{n(l), n(J)-n(M),
whence formula (Hi) follows.
(iv) Denote by M the ideal of R such that n(M) = Max (n(I), 11(/». It
follows from condition (ii) that M c I, M c J, and so M c: Ir'\I. We also
have n(I)  n(In/), n(J)  n(InJ), and consequently n(M)  n(/nJ), whence
it is seen that M ::) I(")J, and finally M = InJ. 0

122
Integral Extensions and Dedekind Domains
[Ch.
'Theorem 3.3.6
If R is a Dedekind domain and I c: R a non-zero ideal, then RII is a principal
ideal ring.
Proof
A non-zero ideal I can be represented as I = Pl ... P:s, where PI' ..., Ps
are distinct prime ideals, nl > 0, ..., ns > O. It follows from Theorem 1.1.9
that RII  RIPl x ... x RIP:s. It is easily seen that every ideal of this product
of rings is of the form 11 x ... x Is, where J , is an ideal of RIP:', i = I, ...,. s
Thus it is enough to show that the ideals of each of the rings RIP" are principal
if P is a non-zero prime ideal and n > O.
The ideals of the ring Rlpn have the form lIP", where I :::> pn is an ideal
of R, whence, using Theorem 3.3.5 (ii), we get I = pili for some m  n, and
therefore J/P" = (P/p n )",. If p E p"p2, then Rp = pp ... P: , where P i= P;,
j = 1, ..., t. Theorem 3.3.5 (Hi) implies Rp+P" = PP ... P: +pn = P.
Accordingly, the ideal PIP" is principal and the ideal J/pn is principal as well.
D
Corollary 3.3.7
If P is a non-zero prime ideal of a Dedekind domain R, then the R-modules
pn /pn+l, n = 0, 1, ..., are isomorphic.
Proof
It follows from Theorem 3.3.5 that P" -:F pn+l, and Theorem 3.3.6 implies
that the R-module pn I pn+ 1 is cyclic. The ideal P annihilates that module,
and therefore pn/pn+l  RIP by the maximality of the ideal P. 0
orollary 3.3.8
If I is an ideal of a Dedekind domain and 0 i= a E I, then there exists an
element bel such that I = Ra + Rb.
Proof
The corollary follows immediately from the theorem applied to the ideal
Ra. D
Theorem 3.3.9
Every ring of fractions Rs of a Dedekind domain R is a Dedekind domain.
For every prime ideal P c: R, the ring R p is a principal ideal domain.
Proof
The first part of the theorem follows from Theorem 3.1.11. The ring R p is
a Dedekind domain with a unique non-zero prime ideal P R p , whence every
non-zero ideal is of the form (PR p )". If p E p"p2 then pR p = PR p , and so
all ideals are principal. D
The theorem which follows is converse to Theorem 3.3.5.

III]
Dedekind Domains
123
1nmeoremm 3.3.10
If R is a domain and if every non-zero ideal of R is a product of prime ideals,
then R is a Dedekind domain.
Proof
Note that the representation of an ideal as a product of invertible prime ideals
is unique-we established this in the second part of' the proof of Theorem
3.3.5 for any ring.
We shall show that under our assumptions every non-zero prime ideal P
is maximal and invertible. Let 0 '#: r E P; then P ::> Rr = P ... P for certain
prime ideals P, ...,. P;, which are invertible since the ideal Rr is invertible.
At least one of these ideals, say P, is contained in P. It is sufficient to show
that P is maximal since then P = P and therefore P is maximal and inver-
tible.
Thus we may assume additionally that P is invertible. Let t E R",P. Then
there exist factorizations of the ideals P+Rt = PI... Prn, P+Rt 2 = Ql ... Qq
into products of prime ideals. Obviously we have P;, ::> P, QJ ::> P, and in the
factor ring R' = RIP these factorizations induce the factorizations R't'
= (P I IP) ... (PmIP), R't'2 = (QIIP) ... (QqIP) = {P 1 IP)2 ... {PmIP)2. The
ideals P,IP, QJIP are invertible as factors of principal ideals, whence the first
part of the proof implies that, renumbering the ideals QJ if necessary, we have
q = 2m, Ql = Q2 = PI' ...,Q2m-l = Q2m = Pm. Thus {P+Rt)2 = P+Rt 2 ,
which shows that P c: P+Rt 2 = (P+Rt)2 c: P2+Rt. For each element
pEP, there exist elements U E p2, ,.' E R such that p = u+,.'t. Accordingly
r't = p - u, and therefore r' E P because t  P. If follows that P c: p2 + Pt
and since the opposite inclusion is obvious we get
P = p2+Pt . P(P+Rt).
As the ideal P is invertible we have R = P + Rt, and consequently P is maximal.
We claim that R is Noetherian. Let 0 =1= I c J be ideals in R. From the
previous part of the proof it follows that I is invertible. If II' = R then IJ'
c: II' = R. There exist distinct prime ideals P l' ... , Pm and non-negative in-
tegers k" I" ni such that I = Pl ... p::n, I = Pit... p:nm, II' = Pl ... p:m.
Since I = (IJ')J, we have, in view of the uniqueness of factorization,
k, = I, + ni, and so k i  Ii, i = 1, ..., m. Therefore, there are only finitely
many ideals containing I, whence R is Noetherian.
We shall show that the domain R is normal. Let x E Ro be an element
integral over R; then the R-module M = R[x] c: Ro is finitely generated.
Therefore there exists an element (the product of denominators of gellerators)
r E R such that I'M c R. Since M is also a subring, we have M = M 2 , and
hence (rM)(rM) = r 2 M2 = ,2M = Rr. rM. The ideal rM is invertible, and
so rM = rR, i.e., M = R. Thus x E R, and R is normal. D
The following theoreln is converse to Corollary 3.3.4:

124
Integral Extensions and Dedekind Domains
[Ch.
eoremm 3.3.11
If R is a domain and if every non-zero ideal of R is invertible, then R is a
Dedekind domain.
Proof
By Theorem 3.3.10 it is enough to prove that every non-zero ideal of R is
a product of prime ideals.
Every ideal in R is finitely generated as an invertible ideal, and so R is
a Noetherian domain.
Denote by f the family of those non-zero ideals of R which are not repre-
sentable as products of prime ideals, and suppose that J is non-empty. Then
there exists an ideal I maximal in the family f. I is not maximal, whence it is
properly contained in a maximal ideal m ::> 1. There exists a submodule
M c: Ro such that mM = R. Then 1M c: mM = Rand M = RM ::> mM
= R, whence M;;2 R. Furthermore, we have I c: 1M and I:F 1M because
otherwise the equality I = 1M (in view of the invertibility of 1) would con-
tradict the condition M :F R. The ideal 1M does not belong to the family .1",
and therefore it is a product of prime ideals. Consequently, the ideal I = (lM)m
is also a product of prime ideals. This contradiction shows that the family J
is empty, and the proof of the theorm is complete. 0
Theorem 3.3.11 enables us to derive the following theorem:
Theorem 3.3.12
If R is a Dedekind domain and Ro c K a finite field extension, then the integral
closure of R in K is a Dedekind domain.
Proof
Let T denote the integral closure of R in K. It is known that there exists a field
L such that Ro c L c K, the extension Ro c: L is separable, and the extension
L c: K is purely inseparable (see [L, p. 187]).
We shall show that the integral closure R of R in L is a Dedekind domain.
Since R is Noetherian, it follows from Theorem 3.2.5 that R is a finitely gener-
ated R-module. Therefore each ideal of R is finitely generated and R is Noe-
therian. Every non-zero prime ideal of R is maximal by Theorem 3.1.15,
and R is integrally closed in its field of fractions since (R)o = L.
The ring T is an integral closure of if in K, so it remains to prove the
theorem under the additional assumption that the extension Ro c: K is purely
inseparable. Let p > 0 be the characteristic of the field Ro and let q = p.
= (K: Ro). For every element t e T, we have t,e e Ro and this element is in-
tegral over R; so, by the normality of R, the ring T consists of elements x e K
such that xpl! e R. Denote by qJ: K -+ K the Frobenius automorphism of the
algebraic closure K of the field K (i.e., q;>(y) = Y') and define a subfield K '
of K by
K' = q;-e(K) = {y E K ; y"re K}

III]
Dedekind Domains
125
and set
U = lp-e(R) = {y E K'; y"ee R}.
Then we have T = UnK. Since fPe is an automorphism and fPe(u) = R,
U is a Dedekind domain.
By Theorem 3.3.11 it remains to prove that every non-zero ideal I of T
is invertible. The ideal IU of U is invertible, whence there is a U-submodule
M c: U o such that (IU)M = U. It is possible to choose such elements at, ...
..., ak e I, ml' ..., mk e M that L aim, = 1. Since (m,)q e K, then by setting
b i = «:-1 · m, i = 1, ..., k we obtain elements belonging to K, L a,b i = 1
and hi! = af-l · m11 c: m11" c U, because mil c: U. Moreover, we have
b,l c K, which implies b,! c: UnK = T, and this inclusion together with
the equality L aibi = 1 yields (Tb 1 + ... + Tbk)I = T. Thus the ideal I is
invertible. 0
Exercises
1. Prove that, if R is a Dedekind dOlnain and 1 1 , 12, are ideals in R, then there exist,
an ideal J c R such that the ideal/1J is principal and 1 2 +J = R.
2. Prove the following Krull-Akizuki Theorem: Let R be a Noetherian domain whose
non-zero prime ideals are maximal. Further, let Ro c: L be a finite field extension and T
a ring such that R c TeL. Then Tis Noetherian, non-zero prime ideals of Tare maximal
and, for every non-zero ideal I c T, the R-module T/I is finitely generated. [Prove success-
ively that
(i) there is an element a with the property O:F a e I(')R;
(ii) if elements h 1 , ..., b m E T form a basis of To over Ro and F = Rbi + ... + Rb". ,
then, for every finitely generated R-module M satisfying the condition Fe MeT, the
R-module MIF is a torsion module (that is, the annihilator of each element is non-zero)
and I(MIF) < co;
(iii) for every positive integer n the following inequality holds:
1 1
I(MlaM) = -/(MlanM)  - [1(F/aflF)+I(M/F)] =
n n
1
= ml(R/Ra)+-/(MIF) < 00.
n
This implies 1«M+Ta)ITa)  I(MlaM)  ml(R/Ra) < 00;
(iv) if Mo is a finitely generated R-submodule of T for which the number I( (M o + Ta)/Ta)
takes the maximal value, then Mo + Ta = T, and therefore I(TITa) < co;
(v) the ideal I is finitely generated;
(vi) if peT is a non-zero prime ideal, then IT(T/P) < co, whence P is a maximal
ideal].
3. Derive Theorem 3.3.12 from the Krull-Akizuki Theorem.
4. Let R be a Dedekind domain, peR a non-zero prime ideal, Ro c K a finite field
extension, and if c K the integral closure of R in K. Let RP = Qt 1 ... Q:s for distinct prime
ideals Q1' ...,Q, of R. Prove that the numbers Ii = I( R/ Q,) = [ RI Q,:R/P], i = 1, ...,s,
s
are finite and I (RI RP ) = l: e,li. If R is a princIpal ideal domain and the extension Ro c K
i= 1
is separable, then l: e,f, = [K:R o ].

126
Integral Extensions and Dedekind Domains
[Ch.
5. Rings T, U contained in the field Q(X) of rational functions are defined as follows:
T = Z[X, 2/(X-2), ..., 2 n /(X-2 n ), ...],
U = {Xm(f/g); f, g e Z[X],(X), m> 0 or nl = 0 andf(O)/g(O) E Z(2)}.
Prove that the rings T and U are not Noetherian.
3.4 IDEAL CLASS GROUP AND PICARD GROUP
In the preceding sections we discussed the reasons why the role played by the
rings of algebraic integers of finite field extensions K:::> Q is so important.
These rings are the prototype of Dedekind domains, and, as we have already
mentioned, are not unique factorization domains in general. In the present
section we shall give the construction of the ideal class group CI(R) of a Dede-
kind domain R, which is, in a sense, a measure of the extent to which factoriza-
tion in R is not unique. We shall prove that for rings of algebraic integers in
finite field extensions K :::> Q ideal class groups are finite; not all Dedekind
domains have this property. In Section 4.6 we shall establish that every Abelian
group is isomorphic to the ideal class group of some Dedekind domain.
The construction of the ideal class group will be generalized in Section 4.1
to the case of arbitrary domains. In this section we construct another general-
ization of the idal class group-the Picard group Pic(R) of a ring R. In the
case of a Dedekind domain the groups CI(R) and Pic(R) are isomorphic.
We begin by considering an example which shows that the ring of integers
of a finite field extension K ::> Q need not be a unique factorization domain.
Example 3.4.1
We shall give an example of a Dedekind domain that is not a unique factori-
zation domain. Let d =t 0 be an integer not divisible by a square of a prime
number. In the field Q(a! d ), the integral closure of the ring Z is the ring
z[ yd] if d:F 1 (mod 4), and the ring Zlf( 1 + y a)J if d == 1 (mod 4) (see
[S, vol. I, p. 312], and Exercise 2, Section 3.1). Taking d = -5, we thus obtain
a Dedekind domain R = Z[y - 5]. Using the representation of the number
21 in the form 21 = 3. 7 = 1 +20 = 12- (2y' -5) 2 = 16+5 = 4 2 -( 0 )2
we get factorizations in R
21 = 3.7 = (1+2y -5 )(1-2 y=s ) = (4+ V =5) (4-y -5) .
All the numbers occurring here, except 21, are irreducible in R, as can easily
be seen with the aid of the norm function N: R -+ Z defined by the formula
N(a+by -5) = (a+by -5) (a-by' -5) = a 2 +5b 2 for a, b E Z. The norm
satisfies the condition N(rs) = N(r)N(s) with r, S E R. If we had, e.g., 1 +
+2V -5 = (a+bV-=-S). r, then the norm N(l +2y -='5 ) = 21 would be divi-
sible in Z by the number N(a+by -5 ) = a 2 +5b 2 , which, as can easily be
verified, is possible only in the case where b = 0, a = + 1, or r = + 1. Like-

III]
Ideal aass Group and Picard Group
127
wise it can be checked that the only invertible elements in Rare + 1. It follows
that the above factorizations are factorizations into products of irreducible
and not associated factors, and so R is not unique factorization domain.
Put 8 = V - 5 and consider the ideals, in the ring R, P 2 = (2,1 + 8),
P3 = (3,1 + e), p = (3,I-e), P 7 = (7,3+8), p = (7,3-e). All these ideals
are prime, which we can verify by representing R in the form of the factor
ring Z[X]/(X 2 +S). Then the ideal (2,I+X)/(X2+5) corresponds to P2'
which implies R/P2  Z/2Z. We treat the remaining ideals in a similar
way. The extensions of the prime ideals (3), (7), (2) of Z to ideals of Rare
reducible, R(3) = P3P, R(7) = P7P, R(2) = P 2 P 2 , which is easy to
verify. Moreover, we have P3 P 7 = (21,7+7e, 9+3e, -2+48), and from
the equality 21-2(7+7e)+3(2+48) = 1-2eeP 3 P 7 we easily infer that
P3 P 7 = R(1-28). Likewise P3P = (21,7+78, 9-3e, 8+28), and from the
equality 21 +(7+78)-3(8+2e) = 4+8 e P3P it readily follows that P3P
= R(4+8). Analogously we find PP7 = R(4-e), PP = R(I+2e), whence
it can be seen that the last three distinct factorizations of 21 into irreducible
numbers arise from grouping suitably in pairs the factors in the factorization
of the prime ideal R(21) = P3PP7P.
In order to define the ideal class group of a domain, we have to extend
the range of ideals by introducing fractional ideals.
Definition 3.4.2
Let R be a domain and Ro its field of fractions. We call an R-submodule I
of Ro a fractional ideal of R if and only if there exists a non-zero element
a e R such that 01 c: R. The set of all non-zero fractional ideals of R is denoted
by Id(R).
Clearly, every ideal of R is its fractional ideal, and every fractional ideal
contained in R is an ideal of R. These ideals are also called integral ideals.
An R-submodule of a fractional ideal is a fractional ideal. Every element
x = r Is e R (r, S E R) determines a principal fractional ideal Rx since s(Rx)
= Rr c: R. Every fractional ideal I such that 01 c: R(a :F 0) is contained in
the principal fractional ideal Ra- 1 . Finitely generated R-submodules of Ro
are fractional ideals.
We define multiplication in the set Id(R) by associating with fractional
ideals I, I the fractional ideal II, which consists of elements of the form
L akbk, where Ok E I, b" e I; II is indeed a fractional ideal since if aI c: R,
bl c: R then abIJ c: R.
Extending the definition of the quotient of ideals to non-zero fractional
deals in the following way:
1:/= {xeR o ; xlc: I},
we obtain a non-zero fractional ideal. Indeed, I: I is a submodule of. Ro;
if an element 0 ih a E R satisfies the conditions aI c: R, aJ c: R, and 0 :F ex

128
Integral Extensions and Dedekind Domains
[Ch.
E I, 0 :f.: P E J, then (a{3) (I:J) c: al c Rand 0 #= (afX)J c: a.R c: 1, whence
o #= afX E I: J.
Let us remark further that for non-zero fractional ideals I, J there exists
an isomorphism Hom(J, I)  I:J. We shall show a little more, namely:
if I, J are non-zero R-submodules of the field Ro, then the homomorphisms
f: J -+ I are of the form f(x) = ax for x e J and for some fX e I:J. Indeed,
if 0 ¥= x, y e J then there exist 0 ::/= a, b E R such that ax = by. Hence we
obtain af(x) = bf(y) and therefore the element f(x)/x = af(x)/ax = bf(y)/by
= f(y)/y = fX does not depend on x, which shows thatf(x) = ax for all x E J
and (X e I:J.
Now we assume R to be a Dedekind domain. From Theorems 3.3.5 and
3.3.2 we deduce that Id(R) is a Z-module with a set of free generators
Spec(R)' {O}. Indeed, if 1 c: Ro is a non-zero fractional ideal and al c R
n II
with a non-zero a e R, then (a) = n P:', al = n P" for distinct prime
1= 1 1=1
ideals P l' ..., Pn and non-negative integers k i  0, I,  O. Setting P,-1 = R: Pi
= {x E Ro; XP, c R}, we have P , P,1 = R, from which one readily deduces
n
that I = n P't", where m, = I,-k " and the exponents m, are uniquely
1=1
determined.
Observe that Theorem 3.3.11 implies that Id(R) is a group only in the case
where R is a Dedekind domain.
The non-zero principal fractional ideals form a subgroup Prin(R) of the
group Id(R). The factor group Id(R)/Prin(R) is called the ideal class group
of the Dedekind domain R and is denoted by CI(R).
We note that two fractional ideals I, J determine the same class in the
group CI(R) if and only if there exists an element a e Ro, a ¥= 0, such that
J = al. The condition is fulfilled precisely if the ideals I, J are isomorphic as
R-modules.
The fundamental property of the ideal class group is stated in the following
theorem:
Theorem 3.4.3
For a Dedekind domain R, Cl(R) = 0 if and only if R is a unique factorization
domain.
Proof
Assume that CI(R) = 0 and let I c R be an ideal. Since Id(R) = Prin(R),
the ideal I is principal and therefore R is a unique factorization domain [L f
p. 71].
Assume R to be a unique factorization domain. Let PeR be a non-zero
prime ideal. P contains some irreducible element p, whence Rp c P. If a, b E R
and ab e Rp, it follows from the properties of R that either p divides a or p
divides b, and so the ideal Rp is prime. Thus Rp = P, since the prime ideal Rp

III]
Ideal Class Group and Picard Group
]29
is maximal. Theorem 3.3.3 implies that each ideal of R is principal and con-
sequently CI(R) = O. D
We begin the proof of the finiteness of the ideal class group of rings of
algebraic integers with a lemma.
Lemma 3.4.4
Let a field K be a finite extension of the field of rational numbers Q and let R
be the integral closure of the ring Z in K. Then there exists a positive integer q
such that, given an arbitrary element x E K, there exist an a E R and a positive
integer mo, 1  mo  q, such that /NK(mox-a)/ < 1, where Ng,(y) is the
"
norm of the element y E K (that is to say, NK(y) = n l1J(Y), where (/J: K -+ C,
j=1
j = 1, ..., n, are all the embeddings of the field K into the field of complex
numbers, n = (K: Q».
Proof
Denote by Xl' ..., X n elements of R such that R = Zx 1 + ... + Zx n ; their
existence follows from Theorem 3.2.5. For every element Y E K, there exist
an element b E R and numbers ai' ..., n E Q satisfying the conditions y - b
n
= L ,Xi, 0  rt, < 1, i = 1, ..., n. Fix a positive integer s > 1 and denote
;=1
q = sri; for each m = 0, 1, ..., q, there exist b m E R and numbers mj E Q,
n i
i = ], ..., n such that 0  ml < 1 and mx-b m = L amlxi. The product
j=J
[0, 1) x ... x [0, 1) of n copies of the closed-open interval [0, 1) is the union
of q = sft products of intervals of the form [(k-l)/s, kls), k = 1, ..., s.
Accordingly, there exist integers 0  m' < mil  q such that r«m'l- CXml/il
< lis for all i = 1, ..., n. Setting mo = m"-m', a = bm"-b m ,, we obtain
INK(mox-a)/ = INK«m"x-bm,,)-(m'x-b m ,»/
n n n
= iNK ( L (ocm/lt- OCm,t)xl)1 = In qJ (L (OCmllt- OCm,t)Xt)1
1=1 j=-1 ;=1
11 "
 n ( L IOCmlll- OCm,tl · /qiXt) I )
= 1 1= 1
"" n n
< D +  IqiXt)/ =  J]  IqiXt)l.
The lemma follows if \ve take for s a number satisfying the inequality
n "
sft > n L IqiXt)l.
J= 1 1= 1
o

130
Integral Extensions and Dedekind Domains
[Ch.
Theorem 3.4.5
Let a field K be a finite extension of the field of rational numbers and let a ring
R be the integral closure of the ring Z in K. Then the group CI(R) is finite.
Proof
We preserve the notation of the preceding lemma. Denote by ql the least
common multiple of the numbers 1, 2, ..., q. Let I be a non-zero ideal in R;
we shall show that there exists an ideal J in R determining in the group CI(R)
the same element as the ideal I and such that ql E J.
We know that the values of the norm N K are rational numbers, which
implies that NK(y) E Z for Y E R. Hence there exists a non-zero element C E I
such that INK (C) I  INk{y) I for Y E I, Y ¥= O. By Lemma 3.4.4, for each element
Y E I there exist an element a E R and an integer m, 1  m  q, such that
INK (m(y/c) -Q)I < 1, i.e., INK {my- ca)1 < INK(c)l. Since my- CQ E I, it follows
from the minimaIity of the norm of c that mY-CQ = O. This, in view of mfql,
yields ql I c: Rc, whence, by Theorem 3.3.5, there exists an ideal J c R such
that ql I = cl. Since c E I, this relation shows that qt C E cJ, and consequently
q 1 E J.
If J = Pl ... P;k, Rql = PT1... P:k, where PI' ..., PIc are distinct prime
ideals, then it follows from the condition Rql c J that 0  nl  ml, ..., 0
 n"  m". The numbers ml, ..., mk do not depend on the choice of the ideal I.
By the relation ql1 = cJ the ideal I determines in Cl(R) the same element
as the ideal J, and therefore the group CI(R) has at most (ml + 1) ... (mk + 1)
elements. 0
An incomparably better estimate of the cardinality of the group CI(R) will
be found in [C].
To construct the Picard group of a ring, one uses invertible modules.
We proceed to discuss this concept.
Definition 3.4.6
Let R be any ring. An R-module M is said to be invertible if and only if'there
exists an R-module N such that M @ N  R.
The following lemmas concern the structure of invertible modules:
Lemma 3.4.7
If M, N are R-modules and M(8)N is a non-zero free R-module, then M, N
are projective.
Proof
Let f: F -+ M be an epimorphism of a free module F. Then the epimorphism
f@ IN: F@N -+ M@N splits, and hence the epimorphismf@ IN (8) 1 M : F@
@N@M -+ M@NM also splits. It follows that the module M, as a mod-
ule isomorphic to a direct summand of the module M (8) N @ M  R'fi)
@M(s > 0), is also isomorphic to a direct summand of the moduleF@N@M
z F@ R S , which is free. Thus M is projective. 0

III]
Ideal Class Group and Picard Group
131
Lemma 3.4.8
An invertible R-module M is finitely generated and projective, and for every
P e Spec(R) the Rp-module M p is free of rank 1. If M@N  R, then N
 Hom(M, R).
Proof
k
Let f: M @N -+ R be an isomorphism and let f ( L mi <8> n,) = 1. Denote
1= 1
by Mo the submodule of M generated by elements ml, ..., mk. Let w: Mo
-+ M be an injection. The homomorphism w @ 1: M 0 @ N  M @ N is an
epimorphism and, by the projectivity of N, it is also a monomorphism. This
shows that (M/Mo)@N = 0, whence M/Mo  (M/Mo)@N@M = 0, so
that M = Mo. Therefore M is a finitely generated module; the same applies
to N. The finitely generated Rp-modules M p , N p are projective and so free
(see Exercise 8, Section 1.3). Moreover, we have M p @N p  R p and so the
rank of M p is equal to I.
Denote by l'the homomorphism N -+ Hom(M, R) given by the formula
(f'(n»(m) =/(m@n) for meM, nEN. For each ideal PESpec(R), we
have M p  R p  N p , and the homomorphism I;: N p  Hom(M, R)p
 Hom(M p , R p ) is an isomorphism. Hence we can easily conclude that I'
is an isomorphism (see Exercise 6, Section 1.4). 0
In the case of a domain, the connection between invertible modules and
invertible ideals is expressed by the following lemma.
Lemma 3.4.9
(i) Let S be a multiplicative subset of a d<?main R. If M, N are invertible
R-submodules of the ring Rs, then the mapping p: M@N -+ MN c Rs,
for which p(m@n) = mn, is a monomorphism,
(ii) If R is a domain, then every invertible R-module is isomorphic to an
R-submodule of the field of fractions Ro, which is an invertible fractional
ideal.
Proof
(i) The R-submodule N is projective, whence the inclusions M ( Rs,
N ( Rs induce a monomorphism
M@N -+ Rsfi;N -" R S fi9 R R S  Rs
which maps m <8> n on mn.
(ii) Let M be an invertible R-module and M @ N  R: then
(M@RRo)@Ro(N<8>RRo)  (M(8)RN)fi9 R R o @RR o  Ro.
Since Ro is a field, the above formula shows that M @ R Ro  Ro. The module
M is projective, and so the inclusion R C Ro induces a lllonomorphism
M  M@R M@Ro  Ro. 0

132
Integral Extensions and Dedekind Domains
[Ch.
Lemma 3.4.9 indicates the possibility of finding an analogue of the ideal
class group (which we have defined for a Dedekind domain) for an arbitrary
ring by replacing invertible ideals with invertible modules, and the multi plica.
tion of ideals with the tensor product of modules. We present one more con-
struction of this kind in Section 4.1.
Now we proceed to define the Picard group of a ring R. For any invertible
R-module M, we denote by [M] the isomorphism class of the module M.
It follows immediately from the definition of an invertible module that the
tensor product of two invertible R-modules is invertible. The operation of
multiplication of isomorphism classes is given by the formula
[M] · [M 1 ] = [M@M 1 ].
The unity element is [R], and, by Lemma 3.4.8, the inverse for [M]
is [Hom(M, R)]. The isomorphism classes of invertible modules with the
multiplication defined above form a group, which is called the group of classes
of invertible R-modules, or the Picard group of R, and is denoted by Pic(R).
We show that a homomorphism of rings f: R -. T induces in a natural
way a homomorphism Pic(j): Pic(R) -. Pic(T). In fact, for any invertible
R-module M, the T-module M@R T is invertible, and we set Pic(f) [M]
= [M @ R T]. It is readily seen that the definition is correct and that Pic(f)
is a homomorphism. In this way Pic becomes a covariant functor.
Theorem 3.4.10
If R is a Dedekind domain, then the group Cl(R) of classes of invertible ideals
is isomorphic to the group Pic(R) of classes of invertible modules.
Proof
By Lemma 3.4.9 (i), associating with a fractional ideal 1 c Ro its isomorphism
class [I] induces a homomorphism Id(R) -+ Pic(R), which is an epimorphisln
by part (ii) of the above lemma. The kernel of the homomorphism is of course
the subgroup Prin(R), and the theorem follows. 0
To investigate the properties of the functor Pic we use another group
Pic(R, S), defined in the case when S is a multiplicative subset of Rand S
does not contain zero divisors. In order to define it, let us call an R-module M
contained in the ring Rs S-invertible if there exists an R-module N contained
in Rs such that M. N = R. The module N with this property is determined
uniquely: N = {x e Rs; xM c R}. The group Pic(R, S) is defined to be the
group (under multiplication) of S-invertible R-submodules of the ring Rs.
Let us observe that isomorphic but different S-invertible modules give rise to
different elements of Pic(R, S). By Lemma 3.4.9, there exists a natural homo-
morphism [ ]: Pic(R, S) -+ Pic(R) which sends a S-invertible module M c Rs
onto its isomorphism class [M] e Pic(R).
If R is a domain, then Pic(R, R" {O}) is the group of invertible fractional
ideals. '

Ill]
Ideal Class Group and Picard Group
133
Theorem 3.4.11
Let S be a multiplicative subset of a ring R and let S contain no zero-divisors.
The sequence
o  R*  (Rs)* -+Pic(R, S)i_! Pic(R) Pic(}) -+ Pic(Rs),
where j: R ( Rs; j* is an induced map of groups of units, and e(a) = Ra
for a e (R s )* is exact.
Proof
(1) The homomorphism j* is obviously a monomorphism.
(2) An element a e (Rs)* belongs to Ker(e) if and only if Ra = R, i.e.,
if a E R*; this means that Ker(e) = Im(j*).
(3) Of course, we have rea] = [Ra] = [R] for a E (R s )*, hence Ker([ ])
 Im(e). If M c Rs is an S-invertible module and [M] E Ker([ ]), then
there is an element mo e M such that M = Rmo  R. Let MN = R; then
Rsmo = Rs(Rmo) = RsM = RsRsM ::> RsNM = Rs,
which shows that mo e (Rs)* and M = e(mo) E Im(e).
(4) If M, N c Rs are R-submodules and MN = R, then PicU)[M] = M(8)
(8)RR s  MRs = MRsRs ::> MNRs = Rs, and therefore MRs = Rs. Thus
the composition of the last two homomorphisms in the sequence is zero, i.e.,
Ker(Pic(j»)  Im([ ]).
To prove the opposite inclusion, consider an invertible R-module M such
that [M] E Ker(Pic(j»), which means that M(8)RR s  Rs. We have to show
that M is isomorphic to an S-invertible sub module of Rs.
Clearly, there exists a monomorphism M  M@R -+ M(8)R s  Rs,
and so, without loss of generality, we may assume that M c Rs and M s = Rs.
Let h: F -+ M be an epimorphism of a free R-module with a basis el, ..., ek.
The module M is projective, whence there exists a homomorphism g: M -. F
such that hg = 1 M - Let Pi: M -+ R be the composition of g with the i-th
projection mapping F -+ R, i = I, .." k. Then, for x EM, we have
x = L Pi(x)h(e,). Observe that the homomorphisms f: M -+ R are of the
i
form f(x) = ax for some a e R: M. Indeed, a homomorphism M 4 R -+ Rs
can be extended to a homomorphism Ms = Rs  Rs in which x H ax. It
follows thatf(x) = ax and a e R:M.
Therefore, there exist elements a, e R:M such that p,(x) = a,x, and con-
sequently x = 1: xa,h(e,) for all x e M. This in1plies that 1-1: a,h(e;)
EAnn(M) c Ann(M@N) = o provided thatM@N z R. Thus 1 = 1:a,h(e,)
,
eM. (R:M), hence M. (R:M) = R, and M is an S-invertible module. We
have proved that Ker(PicU») = Im([ ]). D
Exercises
1. Let a field K be a finite extension of the field of rational numbers, and let a ring R
be the integral closure of the ring Z in K. Prove that there exists a finite field extension L  K

134
Integral Extensions and Dedekind Domains
[Ch.
such that for each ideall c: R the ideal TI is principal, T being the integral closure of Z
in L. [Use the fact that the group Cl(R) is a direct sum of finite cyclic groups.]
2. Let R be a Dedekind domain. Prove that the group CI(R) is a torsion group if and
only if every ring T such that R c T c: Ro is a ring of fractions of R.
3. Prove that, if M is a projective finitely generated R-module and the ranks of the
Rm-modules M m are equal to 1 for all m e Max(R), then M is invertible.
3.5 MODULES OVER DEDEKIND DOMAINS
We begin the study of modules over Dedekind domains by proving a theorem
which characterizes these domains in terms of modules. We recall that an
R-module M is said to be divisible if and only if for every non-zero r E R
we have rM = M. It follows immediately from the definition of an injective
module (see Section 1.3) that each module of this kind is divisible; an
R-module M is called torsion free if Ann(m) = 0 for every element 0 ¥= m e M.
If R is a domain, then the set t(M) = {m EM; Ann(m) :F O} is a submodule
called the torsion submodule of M.
Lemma 3.5.1
Let R be a domain. A non-zero ideal I c R is a projective R-module if and
only if I is invertible, that is, if there exists an R-submodule I' c Ro such
that II' = R.
Proof
Assume that I :f.: 0 is a projective R-module. Accordingly there exists a free
R-module F =  Re, with basis {e,} such that Ie M = F for some module
M. Denote by n,: F -+ R the projection mappings and let a, + mi = e"
where ai E I, mi e M. For each element x E I, we have
x = 2: n t(x)et = 2:nt(x){at+mt) = ) nt(x)at.
If 0 #= x, Y E I then n,(xy) = xn,(y) = ynt(x), and therefore the elements
b z = ltl(x)lx E Ro do not depend on x, are almost all equal to 0, and b,I c R.
For every x E I, we have x = L ni(x)ai = L (xb,)ai' and so 1 = L b,a,.
Setting I' = L Rbi' we conclude that II' = R. Thus the ideal I is invertible.
n
Assume now that the ideal I is invertible and II' = R. Let 1 = L aJb J
J=1
for some aJ E I, b j E /', and let Re e... e Re be a free module. The mappings
I  Re  ... eRe  Igiven by w(x) = L (b j x)e,n(e) = aj,j = 1, ..., 1l,
satisfy the condition nw(x) = L (bjx)ai = x for x E I. Thus the mapping
w determines an isomorphism between I and a direct summand of a free module,
and consequently the ideal I is projective. 0
Theorem 3.5.2
A domain R is a Dedekind domain if and only if every divisible R-module is
injective.

III]
Modules over Dedekind Domains
135
Proof
It is known (see [P], p. 31), that a module M is injective if and only jf for every
homomorphism f: I -+ M of an ideal I there exists an extension /: R -. M
off.
Now assume that R is a Dedekind domain, that M is a divisible R-module,
and that f: I -+ M is a homomorphism of an ideal R. It is sufficient to prove
that there exists an extension f: R -+ M. Since I is invertible, II' = R for some
k
ideall'. Let L aibi = 1, where at, ..., ak E I, b i , ..., b k e I'. The module M
;==1
is divisible, thus there exist elements mi E M such that f(aa = at mi. Let
k
the homomorphism f: R -+ M"be such that f(x ) = x L (aib,)mi. If x E I
1=1
then xb t E R and x = L (xbi)ai, whence
f(x) = L (xb,)f(a,) = L (xb,)(a,m,) = j(x),
and thus the module M is injective.
Assume now that every divisible R-module is injective. By Lemma 3.5.1
and Theorem 3.3.11 it is enough to show that every non-zero ideal I c R
is projective. Let 0 -+ M' -+ M 1.-+ M" -+ 0 be an exact sequence of R-modules
and let f: I -+ M" be an R-homomorphism. Consider the commutative
diagram of R-modules with exact rows
.. M'
y
1 -
!I
. M"
!p
o
R C
o
:.- 0
 M
!a
l'
0-
.. M'
 E
 ElM'
 o
in which E is an injective module containing the module M, 0: is an injection,
P is the homomorphism induced by 0:, and v is the natural homomorphism.
The module ElM' is a homomorphic image of the divisible module E, so it is
divisible and hence injective. Thus there exists an extension f: R -+ ElM'
of the homomorphism Pf. Obviously, there exists a homomorphism g: R -+ E
such that f = "g. If x e I then "g(x) = f(x ) = Pf(x) e Im(p) = Im(py) = v(M),
and therefore g(x) e M and yg(x) = f(x). It follows that the ideal I is projec-
ti 0
Now we proceed to describe finitely generated modules over Dedckind
domains. We first concentrate on torsion free modules.

136
Integral Extensions and Dedekind Domains
[Ch.
If M is a torsion free module, then M -+ M@RRo = Mo is an injection.
Generalizing the definition of the rank of a free module, we define the rank
of a module M to be the dimension of the vector space M (8) R Ro over the field
Ro, and denote it by rk(M). If I is a non-zero ideal of R, then rk(I) = 1.
Theorem 3.5.3
If R is a Dedekind domain and M a finitely generated torsion free R-module,
then M  1 1 E9 ... elk, where 11' ..., It are non-zero ideals of Rand k
= . rk(M).
If 1 1 , ..., J" are also non-zero ideals of R, the modules 1 1 EB ... <3:J I" and
J 1 Et> ... Ea J k are isomorphic if and only if the ideals 1 1 ... I k , J 1 ... J k are
isomorphic as R-modules. Moreover
M  IlEa ... Et>IA  Rk- 1 EaI1 ...I k .
Proof
Since rk(I i ) = 1, i = 1, ..., k, we have rk(M) = k when M  11 <3:J ... <3:J I".
The rest of the theorem will be proved by induction on rk(M).
If rk(M) = 1, then M is isomorphic to a finitely generated submodule M'
of the field Ro. Denote by s the product of the denominators of generators
of M'; we then have M  M'  sM' = 1 1 c Rand 11 is an ideal of R.
Assume that all torsion free modules of rank < rk(M) are isomorphic
to direct sums of ideals and rk(M) > 1. We identify M with a submodule
of M(8)Ro, choose any non-zero element mo EM and set N = RomonM.
Then N is a submodule of M and mo EN. We shall show that M/N is a torsion
free R-module and rk(M/N) < rk(M). If x+N E M/N and r(x+N) = 0,
r i= 0, then rx EN, and hence rx = (t/s)mo for some t, S E R. Thus
x = (t/rs)mo EN, and therefore x+N = 0 and M/N is a torsion free module.
We have an exact sequence
o -" N(8)R o -+ M(i!)R o -+ (MIN)fg)Ro -+ 0,
whence rk(M/N) < rk(M), rk(N) = 1. Consequently the modules M/N,
N are direct sums of ideals, and it follows from Corollary 3.3.4 and Lemma
3.5.1 that M IN is a projective module. This shows that M  M IN eN, proving
the first part of the theorem.
Assume that there exists an isomorphism f: II Et> ... e Ik  J 1 e... e J",
where / 1 , J 1 ... are non-zero ideals. Choose elements 0 ¥= rti E Ii, 0 ::/= Pi E J i ,
i = 1, ..., k. By replacing the ideals Ii, I, with the fractional ideals (XlI"
p,ll" we may assume additionally that ideals Ii' J , contain the ring R. Let
1(0, ..., 0, 1,0, ..., 0) = (a'l' ..., aik) for some au e J} (1 stands in the i-th
place). Using the description of the homomorphisms Ii '-+ I j (see Section 3.4),
we obtain
1(0, ...,0, x" 0, ...,0) = (ail x" ..., a'k X ') Ell e... EaJ k
for x, EI,. ThusalJli c Ijandforeverypermutation i 1 , ..., i k of the sequence
1, .,., k, we have
11 ... J"  0'1 1 ... a,,,,,I'1." I'k = 0'1 1 ... a't."I! ...1",

III]
Modeles over Dedekind Domains
137
which yields 1 1 ... J" ::> det(aiJ)I 1 ... It. In the same way the homomorphism
1- 1 determines a matrix (a'J)-l, and 1 1 ... 1" ::> det(a'J)-1/ 1 ... J t . The two
relations imply J 1 ... J t = det(a'j)I 1 ... It, and thus the ideals II ... lie, 1 1 ... J t
are isomorphic.
Before undertaking the last part of the proof, we shall establish a lemma
which is a particular case of a more general fact (see Theorem 4.4.10).
Lemma 3.5.4
Let P , PI' ..., Ps be distinct non-zero prime ideals of a Dedekind domain R.
Then there exists an element a E R and an ideal I c R such that Ra = p. I,
and I is prime with respect to the ideals P, P 1, ..., Ps.
Proof
Pick an element c e p,p2; renumbering, if necessary, the ideals PI' ..., Ps,
we may assume additionally that C E PI' ..., c e P", C  P k + 1 , ..., c  p.
(0  k  s). It follows from Corollary 1.1.8 that there exist elements b
E P2'(PI U ... uP,,), b , E P i ,{P 1 U ... uP t ) for i = k+ 1, ..., s. Hence we get
bb"+l ... b s e p2Pt+l ... P S ,,{P 1 U ... uP,,). Since C E PP 1 ... p",p2, we thus
find that the element a = c+bb"+l ... b s belongs to the set P'{P 2 UPIU ...
... uP.). Accordingly the ideal Ra takes the form Ra = PI, I being a product
of prime ideals distinct from P, PI' ..., P8. The element a satisfies the condi-
tions of the lemma. 0
We return to the proof of Theorem 3.5.3; it remains to show that the
isomorphism 11 ... Ik  J 1 ... J t of non-zero ideals implies an isomorphism
1 1 €a ... Eelk  J 1  ...  J t of modules. To this end it is sufficient to prove
that II e ... Eel"  R,,-1 e 1 1 ... It, whereas the obvious induction reduces
the proof of the last formula to the case k = 2.
Let 1 1 , 1 2 be non-zero ideals of R. There exists an element r E R, r #= 0,
such that rIll c R. We have factorizations
,1 - 1 - P ili P n, I - P III1 P '
1 - 1 ... ., 2 - 1 ... S'
where P 1, ..., Ps are distinct prime ideals and the exponents are non-negative.
By Lemma 3.5.4, there exist elements aj E R such that Ra, = Pi · J; , i = 1, ...
..., s, and the ideals J; are prime with respect to the ideals PI' ..., Ps. Let
a = a1 ... a:'; then Ra = pr 1 ... P:. · J = rIll J, and the ideal J = 1"1 ... J;n.
is prime with respect to the ideals P1, ..., Ps. Thus ar- 1 I 1 = I and 1+1 2 = R.
It follows that 1 1 e 1 2  a,-lI 1 EB 1 2 = J  1 2 . By composing this isomorphism
with the epimorphism J €a 1 2  J + 1 2 = R, which associates with a pair
(x, y) the element x+y, we obtain an epimorphism 1 1 €a 1 2  R. Therefore
there exists a decomposition II  1 2  R Eel, where I is a finitely generated,
torsion free module of rank 1, and hence it may be identified with an idea1
of R. The previous part of the proof yields II .1 2  R · I = I, and consequently
we have 1 1 I2  REall .1 2 . 0

138
Integral Extensions and Dedeldnd Domains
[Ch.
Corollary 3.5.5
A finitely generated, torsion free module over a Dedekind domain is projective.
Theorem 3.5.6
If R is a Dedekind domain and M is a finitely generated R-module, then
M  R"-1Iffi  R/Pl,
(Pit nl)
where k = dimRo (M @ R R o ), I c: R is an ideal determined uniquely up to
isomorphism, Pi are non-zero prime ideals of R, n, > 0, and the finite set
of pairs (Pi, n,) is uniquely determined.
Proof
The R-module Mft(M) is torsion free, and hence projective, which shows that
there exists a decomposition M = M' t(M), where M'  Mft(M) (see
Exercise 2, Section 1.3). Applying Theorem 3.5.3, we may confine ourselves
to the case of a torsion module.
Thus, let M be a finitely generated torsion module. The ideal Ann(M)
is then different from zero and has a factorization Ann(M) = Pi 1 ... P:",
where PI' ..., Ps are distinct prime ideals, ml > 0, ..., ms > o. We claim
that M = Mt  ... E9 Ms, where MJ = {x E M; P'jJX = OJ, j = 1, ..., s. It,
follows from Theorem 3.3.5 that if we denote I j = PT1 ... P;'!.i 1 Pj+lt ... P':',
then L IJ = R. Accordingly M = L (IJM) and clearly PjJ(IJM) = 0, whence
M J :::> IJM. On the other hand, if x e Mtt1(M2 + ... +Ms), then Pi 1 x = 0,
Pi" ... P:'x = 0, whence the equality Pi 1 +Pi 2 ... P:& = R implies x = O.
In the same way we establish the equality M J t1(M I + ... +M J - 1 --t-M J + 1 +
+ ... +M.'l) = O. Thus M = L MJ is the direct sum  MJ.
1
There remains to be considered the case where Ann(M) = pm, m > O.
We may regard the module M as an Rp-module since if s  P then pm + Rs = R,
which shows that M ...!-. M is an isomorphism. The ring R p is a principal ideal
domain by Theorem 3.3.9, and therefore M is a direct sum of cyclic modules
of the form Rp/(PRp)n  R/p n , where the representation of M in this form is
unique (see [L, p. 390]). This ends the proof of the theorem. 0
Exercises
1. Prove that if R is a principal ideal domain then a submodule of a free R-module
is free.
2. Let R be a Dedekind domain. Show that:
(i) a submodule of a free R-module is isomorphic to a direct sum of ideals of R, whence
it is a projective module.
(ii) a projective R-module which is not finitely generated is free.
3.6 VALUATIONS
In this section we discuss the basic properties of valuations. A valuation is
a generalization of the notion of multiplicity of a root of a polynomial. We

III]
Valuations
139
first concentrate on valuations with values in an arbitrary ordered group
and then confine our attention to discrete valuations, i.e., valuations with
values in the group of integers. We define and characterize some rings associ-
ated with valuations, called valuation rings. Using Dedekind domains as an
example we outline the connection between discrete valuations and the theory
of ideals. Finally, we discuss completions of fields with respect to discrete
valuations. For a more extensive treatment of valuation theory the reader is
referred to [L], [8, vol. II].
Let R be a unique factorization domain and pER and irreducible element.
Each element 0 :/:. ,. E R has a unique representation ill the form r = p"p(r) r',
where r' e R is not divisible by p, and v,(r) is an integer  O. The function vI'
obtained in this way can be extended to R = Ro"-. {O} by setting v,(rls) = m'
if rls = pm. r'ls', where r', Sf e R are not divisible by p and m is an integer.
We thus have v,(rls) = v,(r)-v,(s) and the values v,(rls) are integers. The
function v,: R -+ Z satisfies of course the following conditions:
vp(x. y) = vp(x)+vp(y), vp(x+y)  min(vp(x), vp(y».
Observe that the set {x E R; v,(x)  O}u {o} is the ring of fractions of R
with respect to the prime ideal (p). The function 'lJ, depends only on the ideal
(p) and does not depend on the choice of the generator p of this ideal. The
function v, is called the p-adic valuation of the field Ro.
The most important examples of p-adic valuations are the valuations of the
field of rational numbers, determined in the above way by prime numbers p.
Of equal importance are the valuations of the field K(X) of rational functions
in one indeterminate, determined by irreducible polynomials p e K[X]. If the
polynomial p is of the form X-a for a e K, the value of the appropriate
valuation on a rational function f E K(X) is the multiplicity of a as a root
or as a pole of the function/.
We are going to generalize the notion of a padic valuation. Observe
first that in the statement of the basic property v,(x+y)  min(v,(x), v,(y»
an essential role is played by the order relation  in the group Z; all the sub..
groups of the group R of real numbers admit this relation. These groups are
the prototypes for ordered abelian groups.
An ordered Abelian group is an Abelian group G together with the (linear)
order relation  which satisfies the following consistency requirement: if
ex, P, reG and tX  p, then ex+r  P+r. An ordered group has no non-zero
elements of finite order: indeed, if n > 0 and tX > 0 then nex = ex + ... + ex > 0,
and hence nex :/:. O.
Definition 3.6.1
A valuation of a field K with values in an ordered Abelian group G is a function
v: K"- {O} -+ G which fulfils the conditions:
(i) v(x. y) = v(x)+v(y),
(ii) v(x+y)  min(v(x), v(y») provided that x+y :F O.

140
Integral Extensions and Dedekind Domains
[Ch.
A valuation determines a homomorphism of the multiplicative group
K' {O} into the group G. Note that in the above definition we can replace
the field K by a domain R because a valuation w: R' {O} -+ G of R can be
uniquely extended to a valuation of the field Ro defined by the formula v(/.ls)
= w(r)-w(s).
The most important role in ring theory is played by valuations with values
in the group of integers; they will be discussed in this section.
Example 3.6.2
We shall show that every ordered Abelian group Gis the value group of some
valuation of a field. Let Z[G] be the group ring of G; its elements are sums
L no · g, where ng E Z and almost all the coefficients ng are equal to O. The
geG
multiplication of elements is an extension of the n1ultiplication of monomials
(ngg) (ng,g') = (ngng,) (gg'). The ring Z[G] is a domain because of the existence
of an order relation in G. Each non-zero element of Z[GJ can be uniquely
written as ntgt +n2g2 + ... +nkg", where gl < g2 < ... < g,,, and the
coefficients nl, ..., n" are all non-zero. Putting v(nlgt + ... +n"gk) = gl,
we obtain, as is easy to verify, a valuation of the ring Z[G] and hence a valua-
tion of the field (Z[G])o. Every element of the group G is a value of'lJ on some
element of the ring Z[G].
We shall establish the simplest properties of valuations:
v(l) = 0, v(x- 1 ) = -vex), v( -x) = vex),
v(x + y) = v(x) whenever vex) < v(y).
The first three formulae follow from the equalities 1 · 1 = 1, x- 1 · X = 1,
(-1)( -1) = 1. The last formula is a consequence of the inequalityv(y) > vex)
= v«x+y)-y)  min(v(x+y),v(y», which implies that min(v(x+y), v(y»
= vex + y), and, because of the fact that v(x + y)  min (v(x), v(y») = vex),
we finally get v(x + y) = vex).
Definition 3.6.3
If v: K* =  {O} -+ G is a valuation of a field K, then the set
Rv = {x e K*; v(x)  O}u {OJ
s a ring which we call the ring of the valuation '0.
Our first goal is to characterize rings of the form Rf) in the set of subrings
of the field K and to investigate their properties. The characterization is very
simple if we impose no restrictions on the ordered group G.
Theorem 3.6.4
Let K be a field. A subring R c K is the ring of some valuation v of K if and
only if, for every element x E K*, at least one of the elements x, x- t belongs
toR.

lIT]
Vatoations
141
Proof
Let v: K*  G be a valuation such that R = R". Since the group G is ordered,
for each element x E K* we have either vex)  0 or v(x- 1 ) = -v(x)  0,
and therefore either x E R" or x- 1 E Rf).
Before we pass to the second part of the proof, let us observe that Im(v)
c: G is a subgroup of G, the valuation v is a homomorphism of the group
K* onto Im(v), and its kernel is
{xEK*; v() = O} = {xEK*; vex) = v(x- 1 ) = O} = R;.
Thus the group Im(v) is isomorphic to the group K*/R:, and the ordering
in this group, which corresponds to the ordering in the group Im(v), is x  y
<=> yx- 1 E R" for x, y E K*(X, Y denote the respective residue classes).
This observation allows the reader to check easily that if a ring R c: K
satisfies the assumption of the theorem, G = K* / R*, f): K*  G is the natural
homomorphism and we put x  y <=> yx- 1 E R then G becomes an ordered
group, v is a valuation and R" = R. 0
Definition 3.6.5
Valuations v: K*  G, v': K*  G' are said to be equivalent if R" = Rv,.
Following the proof of the previous theorem, the reader will verify without
difficulty that valuations v, v' are equivalent if and only if there exists an
isomorphism of ordered groups lfJ: Im(v)  Im(f)') such that lp v = v'.
Definition 3.6.6
A subring R of a field K is called a valuation ring of K if and only if given any
element x E K*, at least one of the elements x, x- 1 belongs to R. A domain
R is called a valuation ring (the field not being specified) if R is a valuation ring
of the field Ro.
Valuation rings have an obvious intrinsic characterization.
Corollary 3.6.7
A domain R is a valuation ring if and only if for all elements r, S E R, \ve
have either Rr c: Rs or Rs c: R,..
A valuation ring ned not be Noetherian (Exercise 2).
The fundamental properties of valuation rings are stated in the following
theorem:
Theorem 3.6.8
Let R be a valuation ring of a field K. Then:
(i) the ring R is quasi-local, i.e., it has only one maximal ideal,
(ii) every finitely generated ideal of R is principal,

142
Integral Extensions and Dedekind Domains
[Ch.
(Hi) if R c: R' c K and R' is a ring, then R' is a valuation ring, and R' = R p
for some prime ideal P c: R,
(iv) the ring R is normal.
Proof
According to Theorem 3.6.4, there is a valuation v: K* -+ G such that R = Rv.
(i) An element 0 :/:. x e R is invertible in R <=> vex) = 0, whence every
proper ideal of R is contained in the set {x E R; v(x) > O}u {O}, which is an
ideal. This ideal is therefore the only maximal ideal of R.
(ii) Let I = Rr 1 + ... + Rrk; one of the elements vCr 1), ..., vert) is the least,
assume it is v(rl). Thus v(ri/'I) = v(ri)-v(r 1 )  0, and consequently rift 1
E R, which shows that r, e Rr1, i = 1, ..., k, and finally I = Rr1.
(Hi) The ring R' is of course a valuation ring; let m' c: R' be its only
maximal ideal and set P = m'raR. The ideal P is prime; we claim that R' = R p .
Indeed, if S E R"P, then s m', and so 8- 1 E R', which yields R p c: R'.
If x ER'''R then X-I ER, hence x, X-I E R' and x- 1 m' therefore X-I ER"P
This implies x E R p and finally R' = R p .
(iv) Let an element 0 :/:. x E K satisfy the equality x", + r m-1 xm-1 + ... + to
= 0 for some ro, ..., r m -1 e R. If we had x  R, then X-I E R, and from the
above equality we would get -x = r"'-1 +r m _2 x - 1 + ... +rox-(.n-1) E R,
which contradicts the condition x  R. This shows that x e R. 0
In Theorem 3.6.9 we give another characterization of valuation rings
of a field K, which will be used for representing the integral closure of a 1.ing
as an intersection of valuation rings.
Observe first that if V is a valuation ring of a field K, if m c: V is a maximal
ideal, and if L is an arbitrary field, then no homomorphism of the form V
 VIm -+ L can be extended to a subring of the field K properly containing
V: if x e K,Vthen x- 1 E V and x- 1 em, and every homomorphism V[x] -+ L
which is an extension of the homomorphism V  Vft1t -+ L sends X-I into 0,
contrary to the relation xx- 1 = 1.
Theorem 3.6.9
Let K be a field and L an algebraically closed field. Denote by R4 the partially
ordered set whose elements are pairs (R,f), where R is a subring of K, the
mapping f: R -+ L is a homomorphism, and (R,f)  (R',f') <=> R c R'
andf'IR = f.
A pair (R,f) is a maximal element in the set f1A if and only if R is a valuation
ring of K and Ker(/) is the unique maximal ideal of R.
Proof
We gave the proof of the implication <= before stating the theorem. Let (R,f)
be a maximal lement in R4. First \ve prove that m = Ker(/) is the unique
maximal ideal of R. It is clear that the ideal m is prime, and for each S E R"m

III]
Valuations
143
the element 0 :f:. 1(8) E L is invertible, whence the homomorphism f: R  L
can be extended to a homomorphism Rm -+ L. It follows from the maximality
of the pair (R,f) that Rm = R, and therefore every element not belonging
to m is invertible, which implies that m is the unique maximal ideal.
. Let 0 :/:. x e K. We shall show that the equalities mR[x] = R[x], mR[x- 1 ]
= R[x- 1 ] cannot both hold. Indeed, if they did, then for some elements
'0' ..., r" ,., ..., r Em, we would have
rJJxP+ ... +rlx+rO = 1,
, -t + + ' -1 + ' 1
rqx ... r1X ro = .
(8)
(9)
We may additionally assume that the numbers p, q are nlinimal and, in view
of the symmetry of the assumptions about x and x- 1 , that q  p. Multiplying
equality (8) by x- t , we get
rpx P - t + ... +r1x1-t = (l-ro)x- t . (10)
The element l-ro is invertible in R, and so, computing x- t from (10) and
substituting the result in (9), we obtain an equality of the form
r 1 x- t+ 1 + ... + r' x- 1 + r = 1
for some r;{, ..., r: 1 Em, contradicting the minimality of the number q.
Let 0 :j:. x E K; we claim that either x E R or x- 1 e R. By replacing, if
necessary, x with x- 1 , we may assume, on account of the previous part of the
proof, that mR[x] :F R[x]. There exists a maximal ideal m' of the ring R[x]
containing the ideal mR[x]. We then have m' nR = m, and the embedding
R (R[x] induces a homomorphism R -+ RIm -+ R[x]/m'. Since the field
R[x]/m'is a homomorphic image of the polynomial ring (RIm) [Y], the residue
class x+m' is an element algebraic over RIm. The field L is algebraically
closed, whence there exists an extension of the homomorphism I', where
f: R  RIm  L, to the field R[x]/m', and consequently also an extension
of the homomorphismfto the ring R[x]. Thus it follows from the maximality
of (R,f) that R[x] = R, which shows that x e Rand R is a valuation ring. 0
Theorem 3.6.10
If R is a domain, then the normalization of R (i.e., the integral closure of R
in Ro) is the intersection of all valuation rings of the field Ro that contain R.
Proof
Let V :::> R be a valuation ring of the field Ro. The ring V is normal, whence
V ::;) R, and therefore the normalization R. is contained in the intersection
of all valuation rings which contain R.
Let x e Ro "- R ; then it immediately follows from the definition of an integral
element that x  R[x- 1 ]. Hence the element x- 1 is not invertible in the ring
R[x- 1 ], and therefore there exists a homomorphismf: R[x- 1 ] -+ L such that
f(x- 1 ) = 0, where L denotes an algebraically closed field. The set BI described

144
Integral Extensions and Dedekind Domains
[Ch.
in Theorem 3.6.9 (we take K = Ro) satisfies the hypotheses of the Kuratowski-
Zorn Lemma, whence there exists a maximal element (T, g)  (R[x- 1 J,f)
in fIl. The ring T is a valuation ring; if we had x E T, then, in view of the fact
that x- 1 e T, we would obtain 1 = g(I) = g(xx- 1 ) = g(x)g(x- 1 ) = g(x)f(x- 1 )
= 0, which is not true. Consequently, the element x does not belong to the
valuation ring T, and the theorem follows. 0
The following extension theorem for valuations is if great importance:.
1rheoremm 3.6.11
Let v: K* --) G be a valuation of a field K, v(K*) = G, and let K c: L be a field
extension. There exist an ordered group G' => G and a valuation v': L *  G'
which is an extension of v.
Proof
There exists a homomorphism It: R"  n into an algebraically closed field D.
In the partially ordered set of pairs (T, g), where T is a subring of the field L,
g: T -+ D, there exists a maximal element (R,f)  (Rf)' h). We have of course
R*(')K* => R:, and if x e R*nK* then also x- 1 E R*nK* and at least one
of the elements x, x- 1 belongs to the ring R". If x- 1 E R" then h(x- 1 ) = f(x- 1 )
:F- 0, and consequently x- 1 is invertible in Rv, thus x E R". As a result we get
R*nK* = R* and
.,
G = v(K*)  K*fR: = K*f(R*r.K*) c: L*/R*.
In the group G' = L*fR* one can introduce an ordering relation compatible
with the group operation, just as in the proof of Theorem 3.6.4; then the above
monomorphism preserves ordering. The valuation v': L*  L*fR* = G'
is an extension of the valuation v if we identify G with a subgroup of the
group G'. 0 .
Let us observe that in proving the theorem the possibility of extending
the value group of a valuation was essential. For the study of valuations with
values in the group Z we shall need a more detailed investigation of the connec-
tion between the groups G' => G and the degree of the field extension K c: L.
1rheorem 3.6.12
Let K c: L be a field extension and let v: K*  G, v': L* -+ G' be valuations
satisfying the conditions: Im(v) = G, Im(v') = G'. If G c: G' and the valuation
v' is an extension of the valuation v, then
(G':G)[R",fm':R,,/m]  [L:K],
where (G': G) is the index of the group G in the group G' (i.e., the number
of elements of the factor group G' /G), and m', m = m' nRu are maximal
ideals of the rings Ru" R".

III]
Valuations
145
Proof
Choose elements Y 1, .." Yk E L such that 'I)' (Yi) - v' (y J)  G for i:/:. j, and
elements r, ..., r; E R such that the residue classes 1" +m', ..., r; +th' E Rv,/m'
are linearly independent over the field Rim. Even though we do not assume
that the numbers occurring in the formula are finite, it is sufficient to establish
that the elements rYJ E L*, i = 1, ..., l,j = 1, ..., k, are linearly independent
over K.
Suppose there exist elements CXii E K, not all equal to 0, such that
L: O(IJr:YJ = O.
I.i
If PJ = 2: (XiJr;, j = 1, ..., k, then 2: PJYJ = O. Fix an index j such
i j
that not all of the coefficients CXl), ..., CXl} are equal to 0; there is an index
i(j) such that the valuation of the element CXI(J)J is the least of the valuation
of these elements among CXl}'''.' CXl} which are non-zero. We then have
P J = CXi(j)} 2: cxJ"' where cx},..., cx;J e Rv and cx(})j = 1. Froln the linear
i
independence of the residue classes r +m', ..., r; +m' over the field Rvlm
it follows that 2: cxJr: e Rv,,,-m', and hence PJ =1= 0 and v'({JJ) = V(CXI(J)J). By
ij
renumbering the elements Y 1, ..., Yk if necessary we may assume moreover
that the elements Pl, ..., pq are non-zero for a certain q  k, (Jq+l = ... = Pic
= 0, and V'(PlYl)  V'(P2Y2)  ...  v'(pqYq). If we had q  2, then V'(PlYl)
q
= V'(-PlYl) = v' (2: PJY})  V'(P2Y2), and consequently V'(PIYl) = V'(P2Y2)
j=2
and V'(Yl)-V'(Y2) = V'(P2)-V'(Pl) = V(CXi(2)t2)-V(i(1)tl) eIm(v), contrary
to the assumption. The condition q = 1 means that PlYl = 0 and also implies
a contradiction. Thus the set {r;YJ} is linearly independent over K. 0
As already mentioned, the most frequently used valuations are those with
values in the ordered group Z of integers. Such valuations v: K* -+ Z are
called discrete valuations of the field K. We can distinguish discrete valuations
among all valuations by properties of their valuation rings in the following
manner.
1rheoremm 3.6.13
Let 'v: K* -+ G be a non-trivial valuation of a field K. The ring Rv of the
valuation v is Noetherian if and only if the group Im(v) c: G is isomorphic
to the ordered group of integers.
Proof
We assume the ring Rv is Noetherian. It follows from Theorem 3.6.8 that
every ideal in R" is principal and therefore, R" being a local ring, there exists

146
Integral Extensions and Dedekind Domains
[Ch.
in Rv an irreducible element 0 :/:. p e Rv. If 0 i= r e Rv then Rv I. = Rvp" for
some non-negative integer k, whence we deduce that
Im(v) = {O, + v(p), + 2v(p), ...}  z.
Suppose, that the group Im(v) is isomorphic to the group Z and let 1 E Z
be the image of some element v(p) with p E K. Obviously v(p) > 0 in the
group G, and if 0 :/:. r E R; then vCr) = kv(p) = v(P") for some non-negative
integer k. Since vCr/pIc) = 0, rlpk E R and Rvr = Rup". If 0 :/:. I c: Rv is an
ideal then
I = U Rv r = U Rvpv(r) = Rvpv(r o )
O:prel O:#=t'el
provided that 0 :f: ro E I, and v(ro) = min {v(r); 0 :/:. rEI}, and so Rv is a
principal ideal ring. 0
Theorem 3.6.14
A domain R is a ring of some discrete valuation of the field Ro if and only if R
is a principal ideal domain \vith only one non-zero prime ideal.
Proof
The implication => was established in the proof of the preceding theorem.
Assume that R is a principal ideal domain with only one non-zero prime ideal
Rp. The only irreducible element in R is p, and so every element 0 :/:. r E R
has the form I. = up", where u e R*. The formula vCr) = k defines a valuation
of R, which can be uniquely extended to a valuation of the field Ro, and we
clearly have Ru = R.
In the case of an arbitrary valuation, the intrinsic properties of the ring Rv
lead us to the definition of a valuation ring; similarly, the above theorem leads
to a definition of a discrete valuation ring.
Definition 3.6.15
A domain R is said to be a discrete valuation ring if R is a principal ideal
domain with only one non-zero prime ideal.
A discrete valuation ring R determines a unique discrete valuation v:
Ro -+ Z such that R = Rv and Im(v) = Z.
There are many characteristic properties of discrete valuation rings. Some
of them are given in the following theorem:
Theorem 3.6.16
Let R be a local (and hence Noetherian) domain with the maximal ideal
m :F O. The following properties are equivalent:
(i) R is the ring of a discrete valuation,
(ii) R is normal and m is the only non-zero prime ideal,

III]
Valuations
]47
(iii) R is normal and there exists a principal m-primary ideal,
(iv) m is a principal ideal,
(v) dimRl m m/m 2 = 1, 
(vi) there exists an element 0 i= pER such that every non-zero ideal is of
the form (p"), k = 0, 1, ...,
(vii) every non-zero ideal has the form nt k , k = 0, 1, ...
Proof
(i) => (ii). This is obvious.
(ii) => (iii). The only prime ideals of Rare 0 and m, whence every non-zero
proper ideal is m-primary.
(iii) => (i). Let the ideal Rr be m-primary, and let q be the least positive
integer such that Rr:::> mil. and s em 4J - 1 "Rr. Then am c: m 4 c: Rr and
therefore (alr)m c R. The element a/,. e Ro does not beloJ:1.g to R, whence
it is not integral over R, which by Theorem 3.1.6, yields (a/r)m cJ: m. Thus
(a Ir)m = R, and there exists an element p Em such that (a /r)p = 1. For any
element x em, we have x = (alr)xp, (alr)x E R, and consequently m = Rp.
Let I be an arbitrary ideal; if 0 :/:. Y E I, then y = tp", t  m, whence Ry = Rp",
which shows that I = Rp"o, where ko is the least number with the property
p"o e I.
(i) => (iv). This implication is obvious.
(iv) => (v). The ideal m is principal, and hence dimm/m 2  I; dimm/m 2
> 0, since the Nakayama Lemma shows that m i= m 2 .
(v) => (vi). If p em"m 2 then (v) implies that Rp+m 2 = m thus m(m/Rp)
= (m 2 +Rp)IRp = m/Rp and by the Nakayama Lemma it follows that
m = Rp. If I c: R is a non-zero ideal, then there exists a largest number q
such that I c: m 4 . For an element r em4J+l, we therefore have r = p 4 t,
t e R"m. Thus I:::> Rr = Rp4 = m 4J :::> I, and I = Rp4.
(vi) => (vii). Obvious.
(vii) => (i). If p Em"m 2 then m = Rp, whence every ideal is principal,
and m is a unique non-zero prime ideal. 0
Let us state one more simple property of discrete valuation rings.
Theorem 3.6.17
If R is a discrete valuation ring and if R' is a ring such that R c R' c: Ro,
then either R' = R or R' = Ro.
Proof
There exists a valuation v: R  Z such that R" = R. Suppose that R' #= R;
thus there exists an element x E R'''R, and consequently v(x) = -m < O.
If y e Ro "R, then v(y) = -Ie < 0 and, taking q to be a positive integer such
that k  qm, we get v(y/x q ) = -k+qm  0, and therefore ylx q E R thus
y e R[x] c: R', which yields R' = Ro. 0

148
Integral Extensions and Dedekind Domains
[Ch.
Now we shall concentrate on extensions of discrete valuations. Let a field L
be a finite extension of a field K and let v: K*  Z be a discrete valuation of K.
It follows from Theorem 3.6.11 that there exists an extension Vi: L* -+ G
of the valuation v, where G is an ordered group containing the group Z.
Theorem 3.6.12 shows that (G:Z)  [L:K], whence the group GIZ is finite.
Accordingly the group G is finitely generated and, as any ordered group is
torsion free, it is a free Z-module. The group G I Z is finite only if G  Z. The
extension Z eGis therefore isomorphic to an extension Ze c: Z, where
e > O. Thus \ve have established
Corollary 3.6.18
For every finite field extension K c: L and a discrete valuation v: K* -. Z,
there exists a discrete valuation Vi: L*  Z and a positive integer e such that
v'(x) = ev(x) for x E K*. Every extension of the valuation v with values in an
ordered group is equivalent to a discrete valuation.
The number e is called the ramification index of the valuation v with respect
to the valuation v'; the number [Rvi 1m': Rvlm] is denoted traditionally by f.
It is known (see [L, p. 308]) that, if the extension K c: L is finite and separable,
then there exists a finite number of non-equivalent extensions of v to discrete
q
valuations Vt, ..., v q of L, and  eJl = [L: K], where e" fl are the natural
;=1
numbers, defined above, corresponding to the valuation Vi, i = 1, ..., q.
If K c L is a Galois extension, then el = ... = e q , 11 = ... = /q.
The connection of discrete valuations with ideals and the role of the ramifica..
tion index will be illustrated by the example of Dedekind domains.
Example 3.6.19
Let R be a Dedekind domain; we shall describe those discrete valuations
v: R  Z for which Rf) :::> R. The intersection mnR = P, where m is a maxi-
mal ideal of Rf)' is a non..zero prime ideal of R. Since every element of the set
R"P is invertible in Rv, we have R p c Rv. It follows from Theorem 3.6.16
(ii) that R p is a discrete valuation ring, and so we have R p = Rv by Theorem
3.6.17. This shows that the valuation v is of the form V p , where P is a non-zero
prime ideal of the ring Rand vp(r) = n <::;- Rr = pn pr 1 ... P:s provided that
o ¥= r E R and the prime ideals P, PI' ..., Ps are all distinct.
Let Ro c: L be a finite field extension. Then the normalization R of R in L
is also a Dedekind domain by Theorem 3.3.12. Let v' be an extension of the
valuation Vp to the field L (i.e., there exists a positive integer e such that
v' (x) = ev"(x) for 0 #: x e R). The ring Rf)I. is normal and RV I :::> Rv :::> R
- p ,
whence RUI ::> R; thus the valuation Vi has the form v' = va, where Q is
a non-zero prime ideal of R .
The ideals P, PI' ..., Ps are all maximal, whence it can readily be deduced
that factorizations of the ideals R P, RP t, ..., R P s into products of prime

III]
Valuations
149
- -
ideals have no common factors. The ideal RP has a factorization RP = Qi J ...
. .,. Q:4, where Q l' ..., Qq are distinct prime ideals, e 1 > 0, ..., e q > O. We
then have for the element r considered above
Rr = R(Rr) = R(P"p7. 1 ... P:') = ( R P)n( RP j )n 1 ... ( R Ps)n s
= Qel ... Q:e q . I,
and the factors Q l' ..., Qq do not occur in the factorization of the ideal I.
Hence it is easily seen that Q is one of the ideals Q l' ..., Qq.
We have proved that, if the extension R P of a prime ideal P c: R has a
factorization R P = Q1 ... Q:q, then the only extensions of the valuation Vp
are the valuations val' ..., Vall and the numbers e 1, ..., e q are the ramification
indices of Vp.
This example indicates how much information about prime ideals and
their factorizations can be obtained by means of valuations.
We shall establish one more theorem about extending valuations, this time
in the case of non-algebraic extensions.
Theorem 3.6.20
If V is a valuation of a field K with values in an ordered group G, and if K(X)
is the field of rational functions in one indeterminate, then the formulae
v (an xn + ... + at X + ao) = min {v(a n ), ..., v(a1)' v(ao) },
where 00' ..., an E K, and
vlf/h) = v(f)-v(h) for f, h E K[X],
define a valuation of K(X) with values in G, and
R; = Rv[X]m(R,,)R.,rXj.
Proof
Let f = anX n + ... + ao, h = blnXnI + ... + ho be non-zero polynomials of
K[X], and let k, I be the least indices such that v(ak) = iJ(f), v(b,) = v (h).
The coefficients Co, ..., C n + In of the polynomialfh can be expressed in the form
c p = L aibJ, whence v(c p )  v(ak)+v(b,), and therefore v(fh)  v(f)+
I+J=p
+v(h). We shall compute V(Ck+'):
(1) if i = 0, 1, ..., k - 1, then
v(a,b,,+,_,) c: v(a,)+v(b k + I - i ) > v(ak)+v(b k + l -,)  v(a,,)+v(b,),
(2) if i = k+ 1, ..., n, then k+l- i < 1, and similarly we obtain
v(a,b k + , - i ) > v(ak)+v(b,).
Using the equality v(a"b,) = v(a,,) + v (b,) and the above inequalities, we
see that fJ(Ck+') = v(a,,) + v (b,), hence v(fh) = v(f) +v(h).
Assume moreover v(ak)  v(b,). Then the inequality v(ai + hi)  min (v(a,),
v(b j ) ) yields

150
Integral Extensions and Dedekind Domains
[Ch.
fl(f+h) = min {v(a, +b,)},  rnin {v(a,), v(b J ) }'J = v(b,)
= v(h) = min«v.n, v(h»,
and v is actually a valuation.
To prove the last assertion, assume that flh e R;, i.e., fllf)  v(Il). We
then have v(aj)  v(b,), v(b J )  v(b,) for all i, j, which shows that fIb, E R,,[X]
and v(h/b,) = O. Hence hlb , m(Rv)R;;[X], which establishes the validity of the
inclusion c:. The opposite inclusion is obvious. D
It is easily seen that the above theorem extends without any difficulty
to the case of the field of rational functions in an arbitrary number of indeter-
minates. The value of the valuation v at a non-zero polynomial should then
be defined as the minimum of its values on the coefficients.
The classical notion of the absolute value of a complex number allows
us to introduce a metric, and hence also a topology, on every field contained
in the field of complex numbers. We shall generalize this situation to the case
of any field in the following definition:
Definition 3.6.21
An absolute value (or norm) defined on a field K is a function I I: K  R, where
R is the field of real numbers satisfying, for x, y e K, the following require-
ments :
(i) Ixl ;?; 0,
(ii) Ixl = 0 <=> x = 0,
(Hi) Ix. yl = Ixllyl,
(iv) Ix+ YI  Ixl + 'yl.
If, in addition, the condition
(iv') Ix+ YI  max(lxl, Iyl),
is satisfied, then the abosulue value I I is said to be nonarchimedeao.
The absolute value determines a metric d on K by the formula d(x, y)
= Ix - yl. Just as in the case of the field of complex numbers, we prove that
the operations +, -, ., : are continuous with respect to the metric. Proceeding
as in Section 2.6 we can construct the completion of the field K with respect
1\
to the metric d, and we call the resulting field K the completion of K with
respect to the absolute value I I. Of particular importance are the nonarchi-
medean absolute values arising from a valuation 'lJ: K* -+ R of the field K
with values in the ordered group of real numbers, especially those arising
froln a discrete valuation. Let 0 < a < 1 be a real number. We define an
absolute value I 'v: K  R by the formula Ixl v = afJ(:t) for x e K*, 101., = O.
"
The completion K of K with respect to I 10 is also called the completion of K
with respect to the valuation v (it is independent of the choice of the number
a).

III]
Valuations
151
Fields conlplete with respect to a valuation play an important role in
algebraic number theory. The most important examples are the completions
"
Qp of the field of rational numbers with respect to the p-adic valuations,
which are called the fields of p-adic numbers. The closure of the ring Z in the
"
field Q, is called the ring of p-adic integers. One easily verifies that this ring
is isomorphic to the completion of the local ring Z(P).
Exercises
1. Let R be a domain and Q the ideal of the polynomial ring T = R[X 1 t ..., X n ], generated
by Xl, ..., XII. Show that the function V: T"" {O} -+ . Z given by the formula vet) == k <=> t
e Q"""-..Q" + 1 for t e T is a valuation of T. Find the corresponding valuation ring of the field
of fractions To. State the corresponding property for graded rings.
2. Denote by Q+ the ordered additive group of rational numbers. Extend the valuation
Z[Q+] -+ Q+ described in Example 3.6.2 to a valuation v of the field of fractions. Prove
that the valuation ring Rv of v is not Noetherian.
3. Prove that if R is a principal ideal domain then every valuation ring of the field Ro
containing R has the form R p , where peR is a prime ideal.
4. Describe all the valuation rings of the field K(X) of rational functions which contain
the field K.
5. Let R be a subring of a field K and peR a prime ideal. Prove that there exists a
valuation ring V of K such that ReV and mJ'r\R = P, where my is the maximal ideal of V.
6. Let R be a domain. Prove that the following conditions are equivalent:
(i) every finitely generated ideal of R is invertible,
(H) for every prime ideal peR, the ring of fractions R,. is a valuation ring,
(Hi) for every maximal ideal m c: R, the ring of fractions Rm is a valuation ring.
7. Let v: K* -+ Z be a discrete valuation of a field K and let I I,,: K -+ R be the corre-
sponding absolute value. Prove that: lOin the completion R" of the metric space (R." d,,),
where du(x,y) = Ix-ylv, there exist (unique) extensions of the operations +, .: Rox
III III
X Rv -+ Rv to continuous operations in Rv; Rv is a ring under these operations; 2 0 the ring
- "
Rv is isomorphic to the completion Rv of the ring R" in the m,,-adic topology.
NOTES AND REFERENCES
The notion of divisibility of positive integers and the notion of a prime number and also
their basic properties were well known already to Euclid. As we mentioned in the first section,
the need for developing a theory of divisibility in sets of complex numbers, substantially
larger than the set of integers in which polynomials under consideration admit a convenient
decompositions, emerged from problems of number theory. The first difficulty consisted in
the problem of selecting an appropriate set of complex numbers. Today, the main object
of study in algebraic number theory is, as a rule, a subfield K of the field of complex numbers,
of finite degree over Q, and the ring R K of integral algebraic numbers of K. Another essential
difficulty, first discerned by Kummer (and traditionally but wrongly ascribed to Dirichlet),
was connected with the nonuniqueness of factorization of numbers in the ring Rx into irre-
ducible factors from this ring. The first important steps towards overcoming this difficulty
were made by Kummer, who established, about 1845, the theory of divisibility in the rings
of integers Z[C] of cyclotomic fields Q(C), CP = 1, for prime p. He gave a description of "ideal
prime numbers" and a method of finding the multiplicity with which they divide a given
number; in modern terminology it is a description of discrete valuations of the rings Z['] (or,

152
Integral Extensions and Dedekind Domains
equivalently, the description of non-zero prime ideals), and finding the values of those
valuations at a given number. A detailed discussion of the theory of algebraic numbers at
that stage is to be found in [B].
It was only about 1870 that Dedekind realized the need to extend the ring Z[,Id] to the
ring Z [ (JI d+ 1 )] in the case where d == 1 (mod 4) in order to obtain a satisfactory theory
of divisibility of "ideal numbers". The basic ideas of algebraic number theory were developing
slowly, along with the gradually increasing information on particular types of field. The
papers by Dedekind, and particularly his famous Supplements to Dirichlet's book [6], contain
a very clear synthesis of the concepts, methods, and contemporary results of the theory.
It is there that the notion of a module occurs for the first time.
A recapitulation of the nineteenth century studies, particularly those by Kummer, and
by Dedekind and Kroneckeli competing with each other, and also new results and directions,
were presented by Hilbert in the so-called Zahlbericht [10]. The term "ring" was introduced
by Hilbert in place of the former "order" (Ordnung). The modern, abstract concept of a field
was worked out by Steinitz [26] under the influence of Hensel's theory of fields of p-adic
numbers. The development of this theory has led to a very useful method of completion
of a field with respect to a given valuation. The theory of valuations is one of the links be-
tween the theory of fields and the theory of algebraic curves and Riemann surfaces. The
continuous development of the theory of algebraic numbers in the 20-th century is connected
mainly with the names of Artin, Hasse, Chevalley, Serre, Tate. The modern methods go
considerably beyond the scope of this book.
The theorems concerning connections between the prime ideals of a ring and those
of its integral extension were proved by Krull [13]. Valuation rings were first defined and
studied by Krull in [12].

Chapter IV
Divisors and Krull Domains
In Chapter III, we indicated how some basic methods of algebraic number
theory \vere transferred to the class of Dedekind domains. However, the
question of the uniqueness of factorization into irreducible elements applies
to any domain. As we have seen (cf. Exercise 7, Section 2.3), every element
of a Noetherian domain is a product of irreducible elements, but, in general,
this factorization is not unique.
The theory of the ideals of Dedekind domains, which has been successfully
employed in number theory, makes it possible to obtain a unique factorization
of an ideal into prime ideals which, of course, are irreducible. Thus, the theorem
on unique factorization of elements can be preserved at the cost of admitting
new factors which are not numbers but ideals (corresponding in the classical
version to so-called 'ideal numbers').
Quite naturally the question arises whether one can construct a similar
theory of factorization of elements or ideals for a class wider than that of
Dedekind domains. One possible answer to this question is given by Theorem
2.3.25, concerning the representation of ideals of a Noetherian ring as intersec-
tions of primary ideals with the uniqueness of non-embedded components
and the above-mentioned theorem on the factorization (in general, not unique)
of an element of a Noetherian domain into a product of irreducible elements.
A generalization of the theory of ideals of Dedekind domains, or, more pre-
cisely, a generalization of the theorem on the representation of an ideal as an
intersection (not a product) of powers of prime ideals, is the theory of divisors,
which will be presented in this chapter. In Section 4.1, we shall describe for
any domain a class of ideals called integral divisors or integral divisorial ideals,
and in Section 4.4, restricting the rings to the class of Krull domains, defined
and studied in Sections 4.2 and 4.3, we shall develop a theory similar to the
theory of ideals of Dedekind domains. The class of Krull domains proves
to be the most convenient for this purpose. In Sections 4.5 and 4.6, we shall
study an important invariant of a Krull domain, namely its divisor class group
which is a kind of measure of the non-uniqueness of the factorization of ele-
ments into a product of irreducible elements. Making use of practically the
whole theory of Krull rings, we shall prove that every Abelian group is isomor-
phic to the ideal class group of some Dedekind domain. This result gives a deep

154
Divisors and Krull Domains
[Ch.
insight into the non-uniqueness of factorization in Dedekind domains and is
the final outcome of our considerations on the subject.
. In Chapter III an important role was played by the property of normality
of a domain, partly because every unique factorization domain is normal
and every domain R has a unique nornlalization R, R c: if c Ro. For a long
time it was not known whether or not the normalization of a Noetherian
domain were also Noetherian. This normalization turned out to be a Krull
domain which may be not Noetherian ([N] and Example 4.7.3). This result, the
possibility of constructing a satisfactory theory of divisors in Krull domains,
and many other good properties of this class of rings (it is closed with respect
to localization and to the construction of polynomial rings in an arbitrary
number of indeterminates, etc.) confirm that it is a very natural class to study.
The class of Krull domains is closely connected with the class of valuation
rings. As we know, the normalization of a domain R is the intersection of all
valuation rings of the field Ro which contain R, and of course every intersection
of valuation rings of a given field is a normal domain. The class of domains
R which are intersections R = n VA of a family {VA} of discrete valuation rings
contained in a :fixed field, such that every non-zero element r e R is invertible
in almost all rings VA' is a class of Krull domains (Section 4.3).
The most important relations between the class of Krull domains and other
familiar classes of domains are the following: a unique factorization domain
is a Krull domain; a Dedekind domain is a Krull domain; a Krull domain
is normal; a normal Noetherian domain is a Krull domain.
There are comparatively few textbooks on commutative algebra which
devote sections of any length to Krull domains; however, a detailed treatment
of the subject seems to be indispensable for a full explanation of the links
between the properties of unique factorization, normality, and the theory
of divisors. A more extensive discussion of the theory of Krull domains wilt
be found in [D] and [F].
Throughout this chapter all rings considered are domains and Ro denotes
the field of fractions of R.
4.1 DIVISORS AND THE SEMIGROUP OF DIVISOR CLASSES
In this section we shall introduce the important notion of a divisor. Its prototype
appeared first in studies of rational functions on non-singular algebraic curves
and was subsequently generalized to the case of algebraic varieties of higher
dimensions. Let us consider the simplest case, where the curve V is a line K 1
whose ring of polynomial functions is the polynomial ring R = K[X]; K is an
algebraically closed :field. Every rational function f e Ro determines a linear
combination div(f) = L vp(f)P, where 'lJp is the valuation corresponding to
PeY

IV]
Divisors and the Semigroup of Divisor Classes
155
the point P on the curve V (in our example, vp(f) is the multiplicity \vith which
the irreducible polynomial X - P occurs as a factor in f). Since almost all
coefficients vp(f) are zero, div(f) is an element of the free Z-module generated
by the points P e V. Every element D = L dpP of this module, called a
divisor in the classical version, determines a set L(D) of those rational functions
f e Ro which satisfy the conditions 'lJp(f)  - d p for all P e V. Obviously,
the set L(D) c: Ro is an R-module, and there exists an element 0 =/:. r E R
such that rL(D) c: R.
This description extends to any smooth algebraic curve. In the case of a
straight line, it is easy to observe that, for every divisor D the R-module L(D)
is cyclic and generated by a function 10 such that div(fo) = - D; however,
this does not hold for an arbitrary curve.
Similar constructions can be applied by replacing a curve with an algebraic
variety of dimension n > 1. Points are then replaced by submanifolds of
dimension n-l (and not by points); which are determined by prime ideals
P c K[V] of height 1. In the case where the ring K[V] is normal there exist
suitable valuations.
In this section, we shall establish an algebraic analogue of the situation
outlined above. Fractional ideals of a domain will correspond to modules
L(D). Divisors will be defined as fractional ideals of a particular type. This
definition does not refer directly to the geometric situation described above;
nevertheless, we shall. show in Section 4.4 that, in the case of Krull domains
(particularly of normal Noetherian domains), a divisor becomes an object
analogous to the L(D)-type modules determined by a linear combination
of prime ideals of height 1 with integer coefficients.
Further on, also following the patterns of the constructions applied in
algebraic geometry and the theory of algebraic numbers, we shall establish
a relation of linear equivalence in the semigroup of divisors, obtaining a semi..
group of divisor classes. We shall prove that in the case of a Dedekind domain
the latter is isomorphic to an ideal class group.
In the class of all fractional ideals, we shall distinguish an important
subclass of ideals which are called divisors (or divisorial ideals).
Definition 4.1.1
A non-zero fractional ideal J c: Ro is called a divisor of a domain R if and only
if it is an intersection of a set of principal fractional ideals. We shall denote
the set of all divisors of R by Div(R).
It follows directly from the definition that a divisor J is the intersection
of all principal fractional ideals Rx, x e Ro, such that Rx => J. A non-zero
principal fractional ideal is obviously a divisor.
Definition 4.1.2
For any non-zero fractional ideal J e Id(R), we denote by D J the set {x
E Ro: J c: Rx}. We define

156
Divisors and Krull Domains
[Ch.
div(J) = n Rx.
xeD,J
Instead of div(Ry) we shall \vrite div(y).
Obviously, div(J) => J; if aJ c: R (a :1= 0) then div(J) c: Ra- 1 , and div(J)
is a divisor; it is the smallest divisor containing the fractional ideal J. Thus
we have defined the mapping div: Id(R)  Div(R). A fractional ideal J e Id(R)
is a divisor provided J = div(J). For every J E Id(R), we have div (div(J) )
= div(J). A non-zero intersection of any family of divisors is also a divisor.
Theorem 4.1.3
For any non-zero fractional ideals I, J of R, the following are satisfied:
(i) I:J = ny-II,
OyeJ
(ii) if I e Div(R), then I: J e Div(R),
(iii) div(l) ::> div(J)  R: I c: R:J,
(iv) div(I) = div(J)  R: I = R: J,
(v) R:I = R: (R:(R:l)},
(vi) div(I) = R:(R:I).
Proof
(i) results from the following equivalences:
x E I:J  xy e I for every 0 ¥= Y E I 
XEy-l I for every 0 =1= YEJ.
Condition (ii) follows from (i) since y-l1 e Div(R) for 0 =1= y e J.
In order to prove (Hi), observe that if div(I) :::> div(J) then, for every
x e D 1 , we have Rx :::> div(I) :::> div(J) :::> J, whence xED J, and so, as can
easily be seen div(l) :::> div(J) <=> Dr c DJ. On the other hand, for an element
y eRa we have
Y E R:l <=> yl c: R <=> 1 c: Ry-l  y-l E D "
whence D, c: DJ <:> R:l c: R:/, which ends the proof of (iii).
Condition (iv) is the consequence of (Hi).
(v) Immediately froln the definition it follows that 1 c: R:(R:I) and so
R:/:::> R:(R:(R:1); replacing I with R:I in the first formula, we obtain
R:I c: R:(R:(R:1)), which yields condition (v).
To prove (vi), observe that (iv) and (v) yield div(/) = div(R:(R:I), and
therefore, as R e Div(R), from (ii) we obtain div(R:(R:J)) = R:(R:I), whence
it follows that div(/) = R:(R:l). 0
We shall now prove that the surjection div: Id(R)  Div(R) and the
multiplication in Id(R) induce a certain operation in Div(R). Namely, we shall
prove the following implication: if div(I) = div(/ 1 ) and div(J) = div(J 1 ),
then div(/. J) = div(11 · 11). Choosing elements 0 =1= x E R: (1. J) and 0 #: fJ

IV]
Divisors and the Semigroup or Divisor Classes
157
e J, we have xfJI c: R, so I C R(xfJ)-l, whence it follows that 1 1 c: div(/1)
= dive!) c R(xfJ)-1. Hence XP1l c R, and therefore xJI! c: Rand
x e R:(I 1 !). Thus we have shown that R:(l. J) c: R:(11 · J), whence, by
symmetry, we obtain R:(I. 1) = .R:(I 1 ..1), which by formula (iv) in Theorem
4.1.3, yields dive!. J) = dive!! · J). This formula implies the equality, dive!. J)
= div(I1 · J 1 ).
The implication just proved enables us to define an operation + in the
set Div(R) by means of the formula
div(I) + div(J) = div(!. J).
The sum of divisors defined above may not be an ideal generated by ideals I
and J, even in the case of I = div(I), J = dive!); we show in Example 4.1.8
that the ideal I. J need not be a divisor.
It is easy to see that Div(R) endowed with the operation + is an associative,
commutative semigroup with zero, which we call the semigroup of divisors of R.
Observe that the operation + preserves inclusion of divisors, i.e., satisfies the
following condition:
if divt1) :::> div(ll) then dive/) + dive!) => div(ll) + div(J).
Indeed, from formula (iii) in Theorem 4.1.3 it follows that R:I c: R:l 1 ,
and so if xeR:(I.J) then xJc: R:lc: R:I 1 , and therefore xeR:(I 1 .J).
This implies that R: (I · J) c: R: (1 1 · I), which yields the desired result. Thus,
the semigroup Div(R) with the relation of inclusion forms a partially ordered
semigroup.
Let us now observe that, if a fractional ideal I is invertible, i.e., if there
exists a fractional ideal I' such that I. I' = R, then the ideals I and [' are
divisors and 1+1' = 0 in the semigroup Div(R). Indeed, if elements ai' ..., ak
e I and a, ..., a e I' satisfy the condition L ala = 1, then, for every x
e R: 1, we have x = L (xaj) a e 1', whence, by the condition I. [' = R, it
follows that I' = R:l. Analogously, we have I = R:I', whence div(I)
= R:(R:I) = R:[' = I, and similarly we obtain div(I') = ['. Consequently,
invertible ideals are identical with invertible divisors in the semigroup Div(R).
In a Dedekind domain R every non-zero fractional ideal is invertible,
whence the semi group Div(R) is a group. The following theorem states necess-
ary and sufficient conditions for the semigroup Div(R) to be a group. Let
us recall that we defined in Chapter III a completely normal domain as a
domain R which satisfies the following condition:
if xeR o and the R-module R[x] = R+Rx+Rx2+ ... is contained in
a finitely generated R-submodule of Ro, then x belongs to R.
Theorem 4.1.4
A semigroup of divisors Div(R) is a group if and only if the domain R is
completely normal.

158
Divisors and Krull Domains
[Ch
Proof
Assume that Div(R) is a group_ Suppose that for some x E Ro the R-module
J = R+Rx+... is contained in a finitely generated R-module I c Ro.
Hence, there exists an element 0 :/= r E R such that R ::> rI ::> rJ, and therefore J
is a fractional ideal. Since xJ c: J, we have div(J) ::> div(xJ) = div(x) + div(J).
By the assumption, there exists a J' E Id(R) such that div(J) + div(J') = div(R)
= 0, and so the previous inclusion yields div(R) ::> div(x). Hence, by Theorem
4_1_3, we conclude that R = R:R c R:Rx, whence it follows that x E R,
and thereby the domain R is completely normal.
Now, assume R is completely normal, and let ]be a divisor. It is sufficient
to prove that div(/-(R:I)) = 0, since it implies that div(1)+div(R:/) = O.
Let Rxo be a principal fractional ideal which contains the ideal I- (R: I).
For every element Y E D 1 , we have Ry::> I, and so y-I E R:I and
Iy-I c ]-(R:I) c Rxo, whence xoI1 cRy. Since] = n Ry, we have XCiII
yeD,
C I. Hence we conclude that for every positive integer n we have (xCi I )"I c I.
Choosing elements 0 1= a e I and Yo E D 1 , we have Ra c ] c Ryo and (xCi 1 )na
E (XCi 1)"] c: ] c Ryo, i.e., (xol)n C R(Yo/a) for n = 0, 1, ... Hence it follows
that the R-module R[xo l ) is contained in the module R(Yo/a), and therefore
XCiI is an almost integral element, which implies that xCi 1 E R. Thus we have
Rxo ::> RXolxo = R, and, since the ideal div(/-(R:/») is an intersection
of ideals of the form Rxo, we have div(/-(R:I))::> R. Since [-(R:I) c: R,
we finally obtain div(/-(R:I)) = div(R) = o. 0
In the semigroup of divisors Div(R) we define a congruence relation ==,
called the relation of linear equivalence of divisors, assuming that
div(I) == div(J) <:> there exists 0 :1= x E Ro such that
div(I) = div(J)+div(x)_
Since div(J) + div(x) = div(xJ) = xdiv(J), we easily check that the relation
== is reflexive, symmetric and transitive. If, moreover, div(I I ) = div(J I )
then, for some element 0:1= Xl E R, we have div(]I) = div(J I )+div(xl)'
whence we obtain
div(I) + div(I I ) = div(J) + div(x) + div(J 1 ) + div(xI)
= div(J) + div(J 1 ) + div(xxl) ;
thus == is indeed a congruence relation_
Definition 4.1.5
The factor semi group Div(R) / == of the semigroup of divisors of the domain R
is called the semigroup of divisor classes of R, and written CI(R)_
Corollary 4.1.6
If a domain R is completely normal, then the semigroup Cl(R) of divisor
classes of R is the factor group of the group Div(R) by the subgroup Prin(R)
of principal divisors_

IV]
Krull Domains
159
As we have mentioned above, if R is a Dedekind domain then every non-
zero fractional ideal is a divisor. The group of divisor classes is then identical
with the ideal class group, Id(R)/Prin(R), also denoted by CI(R), which was
defined in Section 3.4.
In subsequent sections, we shall study the class of Krull domains and prove
that a Krull domain is a unique factorization domain precisely if its divisor
class group is zero.
Example 4.1.7
We now give an example of a prime ideal P which is a divisor such that p2
is not a divisor.
Let K be a field, let T = K[X 1 , ..., Xn] be a polynomial ring with n  2,
and let R = K[Xf,X 1 X 2 , ...,XiX), ...,X;] be the subring of T generated
by the field K and all the products X, XJ. The ring R consists of all polynomials
whose homogeneous components of odd degrees are equal to zero, R = To +
+ T 2 + T4 + .. . We shall show that the prime ideal P = TX 1 nR is a divisor
and the ideal p2 is not a divisor of R. The ideal P is generated by the monomials
Xl, X 1 X 2 , ..., X1X n . Immediate calculation shows that
D p = {xt/X 1 !} with 0 :FIe T 1 +T3+ ...,
D P2 = {Xf/g} with 0 1= geR.
It is easy to verifythatdiv(P) = n R(Xf/X 1 !) = P and div(P2) = nR(Xi/g)
= RXl '¥= P2. The same calculation may be performed in the case of an infinite
number of indeterminates.
Exercises
1. Prove that if R is a normal Noetherian domain, P and Q are prime ideals of R and
p  Q:I= 0, then P is not a divisor.
2. Let R be a subring Z[X, Y 2X] of the field Q(X, Y 2X) , and let P be the ideal of R
generated by the elements X and y 2X. Prove that R is normal, P is a prime ideal, minimal
in the set of non-zero prime ideals and p2 is not a divisor. [Show that X-l p2 is a prime ideal
which properly contains P.]
4.2 KRULL DOMAINS
In this section we shall define Krull domains R imposing conditions on the
rings of fractions R p with respect to some prime ideals PeR. We shall
demonstrate several fundamental properties of the class of Krull domains,
e.g., that it is closed with respect to the formation of rings of fractions and
rings of polynomials in an arbitrary number of indeterminates. We shall
regard rings of fractions of a domain R as subrings of the field of fractions
Ro.

160
Divisors and Krall Domains
[Ch.
In Definition 1.2.12 we introduced the notion of the height, ht(P) , of
a prime ideal P. Here we extend this notion to arbitrary ideals by setting
ht(l) = inf ht(P), where P is running through all prime ideals containing
P)I
the ideal I. In particular, ht(l) = ht(P) for any P-primary ideal I. We
denote by SpecteR) the set of all prime ideals of R of height 1.
Definition 4.2.1
A doain R is called a Krull domaio provided it satisfies the following condi-
tions :
(i) for every prime ideal P e Spec 1 (R), the ring R p is:a discrete valuation
.
nng,
(H) R = n R p ,
P e Spec1(R)
(Hi) for any non-zero ,. E R, there exist only finitely many prime ideals
P e Spec 1 (R) such that P  Rr.
By Theorem 3.5.6, condition (i) is equivalent to each of the following
conditions:
(it) R p is a Noetherian valuation ring,
(i") R p is a local principal ideal domain.
It is easy to see that condition (iii) is equivalent to each of the following condi-
.
tl0ns :
(iii') every non-zero element r e R is invertible in almost all rings R p ,
(Hi") every non-zero element x e Ro belongs to almost all R p and is inver-
tible in almost all R p .
Examlle 4.2.2
Every unique factorization domain is a Krull domain. Indeed, given
P E Spec 1 (R), there exists an irreducible element pEP. Since the ideal Rp
is prime, Rp = P. The ring Rp consists of all the fractions rls e Ro such that
p ,r s, and so it is a discrete valuation ring. Clearly, conditions (ii) and (iii)
are satisfied. In particular, every principal ideal domain and every field are
Krull domains. A polynolnial ring K[X 1 ,X 2 , ...] in a countable (or greater)
number of indeterminates is not Noetherian, but it is a Krull domain (see
Theorem 4.2.9).
Example 4.2.3
Every Dedekind domain is a Krull domain. Indeed, condition (i) follows from
Theorem 3.3.11, condition (ii) results from Theorem 1.4.10, since Spec 1 (R)
is the set of all non-zero maximal ideals of a Dedekind domain R and finally,
Theorem 3.3.5 yields condition (Hi).
The definition given above is stated in terms of properties of rings of frac-
tions R p for prime ideals P of height 1. We shall show that conditions (ii)
and (Hi) can be replaced with equivalent conditions imposed on principal

IV]
Krull Domaias
161
ideals of R. In the sequel we shall prove (see Theorem 4.3.1) that Krull domain
may be defined as an intersection of a family of discrete valuation rings satis-
fying an additional condition.
Theorem 4.2.4
Conditions (ii) and (fii) in Definition 4.2.1 are equivalent to the following
condition:
(H') for every non-zero non-invertible element ,. E R there exist primary
ideals Q1, ..., Qrn such that Rr = Ql n ... nQm and ht(Ql) = ... = ht(Qm)
= 1.
Proof
Suppose that the domain R satisfies conditions (ii) and (iii) from Definition
4.2.1. Let r E R be non-zero, and noninvertible and let PI' ..., P 1ft be all the
prime ideals of height 1 which contain the element r.
m
Setting I = n (rR p nR), we have I ::> Rr. If a prime ideal P of height 1
. 1 '
,=
differs from the ideals P 1 , ..., Pm' then r, P, and thus rR p = R p and I'RpnR
"= R, whence we obtain 1= n(I'RpnR) = (nrRp)nR = r(nR,)"R = ,".R,
where P ranges over Spec 1 (R). The ideals P 1 J ..., Pm are minimal prime ideals
of the ideal Rr whence for each of the rings R p " i = I, ..., m, the maximal ideal
.P,R p , is the radical of the ideal rR p, (cf. Corollary 1.1.5). Thus, the ideal rR p ,
is primary, whence each of the ideals Qt = rRp,nR is also primary and,
clearly, rad(Q,) = Pl. Therefore condition (H') is satisfied.
Suppose that R satisifes (H'). Let 0 1= r E R, and the decomposition Rr
 Qln ... nQ", be that of (H'). If P E SpecteR) and P ::> Rr, then P ::> Qi
for some i = 1, ..., m. Since ht(P) = ht(Q,), it follows that P = rad(Qi)
and condition (iii) is satisfied.
We shall show that condition (ii) is satisfied as well. If T = n RpJ
P e Spec 1 (R)
then T::> R. Assume that a/b E T for some a, b E R, b :1= O. There exists
a decomposition Rb = Q1n ... nQm which satisfies condition (H'). For every
ideal P E SpecteR) we have a/b E R p _ and therefore a E bR p . If P is the radical
of Qi for some i, then bRpnR = RpQinR = Q,. If P differs from the radicals
of the ideals Q1, ..., Qm, then b rt P and consequently bRpnR = Rpr\R = R.
It follows that a E n (bRpnR) = n Qi = Rb, and thus a/b E R, whence
P i
T = R, which is the desired result. 0
Clearly, it follows from conditions (i) and (ii) of the definition that a Krull
domain is normal; we shall now prove a stronger result.
Theorem 4.2.5
A Krull domain is completely normal.

162
Divisors and Krull Domains
[Ch.
Proof
Suppose that R is a Krull domain, and that an element x E Ro and a finitely
generated R-module M satisfy the condition R[x] c: M c Ro. For every ideal
P E Spec 1 (R), we then have Rp[x] c M p c Ro = (Rp)o; thus the element x
is integral over R p , whence x e R p for all ideals P E Spec 1 (R), and consequently
x E R. 0
As we know, all Dedekind domains are Krull domains. The following
theorem clarifies the relation between those classes of domains.
Theorem 4.2.6
A Krull domain R is a Dedekind domain if and only if Spec(R) = Spect(R)u
u {O}, i.e., every non-zero prime ideal is maximal.
Proof
The implication => is obvious. Thus let us suppose that the Krull domain R
satisfies the condition Spec(R) = Spec 1 (R)u {OJ. By the previous theorem,
it remains to prove that R is Noetherian. Let I be a non-zero proper ideal,
o :f:- rEI. There exist primary ideals Q 1, ..., Qm of height 1 such that Rr
= Qltl ... nQm, whence there exiss a monomorphisln R/Rr -+ R/Q1 Ea ...
E9 R/Qm. It is sufficient to prove that the R-moules R/Q 1, ..., R/QlII are
Noetherian because then the module II Rr as a submodule of a Noetherian
module, will be finitely generated (see Corollary 2.1.7), and consequently
the ideal I will also be finitely generated.
The only maximal ideal of RIQl is Pt/Qt, where Pt = rad(Ql). Hence
R/Ql = (RIQ1)P1/QI  Rpt/QtRpl. The ring R p1 is Noetherian, since ht(Pl)
= 1, and consequently R/Ql, being the homomorphic image of R pt , is also
Noetherian. Tberefore the R-module R/Ql and, similarly, the modules R/Q2, ...
..., R/Q". are Noetherian. 0
The next theorem is the first result which links together the properties
of Noetherian domains and those of Krull domains. Let us recall that, though
the normalization of a Noetherian domain is not necessarily Noetherian (see
Example 4.7.3), however it is a Krull domain (Theorem 4.7.2).
Theorem 4.2.7
A normal Noetherian domain is a Krull domain.
Proof
Let R be a normal Noetherian domain. If P E SpecteR) then, by Theorem
3.1.11, the ring R p is normal, and therefore it is a Dedekind domain. By
Theorem 3.3.11, R p is a principal ideal domain, and thus condition (i'), which
is equivalent to condition (i), is satisfied.
We shall prove that condition (ii') of Theorem 4.2.4 is satisfied. Let 0 :1= r
E R be non-zero and noninvertible, and let P be the associated prime ideal

IV]
Krull Domains
163
of the ideal Rr. We shall prove that ht(P) = 1. By Theorem 2.3.15 there exists
an element s e R such that P = (r): (s). It is easy to see that the condition
P :P R implies sIr, R, whence the element sIr is not integral over R. It follows
by Theorem 3.1.6 that (slr)P cf:: P. Since (slr)P c: R there exists Po E P such
that the element '0 = (slr)po belongs to R,,"P. The equality rolpo = sir
implies (Po): (ro) = (r): (s) = P. Since ro rt P, this yields PR p = PoRp, whence,
by the implication (iv) => (ii) of Theorem 3.6.16, we deduce that ht(P) = 1. 0
Theorem 4.2.8
Let R be a Krull domain, and let S be a multiplicative subset of R. The ring
of fractions Rs is then a Krull domain and Rs = nRp, where P ranges ,over
ideals P e Spec 1 (R) such that PnS = 0.
Proof
Let us begin by proving that Rs = nRp.
If Pr.8 = 0 then S c: R,,"P. Hence Rs c R p . If x E nRp then, by condi-
tion (iii"), there exist only finitely many prime ideals PI' ..., Pm E SpecteR)
such that x rt Rp;,j = 1, ..., m. Thus, we have PJnS =F 0, whence there exist
elements sJ E PJnS and, since the rings R pJ are discrete valuation rings, we get
sJJX E RpJ' j = 1, ..., m, for sufficiently large exponents kJ. Finally, setting
s = Sl ... s, we obtain sx E R p for all prime ideals P E Spect(R). Hence
sx E R, therefore x = (sx)ls E Rs, and the equality Rs = nR p is proved.
From Corollary 1.4.8 it follows that prime ideals of height I of Rs are
of the form PRs, where P E SpecteR) and PnS = 0. The ring of fractions
(RS)PR s is the same as R p , as one can easily deduce from the condition PnS
= 0. From the above results we readily conclude that the domain Rs satisfies
conditions (i), (ii), (Hi") in the definition of Krull domain, and thus it is a Krull
domain. 0
The results obtained in Section 4.3 enable us to infer that if R is a Krull
domain then so is every intersection nR p , where P ranges over an arbitrary
subset of SpecteR).
Theorem 4.2.9
If R is a Krull domain, then a polynomial ring R[ {Xy}yer] is a Krull domain
for every set of indeterminates {Xy}yer.
Proof
For brevity we shall write Ro[T], Rp[r], etc., to denote rings of polynomials
in the indeterminates {X"},,er. Moreover, let us write T = R[F].
The proof will be divided into two parts.
1. First, we shall describe prime ideals Q of T of height 1. We consider
separately two cases: QnR = 0 and QnR :1= o.
Suppose that QnR = O. Then, the ring of fractions TR"o = Ro [11 is a
unique factorization domain, and the prime ideal QTR"o, as a prime ideal

164
Divisors and Krull Domains
[Ch.
of height 1, is generated by an irreducible element q E Ro[lj. By Theorem
1.4.7 (v), we obtain Q = QTR,,",o r.T= Ro [ljqn T. Every prime ideal of Ro[rJ
which is of the form Ro[r]q determines in the same way a prime ideal of height
1 of T.
Suppose now that QnR = P =F O. Obviously, the height of Pis 1, and the
prime ideal PT is contained in Q. Since ht(Q) = 1, it follows that Q = PT.
We shall prove that, conversely, for every ideal P E Spec 1 (R), the prime ideal
PT of T is of height 1. Since PTn(R"P) = 0, it is sufficient to prove that
the ideal (pT)Rp[r] = PRp[r] is of height 1; thus, we have reduced the
assertion which we are proving to the following one: if V is a discrete valuation
ring with the maximal ideal m :1= 0, then the ideal m(V[r]) is of height 1. Let
o :F Ql c mV[lj be a prime ideal. IfQ1nV = 0 were true, then, on the basis
ot [he first case considered above, there would exist a polynomial qt E V[11,
irreducible in Vo[F] such that Ql = VO[F]qlnV[lj. The polynomial qt would
be of the form ql == P,"q2, where k  0, p, is a generator of the ideal m, and
q2 e V[l'], q2 ,mV[r]. From the condition QlnV = 0 it follows that p" Ql'
thus q2 e Ql' since Qt is a prime ideal and ql E Qt. However, condition
q" e Qt together with q2,mV[r] contradicts the inclusion Qt c mV[lj,
and we conclude that Qlr.V 1= O. Therefore Qlr.V == m, which implies
Ql :;:) mV[r], whence Q1 = mV[r], and thus it follows that the ideal mV[lj
is of height 1.
This part of the proof leads to tbe following corollary:
Coronary 4.2.10
If R is a Krull domain, then prime ideals Q of the ring of polynomials R[F]
= R[ {X,,}YEr] which are of height 1 are of the form:
1 ° Q = Ro [lj qr.R[lj, where Ro [lj q = QRo [r] is a prime ideal of height
I, in the case where Qr.R == 0;
2° Q = PR[r] where P = QnR E Spec 1 (R), in the case where QnR =1= O.
2. We shall prove that the ring T = R[r] satisfies conditions (i), (ii), (iii')
in Definition 4.2.1.
Let us consider ideals Q E Spec 1 (T) which are of type 1°. Writing TO
= Ro[r], we have Q = TOqr.T. We shall prove the identity T Q = T¥oq. Since
T'Q c 'J'O"TOq, we have T Q C Toq; ift, U E TO and u, TOq, then tlu E TOf
and there exists a non-zero element r E R such that rt, ru E T. The element rll
does not belong to the ideal Q, since otherwise, because of u rt TOq, we would
have r E TOqr.R = 0, contrary to the assumption. Hence, tlu = (rt)/(ru) ETa,
and we have shown that To = Toq. The ring TO = Ro[r], being a unique
factorization domain, is a Krull domain, and the ideals of the form r'q with
irreducible q are the only ideals in Spec 1 (TO). Hence it follows that TO = n T:.o q ,
q
which yields
n Ta = Ro[l'],
Q
(1)

IV]
Kroll DOlpais as Intersections of Discrete Valuation Rings
165
where Q ranges over all ideals of type 1°. Let us note that every non-zero
element of Tis invertible in almost all rings To, and that these rings are discrete
valuation rings.
Let us consider the ideals Q E Spec 1 (T) of type 2°. We have Q = PT,
where P E Spec 1 (R). We shall prove that
TpTtiRo[F] = Rp[F], (2)
i.e., that T Q r1To = Rp[F]. Since R"-.P c T'Q, we have R p c: To, and
consequently Rp[r] c TanTo. We shall prove an opposite inclusion. Observe
that (Rp)o = .Ro. Since PR p is a principal ideal, every element of the ring
TO = Ro[T] can be represented in the form tIs, where t E Rp[r], s E R p ,
'Op(s) = 0, or 'Op(s) > 0 and t  PRp[F] ('Op denotes the discrete valuation
of the field R o , its valuation ring being R p ). If such an element tIs belongs
to To, then tIs = u/'O, where u, '0 e T, '0 ,Q. Considering the relation tf) = SU,
and applying the foregoing conditions, we obtain 'Op(s) = O. Hence tIs E R p ,
and equation (2) has been proved.
From (1) and (2) we deduce
n To = R o [l1ti n T pT = n Rp[r] = R[l1,
QeSpecl(T) PeSpecl(R) PeSpec 1 (R)
and therefore condition (H) is satisfied.
A non-zero element t E T belongs to an ideal of the form PT where P
E Spec 1 (R) if and only if all its coefficients belong to P. There is only a finite
number of such ideals and thus t is an invertible element in almost all rings
T pT . The above remark implies that condition (Hi') is satisfied.
It remains to prove that if P E Spec! (R) then T pT is a discrete valuation
ring. To this end, observe that T pT = (R p [l1)PR p [rJ; it is therefore sufficient
to prove that if V is a discrete valuation ring with the maximal ideal m, then
the ring (V[r])mJ'[r] is also a discrete valuation ring. This folIows immediately
from Theorem 3.6.20 which holds also in the case of an arbitrary number
of indeterminates. 0
Exercises
1. Prove that a Noetherian domain is normal if and only if it satisfies the following
conditions:
(i) for every P e Spec 1 (R), the ring R p is a discrete valuation ring,
(ii) every associated prime ideal of any non-zero proper principal ideal is of height 1.
4.3 KRULL DOMAINS AS INTERSECTIONS OF DISCRETE
VALUATION RINGS
The definition of a Krull domain includes the requirement that the ring R
should be an intersection of discrete valuation rings R p for P E Spec l (R).
In order to verify whether a domain is a Krull domain. we need a description

166
Divisors and Krull Domains
[Ch.
of all its prime ideals of height 1. We obtained such a description when proving
that the class of Krull domains is closed under the formation of rings of frac-
tions and of polynomial rings. An attempt to prove that an intersection of two
Krull domains contained in a given field is also a Krull domain shows that
finding the description of all prime ideals of height 1 in an intersection of rings
is very troublesome. The difficulty can be overcome by adopting a new charac-
terization (often regarded as a definition) of a Krull domain as an intersection
of discrete valuation rings, which will be given in Theorem 4.3.1. The represen-
tation of a Krull domain in the form of an irreducible intersection of discrete
valuation rings permits us also to describe the prime ideals of height 1 of that
ring (see Corollary 4.3.9). In the sequel we shall use this characterization in
order to prove that the normalization of a Krull domain in a finite extension
of its field of fractions is a Krull domain and that a ring of power series with
coefficients in a Krull domain is also a Krull domain.
Theorem 4.3.1
Let K be a field, and let {VA1eA be a family of discrete valuation rings of K.
If the ring R = n VA satisfies the condition that every non-zero element
AeA
rE R is invertible in almost all rings VA' then R is a Krull domain.
To facilitate the understanding of the circumstances occurring in the proof
we shall begin by discussing the following example:
Example 4.3.2
Let K be an algebraically closed field. In the field of fractions Ro of the poly-
nomial ring R = K[X, Y] we define the valuation 'lJ as follows: iff, g E Rare
D;on-zero polynomials and I = /,,. + fm+ 1 + ... + Is, g = gll + g,.+ 1 + ... + gq,
where fi, g, are homogeneous polynomials of degree i, and 1m 1= 0, gn :/: 0,
we set v(f/g) = m-n. For any elements ex, (1 E K, we define the valuation
va.,{J of Ro by the formula va.,p(f/g) = v (flJa.,p(flg)), where CfJa.,{J is the K-auto-
morphism of Ro which satisfies the conditions lfJrt.,p(X) = X - «, qJa.p(Y)
= Y - (J. Hence 'lJ = Vo,o, and for fER we have va.,p(f) > 0 <:> I(rx, (J) = O.
We shall prove that R = n Ru fJ . The inclusion c is obvious. Let fIg
R a.,
a.",
E Ro "R, and suppose that the polynomials f, g are relatively prime. Then,
for n = 1, 2, ..., we have g,ffn, i.e., frt rad(Rg), whence rad(Rf) cI: rad(Rg);
thus, by Corollary 1.2.5, there exist ex, (J E K such that g(<<, (1) = O,f(<<, (J) -:p o.
It follows lIenee that 'Oa.,fJ(f/g) < 0, and therefore fig, Rva.,fJ' which yields
the desired equality R = n Rv R.
P a.,,.,
OCt
Each of the rings Ru fJ is a discrete valuation ring and their intersection,
oc,
as a unique factorization domain is a Krull domain. We shall show that none
of the rings Rva.. fJ is a ring of fractions of R. Since CfJa., p(Rva.,p) = Ru, qJa..{J(R)
= R, it is sufficient to prove that RI1 is not a ring of fractions of R. The ideal

IV]
Krull Domais as Intersections of Discrete ValuatioD Rings
167
RX + RY is a maximal ideal of R, and we have RX + RY c Rnmv, where mu
is the maximal ideal of Rv, and thus RX+RY = Rnmv. This implies ReX,f)
* Rv in view of X/Y E Rv"Rex,y). If for some multiplicative set S c R, we
had Rv = Rs, then, by the above inclusion the set S would have to contain
an element U E RX +RY = (X, Y); thus u- 1 E Rv in spite of v(u) > O. Conse-
quently, the rings Rv fJ are not rings of fractions of R.
ex,
To sum up, the representation of the Krull domain R = K[X, Y] in the
form of the intersection of the discrete valuation rings R = n Rv P has the
P oc,
«,
following properties:
(a) the prime ideals Rnm(R"oc.fJ) are not of height 1, and the localization
of R with respect to these ideals differs from Rvex,fJ'
(b) a non-zero irreducible element fER is not invertible in infinitely many
rings Rv fJ (indeed, for infinitely many pairs lX, (J E K, we have 1(0=, (J) = 0,
ex,
and therefore vex,fJ(f) > 0).
Before proceeding to the proof of Theorem 4.3.1, we can without loss
of generality assume that K = Ro. In fact, it follows directly from the definition
of a discrete valuation ring that the rings VAr\R o are discrete valuation rings
of the field Ro and satisfy tbe hypotheses of the theorem. In. proving the
theorem, we shall demonstrate that if P E Spec 1 (R) then the ring R p is one of the
rings VA provided K = Ro.
We shall preface the proof of Theorem 4.3.1 with a few lemmas.
Lemma 4.3.3
Under the hypothesis of Theorem 4.3.1 the domain R is completely normal,
and every increasing sequence of ideals of R which are divisors becomes stable.
Proof
The rings VA are Noetherian and normal, and thus they are completely normal.
If 0 =1= x E Ro and the R-module R[x] is contained in a finitely generated
R-module, then the VA-module VA [x] is finitely generated, and therefore the
element x, as an element which is integral in (VA)o = Ro, belongs to the ring
VA. Thus x belongs to n VA = R, whence R is completely normal.
1
Let I c Ro be a non-zero fractional ideal of R. Then we have R: 1
= (nVA):1 = n(VA:I) = n(V A : VAI). Applying this formula to the ideal R:/,
we obtain
divI = R:(R:I) = n(fA:V,\(R:I)),
(3)
let 0 :F 11 c: 1 2 c... be an increasing sequence of divisors contained in R.
Then R:/ 1 ::> R:/ 2  ... and R:/" ::> R, k = 1, 2, ... For every A e A we
obtain an increasing sequence of ideals
VA: V A (R:l 1 ) c: VA: V A (R:/ 2 ) c ...

168
Divisors and Krull Domains
[Ch.
of VA because VA(R:J k ) ::> VA' k = 1, 2, ... The ring VA is Noetherian, so
each of these increasing sequences becomes stable. The condition imposed
in Theorem 4.3.1 implies that V}.1 1 = VA for almost all A EA. Thus from the
inclusion VA c V,,(R:/ t ) C V;.: V"/ l we infer that V;.: V;.(R:1 1 ) = VA for almost
all A. Consequently, by (3), the sequence div(1 1 ) c: div(1 2 ) c: ..., being identical
with the sequence 1 1 c: 1 2 C ..., becomes stable. D
Lemma 4.3.4
A domain R is completely normal if and only if, for every non-zero fractional
ideal 1 c Ro, we have 1:1 = R.
Proof
Suppose R is completely normal and I :f:. 0 is a fractional ideal. Let 0 1= ,. E 1
and 0 i= de R satisfy the condition dl c R. If x E 1:1, then xl c I, x 2 1
C I, ..., whence rR[x] c I and R[x] c r- 1 1 c (rd)-lR c: Ro. Hence x e R,
and consequently 1:1 = R.
Suppose 1:1 = R for every non-zero fractional ideal I c Ro. Let x E R
be such that the R-module 10 = R[x] is contained in a finitely generated R-
submodule of Ro. The module 10 is thus a fractional ideal and x e 10: 10 = R.
Hence the ring R is completely normal. 0
Lemma 4.3.5
If a domain R is completely normal and a divisor I c R is maximal in the family
of all divisors properly contained in R, then I is a prime ideal of height 1.
Proof
We shall show that I is prime. Let the elements r, s e R satisfy the conditions
rs E I and r rt I. By Lemma 4.3.4, we have I c 1+10- c: 1:(1+&) c 1:1 = R.
By Theorem 4.1.3 (ii), the ideal 1:(1+&) is a divisor, and thus the maxiInality
of the divisor [implies 1:(I+Rs) = R, whence we obtain s e I. Thus the ideal 1
is prime. .
Suppose that there exists a prime ideal PeR such that 0  P * J. For
any element r e P we have (R: I) r c R, whence it follows that (R: 1) Pr c: P;
since r rp P, we conclude that (R:/)P c P. By Lemma 4.3.4, the last inclusion
yields R:I c P:P = R, whence R:I = R, contrary to the assumption I
== R:(R:I). 0
Lemma 4.3.6
If a domain R is completely normal and every increasing sequence of ideals of R
which are divisors becomes stable, then every prime ideal of height 1 is maximal
in the family of all divisors properly contained in R.
Proof
If P e Spec 1 (R) and 0 :/: pEP, then Rr.Rp c P. Hence the family SIt of all
divisors of the form RnRx-1, 0  X E Ro, which are contained in P is not

IV]
Krull Domais as Intersections of Discrete Valuation Rings
169
empty. Let I = R()Rxol be a maximal divisor in the family d. We shall
prove below that I is prime. Since 0 :F I c: P and ht(P) = 1, it follows that
I = P, and thus that P is a divisor. Hence, by Lemma 4.3.5, we conclude
that P is a maximal divisor in the family of divisors properly contained in R.
It remains to prove that the ideal I = RnRxc;l is prime. Let r, S E Rand
rs E I.
Suppose RnR(xor)-l c P. Since RXc;l c R(xor)-I, we have I = Rn.Rxc;l
c RnR(xor)-1 c P, the maximaIity of I i.l the family d implies that I
= RnR(xor)-I, and the condition rs E InRxol yields S E RnR(xor)-l = I.
Suppose now that RnR(xor)-l q: P. Then there exists an element tERn
nR(xor)-1'P. Thus rt E RnRxol = I, and consequently t(RnR(x o t)-l)
= RtnRxo1 c RnRxol = Ie P. We have t rt P, hence Rr\R(x o t)-l c P.
Since 1 = RnRxOI c RnR(xo t)-1 c: P, the maximality of I in the family .91
implies that I = RnR(xo t)-I. The condition t E R(xo r)-1 yields r E R(xo t)-I,
whence rEI.
We have shown that the condition rs E I implies either rEI or S E I, i.e.,
that the ideal I is prime. 0
Corollary 4.3.7
If a domain R is completely normal and every increasing sequence of ideals
of R, which are divisors, becomes stable, then the set of all maximal divisors
contained properly in R is identical \vith the set Spec 1 (R).
Lemma 4.3.8
If a domain R is completely normal and every increasing sequence of ideals
of R which are divisors becomes stable, then for every ideal P e Spec. (R)
the ring of fractions R p is a discrete valuation ring.
Pro.of
Suppose that R p is not a valuation ring. As we have already seen (Theorem
1.4.11), R p = {yeRo: RnRy-I cI: P}u{O}. Hence there exists a non-zero
element x E Ro such that x, X-I rt R p , i.e., RnRx- 1 c: P and RnRx c P.
By the formula R:(Rx-l n R) = Rx:(.RnRx), and by Lemma 4.3.4 the above
inclusions yield
R:P c [R:(RnRx- 1 )]n[R:(RnRx)]
= [Rx : (RnRx)]n [R : (RnRx)]
= (RxnR):(RnRx) = R,
whence we have div(P) = R:(R:P) ::> R:R = R. Hence div(P)  P in spite
of Corollary 4.3.7. Thus we have proved that R p is a valuation ring.
In order to prove that R p is a discrete valuation ring, \ve shall :first show
that the maximal ideal PR p is principal. Choose an element y-l E R:.P'R.
Then y-l fP:P == R, and therefore y-1P cI: P and y-IP c R, whence it
follows that y-1PR p = R p , i.e., PR p = yR p , and of course y E R p ; thus the
ideal P R p is principal.

170
Divisors and Krull Domains
[Ch.
Let v:R -+ G be a valuation such that Rf) = R p . Set I = n (PR p )"
n
= nynR p and let a, b e Rp"I. If a eY'R p "Y'+lR p , then a = yka', where
n
a' rp yR p . Hence a' is invertible in R p and v(a') = 0; so v(a) = kv(y), k  0
and similarly v(b) = Iv(y), I  O. Hence we obtain v(ab) = (k+l) v(y) and
thus we see that ab rp I, and I is prime. Since ht(PR p ) = 1, we have either
1=0 or 1= PR p . The latter identity, however, yields yR p = y 2 R p , which
implies that y is invertible in R p , contrary to the condition ht(PR p ) = 1. Thus
we have I = 0, whence Im(v) is the group generated by v(y), and R p is a discrete
valuation ring. 0
Proof of Theorem 4.3.1
We have shown that, without loss of generality, it may be assumed additionally
that K = Ro. Disregarding the case where R is a field-the theorem is then
obvious-we may also suppose that all the rings VA .are different from the
field K = Ro.
It follows from Lemma 4.3.3 and Lemma 4.3.8 that for every P e Spec 1 (R)
the ring R p is a discrete valuation ring; thus condition (i) in the definition
of a Krull domain (Definition 4.2.1) is satisfied.
We shall show that R = n R p . Suppose 0 #= x e R p for all P
Pe Spec l (R)
e Spec 1 (R). Since R p = {y e Ro: RnRy-l q: P}u {OJ, we have Rr.Rx-l cf: P.
Hence, by Corollary 4.3.7 and Lemma 4.3.3, the integral divisor Rr.Rx-l
is not contained in any maximal divisor properly contained in R. This divisor,
therefore, is not proper, i.e., RnRx- 1 = R, whence it follows that 1 e Rx-1
and x e R. Thus condition (ii) in the definition of a Krull domain is satisfied.
It remains to prove condition (Hi). For any ideal P e Spec 1 (R), write
A p = {A E A: m(VA)nR c P}. We shall prove that the sets A p are not
empty. Suppose A po = 0 for some Po E Spec 1 (R). From the requirement
that every non-zero element of R should be invertible in almost all rings VA
it is easy to deduce that if 0  x e Ro then there exist elements AI, ..., Am
such that x f VAl' ..., X rp V Am and x E VA for A 1= AI, ..., Am. Since AI, ...
..., Am e A po = 0, there exist elements Sj e m(VAJ)nR such that sJ rp Po, j
= 1, ..., m. For sufficiently large exponents k J , we have sJJ x e V AJ , j = 1, ...
... , m. Setting s = Sl ... sn, we have S e R"P 0 and sx e VA for every I" eA.
Hence sx e R and x = (sx)/s e Rpo. It follows that R po = Ro, contrary to the
assumption Po :p O. Thus the sets A p are not empty.
If A e A p , then R"P c VA"m(V A ). It follows that the elements of the set
R"P are invertible in VA' whence R p c VA $ Ro. The ring R p , as a discrete
valuation ring is a maximal subring of the field Ro (see Theorem 3.6.17),
and therefore R p = VA for A e A p . Every non-zero element r e R is invertible
in almost all rings VA' and m(VA)nR = P for A e A p ; thus r is invertible in
almost all rings R p , whence condition (Hi) in the definition of a Krull domain
is satisfied. 0

IV]
Kru1l Domais as Intersections of Discrete Valuation Rings
171
The following corollary will be used later to describe explicitly prime ideals
of height 1 in certain Krull domains.
Corollary 4.3.9
If {VA},\eA is a family of discrete valuation rings of the field Ro, where R = n VA
AeA
and the following conditions are satisfied:
(i) every non-zero element r E R is invertible in almost all rings VA,
(ii) the representation of R in the form of the intersection n VA is irredun-
AeA
dant (i.e., if A' * A, then n VA 1= n VA), then the function cp defined by the
AeA' AeA
formula cp(A) = m(VJi1R establishes a bijective correspondence A  Spec 1 (R).
Proof
Indeed, given an ideal P E Spec 1 (R), choose an element 1p(P) E A p . The con-
siderations in the latter part of the proof of Theorem 4.3.1 show that R
= n VA and Vtp(P) = R p , thus condition (ii) implies that "p is injective and,
AeIm(1p)
moreover, Im(lJ') = A. Furthermore, we have m(Vy(p»nR = m(Rp)nR
= PRp()R = P, whence it follows that the function "p is the inverse of cp. 0
Corollary 4.3.10
If R is a Krull domain and A c Spec l (R), then the ring R(A) = n R p is a
PeA
Krull domain, the representation of R(A) in the form n R p is irredundant
PeA.
and the inclusionj: R c R(A) induces an injectionj1: Spec 1 (R(A») -+ Spec 1 (R)
which has the set A as its image.
Proof
Indeed, R(A) is a Krull domain by Theorem 4.3.1 since it satisfies condition
(iii'). If the representation R(A) = n R p were redundant, then for some
PeA
Po e A, we would have Rpo::> n R p , whence R =:= n R p . From the
Po:l:PeA P:l:Po
latter part of the proof of the theorem we deduce that R p = R po for some
P  Po, which is not true. Thus the intersection n R p is irredundant. The
PeA
mappingj: Spec (R(A) ) -+- Spec(R) induced by the inclusionj is defined by the
N
formulaj(P') = P'nR. If P' e Spec 1 (R(A», then by the preceding corollary,
there exists a unique ideal PEA such that P' = m(Rp)nR(A). We then have
".,
j(P') = P'i1R = m(Rp)nR(A)nR = m(Rp)f'\R = PRp()R = P, whence f<P')
EA. 0
eorem 4.3.11
If {R)'}AeA is a family of Krull domains which are contained in a fixed field
and every non-zero element r e n R). is invertible in almost all rings RA' then
AeA
n R A is a Krull domain.
AeA

172
Divisors and Krull Domains
[Ch.
Proof
Since, for every A E A, RJ. = n (R,)p, it follows that R = n R).
P E Specl(R).) A e A
= n n (R,)p. The rings (R,)p are discrete valuation rings of the field
J. e .J PeSpecl(R,\)
(R})o. Hence (R})pnR o are discrete valuation rings of the field Ro and, clearly,
R = n [(R.t)pnR o ]. If a non-zero element r e R is invertible in the rings R).
At P
for A =F A 1 , ...,Am' then it is invertible in the rings (R).)p. For every AJ, j
= 1, ..., In, the element r is invertible in almost all rings (RAJ)P nRo for P
E Spec 1 (R;.}). Thus the intersection n (Ri)p satisfies the conditions in Theorem
).,P
4.3.1, and consequently R is a Krull domain. 0
Corollary 4.3.12
If a Krull domain R is contained in a field K, and if L is a subfield of K, then
RnL is a Krull domain.
1nbeorem 4.3.13
If R is. a Krull domain, and a field K is a finite extension of the :field Ro, then
the integral closure of R in K is also a Krull domain.
Proof
For every ideal P E Spec 1 (R), the ring R p is a discrete valuation ring. Hence,
by Theorem 3.3.12, it follo,vs that the integral closure R p of R p in K is a Dede-
kind domain and therefore a Krull domain. Since R = n R p , we see that the.
p
integral closure R of R in K satisfies the condition if c: n R p. Now, we shall
prove the opposite inclusion.
Let x E n R p and IE Ro[X] be the minimal polynomial of x. Applying
p
Lemma 3.2.3 to the rings R p and R p , we infer that the coefficients of the
polynomial f belong to R p , and consequently to R, which means that x e R,
and the desired inclusion is proved.
Thus we have R = n R p . Since all the rings R p are Krull domains, it
remains to prove, in virtue of Theorem 4.3.1, that every non-zero element
t e R is invertible in almost all rings R p . Let t m +'m_l t m - 1 + ... +ro = 0,
where '0, ..., , m-l e R, be the minimal equation for t over R. If ro 1 e R p for
some ideal P E Spec 1 (R), then the equation t- 1 = _'.01(t m - 1 + ... +rl)
yields t- 1 e R p . Since the condition rol E R p is satisfied for almost all ideals
P e SpecteR), we also have t- 1 e R p for almost all P. 0
eoremm 4.3.14
If R is a Krull domain, then the ring of formal power series R[[X 1 , ..., X,,]]
is also a Krull domain.

IV]
Krull Domais as IntersectioDS of Discrete Valuation Rings
173
Proof
It is sufficient to show that the ring of formal power series in one indeterminate
T = R[[X]] is a Krull domain. For any prime ideal P e Spec 1 (R), denote
by Vp the discrete valuation of the field Ro whose ring is R p . We define its
extension to a valuation vp of the field To setting vp(ro+rIX+ ...) = vp(r m ),
where m is the least index such that r m #= 0, and extending it uniquely to To.
We assign to the zero ideal of R the valuation vo, setting vO(rO+rlX+ ...)
= m, with m as above, and extending it to To. Clearly, we have T c nRijp'
p
where P ranges over prime ideals of height  1. We shall prove the opposite
inclusion.
Denote T' = n Rvp. Observe that T' contains the ring R[X].
ht{P)  1
Every non-zero element t' E To can be represented in the form t' = X"(a/b),
where a = ao+a 1 X+ ..., b = bo+btX+ ... E T, ao =F 0, b o =F 0 and k
= vo(t'). Suppose the element t' belongs to T'. Then k  0 and, setting t = a/b,
we have vp(t') = vp(t) for P E Spec 1 (R). It is sufficient to show that t e T.
Since for all P E SpecteR) we have 0  vp(t) = vp(a)-ifJp(b) = 'Vp(ao)-
-vp(b o ) = vp(ao/b o ), the element to = ao/b o belongs to n R p = R. Hence
p
the element
t-to = [(ao-tobo)+(at-tobl)X+(a2-tob2)X2+ ...]b- t
= X[a +aX+ ... ]b- 1 ,
where a = an - to b", n = 1, 2, ..., belongs to T'. It f01lows from the definition
of T' that the element [a +aX + ...]b- 1 belongs to T', and, as before, we can
find an element tiE R such that
t-fo-ttX = X2[a +a + ... ]b- I E T',
where a' = a-ttbll-l = a,.-tob,.-tlb n - t , n = 2,3,... Proceeding in this
way, we obtain a sequence of elements to, t 1 , ... e R which satisfy the conditions
Do = to b o , a1 = to b l + t 1 b o , ...,
an = tobn+tlbn-l + ... +t,.b o , ...
This means that a = (t o +t 1 X+ ...)b, i.e., t = fo+ttX+ ... Thus it follows
that t e R [X] , and we have proved that R[ [X]] = n R.up.
ht(P)  1
A non-zero element t of R[[X]] has the form t = Xm(am+ana+tX+ ...),
where am #= 0 and vp(t) = vp(a m ). The element am is invertible in almost all
rings R p = Rvp, whence the element t is invertible in almost all rings Rvp.
Thus, by Theorem 4.3.1, we have concluded the proof. 0
The representation of R[ [X]] ill the form of the intersection nR Up is irredun-
dant; since v p (I/X) = 0 for P =F 0, the component RVG cannot be omitted,
and it is sufficient to make use of the fact that the representation nR p = R
is irredundant because R-vp()R = R p for P =F 0, and R-t";onR = R.
Corollary 4.3.9 yields

174
Divisors and Krull Domains
[Ch.
Corollary 4.3.15
If R is a Krull domain, then the prime ideals.Q of height 1 of R[[X]] have
the form
1 0 Q = XR[[X]] in the case when Qr.R = 0,
00
2 0 Q = {L r,.xn; '0, '1' ... eP} in the case when P = QnR eSpec 1 (R).
11-0
The ideal Q described in 2 0 need not be identical with p. R [[X]] (see
Exercise 1).
Exercises
1. Let K be a field, let T = K[X 1 , X2, ... ,] be a polynomial ring, and let R = K[Xf,
X 1 X 2 , ...,X,X}, ...] be the subring of Tgenerated by Kand all the monomialsX,X}. Denote
P = TX 1 nR. Prove that
(i) R is a Krull domain,
(ii) P is a prime ideal of height 1,
(iii) P is not finitely generated,
(iv) the ideal PR[[X]] is not prime.
00
[Consider the element L X 1 X n X n .]
nal
4.4 DMSORS IN KRULL DOMAINS
In this section, we continue our presentation of the theory of divisors, which
was begun in Section 4.1, concentrating on Krull domains. The basic Theorem
4.4.7 states that the group of divisors Div(R) of a Krull domain R is a free
Z-module with the basis Spec 1 (R). Consequently we can identify divisors with
linear combinations  np · P of ideals p e Spec 1 (R) with integer coefficients
p
and prove that a Krull domain R is a unique factorization domain if and only
if the ideal class group CI(R) is zero. We shall also prove an approximation
theorem for valuations (Theorem 4.4.10) and show that every divisor of a Krull
domain is an intersection of two principal divisors.
Throughout this section R denotes a Krull domain.
Theorem 4.4.1
If R is a Krull domain and I is a fractional ideal of R, then
div(I) = n IR p .
PeSpec 1 (R)
First, we shall prove two lemmas.
Lemma 4.4.2
If I, J E Id(R), then IR p = JR p for almost all P e Spec 1 (R).

IV]
Divisors in Krull Domains
175
ho .
It is sufficient to prove that IR p = R p for almost all P e Spec. (R). There exists
a non-zero element r e R such that rI c R. Moreover, we choose a non-zero
element a E I. Each of the elements rand ra belongs to a finite number of ideals
from Spec 1 (R). If P e Spec 1 (R) and r f P, ra  P, then the elements rand ra
are invertible in R p , whence R p ::> rIR p = IR p  raR p = R p . Thus the lemma
is proved. 0
Lemma 4.4.3
If Ie Id(R), then (R:I)R p = Rp:IR p for almost all P e Spec 1 (R).
Proof
Since R: I c R p : IR p , we have the inclusion c:. Let 0  x e R p : IR p . Then
xIR p c R p , and by the preceding lemma, there exist finitely many ideals,
say PI' ..., Pm e Spec 1 (R), such that xIR p , cI: R p " i = 1, ..., m, and for each
ideal P:f:. PI' ..., Pili of Spec 1 (R) we have xIR p c R p . There exist elements
at e P,,,-P and positive integers k" i = 1, ..., m, such that ak'xIR p , c R p ,.
Setting a = a1 ... am, we obtain an element a eR" P which satisfies axIR p ,
c Rpi and hence axI c Rpl for all P' e Spec 1 (R). It follows that axI c:: n R p ,
p'
= R, and therefore ax e R:I and x = (ax)/a E (R:1)R p . Hence the inclusion
::> is proved. 0
Proof of Theorem 4.4.1
We may suppose that I :p o. Then, by Lemma 4.4.3, for every P e SpecteR),
we have
(divI)R p = (R:(R:I)}R p = Rp:(R:/)R p = Rp:(Rp:IR p ) = IR p
because every ideal of R p is a divisor. Thus dive/) c (div(1) }R p = IR p and
dive/) c:: nIRp. We shall prove the opposite inclusion. If y e Ro and Ry ::> I,
then RpY ::> IR p . Hence Ry = (nRp)y = n (RpY) ::> n IR p , which yields
p p p
div(I) ::> n IR p . This ends the proof. 0
p
Corollary 4.4.4
All prime ideals of height 1 in a Krull domain are divisors.
Proof
Indeed, if Po E Spec 1 (R), then, for every ideal P e Spec 1 (R) different from Po,
we have PoRp = R p , whence
div(Po) = PoRpon n R p
PPo
= PoRponRpor\ n R p = PoRponR = Po,
pp.
and therefore Po is a divisor. 0
In the following theorem we give a description of divisors.

176
Divisors and Krull Domains
[Ch.
Theorem 4.4.5
A non-zero fractional ideal I of a Krull domain R is a divisor if and only if
there exists a function p: Spec 1 (R) -+> Z which satisfies the following conditions:
(i) p,(P) = 0 for almost aU P E Spec! (R),
(ii) I = {x E Ro; vp(x)  p,(P) for all P E Spec 1 (R) }u {O}, where 'lJp: R:
 z is a discrete valuation whose valuation ring is R p .
The function p is determined uniquely by the ideal 1.
Proof
Let I be a divisor. By Theorem 4.4.1, 1= div(l) = n IR p . The ring R p
p
is a principal ideal domain; if PR p = T&pR p for some np E .R p , then the frac-
tional ideal IR p is of the form
IR p = n'P(P)R p = (3t p R p )ll(P) = (PRp)J.t(P),
where p,(P) is an integer. Thus we have IR p = {x e Ro: vp(x)  Jt(P)}u {O},
which implies condition (ii). Condition (i) follows from Lemn1a 4.4.2.
Suppose now that the function p,: Spec 1 (R)  Z satisfies condition (i).
Let I = {x E Ro; vp(x)  /J(P) for P e Spec 1 (R) }u {O} Clearly, the ideal .1
is a submodule and, as can easily be deduced from conition (i), a fractional
ideal. We have to prove that it is a divisor. From the definition of I it follows
that I = n (PRp)I&(P). Thus, by Tbeorem 4.4.1, it is sufficient to prove that
p
IR p = (PRp)"(P) for all P. This formula ensures the uniqueness of the function
1'.
Suppose first that p,  0, fix an ideal Po E Spec 1 (R), and let p,(P) = 0
for P ¥: Po, P1, ..., Pm. Choose elements '0'''1'...' rill E R satisfying the
conditions r, EP"P o for i = 1, ..., m and roRpo = PoRpo' and set
r = rS(Po) ... (Pm). We then have vpo(r) = It(Po), 'Vp,(r)  fl(P i ), i = 1, ..., m,
and therefore rEI. Moreover, {PORpo)IA(Po> = rR po C IR po c: (PORpOYl(Pf)
whence IR po = (Po Rpo'f<Po>.
. Suppose now that (P 1), ..., p,(P k ) < 0, /J(P k + 1), ... , p(Pna) > 0 and p,(P)
= 0 for P ¥: P 1, ... , Pm. Choose elements '1' ..., rk E R such that vp;{r,) = 1,
i = I, ..., k, and set r = r1/.t(Pl) ... 'kl'(PIc). We have r E Rand rI = {rx E Ro;
vp(x)  p,(P) for all P}U{O} = {y E Ro; vp(Y)  1J(P) +vp(r) }U{O}. All the
numbers p,(P) +vp (r) are nonnegative and almost all of them are equal to
zero. Hence, by the preceding part of the proof, we obtain
(PRp)up(r>(IR p ) = (rR p ) (IR p ) == rlR p = (PR p )P<P)+l1 p (r),
whence it follows that IR p = (PRp)"(P), which ends the proof. 0
Corollary 4.4.6
Let us associate with a function p,:Spec 1 (R)  Z almost everywhere equal
to zero, the divisor
1 = {x E Ro; fJp(x) ;;, p,(P) for all P E Spec! (R)}u {O}.

IV]
Divisors in Kroll Domains
177
In this way we obtain a bijective correspondence between the free Z-module
with the basis Spec 1 (R) and the set of all divisors of the domain R.
We shall prove below that this correspondence determines an isomorphism
between the free Z-module Z {Spec 1 (R)} and the group Div(R).
Theorem 4.4.7
The assignment associating with a divisor I of a Krull domain R an element
/AI which belongs to the free Z-moduIe Z {Spec 1 (R) } and satisfies the condition
I = {x E Ro: vp(x)  PI(P)}U {O} is an isomorphism of the groups Div(R)
and Z{Spec 1 (R)}, and sasfies the following conditions:
(i) Pdlv(IJ) = PI + /JJ,
(ii) J c J <=> /AJ  PI'
(Hi) PI:J = /JI-P,J,
(iv) Pdlv(I+J) = min{J.tI' !JJ), where I+J is the fractional ideal generated
by I and J,
(v) /AinJ = max(/JI, PJ).
Proof
We know that the correspondence 11-+ PI is one-to-one. Formula (i) states
that this correspondence is a homomorphism of groups.
(i) The formulae IR p = (PRp)fJr(P) and JR p = (PRp)fJJ(P) yield (IJ)R p
= (JRp)(IR p ) = (PRp»).(P), where ,t(P) = /JI(P)+PJ(P). By Theorem 4.4.1,
A = !Jdlv(IJ), and thus condition (i) is satisfied.
Condition (ii) follows directly from the definitions of PI and /JJ and from
the equivalence
(PRp)k c (PRp)m  k  m.
Condition (iii) results from the following equivalences:
x e 1:1 <=> xl c I <=> V (xJ c IR p ) <=> V (xJR p c IR p )
p p
 V (vp(x)  PI(P) - PJ(P) ).
p
Condition (iv) follows from the identity div(l + I) = n (1+ J) R p and the
p
formula
(I+J)R p = IRp+IR p = (PRp)/ll(P) + (PRp)/lJ(P) = (PRp»).(P),
where ,t(P) = min (PI(P), PJ(P».
To prove (v) observe that there exists a divisor I' such that PI'
= max{J.tl, PJ). Thus, by condition (ii), we obtain I::> I', I:::> I', whence
Jr.J ::> I'. Since pInl  PI and pinJ  PI' it follows that PIn}  #1' and
I' ::> Jr.J, and consequently, I' = In/; thus condition (v) is satisfied. 0
Observe that if 0 :p x e Ro then to the principal fractional ideal Rx corre-
sponds the function PRx, which is defined by the formula PRX(P) = vp(x).

178
Divisors and Kroll Domains
[Ch.
Theorem 4.4.8
A domain R is a unique factorization domain if and only if R is a Krull domain
and Cl(R) = O.
Proof
=>. It has been proved in Example 4.2.2 that R is a Krull domain and that
every divisor P e Spec 1 (R) is principal. The assertion follows from Theorem
4.4.7 and Corollary 4.1.7.
<=. By Corollary 4.1.6, the equality CI(R) = 0 implies that every divisor
is principal. Hence, for every ideal P e Spec 1 (R) there exists an element np e R
such that vp(np,) is equal to 1 if pi = P and equal to 0 if pi  P. The elements
np are irreducible since if np = ab, where a, b E R, then either vp,(a) = 0
for all P' or vp,(b) = 0 for all pi, whence either a or b is invertible in R. Given
a non-zero element ,. e R, we set r' = 11 np(r). Since vp(rlr') = vp(r) -fJp(r')
p
= 0, it follows that u = r Ir' is invertible in Rand ,. = u 11 np(r). Hence we
p
deduce that the only irreducible elements in R are, up to invertible ones, ele-
ments of the form np. If r = Il1ij.p, then vpo(r) = Lvpo(nP) = (XPo. Thus
p
the decomposition of an element into a product of irreducibles is unique. 0
C:orollary 4.4.9
A Krull domain is a unique factorization domain if and only if for every
P E Spec 1 (R) there exists an element np e R such that vp,(np) = 0 for pi  P,
fJp(np) = 1.
1rheorem4.4.10
Let R be a Krull domain, let ideals P 1, ... , Pm e Spec 1 (R) be distinct, and let
k 1 , ..., k m be integers. Then there exists a non-zero element x e Ro such that
VPt(x) = k 1 , ..., vPm(x) = k m ,
vp(x)  0 for P e Spec 1 (R)" {PI' ... , Pm}.
Proof
To begin with let us prove that, if the ideals P, P, ..., P; e Spec 1 (R) are
distinct then there exists a non-zero element n e R such that vp(n) = 1,
vpr(n) = ... = vp;(n) = O. There exists an element r e R such that rR p = PR p ;
renumbering the ideals P, ..., P; , if necessary, we may suppose that r e P, ...
..., r E Pi, r  Pi+l, ..., r rp P;, where 0  j  s. By Corollary 1.1.8, there
exist elements
J j
'0 e p2"U P;"h e P"U P, for h = j+ 1, ..., s.
i=1 1-1
Then
'O'J+l ...'s Ep2Pi+1 ... P"UP;
i-I

IV]
Divisors in Krull Domains
179
and the element n = r-rOrJ+l ... rs satisfies the conditions n e P, n, p2,
'1'&  p, ..., n , P;. Hence vp('1'&) = 1, fJp(n) = ... = vp(:n) = O.
From the preceding considerations it follows that there exist elements
nl, ..., '1'&m e R such that fJp,('1'&J) = 'J for i,j = 1, ..., m. Setting Y = n1 ...
... nm e Ro, we have VPl (y) = k 1 , ..., vP", (y) = k m . Let P', ..., P' be all
ideals P" from Spec 1 (R)' {P 1 , ..., Pm} such that vp,,(y) < O. Applying once
more the result of the first part of the proof, we obtain elements '1'&', ..., n' e R
such that fJp,,(n,') = 1 and vp('1'&') = 0 for P = P 1 , ..., P lH P',..., P;l'
P " P I'. I Th I t ( " ) nl ( " ) "q h
i+ 1, ..., q, I = ,... , q. e e emen x = Y '1'&1 ...:n q , were ni
=fJp,'(Y), i = 1, ..., q, satisfies the conditions of the theorem. 0
The above theorem permits a more precise description of divisors in a Krull
domain.
1rheoremo 4.4.11
If R is a Krull domain and I is a divisor of R, then there exist elements Xo, Xl'
Yo, Yl E Ro such that
I = RxonRxl = div(Ryo + RY1).
Proof
The divisor I determines a function Ill: Spec 1 (R) -. Z which satisfies the
conditions of Theorem 4.4.5. Let Ill(P) = 0 for P -:F P 1 , ..., Pm. By Theorem
4.4.10, there exists a non-zero element Yo e Ro such that vp(Yo) = Ill(P) for
P = PI' ..., Pm. Denote by P m + 1 , ..., Pq all ideals P E Spec 1 (R)' {P 1 , ...
..., Pm} such that vp(Yo) > O. There exists a non-zero element Yl e Ro such
that
Vp(Yl) = IlI(P) for P = PI' ..., P n "
Vp(Yl) = 0 for P = P m+ 1, ..., P tl'
Vp(Yl)  0 for P  P 1 , ..., Pq.
It is easy to see that III = min (v (Yo) , V(Yl») = min(,uRYo' IlRY1) where v(yo)
denotes the function P 1-+ vp(Yo). By Theorem 4.4.7 (iv), we have I = div(Ryo +
+ RY1).
It follows from Theorem 4.1.4 (ii) that the ideal R: I is a divisor; thus, by
the previous result, there exist non-zero elements Xo, Xl e Ro such that R:I
= div(Rxol+Rxl1) = R:(R:(Rxo1+Rxl1»). Using Theorem 4.1.4 (iv), we
obtain 1= R:(RxOl+Rxl1) = (R:Rxol)n(R:Rxl1) = RxonRxl. D
Exercises
1. Prove that if a Noetherian domain R is normal, then:
(i) for every ideal P e Spec 1 (R) and for every positive integer n, the ideal p<n) = pnRpnR
is P-primary,
(ii) for every divisor Ie R we have 1== n p<np), where np = Jtl(P); and this
PeSpec 1 (R)
is an irredundant primary decomposition.

180
Divisors and Krull Domains
(Ch.
2. Let R be a Krull domain. Prove that CI(R) is a torsion group if and only if for every
subset A c Spec! (R) the ring n R p is a ring ,of fractions of R.
PeA
3. Let R be a Krull domain. Prove that the following conditions are equivalent:
(i) R is a Dedekind domain and CI(R) is a torsion group
(ii) if P and Pu.' (X e A, are prime ideals of Rand pen Pu., then there exists(Xo E A
cxeA
such that P C P«o.
In Exercises 4-11 we give preliminary information about the theory of Cartier divisors;
a more comprehensive presentation of that theory will be found in textbooks on algebraic
geometry, for instance in [0].
Given an element h of an arbitrary ring R, we denote by RII the localization of R with
respect to the multiplicative set {I , h, ,,2, ...} (including, exceptionally, the case of a nilpotent
h) and by K" the total ring of fractions of R". We write also D(h) =- {P E Spec(R); h rp P}.
4. We consider finite families {D(h,) , I,}, i = 1 t ..., s, where h. e R,ji e (K",)", such that
(i) Rh 1 + ... +Rh, = R,
(ll) R"'''J!i = R",II J Ii, i, j = 1, ..., s.
More precisely, condition (ii) states that
R","i1IlUi) == R"'''Je.}(h) ,
where e'J: K", -. K"'''J is the homomorphism induced by the natural homomorphism &t
-. R",,,} (show the existence of ell).
Two families {D(h,),!i}, {D(h),f} which satisfy conditions (i) and (ii) are said to be
equivalent if we have R"'''"fi = R",,,, f, for every pair i, k. Verify that the relation is indeed
. Ie
an equivalence relation. Show that for a finite number m of families there exist equivalent
families of the form {D(h,) , 1,<Ie)}, k = 1, ,.., m.
5. The equivalence class of a family {D(h,) , Ii} which satisfies the conditions in Exercise
4 is called a Cartier divisor of R. Prove that Cartier divisors form an Abelian group with
respect to the operation of addition induced by +: {D(h,),/i}+ {D(h,),r;} = {D(h,),JiI:}
(this sum is not defined for all pairs of families). Prove that the above operation is well
defined. The group of Cartier divisors is denoted by Ca Div(R). Divisors determined by fami-
lies of the form {D(h,),fi}, where Ii = ffor i = 1, ..., s, are called principal. AIl principa
divisors form a subgroup Ca Prin(R) of the group Ca Div(R). The factor group Ca Div(R)1
ICa Prin(R) is denoted by Ca Cl(R) and called a Cartier divisor class groop.
6. Assume that a homomorphism of rings g: R -. T is flat, i.e., that the ring T is flat
as an R-moduIe, which means that every monomorphism of R-modules qJ: M' -+ M induces
a monomorphism qJ  1: M' (8) R T -. M  R T. Prove that if an element a E R is not a zero-
divisor, then the element g(3) E Tis not a zero-divisor either. Deduce that the homomorphism
g induces a homomorphism of the total ring of fractions of R into the total ring of fractions
of T. Show that for hER the induced homomorphism c,,: R" -. T,(II) is also flat, and there-
fore induces a homomorphism i" of total rings of fractions.
7. Using the results in the previous exercise prove that a flat homomorphism of rings
g: R -+ T induces a homomorphism Ca Div(g):Ca Div(R) -. Ca Div(T) which assigns
to the divisor determined by the family {D(h,), Ii } the divisor determined by the
family {D(g(h,»), K",(fi)}. Prove that this homomorphism induces the homomorphism
of the appropriate Cartier divisor class groups. Examine the homomorphism induced by
a composition of two flat homomorphisms.
8. Prove that if R is a Noetherian ring then the group Ca Prin(R) is isomorphic to the
group (Rs)*1 R*, where S denotes the set of all elements which are not zero divisors of R.
[Let (X: (R s )* -+ Ca Prin(R) denote a homomorphism which maps an element rIa e (Rs).
onto the divisor determined by the family {D(I), rls}; thus (X induces a monomorphism
jj: (R s )* I R* -. Ca Prin(R). To prove that iX is an epimorphism, observe that every principal
divisor D is determined by a family {D(h,), r,ls,}, r" a, e R, which satisfies the conditions

IV]
Induced Homomorphisms of Class Groups
181
"SJ == rJI,. Then show that there exists an element of the form L IJIJ belonging to S. Prove
j
that the family {D(h,), (tJrJ)/(i= tJsJ)} determines the divisor D.]
9. Let R be a Krull domain. Prove that if a family {D(h,) , fi)} determines a Cartier
divisor D of R then the formula rp(D) ==  t}p(fi(p»P, where 1  i(P)  s is an arbit-
p e Specl(R)
rary index such that P E D(h,(p» and fJp is the valuation corresponding to the ideal P, deter-
mines a homomorphism 9': Ca Div(R) -+ We Div(R) == Z {Spec1 (R)}. Prove that qJ is an
isomorphism.
10. Let R be a Noetherian domain. Prove that, if all rings Ro for Q E Spec(R) are unique
factorization domains, then R is normal, thus a Krull domain and the homomorphism
qJ: Ca Div(R) -+ We Div(R) is an isomorphism which maps Ca Prin(R) onto a subgroup
We Prin(R) generated by all elements of the form div(x), x E Ro. [In order to construct
a homomorphism tp inverse to qJ, fix a divisor W = L npP e We Div(R) and an ideal Q
e Spec(R). The divisor W induces the divisor W o = L npP of Ro. The divisor Wo is
pcQ
principal, Wo = div(fo), for some 10 E Ro. Prove that there are only finitely many prime
ideals of height 1, P 1 , ..., Pm, such that Pl, ..., Pm cI: Q and fJp,(fo) =F np. Choose ho
e PI" ... nP.."Q and show that div(fo) ==  npP in Ro. Next prove that if (ho 1 ,...
PeDt.ho)
..., h o .) = R then lo,lfo J e (RhO/'o).' i.e., the family {D(ho,),/o,} determines a Cartier
divisor ,,(D) of R. Show that tp(D) is well-defined. To prove the latter part of the assertion,
fix a non-zero ideal Q e Spec(R) and let Po c Q be a prime ideal of height 1. Let a family
{D(h,),fi} determine a Cartier divisor such that feD) == 1. Po. Prove that if 11 ; Q then
we have an equality div(Ji) == Po of divisors of the ring Ro; then apply Corollary 4.4.9.]
11. Let R be a domain and let D be a Cartier divisor determined by a family {D(h,) , fi }
i== 1, ...,S. Write
M, = {x E Ro; x E R p l,l for P e D(h,)},
8
and (j)(D) == n M.. Prove that the function (j) is well defined, establishes an isomorphism
lal
of groups
(j): Ca Div(R) -+ Pic(R, R" {O})
"-I
and induces an isomorphism Ca CI(R) .:+ Pic(R). [Prove that if M == n M, then the module
M, is the localization of M with respect to {I, h., hf, ...}. To prove that M is an invertible
module, denote by N" N the modules assigned in a similar way to the family {D(h,),!,1}.
There exist representations Ji == n,lh',/f l == m,lh7, for some n, EN" m, eM.; deduce that
R c MN. When proving that (j) is an epimorphism note that the Ro-modules Mo are free
of rank 1 for Q E Spec(R»).
4.5 INDUCED HOMOMORPHISMS OF CLASS GROUPS
In Sections 4.1 and 4.2 we have assigned to a Krull domain R the divisor
class group Cl(R) = Div(R)/Prin(R). Quite naturally the question arises
whether this assignment is functorial, i.e., whether homomorphisms of Krull
domains f: R -+ R' induce homomorphisms of groups Cl(f): Cl(R) -+ CI(R')
which satisfy the conditions CI(gf) = Cl(g)CI(f) and Cl(I R ) = lC1(R), where
g: R'  R". An analogous question can be asked about the group of divisors
Div(R).

182
Divisors and Krull Domains
[Chi
It turns out that the homomorphisms Cl(f) with the expected properties
can be defined for a relatively small class of homomorphisms f. In certain
cases we give a description of the kernel of a homomorphisms Cl(f), and
obtain as a result an isomorphism between class groups of a Krull domain R
and that of a ring of polynomials in arbitrarily many indeterminates with
coefficients in R.
To begin with let us illustrate the difficulties involved.
Example 4.5.1
Consider the rings T = K[X 1 , X 2 , ..., X,,] and R = K[Xf, XI X 2 , ..., X;]
which were described in Example 4.1.7. The ideal P = TX 1 nR is a divisor
and is generated by the elements X;,X 1 X 2 , ...,X 1 X". The ideal PTis equal
to XI · I where I = TX 1 + ... + TX n , and so we have div(PT) = div(X I · I)
= XI div(l) = XI T  PT since div(I) = T. We see that the inclusion j:R
c: T, which induces the mapping of the sets of fractional ideals IdU): Id(R)
 Id(T), Id(j) (J) = JT does not carry divisors into divisors.
Suppose that j:R c: R' is an extension of Krull domains. We define the
mapping of groups of divisors Div(j): Div(R)  Div(R') by the formula
Div(j) (I) = div(R'J) = R':(R':(R'I) for IeDiv(R).
We obviously have Div(j) (Rx) = R'x, and thus Div(j) (Prin(R» c: Prin(R').
We shall find conditions for the mapping Div(j) to be a homomorphism
of groups.
Write /A = /AI and /1/ = PdiV(IR') for the functions which correspond to the
divisors land div(IR') as indicated in Theorem 4.4.5. Denote by j: Z {Spec 1 (R)}
-+ Z {Spec 1 (R')} a function such that j(P, I) = Pdiv(IR') = p,' for every divisor
IE Div(R). Since R"-(P' nR) c R'""P', we have RP'nR C R p , and by Theorem
4.4.1
(P'R')/A'(P') = (IR')R p , = IR, = (IRp'nrJR,
for every ideal P' E Spec 1 (R'). Thus the number p'(P') depends only on the
extension of the ideal IRP'nR to the ideal (IRp'nR)R p ,. We may represent
the ideal IR p ' nR in terms of the function p, if the height of P' nR is not greater
then 1. Indeed if P' nR = 0, then IR p ' nR = Ro since I  0 and p,' (P') = 0,
and if P' nR = P is of height 1, then IR p = (P R p )fJ(P). In the latter case,
there exists a positive integer e(P' / P)  1 such that
(PRp)R p , = (p'R,)e(P'IP) where P = P'nR and ht(P) = 1,
and finally we have
(P'R,)fJ'(P') = (PRp)/'(P)R, = (PRpR,)f.I(P) = (p'R'p,)e(p'/P)/J(P>,
i.e., p,(P') = e(P' I P) p,(P) for P' e Spec l (R') and P = P' nR e Spec 1 (R). Thus
we have proved.

IV]
Induced Homomorphisms of Class Groups
183
Lemma 4.5.2
If for every ideal pi E Spec 1 (R'), the prime ideal P = p'nR is of height not
greater than 1, then
{ e(p i / P) f.t,(P), if ht(P) = 1,
P,div('R')(P') = 0 , f P 0
i =,
or every divisor I e Div(R).
Corollary 4.5.3
Let j: R c: R' be an extension of Krull domains. If for every ideal
pi E Spec 1 (R') the height of the ideal P = P' nR is not greater than 1, then
the mapping Div(j): Div(R)  Div(R') given by the formula DivU) 1
= div(IR'), where I E Div(R), is a homomorphism of divisor groups.
Proof
The previous lemma yields
j(P) (P') = { e(p, / P) p,(P),
0,
if ht(P) = 1, P = P'nR,
if P'nR = o.
From this formula we immediately obtain j(p + Ili) = j(P) + j(P 1) for aU
p" P,1 E Z(Spec 1 (R»). Hence j, and therefore Div(j), are homomorphisms. 0
C:orollary 4.5.4
Under the hypotheses of Corollary 4.5.3 we have the identity
Div(j)(P) = Le(P'/P)P'
P'
for P E Spec 1 (R), P' ranging over ideals from Spec 1 (R') such that P'nR = P.
Since the group Div(R) is a free Z-module generated by Spec 1 (R), the above
identity might be taken as a definition of the homomorphism Div(j) in the case
of arbitrary extensions of Krull domains j: R c R ' . In the general case,
however, this homomorphism does 110t carry the group Prin(R) into the
group Prin(R'), and consequently it does not determine a homomorphism
of divisor class groups.
We shall now prove a theorem on induced homomorphisms of class groups.
Theorem 4.5.5
Suppose that an extension of Krull domains j: R c R' satisfies one of the
following conditions:
(i) the extension R c: R' is integral,
(ii) the extension R c: R' is flat, i.e., R' is a flat R-module,
(Hi) the ring R' is a ring of fractions Rs with respect to a multiplicative
subset S c: R,
(iv) The ring R' has the form R' = n R p , where A c: Spec 1 (R).
PeA

184
Divisors and Krull Domains
[Ch.
Then, for every ideal p' e Spec 1 (R'), the height of the ideal P' nR is not
greater than 1. Also the mapping DivU): Div(R)  Div(R'), given by the
formula DivU)I = div(IR'), is a homomorphism which maps principal
divisors to principal divisors, Div(j)(Prin(R») c Prin(R'); thus it induces the
homomorphism
CI(j): Cl(R)  CI(R').
In proving this theorem we shall make use of the following
Lemma 4.5.6
If a ring extension R c: R' is flat and I, J are fractional ideals of R, then
R'(InJ) = R'/nR'J.
Proof
Replacing the ideals I and J by rl and rJ, if necessary, we Inay suppose in
addition that I, J c R. We have the exact sequence
h
o -. In! -+ R -+ R/lfd;)R/J,
where h(r) == (r+I, r+J); since R' is flat as an R-module, tensoring by R'
induces the exact sequence
0-.. (InJ)@RR' -. R@RR' -. (RII0 R R')ff)(RIJ@RR').
Replacing the modules with isomorphic ones, we obtain the exact sequence
hi
0-. R'(/nJ) -+ R'  R'IR'I$R'/R'J,
where h' (r') = (r' + R'I, r' + R' J). Since Ker(h') = R' InR' J, the lemma is
proved. D
Corollary 4.5.7
If an extension R c: R' of Krull domains is fiat and I is a divisor of R, then
R'I is a divisor of R'.
Indeed, by Theorem 4.4.11 the divisor 1 is an intersection of two principal
divisors; thus, by Lemma 4.5.6, R'I is also an intersection of two principal
divisors, and therefore it is actually a divisor.
Proof of Theorem 4.5.5
By Corollary 4.5.3 it is sufficient to prove that for every ideal P' E Spec 1 (R'),
the ideal P' nR is of l1eight not greater than I.
(i) Suppose that the extension R c R' is integral. As the domain R is
normal, Theorem 3.2.4 yields ht(P' nR)  1. Since by Theorem 3.1.13 P' nR
:F 0, we infer that ht(P' nR) = 1.
(ii) Suppose that the extension R c: R' is flat and P e Spec 1 (R'). Assume
that ht(PnR) > 1, and choose a non-zero element r e PnR. Let PI' ..., Pm
be all ideals in Spec 1 (R) which contain the element r. Since ht(PnR) > 1,

IV]
Induced Homomorphisms of Class Groups
185
we have P 1 U... UPm:P PnR. Hence there exists an element sePnR
such that S  PIU ... uP".. It follows that
'lJp,(rs) = 'lJp,(r) for i = 1, ..., m
'lJp(rs) = vp(s) for P 1:. Pi' ... , Pm'
whence we obtain the identity vp(rs) = max (fJp(r), 'lJp(s») for all P e Spec 1 (R).
By Theorem 4.4.7 (v) this identity yields RrnRs = Rrs. It follows from
Lemma 4.5.6 that R'rs = R'(Rrs) = R'(RrnRs) = R'rnR's. Hence for all
P'eSpec 1 (R'), we have 'lJp,(rs) = max('lJp,(r), 'lJp'(s»); for P, however, we
have r e p and s e P, and therefore fJpo(rs) = 'lJpo(r)+fJpo(s) > max('lJpo(r),
'lJ Po (s»). The resulting contradiction proves the assertion.
(Hi) Since the R-module Rs is flat, condition (Hi) follows from the preceding
one.
(iv) This condition is implied by Corollary 4.3.10. 0
In the case of extensions of the formj: R c Rs, we are able to give a full
description of the kernel of the homomorphism CIU).
Theorem 4.5.8 (Nagata [NJ)
If S is a multiplicative subset of a Krull domain R, then we have the following
exact sequence:
f CIOO
F -. CI(R) -.. CI(R s ) -. 0,
where F is the subgroup of Div(R) with the basis {P e Spec 1 (R),PnS 1= 0},
the homomorphismfis the composition F c Div(R) --.. Cl(R) and moreover
Ker(CI(j») = (F+Prin(R»)/Prin(R)  F/(FnPrin(R»).
Proof
By Theorems 4.4.7 and 1.4.8, we can identify the groups Div(R) and Div(Rs) €a
G) F. The homomorphism DivU) is then a projection onto the first s umm and,
and hence CI(j) is an epimorphism. Since DivU) (Prin(R») = Prin(Rs), it
follows that Ker(Cl(j») = (Ker(DivU»)+Prin(R»/Prin(R) = (F+Prin(R»)
{prin(R), which completes the proof. D
Corollary 4.5.9
Let R be a Krull domain and assume that {py}y er is a family of elements of R
such that ideals Rpv are prime. Denote by S a multiplicative subset of R
consisting of all products of p;,s. Then the extension j:R c: Rs induces an
isomorphism CI(j): Cl(R)  CI(R s ).
Proof
If P e Spec 1 (R) and PnS rF 0, then there exists r e r such that Pv e P. Since
ht(P) = 1 and Rpy c P, it follows that P = Rp". Using the notation of
Theorem 4.5.8, we can write F c. Prin(R), hence Ker(CIU») = o. 0

186
Divisors and Krull Domains
[Ch.
1nmeoremo 4.5.10
If R is a Krull domain, then the extension j: R c: R[ {Xv}yer] of R to a poly-
nomial ring induces an isomorphism
CIU): CI(R)  Cl(R[ {Xv }yer]).
Proof
In Corollary 4.2.10, we described the prime ideals Q of height 1 of the ring
T = R[ {X Y }l'er] as follows: 1° if QnR = 0, then Q  Ro[F]qnT for some
irreducible polynomial q e Ro [F], 2° if QnR = P  0, then Q = PT and
P e Spec. (R).
The domain Ro [r] = TR",o is a unique factorization domain. Thus CI (TR"'o)
= 0, and, by Theorem 4.5.8, we have CI(T) = Ker«CI(T» -+ CI(TR",o»
= (F+Prin(T»/Prin(T), where F is a subgroup of Div(T) with the basis
{Q e Spec 1 (T): QnR  O}. Since Div(j)P = PT for P e Specl(R), it follows
from the above description of the ideals Q that the homomorphism Div(j):
Div(R)  Div(T) is a monomorphism and 1m (Div(j» = F, whence CI(j) is an
epimorphism.
Now, applying Corollary 4.5.7, we see that the kernel of the homomorphism
CIU) consists of all residue classes with respect to Prin(R), determined by
divisors Ie Div(R) such that DivU)I = div(IT) = IT is a principal divisor.
Replacing I with a divisor of the form rI, we can assume in addition that
I c R. If IT = Tt for some t e T, then TtnR = I  0, whence t is a polynomial
of degree 0, i.e., t e R. Hence we have I = TtnR = Rt e Prin(R), and conse-
quently CIU) is an isomorphism. 0
Corollary 4.5.11
A domain R is a unique factorization domain if and only if for every set of
indeterminates {X" }l'er the polynomial ring R[ {Xv }yer] is a unique factorization
domain.
The implication  follows from Theorems 4.4.8 and 4.5.10. The converse
implication is an immediate consequence of the fact that the factorization
of an element of R in the polynomial ring is a factorization in R.
Let R = n R p be a Krull domain and let A c Spec l (R). As we
Pe Spec 1 (R)
know, the ring R(A) = n R p is also a Krull domain. Denote by j the embed-
PeA
dingj: R c R(A). Proceeding as in the proof of Theorem 4.5.8, we can show
that if F is the subgroup of Div(R) with the basis Spec! (R)"A then, with the
groups Div(R) and Div (R(A) ) $ F identified, the homomorphism Div(j) is
the projection on the first summand, and consequently Ker(CI(j» = (F+
+ Prin(R) )/Prin(R).

IV]
A Representation of an Abelian Group as a Divisor Class Group
187
Exercises
1. Let R be a unique factorization domain in which the element 2 is invertible. Prove
that, if r e R is a non-invertible element which is not divisible by any square of a prime
element, then the ring R[XJ/(X 2 - r) is normal.
2. Let K be a field of characteristic different from 2. Prove that if F is a non-degenerate
quadratic form in the ring K[X 1 , ..., XII], n  3, then the ring K[X 1 , ..., Xn]/(F) is normal.
3. Prove that the ring C[X, Y, ZJ/(X 2 +y2+Z2) is normal, and its divisor class group
s cyclic of order 2. [Hint: set t = x+iy, U = x-iy, where x, y, z are the residue classes
of X, Y, Z. Then R == C[x, y, z], tu+z 2 = 0, and RII = C[z] [u, u- 1 ] is a unique factorization
domain. Apply Theorem 4.5.8 to prove that P = (u, z) is the only prime ideal of height 1
of R. Show that 2 div(P) = div(u ).]
4. Let R be a Krull domain. Prove that the map which associates with an invertible
fractional ideal I c R the class of its divisor div(I) in Cl(R) induces a monomorphism
(J(R): Pic(R)  Cl(R). Show that if g: R -+ T is a flat extension of Krull domains then the
diagram
Pic(R) Pic (g) Pic(T)

P(! CI (g) ! P(T)
CI (R) . CI (T)
commutes. .
5. Prove that if R is a Krull domain then the embedding R -+ R[X] induces an iso-
morphism Pic(R) xPic(R[X]). [Hint: Apply the preceding result and Theorem 4.5.10.]
4.6 A REPRESENTATION OF AN ABELIAN GROUP AS
A DIVISOR CLASS GROUP
We have assigned to every Krull domain R an Abelian group CI(R) = Div(R)
/Prin(R) and proved that it is zero if and only if R is a unique factorization
domain. The group Cl(R) can be regarded as a measure of the non-uniqueness
of factorization of elements onto irreducibles. In Section 2.4 we proved that
if R is a Dedekind domain of algebraic integers of a finite extension of the
field of rationals then the group CI(R) is finite. Naturally the question arises
as to which Abelian groups are isomorphic to groups of the form CI(R) where R
is a pedekind domain, or, more generally, a Krull domain. A complete answer
is due to Claborn [4], who proved that for every Abelian group G there exists
a Dedekind domain R such that Cl(R)  G. This result is very important since
it clarifies relations between classes of unique factorization domains and the
class of Dedekind domains, stating that there exist Dedekind domains with
factorizations which are non-unique and arbitrarily complex with respect
to the classification in terms of the group Cl. It should also be mentioned that
in Claborn's construction, although it yields a Dedekind domain, a very
important role is played by Krull domains, and use is made of all their prop-
erties previously discussed here.

188
Divisors and Krull Domains
[Ch.
Let G be an arbitrary abelian group. We wish to represent G as an image
of the homomorphism 1p: Z {r} --. G of a free Z-module with a basis r.
In order to obtain a Dedekind domain whose class group is isomorphic to the
group G we construct successively:
(I) a Krull domain R 1 whose class group CI(R 1 ) is isomorphic to the group
Z {r},
(II) a Krull domain R 2 whose class group CI(R 2 ) is isomorphic to the group
G,
(III) a Dedekind domain R3 whose class group CI(R 3 ) is isomorphic to the
group G.
I. The definition of the ring R 1 will be preceded, by remarks which may help
in understanding the general idea behind the construction, although these
remarks are not indispensable for the proof of the properties of the ring R t .
Let K be a field; the rings and homomorphisms under consideration will
be K-algebras and K-homomorphisms of algebras respectively. The simplest
cause of disturbing uniqueness of factorization is the relation ac = bd satisfied
by four non-associated irreducible elements a, b, c, d. The subalgebra generated
by the elements a, h, c, dis a homomorphic image of the algebra. A = K[X, Y,
T, UJ/(XT- YU), in which the residue classes x, y, t, u of X, Y, T, U are non-
associated and irreducible, and xt = yu. The reader will readily verify that the
homomorphism (! of the algebra of polynomials [X, Y, T, U] into the algebra
of polynomials K[X, Y, Z] given by the conditions (!(X) = X, (!(Y) = Y,
e(T) = YZ, (!(U) = XZ, has the kernel (XT- YU), and consequently the
algebra A is isomorphic to the subalgebra R = K[X, Y, XZ, YZ] of K[X, Y,
Z]. It is easy to represent A as an intersection of discrete valuation rings. The
non-unique factorization in R is realized with X(YZ) = Y(XZ). This relation
will serve as a generator of a suitable class group. We wish to obtain a ring
whose class group is isomorphic with Z {r}, and thus we should assign a rela-
tion of this type to every element i' e r. To this end, we map the algebra
Al = K[ {Xy, Yy, Ty, Uy}yer]/{ {XyTy- Y y Uy}yer) isomorphically onto the sub-
algebra R 1 = K[ {Xy, Yy, XyZy, Y"Zy}"er] of the algebra of polynomials
K[ {Xy, Yy, Zy}yer]. The description given above of the algebra R 1 relies on
intuition, but it is not convenient for proving that R 1 is a Krull domain and
O(Rt)  Z {r}. We shall give a different but equivalent description of the
algebra R 1 .
Let W = K[ {Xy, Y y ,Zy}yer] be an algebra of polynomials with coefficients
in the field K. For a fixed index )'0 e r, we write W Yo = K[ {Xy, Yy, Zy}yo"er]
and Ayo = W"o[X Yo ' Y Yo ' XYoZYo] = WYo[XYoZYo] [X Yo ' YYo] c: W. The algebra
Ayo is an algebra of polynomials in X"ot Y Yo with coefficients in WYo[XYoZYo].
In Ayo' the ideal Q)'o' generated by the variables XYo' Y Yo ' determines a
discrete valuation 'Vo given by the formula vYo(a) = q  a E Qo"Q:1,
extended onto the field (A)'o)o = W o (cf. Exercise 1, Section 3.6).
We define the ring R 1 by the formula R 1 = n Rv n W, where R" is the
yer y y

IV] A Representation of an AbeUan Group as a Divisor Class Group 189
valuation ring of v". By Theorem 4.2.9, the ring W is a Krull domain, whence
W = n W p , and we obtain the representation of the domain R 1 as an
PeSpec 1 (W)
intersection of discrete valuation rings
R 1 = n R" n n W p . (4)
"er V PeSpec1(W)
(The reader will easily prove that Rvyn W is an algebra generated by W y
and by the elements X"' Yl" X"Z", Y"Z", and hence that R 1 is an algebra
generated by all the monomials X"' Y v ' XyZ", Y"Z", ,., e r, i.e., the algebra
described above). Every non-zero element r e R 1 is a polynomial in the indeter..
minates X"' Y", Z" and consequntly depends only on a finite number of those
indeterminates. If the indeterminates XYo' Y"o' Zvo do not enter into the poly-
nomial r, then'Vvo(r) = O. Hence,. is an invertible element in almost all rings
RV)l and in almost all rings W p . Thus, by Theorem 4.3.1, R 1 is a Krull domain.
To describe Spec 1 (R 1 ) we shall use Corollary 4.3.9, proving first that the
representation (4) is irredundant. The reader will easily verify what follows:
(1) Zvo  Rvvo and Z"o belongs to all the remaining factors in (4);
(2) Y"ol Xvo  W wX vo and YYo/ Xvo belongs to all the remaining factors in
(4);
(3) given an ideal P = Wfe Spec 1 (W) which is not of the form WX y ,
if "1'. . . , "" E r are all the elements ,., e r such that 'Vv(f):F 0, and
g = f-l X';,,1 ... X::C", where ml = max(O, V"l (f)), ..., m" = max(O, !I)I" (f»),
then g  W p and g belongs to all the remaining factors in (4).
It follows from Corollary 4.3.9 that Spec 1 (R 1 ) consists of all ideals Py
= m(R 17 )nR 1 for" e r and of all ideals PW p nR 1 = PWpn WnR 1 = PnR 1
for P e Spec 1 (W). Accordingly, the group Div(R 1 ) is isomorphic with the
group H = Z {r}  Z {Spec 1 (W)} and the isomorphism assigns to div(R1 x)
e Prin(R 1 ) c Div(R 1 ) the element h = L 'Vv(x)Y + L vp(x) P for O:F x
" P
e (R 1 )o = WOe Clearly, h X1 = h+h, and CI(R 1 )  H/H', where H'denotes
the group of all elements of the form h.
Observe that if P = Wfp, fp e W, then fp e (R 1 )o and
hi" = L 'lJ"ifp)y+ 1 · P,
v
whence we deduce that H = Z {r}  H", where H" c H' is the group gener-
ated by all elements of the form h f p for P e Spec 1 ( W). We claim that H" = H'.
Suppose that 0 :F x e (R 1 )o and consider the element y = x DfiVp(X). Clearly,
P
we have'Vp(Y) = 0 for all P e Spec 1 (W). Thus yeW and y is invertible in W;
hence y E K and therefore y is invertible in R 1 . Accordingly, h, = 0 and h
= Lvp(x)h fp eH", whence H" = H', and consequently CI(R 1 )  H/H'
p
= BIH"  Z{r}.

190
Divisors and Krull Domains
[Ch.
II. We have represented the free Z-module Z {r} as the class group of some
Krull domain R 1 . In the sequel, we shall represent the kernel of the epimor-
phism 1p: Z{r} -+ G as the kernel of an epimorphism of the form CI(R 1 )
 Cl(R 1 [X]) CI(l) ) CI (Rl [X] (A) ), where A is a suitable subset of Spec 1 (R 1 [X]).
This construction and the reason for replacing the ring R 1 by the ring of
polynomials R 1 [X] are based on the possibility of representing every element
of the group CI(R 1 [X]) as the residue class of a prime ideal of R 1 [X]. Strictly
speaking, we have the following
Lemma 4.6.1
If R is a Krull domain then for every element (t e CI(R[X]) there exists an ideal
P e Spec 1 (R[X]) such that (t = P+Prin(R[X]).
Here, and in the third part of our construction, we shall use
Lemma 4.6.1
Let R be a normal domain, 0 :F r, s e R, and I = Rr- 1 r.Rs-l. Then (r X -
-s)Ro[X]r.R[X] = (rX-s)IR[X].
Proof
If y e I then (rX -s)y e R[X] and y E Ro. Thus the inclusion => is proved.
To prove that the converse inclusion holds, assume that the polynomial
In
1= (rX-s)(t".X'"+ ... +t o ) = rt m Xm+l+ LUk Xk
k=O
has coefficients in the domain R, 10' ..., t m e Ro, and t m -:f: O. We claim that
the element st m is integral over R and thus belongs to R.
Denote by g e Ro [X] the polynomial given by the formula
In
g(X) = I(X/rtm) = (rtm)-m xm+1 + L uk(X/rt",)k
k=O
Since f(s/,,) = 0, we have g(st m ) = f(st",/rt m ) = 0; moreover, (rt m ) mg (X)
In
= xm+ 1 + 2: uk(rtm)",-kX" e R[X], whence it follows that st m is integral over R.
k=O
The polynomial f-(rX-s)lmX'n = (rX-s)(t m _ 1 X m - 1 + ... + to) also has its
coefficients in the ring R; we proceed as before to show that rt m-l, st m - 1 e R,
and so on.
We have thus proved that the coefficients of the polynomial tmP + ... + to
belong to Rr- 1 r.Rs-l = I, whencefe (rX -s)IR[X], and the lemma is proved.
o
Proof of Lemma 4.6.1
Using Theorem 4.4.1 we can easily prove that every divisor J of R is of the
form t- 1 (Rr- 1 r.Rs-l) for some non-zero elements r, s, t of R. Since Cl(R)
 CI(R[X]), there exists a divisor I of R which has the form I = Rr- 1 r.Rs-l

IV] A Representation of an Abelian Group as a Ditisor Class Group 191
and is such that the element oc is the class of the divisor IR[X]. The element
rX -8 is irreducible in the ring Ro[X]. Hence, by the previous lemma and
Corollary 4.2.10, P = (rX-s)IR[X] is a prime ideal of height 1 of R[X].
Obviously the element oc is the class of the divisor (rX -s)IR[X] = P. This
completes the proof. 0
To accomplish the second part of the construction, let us denote by "P'
the epimorphism "P': CI(R 1 [X])  CI(R 1 )  Z {r} -. G and choose for every
element E e Ker( "P') a prime ideal PeE Spec 1 (R 1 [X]) whose class is E. Set
A = Spec 1 (R l [X])" {P'}feKer('P'). Then Corollary 4.3.10 implies, that
R 2 = n (R 1 [X])p is a Krull domain. It follows from the considerations
PeA
in the latter part of the previous section that the embedding i: R 1 [X] ( R 2
induces an epimorphism CI(i): CI(R 1 [X]) -+ CI(R 2 ) whose kernel is the sub-
group Ker(1p'). Accordingly, the group CI(R 2 ) is isomorphic to the group G.
III. Given an Abelian group G, we have constructed a Krull domain R 2
such that CI(R 2 )  G. We shall now construct a Dedekind domain R3 such
that CI(R 3 ) = G; R3 will be a ring of fractions of the ring T == R 2 [Xl' X 2 , ...,]
with respect to some multiplicative subset SeT. We know by Theorem 4.5.10
that the embedding R 2 c T induces the isomorphism Cl(R 2 )  CI(T), and
thus CI(T)  G. We shall choose such a set S that it will satisfy the following
two conditions:
(1) the ring Ts is a Dedekind domain,
(2) the embedding T c Ts induces an isomorphism of class groups.
By Theorem 4.2.6 and Corollary 4.5.9, we shall satisfy both conditions if
we succeed in finding for every prime ideal Q c T of height > 1 a prime element
fQ e Q (i.e., an element such that the ideal TfQ is prime) because the set S
of all products of elements of the form fQ satisfies (1) and (2).
Let Q c T be a prime ideal of height greater than 1. Choose a non-zero
element r e Q, and denote by PI' ..., P" all the ideals in Spec 1 (T) which contain
the element r. Since ht(Q) > 1, we have Q cI= P 1 u ... uP k . Hence there exists
an element s e Q such that s  PI U ... uP k . Furthermore, there exists a number
m such that r, s e R 2 [X 1 , ..., X m - 1 ]. SettingfQ = rXm-s, we obviously have
o :F fQ E Q. Let A = R 2 [Xl' ..., X m - 1 ]. It is sufficient to prove that fQA [X m ]
is a prime ideal of the ring A [X m] because this implies that the ideal fQ T
= fQA[Xm, X m + 1 , ...] of T is also prime. The elements 1', seA c T have
been chosen so that fJp(rs) = max (vp(r), fJp(s) for all ideals P e Spec 1 (T).
Thus this condition is also satisfied by all ideals P' e Spec 1 (A) by Lemma
4.6.2, whence Ars = ArnAs, and consequently A = Ar- l nAs-l. It follows
from Lemma 4.6.2 thatfQA[X m ] = (rXm-s)A[Xm] = (rXm-s)Ao[Xm]nA[X m ],
whence the idealfQA[X m ] is prime.
We define R3 = (R 2 [X 1 , X 2 , ...])s, where S denotes the set of all products
of elements of the form fQ. The ring R3 is actually a Dedekind domain and
Cl(R 3 )  G. Thus we have proved

192
Divisors and Kroll Domains
[Ch.
Theorem 4.6.3 (Claborn [4])
Every Abelian group is isomorphic to the divisor class group of some Dedekind
domain.
An entirely different proof may be found in [16].
4.7 NORMALIZATION OF NOETHERIAN DOMAINS
The normalization R of a Noetherian domain R (i.e., the integral closure
of the domain R in the field of fractions Ro) need not to be a Noetherian
domain (see Example 4.7.3). In view of the fundamental rOle played by Noe-
therian domains in the whole theory of rings it is natural to ask about the
structure of rings R . This section is devoted to the proof of the theorem stating
that R is a Krull domain. The proof is long and complicated (other proofs
are to be found in [21], [23], [24]). Nevertheless, it is a good illustration of the
application of complete local rings and of integral extensions techniques.
We regard this theorem as essential for the linking together of ideas discussed
in this book, and that is why we present its proof in full. The first step is to
prove a theorem analogous to Theorem 3.2.5, which states that the integral
closure of a complete local domain R in a finite extension of the field of frac-
tions Ro is a finitely generated R-module.
Theorem 4.7.1
Let R be a complete local domain. If Ro c:: K is a finite field extension, then
the integral closure of R in K is a finitely generated R-module.
Proof
Denote by R the integral closure of R in K. We shall construct a diagram in
which all the mappings are injections.
T
 To
 Ro
T"
K"
T=R
T' /
 K

:. K'

IV]
NormaUzation of Noetherian Domains
193
. In [B] we prove (see Theorem 2.3.18) that R contains a subring T which
is local and complete, and satisfies the following conditions:
(i) the ring T is normal,
(ii) the ring R is a finitely generated T-module.
In the case where the characteristic p of Ro is positive, we may take for T
a ring of the form T = k[[X 1 , ..., X n ]], where k is a field. The extension
To c Ro is then finite, and the integral closure T of T in the field K is equal
to R. It is easily seen that the properties of purely inseparable extensions
([L], Chap. VII,  7) imply the existence of finite field extensions K c K',
To c K" c K' such that To c K" is purely inseparable, and K" c K' is
separable. Denote by T', T" the integral closures of T in the fields K' and K",
respectively.
Observe that it is sufficient to prove that T" is a finitely generated T-module.
Indeed, by Theorem 3.2.5, the ring T', as the integral closure of T" in K',
is contained in a finitely generated T" -module which, regarded as aT-module,
is also finitely generated by our assumption. Now, since T is a Noetherian
ring, it follows from R c T' that R is a finitely generated T-module. Hence, R
is a finitely generated R-module.
We shall prove that the ring T" is a finitely generated T-module. If the
characteristic of Ro is 0, then the purely inseparable extension To c K" is
trivial, To = K", whence, by the normality of T we have T" = T. Suppose
that the characteristic p of Ro is positive. As we have mentioned before, we
may assume that T = k[[X 1 , ..., X n ]], where k is a field. The degree of the
extension q = (K": To) is a power of the characteristic, q = pee As in the proof
of Theorem 3.3.12, write q;: To -+ To for the Frobenius automorphism tp(y)
= y' of the algebraic closure To of To which contains K", and assume
T = tp-e(T), k = rp-e(k), Y 1 = tp-e(x 1 ), ..., Y n = tp-(Xn).
Since cpe(T o ) c To, it follows that To c: tp-e(T o ), and since the extension
To c K" is purely inseparable of degree q = pe, we also have K" c qJ-e(T o ).
If t" E T", then (t,,)q E To, whence also (t")q E T since the ring T is normal.
Thus T" c q;-e(T) = T (in fact, even T" = TnK"). Consequently, we obtain
a diagram where all the mappings are inclusions
7'- . 'fo
+ +
1""  K"
+ +
lOtI :. cp-e(T 0)
T == q;-'(T)

194
Divisors and Krull Domains
[Ch.
II>' ,., ,.,
The ring T is the ring of formal power series T = k [[Yl' ..., Y n ]], and the
inclusion T c T associates with the series L IX/I...IIIX1i ... XII the series
L a.1t ...III Yf 'l ... Y'n. Clearly, the extension T c: T is integral.
Every element a E T" can be regarded as a power series in Yl, ..., Y n
with coefficients in the field k; for a :oF 0, we denote by ).(a) the homogeneous
component of lowest degree (the initial form) of the series a. If a e T, then
also ).(a) e T, and coefficients of the form ).(a) belong to the field k.
Observe tht if al, ..., am e T" and the forms ).(al)' ..., ).(a m ) are linearly
independent over T then al, ..., am are also linearly independent over T.
Indeed, if b 1, ..., b m e T were not all zero and L h, a, = 0, then the sum
of those of the forms )'(h,) ).(a,) which are of the lowest degree would be equal
to zero, contrary to the linear independence of the forms A(al)' ..., A(a",).
This remark implies, that if t 1, ..., tk e T" and initial forms ).(t 1), ..., A(t k )
are linearly independent over T, then k  q = (K": To). Consequently for
some al, ..., am e T", m  q, initial forms A(al)' ..., A(a m ) constitute a maxi-
mal linearly independent (over T) subset of the set of all forms A(a), a e T " .
AI
Denote by k' c /, the field which is formed by adjoining all coefficients of the
forms A(al)' ..., A(a m ) to the field k; let Po = 1, Pl, ..., P, be a basis of the
field k' over k. Further, denote by 'fJo = 1, 'Y/l, ..., 'fJr the set of all monomials
in Yl, ..., Y" of degree < q with respect to each of the indeterminates. For
every non-zero element a e T" the forms A(al)' ..., A(a m ), ).(a) are linearly
dependent over T. Hence there exist elements b, b l , ..., b m e T, h  0, such
that bA(a) = hi ).(a 1 ) + ... +b".).(a m ). This equality implies that the form
)'(b) A(a) is a sum of forms of type A(h,) A(a,), whence it is easily deduced
(by using the basis for k over k') that all coefficients of the form A(a) belong
to k'.
Every monomial in Yl,'..' Y" has a representation of type (Yf)k 1 ...
... (yg)k n 'Y}I; thus, every form in Yl' ..., y" with coefficients in k belongs to the
r
module L T1J,; consequently, every form '\(a), a e T", belongs to the module
1..0
L T'Y},PJ. Since the ring Tis Noetherian, the T-module M c: L T'Y/,PJ gener-
tJ tJ
ated by forms ).(a), a E T " , is finitely generated. Let A(CI)' ..., ).(c,) for some
c 1, ..., C, e T" be generators of M. Since the elements c 1, ..., c, are integral
over T, the ring V = T[Cl' ..., c,] c T" is a finitely generated T-module.
Clearly, the extension V c: T is integral, and since T is a local ring it follows
by Theorem 3.1.12, that the ring V has a unique maximal ideal. The ring V
is Noetherian, and hence local. It follows from Theorem 3.1.18 that the natural
topology of V is identical with that of V as a T-module. By Corollary 2.6.23,
the ring V, as a finitely generated T-module, is complete. Thus it is a complete
local ring.
We shall prove that the natural topology of V is identical with that induced
,., ,., ,.,
by the natural topology of T. Write m and m for the maximal ideals of T

IV]
Normalization of Noetherian Domains
195
and V, respectively. Since the extension V c: T is integral, we have m = mnV.
The induced topology of V is determined by the decreasing sequence of ideals
In = mnnV, n = 1, 2, ... Clearly, we have m" c: fftnnV = In; thus, it remains
to prove that for every n = 1, 2, ... there exists a positive integer Zen) such that
I'(n) em,.. Suppose the contrary, i.e., that there exists a positive integer no
such that Ik cJ: m"o for k = 1, 2, ...; then also Ik cJ: m" for n  no and k
= 1, 2, ... The factor rings V 1m" are Artin rings, and so there exist positive
integers ken) such that l"(n) -)-m" = Ik +m" for J,  ken) and we can suppose
in addition that ken) < k(n+ 1). The inclusion [ken) c I 1 (IJ)+m n = Ik(..+I)+
+m Jl implies that we can successively find elements x" E l"(n), for n  no
such that x"o ,m"o and X n - X n + 1 E m". The ring V is complete, and so the
sequence {xn} converges to some x e V. Moreover, for every n  no, elements
X n , X,,+ l' ... belong to the ideal I" since x e In and m" c: In. Every ideal In
is closed (see Exercise 6, Section 2.6); hence
x E n 1" = n (tiinnV) = 0,
II 110 n 110
a11d therefore x = O. Furthermore, we have
xllo-x n = (x no -X no +l)+". +(X n _l- X n)em"o+ ... +m"-1 =m"o
and X n E nt"O for sufficiently large n, contrary to the former assumption xno f: mno.
Thus the natural topology of the ring V is identical with that induced by the
natural topology of the ring T.
We claim that V = T". For every t e T" we shall construct elements
t o ,t 1 ,... of V such that forj=O,l,..., the degree of the form A(t-t})
is  j. Set to = 0, and suppose that the elements to, ..., tJ-l satisfy the above
condition. The form A(t - tj-l) belongs to the module M, and so there exist
elements Ul, ..., umETsuch that A(t-fi_l) = Ul A(C 1 ) + ... +tlmA(c m ). Replac-
ing Ul,"" U m by their homogeneous components of suitable degrees (with
the component of degree deg A(t - tj -1) - deg A( c,) of the element u,), we can
suppose in addition that Ul' ..',U m are forms. Then A(UI C 1+." +umc m )
= Ul A(Cl) + ... +ll,nA(C m ); setting tJ = tJ-l f'Ul Cl + ... +UrnC m , we have
deg A( t - t J)  j. The sequence {t J} is convergent in T to an elenlent t; V is
closed in T" and tJ e V thus t e V. We have proved that T" c V, and, since
V c: T", we have T" = V. Thus the ring T" is finitely generated as aT-module;
this ends the proof of the theorem. 0
Theorem 4.7.2 (Nagata [N])
Let R be a Noetherian domain. Then the normalization R of R (i.e., its integral
closure in the field Ro) is a Krull domain. If P is a prime ideal of R, then there
are only finitely many prime ideals Q in R such that QnR = P.
Proof
First we shall prove, in two steps, a special case of the theorem.

196
Divisors and Krull Domains
[Chi
1. Suppose additionally that R is a local domain with the maximal ideal P.
A
We shall prove the first part of the theorem. The completion R of R may have
A
proper nilpotent elements (see [N, Pi 208]) but the homomorphism R -+ R
A
-+ R/rad(O) is an injection. Thus R can be identified with a subring of the
A
Noetherian ring T:: R/rad(O). We shall prove that the elements of Rare
not zero-divisors in T. Indeed, if 0 f:. , E R, we have an exact sequence 0 -+ R
 R whose tensor product with R is, by Corollary 2.6.20, also an exact se-
A
quence, whence it follows that,. is not a zero-divisor in RI Theorem 2.3.20 easily
implies that r is not a zero-divisor in the ring T either. Write T for the total
ring of fractions of T. According to the above remarks, we can identify the
field Ro with a sub ring of T. Writing T for the integral closure of T in t, we
obtain a diagram whose mappings are inclusions
R  R  Ro
! (5)
V
..,
.r  T .. 7'
We shall prove the formula R = Tn Ro; this will enable us to transfer some
properties of the ring T onto the ring R. Obviously, R c: Tn Ro. If a, b e R
- A A A
and a/b e T, then there exist elements '0, ..., ',.-1 e R such that (alb)"+
+y n _ 1 (alb)n-l + ... +ro e rad(O). Hence, for sufficiently large k, there exist
elelnents u o , . I., Uk-l e R such that d' + Uk-l a"-lb + ... + lIo b k = O. It follows
A
from Theorem 2.6.25 (i) that a k belongs to the ideal R(d'- l b, ..., bk)nR
= R(a k - 1 b, ..., b k ); thus a/b is an integral element over R, i.e., alb e R ,
- -
and we have proved that R = TnRo.
In the ring T there are no non-zero nilpotent elements, and therefore, by
m
Theorem 2.7.16, the total ring of fractions T of T has the form n (T/P,)o,
1=1
where PI' ..., P na denote the minimal prime ideals of T and the element t e T
is identified with the sequence of its residue classes modulo P l' ... , Pm. Denote
by T, the normalization of the ring TIP h i.e., the integral closure of TIP,
m
in the field L, = (T/P,)o, i = 1, ..., m. We shall prove that T = n T"
i1
If an element (t 1, ..., t m ) E i'is a zero of a monic polynomial with coefficients
in T, then every element t 1, . I. , t m is also a zero of that polynomial. Hence
11' ..., 1m are integral over TIP t , ..., T/P nu respectively, and consequently
m
T c IT T, . If 1 1 e T 1 , ..., till E Tm , there exist monic polynomials 11' ..., 1m
1=1
e T[X] such that t t, ..., 1m are zeros of the residues classes of the polynomials

IV]
Normalization of Noetherian Domains
197
11' ... ,1m in the rings TIP 1 [X], ..., TIPm[X]. The element (t 1, ..., t",) is a zero
m
of the monic polynomial /1 ... 1m E T[X], whence n T, c T.
1=1
The local domains TIP, are complete, thus Theoreln 4.7.1 implies that the
rings T, are finitely generated TIp,-modules; consequently the rings T, are
- -
Noetherian and normal, whence T 1 , ..., T". are Krull domains by Theorem
4.2.7.
Finally, we get
In
R = Tn Ro = (IT r;) nRo
1=1
III
= n [(L 1 x ... XL'_l X T , xL , + 1 X... xLm)nR o ],
1=1
and, by Theorem 4.3.11, in order to prove that R is a Krull domain, we only
have to prove that each of the factors in the latter intersection is a Krull
m
domain. The embedding w:R o c n L, is determined by the injections 1Vl: Ro
tal
-+ L 1 , ... , w na : Ro -. Lrn, and clearly
( T 1 xL 2 x... xL",)nR o = ( T 1 xL 2 x... xLIII)nIm(w)
= {(lV 1 (et), ..., wm(et»:« eRo, wl(a) E 7\}  T 1 n!m( 1v l).
The latter ring is a Krull domain by Theorem 4.3.11, and the sanle argumenta-
tion applies to the other factors.
2. We shall prove the second part of the theorem, assuming as before that R
is a local domain with the maximal ideal P. We retain the notation of the first
part of the proof.
Let Ql, ..., Qk be distinct maximal ideals of R. We shall give an upper
estimate of the number k. For i:f: j, i, j = 1, ..., k, there exist elements
tlj E Q,,,Qj. Denote by V the ring generated by R and the elements ti} (it is
not the ring V considered in part one). We have ReV c R, Ii} E V, whence
the maximal ideals Q 1 ('\ V, ..., Qkr. V are pairwise different. The ring V is
finitely generated as an R-module; by Theorem 3.1.18 it is semilocal and
therefore Noetherian. Write V' for the subring of T generated by T and the
elements I,) E R c T. We extended diagram (5) to
R-  v-  R.. ... R
! «! !p
-
T- . V'  T ..- T

198
Divisors and Krull Domains
[Chi
It. I\. AI
Since T = R/rad(O) is an R-module, so are V' and T; the homomorphism oc
A @I A I\.
induces the homomorphism a,/: V (8) R R V' (8) R R -. V' @ R R = V', and simi-
larly for {J. We obtain a commutative diagram
I\. I\. Y A
R :. VRR  Ro0RR
! ! a' !p,
AI
T-  V'  T'
in which the horizontal mappings are monomorphisms because of the flatness
A I\. I\.
of R. Suppose that the element t E V@RR belongs to Ker(a,'). Then, for some
A A "- A
rl E Rand 0 ¥= r e R, we have ,,(t) e Ker(fJ') and ,,(t) = (llr) f8> '1' Hence
"-
o = {l'r(t) = p' «ll r ) (8)r 1 ) = (1/1')(r 1 +rad(O»), which implies '1 e rad(O).
I\.
Thus there exists a positive integer m such that ,n = 0 and so ,,(t m ) = 0 and
"-
t,n = 0, whence Ker(') c: rad(O). It follows that the ring Im(a,') = V' has
"-
as many maximal ideals as the ring V  R.
Denote by P 1, ..., P q all maximal ideals of the ring V; we then have q ;?; k.
Since the natural topology of semi-local ring V is identical with the topology
of V considered as an R-module, \ve have, by Theorem 2.6.19, an isomorphism
I\. "
V  V @ R R. By Theorem 2.6.28, the completion of a semi-local ring is of the
I\. q "-
form V = n V P , and the number of its maximal ideals is equal to q. Hence,
ial
the ritlg V' also has q maximal ideals.
III
The ring T = n T. is Noetherian as Q product of Noetherian rings and
tal
is the integral closure of V' in 1'; its maximal ideals have the form T 1 x ... x
X T' -l X M, X T, + 1 X ... x T m where M L c: T j is a maximal ideal. The rings
T i are semi-local, whence so is the ring T.
Finally, k  q and q is not greater then the number of maximal ideals of the
ring T, whence R is a semi-local ring. By Theorem 3.1.12, there exist finitely
many prime ideals Q of R which satisfy the condition Q()R = P. .
3. We pass to the general case. First, we shall prove the second part of the
assertion. Let P be a prime ideal of R. The ring RR''-P = R p is local with the
maximal ideal PRR'\.p, whence we can apply to it t he al ready proved particular
case of Theorem 4.7.2. The formula ( R) R''-.P = (RR'\.P) follows directly from
the definition of integral closure (cf. Exercise 1, Section 3.1). Hence (R)R'P
is a Krull domain, and there exist finitely many prime ideals Q' of (R) R'\.P
such that Q' nRR"-.p = P R p . If Q c: R is a prime ideal and Q()R = P, the
ideal Q" == Q (R) R'\.P = QR'\.P satisfies by Corollary 1.4.18 the condition

IV]
Normalization of Noetherian Domains
199
Q" t1RR""P = P R p . By the preceding remark, there exist finitely many prime
ideals Q of R such that Qr'\R = P. We have thus proved the second part of the
theorem.
We shall prove that R = n Ro , where Q e Spec 1 ( R) , and Ra are discrete
valuation rings. Indeed, for every Q E Spec 1 ( R) the ideal P = Qr'\R is prime
and R","P c: R" Q; hence RQ is a ring of fractions of the Krull domain
( R) R""P' and, by Theorem 4.2.8, it is a Krull domain. By Theorem 4.2.6 and
Theorem 3.3.9, R Q is a discrete valuation ring. If M c:  is a maximal ideal,
then RM is a ring of fractions of the ring ( R) R'MnR' whence RM is a Krull
- -
domain. From the equality R == n R M , where M ranges over all maximal
ideals, and from the equality RM = (I R a, where Q ranges over ideals Q
E Spec 1 ( R) contained in M, we obtain the required formula R = n RO e
To end the proof, we must show that every ideal Ra , a e R , is contained
in a finite nUDlber of ideals from Spec 1 ( R) .
Observe first that without loss of generality, we may additionally assume
that a e R. Indeed, the ring R[a] is Noetherian and its normalization is equal
to R . Thus R can be replaced with R[a] 3 a without changing the ring if to
which the assertion refers.
Suppose thus that a e R; since R is a Noetherian ring, the ideal Ra has
only a finite number of minimal prime ideals. As we have proved above,
there exist only finitely many prime ideals of R over any of the minimal prime
ideals of the ideal Ra. Thus, it is sufficient to prove that, if Q c: R is a minimal
prime ideal of Ra , then P = Qr'\R is a minimal prime ideal of Ra.
We shall reduce the latter assertion to the case of R local. Denote by I
the ideal Q( R) R'Pr'\RR""P. Thus we have
lraR = Q( R) R""pnR = Q( R) R""pn Rn R = QtiR = P,
whence, applying Theorem 1.4.7 (i) to the localization R c R p , we get
--. --
PR p = Q(R)R""pnR p = Q(Rp)nR p .
The ideal Q R p is a minimal prime ideal of Rp a, whence it is sufficient to prove
the above assertion under the additional hypothesis that R is a local ring with
the maximal ideal Qr'\R = P.
In view of the first part of the proof, concerning local rings, R is a Krull
domain; since there are only finitely many prime ideals over the maximal
ideal P, the properties of integral extensions imply (see Theorem 3.1.12)
that there are finitely many maximal ideals in R. By Theorem 4.2.4, Q is ofheight
1. By Corollary 1.1.8, there exists an element t E Q which does not belong to
any other maximal ideal of R . Write V = R[t], Ql = QnV; the ideal Ql
is maximal, and Q is the only maximal ideal of R lying over Ql. Hence, and
by Theorem 3.1.17 (going up) and the equa1ity ht(Q) = 1, we conclude that

200
Divisors and Krull Domains
[Ch.
ht(Ql) = 1. Since there is a maximal ideal of R over each maximal ideal of V,
the ring V, as a Noetherian ring, is semi-local. By Theorem 2.6.28, there exists
'" A
an isomorphism 1p: A x B -+ V, where A = V Q1 (if Q 1 is the only maximal
ideal in V, the factor B is absent). The natural topology of V induces the natural
A '"
topology on R, and thus R c: V.
The element t is integral over R, whence V is a finitely generated R-sub-
module of Ro. Thus there exists a non-zero element u E R such that uV c: R,
'" A
whence uV c R. If u is invertible, then V = R, Ql = P, whence ht(P) = 1
and the assertion holds. Suppose that u is not invertible; we then have U E P
c: Ql. The maximal ideal Ql A of A is of height 1 and u E R c: A is not a zero-
divisor in A (see Corollary 2.6.20). Hence the ideal Au is Ql A-primary, and
therefore there exists a positive integer k such that
Au ::> Q A ::) pt.
Suppose that P is not a minimal prime ideal of the ideal Ru. Then there
exists a non-zero element pEP which is not a zero-divisor on R/Ru (i.e., is
such that R/Ru!. R/Ru is a monomorphism). Since rad(Ru) = rad(Ru"),
n = 1, 2, ..., the elementp is not a zero-divisor on R/Ru" (by Corollary 2.3.21)
A
Since R is a flat R-module, it follows from the isomorphism
"" '" '"
RIRun@RR  R/Rtf
"- A
that p is not a zero-divisor on R/Ru n . Let x e P" be an arbitrary element;
A A
then x E Au and (x, 0) e 1p(Au x Bu) = Vu c R, and
'" ""
p"k1p(X,O) = 1p(pnkx, 0) E 1p(Atf+ 1 X Bu"+l) = VU,,+l C Run.
"-
Consequently 1p(X, 0) e Ru", n = 1, 2, ..., whence
"" "-
1p(x,O) E n Rtf c: n 1n n = 0, and finally x = o.
It follows that P = 0, and R is a field, the assertion being evident in this case.
Suppose now that P is a minimal prime ideal of Ru. By Theorem 2.3.15,
P = Ru: Rw for some w E R. Hence aw = bu for some b e R. It is easily seen
(cf. Exercise 6, Section 1.1) that Ru:Rv = Ra:Rb; applying again Theorem
2.3.15, we conclude that P is also a minimal prime ideal of the ideal Ra. 0
Example 4.7.3 (Nagata [N])
We shall give an example of a Noetherian domain whose normalization is not
Noetherian.
Let K be a field of characteristic 2 such that its subfield K2 c: K, which
consists of the squares of the elements from K, is identical with K (every finite
field of characteristic 2 satisfies this condition).
Denote by L the field K(X o , Xl' X 2 , ...) of rational functions in the indeter-
minates X 0, Xl , X 2 , ..., and by R' the ring of formal power series L [[Y, Z, T]]
in the indeterminates Y, Z, T. Then, write R" for the subring L2[[y, Z, T]]
of R', and denote by R the ring L2 frY, Z, T]] L2L. It is easily seen that R

IV]
Normalization of Noetherian Domains
201
can be identified with the subring of R' which consists of all series L lilk Y'ZJT: k
such that the linear L2- su bspace of L generated by all the coefficients I'Jk is
finite dimensional. We then have the inclusions R' => R ::) R". We set
00 00
b = Y LX 2n T"+Z LX 2n +1 T" eR',
n=O n=O
In the sequel, we shall prove that the subring R[b] of R' is Noetherian, but its
normalization is not Noetherian.
1. To begin with, we shall show that R[b] is a Noetherian ring. It is of
course sufficient to prove that the ring R is Noetherian.
Let us consider the ideal I of R generated by elements a1, ..., am e R.
We shall prove that R'Ir\R = I. The inclusion R'lnR ::) I is obvious. Let
a e R'Ir\R, and denote by La the subfield of L generated by L2 and the coeffi.
cients of the series a, al, ..., am. The dimension of Lo over L 2 is finite. There
exist elements r, ..., r E R' such that a = r a1 + ... +ram. Considering
a basis of Lover Lo, it is easy to verify that the series ,. , ..., ,. can be chosen
in such a way that all their coefficients belong to Lo. Consequently, r, ..., r 
E R, whence a e I, and R'Ir\R = I.
Suppose that R is not Noetherian. Then there exists an increasing sequence
11 * 1 2 $ ... of finitely generated ideals of R. For the sequence R'1 1 c: R'[2
c: ... of ideals of the Noetherian ring R', there exists no such that R'l n = R'I"
o
for n  no. Thus, for n  no, we obtain
1,1 = R'I"nR = R'I"onR = 1"0'
contrary to the assumption. Thus R is a Noetherian domain.
2. We shall prove that the normalization of the ring R[b] is the ring (R[b])of\
(1 R' .
Before we begin the proof, let us observe that char L = 2 implies
00 CIJ
b 2 = y 2 LXfnT2n+Z2 LXln+1T211.
n=O n=O
The coefficients of this series belong to L2, whence b 2 E R. It follows that
R[b] = R+Rb and (R[b])o = Ro+Rob. Furthermore by the assumption
K" = K, we have L2 = K(X5,X, ...), whence it is easily deduced that the
monomials X A = X A1 X A2 ... X AJ for A = (Al, A2, ..., AJ), 0  At < A2 < ...
< AJ, j  0, forln the basis of Lover L2. Thus the elements of R can be unique.
Iy represented as finite sum L SAX A , where SA E R". If '.1"2 e Rand
'2 :F 0, then rl/r2 = rl r2/r and ,. E R". Hence every element of the field
of fractions Ro has a unique representation as a finite sum L: tAXA' where
tA E (R")o.
Suppose that t e (R[b])onR'; then t 2 E R'2 c: R" c: R c: R[b], Mrtl.
integral over R[b). Thus we have proved that the ring (R[b])onR'(fon],?
in the normalization of R[b]. '

202
Divisors and Krull Domains
[Ch.
Suppose now that the element t = (p+qb)/s (where p, q, e R, s 'f:. 0)
of the field of fractions (R[b])o in integral over R[b] and satisfies the equation
t m +rm_ 1 t m - 1 + ... +ro = 0, "0, ..., r m - 1 e R[b].
Then (t2yn+r_1(t2)m-l+ ... +r = 0 and r, ..., r-l E (R[b])2 c: R", and
so the element t 2 is integral over R". Since t 2 = (p2+q 2 b 2 )/S2 and p2, q2,
b 2 , S2 e R", we have t 2 e (R")o, whence, by the normality of the ring R",
it follows that t 2 e R". This means that S2 divides (p+qb)2 in R", thus a fortiori
in R'. Since R' is a unique factorization domain, s divides p+qb in R', i.e.,
t e R'. We have proved the opposite inclusion, and thus also the assertion.
3. We shall prove that the normalization of R[b] is identical with the ring B
generated by R and the sequence of elements b o = b, hi, ..., where
00 00
b k = Y  X 2n T"-k+Z LX 2n +1 rn-k, k = 0, 1, 2, ."
 nk
Observe first that the following equalities are satisfied
b" = Tb"+l + YX 2 " + ZX 2k + 1 , k = 0, 1, 2, ..., (6)
and consequently all the elements b o , b 1 , b 2 , ... belong to the field (R[b])o
since T, YX 21c + ZX 2k + 1 e R. Obviously b e R, and therefore the ring B is
contained in the normalization of R[b].
In order to prove the opposite inclusion, consider an element t of (R[b])o
integral over R[b]. It follows from Part 2 that t e R' and the element t can be
represented as a finite sum
t = (l/s) L: (PA+qAb)X A ,
"eA
where PA, qA, s c: R". Hence
st = L(PA+qAb)X A ,
leA
(7)
Let m be a positive integer greater than all components AI'...' A J of
elements A e A. Since t is a series with coefficients in L, (7) is an equality of
series with coefficients in L. We shall find the coefficients (which are series
with coefficients in L2) of the monomials X).oX 2rn and XAoX2m+ l' for a fixed
Ao e A, on the right-hand side of (7). The summands p"X)., A E A, do not
contain such monomials since they contain neither X 2m nor X 2m + 1. It
follows from the definition of b that the corresponding coefficients are equal
to qJ. o YT m X 2m X).o and q).oZTmX2m+lXAo. From (7) we deduce that sfqAoYTm
and SfqA o ZT'" in the ring R". Hence it follows that SfqA o T m , and so there exist
I  0 and S1 e R" such that s = T's l and slfq" for all A EA. The formula
T ' s 1 t = L(PA+q,tb)X A and the relation sllq" imply sll LPAX,t in the ring R,
whence also slIp" in the ring R". Consequently, we obtain a representation of t
in the form t = (jJ+qb)jT ' , where p, q e R. From (6) we deduce that
b = b o = b+T'b"

IV]
NormaUzatioD of Noetherian Domains
203
where
b = Y(X o +X 2 T+ ... +X 2 '_l T '-1)+Z(X I +X 3 T+
+ ... +X2'_lT-1) e R,
whence
t = T-'(p+qb+qT'b,) = qb,+i/T', where t= p+qbNeR.
We know that t and qb, belong to the normalization of R[b]; thus the ele-
ment tIT' also belongs to the normalization of R[b]. Hence the element tIT'
N N
is a series with tbe coefficients in L. Since t E R, the coefficients of the series t
generate an L 2 -subspace of finite dimension in L. The coefficients of the series
 N
tIT' have the same property, whence tIT' E R. Thus we have proved that
t e R [b, b,] c: B, which is the required result.
4. We shall prove that the ring B is not Noetherian by showing that the
sequence of its ideals
B(b o ) c B(b o , b 1 ) c B(b o , b 1 , b 2 ) c: ...
is strictly increasing.
Suppose the contrary. Thus there exists a positive integer k such that
b" e B(b o , ..., b"-I). It follows that for some elements co, ..., Ck-l E B we
have
b" = cob o + ... +Ctc-lb"-I. (8)
Since for every j we have bJ E R, every element of B can be represented in the
form r+ L: r'l to . i ,b i1 ... b ,s ' where r, ";1".;3 E R. lIenee comparing in
Ii < ... <I,
formula (8) the coefficients at YZo which are series belonging to L[T] , we
conclude that
X2,,+X2k+2T+ ... = c(T)(Xo+X2T+ ...)+
+c(T)(X2+X4T+ ...)+ ... +C-I(T)(X2k-2+X2kT+ ...) (9)
where c(T), ..., c_l(T)EL[T] and Cj == c(T)mod(Y, Z), i = 0,1, ..., k-
-1. These congruences immediately imply that the series c(T), ..., C-1 (1)
belong also to the ring L2 [T] @ L2 L, whence their coefficients are linear
combinations of a finite number of monomials X A , with coefficients in L 2 .
Let m  k be a positive integer greater than all components of indices A.
mentioned above. Comparing coefficients of X 21n in (9), we obtain
rn- k = c(T)Tm+c'(T)TIn-l+ ... +Cl(T)T'n-k+l
for some series c;f (T), ..., C'-l (T) in L2 [[T]], which is of course impossible.
This contradiction proves that the sequence of ideals under consideration is
strictly increasing.
NOTES AND REFERENCES
Towards the end of the 19th century it was well known that many results in algebraic number
theory have their counterpart in the theory of algebraic curves. The main results of the
former stimulated attempts to transfer its methods to classes of rings occurring in algebaric

204
Divisors and Krull Domains
geometry, i.e., rings corresponding to manifolds of dimension greater than one, and, if
possible, to Noetherian rings or even larger classes of rings. Thus arose the theory of Krull
domains, initiated in [13], and the theory of divisors, which emerged in the 19308.
The divisor class group of cyclotomic fields were studied in detail by Kummer. He
demonstrated the finiteness of their orders, properties of which were of importance in the
examination of the Fermat equation. As we have mentioned, the divisor class group of a given
domain is a kind of measure of deviation from unique factorization into irreducibles. The
Theorem of Claborn (Theorem 4.6.3, see [4]) shows that even in the class of Dedekind do-
mains, so closely related to classical rings of algebraic numbers, non-uniqueness of factoriza-
tion may have a structure of any degree of complexity. The methods of classification which had
been first applied to divisors were developed in many fields and eventually turned into the
modern algebraic K-theory.
It was Dedekind who first realized the importance of the normality of a ring in number
theory; the Zariski Theorem, stating that the codimension of the set of singular points
of a normal algebraic variety is greater than 1, i.e., that this set is "smaller" than it may
be without the assumption of nonnality, pointed out the essential rOle of normal rings in
algebraic geometry. The transition from a ring to its normalization is geometrically equivalent
to the replacement of an algebraic variety by a variety, rationally equivalent to the former
one, with a smaller set of singular points. This explains the importance of the topics we
Dresent in the later sections. The fundamental Theorem 4.7.2 was proved by Mori (18] and
Nagata [19].

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[26] Steinitz, E., Algebraische Theorie der Karper, J. Reine Angew. Math. 137 (1910), 167-
309.

Index
Absolute value 150
nonarchimedean absolute value 1 SO
algebra 12
affine algebra 12
algebraic set 20
irreducible algebraic set 23
reducible algebraic set 23
algebraic variety 23
annihilator 29
Artin-Rees lemma 80
Basis of a module 31
Category of (commutative) rings 12
of graded modules 50
of graded rings 48
of R-modules 28
chain
ascending chain condition 54
composition chain 33
descending chain condition 96
refinement of a chain 33
of submodules 33
characteristic of a ring 12
Chinese remainder theorem 17
change of coefficients 29
completion of a field 150
of a module 86
of a ring 86
cokernel 28
component
homogeneous component 48
irreducible component 62
coproduct of modules 30
decomposition
irredu ndant primary decomposition 66
74
irredundant primary decomposition of an
ideal 66
primariy decompostion of an ideal 66
primary decomposition of a submodule
74
degree of a polynomial 18
direct summand of a module 31
direct sum. of modules 30
divisor
divisor of a domain 135
Cartier divisor 180
Weil divisor 181
zero-divisor 12, 30
pomain 12
completely normal domain 109
Dedekind domain 119
integrally closed domain 108
Krull domain 160
normal domain 108
principal ideal domain 13
unique factorization domain 13
Element
almost integral element 109
an inverse of an element 12
homogeneous element 48
integral element 106
invertible element 12
irreducible element 13
nilpotent element 12
epimorphism 29
extension
integral extension of rings 106
of an ideal 40
of rings 106
factor
module 28
ring 13
field of rational functions 38
filtration of a module 84
fraction 37
function

208
Index
polynomial function 24
rational function 38
"Goind up" theorem 112
"Going down" theorem 11 S
group
Abelian ordered group 139
Cartier divisor class group 180
ideal class group 128
of classes of invertible modulus 132
Picard group 132
Height of a prime ideal 27
Hilbert basis theorem 58
Hilbert's Nullstellensatz 22
hypersurface 21
homomorphism
flat homomorphism of rings 180
natural homomorphism of rings 28
of algebras 12
of graded modules 49
of graded rings 48
of modules 28
of rings 11
Ideal 13
associated isolated prime ideal 69, 78
associated prime ideal 68, 72
associated prime ideal embedded 68, 72
finitely generated ideal 13
fractional ideal 127
generated by a set 13
homogeneous ideal 49
integral ideal 127
irreducible ideal 6S
maximal ideal 14
minimal prime ideal of an ideal 69
P-primary ideal 64
primary ideal 63
prime ideal 14
principal ideal 13
idempotent 27
image of a homomorphim
of modules 28
of rings 13
injection 13, 28
integral closure 108
inverse of an element 12
isomorphism
of modules 28
of rings 12
JordanH6lder theorem 34
Kernel
of a homomorphism of modules 28
of a homomorphism of rings 13
Krull intersection theorem 82
Lemma
Artin-Rees lemma 80
Nakayama lemma 33
length
of a chain 27, 33
of a module 34
limit of an inverse system 88
localization
of a module 43
of a ring 38
Maximum condition S4
minimum condition 96
Module 28
Artin module 97
complete module 86
divisible module 134
factor module 28
flat module 44
free module 31
graded module 44
injective module 32
invertible module 119, 130
Noetherian module 5S
of fractions 43
projective module 32
semi-simple module 36
simple module 29
S-invertible module 132
topological module 84
torsion-free module 134
torsion module 134
monomorphism 28
multiplicative subset of a dng 15
Nakayama lemma 33
nilradical 14
norm 1 SO
normalization of a ring 108
Polynomial function 24
product
of a family of modules 31
of a family of rings 17
of ideals 13
projection 31

quasi-compact space 26
quotient of ideals 14
Radical
Jacobson radical 15
of an ideal 14
ramification index of a valuation 148
rank of a free module 32, 136
rational function 38
rafinement of a chain 33
restriction of an ideal 40
ring 11
Artiti ring 97
commutative ring 11
discrete valuation ring 146
factor ring 13
graded ring 48
graded ring associated with an ideal
local ring 57
local ring of a point 39
local ring of a subvariety 39
Noetherian ring S5
of algebraic integers 110
of a valuation 140
of coefficients 28
of formal power series 18
of fractions 38
of p-adic integers 151
of polynomial functions 24
of polynomials 18
quasi-local ring 39
reduced ring 13
semi-local ring 50
topological ring 85
total ring of fractions 38
valuation ring 141
Saturation of a submodule 77
Schreier theorem 33
Index
so
semi-group od divisors 157
semi-group of divisor classes
sequence
Cauchy sequence 85
exact sequence 29
sequence splits 31
short exact sequence 29
spectrum
maximal spectrum 25
prime spectrum 25
submodule 28
cyclic submodule 29
finitely generated submodule
generated by a set 29
primary submodule 74
subring 12
support of a module 79
surjection 13, 29
Topology
I-adic topology
linear topology
natural topology
Zariski topology
torsion submodule
trace 116
91
84
95
20, 26
125
Unit 12
unity 11
Valuation 139
discrete valuation 145
equivalent valuations 141
p-adic valuation 139
ring 141
ring of a valuation 140
Zero-divisor 12, 30
209
157
29

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