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Sfanisaw Balcerzyk and T adeusz J6zefiak
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Ellis Horwood Series in
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COMMUTATIVE RINGS: Dimension, MultiplicitY
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STANISlAW BALCERZYK and TADEUSZ JOZEFIAK,
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COMMUTATIVE RINGS
DIMENSION, MULTIPLICITY AND HOMOLOGICAL METHODS
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Commutative Rings
Dimension, Multiplicity
and Homological Methods
STANISLAW BALCERZYK
Professor, Institute of Mathematics
Polish Academy of Sciences
TADEUSZ J6ZEFIAK
Professor, Institute of Mathematics
Polish Acadelny of Sciences
Translation Editor
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1. Commutative rings
I. Title II. J6zefiak, Tadeusz
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Table of Contents
Preface
Chapter I - DIMENSION AND MULTIPLICITY
1.1 The Dimension of Affine Algebras
1.2 The Krull Dimension. . .. ... ....
1.3 Polynomial Rings . . . . . . . . . . . . . .
1.4 Sets of Parameters . . . . . . . .. ...
1.5 Hilbert-Samuel Polynomials and Multiplicity . . . .
1.6 Intersection Multiplicity of a Pair of Modules; Geometric
Applications . . . . . . . . . . . . . . . . . . . . .
Chapter II - REGULAR LOCAL RINGS
2.1 Homological Characterization . . . . . . .
2.2 Uniqueness of Factorization. . . .
2.3 Structure of Complete Local Rings . . . .
Chapter III - COHEN-MACAULAY RINGS ....
3.1 Regular Sequences and the Depth of a Module.
3.2 Regular Ideals . . . . . . . . . . . . . .
3.3 Characterizations of Cohen-Macaulay Rings
3.4 Basic Properties of Cohen-Macaulay Rings . . .
3.5 Perfect Ideals . . . . . . . . . . . .
3.6 Structure of Perfect Ideals of Depth 2 .
Chapter IV - GORENSTEIN RINGS ......
4.1 The Type of Local Cohen-Macaulay Rings.
4.2 Zero-dimensional Gorenstein Rings ..........
4.3 Homological Characterization and Properties of Arbitrary
Gorenstein Rings. . .. .............
4.4 Gorenstein Ideals ..................
Appendix - HOMOLOGICAL BACKGROUND
A.I Projective and Injective Dimension
A.2 Derived Functors . . . . . . . . . .
VII
1
1
12
26
34
46
60
79
79
81
85
101
103
107
113
118
124
132
137
137
142
]46
151
157
157
160

VI
Contents
A.3 Functors Ext" and Torn .............. 164
A.4 Injective, Projective and Global Dimension in Terlns of the
Functors Ext and Tor: . . . . . . . . . . . . . .. 168
A.5 Injective, Projective and Global Ditnension over Local Rings 171
A.6 The Koszul Complex. . . . . . . .. ....... 175
References . . . . . . . . 179
Index of Theorems and Definitions quoted frool COlnmutative Noet/ler-
ian and Krull Rings ............ . .. 183
Index
191

Preface
This book constitutes, with minor changes, a translation of the last four
chapters of our book Commutative Rings (Pierscienie Przemienne)
published in Polish in 1985. The first four chapters of the book have recently
appeared in English in this series under the title Commutative Noetherian
and Krull Rings. The two books, treated as a whole, form a self-contained
presentation of the fundamentals of the theory of commutative rings.
For the convenience of the reader we enclose at the end of this book
the list of all results from the first book which are cited in the present
volume.
The COlnn1on feature of all the chapters of the present book is a wide
application of homological algebra to purely ring-theoretic problems. The
most illustrious example is the homological characterization of regular
local rings as local rings of finite global dimension in Chapter II. As a simple
consequence one gets the proof of the fact that a localization of a regular
local ring is again regular. It had been an open problem for many years
until it was solved by homological methods in the mid-fifties. To nanle
some other examples of the impact of these methods on the theory of com-
mutative rings we mention a description of intersection multiplicity in
terms of Tor functors in Chapter I, a characterization of the depth of an
ideal-a very useful and effective invariant of an ideal-in terms of Ext
functors in Chapter III, and a characterization of local Gorenstein rings
as those of finite injective dimension in Chapter IV.
For the convenience of the reader we summarize those basic facts
from hOlnological algebra needed in this book in the Appendix, and pro-
vide detailed references for all unproved theorems in D. G. Northcott's
books [8], [U]. The only exception concerns spectral sequences which
we use twice in Section 1.6. The necessary background can be found, for
instance, in S. MacLane's book [P].
The list of contents gives a fairly detailed picture of the main topics
treated, but a few additional comments may be of some help.
We start in Chapter I with two important notions of geometric ori-
gin-dimension and multiplicity. The dimension is defined first for an
affine algebra which is a domain as the transcendence degree of its field

VIII
Preface
of fractions, and then extended to an arbitrary Noetherian ring using
Krull's characterization in terms of chains of prime ideals. In Chapter I
we also discuss the related notion of the height of an ideal, and properties
of dimension and height in polynomial rings (Section 1.3) and local rings
(Section 1.4). In Section 1.4 we introduce sets of paran1eters and use them
to define the important class of regular local rings. We also indicate a con-
nection between regular local rings and simple points of algebraic varieties.
The Hilbert-Samuel polynomial, studied in Section 1.5, allows us to give
one more interpretation of dimension, as well as to introduce the multi-
plicity of a module with respect to an ideal. The homological description
of multiplicity leads to a more general notion of the intersection multiplicity
of any pair of modules over a regular local ring. We study this notion in
Section 1.6 where we also give the necessary algebraic background for a
construction of the Chow ring of an algebraic variety. III this section we
follow closely Serre's lecture notes [X].
Chapter II is devoted entirely to regular local rings. In Section 2.1 we
present the homological characterization already mentioned, in Section 2.2
we prove that regular local rings are unique factorization domains, and
Section 2.3 contains the Cohen structure theorem for complete regular
local rings. In Section 2.3 we follow Nagata's exposition in [R].
Chapters III and IV are concerned with two classes of rings (both con-
taining the class of regular local rings): Cohen-Macaulay and Gorellstein
rings, respectively. Both notions are defined in purely ring-theoretic terms.
The definition of a Cohen-Macaulay ring has a geometric flavour and is
motivated by an old Theorem 1.3.7 of Macaulay, whereas local Gorenstein
rings are distinguished among Cohen-Macaulay rings by the property
that any ideal generated by a set of parameters is irreducible. Chapters III
and IV bring together various properties and characterizations of the rings
in question. The most important tool in studying them is the notion of
a regular sequence and the related concept of the depth of a module
(Section 3.1).
Exercises are included at the end of almost every section, the more
difficult being furnished with hints in brackets [ ]. With few exceptions
the material contained in the exercises is not used in the text.
Each chapter is followed by short Notes and References, in which
we make a few comments and try to trace the methods and theorems of
the chapter to their sources.
Among numerous books in English devoted to commutative rings
(some of them listed itl the Bibliography at the end of the book) we should
mention here I. Kaplansky's Commutative Rings and M. Nagata's Local
Rings which were very helpful to us when preparing this book.
We wish to express our gratitude to Professor Andrzej Bialynicki-Birula
for the encouragement to write the book, for detailed examination of an
early version of the manuscript and for his helpful criticism. We are indebted

Preface
IX
to Dr. Andrzej Pr6szynski who made accessible to us his notes from the
lectures by one of the authors. They were very useful in writing C.hap-
ter II. We are also grateful to Dr. Jerzy Weyman for the permission to in-
clude his Lemma 3.3.3 and for the preparation of the proof of Theoreltl
4.4.7.
Stanislaw Balcerzyk
Tadeusz Jozefiak

Chapter I
Dimension and Multiplicity
In algebraic geometry there exists a fundamental correspondence which
associates with an algebraic set V c Kn (i.e., the set of solutions in K n of
a system of polynomial equations in n indeterminates over a field K) a K-al-
gebra K[V] of polynomial functions on V. Any such K[V] is a finitely
generated K-algebra: hence it is also a Noetherian ring (as a homonlorphic
image of a ring of polynomials K[X 1 , ..., X n ]). It is possible to transfer
to the class of Noetherian rings many classical theorems proved in their
original versions for algebras of the form K[V]. The advantage of this gen-
eralization lies not only in the extension of the domain of validity of the
theorems, but also in a radical simplification of the proofs. We are thus faced
with the natural problem of transferring the largest possible range of
concepts and theorems of a geometric and algebraic character to the theory
of Noetherian rings. We shall see that some intuitive geometric notions
(e.g., the non-singularity of a point, or the multiplicity of a point on
a subvariety) can be defined precisely only in terms of the theory of rings.
Our principal aim in this chapter is to define and investigate the notions
of dimension and multiplicity for a Noetherian ring. These t\VO notions
will be introduced in different ways: we shall first define dimension for
finitely generated algebras in terms of transcendence degree, and then
prove those properties of dimension which will permit us to give, in a quite
natural way, a new definition in terms of chains of prime ideals for any
commutative ring. The concept of multiplicity will be presented at once
in the full generality for modules over local rings.
1.1 THE DIMENSION O:F AFFINE ALGEBRAS
We begin by recalling some fundamental concepts of algebraic geometry
which will allow us to exhibit geometric aspects of the definition of Krull
dimension. A slightly more detailed exposition is to be found in [B].
Let K be an arbitrary field. Any ideal I of the ring of polynomials
K[X 1 , ..., Xn] determines an (affine) algebraic set V(I) c K!', which consists
of all elements (t 1, ..., I,,) E K" such that

2
Dimension and Multiplicity
[Ch,
f(t 1 , ..., t ll ) = 0 for all fE I.
One can assume that this equation is satisfied only for the elements of any
finite set 11' ... ,I p of generators of I, Le., V(l) = V((fl»)n ... n V ((fp) ).
We know that I admits a finite set of generators, because K[X 1 , ..., I]
is a Noetherian ring (see [B], Corollary 2.2.2).
Any subset E c: K n determines an ideal ICE) of the ring K[X 1 , ..., X n ],
which consists of all polynomials which take zero value at all points of E.
An algebraic set W is called irreducible (or an algebraic variety) if it is
not a union of two algebraic sets distinct from W. It is shown in Defini-
tion 1.2.7 of [B] that an algebraic set W is an algebraic variety if and only
if the ideal I(W) is prime, or equivalently W = V(P) for some prime
ideal P. Any algebraic set Wadmits a presentation W = WI U ... u W p as a
union of algebraic varieties WI' ... , W, which satisfy conditions Wi t U W j
):1=;
for all i (Le., no component can be omitted); the varieties WI'. to, W p
are uniquely determined and we call them irreducible components of the
algebraic set W (see Theorem 2.3.1 in [B]). If W = Vel) then Wi = V(P i ),
i = 1, ..., p, where PI' ..., P p are minimal prime ideals of the ideal I.
Any polynomial g e K[X 1 , ..., Xn] determines a polynomial function
g': Kn -)- K defined by the formula g'(t 1 , ..., tn) = g(t 1 , ..., tn) for t 1 , ...
. . . ,t n e K (distinct polynomials may determine the saine polynomial function,
e.g., in the case when K is a finite field). A restriction g of g' to an algebraic
set W is called a polynomial function defined on W. All polynomial functions
defined on W form a K-algebra K[W] and we call it the K-algebra of poly-
nomial functions of the algebraic set W (or the K-a1gebra of affine coordi-
nates of W). Since I( W) is the ideal of all polynomials in K[X 1 , ..., X,,]
which vanish on Wthen K[W] = K[X 1 , ...,XnJ/I(W). If Vis an algebraic
variety and V = V(P) for some prime ideal P then the algebra K[V]
= K[X 1 , ..., X,,]/ P is a domain; the converse implication also holds.
Let a = (al, ..., an) be a point of an algebraic variety V c: K". Let us
denote by ma the maximal ideal of a K-algebra K[V] generated by residue
classes of elements Xl - ai' ..., X" - an. It is easy to see that the localization
of K[V] (see Section 1.4 of [B])
()(a, V) = (K[V)m.
which is called the local ring of the point a oo'the algebraic variety V, consists
of an elements IlK of the field of fractions of K[V] such that g(o) 1= 0;
any such element f(g takes a definite value [(a){g(a) at the point a. The
maximal ideal m(a, V) of the ring (a, V) consists of all elements which
take zero value at a.
Let U be the subvariety of the aJgebraic variety V = V(P) determined
by the prime ideal P' * P and let Q be the prime ideal P' / P of K[V]; the
local ring

I]
The Dimension of Affme Algebras
3
(!J(U, V) = (K[V])Q
is called the local ring of the subvariety U on the algebraic variety V.
Now,ve pass to the definition ofa dimension of a K-algebra. Algebraic
varieties are studied mainly by means of their algebras of polynomial
functions. If we want to model the definition of the dimension of an alge-
braic variety on that of a differentiable (or topological) manifold, we should
express its dimension in terms of the algebra of differentiable functions
defined on it. Every point of n-dimensional, connected, real differentiable
manifold has a neighbourhood isomorphic with the Euclidean space Rn;
the differentiable functions defined on such a neighbourhood form an
algebra which is isomorphic to the algebra All of differentiable functions
in n independent variables over R. The number n can thus be defined as the
maximal number m such that Am is a subalgebra of functions which are
differentiable in a certain neighbourhood of any point of the manifold.
For an algebraic variety V of dimension n defined over a field K, the
condition imposed on the algebra K[V] of polynomial functions is analogous
to that described above, namely that in the algebra K[V] there exist n
functions algebraically independent over K, the number n being maximal
with respect to this property. In the sequel, we shall adopt this number as
the dimension of the algebra K[V]. The dimension will be also described
in terms of the field of rational functions K(V). To this end, let us recall
that if a field L is an extension of a field K then there exists a maximal
subset BeL algebraically independent over K. The cardinality of B
depends only on the extension K c L (see [N], p. 254); it is called the
transcendence degree of the field Lover K and denoted by trdegKL. The
extension K c L is algebraic if and only if trdegKL = O. If L is finitely
generated over K, then trdegKL < 00 ([N], p. 254).
In these terms, we may say that the dimension of a variety V is equal to
trdegKK(V).
These remarks provide a geometric and algebraic motivation for the
definition of the dimension of an algebra over a field. We shall begin this
section with that definition. We shall then study the properties of dimension.
The most significant result in this respect is the description of the dimension
of an algebra in terms of chains of prime ideals; it permits us to extend the
definition of dimension from the class of algebras to the class of all rings.
Unfortunately not all theorems concerning the dimension of algebras can
be extended to rings.
Throughout this section K will denote a field, and the K-algebras under
consideration will (usually) be finitely generated over K, Le., they will be
homomorphic images of K-algebras of polynomials in a finite number
of indeterminates. Such algebras will be called affine because of their
connection (described in [B], Chap. I) with affine algebraic sets. Affine
algebras are Noetherian rings.

4
Dimension and Multiplicity
[Ch.
Definition 1.1.1
IJet a domain A be a K-algebra, and let L be the field of fractions of A.
The dimension dinlA of the algebra A is the number trdegKL.
Example 1.1.2
It follows directly from the definition that the dimension of the K-algebra
of polynomials K[X 1 , ..., Xn] is equal to n.
Example 1.1.3
If K and L are fields and A is both a K-algebra and an L-algebra, then the
dimension of A as a K-algebra may be different from its dimension as an
L-algebra. A simple example is provided by the algebra A = K(X) [Y],
where X, Yare algebraically independent over K. When L = K(X), the
dimension of A as a K-algebra is equal to 2, whereas the dimension of A
as an L-algebra is equal to 1. Let us observe that A is not an affine K-algebra,
this, however, is not required in Definition 1.1.1.
Definition 1.1.1 cannot be automatically extended to algebras with
zero-divisors. However also in this case a geometric interpretation suggests
the proper definition.
When V 1 , ..., V s are the irreducible components of an algebraic set V,
our geometric intuition suggests that we should define the dimension of V
as the maximum of the dimensions of the varieties V 1 , ..., V s . If K[V] is
the algebra of polynomial functions on V and the prime ideals P l' ..., P s
in K[V] correspond to the varieties V 1 , ..., V, then P 1, ..., P s are prime
ideals associated with the zero ideal in K[V] and the foregoing geometrical
interpretation leads to
dimK[V] = maxdimK[V]/P i .
I
In the above equality we may of course confine ourselves to minimal prime
ideals of the algebra K[V]. Thus, we adopt the foJIowing definition:
Definition 1.1.4
Let A be a K-algebra, and let {Pi} denote the set. of minimal prime ideals
of A. The dimension dimA of the algebra A, is the number
supdimA/P i .
i
Remark 1.1.5
If A is an affine K-algebra, then it is also a Noetherian ring; hence, by [B
Definition 2.4.20, the set of minimal prime ideals in A is finite.
The following theorem will enable us to investigate the relation between
the dimension of an algebra and the lengths of the chains of prime ideals
which it contains:

I]
The Dimension of Affine Algebras
5
Normalization Theorem 1.1.6 (E. Noether)
If K is a field and A is an affine K-algebra, then there exist elelnents Z 1, ...
. . . , Zd E A algebraically independent over K such that A is an integral
extension of the algebra K[Z1' ..., Zd].
Proof
Consider the family d consisting of all finite subsets {U1, ..., u,} of A such
that A is an integral extension of the algebra K[U1' ..., u,]. The family d
is non-empty since the algebra A is finitely generated over K. Let {z 1, ..., Zd}
be an element of d of minimal cardinality. To end the proof, we shall
show that the elements Z 1, ..., Zd are algebraically independen.t over K.
Suppose that, on the contrary, there exists a polynomial f in d indeter-
minates such that f(z 1, ..., Zd) = o.
Take a positive integer e greater than any of the exponents of the powers
occurring in f and set z; = Z, - Zi -1, i = 2, 3, ..., d. Mter the substitu-
tion Zj = zi + z1 i -1, the monomial az 1 ... Zd takes the form
az n 1 +"2 e + ...+nd ed - 1 +g(z Z ' Z ' )
1 1,2,.", d,
the degree of the polynomial g with respect to Zl being less than nl +n2e+
+ 4-1
· . · nd e .
We order the sequences of exponents of monomials occurring in f lexi-
cographically: (n1, ...,nd) < (m1, ...,md) if and only if nj < mj, where
i is the least of the numbers 1, ..., d for which nj '1= mi. Since the nwnber
e has been chosen greater than nJ, mJ, j = 1, ..., d, we have (n1, ..., nd)
< (m1, ..., md) if and only if
n1 +n2e+ ... +nd ed - 1 < nt 1 -I-m2e+ ... +n1ded-l.
Among the sequences of exponents of the monomials occurring in f
there exists a greatest one (under the lexicographical order), from which,
after expressing in terms of Z 1 , Z, ..., z, \ve can isolate the monomial
bzf so that the equation f(z 1, ..., Zd) = 0 takes the form
bz1+h(Zl'Z, ...,z) = 0,
where the cofficient b is a nonzero element of K and the degree of the poly-
nomial h with respect to Z1 is less than N.
This yields the integral dependence of Z 1 over the ring K[z, ..., z].
Applying Corollary 3.1.9 from [B] to the sequence of ring extensions
K[z, ..., z] c: K[Z1, ..., Zd] = K[Z1' z, ..., z] c A,
we infer that A is an integral extension of the algebra K[z, ..., Zd], i.e.,
that the set {z, ..., z} of d -1 elements belongs to the family d, contrary
to the choice of the number d. This ends the proof. 0

6
Dimension and Multiplicity
[Ch.
Theorem 1.1.7
If A is an affine K-algebra, then dimA is equal to the maximum length
of the chains of prime ideals in A (for the notion of. length of a chain of
prime ideals, see the beginning of Section 1.2).
Proof
Denote by d(A) the maximum length of the chains of prime ideals in the
algebra A. Then d(A) = maxd(A/P), where P ranges over the prime ideals
of A. Since Definition 1.1.4 implies directly that dim A = maxdim(A/P),
the proof reduces to the case of algebras which are domains.
Thus suppose that the algebra A is a domain. By the Normalization
Theorem 1.1.6, there exists in A a subalgebra B, isomorphic to the algebra
of polynomials K[X 1 , ..., X d ] and such that the extension B c: A is integral.
Since the fields of fractions of the algebras B and A form an algebraic
extension, we have d = dimA. Let us now compare the numbers d(A) and
deB). By Theorem 3.1.13 in [B], for any chain
Poc Ptc ...c Ps
of prime ideals of A, the sequence
PonBc PtnBc ...c PanB
is also a chain of prilne ideals of A; hence d(A)  deB). Theorem 3.1.17
in [B] (going up) yields the opposite inequality, whence d(A) = deB).
In Section 1.3 we shall show, independently of the considerations of
this section, that deB) = d (Corollary 1.3.5).
Finaly, we get
dimA = d = deB) = d(A).
o
The preceding theorem enables us to extend the concept of dimension
to any commutative ring. We shall do so in the next section. Now we give
the more important corollaries to Theorem 1.1.7.
Corollary 1.1.8
If an affine K-algebra A is a field, then the field extension K c A is algebraic.
Proof
Since the zero ideal is the only proper prime ideal of A, we have by Theorem
1.1.7, tr degx A = O. 0
Directly from Corollary 1.1.8 we obtain
Corollary 1.1.9
If m is a maximal ideal of an affine K-algebra A, then the field Aim is an
algebraic extension of the field K.

I]
The Dimension of Affine Algebras
7
Corollary 1.1.10
If K is an algebraically closed field, then any maximal ideal of the poly-
nomial ring K[X l , ..., X,,] is of the form (Xl - eXl , X 2 - OC2, ..., X n - Ct n ) for
some c'X 1, ..., Ct ll E K.
Proof
It follows from Corollary 1.1.9 that if m is a tnaximal ideal in K[X t , ..., XII]'
then K[X l , ..., X,,]/m = K since the field K has no proper algebraic exten-
sions. Thus, there exist elements (Xl' ..., OC" E K such that Xl - (Xl' ...
..., XII-CX" Em, Le., m = (Xl -OC I , ..., Xn-Ctn). 0
Corollary 1.1.11
If K is an algebraically closed field and I is a proper ideal in K[X l , ..., Kn]
then the set V(I) is non-empty.
Proof
The proper ideal I is contained in some Dlaximal ideal which, by Corollary
1.1.10, has the form (Xl - OCl, ..., X n - an), where al,..., an E K. Hence
V(I) contains the point (a I , ..., an). 0
The next result is known as Hilbert's Nullstellensatz.
Corollary 1.1.12
If K is an algebraically closed field and a polynonlial f E K[X l , ..., Xn]
vanishes at all the points of the algebraic set V(I) determined by an ideal
I c K[X 1 , ..., X n ], then f belongs to the radical of I.
Proof
Suppose I = (gl, ..., Cm), and denote by J the ideal of the polynomial
ring A = K[X l , ..., X n , X n + 1] generated by gl, ..., gm, 1- fX II + 1. We
claim that J = A. Indeed, if J were a proper ideal, then, by Corollary
1.1.11, there would exist a common zero of the polynomials g 1, ..., gm,
1-.fX n + 1. This leads to a contradiction, since by the assumption, this
con1mon zero would be a zero of the polynomial f Since J = A, there
exist polynomials 11 1, ..., /1"., h E A such that
hI gt -t- ... +hmg". +11(1- !X n + 1 ) = 1.
(1)
Denote by oc: A  K(X 1 , ..., X n ) a homomorphisll1 of K-algebras such
that a(X;) = X, for i = 1, ... n, and a(.+ 1) = 1 If (we exclude the trivial
case off = 0). Under the homomorphism a, equation (1) yields
h 1 (Xl' ..., " 1 If) g 1 + ... + hili (X t , ..., X n , 1 If) g m = 1.
Multiplying both sides of the above equation by a sufficiently high power
off, we get thatfbelongs to the radical of I. 0

8
Dimension and Multiplicity
[Ch.
We aim now at a more precise examination of the lengths of chains of
prime ideals in affine algebras. To this end we shall need a stronger version
of the Normalization Theorem. However, we shall begin with an auxiliary
lemma.
Lemma 1.1.13
For every polynomial f E A = K[X 1 , ..., Xn] of positive degree, there exist
elements 12, ...,f" E A such that the extension K[f,f2' ... ,.h,] c A is
integral.
Proof
As in the proof of the Normalization Theorem by substituting Zi
= X,-Xr'- l , i  2, for a sufficiently large number e we can \vritefin the
form
f(X 1 , ..., X n ) = bXr + g(X 1 , Z2, ..., Zn),
where b is a non-zero element of K, and the degree of g with respect to Xl
is less than N. Write A' = K[f, Z2, ..., Zn]. In view of the above, A' c A
is an integral ring extension, hence Z2, ..., Zn are the required elements. D
Theorem 1.1.14
Let A be an affine K-algebra, and let / be an ideal of A. If dimA = d,
then there exists a sequence Z 1, ..., z" e A of elements algebraically inde-
pendent over K such that
(i) the extension K[z 1, ..., z,,] c: A is integral,
(ii) If1K[Zl' ..., Zd] = (Zl, ..., zp) for some p.
Proof
Consider the family  of d-element subsets {Ul"'" u,,} of A satisfying
condition (i). By the Normalization Theorem 1.1.6 the family  is not
empty. Ifu = {Ul, ..., Ud} E we may suppose that Ul, ..., up e/, U p +l, ...
... , Uti rp /, for some p. In this way we have associated with every U E  the
number p = p(u). Select in  an element Z = {z 1, ..., ztJ} for which
p = p(z) is maximal. We claim that /f1K[Zl' ..., Zd] = (Zl' ..., zp).
Write B = K[Zl' ..., Zd]; we have
(/flB)/(Zl' ..., zp)Bc: B/(Zl, ..., zp)B = K[ZP+l' ..., Zd]'
If the ideal on the left-hand side is non-zero, then there exists a polynomial
rEI nB, f '(Z 1, ..., Z p) B. Denote by f the residue class it determines in
K[ZP+l' ..., Zd]. By virtue of Lemma 1.1.13, there exist polynomials
h+2, ...,jd eBsuch that the extension K[ f,h +2' ..., Jd ] c K[ZP+l' ,,,,Zd]
is integral. Hence the extension K[Zl' ..., zp,f,fp+2' ... ,Jd] c B is also
integral Le., y = {Zl, ..., zl1,l,h+2, ... ,fd} E CC. Since IE I, we have
p(y) > p(z), contrary to the choice of z. This contradiction ends the proof. 0

I]
The Dimension of Affine Algebras
9
To state an important property of affine algebras we recall that a chain
of prime ideals
Poc P 1 c ...c P d
is said to be saturated if Pi c Q c: P i + 1 implies Q = Pi or Q = P i + 1
for a prime ideal Q and any i = 0, 1, ..., d-l.
Theorem 1.1.15
Let a domain A be an affine K-algebra and let P be a prime ideal in A.
Then the length of any saturated chain of prime ideals of A, beginning
at (0) and ending at P is equal to dimA-dim(A/P).
Proof
It is sufficient to prove that if P is a minimal non-zero prime ideal of A,
then dimA = dim(A/ P) + 1. Indeed, let (0) c: P 1 c: ... c: Ps = P be
a saturated chain of prime ideals of A between (0) and P; applying the
above assertion to the ideal PI we obtain dimA = dimA/P 1 + 1. Applying
the same assertion to the algebra AlP 1 and the minimal ideal P 2/ P 1,
we obtain
dimAIP t = dimAIP 2 + 1.
Proceeding this way, we arrive at the identity
dimAIP s - 1 = dimA/Ps+ 1.
Summing these equalities, we obtain the formula or the theorem. To estab-
lish the statement formulated at the beginning of the proof we apply
Theorem 1.1.14 to the algebra A and the ideal P. If d = dimA, then there
exist elements Z 1, ..., ZtJ E A such that the extension B = K[z 1, ..., Zd] c: A
is integral and Pf1B = (z 1, ..., zp) for some p  d. Since P is a minimal
prime ideal in A, it follows from Theorem 3.2.4 in [B] (going down), that
p = 1. Moreover, the extension B/(BnP) c: A/P is integral, and Bj(Bf1P)
 K[Z2' ..., Zd] is a ring of polynomials in d-l variables. Hence
dimA/P = dimB/(BnP) = d-l = dim A-I.
D
Corollary 1.1.16
If a domain A is an affine K-algebra, then every saturated chain of prime
ideals in A has the same length, equal to dimA.
Example 1.1.17
There exist affine algebras (with zero-divisors) for which the conclusion
of Corollary 1.1.16 is not valid. An example is furnished by the algebra
A = K[X, Y, Z]j(XY, XZ). If we denote by x, y, Z the residue classes of
the corresponding indeterminates X, Y, Z in the algebra A, then the two
chains of prime ideals (x) c: (x, y) c: (x, y, z) and (y, z) c: (x, y, z) are
both saturated, but of different lengths. The simple verification is left to
the reader.

10
Dimension and Multiplicity
[Ch.
Corollary 1.1.18
If A is an affine K-algebra, and .P, Q are two prinle ideals in A such that
p c: Q, then all saturated chains of prime ideals between P and Q have
the same length equal to dimA/P-dimA/Q.
The proof follows froln Corollary 1.1.16, applied to the affine algebras
AlP and AIQ. 0
Remark 1.1.19
The conclusion of Theorem 1.1.15 does not hold for arbitrary Noetherian
domains (see Example 1.2.22). Nevertheless, the upper bound of the lengths
of descending chains of prime ideals beginning at P is a very important
invariant of the ideal P, called its height. The height of an ideal will be
studied more closely in the next section.
We shall now give another version, a homogeneous one, of the Normal-
ization Theorem which will be employed in Section 1.4.
Theorem 1.1.20
co
If a domain A = E9 A" is a graded K-algebra generated by a finite number
n=O
of elements of degree 1, Ao = K and Xl is a non-zero honlogeneous element
of positive degree, then there exist homogeneous elements X2, ..., Xd of
positive degrees such that the algebra A is an integral extension of the
algebra K[x 1, ..., Xd], where d = dim A.
Proof
We shall find homogeneous elements x 2, ..., Xd such that the ideals 10 = 0,
/ 1 = (Xl), ..., 14 = (x t, ..., Xd) satisfy the condition
dimA/Io > dimA/lt > ... > dimAll d ,
and then we shall show that A is an integral extension of the algebra
K[x 1, ..., x,,].
It follows from Theorem 1.1.7 that dimA/lo > dimAll t , since / 1 i= O.
Suppose that we have constructed elements X2, ..., Xi, d > i  1, which
satisfy the above conditions. Denote by P t, ..., P, the Ininimal prime
ideals of the ideal Ii. By Corollary 1.5.9 in [B], the PJ are homogeneous.
As the intersection P 1 () ... () p.t; is irredundant, there exist homogeneous
elements Zt rp P l , ..., Zs ft Ps such that Z1 E P2() ... ()ps, ..., Zs E Pt n ...
· · · ()p s - t · Replacing the elements Z t, ..., Z s by suitable powers of themselves
we may assume that they have the same degree. Setting Xi+1 = Zt + ... +z"
we get Xi+ t  P 1, ..., Ps. Furthermore, if P is a minimal prime ideal of
the ideal /1+1 = (Xl' ..., Xi+t), then P :/= Pi' ..., Ps and P ::> PJ for some
j, 1  j  s. Hence dimAlli+l < dim A/I;.

I]
The Dimension of Affine AJgebras
11
Clearly dimA/ld = 0, and thus the minimal prime ideals of Id are
homogeneous maximal ideals. The only such ideal is J = (f) An (see
n>O
Theorem 1.5.9 in [B]), hence Id is J-primary, and therefore Id :::> Jk for
some positive integer k.
Write B = K[x 1, ..., Xd], and let Y 1, ..., y, be homogeneous generators
of degree 1 of the K-algebra A. Denote also by 'YJo = 1, 'YJ1, ..., 'YJq all the
monomials in YJ, ...,y, of degree < k. We shall show inductively that
All c C, where C = B'YJo + ... + B'YJq. Clearly Ao, ..., A k - 1 c: C; suppose
that Ao, ..., An-1 c C for some 1l, n  k. Let'YJ be a monomial in Yl, ..., )'r,
of degree n. Then 'YJ EJ n c Jk c I d , and 'YJ = r1x1+ ... +r"xd for some
r 1, ..., r d EA. Since the elements 'YJ, x 1, ..., Xd are homogeneous, we can
assume the same of r 1, ... , rd. The degrees of r 1, ..., r d do not exceed n - 1 ,
hence r1, ..., rd E C. Since C is a B-nl0dule and Xl' .oo, Xd E B, we have
'I} E C. Thus we have proved that A = B'YJo + ... + B1'Jq. Hence, by Theorem
3.1.6 in [B], A is an integral extension of B. D
Exercises
1. Is the localization of an affine algebra with respect to a prime ideal always an
affine algebra?
2. Let A c: B be an integral extension of K-algebras. Prove that if B is an affine
K-aIgebra, so is the K-algebra A. [Write b 1 , ..., b s for the generators of the K-algebra B
and consider an affine subalgebra C of A, generated by the coefficients occurring
in integral dependence relations of b 1 , ..., b , over A.]
3. Let G be a finite group of automorphisms of an affine X-algebra B. Prove that
the algebra B G = {b e B: g (b) = b for all g E G} is also an affine K-algebra. [Deduce
from the identity II (b-g(b» = 0 that an element b E B is integral over B G .]
geG
Let A, B be K-algebras; the tensor product A @xB becolnes a K-algebra when we
define the multiplication on generators by (a@b) (al@b 1 ) == aal@bb h where 0, a1 E A,
b, b 1 e B.
4. Prove that
(i) "if L * K is an algebraic field extension, then the K-algebra L@ xL has zero-
divisors. [If ct e L""K is an element algebraic over K, then «(8)1-1 Ot is a zero-divisor.]
(ii) if K is a field of characteristic p > 0, and the element a E X is not the p-th power
of an element of K, then the K-algebra K(ya)@xK(ya) has non-trivial nilpotent
elements,
(Hi) if K(Ot) ::> K is a separable algebraic field extension, and L ::> K is an arbitrary
field extension, then the K-algebra K(Ot)@xL has no non-trivial nilpotent elements.
5. Prove that if a field K is algebraically closed and the domains A, Bare K-algebras,
then A@xB is also a domain. To this end, prove the foHowing statements:
(i) Without loss of generality, we can additionally assume that A, B are finitely
generated algebras.
(H) If a field L is algebraically closed in a field F (i.e., the only algebraic elements
of F over L are those of L), then the field L(X 1 , ..., X n ) is algebraically closed in F(X 1 , ...
... J X n ). [Prove first that, if u, v e F[X 1 ] are monic polynomials, and uv E L[X 1 ], then
U, v e L[X 1 ]; to this end, observe that the zeros of the polynomials u, v are algebraic
over L. Next, show tbat if w E F(X 1 ) is algebraic over L(X 1 ), then there exists an element

12
Dimension and Multiplicity
[Ch.
f E L[X 1 ], / ¥: 0, such that g = Iw is a polynomial, with coefficients in F, integral over
L[X 1 ]. If g'"+alg m - 1 + ... +am = 0, al, ..., a," e L[X 1 ], and if a positive integer s is
greater than the degrees of the polynomials g, aI, ..., am, then the polynomial h = g- X'
satisfies an equation of the form h rn +b 1 h m - 1 + ... +b m = 0, where b i . ..., b", e L[XtJ,
and b m = (Xf)m +a1(Xl)m-l + ... +a". is a monic polynomial. Now, apply the first
assertion to the polynomials -h, hm-l+b 1 hrn-2+ ... +b m - 1 .]
(iii) If L c: F is a finite field extension, and elements Xl' 0'0' Xu (contained, together
with F, in some field) constitute a set algebraically independent over L, then this set
is also algebraically independent over F, and [F(X 1 , . 0 . , X n ): L(X 1 , ..., XII)] = [F: L).
(iv) Note that every field Fwhich is a finitely generated extension od an algebraically
closed field K is of the form K(T 1 , .oo, T q , y), where the set T l , ..., Ta is algebraically
independent over K and y is an element algebraic over K(T 1 , ..., Tq) (see [N], p. 185
and p. 265). Deduce from this that there exists an algebraically closed field {J which is
an extension of K, elements Xl, ..., XII' Y I ..., Y m e {J which form a set algebraically
independent over K, and elements ex, {J e (J, algebraic over K(X) = K(X 1 , ..., XII) and
K(Y) = K(Y!, .oo, Y m ), respectively, and such that there exist isomorphisms qJ: Ao
 K(X, ex), VJ: Bo  K(Y, (J) of the fields of fractions Ao, Bo of the algebras A, B.
(v) Prove that [K(X, Y) (ex, (J): K(X, Y)] == [K(X, ): K(X)] [K(Y, (J): K(Y)]. [Show
that [K(X, Y) (ex, (3): K(X, Y) ()] = [K(Y, (3): K(Y)]; to this end, consider the minimal
polynomials 11,12 of the element (3 over the fields K(Y) and K(X, Y) (ex), respectively.
It follows that 12111 , and that all the zeros (3 = (31' .oo, fJ, of/ 2 in the field (J, are algebraic
over K(Y). Applying (ii) to the extension K C K(X, ex), deduce that the coefficients of
12 belong to K(Y), i.e., that 11 = /2']
(vi) Prove that the homomorphism p :Ao C KBO -+ (J, satisfying the condition
p(a@b) == qJ(a)lJl(b) for a e Ao, b e Bo, is a monomorphism. [Observe that p, is a mono-
morphism on qJ-1(K[XJ)1p-l (K[Y]), and then apply (v).)
6. Deduce from the results of Exercise 5 that if K is an algebraically closed field,
V c: Kit, W c: K m are K-varieties, then the set Vx We K,,+m is also a K-variety, and
K[Vx W]  K[V]@KK[W].
1.2 THE KRULL DIMENSION
In this section, we shall apply Theorem 1.1.7, proved in Section 1.1, which
states that the dimension of an affine algebra is equal to the maximal length
of a chain of prime ideals of this algebra. We shall also extend the defini-
tion of dimension to arbitrary rings so introducing the concept of the
Krull dimension of a ring. We prove some basic theorems on dimension
and give a number of examples which show that not all the theorems on
the dimension of algebras which were proved in Section 1.1 can be extended
to the general case.
We recall that the length of a chain of (distinct I) prime ideals Po c Pt
c: ... C P d is the number d. We say that this chain connects the ideals Po
and Pd. Theorem 1.1.7 justifies the following generalization of the definition
of the dimension of an affine algebra to the case of arbitrary rings:
Definition 1.2.1
The Krull dimension of a ring R, dim R, is the upper bound of the lengths
of chains of prime ideals

I]
The Krull Dimension
13
PaC PtC ...C P d
(2)
of R.
The height, ht(P), of a prime ideal P of a ring R is the upper bound
of the lengths of those chains of prime ideals of form (2) for which P d = P.
The height, ht(I), of any ideal I of the ring R is the number
ht (1) = inf ht (P),
P::>]
where P ranges over all the prime ideals containing I.
Dimensions and heights take the values 0, 1, 2, ..., 00.
Example 1.2.2
If R is a Dedekind domain and is not a field, then dimR = 1. If R c: T
is an integral ring extension, then, by Theorems 3.1.17 and 3.1.13 in [B],
dimR = dimT.
The following statements result directly from the definition:
(1.2.3) If I c J are ideals in R, then
dimRjJ  dimR/I, ht(I)  ht(J).
(1.2.4) If Pt, ..., P:t are the minimal prime ideals of an ideal I, then
dimR/I= max(dimR/P t , ...,dimR/P s ),
ht(I) = min(ht(Pt), ..., ht(P s »).
Making use of the correspondence between prime ideals of a ring Rand
those of its ring of fractions (Theorem 1.4.7 in [B]), we get
(1.2.5) If P is a prime ideal in R, then dimR p = ht(PR p ) = ht(P).
(1.2.6) If S is a multiplicative subset of R, and P is a prime ideal in R
such that Pf1S = 0, then ht(PRs) = ht(P), whence ditnRs  dimR.
(1.2.7) dimR = sup dimR p = sup dimRm.
Pe Spec(R) me Max(R)
(1.2.8) dimR = sup ht(P) = sup ht(m).
P e Spec(R) me Max(R)
(1.2.9) If a domain A is an affine algebra then Theorem 1.1.15 implies
that, for every prime ideal P in A, ht(P)+dimA/P = dimA. Whence ft is
readily deduced that also for an arbitrary ideal I in A we have ht(I)+
+dimA/I = dimA. This however, is not true for every Noetherian do-
main* (see Example 1.2.22).
One of the most important theorems in the theory of rings is the follow..
ing theorem of Krull.
Theoem 1.2.10 (Krull)
Let R be a Noetherian ring, and let P be a minimal prime ideal of an ideal
generated by n elements. Then ht(P)  n.
* These formulae have essentially contributed to the diversity of the terminology; at
times, the height of an ideal is termed ranI" and its dimension is termed co-height or
co-rank.

]4
Dimension and Multiplicity
[Ch.
In the case of n = 1, this theorem is called the Krull Theorem on Princi-
pal Ideals.
Proof
1. We shall first prove the theorem for n = 1.
Suppose x E R generates the proper ideal (x). Let P ;:) (x) be a Dlinimal
prime ideal of (x).
Suppose that ht(P) > 1; thus, there exists a chain of prime ideals
P :::> P 1 :::> Po. Replacing R with R/ P 0, we reduce the proof to the case
where R is a domain. Since the height of the prinle ideal P remains un-
changed under the localization with respect to P (see (1.2.5), we may assume
that, in addition, R is a local domain whose maximal ideal m is a minimal
prime ideal of (x). Note that now to the ideal PI corresponds a non-zero
prime ideal Q  m; clearly x  Q.
Take a non-zero element y E Q; we shall prove that the decreasing
sequence of ideals
(x, y) :::> (x 2 , y) ;:) (x 3 , y) :::> ...
(3)
becomes stable. To begin with let us observe that, by Theorem 2.7.12 of
[B], Rj (xI') is an Artin ring for k  1. According to Theorem 2.7.11 in [B],
it has finite length. Using additivity of lengths for exact sequences, and
applying the isomorphisms
. Rj(xm):(y)  (x m , y)j(x m ),
(x nl ) : (y)/(x Pn )  (y) : (xm)j(y) ,
we shall calculate the (finite) length of the module R/(X", y). We have
I (R/(x"' , y») = l(R/(xm»)-/«x m , y)j(x m »)
= 1 (R/(x m ) )-l(R/(x m ): (y») = 1 «x m ) : (y)/(x nI ) )
I:: / ( (y) : (x"') / (y) ) .
Since the sequence of ideals {(y) : (x m ) }, n1 = 0, 1, ... is increasing and the
ring R is Noetherian we infer that the sequence of lengths we have calculated
becomes stable, hence the sequence (3) also becomes stable.
Let (X', y) = (x"+1, y) for some k. Accordingly, there exists t E R such
that )(' - t+ 1 E (y) c Q, whence xk(l- tx) E Q. By the invertibility of
1- tx, we have x E Q, contrary to the assumption. Thus we have proved
the theorem for n = 1.
2. Assume the theorem to be valid for ideals generated by less than n
elements, n > 1. Let I be a proper ideal generated by n elements, say
Xl' ..., x n , and let P be a minimal prime ideal of I. It is our purpose to
show that ht (P)  n.
Suppose the contrary, ht(P) > n; thus, there exists a chain of prime
ideals Po c: PI c: ... c: Pn C P. We may assume, as before that R is

I]
The Krull Dimension
15
a local ring with the n1aximal ideal P. Replacing Pn by the ideal which
is maximal in the family {P'} of the prime ideals satisfying the condition
Pre C P' $ P, we can assume additionally that there exists no prime ideal
P" such that Pn $ P" $ P. The ideal P is a minimal prime ideal of I,
hence I cJ: Pn. Accordingly, at least one of the elements Xl' ..., Xn' say X n ,
does not belong to Pn. Now, it follows that P is a minimal prime ideal
of the ideal Pn + (x,,); hence, by Lemmas 2.3.20 and 2.3.5 in [B], Pn + (xn)
is a P-primary ideal. Thus there exists a positive integer q such that pq
C Pn + (XII)' and consequently, for some t l' ..., t n - 1 E Pn, Ul , ... , U n - 1 E R,
we have
xf = t i + u, x," i = 1, ..., n - l.
(4)
Write J for the ideal generated y t 1, ..., t n - l' Clearly J c P", but P II is
not a minimal prime ideal containing J in virtue of the inductive hypo-
thesis since ht(Pn)  n. Thus there exists a prime ideal Q such that J c Q
$ Pn. It follows from (4) that the ideal Q + (x n ) contains a power of I,
whence P is a minimal prime ideal of Q + (x,,). The ideal P /Q of R/Q is
a minimal prime ideal of a proper principal ideal, and thus the first part
of the proof yields ht (P /Q)  I. However, the ring R/Q contains the chain
of prime ideals (0) c Pn/Q c: P /Q, contradicting ht (P /Q)  1. This ends
the proof. 0
Directly from the Krull theorem we deduce the following corollaries:
Corollary 1.2.11
Every ideal of a Noetherian ring is of finite height.
Corollary 1.2.12
The Krull dimension of a local ring with the maximal ideal m is finite and
does not exceed the number of generators of any m-primary ideal.
Corollary 1.2.13
If (R, m) is a local ring and K = R/m, then dimR  dim K m/m 2 .
Proof
If the residue classes of Xt, ..., x q Em modulo m 2 form a basis for m/m 2
over K, then Xl' ..., X q generate m by Lemma A.S.l. Theorem 7.2.10
yields dimR  q. D
Corollary 1.2.14
A Noetherian ring does not contain infinite descending chains of prime
ideals.
As a geometrical consequence of the Krull theorem, let us add the
following:

16
Dimension and Multiplicity
[ChI
Corollary 1.2.15
If an algebraic set V in an affine space Kn is determined by d polynomials,
i.e., V = V(I), where I is an ideal in K[X I , ..., Xn] which is generated by d
elements, then dim V  n - d.
Proof
Let R = K[X 1 , ..., X n ]. As we know from Section 1.2, dim V(I) = dimR/I.
By Example 1.1.2 and by 1.2.9, dimR/I = n-ht(I); hence, the Krull
theorem yields dimR/I  n-d. 0
Let us note that the Krull theorem cannot be strengthened by asserting
that ht(P)  n for every prime ideal associated with an ideal generated by
n elements (see Exercise 2, Section 1.3). Although chains of prime ideals
in a Noetherian ring R are finite, the dimensions of local rings Rm are also
finite, and dimR = supdimR m , where m ranges over all the maximal ideals
of R, nevertheless this bound may be infinite; a suitable example will be
given in Example 1.2.20. Let us also note that a non-Noetherian ring may
have a finite Krull dimension (Exercise 1).
A ring of polynomials in a countable number of indeterminates pro-
vides a simple example of a non-Noetherian ring of infinite Krull dimension.
We shall quote another theorem related to the Krull theorem.
Theorem 1.2.16
Let P be a height d prime ideal of a Noetherian ring R. Then there exist
elements Xl' ..., Xd E P such that the ideals It = (Xl' ..., Xk), 1  k  d,
and 10 = 0 satisfy the following conditions:
(i) P is a minimal prime ideal of I d ,
(ii) ht(lk) = k for 0  k  d.
Proof
We shall first use induction on k to prove (ii). The case k = 0 is trivially
valid. Take k < d, and suppose we have constructed Xl'.'" Xt which
satisfy (ii). Let PI' ..., P:t be the minimal prime ideals of Ik which are of
height k. Since k < d = ht (P), none of the ideals PI' ..., P II contains P.
Hence P <t: PI u ... u P:t by Corollary 1.1.8 in [B]. Thus there exists an
element Xk+l eP,",(Pl u ... uP:t). Let Q be a minimal prime ideal of 1,,+1.
By Theorem 1.2.10, ht (Q)  k + 1; on the other hand, Q contains a certain
minimal prime ideal P' of Ik. Hence ht(P')  k. If ht (P') > k, then ht (Q)
 k + 1. If, on the other hand, ht (P') = k, then P' = P, for some i, 1  j
 s; thus Xk+l rp P', but Xk+l e Q, whence P':f:. Q, and finally ht(Q)
> ht(P') = k. Thus, it follows that ht(Q) = k+ 1.
Property (i) follows from (ii). Indeed, if P were not a minimal prime
ideal of I d , (ii) would imply ht (P) > d, contrary to the assumption. 0

I]
The Krull Dimension
17
Corollary 1.2.17
If I is an ideal of a Noetherian ring, and x E I is not a zero-divisor, then
ht(II(x») = ht(1)-I.
Proof
Suppose first that I is a prime ideal, and let ht(ll(x)) = d. A descending
chain of prime ideals in RI(x) of length d beginning at II(x) induces a chain
of prime ideals I = P d :::> P d - 1 :::> ... :::> Po in R, where x e Po. Since x is
not a zero-divisor, by Theorem 2.4.22 in [B], Po is not a minimal prime
ideal in R, whence ht(I)  d+ I. On the other hand, applying Theorem
1.2.16 to the ideal II(x), we infer that I is a minimal prime ideal of an ideal
generated by d+ I elements, Le., ht(I)  d+ 1 by the Krull theorem.
We leave to the reader the simple arguments in the case where I is not
. 0
Corollary (1.2.17), immediately implies
Corollary 1.2.18
If (R, m) is a local ring and x Em is not a zero-divisor, then
dimR/(x) = dimR-I.
The following theorem is complementary to the Krull theorem on
principal ideals:
Theorem 1.2.19
Let R be a normal Noetherian domain. Then every associated prime ideal
of a non-zero proper principal ideal of R is of height 1.
Proof
Let P be an associated prime ideal of a proper principal ideal (x) ¥= o.
By Theorem 2.3.15 of [B] there exists an element Y E R such that P
= (x):(y). It is easy to see that condition P :f:. R implies ylx rp R. Thus
ylx is not integral over R in view of normality of R. By Theorem 3.1.6 of
[B] we deduce that (ylx) P q: P; since (ylx) PeR there exists Xo E P
such that the element Yo = (ylx) Xo belongs to R"",P. The equation Yolxo
= ylx implies (xo):(Yo) = (x):(y) = P; therefore from the condition
yorpP we obtain PR p = {(xo):(yo))R p = (xo)R p . By the Krull theorem
on principal ideals we finally get htP = ht(PR p ) =. ht{(xo) R p ) = I. 0
We conclude this section with three examples showing that by extending
the notion of dimension from affine algebras to arbitrary Noetherian rings
we lose in general the good properties of dimension which we proved in
Section 1.1.

18
Dimension and Multiplicity
[Ch.
Example 1.2.20
We shall describe the construction, due to M. Nagata (see [R]), of a Noe...
therian domain of infinite Krull dimension.
Let K be a field, and let T = K[X 1 , X, ...] be a ring of polynomials
in a countable number of indeterminates Xl' X 2 , ... Let us also take
a sequence of positive integers ml, m2" m3, ... satisfying the condition
0< m2-m1 < m3-m2 < ... Write Pn, n = 1,2, ..., for the prime ideal
00 co
(X m +1' ..., X m +) of T. Then the set S = n (T"P n ) =  U P II
'1 n 1 1 1
n= n=
is multiplicative. It is our purpose to show that the dOlnain R = Ts is
Noetherian and has infinite Krull dimension.
The inclusion S c: Pn implies that Rpn R = RT"",Pn = TT'.Pn = T Pn
and the rin g .Tp = K[X 1 , X2" ...]p has the form ( Kn[" + 1, ..., X m +1 ]);; ,
PI _ PI PI n n
where K,. is a field and Pn is the extension of Pn (see Ex. 8, Section 1.4 in
[B]). Thus T p is a Noetherian ring, and dinlT p  111n+1 -m n (it follows
PI n
from Theorem 1.1.7, that actually dimT Pn = mn+1-mn). By (1.2.7),
dimR  dimR p R = dimT p  mn+1-mn. Hence dimR = 00
n n
It remains to prove that the ring R is Noetherian. We shall employ the
fol1owing lemma, which will be proved in the sequel:
Lemma 1.2.21
If a ring R satisfies the conditions:
(i) for every maximal ideal m c: R the ring Rnt is Noetherian,
(ii) every non-zero element of R belongs to a finite number of maximal
ideals of R,
then R is a Noetherian ring.
Let us show that the ring R which we have constructed above satisfies
these two conditions.
Let m c: R be a maximal ideal of R. Then P = mnT is a prime ideal
and, by Theorem 1.4.7 (i) of [B], we get 111 = PRo Clearly, p()S = 0,
00
hence P c: U P n ; we claim that P is contained in one of the ideals Pn.
n=l
m
By Theorem 1.1.8 of [B], it is sufficient to prove that P c: U Pn for some
11=1
m
m  1. Suppose the contrary, Le., P q: U Pn for all m. Take an arbit-
n=l
rary elen1ent 0 i= pEP; then PEP" for a certain k, and p is a linear com-
bination, with the coefficients in K, of monomials containing one of the
indeterminates X m" + l' ..., X mA;+ 1 as a factor. There exists an index q > /,
such tbat p does not depend on any of the indeterminates X m + 1, X m +2, ...
q II
q q
Since P q: U Pn, there exists pi E P such that pi E U Pn. For some
n=l n=1
I > q, we have pi E Ph hence pi is a linear combination of monomials
containing one of the indeterminates X m ,+ 1, ..., X ml + 1 as a factor. Thus

I]
The Krull Dimension
19
the monomials occurring in p and pi are distinct. The elelnent p + pi belongs
to P, and hence to PJ for a certainj. Consequently, it is a linear combina..
tion of monomials containing one of the indeterminates XmJ+ 1, ..., X mJ + 1
as a factor, and therefore p E PJ. The element p is independent of the
indeterminates X m4 + 1 ' X m .+ 2 , ..., whence j+ 1  q. Now, P E Pj, p+p' E PJ
q
imply pi E Pi; we have assulned, however that p' rt U Pn.
1Z=1
This contradiction proves that P is contained in one of the ideals
Pn, n = 1, 2, ... Since P "nS = 0, the ideal P,. R is a proper one, and
contains the maximal ideal m = PR; thus P = PIt. Since P = mnT, the
condition tIs E tn, for t E T, S E S, is equivalent to t  P. Hence, in view
of S c: p and P = Pn, it follows that Ran = (Ts)m = (1:)T""-P = TT""-P
= T p , and so Rm is a Noetherian ring.
n
An element 1" = tis of the ring R = Ts, where t E T, S E S, belongs to
the maximal ideal m if and only if t belongs to the ideal mnT, which is
one of the ideals P", n = 1, 2, ... The polynomial t depends only on a finite
number of indeterminates, and so it belongs to a finite number of ideals Pn.
Accordingly, r belongs to a finite number of the n1aximal ideals of the
ring R.
Thus, we have proved that both conditions of the lemma are satisfied
by the ring R. It remains to prove the lemma.
Proof of Lemma 1.2.21
If I is a non..zero ideal of R, it follows from (ii) that there exist only finitely
many maximal ideals, ml, ..., mh, containing I. For any non-zero r € R,
write M(r) for the finite set of all the maximal ideals which contain r.
We then have {ml"'" m,,} = n M(r) , and, the sets M(r) being finite,
reI
q
there exist '1, ..., rq E I such that {ml"'" mh} = n M(r,). It follows
;=1
from (i) that there exist a finite number of elements r q + 1, ..., r n E I which
generate all the ideals 1Rm, c: Rani' i = 1, ..., h.
n
Obviously 2: r, R c: I. Let nt be a maximal ideal of R. If I c: m,
1= t
n n
then m = ttt) for some j, 1  j  h, and 2: ri Rut :::> 2: 1', Rm = IRtn ) '
. 1 J. 1 J
.= .=q+
If I cf: m, then 1Rm = Ran and m =F ml, ..., mh, whence there exists i,
1  i  q, such that m rt M(ri). This implies ri  m so that ri Rm = R.
n
Accordingly, for all maximal ideals m, we have (2: riR) R n1 = IRm,
i= 1
11
and therefore 2: ri R = I. Thus R is a N°oetherian ring.
;= 1
o

20
Dimension and Multiplicity
[Ch.
Example 1.2.22
Now we shall give an example of a Noetherian domain which contains
saturated chains of prime ideals of different lengths.
Let K be a field, and let T = K[[Y]] [X] be the ring of polynomials in
one indeterminate X with coefficients in the ring of formal power series
R = K[[Y]]. Let m = (XY -1), m' = (X, Y) be ideals of T. We shall show
that m and m' are maximal ideals of heights 1 and 2, respectively. To this
end we use Lemma 1.3.3. Since mt1R = 0, it follows that ht(m) = 1.
Similarly, m't1R = RY is of height 1 in R = K[[Y]], whence, by Lemma
1.3.3, ht(m') = 2. It is evident that the ideals m, m' are maximal since
TIm'  K and TIm  K[[Y]] [1/Yj is the field of fractions of the ring
K[[Y]].
It is a great deal more difficult to find a Noetherian domain together
with a pair of prime ideals between which there are saturated chains of
prime ideals of different lengths. This will be done in the example below.
Example 1.2.23
Following M. Nagata (see [R]), we shall construct a local domain R in
which there exists a saturated chain of prime ideals of length less than
dimR.
1. Let K be a field; we shall show tbat for every positive integer r there
exist, in the ring of formal power series K [[X]], elements Zt, ..., z, alge..
braically independent over K[X]. Indeed, if k c: K is a simple subfield,
then Ikl  o and Ik [[x]]1 = Ikl No = 2 Mo > o = Ik(X)/. Hence the trans-
cedence degree of k [[X]]o over k(X) is equal to 2 No , and consequently,
there exist elements Z 1, ..., Z, E k [[X]] algebraically independent over
k [X]. It is easily seen that Z 1, ..., z, are also algebraically independent
over K[X].
2. Let m  1 be an integer. We shall construct a domain R 2 and two
maximal ideals m, 11 of that domain such that the rings (R 2 )m, (R 2 )n will
be Noetherian (hence local), regular, of dimensions m+ 1 and r+m+ 1,
respectively (for the notion of regularity of a local ring, cf. Definition
1.4.12, and also Chap. II).
The elen1ents z 1, ..., Zr constructed in 1. are of the form Zt = L a;r,X k ;
k
we set
Z,j = QIjX+a ,J + 1 X 2 + ...
j-I
= (z,- L a,.X 1c )/X J - 1 for i = 1, ..., r, j = 1, 2, ,..,
k=O
and write R 1 = K[X, Zij] for the K-subalgebra of K[[X]] generated by the
elements X, Z'j, i = 1, ..., r,j = 1,2,... Let R 2 = R 1 [Y 1 , ..., Y m ] be

I]
The Krull Dimension
21
the ring of polynomials in indeterminates Yl, ..., Y m ; thus, R 2 = Rt [Y t , ...
..., Y m ] == K[X, Zij, Y t , ..., Y m ]. We set
m = (X, Y t , ..., ,,), n = (X - 1, Z t, ..., Z r, Y 1 , ..., Y m) ·
We claim that the ring R 2 and the two its ideals m, n have the desired
.
properties.
The equations XZ ij + t = zlJ-aIjX, which hold in R 1 , yield Z'j ER I X,
whence it follows that Rt/RtX  K. Accordingly,
R 2 /R 2 (X, Y t , ..., 1';)  K[Yi+l, ..., Y m ], I = 0, 1, ..., m',
and consequently we obtain in R 2 the chain of prime ideals of length m + 1,
o c: (X) c (X, Y t ) c: ... c: (X, Yl, ..., Y m ) = m.
The ideal m is maximal, and ht(m)  m+ 1. From R 1 t1m = R 1 Xit follows
that R 1 "Rt X c: R 2 "m. Hence (R 2 )m is a ring of fractions of the ring
11 [Y 1 , ..., Y m ]R 1 'R 1 K = (R t )R 1 X[YI, ..., Y",], which is Noetherian (by
virtue of Corollaries 2.2.2 and 2.1.12 of [BJ) whence (R 2 )m is also Noether-
ian. The ideal m is generated by m + 1 elements, and hence, by the Krull
theorem, ht (nt(R 2 )m)  m + 1. Thus (R 2 )tn is a regular local ring of dimen..
sion m + 1 (because m + 1 generators of m form a set of parameters by
Theorem 1.4.11 (i»).
We study in a similar way the ring (R 2 ) . The natural homomorphism
P: R 1 -+ Rt/Rl(X -1) can be uniquely extended to the homomorphism
P: (R t ){X,X2, ...} = R 1 [1/X] = K[X, zu][l/X]
= K[X, I/X, Zt, ..., z,] = K[X, l/X][zt, ..., zr] -+ R 1 /R t (X-l),
Ker(v) being the principal ideal generated by X-I. Hence
Rt/Rt (X -1)  K[X, 1 /X][Zt, ..., zr]/R t [1 /X](X -1)
 K[zt, ..., zr]'
Thus we obtain in the ring Rl the chain of prime ideals
o c: R 1 (X -1) c: R I (X -1 , Zt) c: ... c: R I (X -1, Zt, ..., zr)
and in the ring R 2 the chain of prime ideals of length l' + n'l + 1
o c: (X-I) c: (X -1  Zt) C ... C (X -1 , Zt, ..., zr)
c (X - 1 , Z t, ... , Zr, Y 1 ) c: ... c: (X - 1 , Zt, ... , Zr, Y t , ..., Y",) = n.
The ideal n is maximal, and ht(n)  r+l11+1. Let us set n 1 = Rt(X-l,
Zl, ..., z,); thus we have Rtf11t = nt, whence Rt"nt c: R2"n and (R 2 )"
is a ring of fractions of
R 1 [Y t , ..., Y m ]R 1 '"1 = (R t )n 1 [Y t , ..., Y m ].
Thus it is sufficient to prove that the ring (R t )n 1 is Noetherian.

22
Dimension aod Multiplicity
[Ch.
Denote by S1 the multiplicative set K[XJ"'-K[X](X -I). The equation
Stnn! = 0 (the proof of which will be given below) implies that S1 c Rl"
"'-nt, and consequently (Rt)n 1 is a ring of fractions of (R 1 )Sl. FrOln
j-I
Z'J = (ZI- I>,k Xk )/XJ-l E K[X, Zl, ..., zrk c (R 1 )s,
"=0
it follows that
(R t )Sl = K[X, Zl, ..., zr]s, = K[XJs 1 [Z1, ..., zr].
Accordingly, (R t )Sl is a Noetherian ring, hence so is (R t )n 1 . The ideal
n is generated by ,. + m + 1 elements, which, in view of the inequality
ht(n)  r+m+ 1, implies that (R 2 )n is a regular local ring of dimension
r+m+ 1.
To prove that Stnn! = 121, suppose that a polynomial }-v E K[X] belongs
to nl; then w = qo(X-I)+q1z1 + ... +qrzr where qo, ..., qr E Rl' and
consequently qo, ..., q, can be expressed in terms of X and a finite number
of the elements Zij. It follows from Xz ,J + 1 = zIJ-aiJX that, for a sufficiently
large N, the elements qo, ..., qr can be expressed in terms of X and C 1
N-1
= ZtN, ..., C, = ZrN. Since X N - 1 Ci = Zi- L aikXk, it follows that
k=O
R 2  R3 = K[X,  t, ..., C,] ::> K[X, Z 1, ..., z,], the elements. X, C 1, ..., C,
being algebraically independent over K. By the choice of N, the element w
belongs to the ideal I c: R3 generated by the elements X-I, X N -1 C 1 +
N-l N-l
+ L: ai"X k ,..., XN-1r + L a'tXk, and consequently the isomorphism
k=O k=O
R 3 /R 3 (X-I)  K[X, C1, ..., C,]/R 3 (X -1)  K[C1, ..., C,] carries the ideal
N-l N-l
I/R 3 (X-I) onto the proper ideal (C1+ L Qik, ..., C,+ L: ark) of the
k=O k=O
rjng K[C 1, ..., Cr], and the residue class of w into the element w(l) E K.
Hence wet) = 0 and (X -I) I w, Le., w  S1. Thus, we have S1nn! = 0.
3. Denote by S the multiplicative subset (R 2 "'-m)()(R 2 "'-n) of the ring
R 2 . The ideals of the ring R' = (R 2 )s are the extensions R'I of those ideals
I of R 2 for which InS = 0. The latter condition is equivalent to I c: mun
which in turn, by Corollary 1.1.8 in [B], is equivalent to I c: m or I c n.
This, however, proves that R'm and R'n are the only maximal ideals ill R'.
Now, the inclusion S c R 2 "'-m c: R"'-.R'm yields RR'm = (R 2 )mt and
similarly we get RR'u = (R 2 )n and, since we have proved that these two
rings are Noetherian, it follows from Len1ma 1.2.21 that R' is also a
Noetherian ring.
We shall show that the subring R = K+R'mnR'n of R' has the desired
property. The basic properties of tllis ring result from the following:

I]
The Krull Dimension
23
Lemma 1.2.24
Let K be a field, and let R' be a semilocal K-algebra \vith the maximal
ideals ml, ..., 1nq. Suppose also that R'/m! = o.. = R'/m q = K, and put
J = m1t1 ... t1m q , .R = K+J. Then R is a K-algebra which satisfies the
following conditions:
(i) the extension R c: R' is integral,
(ii) R' is a finitely generated R-module,
(iii) the algebra R is local.
Proof
We already know that the K-module R' /1 can be embedded in the direct
sum
q q
fP; R'/m j = $ K.
j=O )=0
Thus there exists a finite basis for R'IJ over RIJ = K. Let the residue
classes modulo J of elements r 1 = 1, r 2, ..., tit constitute such a basis.
For every r E R', its residue class r = r+J E R'IJ fulfils an equation
-p -p-1 0
r +t1'-lr +.o. +t o =
for some to, ..., t p - 1 ER. Hence U = r"+I"_1,,,-1+ ... +t o belongs to I,
which implies (i) since to - U E R. There exist elements u l' ..., Un E R such
that r =  u,r" and thus r =  u,r,+u for some u E J, whence we get
I . I
R' =  RI', + J. Since J c: R = Rr 1 , we have R' = L Rr, which proves (ii).
i I
To prove (iii), observe that every element of R which does not belong
to Jis of the form tX(l+u), where 0 i= tXEK, ueJ. Hence l+uml' ...
..., mq, and therefore 1 +u is a unit in R. If Y E R' and (1 +u) Y = 1, then
Y = 1- uy E K + J = R, and tX(1 + u) is a unit in R'. Accordingly, J is
a unique maximal ideal in R. To prove that R is a Noetherian ring, it is
sufficient, in view of Theorem 2.1.15 of [B]t to show that every prime ideal
PeR is finitely generated. It follows from Theorem 3.1.16 in [B] that
there exists a prime ideal Q c: R' such that Qt1R = P. Since P c: J, we
have P = P t1 J = Q t1 R t1 J = Q t1 J. Hence, P is also a prime ideal of R',
and consequently, P = R't 1 + .o. +R't s for some t 1 ,..., t E P. The
elements fJr) belong to P, and therefore the ideal I =  Rt,r) of the
i.)
ring R is contained in P. If pEP. then p =  u, t, for some Ul, ..., Un E R,
I
whence u, =  v,)r) for some v,) E R; thus it follows that p =  u, I,
J i
= LVi) t;r) E I, and therefore 1= P, i.e., the ideal P is finitely generated. 0
I.l
It follows from the lemma that the ring R = K+R'mnR'n is local
with the maximal ideal R'mnR'n = R'(mt1n), and the extension R c: R'
is integral.

24
Dimension and Multiplicity
[Ch,
4. It is our purpose now to show that the ideal R' X r\R of the ring R is
prime, and every saturated chain of prime ideals with this ideal as a member
is of length less than dim R.
Let us set Q = R' X r\R; the ideal R' X is an extension of the prime
ideal R1X c: m, whence R1XnS = 0. Thus R'X is a prime ideal in R',
and Q is a prime ideal in R. We shall show that R'X is the only prime ideal
over Q in R'. Let us first note that Q = R'Xr\R'n; indeed, the inclusions
Q c: R'(mr\n) = R'mr\R'n and R'Xr\R'n c: R'mnR'n c R yield Q = Qr\
R'mr\R'n = R'Xr\Rr\R'mr\R'n = R'XnRnR'n = R'Xr\R'n. According..
Iy, QRR'u c: nRR'n, and ifr En then rX E R'X()R'n = Q, and the condi-
tion X  R'n implies r = (rX)/X E QRR'n, whence QRR'n = nRR'n' Let
P' c: R' be a prime ideal lying over Q, i.e. P'r\R = Q, and suppose that
P' c: R'n. Then P' RR'n  QRR' = nRR'n, whence P'RR'n = nRR'n, and
consequently P' = R'n, Le. R'ltnR = Q. Since R'X c R'm, ht(R'm)
= m+ 1 > 1, ht(R'X) = I, the ideal R'X is not maximal. By Theorem
3.1.12 in [B], Q is not maximal either, and R'n does not lie over Q. In view
of this contradiction the ideal P' is not contained in R'n, i.e., P' c: R'm.
Since XEm, the condition X(X-I) Emr\n c:R'mr\R'n c: R yields
X(X -1) E R'Xr\R = Q c: P' c R'm, whence, in view of X-I rt R'm,
it follows that X E P', and consequently R'X c: P'. Now, Theorem 3.1.13,
in [B] yields P' = R'X. Thus we have proved that R'X is the unique prime
ideal in R' lying over Q.
Let 0 = Qo c: QI c... c: Qs = Q be a chain of prime ideals in R;
by Theorem 3.1.17 in [B], there exists in R' a chain of prime ideals
o = P c: P c: ... c: P = R'X such that Pi()R = Qt. The condition
ht(R'X) = 1 implies s = 1, whence ht(Q) = 1.
- - -- -
Suppose that 0 = Qo C QI c: ... c: Qd, where Ql = Q, is a saturated
chain of prime ideals in R; then there exists a chain of prime ideals 0
= Po c: PI c: ... c: P d in R' such that Pi r\R = Q, . We have already
proved that PI = R'X, and thus, since X  n, each of the ideals P, is con-
tained in R'm. Accordingly, d  ht(R'm) = m+ 1 < r+m+ 1 = dimR'
= dimR by Example 1.2.2, and R has the required property. 0
Exercises
1. We define the rule of multiplication in the Abelian group ZEB Q/Z by the formula
(a, q) (a', q') = (aa', aq' +a'q) for a, a' e Z, q, q' e Q/Z. Find the Krull dimension of
the ring thus obtained. Is it a Noetherian ring?
2. Find the Krull dimension of the factor ring of the polynomial ring R [Xl, X 2 , o. oj,
with the coefficients in R, by the ideal
(i) (Xl' X 2 , ...)2,
(ii) (Xl, X, ...).
3. Prove that, for every ideal 1 of a ring R and every integral extension R/ Ie-+- T,
there exist an integral extension R c: R' and a surjection R' -. T such that the diagram

I]
The Krull Dimension
25
r
Rll
t
R

T
r
t

R'
commutes. Show that if, moreover, Rand T are domains, then R may also be chosen
to be a domain.
4. Prove that if S is a multiplicative subset of a domain Rand Rs c -+0 T is an integral
ring extension, then there exist an integral: extension R c-+ R' and an isomorphism
R;  T such that the diagram
R r
 R' L ).- R;

 T

Rs (
commutes.
5. Prove that if R C-+o T1' R c-+ T 2 are integral extensions and Tl' T 2 are do-
mains, then there exist a domain R' and a commutative diagram
R (
.. T 1
f c
T,.
r
.. R'
in which all extensions are integral. Show that an analogous assertion is not true for
rings with zero-divisiors.
We recall that a saturated chain of prime ideals Po c: ... c P" of a ring R is maximal
if Po is a minimal and Pit. a maximal ideal of R.
The remaining exercises refer to the following four properties of a ring R.
(A) The length of any maximal chain of prime ideals of R is equal to dimR,
(B) For every minimal prime ideal P of R, dim RIP = dimR; furthermore, every
domain R' which is an integral extension of the ring RIP satisfies (A),
(C) For every pair of prime ideals P c: Q of R, the ring RQIPR Q satisfies (B),
(D) For every pair of prime ideals P c: Q of R, the ring RQIPR Q satisfies (A).
The first three conditions are known in the literature as "chain conditions for prime
ideals". A ring with property (D) is called catenary.
In the Exercises 6-16, the Krull dimension of the ring R is assumed to be finite.
We use the results of the Exercises 3-5.
6. Prove that R satisfies (D) if and only if every saturated chain of prime ideals
connecting ideals P c Q is of length ht(Q/ P).
7. Prove that R satisfies (C) if and only if, for every integral extension R C R',
every saturated chain of prime ideals of R' connecting ideals P' c Q' is of length
ht (Q' nR)/(P' nR).
8. Prove the following implications:
(B) => (A) => (D),
(B) => (C) => (D),
(B)  «A) and (C».
Show that (A) does not imply (B). [To prove the latter statement, employ the con-
struction from Example 1.2.23, with m = O. The local domain R thus obtained does
not satisfy (B) since the ring R' does not satisfy (A). To prove that R satisfies (A), use

26
Dimension and Multiplicity
[Chi
Property 8, Sec. 3.4 and Example 4, Sec. 3.3, which imply that the regular local ring
RR'n satisfies (A).]
9. Prove that if P is a prime ideal of Rand R satisfies (A) [(B)], then also the rings
RIP and R p satisfy (A) [(B)].
10. Prove that if I is an ideal of R, S c: R is a multiplicative subset, and R satisfies
(C) [(D)], then also the ring RslIRs satisfies (C) [(D)].
11. Prove that .R satifies (B) if and only if R satisfies (C) and for every mil1inlal
prime ideal P and every maximal ideal tn such that P c m, we have ht(mIP) = dimR.
12. Prove that R satisfies (C) if and only if for every minhnal prime ideal P, and
every maximal ideal 11t such that P c m, the ring Rml P Rm satisfies (B).
13. Let R be a domain. Prove that
(i) R satisfies (A) [(B)] if and only if R satisfies (D) [(C)], and ht (m) = dimR for
every maximal ideal m e Max(R),
(ii) R satisfies (C) if and only if, for every tn e Max(R), the ring Rm satisfies (C) [(B]),
(Hi) R satisfies (D) if and only if, for every tn e Max(R), the ring Rm satisfies (D)
[(A)].
14. Let R c: R' be an integral ring extension. Prove that if R' satiisfies (A) [(B)]
then also R satisfies (A) [(B)].
15. Let R c R' be an integral extension of domains. Suppose that for every
m' e Max(R'), ht (m') = ht (m' nR). Prove that if R' satisfies (C) [(D)] then also R satisfies
(C) [(D)], and ht (Pi) = ht (P' nR) for every prime ideal P' c: R'.
16. Prove that if R c R' is an integral ring extension and R satisfies (C) then so
does R'. [Apply Exercise 7.]
17. Prove that a Noetherian domain R satisfies (A) if and only if every domain
R' containing R finitely generated as an R-module, satisfies (A). [To prove the suffi..
ciency, apply induction on dimR. Let P' c R' be a prime ideal of height 1. Consider
the normalizations R and R' of the rings Rand R', R c R '. By Theorem 4.7.2 of [B],
there exist finitely many prime ideals of R, sa.y Q 1, ..., Q", lying over the ideal P = P' nR,
and there exist elements q2 e Q2"Ql, ..., q" E Q,,"Ql. Prove that the only prime ideal
of R which lies over the ideal Qlr.T of T = R[ql, .00' q,,] is Ql. Deduce from this that
ht (Q 1 n T) == 1, dim RI P = dim R -1. Show that RIP satisfies the hypothesis of the
implication we are proving.]
1.3 POLYNOMIAL RINGS
In this section, we shall prove a number of theorems concerning the Krull
dimension. of polynomial rings. We shall start with an estimation of the
Krull dimellsion of the polynomial ring R[X], and, in the case of a Noether-
ian ring R, we shall calculate its value, dimR+ 1. In proving those results
from which it follows that dim K[X 1 , ..., Xn] = n, where K is a field, we
use only the definition of dimension in terms of chains of prime ideals.
This enables us to complete the proof of Theorem 1.1.7.
The next two theorems are algebraic versions of certain geometric
results. As we already know (Corollary 1.2.15), the set of zeros of d poly...
nomials in n variables is of dimension not less than n-d. We shall prove
a theorem, due to Macaulay, stating that in the case where this set has
the minimal dimension, i.e. n-d, all its irreducible components are of
the same dimension n - d. The second theorem states that if U, U' are

I]
Polynomial Rings
27
algebraic varieties in K n then the dimension of the irreducible cOlnponents
of the algebraic set U t\ U' is not less than dim U + dim V' - n.
The facts we need to calculate the Krull dimension of the polynomial
ring are contained in a few lemmas below.
Lemma 1.3.1
If Q, Q' are two different prime ideals in R[X], and Q c Q' while Qt\R
= Q't\R = P, thenQ = PR[X].
Proof
Replacing the ring R by RIP, we can assume \vithout loss of generality
that P = O. Write S for the multiplicative set R"(O), and observe that
R[XJs = K[X], where K is the field of fractions of R. By the assumption,
Qt\S = Q't\S = 0, hence QR[X]s c: Q'R[X]s are different prime ideals
in R[X]s. Since R[XJs is a principal ideal ring, QR[XJs = 0, hence Q = o. 0
Lemma 1.3.1 directly implies
Corollary 1.3.2
If Qo c: Ql c: Q2 is a chain of prhne ideals of the ring R[X], then Qot\R
i= Q2t\R.
Lemma 1.3.3
Let R be a Noetherian ring, let Q be a prime ideal of the polynomial ring
R[X], and suppose P = Qt\R.
(i) If Q = PR[X], then ht(Q) = ht(P);
(ii) If Q :f; PR[X], then ht(Q) = ht(P) + 1.
Proof
(i) The inequality ht (P)  ht (Q) follo\vs directly from the definition. To
prove the opposite inequality, we apply Theorem 1.2.16, ,vhich asserts
the existence of an ideal I in R, generated by ht (P) elements, such that P is
a minimal prime ideal containing I. We claim that Q = PR[X] is a minimal
prime ideal containing IR[X]. Indeed, if Q' ,vere a prime ideal between
IR[X] and Q, we would have, by the minimaIity of P, Q't\R = Qt\.R = P.
Hence, by Lemma 1.3.1, Q' = Q. Since the ideal IR[X] is generated by
ht (P) elements, the Krull Theorem yields ht (Q)  ht (P).
(ii) Let n = ht (Q), and let Qo c: ... c: Qn = 0 be a chain of prime ideals
in R[X] of length n. Denote also Pi = Qit\R, 0  i  n; if the ideals
Po, ...,P" were all different, we would have ht(PR[X]) = ht(P)  11,
contrary to the assumption PR[X] $ Q. Thus, let k be the greatest number
such that P k = P k + 1 . It follows from Lemma 1.3.1 that Qk = PkR[X],
whence, in view of (i), ht(P k ) = ht(Qk)  k. By the definition of k, we have
the chain P" c: P k + 2 ell. c P n = P of length n - k -- ], and consequently

28
Dimension and Multiplicity
[Ch.
ht(P)  n-l. On the other hand, (i) and the assumption Q i= PR[X]
yield
ht(P) = ht(PR[X])  ht(Q)-1 = n-l.
o
Theorem 1.3.4
For any ring R, the following inequalities hold:
dim R+ I  dim R[X]  2dim R+ I.
If R is a Noetherian ring, then
dim R[X] = dim R+ 1.
Proof
A chain of prime ideals, of length d, in R, Po c: PIC ... c: P 11, determines
a chain of prime ideals
Po.R[X] c: P 1 R[X] c: ...c: PdR[X] c: PdR[X]+XR[X],
of length d+ 1, in the ring R[X]. Thus dimR[X]  dimR+ 1.
Now, let Qo c: Ql c: ... c: Qn be a chain of prime ideals of R[X].
Applying Corollary 1.3.2 to the chains Q2k c Q2k+ 1 c: Q2t+ 2 we infer
that
QonR c: Q2nR c: Q4nR c: ...
is a chain of prime ideals of R. If n = 2q or n = 2q + 1, then the last term
in this chain is Q2Q"R; hence q  dimR. Consequently, we get n  2q+ 1
 2dimR+l, whence dimR[X]  2dimR+l.
Suppose now that R is Noetherian. In this case it is sufficient to prove
that dimR[X]  dimR+ 1; this, however, follows directly from Le mma
1.3.3. 0
By the Hilbert Basis Theorem (Theorem 2.2.1 in [B]), we get the follow..
i ng coroIlary to the foregoing theorem:
C:orollary 1.3.5
If R is a Noetherian ring, then
dimR[X 1 , ...,X n ] = dimR+n.
Remark 1.3.6
For each pair of positive integers m, n such that m+ 1  n  2m+ 1,
there exists a ring R (not Noetherian if n > m+ 1) such that dimR = In
and dimR[X] = n. This result, as well as many other details concerning
the dimension of polynomial rings, can be found in [36], [37] and [J].
Theorem 1.3.7 (Macaulay)
If K is a field, and an ideal I of the polynomial ring K[X 1 , ..., Xn] is gener-
ated by d elements and is of height d, then all the associated prime ideals o.f
1 are also of height d.

I]
Polynomial Rings
29
The property of a ring of polynomials with coefficients in a field, stated
in the above theorem of Macaulay, has led to a new class of rings, so-called
Cohen-Macaulay rings, which will be discussed in Chapter III. Within the
theory of Cohen-Macaulay rings, we shall prove, among other things,
that if a ring R belongs to that class, then so does the polynomial ring
R[X]. Hence it folIows immediately that a ring of polynomials K[Xt, ..., Xn]
with coefficients in a field is a Cohen-Macaulay ring. We thus obtain
a new proof of Theorem 1.3.7. However, in view of the great importance
of this theorem, we give another proof of it in this section.
For the proof of Theorem 1.3.7, we need a lemma.
Lemma 1.3.8
Let Ql, ..., Q" be prime, not maximal, ideals of the ring R = K[X I , ..., X n ].
Then there exist: an index k, 1  k  n, and positive integers m 1, ..., mk- 1 ,
such that the polynomial X k + X;:'2t + ... + XfIc-" is algebraically independ-
ent over K modulo each of the ideals Q 1, ..., Qs (i.e. the residue class
it determines in R/Qi is algebraically independent over K, i = 1, ... , 11).
Proof
In the set {I, 2, II., s}, we define pairwise disjoint sets At, A 2 , II.' An
\vhich satisfy the conditions:
i E Al «> Xl is algebraically i11dependent modulo Qh
i E A 2 <=> i  AI' and X 2 is algebraically independent modulo Q"
i E A3  i  Ai uA 2 , and X 3 is algebraically independent modulo Qh
............................. .
iEAnirtAlu ... uA n - 1 , andX n is algebraically independent mod.
ulo Qt.
For every j, 1  j  n, we have dimR/Q) > 0; hence, denoting by
I the least index such that X, is algebraically independent modulo QJ, we
see that j E A,. Thus the sets AI' ..., An constitute a partition of the set
{I, 2, ..., s}. Let k be the greatest index such that A" :F 0.
If ieA", then iAIU ... UA"_l, and consequently, the elenlents
Xl' ..., X,,_ 1 are algebraic modulo Qi.
If j E A k - 1, then for distinct positive integers m, p, the elements X" +
+ Xf'-l , Xli: + Xf-l are not both algebraic modulo QJ, since otherwise their
difference XI:_1 -XC_1 wauld be algebraic modulo Q}, contrary to the alge-
braic independence of X,,_ 1 modulo QJ. Thus, for every j E At-I, there
exists at most one positive integer m such that Xt +X:'-1 is algebraic modulo
Q). Accordingly, there exists m 1 such that X t +x:'21 is algebraically inde-
pendent modulo Qj for all the j E A"-t. Since X"-1 is algebraic modulo
Q, for i E A"-I, it follows that X k +X:l is algebraically independent modulo
Q, for i E A"_luA". Now, if q E Q"-2, the elements X t +x:'21 +Xk'-2 and
X,,+Xk'21 +Xl- 2 , with distinct positive integers m,p, are not both algebraic
modulo Qq. As before we can find m2 such that Xk+x:'1 +xr2 is alge-
braically independent modulo Qi for i E A"_2UA"_luA". Proceeding in
this way, we find the desired m3, ..., m"-l. 0

30
Dimension and Multiplicity
[Ch.
Proof of Theorem 1.3.7
If d = 1, then the ideal I is principal, I = (fl) and, since R = K[X 1 , ..., Xn]
is a unique factorizatiol1 dOlnain, the prime ideals associated with I are
principal ideals, generated by the irreducible factors of the polynomial fl'
Thus, those prime ideals are of height 1. Assume that the assertion holds
for numbers less than d (and for arbitrary n), where d  1, and let I
= (fl, ... , fd) be an ideal of height d. Let us consider a prime ideal P :::> I,
associated with I, and denote p = ht(P). Replacing R by a suitable local-
ization, we may assume in addition that p = n, i.e. that P is a maximal
ideal. Indeed, by Theorem 1.1.15, dimR/P = n-p, hence, changing if
necessary the numbering of the indeterminates, we may assume that the
Xl' ..., Xn- p are algebraically independent modulo P (over the field K).
Denote by Sthe multiplicative setK[X 1 , ..., ,_p]'" {O}; we have Pr\S = 0.
The ring R = K(X 1 , ..., Xn- p) [X,,_ p+ 1, ..., Xn] is the ring of polynomials
in p indeterminates with coefficients in the field K(X 1 , . "' X n - l ,); the ideal
IRs is generated by d elements. Furthermore, the ideal P Rs is associated,
by Lemma 2.4.14 of [B], with the ideal IRs, and by (1.2.6) ht(PRs) = ht(P)
= p. Hence PRs is a maximal ideal of Rs. Every associated prime ideal
of IRs is of the form QRs, where Q is an associated prime ideal of I such
that Qn8 = 0 (see Lemma 2.4.14 in [B]); thus we have ht(IR s )  ht(I) = d.
By the Krull theorem (Theorem 1.2.10), ht (IRs) = d.
Thus we have reduced the assertion to the case where the ideal I of the
ring R = K[X 1 , ... , Xn] is generated by elements iI' ... , fd and is of height d,
and some associated prime ideal P of I is maximal, hence ht (P) = n by
(1.2.9). We have to prove that d = n. Suppose then that d < n, and write
J = (fl, ... ,fd-l). We deduce from the Krull theorem (Theorem 1.2.10)
that ht(J) 'd-1. If ht(J) < d-l, there would exist a mininlal prime
ideal Q of the ideal J of height < d-l; then, by the KrulI theorem on
principal ideals, the minimal prime ideals of the ideal Q + Rid would be of
height < d, i.e. they could not contain the ideal I of height d. This contra-
diction shows that actually ht(J) = d-I.
The ideal J being generated by d-l elements, the inductive hypothesis
applies to it. Hence all the prime ideals Q 1, ..., Qm associated with J are
of height d - I. Write Q",+ 1, ..., Qs for those associated prime ideals of I
\vhich are of height d. According to the assumption d < 11, none of the
ideals Ql, ..., Qm, ..., Qs is maximal. Applying Lenuna 1.3.8 to the ideals
Ql, ..., Qs, we find an element Y k = Xk+X:l + ... +Xr k - 1 , which is
not algebraic over K modulo Ql, ..., Qs. Obviously R = K[X 1 , ..., .IY,,]
= K[ Xl' ..., X" - 1, Y", X k + l' ..., XII]'
Since P is maximal, it follows from Corollary 1.1.9 that the residue
class y" of Y" modulo P is algebraic over K. Let g E K[T] be a minimal
polynomial of Yk over K; then b = g(Y k ) belongs to P whereas b  Ql, ...
... , b  Qs since Y" is not algebraic modulo Q l' ..., Q,'). Being isomorphic

I]
Polynomial Rings
31
with K(Yk) [Xl' ..., X k - 1 , X,,+ 1, ..., X n ], the ring R/ Rb is a ring of poly-
nomials in n - 1 variables with coefficients in a field.
We shall derive the identity I:P = I, obtaining by Theorem 2.3.15 of [B]
a contradiction with the assumption that P is an associated prime ideal of I.
Suppose II E I: P; it follo\vs that ub E I = J + RJd, whence ub + l'J'h E J for
some W E R, i.e. lV};' E J + Rb. Let Q be an associated prime ideal of the
ideal J+Rb. We shall show that fd  Q. The ideal (J+Rb)/Rb of R/Rb
is generated by d - 1 elements, and is of height d - 1, since b does not
belong to any of the associated prime ideals of J (which are Q 1, ..., Qm
and each of them is of height d-l); hence every Ininimal prime ideal
Q'/Rb of the ideal (J+Rb)/Rb contains one of the ideals (Qj+Rb)/Rb,
1 .i  n1. This, by the Krull theorem, implies ht (Q/) = ht (Qj) + 1 = d and
ht(Q' /Rb) = ht(Q')-l = d-l.
By the inductive hypothesis applied to the polynomial ring R/Rb and
the ideal (J + Rb)/Rb, all the associated prime ideals of the latter are of
height d - I. If fd belonged to a certain prime ideal Q" associated with
J + Rb, we would have ht (Q") = d and, by the inclusion Q" ::J I, Q"
\vould be one of the ideals Qm+ 1, ..., Qs, contrary to the assunlption
b  Qrn+ 1, ..., b  Qs. Thus, in view of Theorem 2.3.22 in [B], the condition
wJd E J + Rb yields lV E J + Rb. Accordingly there exists hER such that
w + hb E J, and, since ub + wid E J, we obtain /ifd b - ub = (/1/d - u) b E J.
The element b does not belong to any ideal associated with J, hence hid-
- U E J. Finally, U E J + Rid = I; thus \ve have proved that I: P = I, which
completes the proof. D
Corollary t .3.9
If an (n - d)-dimensional algebraic set in K" is determined by d polynomials,
then all its irreducible components are of dimension n - d.
Theorem 1.3.10
If PI' P 2 are prime ideals of the ring of polynon1ials K[X 1 , ..., Xn] with
coefficients in a field K and P is a minimal prime ideal of PI +P 2 , then
ht(P)  ht(P 1 )+ht(P2)'
Proof
Write R = K[X 1 , ..., X n ], R 1 = RIEl, R 2 = RIP 2 . We shall use the
diagraln
I L l  RKR
!p
v
> R 1 @KR 2
R

32
Dimension and Multiplicity
[Ch.
where 'J1 is the tensor product of the natural homomorphisms R -+ Rl'
R -4 R 2 , I = Ker(v) = Im(P t @R + R@P 2 -+ R(g)R), and # is the multi-
plication mapping, i.e. ,u(Er,(g)r;) = Erir for r;, r; E R. The kernel Ker(,u)
i I
is generated, as an ideal of R(g)R, by elenlents of the form r@ I-I r,
rER, because Lrjr = 0 implies EriQ5)r = E(rj@l) (l@r-r,@I).
I I i
Since R = K[X 1 , ..., Xn] is a polynomial ring, it is easily seen that Ker(,u)
'is generated by the elements X t (8)1-1@X I , ",,,(g)I-l@Xn'
The ideal p-I(P) is a minimal prime ideal of p,-1(Pt +P2), and, since
Im(,uA) = p,Im(Pl (g)R+R(g)P 2 -+R@R) = PI +P2, it follows that p-l(PI +
+P2) = I+Ker(,u). Hence Vp,-I(P) is a minimal prime ideal of the ideal
V,u-I(PI +P 2 ) = vKer(,u), generated by the elements veX t @I-I@X 1 ), ...
... , v(Xn(g) 1- 1 @X n ). By the Krull theorem, we have ht ('Pp,-1 (P»)  n.
Let Q c: Vp-l(P) be a minimal prime ideal of the algebra R 1 (g)R 2 . By
Theorem 1.1.15, we get
n  ht(vp,-I(P)/Q)
= dim(R I @R 2 /Q)-dim(R I (8)R 2 /vp,-I(P»
= dim(R I (8)R 2 /Q)-dim(R/P),
since RIR2/Vp,-1(P)  R@R/p-l(P)  R/P; hence dimR/P = n-ht(P)
}' ields
ht(P)  2n-dim«RlR2)/Q). (5)
We shall prove that, for any minimal prime ideal Q of R 1 (gJR 2 , ,ve have
dim«R 1 @R 2 )/Q) = dimR 1 +dimR 2 . (6)
Formulae (5), (6), and Theorem 1.1.15 yield the theorem. By the
Normalization Theorem 1.1.6, the K-algebras Rt, R 2 are integral exten-
sions At c: R 1 , A 2 c: R 2 of polynomial algebras. As can easily be deduced,
the algebra R 1 R2 is an integral extension of the algebra Al @A 2 , which is
also a polynomial algebra. Hence (R t @R 2 )/Q is an integral extension
of the algebra (AI @A 2 )/(Qr.(A l (8)A 2 ), and consequently we get
dimR 1 +dimR 2 = dimA l +dimA 2 = dim(AlA2)'
dim(R 1 (g)R 2 /Q) = dim(A t f6)A 2 )/(Qn(A t (8)A 2 »).
Thus, to prove (6), it is sufficient to show that Qn(At @A 2 ) = O. Since
Q is a minimal prime ideal of the algebra RlR2 and the algebra AlA.2
has no zero-divisors, in order to derive the formula Qn(At (g)A 2 ) = 0
from Theorem 2.3.22 in [B], we must show that no element of Al (i9A 2
is zero-divisor in R 1 f8>R 2 .
We use an obvious observation: if T c: T'is a ring extension, and every
element of T' is contained in a free T-submodule of T', then an element of
T which is not a zero-divisor in T is not a zero-divisor in T'. The field of
fractions K, of the ring Rt contains the field of fractions L, of the ring Ah

I]
Polynomial Rings
33
i = 1, 2; hence the L 1 @L 2 -module K 1 @K 2 is free. In the field of fractions
of a domain, every finitely generated submodule is contained in a free
submodule with one generator; the Al (8)A 2 -module L 1 @L 2 has this
property, and consequently no element of At (8)A 2 is a zero-divisor in
£1 (l;)L 2 , thus nor in K 1 @K 2 . 0
Theorem 1.3.10 can also be formulated in terms of the dimensions of
varieties.
Corollary 1.3.11
If U 1 , U 2 are subvarieties in Kn and a variety U is an irreducible compo-
nent of the algebraic set U 1 f1U 2 , then dinlU  dimU 1 +dimU 2 -n.
It is worth stressing that the proof of Theorem 1.3.10 is inspired by the
isomorphism U 1 "U 2  (U 1 x U 2 )f1LJ, where U 1 x U 2 c: K2" is the product
of varieties, with K[Ul]K[U2] as its ring of polynomial functions, and L1
is the diagonal in Kn x K n  K2n.
The following theorem is related to Corollary 1.1.10.
Theorem 1.3.12
In the ring K[X 1 , ..., Xn] of polynomials with coefficients in a field K,
for every maximal ideal m there exist polynomials /1 (X 1 ),/2(X 1 , X 2 ), ...
. . . , f,,(X 1 , ..., Xn) which generate m.
Proof
Let R = K[X 1 , ..., X n ]; since nt is maximal, it follows froln Corollary
1.1.9 that trde8kRlm = o. Writing Xl' ..., x" for the residue classes of
Xl' ..., X n in Rim, we obtain a sequence of finite field extensions
K c: K[XtJ c: K[Xl' X2] C ... c: K[x 1 , ..., XII].
Let 11 (X 1 ), 12 (Xl , X 2 ), ... , In (Xl , X 2 , ..., Xn) be polynomials such that
Jj(x 1, ..., x j _ 1 , X) is the minimal polynomial of the element x J over the
field K[Xl' ..., XJ-l],j = 1, ..., n.
Applying induction on n, we shall show that 11' ..., In generate the
ideal m. For n = 1, the ideal m is principal and is generated by fl. Assume
that the theorem holds for maximal ideals of rings of polynomials in less
than n indeterminates for the polynomials constructed above. Obviously,
the ideal (f1) c: R is prime, R/(fl)  K[x 1 , X 2 , ..., Xn], and ttt determines
in the ring RI(f1) the maximal ideal m = tnl(fl). It follows from the induc-
tive hypothesis that the polynomials 12 (Xl , X 2 ), ... ,In(xl, X 2 , . It' X,,)
generate m , whence the polynomials It, ..., f" generate In.
Exercises
1. Let K be a field. We define a homomorphism g: K[X, Y)(x) -+ K(Y) of the loeal-
ization of the polynomial ring by the conditions g(X) = 0, g(Y) = Y, and denote R
= g-1 (K), h = gl R: R  K, P = Ker(h). Prove the following assertions:

34
Dimension and Multiplicity
[Ch.
(i) The ring R consists of rational functions of the form (au+Xv)f(u+Xw), where
a E K, 0 ::p. u e K[Y], v, IV E K[X, Y].
(ii) The ideal P is not finitely generated.
(Hi) P is the only maxin1al ideal of R.
(iv) If I' = Xlnvf(u+Xw), m > 0 and u, v, w satisfy the conditions III (i) then
P'"+ 1 eRr, and, if X does not divide v in K[X, Y], then pm ct: R,..
(v) P is the only non-zero prhne ideal of R, whence dimR = 1.
(vi) The Krull dimension of the polynomial ring R[T] is equal to 3. [Define the
homomorphism f: R[T].... K(X, Y) by the conditions fer) = r for r E R,f(T) = Y, and
then show that the kernel of I, generated by the elements XY"(Y - T)f(u+Xw), "
= 0, 1, ..., is properly contained in the ideal PR[T].]
(vii) The ring R is normal.
2. Let R = K[X 2 , Xy2, y2, X 3 ] be a subring of the polynomial ring K[X, Y] with
coeffiients in a neld K. Prove that the associated prime ideals of (X 3 ) c R are the follow-
ing: th n1inimal prime ideal P 1 = (X 2 , XY 2 , X 3 ) of the ideal (X3), to which corresponds
the primary con1nent Ql = (X 3 , X4), and the etnbedded prime ideal P 2 = (X 2 , XY2,
y2, X 3 ). The ideals Q2,nJ = (X 3 , xy 2 , y2n1), 111 = 1, 2, ..., are P 2-pri111ary and (X 3)
= Ql n Qz,m.
1.4 SETS OF PARAMETERS
In this section, we study the dimension of a local ring. We shall prove that
this dimension is equal to the minimal cardinality of sets of generators of
m-primary ideals, where m is the maximal ideal. A set for which this mini-
mum is attained is called a set of parameters of a ring. We shall prove a the-
orem on the independence of a set of parameters from which it follows that
if a ring contains a field K then every set of parameters is algebraically
independent over K; one can draw an analogy between this result and
those in Section 1.1. Of particular importance are local rings for which
there exists a set of parameters generating a maximal ideal. We call such
rings regular, and the corresponding set of parameters-a regular one.
We shall show that a regular ring is a domain. A more detailed study of
regular rings will be found in Chapter II; here we shall prove some of
their basic properties. We give a geometric interpretation of the concepts
of a set of parameters, regular ring, and regular set of parameters; it turns
out that the local ring of a point on a variety is regular exactly when the
point is not singular.
The theorems proved at the end of Section 1.2 enable us to give a new
characterization of the dimension of a local ring.
Theorem 1.4.1
If (R, ttt) is a local ring of dimension d, then no m-prinlary ideal is generated
by less than d elements, and there exists an t1t-primary ideal which is gen-
erated by exactly d elements.

I]
Sets of Parameters
35
Proof
Let Q be an m..primary ideal generated by s elements. Then m is a unique
minitnal prime ideal of Q, whence it follows froln the Krull theorenl that
d = ht(m)  s.
By Theorem 1.2.16, there exists an ideal I generated by d elements such
that m is its minimal prime ideal. Since m is the only prime ideal associated
with I, by Corollary 2.3.18 in [B] I is m-primary. 0
Definition 1.4.2
Let (R, m) be a local ring of dimension d = dimR. Any d-element set of
generators of an m-primary ideal is called a set of parameters of the local
ring (R, m).
Corollary 1.2.18 yields
Corollary 1.4.3
If (R, m) is a local ring and x em is not a zero-divisor, then there exists
a set of parameters of R containing x.
Proof
Write d = dimR/(x). If the residue classes of elements Xl' .11' Xd modulo
(x) generate an m/(x)-primary ideal of R/(x), then x, Xl' ..., Xd generate
an m-primary ideal. By Corollary 1.2.18, dimR = d+ 1; thus X, Xl' ..., Xd
is a set of parameters of R. D
Example 1.4.4
Let K be a field and let R be the localization of the polynomial ring
K[X l , ..., Xn] with respect to the ideal (Xl' ..., X n ). Thus R is the local
ring of the point (0, ..., 0) on the variety Kn. As we know, dimR = n; the
elements Xl' ..., X n generate a maximal ideal, and thus constitute a set
of parameters. Every set of the form Xl, X2, ..., X:n, where k 1  1, ...
. . . , k n  1, is also a set of parameters.
Example 1.4.5
Let (R, m) be the local ring of the point (0, 0) on the curve defined by the
equation X2 = y3; thus R is the localization of the ring K[x, y] = K[X, Y]/
/(X2 - y3) with respect to the prime ideal (x, y). Since the polynomial
X 2 - y3 is irreducible, R is a domain and dim R = 1 (by Corollary 1.2.18).
The maximal ideal m = (x, y) admits two generators; since, in view of
x 2 = y3, m is the radical of the ideal (x), the latter is m-primary. Hence
{x} is a set of parameters, and so is {y}.
Note that m is not generated by one element; indeed, it is easily seen
that
m/m 2  (X, y)/[(X, Y)2+(X 2 _y3)]
= (X, Y)j(X, y)2  KXKY,
whence dimKm/m 2 = 2. and tTI is not a principal ideal.

36
Dimension and Multiplicity
[Ch.
Theorem 1.4.6
If {u 1, ... , Ud} is a set of parameters of a local ring R, then
dim R / (U1, ..., Uk) = dim R - k, k = 1, 2, ..., d,
and the residue classes of the elements Uk + 1, ..., Ud form a set of parameters
of the ring R/(Ul, ..., Uk)'
Proof
Let us denote R = R/(Ul, ..., Ut); write also r for the residue class in R
of r E R. The ideal (Uk + 1, ..., Ud) is m -primary, where m denotes the maximal
ideal of R . Hence, according to Theorem 1.4.1, dim R  d-k.
Suppose the ideal (X l"'" X 3 ) is m- primary. It is easily seen that
the ideal (Xl' ..., X" U 1 , ..., Uk) is m-primary, ,vhence s+k  d. Thus
dimR  d-k, which yields the desired result.
In Section 1.1 we introduced the concept of the dimension of a K-algebra
which is a domain, as the maximal number of elements algebraically inde-
pendent over the field K. It turns out that a set of parameters of a local
ring is an analogue of a maximal algebraically independent set. We shall
now try to exhibit this analogy more precisely. A considerable difficulty
lies in the fact that a local ring may contain no field.
The first, weakest, type of independence of sets of parameters in geo-
metric local rings will be stated as follows.
Let aJ(a, V) be the local ring of a point a on the variety V = V(P) c Kn
determined by a prime ideal P c: K[X 1 , ... , Xn], and let elements Ul = WI +
+P, ..., Uti = Wd+P E K[V]m a = {)(a, V) constitute a set of parameters
of {)(a, V); ma denotes the maximal ideal (Xl -a I , ..., xn-an) of the ring
K[V], corresponding to the point a = (a 1, ..., a,.). We shall show that the
point a forms a component of the algebraic set determined by the ideal
p + (WI' ..., W d) and does not belong to the remaining components, or,
in other words, that a is the only point of a certain neighbourhood of point
a which is a common zero of all the parameters regarded as functions
defined on the respective subsets of V.
Indeed, let I = P+ (WI' ..., w,,) = Qlf1 ... n Q, be an irredundant
primary decomposition of a primary ideal I c: K[X 1 , ..., X n ]. It is easily
seen that also I = Ql (\ ... n Qs is an irredundant primary decomposition
( I, Q 1, ..., Qs are the images of 1, Q 1, ..., Qs under the natural homo-
morphism K[X I , ..., Xn] -+ K[V]). Applying Theorem 1.4.7 (vii) in [B],
to the localization K[V] -+ K[V]m G , we infer that, since the ideal (Ul, ..., Ud)
is ma-primary, one of the ideals Ql, ..., Qs, say Ql, is m-primary, where
m = (X 1 -a 1 , ..., Xn-a,.), and Q2 ct: m, ..., Qs cf: m. Thus V(Ql)
= {a}, a  V(Q2)U... u V(Q,), which is the desired result. D
In the case of an arbitrary local ring, the independence of a set of
parameters is expressed in the following theorem and its corollaries:

I]
Sets of Parameters
37
Theorem 1.4.7
Let Q be an m-primary ideal of a local ring (R, m). Let us also write
K = RIm.
(i) If the ideal Q is generated by a set of parameters of R, then a set
of generators {Ul'"'' Ud} of Q is a set of parameters if and only if the
residue classes "1' II" Ud of U 1, ..., Uti modulo mQ form a basis of the
linear space Q ImQ over the field K.
(ii) If the ideal Q is generated by a set of parameters {U1, ..., Uti}, then
00
the elements u 1, ..., Ud of the graded K-algebra (f) Qn ImQn are gener-
n=O
ators of this algebra and are algebraically independent over K.
The independence of a set of parameters can also be expressed intrin-
sically in terms of the ring R.
Corollary 1.4.8
If a set of parameters Ul, . II , Uti of a local ring (R, m) generates an m-primary
ideal Q, and I(X 1 , ..., Xd) is a homogeneous polynomial of degree s with
coefficients in R, such that f(U1, ..., Ud) E mQs, then all coefficients of the
polynomialfbelong to m.
In the case where the ring R contains a field L (which, by means of the
natural homomorphism R -+ RIm = K, can be identified with a sub field
of K), the above corollary implies the algebraic independence of a set of
parameters over the field L.
Corollary 1.4.9
If a local ring R contains a field L, then any set of parameters is algebraic-
ally independent over L.
Proof
Let {Ul, ..., Ud} be a set of parameters of R. Suppose there exists a non-zero
polynomialfE L[X 1 , II" X d ] such thatf(u1' '11' Ud) = O. Let I = !s+1s+1 +
+ ... +Is+" where h,h+l, ...,Is+t are homogeneous polynomials of
degrees s, s+ 1, ..., s+ t, respectively, and h :P o. Thus we have
!s(U1, ..., u,,)
= -!S+1(U1, ..., u,,)- II' -fs+t(U1' ..., Ud) EQS+l c mQs,
and it follows from the foregoing corollary that all the coefficients of Is
belong to m. Yet, by the assumption, they also belong to L, and Lnm = 0;
this yields f, = 0, a contradiction. D
In just the same way, we prove the next corollary which exhibits the
so-called analytical independence of parameters.

38
Dimension and Multiplicity
[Chi
Corollary 1.4.10
If a local ring R is complete in the m-adic topology and contains a field L,
then a set of parameters {u 1, ..., u,,} of R is. analytically independent over
L, i.e. the continuous homomorphism h: L [[Xl' ..., X d ]] -+ R given by
the conditions h(X 1 ) = Ul, ..., h(X,,) = Ud is an injection.
Proolof Theorem 1.4.7
To prove (i), observe that if the ideal Q is generated by a set of parameters,
then by Theorem 1.4.1, the number d = dimR is equal to the minimal
cardinality of sets of generators of Q. Thus (i) follows from (A.S.l).
We shall now prove (ii); to begin with we define a surjection of graded
00
K-algebras f: K[X 1 , ..., X d ] -+ €a Qrt /mQn by the conditions I(X t ) = Ul +
n=O
+mQ, ...,/(X d ) = u,,+mQ. We shall prove below that if we had I
= Ker(f) :/= 0 there would exist an m-primary ideal generated by less than
d elements; this is impossible, and therefore it will be proved that I is an
isomorphism.
Since the ideal I is homogeneous, there exists a nonzero homogeneous
element WI E I of a positive degree. By Theorem 1.1.20, there exist homo-
geneous polynomials of positive degree W2, ..., w" such that the algebra
K[X 1 , ..., X d ] is integral over the algebra A = K[Wl, ..., }Vd]' Replacing
WI' ..., w" by their suitable powers, we can additionally assume that the
degrees of these polynomials are all equal to p.
The elements, Xl' ..., X d are integral over A, hence, for a sufficiently
large exponent k, there exist homogeneous polynomials g'J = g'j(Wl' ..., w,,)
such that
XJ + g k-l t j XJ -1 + ... + go, j = 0, j = 1, 2, ..., d.
The polynomials W2, ..., w" are of degree p. and I(X,) = ui+mQ E Q/mQ,
consequently I( W 2), ..., f( w d) E QP /mQP , and there exist elements v 2, ...
..., Vd E QP such that few}) = vj+mQP, j = 2, ..., d. Let us write J
= (v 2, ..., 'V,,); since the homogeneous polynomial g 'J is of degree k - i  1,
we have
f(g'j( Wl, ..., W d) X})
= gij(O, 'lJ2, ..., Vd)u}+mQk E (J +mQk)/mQ".
This and the relations of integral dependence above yield jointly u} E J +
+mQk,j = 1, 2, ..., d. Suppose that n  dk; then at least one of the
exponents of the monomial Ul ... u of degree n is not less than k. If i 1  k,
then
U1 ... u = u"(U1-kU2 ... u) E (J+mQk)Q"-k c J+mQ",
and consequently Q" c J +mQ" for sufficiently large n. Thus
m(J+Q")/J) = {mJ+mQn+J)/J
= (J+mQ")/J-::J (J+Q")/J,

I]
Sets of Pa.'ameters
39
and the Nakayanla lemlna yields J + Qn = J. Consequently Q" c J C Ql'
whence it foHows that the ideal J is m-prilnary and has d-l generators. 0
Theorem 1.4.11
If (R, m) is a local ring, then the following conditions are equivalent:
(i) m is generated by a set of parameters,
(ii) dimRltn m/m 2 = dimR,
00
(Hi) the graded Rltlt-algebra E9 rn"/tn"+1 is isomorphic to an Rlm-al-
n=O
gebra of polynomials.
Proof
The implication (i)  (iii) follows from Theorem 1.4.7; we shall give the
proof of the opposite implication in Section 1.5 using quite different tech-
niques. The equivalence of (i) and (ii) follows directly from Lemma A.S,l
and Theorem 1.4.7 (i). 0
Definition t .4.12
A local ring (R, 111) is said to be regular if and only if it satisfies the three
equivalent conditions (i), (ii), (iii) in Theoren1 1.4. t 1. A set of parameters
which generates t11 is called a regular set of parameters.
Example 1.4.13
In algebraic geoJnetry a point a" of a variety V, or more generally a sub...
variety W c V, is said to be regular when the local ring (1)(a, V), or lV( W, V),
is regular. Example 1.4.4 shows that all the points of the variety K" are
regular, while the point (0, 0) of the curve given by the equation X 2 = yJ
is not regular (see Example 1.4.5). We shall give a geometrical interpreta..
tion of the regularity of the ring lV(a, V) at the end of this section.
Example 1.4.14
The ring of formal po\ver series K [[X t , ..., Xn]] is a regular local ring
Indeed, we can easily deduce from Corollary 1.2.18 that its dimension is
equal to 11; its maximal ideal is generated by 11 elements Xl' ..., Xli' Thus
condition (i) in Theorem t .4.11 is satisfied.
Theorem 1.4.15
A regular local ring has no zero-divisors.
Proof
Let x, y be two nonzero elenlents of a regular local ring R. Since, by
00
Corollary 2.5.5 in [B], \ve have n ln n = 0, there exist positive integers
11=0
p, q such that x Em P "nt P + 1 , y Em q "m q + 1 . Consequently, the residue
classes x+ntP+t, y+nt Q + 1 are nonzero elements of Rim-algebra of poly-

40
Dimension and Multiplicity
[ChI
co
nomials  m"/m"+l. Thus the product xy is different from zero because
11=0
xy+m P + f + 1 = (x+m P + 1 )(y+m t + t ) =/:. O.
Let us note that according to the foregoing arguments the assignment
x H p, for x E m P "m P + 1 , is a discrete valuation. In general, R is not a
valuation ring. D
A regular set of parameters fulfils certain conditions which are also
independence-like.
Definition 1.4.16
A sequence Xl' X2, ..., X:J of elements of a ring R is called regular if the
following conditions are satisfied
(O):(Xl) = (0),
(Xl' ..., Xk -1) : (X k ) = (Xl' ..., X k - t) , k = 2, ..., s,
and the ideal (Xl' ..., xs) is proper.
These conditions mean that the residue class of Xk in the ring R/(xt, ...
..., X1c-t), k = 1, 2, ..., s, is not a zero-divisor.
Example 1.4.17
In the ring R = T[X 1 , ..., X,,] of polynomials with coefficients in a ring T,
the sequence XI' X 2 , ..., XII is regular.
Before proceeding to prove that elements of a regular set of parameters
form a regular sequence we shall prove the following
Theorem 1.4.18
Let (R, m) be a local ring. The following conditions are equivalent:
(i) a sequence U t, ..., u" of generators of m is regular,
(ii) a sequence
(0) c (Ul) C (Ul, U2) c: ... C (Ut, ..., u.) = tn
is a chain of prime ideals.
Proof
(i) => (ii). We apply induction on d.
If d = 0, then m = 0, and (0) is a prime ideal. Suppose the implication
(i) => (ii) holds for all rings of dimension less than d, d  1, and consider
the factor ring R = R/(Ut) and the residue classes U2, ..., Ud of U2, ..., Ud.
The sequence U2' ... , Ud is regular; indeed, if Mike (U2, ..., Uk-I), 2  k  d,
then ru" e (U1, ... , U"-l), whence r E (u 1 , ..., Uk- J)' and therefore r
E (U 2, ..., ii t - 1 ).

I]
Sets of Parameters
41
By the inductive hypothesis, the chain CO) c (U2) c ... c: (U2, ..., Ud )
is a chain of prime ideals of R. Hence (Ut) C (UI, U2) C ... c: (Ul' ..., Ud)
is a chain of prime ideals in R.
The condition (0):{U1) = CO) yields Ut :/= 0, thus it remains to prove
that (0) is a prime ideal. Suppose that xy = 0 for some nonzero x, Y E R.
00
According to Corollary 2.6.5 in [B], n (Ul)n = (0); hence there exist
n=O
positive integers p, q such that x = Xt uf, y = Yl ut for some X t , Yt ,p (Ul)'
and consequently 0 = xy = XtYt u+q, which, in view of the condition
(O):(Ul) = (0), implies XtYt = O. Now, (Ut) being prime, we have either
Xl E (Ut) or Yl E (U t ), contrary to the assumption.
(ii) => (i). If r e (Ul, ..., Uk-t) : (Uk), then rUk E (u I , ..., Uk-I)' Since Uk
does not belong to the prime ideal (Ul, ..., Uk-t), it follows that r E (Ul, ...
... , Uk-I), i.e. the sequence Ul, ..., u" is regular. 0
The above theorem allows us to give a new characterization of regular
sets of parameters. ·
Theorem 1.4.19
Elements Ul, ..., Ud of a local ring (R, m) fornl a regular set of parameters
jf and only if Ul, ... , Ud form a regular sequence of generators of the maximal
ideal m.
Proof
=> . Suppose that {u 1, ..., Ud} is a regular set of parameters of R. It
foHows that dimR = d, and R is a regular ring.
If d = 1, then the element U t is not a zero-divisor (see Theorem 1.4.15),
and the one-element sequence Ut is regular.
Suppose that regular sets of parameters form regular sequences in
all rings of dimension less than d, d  1. Let {U 1, ..., Ud} be a regular set
of parameters of the local ring R. The ring R is of dimension d; by Corol-
lary 1.2.18, the ring R = R/(ut) is of dimension d-l and is regular
since the residue classes U2, ..., Ud of the elements U2, ..., Ud form a set of
generators of the maximal ideal. Thus U2, ..., U4 is a regular set of par-
ameters, which, by the inductive hypothesis, forms a regular sequence.
Consequently the sequence u 1, ..., Ud is also regular because, by Theorem
1.4.15, u 1 is not a zero-divisor.
<=. Suppose now that u 1, ..., Ud is a regular sequence of generators
of the maximal ideal. By Theorem 1.4.1 we have d  dimR, and by The-
orem 1.4.18 the ideals (0) c= (U1) c: ... C (Ul, ... , Ud) form a chain of prime
ideals of length d, whence d  dimR. Thus \ve get d = dimR, and con-
sequently the elements Ul, ..., Ud form a regular set of paralneters. 0

42
Dimension and Multiplicity
[Clt.
Remark 1.4.20
From the preceding theorem it follows that any permutation of a regular
sequence of generators of the maximal ideal in a local ring is also a regular
sequence. In Chapter III, we shall show that this property is valid for all
regular sequences in a local ring.
Corollary 1.4.21
If {u 1, ..., Ud} is a regular set of paran1eters of a local ring R, then for
any k, 1  k  d, the local ring R/(Ut, ..., Uk) is regular, and the residue
classes of the elelnents Uk+ l' ..., Uti form a regular set of parameters of
this ring.
Corollary 1.4.22
If (R, m) is a local ring, x Em is not a zero-divisor, and the ring R/(x) is
regular, then the ring .R is also regular.
PI.oof
Suppose the residue classes of U2, ..., Ud E lfr form a regular set u 2 , ..., il ll
of parameters of R/{x). If we take Ul = x, then, by Theorem 1.4.19. the
elements Ut, ..., Uti form a regular sequence of generators of the maximal
ideal of R. Thus, it follows from Theorem 1.4.19 that R is a regular ring. 0
Theorem 1.4.23
If (R, m) is a regular local ring and x is a nonzero element of m, then there
exists a regular set of parameters Ul, ..., Ud such that x, U2, ..., Ud is a set
of parameters.
Proof
The theorem is evident for dimR = ° or 1. Suppose that the assertion i.
true for the rings of dimension less than dimR = d, d > I. Let us denote
by PI' ..., P Il the minimal prime ideals of (x). By Theorem 1.2.10, they
are of height I, hence none of them is the maximal ideal. From Theorem
1.1.7 (iii) in (B], it follows that m ''In 2 ct: PI U ... u P, and therefore
there exists an element U2 em'\.m 2 with U2  PIU ... uPs. Thus we have
(x) ct: (U2)' since otherwise a minimal prime ideal P of the ideal (U2)
would be also a minimal prime ideal of (x), contrary to the choice of U2'
By Corollary 1.2.18, the dimension of the ring R = R/(U2) is equal d-I;
when tn c: R denotes the maximal ideal, it is easily seen that m / m 2 is
generated by d-l elements, and therefore the ring R is regular. Since
x = x + (U2) :f: 0, there exists, by the inductive hypothesis, a regulai. set
of parameters tit, it 3 , ..., Ud of R, such that the elelnents x , U3, ..., lid
form a set of parameters of R If we set Q = (x, U2' U3, ..., tld), then the
ideal Q = Q/{u,.) is m-primary, whence Q is m- primary, and consequently
x, U2, ..., lid is a set of parameters of R; the elements Ul, ..., Ud generte
the ideal m, and so they constitute a regular set of parameters. 0

I]
Sets of Parameters
43
Regular sequences are a very effective tool in the study of ideals. We
shall develop this method in Chapter III.
In the definition of a regular ring there is no apparent geometric intu-
ition. We shall now give a geometric interpretation of the regularity of the
local ring of a point on a variety.
Theorem 1.4.24
Let ((9(a, V), m n ) be the local ring of a point a = (a 1, ..., an) on a variety
V c: Kft determined by a prime ideal P c: K[X 1 , ..., J] = K[X], where
P is generated by polynomials 11' ..., f,.. Then
dimK(m../m;) + rank [ , (a)] = n. (7)
Proof
Let 9J1 = (Xl -aI' ..., X n - an) be the maximal ideal of K[X], corresponding
to the point a. Then, we have (9(a, V)  (K[X]/P)IR/P, which, according to
Corollary 1.4.20 in [B], yields
maIm;  (WlIP)/(9RIP)2 = (rolIP)/(9]l2+P)jP  IDl/(ft)12+P).
We thus have an exact sequence of K[X]-modules
D
0-+ (9J1 2 +p)lm 2 -+ 9J1;9J12  Wl/(9Jl2+P) --. 0 (8)
annihilated by the ideal 9J1. So we can regard these modules as modules
over K[x]/9J1 = K. Clearly, 9Jl19Jl 2  K(X l - a l)EE> ... Et)K(Xn-a n ). The
module Im(D) is generated by the elements D(w+9J1 2 ) for w E P. Since
"
w(a) = 0, it follows that w = 2: ;; (a) (Xj-ai)+W", where w" E 9JP,
i= 1 f
and we obtain an exact sequence
D n v
0--. (9J12+P)f9J12  Et) K(Xi-a,) -+ 9R1(9R2+P) --. 0,
;=1
in \vhich v is induced by the natural homomorphism, and
n
D(w+9JP) = L :; (a)(Xj-aj)'
. I i
1=
The ideal P is generated by the polynomials /1' ..., f,; accordingly , for
W E P, there exist g 1, ..., g, E K[X] such that w = g Ifl + ... + g,f" whence,
because Jj(a) = 0, we have
r n
D(w+9JP) = ))  a(gjjj) (a)(X,-aa
ax,
j= 1 1= 1
r
= Lgi a )D(JJ+9Jl 2 ).
Jf: 1

44
Dimension and Multiplicity
[Ch,
Thus the module Im(D) is gel\erated by D(fl + 9)12), ..., D(f, + 9)12). Since
the rank. of the matrix [:, (a)] is, by definition, equal to the dimension
of the space 1m (D), we have actually proved (7) because 9.1l/(9)l2+P)
 malm. 0
The space tangent to the variety Vat the point a is defined by the system
of equations
n
" afj (
L...J ax (a) X, - a,) = 0, j = 1, ..., r,
1=1 I
which, as can easily be deduced from the foregoing calculations, is equiv-
alent to the system
(9)
"
"'" aw (a)(X, - a,) = 0 for aU w E P .
L...J ax,
1=1
Thus, the space described by the system (9) does not depend on the choice
of generators 11' ..., f, of the ideal P.
The dimension of the tangent space is equal to n-rank [, (a)]
= dimma/m. By Theorem 1.4.24 we have given the number dimma/m;
a geometrical interpretation as the dimension of the space tangent to the
variety V at the point a. Thus, Theorem 1.4.11 yields the following
Corollary 1.4.25
The local ring (9(a, V) of a point a on a variety V is regular if an only if
the dimension of the tangent space to V at the point a is equal to the dimen-
sion of V.
To end off the section we shall discuss a geometric property of regular
sets of parameters of the local ring C!J(a, V).
Assume that a is a regular point of a variety V, i.e. such that the ring
(J) (a, V) is regular; let U I' ...,"" be a regular set of parameters. We can
additionally assume that U I' ..., Ud E K[V] are the residue classes of poly-
nomials WI' ..., W" E K[X]. Because of the regularity of the ring (9(a, V),
the residue classes of U 1 , ..., u" modulo m form a basis for malm. It
follows from the exactness of sequence (8), that 9J1/9)12 = K(wl + 9J12)Ea...
... Et> K(Wd+9J12)Im(D). Thus we deduce from the identity Wk+IDl2
n
= 2: i (a) (X,-a,) + IDP that the rank of the n x (d+r)-matrix
1=1 t
r aWl aWd afl aj',. ]
J a = ax, (a),..., ax, (a), ax, (a), ..., ax, (a)
is equal to n.

I]
Sets of Parameters
45
The above fact has a direct geometric interpretation, at least in the
case of K = C. Let us define a mapping rp: C" --. C d + r by the formula
rp(x) = (WI (X), ..., wd(x),fl(X), ...,!r(x»). The matrix J a is then the value
of the Jacobian of rp at the point a (i.e. J a is the matrix of the tangent
mapping at the point a). The rank of J a is equal to n, whence rp is a homeo-
morphism of a neighbourhood N c cn of a onto rp(N). Since rp(V) c c;4
C Cd+r, and the rank of the matrix [  (0), ..., : (0)] is equal to
d, rplV is a homeomorphism of the neighbourhood NnV onto a neighbour-
hood of 0 E Cd. The mapping rplV: V --. Cd is given by the formula
(rpl V) (x) = (WI (x), ..., Wd(X»,
It is easily seen that non-regular sets of parameters do not have those
properties. Indeed, let us take V = K n , a == (0, ..., 0), and let nl t ...
..., nn: Kn -.. K be the projections n,(xI' ..., x,,) = Xi. Then U1 = ni, ...
..., U II = n; is not a regular set of parameters, the mapping rplV has the
form (Xl' ..., XII) H (x, ..., X;), and it is not a homeomorphism of any
neighbourhood of O.
The importance of regular rings lies mainly in the fact that almost all
points of an algebraic variety are regular. A precise formulation of this
result is given below (the proof can be found for instance in [Y], p. 113).
Theorem 1.4.26
The set of regular points of an algebraic variety V is nonempty and open
in the Zariski topology, i.e. the set of non-regular points of V form an
algebraic subset of dimension less than the dimension of V.
This theorem suggests a plausible direction for the study of algebraic
varieties; non-regular (singular) points can be studied through the analysis
of algebraic properties of the corresponding local rings, i.e. through the
study of non-regular local rings.
In this book we introduce two classes of rings containing regular local
rings, namely Cohen-Macaulay rings (Chapter III) and Gorenstein rings
(Chapter IV). Thus the theories here presented can be regarded as an
attempt to classify algebraically the singularities of algebraic varieties.
Exel'cises
1. Prove that a regular local ring is a normal ring (cf. Theorem 2.2.5). [Let K be
a field of fractions of a regular local ring R. Prove inductively that if an element rls e K
is integral over R then r e Rs+m ft for every n. To this end, observe that if r = as+b,
a e R, b E m ft , then the element b Isis integral over R, and there eixst elements d, C1,
C.z, ... e R such that db" = c"s", k = 1,2,... Put G(x) = x+m" for x e m"""m t + 1 ,

46
Dimension and Multiplicity
[Ch.
and show that G(x, y) = G(x) G(y). Deduce from the former equation that G(.v)/G(b),
i.e. b-a's e m,,+1 for some a' e R.]
2. Let K be a field, R = K[X, Y, ZJ/(X, Y) n(Z), and let x, y, z denote the residue
classes of the indeterminates X, Y, Z in R. Prove that ht (x, y+ z) = 2, but ht (x) = o.
Compare this result with Theorem 1.4.6.
1.5 HILBERT-SAMUEL POLYNOMIALS AND MULTIPLICITY
In this section, we define the Hilbert-Samuel function of a finitely generated
module M over a Noetherian ring R with respect to an ideal Q c: R under
the assumption I(MjQM) < co. This function associates with a positive
integer n the length of the module MjQ"M; it turns out that there exists
a polynomial PQ(X, M), called the Hilbert-Samuel polynomial, such that
PQ(n, M) = /(M/Q"M) for sufficiently large n. Its term of the highest
degree has the form eQ(M) {; , where eQ(M) > 0 is a positive integer;
we shall sho\v that the degree d is equal to the Krull dimension of the ring
RjAnn(M). The coefficient eQ(M) is called the multiplicity of the Inodule
M with respect to the ideal Q. The rest of the section is devoted to the
computation of multiplicity in terms of a suitable Koszul complex under
certain additional assumptions about the ideal Q. The resulting formula
turns out to be of great importance for geometric applications which will
be discussed in the next section.
For any integer-valued functionf defined on the set of positive integers,
we denote by .Af the function given by the formula (L1!) (n) = f(n + 1) - f(n)
and by .Ak[ the result of k-fold application of the operator L1. Iff is a poly-
nomial in one indeterminate its monomial of the highest degree is called
the leading form off. We sometimes write f(n) = and + ..., or f(n) = g(n) +
+ ... to suggest that and is the leading form of f, or that f and g have the
same leading forms.
In the sequel we shall deal with functions f: N --. Z for which there
exists a polynomial g with rational coefficients such that f(n) = g(n) for
sufficiently large n. In this case we shall write briefly that f is a poly-
nomial for large n.
The following simple fact will be repeatedly used throughout this
section:
Lemma 1.5.1
If f: N --. Z is a function such that, for large n, Jf is a polynomial with
rational coefficients, then, for large n, f is also a polynomial with rational
coefficients. Moreover, iff -:F 0 for large n, we have degf = degL1f + 1.
Proof
We assume that f:F 0 for large n, and apply induction on degL1j: If degL1f
= -1, i.e. Jf = 0 for large n, thenfis constant for large n.

I]
Hllbert-Samael Polymonials and Multiplicity
47
If degLJf = s  0, then
(LI.D(n) = Qsn"+a,_t nS - 1 + ... +ao, a,EQ
for large n. We define
h(n) = a s slC.Zl) = S;1 n(n-l)... (n-s+l)(n-s).
Note that h is a polynomial with rational coefficients which takes integral
values, and degh = s+ 1. Furthermore,
(L1h)(n) = assl [( ;:: ) - (S:I)]
\
= assl() = a s n(n-l) ... (n-s+ 1) = Qsn s + ...
Let g = f-h; then Ltg = Llf-Lth, and degLlg < s. By the inductive hypoth-
esis, g is a poJynomial of degree  s for large n, whence f = g+h is
a polynomial of degree s + 1 for large n. D
The above proof implies directly the following
Corollary 1.5.2
n
If f(1I) = L i k , then / is a polynomial of degree k+ 1, and f(n) = n"+l/
1=0
/ (k + I) + ...
We shall begin our discussion of the main subject of this section with
an example.
Example 1.5.3
The ring of polynomials R = K[X 1 , ..., XII] with coefficients in a field
00
K is endowed with the natural structure of a graded ring, R = Ea RIf'
,,=0
and the basis of RII is the set of monomials of degree n in the indetermi-
nates Xl' ..., X d . The number of such monomials X{1X2 ... xjcr is equal to
( n1 ) = dim.ll:Rn = l,,(R II ); indeed, the choice of d-l elements
jt < j2 < ... < jd-l in the segment {I, 2, ..., n+d-l} of positive inte-
gers determines d segments {I,... ,jt -I}, {jt + I, ... ,j2 -I}, ..., {jd-t +
+1, ...,n+d-I}, whose lengths i 1 =jl-I,i2 =J2-jl-1, ...,i d = n+
+ d-l- jd-t determine a monomial of degree n. The function n H lK(R n )
_ ( It+d-l ) _ (n+d-I) ... (n+I) _ nd · I · I f
- (d-l)! - (d-l)! - (d-l)! + ... IS a poynomla 0
degreed-l. By virtue of the isomorphismR/(X 1 , ...,X d )  K and Corol-
lary 1.5.2, we get

48
Dimension and Multiplicity
[ChI
n-l
lR(R/(X t . ....XJ") = I>R«X lo ....XJ'/(X t . ....XJ'+1)
1=0
  ( i+d-l )  j4-1
= L/K(R , ) = L.J d-I = L.J (d-l)' -+ ,,'
1=0 1=0 1=0
(n-I)d n d
= d! +. . = d! + ...
The connection, outlined in the above example, between the Krull dimension
of a ring R and the degree of the polynomial lR( R/(X t , ..., Xcr)ft), will
be studied in details in the present section.
Theorem 1.5.4 (Hilbert)
ex>
Suppose that R =  Rn is a graded ring, and Ro is an Artin ring. Suppose
n-O
co
also that R, as an Ro-algebra, is generated by s elements of Rt. If M = E9 M n
n=O
is a finitely generated graded R-module, then
(i) the function A(n, M) = lRo (M n ), n = 0, 1, o. 0 takes finite values and,
for large n, it is a polynomial of degree less than s,
(ii) if, moreover, the R-module M is generated by Mo, then L1s-1 A(n, M)
 lRo(M o ).
Proof
The ring R is Noetherian as the homomorphic image of a ring of poly-
nomials in n indeterminates with coefficients in Ro.
(i) If 8 = 0, then R = Ro, and M, as a finitely generated Ro.module,
has finite length by Corollary 2.7.13 of [B]. Therefore M,. =.0 for large n,
i.e. A(n, M) = 0 for large n, and we have proved the theorem for s = O.
Suppose that the theorem holds for numbers < s, s  I. Let the ele-
ments at, ..., as e Rt be generators of the Ro.algebra R. Denote by h:
M -+ M the mapping given by the formula h(x) = as x for x EM. Thus
we have h(M n - t ) c: M,., and there exist exact sequences of Ro.modules
o -+ Kn-t -+ M n - t  M n -+ C n -+ 0,
where h n = hIM n - 1 , Kn-l = Ker(h n ), C n = Coker hno Since R is a Noether-
00 ex>
ian ring, K =  K,., C = (i) C" are finitely generated graded mod-
n=O n=zO
ules. Since as annihilates modules K and C, they are graded modules over
the graded ring R/Ra s , which is generated, as an Ro-algebra by 8-1 ele-
ments. From the additivity of the length function (Theorem 1.3.7 of [B]),
we infer that I(M,.) -1(M n - t ) = l( C n ) -1(Kn-l). Accordingly, it follows
from the inductive hypothesis that there exists a polynomial f e Q[X], of

I]
Hilbert-Samuel Polynomials and Multiplicity
49
degree < s-1 such that for large n we have A(n, M)- A(n-l, M) = f(n).
By Lemma 1.5.1, the function A(-, M) is a polynomial of degree < s for
large n.
co
(ii) Denote by T = €a Tn the polynomial ring Ro [Xl' ..., X s ], endowed
n=O
with the natural grading. By the assumption, there exist epimorphisms
T"@RoMo -+ M", whence
( n+S-l ) nS-l
I(M n )  1(T,,(8)R.M o ) = I(Mo) 8-1 = I(Mo) (s-I)1 + ...
nfJ - 1
If A(n, M) = e (s-l) 1 + ..., then LlS-1 A (n, M) = e, and the preceding
inequality yields e  I(M o ). 0
Theorem 1.5.5
Let Q be an ideal of a Noetherian ring R, and let M be a finitely generated
R-module such that l(M/QM) < 00. Then l(M/QnM) < 00 for every
n = 1, 2, ..., and the function n H-l(M /QnM) is for large n a polynomial
of degree  s, where s denotes the cardinality of any set of generators of
the ideal (Q+Ann(M»/Ann(M) of the ring R/Ann(M).
Proof
Let us write J = Ann(M /QM); it follows from the assumption and Corollary
2.7.14 in [B] that R/J is an Artin ring. Referring again to Corollary 2.7.14
in [B], we see that /(Q"M/Qn+1M) < 00, since for any n, QnM/Qn+1M is
a finitely generated R/J-module. Accordingly, I(M/Q"M) < 00 for an
arbitrary n.
In the sequel, we shall need the inclusions
Q+Ann(M) c: J c: rad(Q+Ann(M».
(10)
The first inclusion is obvious. To prove the second one, assume x E J, i.e.
xM c: QM; Lemma 1.3.2 in [B] yields xm+Qm_tXm-1+ ... +qo E Ann(M)
for some qo, ..., qm-l E Q, whence x E rad (Q+Ann(M»). Since R/J is an
Artin ring, it follows in view of (10) and Theorem 2.7.12 in [B] that R/(Q+
+Ann(M») is also an Artin ring.
Let us observe that R/(Q+Ann(M»)  R/I , where R = R/Ann , (M)
and 1= (Q+Ann(M»)/Ann(M).
00
The module Gra(M) = <i> Q"M/Qn+1M is a finitely generated graded
n=O
_ co
module over the ring GrI(R) = <i> I"/In+t. Since R/I is an Artin ring,
n=O
and GrI( R) , regarded as an R/I-algebra, is generated by s elements, it
follows from the Hilbert Theorem (Theorem 1.5.4) that for large n the

50
Dimension and Multiplicity
[Ch.
function n H /(M /Q" M) is a polynomial of degree < s. Applying the
relation
I(Q"M/Q,,+lM) = I(M/Q"+lM)-/(M/Q R M)
and Lemma 1.5.1, we see that the function n H /(M/Q"M) is, for large n,
a polynomial of degree  s. 0
Definition 1.5.6
It follows from Theorem 1.5.5 that if R is a Noetherian ring, M is a finitely
generated R-module, Q c: R is an ideal, and /(M/QM) < 00, then there
exists exactly one polynomial Pa(X, M) E Q[X] which satisfies the condition
PQ(n, M) = I(M/Q"M)
for large n. The polynomial PQ(X, M), written also PQ(M), is called the
Hilbert-Samuel polynomial of the module M with respect to the ideal Q.
The connections between the Hilbert-Samuel polynomials of the mod-
ules in a short exact sequence are stated in the following lemma:
Lemma 1.5.7
Let R be a Noetherian ring, and let
o --.. M' -+ M --.. M" -+ 0
be an exact sequence of finitely generated R-modules. If Q is an ideal of
R such that lR(M/QM) < 00, then also /R(M' /QM') < 00, lR(M" /Q!vl")
< 00, and
poeM) = PQ(M')+PQ(M")-g,
where either g is a polynomial with rational coefficients of degree less than
degPQ(M'), whose leading form has positive coefficient, or g = O.
Proof
Since Ann(M') => Ann(M), we have dimR/ (Q + Ann(M') = 0, whence,
in view of Corollary 2.7.14 in [B] it follows that l(M' /QM') < 00. The
inequality l(M" /QM")  I(M/QM) yields l(M" /QM") < 00. For every
n, we have the exact sequence
o -+ M'/M --.. M/QRM -+ M"/QRM" -.0,
where M = M'nQ"Mc:Q"M'. This sequence implies the following rela-
tion among lengths:
l(M /Qn M) = l(M" /Q" M") + l(M' / M) .
The submodules M of M' satisfy the following conditions:
(i) M' = M => M ::) ...,
(ii) QM c: M+ l' n = 0, 1, ...,
(Hi) there exists a number no such that, for n  no, M = Qn-"oMo.
(11)

I]
Hilbert-Samuel Polynomials and Multiplicity
51
The last condition results from the Artill-Ress Lemma (LeITIIUa 2.5.1
of [B]).
The inclusion Q"M' c: M c: Q"-ISo.M' for n  110 yields
I(M'/Q,,-noM')  I(M'/M:')  l(M'/QnM').
(12)
00
In view of (ii), the module EB M:'/ M, + 1 has the structure of a graded
,,=0
co
module over the graded ring €a Q"/Q"+l and is generated by M/MffJ
n::aO
$ ... ff).Mo/Mo+l' We can apply the Hilbert theoreln and deduce that
the function defined by the formula fen) = l(M' / M) is a polynomial for
large n. From inequality (12) it follows that, for large 11, P o(n - no, M')
 fen)  P Q(n, M'). These inequalities imply that the leading forms of the
polynomials f and P o (M') are identical and the polynomial g = P oeM') - f
(more exactly: the polynomial which is equal to it for large n) has a
non-negative leading form. Thus, formula (11) implies that of the lemma. 0
We can now present a characterization of the dimension of a local
ring in terms of the Hilbert-Samuel polynolnial.
l'beorem 1.5.8
If (R, tn) is a local ring, the follo\ving numbers are equal:
(i) the Krull dimension of R,
(H) the degree of the Hilbert-Samuel polyn.omial .Pm(R),
(Hi) the mini1nal cardinality of sets of generators of m-primary ideals.
Proof
Let us denote the successive nUDlbers specified in the theorem by dim R,
d(R), s(R) respectively.
By Theorem 1.4.1, we have dimR = s(R). We shall prove that dimR
 d(R) and d(R)  s(R).
To prove the first inequality, let us consider a chain Po c: P 1 c: ... c: Pr
of prime ideals in R of length r. Directly from the definition it follows
that
o  degPm(R/Pr)  ...  degPm(R/Po)  degPm(R).
Thus it is sufficient to prove that
degP. n (R/P , + 1 ) < degPm(R/P,), i = 0, 1, ..., ,'-1.
IJet x E Pi + 1 "Pi; applying Lemma 1.5.7 to the short exact sequence
x
o  R/P i  R/P i  R/(x)+P,) -+ 0,
we see that
degPm (R/ «x) + P,») < degPm(R/P,).

52
Dimension and Multiplicity
[Ch.
Consequently, the desired result fonows from the evident inequality
degPm(R/P'+l)  degPm(R/{(x)+P,»).
To prove that d(R)  s(R), denote by Q an m-primary ideal generated
by s(R) elements. By Theorem 1.5.5, we have degPa(R)  s(R). Since Q
is m-primary, there exists, by Coronary 2.4.4 in [B], a number k such
that m t c Q c: m. Hence
Pm(i, R)  Pa(i, R)  Pm(ki, R)
for sufficiently large i. Accordingly, degPo(R) = degPm{R), whence
d(R)  s(R). D
The last part of the proof, and Theorem 1.5.8 ilnply
Corollary 1.5.9
If (R, m) is a local ring, and Q is an m-primary ideal, then degPa(R)
= dim R.
Corollary 1.5.10
A A
If (R, m) is a local ring then dimR = dimR, where R denotes the comple-
tion of R in the m-adic topology (see Section 2.6 in [B]).
Proof
The corollary follows from Theorem 1.5.8 in view of the isomorphism
1\ A
RIm"  RIm" proved in Corollary 2.6.21 of [B]. 0
Using the result of Theorem 1.5.8, we can now complete the proof o.
Theorem 1.4.11.
Proof of TheorenJ 1.4.11 (conclusion)
It is our purpose to prove the implication (iii) => (i). Let elements u 1, .." Ud
form a minimal set of generators of the ideal m. The residue classes Ut, ..., lid
modulo m 2 are then linearly independent over K = Rim (cf. Lemma A.5.1)f
00
By the assumption, the algebra $ m"/m n + 1 is a polynon1ial algebra
n=O
K[Ul, ..., u d ]. Thus, we infer from Example 1.5.3 that
I( n / 1'+1 ) = ( n+d-l ) = n d - t
m ltt d-l (d-l)! + ...,
nil
!(R/m") = d! + ... = Pln(n, R).
Therefore Pm(R) is a polynomial of degree d. By Theorem 1.5.8, d = dimR,
hence U t , ..., Ud is a regular set of parameters of R. 0
As we shall see ill the sequel, it is convenient to extend the definition
of Krull dimension to modules.

I]
Hllbert-Samuel Polynomials and Multiplicity
53
Definition 1.5.11
Let R be a Noetherian ring, and let M be a finitely generated R-module. The
Krull dimension of the R-module M, written dimM, is the Krull dimension
of the ring R/Ann(M).
Since Supp(M) = V(Ann(M)) and every minimal ideal from Supp(M)
belongs to Ass(M) (see Theorem 2.4.22 in [B]), it follows that
dimM = sup dimR/P.
Pe Ass(M)
For every ideal I c: R, the Krull dimension of the R-module RII is
equal to the Krull dimension of the ring RII.
There is also an interpretation of the dimension of a module in terms
of the Hilbert-Samuel polynomial.
Theorem 1.5.12
If (R, m) is a local ring and M is a finitely generated R-module, then the
degree of the polynomial Pm{M) is equal to the Krull dimension of M.
Proof
Denote the degree of Pm(M) by d(M). By Lelnma 2.4.12 in [B], there exists
a filtration of M
0.= MoC: M 1 c ...e: M, = M,
such that M J IM J - 1  R/PJ,j = 1, ..., r, where Pj are prime ideals. Lemma
1.5.7 yields
degPtn(M) = supdegPm(RfPJ).
J
Observe that PJ :::> Ann(M) for every j; by Lemma 2.4.12 in [B] Ass(M)
c {PI' ..., P,}, and thus the sets Ass{M) and {P 1, ..., P,} have the same
minimal elements; consequently
degPm(M) = sup degPm(RfP).
Pe Ass(M)
According to Theorem 1.5.8, we have degPm{RfP) = dimRfP, whence
we finally get d(M) = dimM. 0
Arguments similar to those which preceded Corollary 1.5.9 lead to
the following
Corollary 1.5.13
If (R,m) is a local ring, M a finitely generated R-module, and Q an m-pri-
mary ideal, then
degPa(M) = dimM.
The corollary below is a generalization of Corollary 1.5.10.

54
Dimension and Multiplicity
[Ch.
Corollary 1.5.14
1\
If (R, m) is a local ring and M a finitely generated R-lnodule, then dim M
'"
= dimM, where M denotes the completion of M in nt-adic topology.
The Hilbert-Samuel polynomial P Q(M) determines another important
invariant of a module M, namely, its multiplicity with respect to the ideal
Q. In order to define this invariant, consider a polynomialf(X) = asX'+ ...
... +ao E Q[X], of degree s which, for positive integers, takes integer
values. Since
(L1f)(X) = as [(X+ l)S_X S ] + ... = sa s X. Y - 1 -f- ...,
we have (Ll1) (X) = s! as. The number e = (LJ'l") (X) is an integer, and
as = ejs!. If the polynomial ftakes positive values for large n, then e > O.
Definition 1.5.15
Let Q be an ideal of a Noetherian ring R, and let M be a finitely generated
R-module such that 0 < I(MIQM) < Cfj. The I-lilbert-Samuel polynomial
of the module M with respect to the ideal Q is of the form
eQ(M) "
P Q(X, .M) = d( .-- X + ...,
where d is the degree of the polynomial, and eQ(M) > O. The positive
integer eQ(M) is called the multiplicity of the module M with respect to the
ideal Q.
Example 1.5.16
co
If (R, m) is a regular local ring of dimension d, then EB m,,/mn+l is an
11=0
RIm-algebra of polynomials, and as we have sho\vn, I( Rjnt n ) = "'Id! + ...,
whence em(R) = 1.
Example 1.5.17
Let (R, m) be a local ring, and let Q be an m-primary ideal generated by
a d-element set of parameters. By Corollary 1.5.9, we have degPa(R)
= degPm(R) = d; let us apply the Hilbert Theorem (Theorem 1.5.4 (ii»
GO
to the ring a;> QII IQn+ 1, considered as a module over itself. In view of
11=0
A(n, R) = I R / Q (QrajQrJ+l) = (L1P Q )(n, R),
(Lld-l A)(n, R) = (Lid P Q)(n, R) = e Q(R) ,
we conclude that eQ(R)  I(R/Q).
The remaining part of this section is devoted to the proof of a formula
which expresses the multiplicity of a module as the Euler-Poincare charac-

I]
HUb ert-Samuel Polynomials and Multiplicity
55
teristic of a certain Koszul complex. This formula, contained in Theorem
1.5.20, will playa vital role in the following section.
To begin with, we shall give a necessary definition and a simple
lemma.
Definition 1.5.18
Let G be a finite complex of R-modules (see Appendix, Sec. A.t), and let
its homology modules H,(G) have finite length. The Euler-Poincare char-
co
acteristic X(G) of the complex G is the number L (-l)'l(HI(G».
I :::& 0
Lemma 1.5.19
If the components G, of a finite complex of R-modules
G . 0 G dp G dp_1 G 0
·  p --=--+ p-1 --. ... -.. 0......
have finite lengths, then
00
x(G) = L (-I)'/(G,).
"'=0
Proof
By the assumption, we have /(Im(d,») < 00, /(Ker(d,» < 00 for any i.
The equations
/(G,) = I{Im(d,»+I(Ker(d,»,
/(H,(G» = I{Ker(d,»)-/{Im(d'+t»),
yield
x(G) = L( -1)'/(H,(G»
= L « -1)'/(Ker(d,»- (-I)'[(Im(d'+l»]
= 2)( -1)'/(Ker(d,»+( -1)'/(Im(d,»]
= L( -1)'/(G,).
o
If R is a Noetherian ring, M is a finitely generated R-module, and
elements Xt, ..., x" generate an ideal Q c: R such that I(M/QM) < 00,
then in view of Theorem A.6.2 and Corollaries 2.7.14 and 2.7.13 in [B],
we may speak of the Euler-Poincare characteristic of the Koszul complex
K(x; M), where x denotes the sequence x t, ..., x" (see Appendix, Section
A.6).
Theorem 1.5.20
Let (R, m) be a local ring, let Q be an ideal of R generated by a sequence
x of elements x t, ..., x", and let M be a finitely generated R-module such
that I(M/QM) < 00. Then

56
Dimension and Multiplicity
[Ch.
(i) dimM  k,
(ii) if dimM = k, then X(K(x; M)) = eQ(M), where K(x; M) is the
Koszul complex of the sequence x with the coefficients in M,
(iii) if dimM < k, then X(K(x; M» = O.
Proof
Let us consider the subring T of the polynomial ring R[X] consisting of
those polynomials whose coefficients at x n belong to Q" for every n. We
write symbolically
T = R+QX +Q 2 X 2 + ... +Q"X n + ...
This is a graded ring, the elements of Qnxn being homogeneous of degree n.
This ring has been of use in proving the Artin-Rees Lemma (Lemma 2.5.1)
in [B]. We write the R-module M[X] = M<?9RR[X] in the form
M[X] = M+MX+MX 2 + ...
Consider its graded submodule
E = M+QMX+Q 2 MX2+ ...,
whose homogeneous elements of degree n form a submodule En = Q"MXIJ
According to Lemma 2.5.2 in [B], T is a Noetherian ring and E is a finitely
generated T-module.
Let us observe that the elements XIX, ..., x"X of T generate the ideal
T + containing all the homogeneous elements of T of positive degree.
Now, consider two Koszul complexes over T, associated with the
sequence X1X, ..., x"X and the modules E, M[X]:
A = K(x 1 X, ..., x"X; E),
B = K(x 1 X, ..., x"X; M[X]).
The modules E and M[X] are graded T..modules; hence, denoting by
T", R k the free modules of rank k, and using the formulae
AfJ(T k )f8J T E  (fJf\P(Rk)@REq
q
and the fact that dp(!\P(Rk)@RE q ) C !\P-1(Rk)(8)RE q + 1 , we obtain a de-
composition of the complexes If and B, as complexes of R-module, into
the direct sum of complexes isomorphic to A(ft), Bft), - 00 < n < + 00,
respectively, where
A(n): 0 -+ 1\"(R k )(8)RE n 4> 1\"-1(R k )@RE n + 1 -+ ...
... -+ Rk@RE,.+k_l  En+k  0,
B(ft): 0 -+ Ak(Rk)@RMX"  A k-1(R k )@RMxn+l -+ ...
... -+ R k @RMX"+k-l -+ MX n + t -+ o.
We put En = 0 and xn = 0 for n < O. Let us note that the complex B{ft)
for n  0 is isomorphic to the Koszul complex K(x; M).

I]
Hilbert-Samuel Polynomials and Multiplicity
57
The embedding E  M[X] induces the monomorphism of complexes
A(R) -+ B(IJ). Write C<R) for its cokernel; it has the form
CCII): 0  /\k(R")@R(M/QftM)X"
 !\k-l(R")@R(M/Qft+lM)X IJ + 1  ...
...  R"@R(M/QR+k- 1 M)X IJ + k - 1 -+ (M/Q"+kM)X"+k  0,
whence its components are modules of "finite length. Lemma 1.5.19 yields
k
X(c(n») =  (-l)k-P(k k p)l(M/Qn+PM), for n  O.
pr:::O
On the other hand, for large n, the function 1J Ho /(M /Qft M) is determined
by the Hilbert-Samuel polynomial. From the definition of the operator
LJ it is easy to deduce that, for any function};
k
(LlkJ)(n) = L (-It- p ( k k p )f(n+ p ).
p=O
Thus, for large n, we have
X(c(n» = (LlkPa(M»(n). (13)
It is our purpose now to show that for sufficiently large n the complexes
A(ft) are exact.
Indeed, the module Hj(A) is a finitely generated T-module since T is
a Noetherian ring and E is a finitely generated T-module. Moreover, by
Theorem A.6.2, the module Hi(A) is annihilated by the ideal T+ gener-
ated by all the positive elements in T. Hence, in view of T = R + T +,
it follows that Hi(A) is also finitely generated as an R-module. Moreover,
we have an isomorphism of finitely generated R-modules
00
Hi(A)  Ee H,(A<"»;
n= -00
consequently Hi(A(n» = 0 for sufficiently large n. The complex A being
finite, the complexes A(II) are exact for large n.
Using the long exact homology sequence (see (A.2.1»), associated with
the sequence of complexes
o -+ A(")  B(R)  c(n)  0,
we conclude that X(B("») = X(C(ft» for large n. Applying (13), we obtain
the equation
X(K(x; M) = (!1 k P a (M)(n)
valid for large n.
Since l(M/QM) < 00, it follows from Theorems 2.7.14 and 2.7.12 and
Lemma 2.3.5 in [B] that the ideal Ql = Q+Ann(M) is m-primary. Since,

58
Dimension and Multiplicity
[Ch.
clearly, P Q1 (M) = PQ(M), it follows frODl Corollary 1.5.13 that the degree
of the polynolnial P Q(M) is equal to dim M. On the other hand, the ideal
Ql/ Ann (M) of the ring R/Ann(M) is generated by k elements, whence,
in view of Theorem 1.5.5, dimM = diInR/Ann(M)  k. Thus \ve have
proved (i).
It follows that LJkPa(M) = 0 exactly if dimM < k, and X(K(x; M»
= LJkPQ(M) = eQ(M) if and only if dimM = k. 0
Theorem 1.5.20 yields a corollary of particular interest in the case
where the ideal Q is generated by a regular sequence Xl' ..., x". By (A.6.2)
the Koszul complex determined by such a sequence is a free resolution of
the R-module RIQ, and consequently Hi(K(x; M») = Torf(RIQ, M).
Corollary 1.5.21
Let P, Q be ideals of a local ring (R, m) such that the ideal P+Q is m-pri..
Mary and Q is generated by a regular sequence. Then
ex)
eQ(RIP) = L ( -l)'l(Torf(RIQ, RIP)).
;=0
If R = (!)(a, V) is the local ring of a point a on a variety V, and prime
ideals P, Q satisfy the conditions of Corollary 1.5.21, then a is the only
point of intersection of the subvarieties V t , V 2 corresponding to P and Q;
in algebraic geometry, the number L( -1)'I(Torf(R/Q, RIP») is called the
intersection multiplicity of the varieties V t and V 2 at the point a. It follows
from Corollary 1.5.21 that the multiplicity is a positive number (as re-
quired by the geometric interpretation) provided that Q is generated by a
regular sequence. As \ve shall show in the following section, this condi-
tion is superfluous.
Exercises
1. Let :F be a family of subsets of an m-element set {a 1 , ..., am} which satisfies tho
condition: if FE S; and Fo c: F, then Fo E!F. Let us denote by I() the ideal of the
ring of polynomials K[X 1 , ..., Xm] with coefficients in a :field K, generated by monomials
X,! ... K't' 1  i 1 < ... < it  m, such that {a'l' ..., a"J !F. Prove that the function
A(n, R) = lx(R,,) of the graded ring R = K[X 1 , ..., Xm]/I(F) is given by the formula
m
A(n,R) = L.Ik ( n-l ) ,
k=l k-l
where fk denotes the number of those k-element sets which are members of :F.
00
Throughout Exercises 2-5, R = E9 Rn denotes a graded ring, where Ro is an Artin
n=O
ring, and R, regarded as an Ro-algebra, is generated by homogeneous elements x 1, ... , X m
of positive degrees d 1, ..., d m .

I]
Hllbert-Samuel Polynomials and Multiplicity
59
co
2. Prove that if M = E9 M n is a finitely generated graded R-module, then
n=O
(i) the function A(n, M) = IR o (M ), n = 0, 1, ... takes finite values,
(ii) the Poincare series A(M) E Z[[t]] of M, given by the formula
00
)(M) = L A(II, M)t n ,
n=O
is a rational function of the form 1/(I-t'l) ... (1- t d ",), where Ie Z[t],
(Hi) deduce from (ii) that if d 1 = ... = d m = 1, then the function i..(-, M) is a poly-
nomial of degree < m for large n; give an example showing that this assertion does not
hold in the general case. [Proceeding as in the proof of Theorem 1.5.4, compute first
A(n, M)-A(n-d m , M), and then, A(M)-tdml(M).].
3. Prove that if x e Rd then
l(R/(x» = (1- t d ) A(R)- t d A(Ann(x».
4. Suppose in addition that Ro is a field. Prove that the elements XI, ..., x'" are
algebraically independent if and only if A(R) = [(I-td1)... (l-t dm )]-I.
5. Suppose in addition that Ro is a field, and the elenlents XI, ..., X m are algebraically
independent and are of degree 1. Prove that if 11, ..., Is are homogeneous elements of
degrees k 1, ..., ks and the sequence 11' ..., Is is regular, then the Poincare series of the
8
graded ring R/(!1, ..., Is) is equal to (1- t)s-,n II (1 + t+ ... + t k ,-I).
;= 1
6. Prove that the coefficients ao = 1, ai' ... of the formal power series in Z [[t]]
00
L ant" = (1 +t+t z + ... ) (1 +t 2 +t 4 + ...) = [(1- t) (l-t 2 )]-1
/1=0
determine a function 11 -. an which is not a pO]ynoll1ial for large n. Prove an analogous
result for the series [(1- t 2 ) (1- t 3 )]-I.
7. Let R be the graded subring of the ring K[X, Y, Z] of polynon1ials with coeffi-
cients in a field K, which is generated by K and the elements X 2 , y 2 , XY, Z4, and let
S c R be the subring generated by K and the elements X 2 , y 2 , Z4. From the observation
that R = Sa;>SXY, deduce that A(R) = (1+t 2 ) A(S) = (l+t 2 ) (1-12)-2(I-t 4 )-1
= (1- t 2 )- 3. Prove that R is not isomorphic to the polynomial ring U = K[X 1 , %2, X 3 ]
in three indeterminates of degree 2, in spite of A(R) = A( U).
8. Prove that if (R, m) is a local ring of dimension d, and an ideal Q C R satisfies
the condition ln k C Q c: tn S , for certain positive integers k, s, then em(R) s' eQ(R)
 en1(R) k 4 .
9. Let (R, m) be a regular local ring, and let "1, ..., lid be a regular set of parameters.
Suppose Q = (u1 J, ..., Ud), where Ql, ..., Q4 are positive integers. Prove that eo(R)
= q 1 ... fl. [To prove that the sequence u1 1 , ..., u3. is regular, use the fact that a permu-
tation of a regular sequence is regular (see Exercise 3, Section 3.1)]
10. Find the Hilbert-Samuel polynomial PI11(R) and the multiplicity em(R) of the
ring R which is the localization of the ring K[X, Y](XP - yq), P < Q, with respect to the
maximal ideal generated by the residue classes of X, Y.
11. Prove that the n1ultiplicity of the ring R constructed in Exalnple 1.2.23, with
respect to the maximal ideal is equal to 1. [Using the notation of Example 1.2.23, set
J = R'm()R'n, and observe that if an R-module M is annihilated by J then it is iso-
morphic to M R 'mE9M R 'n regarded as an R'-module. Deduce from this that
I(JR /In+l) = l(n/n+l)+ l{tn1/nt+1),

60
Dimension and Multiplicity
[Ch.
where nb tnt are Inaxhnal ideals of the localization of R' with respect to the ideals R'n
and R'm.]
12. Prove that if M is a finitely generated R-module over a local ring R, and x is
a nonzero-divisor on M, then dim(M/xM) = dimM-1.
1.6 INTERSECTION MULTIPLICITY OF A PAIR OF MODULES;
GEOMETRIC APPLICATIONS
At the end of the preceding section, we proved fr prime ideals P, Q of
a local rin.g (R, m) the formula
00
eQ(RIP) = L ( -l)"/(Tor:(RIQ, RIP»)
n=O
under the assumption that P+Q is m-primary and Q is generated by a regu-
lar sequence. The sum on the right-hand side has an obvious analogue for
a pair of R-moduIes M, N
C()
XR(M,N) = L(-l)"/(Tor:(M,N»)
11==0
if the modules Tor:(M, N) have finite lengths and almost all of theln are
equal to O. The integer XR(M, N) is called the intersection multiplicity of
the pair of modules M, N (or the Euler-Poincare characteristic of M and N).
In the first part of this section, we shall give the conditions for the
existence of the intersection multiplicity of a pair of modules; in the sequel,
we shall prove the main result, namely that for "geometric" local rings
(i.e. for rings R containing a field K which, under the natural homomor-
phism R  RIm, is carried onto the field RIm), we have XR(M, N)  O.
This result allows us to define the intersection multiplicity of two sub-
varieties of an algebraic variety. This concept plays a very important role
in algebraic geometry generalizing the notion of order of tangency for
curves and making possible a subtle analysis of the intersection of subvar-
ieties. By using the intersection multiplicity we can define a certain ring
(the so-called Chow ring of a variety) generated by classes of subvarities.
The operation of multiplication is induced by assigning to two subvarieties
a sum of classes of irreducible components of their set-theoretic intersection
each with the corresponding multiplicity as a coefficient. At the end of
the section we outline an algebraic part of the construction of that ring.
The geometric material needed for its completion (e.g. equivalence of
cycles, Chow Lemma) goes far beyond the scope of this book.
In order to find the conditions which ensure the existence of the inter-
section multiplicity of a pair of modules, let us note first that if the local
ring (R, m) is not regular, then, for infinitely many n, we have Tor:(Rlnt,
RIm) ¥: 0, see Theorem 2.1.5, Corollary A.5.6 and Remark A.5.8; conse-

IJ
Intersection Multiplicity of a Pair of Modules
61
quently we cannot determine the intersection InultipIicity for the simplest
pair of modules, Rim, Rim.
Thus, let us assume that (R, m) is a regular local ring. We shall sho\v
that for finitely generated R-modules M, N, the condition l(M<8JN) < 00
implies I(Tor:(M, N» < 00, n = 0, 1, ... Indeed, by Theorems 2.7.11
and 2.4.21 (iv) in [B], we have {m} ::> Supp(M@N) = Supp(M)nSupp(N).
Theorcln A.3.6 (i) yields
Tor:(M, N»)p  Tor:p(M p , N p ) for Pe Spec(R);
hence Supp (Tor:(M, N») c: {m}, and, in view of Theorems 2.4.22 and
2.7.11 of [B], we obtain I(Tor:(M,N») < 00. Moreover, by Theoren18
2.1.1 and A.4.4, we have Tor:(M, N) = 0 for n > dimR.
On the basis of the above consideration we can adopt the following
Definition 1.6.1
Let .R be a regular local ring, and let M, N be finitely generated R-modules
such that I(M@RN) < 00; thus we have I(Torl(M,N») < 00 for 11 = 0,
1, ..., and Tor:(M, N) = 0 for n > dimR. The integer
co
XR(M, N) = L (-l)n/(Tor!:(M, N))
n=O
is called the intersection multiplicity of the modules M, N (or the Euler-Poin-
care characteristic of the modules M, N).
Our direct purpose is to show that for modules M, N over a geo-
metric ring R containing the field K = Rlttt the intersection multiplicity
is non-negative. The proof consists in expressing this multiplicity as a n1ulti-
A
plicity of a certain module M@KN. The proof of the non-negativity of
XR(M@KN) in a more general case is to be found in [X].
We shall first present the general idea of a proof whose detailed elab-
oration leads to the desired result. Let R be a regular local ring which con-
tains the field K = RIm, and let X, Y be free resolutions of finitely gener-
ated R-modules M, N. The module M@KN has the natural structure of
an R@KR-module and since Hn(X@xY):= Tor(M, N) = 0 for n > 0
(K being a field), the complex X@ K Y is, therefore, a free resolution of
MKN as an .R@KR-module. The mapping t.t: R<8J K R  R, t.t(r<8Jr') = rr',
endows the ring R with the structure of an R@ K R-module, and it is easily
seen that
X@RY  (X@KY) @ R.
Rr;sR
lIenee we obtain
Tor:(M, N) = Hn(X@RY)  Hn«XQS)KY)R@RR)
= Tor:@R(M@xN, R). (14)

62
DimeD$ion and Multiplicity
[Chi
If the following conditions were satisfied: (a) R@ K R is a local ring, (b)
Ker(p) is generated by a regular sequence, then, applying Theorem 1.5.20,
we \vould obtain either XR(M, N) = eKer<,c)(M(8)KN), or XR(M, N) = 0,
i.e. XR(M, N)  O. Unfortunately, conditions (a) and (b) are, in general, not
satisfied. However under the additional assumption of the completeness
of R we shan show that isomorphism (14) takes the form
"- A
Tor:(M, N)  Tor:@R(MC8> K N, R),
A "-
where R@R is a ring to be defined in Lemma 1.6.2, and Ker(R@KR -+ R)
is generated by a regular sequence. This will enable us to infer, without
assuming the completeness of R, that XR(M, N)  0 for all regular rings
R containing the field K = RIm.
In the following lemmas, (R, m) will denote a local ring containing the
field K = RIm, i.e. more precisely, containing the field K which is carried
through the natural homomorphism R  R/m on.to the field R/tn. Thus
R is a K-algebra.
In the sequel, we shall write briefly @ instead of @ K'
Lemma 1.6.2
Let M, Nbe finitely generated R-modules. Denote by 1" = l1t"@R+R@m n ,
n = 0,1, .." and J = m@R+R@m the ideals of R@R. The topologies
determined by the sequences of ideals {L,}, {1 ft } in the R<8>R-module
Mfg)N are equivalent Hausdorff topologies, hence the completions they
,..
determine are identical and denoted by M(8)N:
1\
M(g)N = lim(M@Nlln(M@N») = lim(Mfg)N/ln{M@N»)
-- +-
n n
= lim(M/m"M@N/nt"N).
-
n
A ,.. 1\
The ring R@R is local, and M(8)N, regarded as an R(8)R-module, is
finitely generated.
Proof
It follows directly from the definition of In' that I" :::> In ::> J211, whence
the two topologies are equivalent. We shall prove that the J-adic topology
is Hausdorff. In view of nnt" M = 0 = nmnN, there exist linear
K-spaces Mo, M 1 , ... c: M, No, Nt, ... c: Nsuch that
mnM= MnE9M"+l'''' mnN= NnEeN,,+t$ ...,
and consequently
In(M0N) = L (m P @m Q )(M0N) = L m P M + mqN
p+q=n p+q=n
= L: (E9 M,)0( EaN j ) = <:9 M,@N j .
p+qn ip jq i+jn

I]
Intersection Multiplicity of a Pair of Modules
63
whence it follows that (j/"(M@N) = 0, i.e. that the topologies are both
Hausdorff.
1\
In view of the properties of completions (see Sectioll 2.6 in [B]), M@N
A
has the structure of a finitely generated R@R-module.
A 1\
We shall prove that R@R is a local ring. Write for short T = R@R,
1\ 00 " A
I = TJ, and Gr(T) = (i) In/ln+1. By Corollary 2.6.21 in [B], we have
11=0
1\ A
T/J  R(g)R/J  R/m@Rlm  K, i.e. I is a maximal ideal. If x E T,J,
A A
then there exists 0 i= a E K such that x -« E J, hence y = 1 - (X-I X E I.
It is easily seen that the limit of the sequence {I + y + ... + y"}" is the
1\
inverse of 1- y, and therefore the element x is invertible in T. Hence I is
the only maximal ideal of T.
To complete the proof, we n1ust shovv that the ring T is Noetherian.
A A
Observe that the ring Gr(T) is a K-algebra generated by 111 2 which, in
turn, is finitely generated, and consequently Gr(T) is a Noetherian ring.
00 A A
Let leT be an ideal; then Gr(I) = E9 (/nJ")/(/nJII+I) admits a natural
n=O
A 1\
embedding into Gr(T), induced by the embeddings / nl" c: In; thus we may
identify Gr(I) with an ideal, of course finitely generated, of Gr(T).
Let a 1, ..., Ok be nonzero elements of I whose residue classes in Gr(I)
generate Gr(I) as an ideal of Gr(T). We claim that these elements generate
A A
I. There exist positive integers q 1, ..., qk such that aj E Jq'"lqt+ 1, i = I, ...
..., k. Let F be a free T-module with a basis el, ..., ek, and let cp: F  / be
the homomorphism satisfying the conditions q;( ei) = aj, i = 1, ..., k. In the
module F we introduce a topology in which a basis of neighbourhoods of
k A A
zero is formed by the subn10dules F,. = E9 In-q'ej, where J" = Tfor n < o.
i= 1
A 1\ A
This topology is equivalent to the I-adic topology because J"F c: Fn c: In-q F
for q ;;:: qi, i = 1, ..., k. Thus the module F is complete in this topology.
00
Let us write Gr(F) = Ef) Fn/Fn+l; thus Gr(F) is endowed with the natural
11=0
A
structure of a Gr(T)..module. III view of the inclusion rp(Fn) c: [nJ", there
exists an induced homomorphism Or(9'): Gr(F)  Gr(I), and Gr(rp) is an
epimorphism.
Let us take any 110nzero element a E I; then there exists a maximal
A
positive integer q such that a E Jq. We shall construct a sequence fo, 11' ...
of elements of Fq \vhich satisfy the conditions
(in) In - f,,-1 E F',l+"- 1 ,
A
(ii n ) rp(f,,) - a E Inl q +"
for n = 0, 1, ... (condition (io) means that 10 E Fq-t). We take fo = 0;
since Gr(rp) is an epimorphism, there exists 11 E Fq such that the image

64
Dimension and Multiplicity
[Ch.
of its residue class in Gr(F) is the residue class of a in Gr(I). This means
"-
that q;(fl) - a E I nJq+ 1, and the above conditions are satisfied for n = 0, t.
Suppose we have already found elements 10' ...,1" E F which satisfy these
"-
conditions. The residue class of cp(f,,) - a E I (")Jq+ n in Gr(J) is the image,
by Gr( q;), of the residue class of a certain element Yn E Fq+tt, i.e. q;(f,.) - a -
- q;(y,,) E J(")JQ+n+l, and setting f,,+ 1 = 1" - Yn, \ve have 1,,+ 1 E Fq,f,,+ 1 -
"-
- f" E F q +,., and q;(f,.+ 1) - a E In]q+''+l.
Since F is complete, there exists a limit f of the sequence {f,.}, and
cp(f) = a because of (ii n ). Thus the ideal I is finitely generated, and the
I\.
ring R@R is Noetherian. Ll
Lemma 1.6.3
Let N be a finitely generated .k-module. If
o -. M'  M  M" -+ 0
is an exact sequence of finitely generated R-modules, then the sequence
A A A
o -. M'fl:)N -+ MN -+ M"@N -+ 0
is exact.
Proof
For any positive integer n, we have the exact sequence
o -. M' /(m"MnM') -. Mlm"Nl -+ M" /mnM" -+ 0
which induces the exact sequence of Artin R(8).R-lnodules
0-+ (M'!(m"MnM'»)@(N/m"N)  (M/m n M)<8)(N/l1t"N)
 (M" /mnM")(g) (N/m"N)  0;
accordingly, by Theorem 2.6.12 in [B], we obtain the exact sequence
o -+ lim(M' !(m"MnM')@N/mnN) -+ lim(M/t1t"M(g)N/ln"N)
+--- ---
-. lim(M" /m n M"@N/nt Il N)  o.
--
By the Artin-Rees lemma, there exists a positive integer no such that
for n  no we have mnM' c mnMnM' = mn-11o(m"oMnM') c: mn-noM',
hence the topology of M'(8)N induced by the sequence of submodules
{(m n M nM')@N + M' @m'IN} is equivalent to that induced by the sequence
{I"M'@N}. Now, the assertion follows from Lemma 1.6.2. 0
Lemma 1.6.4
A
Let M, N be finitely generated R-modules. Then dimRRM@N = dimRM +
I\.
+dimRN, eJ(M(g)N) = em (M)e m (N).
Proof
As we know, the lengths f(n) = I (m"M!m n + 1 M) are finite, and for large n
we have

I]
Intersection Multiplicity of a Pair of Modules
65
n d - 1
f(n) = LlP,1t(n, M) = em(M)ld=- l) T+ ..., (15)
where by Corollary 1.5.10 we have d = dimM. The only sinlple R-module
is R/m = K, whence fen) = dimx(m"M!m,,+l M). Similarly, for the module
N and large n, we have
g(n) = L1P m (n, N) = !(m"N!nt'I+IN) = dimK(m"Njm"+lN)
,
nd -1
= em(N) (d'-l)! + ..., (16)
A A 1\
where d' = dimN. We already know that k[(g)NjJ"(M@N)  M@N/J"(Mfl;
@N). Using the notation from the proof of Lemma 1.6.2, we can write
fen) = dimK(m n M/m'J+l M) = dimKM n and g(n) = dimKN" and also
A
In(MfSN) = Ef) Mj@N j . Thus, as the ring Rfg)R is local, we obtain
l+j1I
A A A
I(Mf6)N)/J"(M@N») = dim(M@N/l'I(M@N»)
= dimK.  M,@N J = L: f(i)g(j).
'+J<II i+i<n
First, let us compute the SUln L i'js for positive integers t, s. We have
i+j<n
Il-l 11-1 "
 .r.s  " 'r.s   .' (k . ) S
.LJ I) =.LJ .LJ IJ = .LJ.LJ' -I,
i+j<n k=Oi+i=k k=Oi=O
and a simple calculation gives
"  ( , ) 1 [ n,'+s+2 ]
. i'js = L..J (_l)m ;, r+ m+ T ,: +s+ 2-+ ... ;
'+J<II m=O
to compute the leading coefficient, let us write
.
 ( s ) xr+nr+t
9'r,s(X) = L..J(-I)m m r+m+ f'
m=O
where x ranges over real numbers. We then have tp;,.(x) = x'(l-x)S, and,
integrating by parts, we obtain
1 1
11',..(1) = X'(t-X)SdX = r:C x'+1(l-x)'-tdX
o 0
s!
- -(r+ 1) ... (r+'+ 1) ,
and finally
 °r"s _ r!s! [ ,+s+2 ]
.LJ l J - - ( -- 2) f 1"1 + ... ·
I+j<n' r+s+.

66
Dimension and Multiplicity
[Ch.
H the equations (15) and (16) are satisfied for all n, then
A A A
/(M@N/Jn(M(g)N»)
= .L [em(M) (:)I + ...][e l1t (N) d::), + ...]
,+J<n
,
n d + d
= em(M)em(N)-( d+d')! + ...,
and the theorem fol1ows from Corollary 1.5.10. In the general case, for
large n, the differences f(n)-(L1Pm) (n, M),g(n)-(L1Pm) (Il,N) are equal
to 0, and it is easily proved that the leading forms in the sums 2: f(i) g(j)
i+j<n
and 2: (L1P m ) (i, N) (L1P nt ) (j, N) are the same. This concludes the proof
;+i<n
of the lemma. D
Theorem 1.6.5
If a local ring R contains the field K = RIm and is complete, then the
A.
multiplication mapping p: RQ9 K R  R induces the mapping : R@KR
A.
 R, which endows R with the structure of an R@R-module. For finitely
generated R-modules M, N, we have the isomorphisms
1\ A.
Tor:(M, N)  Tor:fg)R(Mf8JN, R), 1l = 0, 1, ...
If, moreover, the ring R is regular, then Ker{) is generated by ad-element
regular sequence, where d = dimR.
Proof
Since the multiplication mapping p,: R(g)R  R, p,(r@r') = rr', satisfies
the condition ,u(m@R+R@m) = m, ft is continuous and induces the
A A ,..
mapping of the completions f.l: S  R = R, where S = R@R.
Let X, Y be finitely generated free resolutions of M, N. We know that
the complex X@xY is a free resolution of the R(8)R-module M@KN.
"-
Consider a double complex X(8) KY consisting of free S-modules; we define
in it a decreasing filtration by setting Fl' = $  X i @ K Y j . We compute
ip J=O
the spectral sequence of this filtration using Lemma 1.6.2, and the fact
that the homology functor commutes with an exact functor
"
E.q = Xp@KY q ,
1 " "- I 0 for q > 0,
El',q = H q (X p f8J K Y) = Xp@KHq{Y) = ) Xp  KN .t"
l 'CI lor q = 0,
2 A ,, { O for p > 0,
Ep,o = Hp(X(g)KN) = Hp(X)@KN = M  KN t'
'CI lor p = o.
A
The convergence of this sequence implies that the complex X@ KY is a free
"-
resolution of the S-moduJe Mfg)KN. The complexes X, Y being free ones
,

I]
Intersection Multiplicity of a Pair of Modules
67
A
the isomorphism S<8JsR  R yields (X@KY)@sR  X(8)RY; applying
the homology functor we easily obtain the desired formula.
To prove the last assertion, we need a more detailed knowledge of the
structure of a regular ring. We know from Theorem 1.4.1 that the K-algebra
co
€a m n /mn+l is the algebra of polynomials in d indeterminates Xl' ..., X d ,
11=0
where Xl' ..., X d are the residue classes of the parameters. We shall write
briefly K[X], K[[X]], (X) instead of K[X I ,..., X d ], K[[X 1 ,... X d ]],
(Xl' ..., X d ), respectively. Since R is a complete ring, we have
R  limR/m" = Iim(K[X]/(X)n)  K[[X]].
- -
A A.
To compute RQ5) K R  K[[X]]<8>xK[[y]], we apply Lemma 1.6.2. We have
A
R@R = lim(K[[X]]/(x)n@K[[y]]/(y)n)
-
= lim(K[X]/(X)n@K[y]/(Y)JI)
-
= lim(K[X, y]/(X, Y)n) = K[[X, Y]].
(-
A suitable isomorphism carries the mapping it onto the mapping
-p,: K[[X, Y]]  K[[X]] which satisfies the conditions ",, (X , ) = p, (Y,) = Xi,
i = 1, ...,d. From the isomorphism K[[X, Y]]  K[[X1-Y I , ...,Xd-Y d ,
Xl' ..., X d ]] it is readily deduced that the ideal Ker @ ) is generated by the
regular sequence Xl - Y I , ..., X d - Y d , and this ends the proof. 0
We sum up the most important properties of the intersection multi..
plicity of modules in the following
Theorem 1.6.6
Let R be a regular local ring containing the field K = Rim. If M, N are
finitely generated R-modules and I(M@RN) < 00, then dimM +dimN
 dimR, and
(i) the intersection multiplicity of M, N
00
X(M, N) = L (-l)"IR(Tor:(M, N»
n=O
is a non-negative integer,
(ii) X(M, N) = 0  dimM +dimN < dimR,
(Hi) X(M, N) = X(N, M),
(iv) if 0  M'  M  M"  0 is an exact sequence, then
X(M, N) = X(M', N)+ X(M", N).
Proof
We shall show that, without loss of generality, we can assume the ring R
to be complete. Indeed, if X, Yare free resolutions of M, N, then the formula

68
Dimension and Multiplicity
[Ch,
A A
M = M@RR and the exactness of the completion functor imply that
A A A A
X, Yare free resolutions of the R-modules M, N, and
1\ " A 1\
Tor:(M, N) = H..(X@RY) = H,,(X@RR@RY@RR)
A A
 H,,(X@RY@RR)  H n (X@RY)f8J R R
A
= Tor:(M, N)@RR = Tor:(M, N),
where the last equation results from the fact that the R-module Tor:(M, N)
is discrete as a module of finite length.
A A
We know that dimM = dimM, dimR = dimR, and consequently we
may assume that R is a complete ring.
A
By Theorem 1.6.5, the ideal Ker(,u) of S = R@KR is generated by
a d-element regular sequence, where d = dim R; applying the formula
from that theorem, we obtain
M@RN  (M@KN)f8)sR  M@xN/(Ker(p»)(M@KN).
Thus, we can apply Theorem 1.5.20 to the ring S, the ideal Ker(jJ,) and the
A A
module M@KN. By Lemma 1.6.4, we obtain dimM+dimN = dimMN
A
 d, and if d = dimM@N = dimM +dimN, we have
00
X(M,N) = L(-l)"IR(Tor:(M,N»
11=0
00
= L (-l)"/s(for(Mc8>N, R»
n=O
00
= 2: (-l)"ls(H n (K@s(Mc8> K N»)
n=O
A
= eKer()(M@KN) > 0
where K denotes a suitable Koszul complex. Moreover, applying Theorem
1.5.20 (iii), we obtain dimM +dimN < d  X(M, N) = O. Thus, we have
proved Theorem 1.6.6 (i), (ii); formula (Hi) is obvious.
Let 0  M'  M  M" -. 0 be an exact sequence. It induces the
exact sequence
... -. Tor:(M', N) -. Tor:(M, N) -+ Tor:(M", N) -+ ...
... -+ Torf(M", N) -. M'@RN -. M@RN -. M"Q9 R N -. O. (17)
We have of course l(M"@RN) < 00, and consequently also I(Torf(.A-f", N»)
< 00; thus, as the sequence (17) is exact, I(M'@RN) < 00. Accordingly,
all the terms of (17) have finite length, almost all of them being equal to O.
Applying Lemma 1.5.19 to (17), we obtain

I]
Intersection Multiplicity of a Pair of Modules
69
00
2)( - I) 3n l(Tor:(M", N))+( - I)3n+11(Tor:(M, N))+
11=0
+(-1)3n+2/(Tor:(M',N)] = 0,
and this implies Theorem 1.6.6 (iv). D
We shall now present certain geometric applications of the results \ve
have obtained. Let K be an algebraically closed field and let V c Kn be an
algebraic variety without singular points (a smooth variety), i.e. such that
the local ring @(a, V) is regular for every point a E V. As the field K is
algebraically closed, every ring @(a, V) contains the field K = eJ(a, V)/
/m(a, V). The ring R = K[V] is the ring of polynomial functions defined
on V. In Chapter II (see Corollary 2.1.6), we shall prove that the localiz-
ation R p of a regular local ring R is also a regular local ring; this result
implies that for every subvariety U c V, the local ring (!)(U, V) = R 1 (u)
is also regular.
Let U, U' be subvarieties of V determined by prime ideals P, P' c R.
The irreducible components WI' ..., W s of the algebraic set UnU' are
determined by the minimal prime ideals PI' ..., Ps of P+P' = Ann(R/P@R
@RR/P'). In view of Corollary 2.7.14 in [B], we can apply Theorem 1.6.6 to
each of the rings R p , and the modules (R/P)p" (R/P')p,; thus
(a) dim(R/P)p,+dim(RIP')p,  dimR p "
(b) the integers
00
XWI(U, U') = L (- I)nlRI',( (Tor: (RIP, RIP'))p , ),
n=O
i = 1, ..., s, are non-negative,
(c)Xw,(U, U') = O<:>dimU+dimU' < dimV+dimW,.
We derive condition (c) from Theorem 1.6.6 (ii) applying the equations
dim(R/P)p, = dimPR p , = dimP-dimPi = dimU-dimW"
dim(R/P')Pt = dim U' -dimW"
dimR p , = dimV-dimW"
which result from Corollary 1.1.18; from these equations we deduce that
dim U + dim U'  dim V + dim Wi, or equivalently, ht (Pi)  ht (P) + ht (P').
If all the components WI' ..., W of the intersection U f\ U' satisfy the
condition dim V + dim W, = dim U + dim U', then we say that the subvar-
ieties U, U' intersect each other properly.
The above facts enable us to adopt the following
Definition 1.6.7
The multiplicity of an irreducible component W of the intersection Uf\U' of
subvarieties U, U' of a non-singular variety V is the integer

70
Dimension and Multiplicity
[ChI
00
Xw(U, U') = 2: (-l)"IR,.'{ (Tor:(R/P. R/P'»p..).
,,=0
where R = K[V] and P, P', P" are the prime ideals of R which determine
the subvarieties U, U', W. The following conditions are satisfied:
(i) Xw(U, U')  0,
(ii) Xw(U, U.') = Xw(U', U),
(Hi) Xw( U, U') = 0 <:> dim U + dim U' < dim V + dim W.
The problem of finding a satisfactory definition of the intersection
multiplicity of subvarieties has proved to be troublesome. The above
definition, due to Serre, is a generalization of the most suitable earlier
definitions. One great advantage of this definition lies in the possibility of
adopting it to other types of varieties (e.g. analytic varieties). The reader
should be aware of how wide is the scope of the algebraic methods needed
for the very formulation of this definition, and for the proof of the property
(iii) given above in a particular case.
We have seen that the intersection Uf\U' of subvarieties U and U'
determines the irreducible components WI' ..., W s and the corresponding
multiplicities Xw 1 (U, U'), ..., Xw.(U, U'). This could be expressed in alge-
braic terms as
s
UnU' = 2: Xw,(U, U')W,.
1=1
To make precise this intuitive approach and to justify the above "informal"
expression we introduce the following algebraic construction:
Definition 1.6.8
Let V c K" be a d-dimensional smooth algebraic variety and let R = K[V]
be the ring of its polynomial functions. Then the graded Z-module C*(V)
d
= E9 C;(V), whose components are the free Z-modules
1=0
C,(V) = Z{P e Spec(R): ht(P) = i}, i = 0, 1, ..., d
is called the group of cycles on the variety V. The law of multiplication:
C,(V)(g)Cj(V) -+ C,+j(V) is defined for the generators by the formula
p. P' = I: XR,." «R/P)p.., (R/P')p!')p"
pIt
ex>
= L [2: (-l)"I R ,.,,( (Tor: (R/P , R/P') )p..) jP",
P" n = 0
where ht (P) = i, ht (P') = j and P" ranges over those associated ideals of
p + P' which are of height i + j.
We say that cycles z = LcxpP e C,(V), z' = LC(p,P' e CJ(V) intersect
properly if the conditions C(p ¥= 0, (X,p' ¥= 0 imply that the ideals P.. P' inter..

1]
Intersection Multiplicity of a Pair of Modules
71
sect properly, i.e. the miniJnal associated priIne ideals of P + p' are of height
i+j.
Let us note that if pi' is a minin1al prime ideal of P+P' then, as we
have shown, ht (P")  ht (P) + ht (P'); Dloreover ht (P") < ht (P) + ht (Pi)
<=> dimP+dimP' < dim V +dimpN <=> 'lRI',,«R/P)p", (R/P')p") = O.
The generator of the Z-module Co (V) is the zero ideal which corresponds
to the whole variety V, and is the unity element of the operation .; the
operation · is, of course, commutative, and preserves grading.
Example 1.6.9
Let V = C2 be a two-dimensional affine space over the field of complex
numbers C. Then C 2 (V) = E9 Znt a , where ttt a denotes the tnaximal
aeV
ideal corresponding to a point a. We shall compute the product of the
cycles determined by the prime ideals P = (Y-XP), Q = (Y-X t ) of the
ring R = C[X, Y], where p < q. Let nt ::> P+Q be a prime ideal. Since
P =F Q, we have m =F P, and consequently m is a nlaximal ideal and
ht (tn) = 2. Since
q-p
m::) P+Q 3 xq-x p = XP(Xq-p-l) = XP n (X-C k ),
"=1
where C is the primitive (q-p}-th root of I, m turns out to be one of the
ideals mo = (X, Y),mk = (X-C k , Y-ckp), k = 1, ..., q-p. The R-mod...
ule RIP has a free resolution
O R y-X' R 0
 ) ,
and as the polynomials Y - XP, Y - X q are nlutual1y prime, it follows that
Torf(R/P, R/Q) = Ker(R/Q Y-x' -+ RIQ) = 0;
Inoreover, Tor(R/P, -) = 0 for n > 1. \Vriting L = R/P@RR/Q, we
have
L  Rj(P+Q) = R/(Y-X", y-X q )
= R/(Y -XP, Xq-X")  C[X]/(Xf_X')
q-p
 C[X]/XPEa E9 C[X]/X-C").
k;::1
Accordingly, Lillo  CP, Ln1k  C, k = 1, ..., q-p, and we deduce that
q-p
p. Q = ptn o + 6 1tnk '
The proof of the associativity of the operation · is less trivial and
involves modules of a more general type than those of the form RIP. We
shall first present some auxiliary properties and. lemmas.

72
Dimension and Multiplicity
[Ch.
We have defined the dimension of a module M as the dimension of the
ring RjAnn(M); analogously, we now define the height of the module M
as the number
ht(M) = ht(Ann(M») = min{ht(P); P::> Ann(M), P E Spec(R)}
= min {ht (P); P E Supp(M) } .
Since the ring R = K[V] is a finitely generated algebra over a field, we
have ht(M)+dimM = dimR.
Let M be a finitely generated R-module, and let PeR be a prime ideal,
ht (M)  i, ht (P) = i. If M p '¥= 0, then P ::> Ann(M), and because ht (P)
 ht(M), it follows that P is a minimal prime ideal of Ann(M); hence the
inclusions PR p :::> Ann(M p ) :::> (Ann(M»)p imply that Ann(M p ) is "a
PRp-primary ideal, and consequently lRp(M p ) < 00.
To each finitely generated R-1Dodule M, such that ht (M)  i we can
assign a cycle
C,(M) = L: lRp(Mp)P E C,(V),
p
since almost all its coefficients are O. Observe also that Ci(M) = 0 <=> ht(M)
> i, and for every exact sequence 0..... M'  M  M" ..... 0 we have
Ci(M) = Ci(M') + Ci(M").
Lemma 1.6.10
If M, M' are finitely generated R-modules, ht (M)  ;, ht (M')  j, and
ht(M(i9M')  i+j, then
(i) the cycles Ci(M), CJ(M') intersect properly,
ex)
(ii) Ci(M) · Cj(M') = L: (-l)nCi+J(Tor:(M, M'»).
n=O
Proof
(i) Suppose that ht(P) = i, ht(P') = j, M p 1= 0, M, =1= 0, and let P" be a
mi'nimal associated prime ideal of P+P'. We know that ht(P")  ht(P)+
+ht(P') = i+j. On the other hand (Mp")p = M p '¥= 0 implies Mp/l :1= 0,
and similarly we get M;" :F O. It follows from Theorem 2.4.21 (iv) of [B]
that P" E Supp(M@M'), whence ht(P") ;::: ht(M@M')  i+j, and con-
sequently ht (P") = ht (P) + ht (P'); this ends the proof of (i).
(ii) To begin with, let us note that ht(Tor:(M, M')  ht(M@M');
indeed, if P E Supp(Tor:(.M, M'»), then we have 0 =F (Tor:(M, M'»)p
 Tor:p(M p , M), whence P E Supp(M)f\Supp(M') = Supp(M@M').
Thus, the right-hand side of formula (ii) is well-defined.
Let P" be a prime ideal of height i + j. It appears in the cycle on the
right with the coefficient
00
L (-l)"IRpn «(for:(M, M'»p" = XRp..(Mpn, M..).
11=0

I]
Intersection Multiplicity of a Pair of Modules
73
Let us consider the cycle 'i(M). j(M'). If ht(M) > i then C,(M) = 0,
and, since the inclusion Ann(M p ") :::> (Ann(M»)p" yields ht(M p ")  ht(M),
it follows that
ht(Mp,,)+ht(M,,)  ht(M)+ht(M') > i+j = ht(P").
Accordingly, we have XRp,,(M p '" M;,,) = 0, and the coefficient of P"
is equal to 0 on both sides of (ii). We proceed analogously with the case
of ht (M') > j.
Suppose ht (M) = i, ht (M') = j, and consider in M and M' sequences
of submodules, M = M 0 ::> M 1 ::> ... :::) Ms = 0, M' = M ::> M :::> ...
... :::> M; = 0, such that the factor modules Mrl M,+ l' M;I M;+ 1 are of
the form RIQ, where Q is a prime ideal. Now, using the additivity of the
function C, we reduce the proof of (ii) to the case of M = RIP, M' = RIP',
where ht(P) == i, ht(P') = j. We then have Ci(R/P) = P, Cj(R/P') = P',
and, by definition the coefficient of P" in the product Ci(R/P). Cj(RIP')
= p. P' is equal to XR p " (R/P)p", (R/P')p"). This concludes the proof
of (ii). 0
Lemma 1.6.11
Suppose that cycles Z E Ci(V), z' E Cj(V), Z" E C,,(V) have non-negative
coefficients, and the pairs of cycles z', z" and z, z' · z" intersect properly.
Let us set M = $(RjP)a p , where z = L O!pP, the modules M' and M"
p
being defined similarly. Then we have
ht(M'@M")  j+k, ht(M@M'@M")  i+j+k.
Proof
The first inequality follows directly from the assumption that the cycles
z', z" intersect properly. Suppose Q E Supp(M@M'@M") = Supp(M)n
(}Supp(M')(}Supp(M"); since z' and z" intersect properly, there exist
minimal ideals P' in Supp(M') and P" in Supp(M") such that Q contains
a minimal associated prime ideal Po of P+P'. Thus we have ht(Q)  ht(Po)
= ht (P') + ht (P") == ht (M') + ht (M") = j + Ie The ideal Q contains an
ideal P, minimal in Supp(M), which intersects Po properly; hence ht(Q)
ht(P+Po) = ht(P)+-ht(Po) = i+j+k, and the second inequality is
also proved. 0
We can now prove that the operation · is associative. Let Z E Ci(V),
Z' E CJ(V), Z" E Ck(V) be cycles with non-negative coefficients, such that
the pairs of cycles z', z" and z, z'. z" intersect properly, and let M, M',
M" be the modules described in the preceding lemma. Consider free resol-
utions X, X', X" of the modules M, Nf', Mil respectively. In the cODlplex
X(8) R X' @ R X", we define a filtration
Fp = EJ1 (f)nQ9X;(8)X.:'
",p r 3

74
Dimension and Multiplicity
[Ch,
whicl1 determines the spectral sequence whose initial terms are
E,q = XpQS)R( e1 X;@"X s ),
r+s=q
E;,q = X p Q9R Tor: <M', M"),
Ei,q = Tor:(M, Tor:(M', M"»).
The sequence {E;, q},. converges to {E q} which corresponds to a filtration
of the module Hp+q(X@RX'RX"). Applying the preceding lemnlas, and
using the additivity of the function C, we obtain
z. (z' · z") = C,(M) · [Cl(M') · Ck(M")]
00
== CI(M) · [2: ( -1)qCJ+k(Tor:(M', M"»]
q=O
00
= L( -1)qC,(M)' (;J+k(Tor:(M', M"»
1]=0
00 00
= 2:( -1)qL(-I)""'+J+k(Tor:(M, Tor:(M', M"»)
q=O p=O
= 2:( -1)p+qC'+J+t(E.q)
p,q
00
= 2: (-I)" L CI+J+k(E:. q)
n=O p+q="
00
= L (-I)" L CI+J+k(E;'q)
11=0 p+q-=Il
00
= L: (- onC'+J+k(H,,(X@RX'@RX"».
n=O
Assulne that also the pairs of cycles z, z' and z · z', z" intersect properly.
Defining in the complex Xf8jX'@X" a filtration
F = e$ $ XnlX;@X;',
In ,. ap
and proceeding as above, we obtain
00
(z.z').z" = L:(-I)"CI+J+k(H,,(X@RX'@RX"»;
n=O
hence (z. z') · z" = z' (z' · z").
The product of cycles intersecting properly which we have described
above is still not sufficient for the study of varieties. The construction can
be improved by defining a suitable equivalence relation on cycles, similar
to that introduced for divisors; it turns out that for any pair of cycles

I]
Intersection Multiplicity of a Pair of Modules
75
z, z', there exist cycles Zl  Z, z '" z' \vhich intersect properly. Using
this, one can define a ring of classes of cycles for a given variety, called its
Chow ring.
Exercises
1. Let R c T be local rings with the maximal ideals m, 11, and suppose that T, regar-
ded as an R-Inodule, is finitely generated. Suppose also that no element of R is a zero-
divisor in T, and write t 1, ..., tIc for a maximal subset of T, linearly independent over R.
Prove that if Q C R is an m..primary ideal, then Q T is an 1t-primary ideal and keo(R)
= [TIn: RIm] eQT(T). [Write F = Rt 1 + ... + Rt", s = [TIn: RI1U]. Observe that there
exists an element 0 ¥: a e R such that aT C F; furthermore, for aT-module M, we have
IR(M) = SIT(M). Using the exact sequences
FIQ"P-+ (F+QnT)IQnT 0,
0-. (aT+Q"T)IQ"T -+ TIQ"T --) TI(aT+Q"T) -+ 0,
derive the formulae
IR(PIQnF)  IR«aT+Q"T)IQnT) = SP Q (II)-SP Q '(Il),
valid for large II, where Q' = (aT+QF)laT is an ideal of the ring TlaT. Similarly, derive
from the exact sequences
aTlaQnT-. (aT+Qnr")IQ"F-+ 0,
o -. (aT+.QnF)JQnF  FIQ"P) -4 FI(aT+Qnp) -+ 0
the formulae
IR(F/Qnp)  IR(FI(aF+QnF»)+IR(aTlaQnT) = li(FlQ"F)+IR(TIQ"T),
where R = RlaR, F = FlaP, Q = (Q+aR)laR.]
2. Let R be a local ring with the maximal ideal m, and let T::> R be a ring finitely
generated as an R-module. Suppose, moreover, that no element of R is a zero-divisor in
T, and denote by t 1, ..., tIc a maximal subset of T, linearly independent over R. Prove
that if Q c: R is an m-primary ideal, then the minimal prime ideals ml, ..., m q of Q Tare
all the maximal ideals of T. Let Q 1, ..., Qq be the primary components of the ideal
QT with the radicals ml, .... m q . Prove that if all the rings R, Tml' ..., Tmq have the
same dimension, then
q
keo(R) = 2: [TIm,: RIm] eQ,(T).
i= 1
q q
[prove that Q"T = Qn ... f"\Q: and TIQ"T  Et> TIQ1  (f) Tnt,/(Q, T m,)n, then
i= 1 i= 1
proceed as in the foregoing exercise.]
3. Let fields K, L, K C L, be finite extensions of the field of rationals, and let R K c R L
be the rings of integers of the respective algebraic number fields. Prove that if P C Rx
is a nonzero prime ideal, and P R L = Q 1 ... Q:s, where Q 1, ..., Q, are pairwise distinct
prime ideals of R L , then
s
[L: K] = L (RLIQ,: RKIPj e,.
;=1

76
Dimension and Multiplicity
[Ch.
4. Denote by R the closure of the ring K[X 4 ,X 3 y,xy3, Y4] in the ring of formal
power series K[[X, Y]] with coefficients in a field K, and write Q == (X4, Y4) c: R. Prove
that R is a local ring, rad Q is its maximal ideal, eQ(R) = 4, and /(R/Q) = S.
5. Prove that the homomorphism h:K[[x]]0gK[[Y]] -+ K[[X, Y]] given by the
n ft
forlnula h( 2: a,(X)@b,(Y) = L a,eX) b,(Y) for a,eX) e K[[X]], b , (Y) e K[[Y]] is
1=1 i1
not a surjection.
NOTES AND REFERENCES
The interpretation of the dimension of an algebraic variety as the transcendence degree
of its field of rational functions is a classical one. At first, the dimension d of a prime
ideal P in a polynomial ring K[X t , ..., Xn] over a field K was studied as the transcendence
degree of the field of fractions of the factor ring K[X 1 , ..., XnJ/P over K, or as the rank
r = n-dofthe ideal P (Lasker [19], Macaulay [0], Van der Waerden [39]). The normal-
ization theorem (Theoreln 1.1.6) due to E. Noether, comes from [23], and the so-called
Hilbert Nullstellensatz (Corollary 1.1.12) from [13]. In [15] Krull gave an interpretation
of the notion of dimension of affine algebras in terms of chains of prime ideals, which
enabled him to create a theory of dimension in arbitrary Noetherian rings. The funda-
mental theorem (Theorem 1.2.1 0) can also be found there. In proving this, we have
followed Kaplansky [K], with simplifications from [5].
The part of Theorem 1.3.4 concerning the ring of pOlynomials over a Noetherian
coefficient ring is due to Krull [18]. The proof of the remaining part comes from [36].
The famous Macaulay Theorem (Theorem 1.3.7) is to be found in [0].
It was Krull who created the fundamentals of the general theory of local rings. In
his pioneering paper [17] he introduced the concept of a regular local ring and the con-
struction of the graded ring associated with an ideal (called also the ring of forms of
an ideal). He posed the question of finding a structure theorem for all complete, regular
local rings, answered a few years later by Cohen (see Chapter II). Krull's theory was
developed by Chevalley in [6), who also introduced the term "local ring", in general
use today. The concept of the set of parameters of a local ring comes from [6]. The char-
acterization (Corollary 1.4.25) of non-singular points of affine varieties as: those for
\vhich the associated local ring is regular, was used by Zariski in [40] in order to define
non-singular points of abstract algebraic varieties.
In his famous paper [12] on algebraic forms, Hilbert introduced the so-called charac-
teristic function n !-+ I(R/Qft) of an ideal Q of a polynomial ring R over a field, and proved
that for large n it is a polynomial of degree equal to the dimension of Q. In [35] Samuel
applied the method of characteristic functions in his studies of local rings. He proved
that also in this case the characteristic function leads to a polynornial, now generally
called the Hilbert-Samuel polynomial, and expressed the dimension of a local ring and
the multiplicity of an m-primary ideal in terms of this polynomial. Formerly, the notion
of multiplicity of an ideal had been defined in a different way by Chevalley in [6], for
a certain class of complete local rings. The interpretation of multiplicity as the Euler-
Poincare characteristic of the respective Koszul complex (Theorem 1.5.20) is due to
Serre [X]. We have presented the proof based on the approach ofRees [32].
Section 1.6 is based on Serre's lecture notes [X]. The main result, Theorem 1.6.6
(ii), was proved by Serre, more generally, for regular local rings (R, m) whose comple-
tion in the m..adic topology is unramified (see Theorem 2.3.20 for the notion of un ramified
regular local rings). Serre also conjectured that it remains valid for an arbitrary regular
local ring. Recently Roberts [34] has proved the vanishing part of the Serre's conjecture,
i.e. that dimM + dimN < dimR implies X(M, N) = O.

I]
Notes and References
77
One can define the intersection multiplicity X(M, N) for lnodules M, N over an arbit-
rary local ring R by the same formula
00
X(M, N) = L (-1)R1R(Tor:(M, N»
n=O
provided that at least one of the modules is of finite projective dimension and M(j!) RN
is a module of finite length. Roberts' vanishing theorem is valid for more general class
of rings, e.g., for rings that are factor rings of regular local rings by regular ideals t under
the assumption that both modules M and N are of finite projective dimension. This last
condition is essential as is shown by the example given by Dutta, Hochster and
McLaughlin in [8] of a local three-dimensional hypersurface R and two R-modules
M, N, only one of them being of finite projective dimension, for which both implica
tions of the equivalence X(M, N) = 0 <=> dimM+dimN < dimR fail.
The history of the search for a satisfactory definition of the intersection multiplicity
of two subvarieties of an algebraic variety is exciting in itself, and it exposes the funda-
mental role which the intersection theory plays in algebraic geometry. The part of the
algebraic apparatus of the theory presented here is only one small fragment. The reader
interested in intersection theory is directed to the excellent monograph by Fulton [C).

Chapter II
Regular Local Rings
We defined regular local rings in Chapter I. Their significance in geometry
is due to the fact that they give a characterization of non-singular points
of an algebraic variety, in purely algebraic terms. Several quite elementary
questions cotlcerning those rings have remained unanswered for rather
long time (e.g. Krull's problem of whether the localization of a regular
ring is also regular). Adequate methods of research have been missing
until the late fifties, when the works of Auslander and Buchsbaum [1] and
of Serre [38] almost simultaneously led the way to a homological charac-
terization of regular rings and to various consequences thereof. We present
these results in Section 2.1.
It was homological methods that provided also, some three years later,
a (positive) solution to another classical problem, namely, that of the
uniqueness of factorization in regular local rings. We give a proof of this
in Section 2.2.
Section 2.3 contains the results of the fundamental work of Cohen
concerning the structure of complete regular local rings. As we have already
seen (Example 1.4.14), power series rings over a field are regular. These
are actually the only complete regular rings with the property that the
characteristic of the ring equals the characteristic of the residue class
field. In the remaining case, a regular ring is of the form W[[X 1 , ..., Xn]]/(u),
where W is a valuation ring and u is an element outside the square of the
maximal ideal of W[[X 1 , .oo, XIt]].
In Section 2.3 we also give several useful facts concerning complete
local rings (not necessarily regular), e.g. that any complete local ring is
a homomorphic image of a regular ring (this fails for arbitrary local rings).
Throughout this chapter (R, m) denotes a local ring with the maximal
deal m and K = RIm is the residue class field.
2.1 HOMOLOGICAL CHARACTERIZATION
The characterization referred to in the introductory remarks and in the
present title consists in identifying regular local rings as those which are
of finite global dimension (see Definition A.4.6)'

80
Regular Local Rings
[Ch.
Theorem 2.1.1
If (R, m) is a regular local ring then gl. dooR = dimR.
Proof
By Theorem 1.4.19, the ideal m is generated by a regular sequence of
length dimR. Thus the assertion is a direct consequence of Corollary
A.5.6 and Lemma A.6.3. 0
In proving that local rings of finite global dimension are regular we
shall need two auxiliary statements.
Lemma 2.1.2
If x Em "-m 2 then ml Rx is isomorphic to a direct summand of ml Xnt.
Proof
According to the assumption x E m"-m2 there exists a minimal system of
generators ofm containing x; see Lemma A.S.I. Let Ibe the ideal generated
by all elements in that system other than x. Since the relation rx E I implies
r Em (Lemma A.S.I), the embedding I y, m induces an injection
I/Ir.Rx -+ m/xm.
On the other hand, 1+ Rx = m, and the composition
m/Rx = (I+Rx)/Rx  I/InRx -+ m/xm -+ mlRx
is the identity. Hence the statement follows.
o
To state the next result we recall (see [B], Chap. I) that for an R-module
M the set 3(M) consists of all elements of R which are zero divisors on
M, i.e. elements r E R such that rm = 0 for some non-zero element m e M
(in general depending on r).
Lemma 2.1.3
If M is an R-module and x E R is an element such that x  3(R), x 3(M)
then pdR/(x) M/xM  pdR(M).
Proof
Let F be a projective resolution of M. We have an exact sequence of com-
plexes 0 -+ F  F -+ F/xF -+ 0, which induces a homology exact sequence
(Theorem A.2.1). Hence H lI (F/xF) = 0 for n > 1. We also get
x x
H 1 (F/xF) = Ker(Ho(F) -+ Ho(F» = Ker(M -+ M) = 0,
and so F/xF is a resolution of the module
x
Ho(F/xF) = Coker (Ho(F) -+ Ho(F») = M/xM.
Since F/xF is projective as a complex of R/(x)-modules, the lemma is
proved. []

II]
Uniqueness of Factorization
81
Under the hypotheses of Lemma 2.1.3 the two dinlcnsions are in fact
equal; see Lemma 3.5.4 and Exercise 3 in Section 3.6.
Theorem 2.1.4
If the global dimension of a local ring R is finite then R is regular.
Proof
Consider two cases:
1 0 Each element ofm'm 2 is a zero divisor in R;
2 0 Some element ofm'm 2 is not a zero divisor in R.
In case 1°, according to Theorem 1.1.7 in [B], m consists entirely of
zero divisors; hence me Ass(R). We claim that m = O. Assume the con...
trary; since m is not a free R-module, then for n = pdR(K) we have
1  n < 00. Since m E Ass(R), there is an exact sequence 0 -+ K -.. R
-.. L -+ 0, which induces an exact sequence
o -+ Torl(L, K) -+ Tor:(K, K) -+ 0
(because n  1 and pdR(K = n)). By Lemma A.S.S we have Tor:(K, K) =1= 0,
and by Corollary A.5.6 we get Tor:+1 (L, K) = o. The resulting contra..
diction proves the claim.
The argument in case 1 0 shows that, under the assumption of the the-
orem, if dimR = 0, then R is a field.
In case 2 0 there is an x em'm 2 , x  3(R), so that dimR> O. Then
x is not a zero divisor on m either, and we have pdR/(x)(m/xm)  pdR(nt)
< 00, in view of the assumption and Corollary A.S.6. Applying Lemma
2.1.2 we see that pdR/(x)(m/Rx) < 00, and so by Corollary A.5.6 gl. dimR/(x)
= pdR/(x)(K) < 00. By the induction hypothesis R/(x) is regular; thus,
using Corollary 1.4.22, we obtain the regularity of R. 0
Summing up, we see that Theorems 2.1.1 and 2.1.4 jointly result in
Theorem 2.1.5 (Auslander, Buchsbaum, Serre)
A local ring R is regular if.and only if it is of finite global dimension. Then
g1. dimR = dimR.
Corollary 2.1.6
If R is a regular local ring and P is a prime ideal of R then R p is regular.
Proof
Since gl.dimR p  gl.dimR by Theorem A.4.8, the statement follows
from Theorem 2.1.5. 0
2.2 UNIQUENESS OF FACTORIZATION
In this section we show that the elements of a regular local ring admit
unique factorization into irreducible factors.

82
Regular Local Rings
[Ch.
We recall that in a domain an element is irreducible if it is not a prod-
uct of two non-units and is not a unit itself. A domain is called a unique
factorization domain if every non-zero element is a product of a unit and
a finite number of irreducible elements, the representation being unique
up to units and permutation of factors.
In any domain, an element generating a prime ideal (i.e. a prime element)
is automatically irreducible (see [N], p. 70). The converse is characteristic
for unique factorization domains. We now prove this, and we give another
characterizatfon of those rings, which will be of use in the sequel.
Lemma 2.2.1
Let R be a Noetherian domain. The following statements are equivalent:
(i) R is a unique factorization domain:
(ii) Every irreducible element of R is prime, i.e. generates a prime ideal;
(iii) The intersection of any two principal ideals of R is also a principal
ideal.
Proof
(i) => (iii). For any x, y E R the ideal (x)n(y) is generated by the least
common multiple of x and y.
(iii) => (ii). Let x be an irreducible element and suppose ab e (x),
a  (x). By assumption, (a)n(x) = (y) for some y E R. If ax = ocy and
y = ya, then x = ocy and by the irreducibility of x either 0'- or y is a unit.
If y were a unit then (y) = (a) and a E (x), contrary to the supposition.
Thus eX is a unit; hence (y) = (ax) and ab E (ax), and finally, b E (x).
(ii) => (i). Since the ring is Noetherian, every element is a product of
finitely many irreducible elements (see Exercise 7., Section 2.3 in [B]).
Suppose
x = rJ.p 1 ... P" = {J q 1 ... qk, ( I )
where Pi'S and qJ's are irreducible elements of R and eX, p are units in R.
Since pllx, the assumption of Pt being prhne forces qJ E (Pt) for some j.
Renumbering, we may assumej = 1 and ql E (Pt). Consequently, ql being
irreducible, we have ql = YPI, where II is a unit. Substitute this into (1);
division by Pl yields eX'P2 ... Pn = P' q 2 ... q" for some units eX', p'. Applying
induction on the length of factorizations we infer that n = k and the
corresponding irreducible terms are equal up to units. D
The proof of. the basic theorem in this section requires certain auxiliary
facts. The first one of theln is a lemma due to Nagata.
Lemma 2.2.2
Let R be a Noetherian domain. Let t be a prime element of R and let Rt
denote the ring of fractions of R with respect to {t, t 2, ...}. If Rt is a unique
factorization domain then so is R.

II]
Uniqueness of Factorization
83
Proo.f
Call two elements associated if they are equal up to a unit. Let {Pi} be
a maximal set of elements irreducible in Rh pairwise non-associated. We
may assume that all p/s belong t R and are not divisible by t in R, because
the powers of t are units in Rt. We shall show that each Pi is irreducible
also in R and that any element in R admits a unique factorization into
a product of elements from {Pi}V {t }.
Since t is a prime element of R, we can easily verify that every unit in
Rt is of the form rt k with r a unit in Rand k E Z. To show that Pi is irre-
ducible in R, assume Pi = ab with a, b E R. By the irreducibility of Pi in
Rt one of the two factors, say a, is a unit in Rt; then a = rt k , k E Z and I' is
a unit in R. Hence Pi = rtkb, and since }Ji  (1), we have k  O. On the
other hand, we have r = at-k, and so k = 0, ,. being a unit. Consequently
a is a unit in R.
Now let x be any element of R. By assumption, ax = P1 ... Pn in Rh
where (X is a unit in Rh i.e. (X = rt", k e Z, and I' is a unit in R. Suppose
k < 0; then at least one of the Pi'S Inust be divisible by t, because t is prime;
and this is impossible. Therefore k  0, and x is a product of elements
from {Pi}U {t}, up to units in R.
If
rJ..f"Pi 1 ... Pi k = pt"'pjl ... pj:J
(2)
where iX, f3 are units in R, then 11 = m for the same reason as above. Thus
uniqueness of factorization ill .Rt forces uniqueness of factorization (2)
in. R. 0
Lemma 2.2.3
Let R be a Noetherian domain and let I be an ideal of R. Then. I is a pro-
jective R-lnodule if and only if I R p is a principal ideal, for every maxitnal
ideal P.
Proof
Since R is Noetherian, I is finitely generated. Thus, if I is projective then
every localization IR p is a finitely generated projective module over R p .
Hence IR p is free over R p ; see Theorem A.5.4. In a domain every free ideal
has rank 1 (because any two elements are linearly dependent). Consequently
IR p is a principal ideal.
To prove the opposite implication, assume IR p = (tp)R p for every
maximal ideal P. Clearly, it can be assumed that t p E R. For every x e I
there is s  P such that x = (afs) t p , a e R. Since I is finitely generated,
a common denominator can be found for all elelnents x of I (it suffices
to choose a common denominator for the generators); denote it by Spa
Then spI c (tp)R. The ideal generated by all sp's is not contained in
any maximal ideal. Consequently, there exist prime ideals P l' ..., P k and

84
Regular Local Rings
[Ch.
k
elements a1, ..., a" e R such that L a'Sj = 1, where s. = sp,. Denote
;= 1
Ii = t pl and let u. = (alsl)!f" i = 1, ..., k, be elements of the field of
fractions of .R. Note that LUjti = LOis; = 1 and u,l = (a,!f,) (Sf I)
c: (a,/ti) Ii R c: R.
tp"' tp
We define two mappings qJ and 1jJ, I -+ R t -. I, where R k denotes the
free module of rank k with a basis e 1, ... , ek:
k
tp(x) = L>,xe"
;= 1
k k
1J1 2.: x ,e, = 2.: x ,t,.
;= 1 i= 1
It is readily verified that
k
tptp(x) = L (u,x)t, = x;
;:z1
hence I is a direct summand for R k , hence a projective module.
o
Lemma 2.2.4
Let R be a domain and let 1 be an ideal of R such that I$Rn  R"+ 1 .
Then I is a principal ideal.
Proof
We apply the functor of exterior power 1\ n+1 to both sides of the equality
Iff)RIJ  Rn+1. Using the formula for the exterior power of the direct sum
(see [N], p. 430) we get
11+1
1\"+1 (I$R")  , IV (I)@R /\"+1-1 (R")
 /\n+l(R'H-l)  R.
The summand corresponding to i = 0 is zero because 1\ n+l Rn = O. For
i = 1 we get I @RR. If i > 1 thenl\i(I) is a torsion module (every element
has a non-zero annihilator) because any two elements of I are linearly
dependent. Thus the summands corresponding to i > 1, on the one hand,
are torsion modules and, on the other hand, are isoll10rphic to ideals of R.
Hence they are zero since R has no non-zero elements with non-zero anni-
hilators. Consequently 1  I@ R R  R is a principal ideal. 0
'Theorem 2.2.5 (Auslander, Buchsbaull1)
Every regular local ring is a unique factorizatiol1 domain.
Proof (due to Kaplansky)
We know that every regular local ring R is a domain (see Theorem 1.4.15).
We apply induction on the dimension of R. If dimR = 0, R is a field and
the statement is obvious. Let dimR > O. Select an element t belonging

II]
Structure of Complete Local Rings
85
to a regular set of parameters of R. By Theorem 1.4.19 t is a prime element.
On account of Lemma 2.2.2 it is enough to show that Rt, the ring of quo-
tients, is a unique factorization domain. To this effect we appeal to the
characterization given in Lemma 2.2.1 and we prove that the intersection
of any 1:\vo principal ideals in T = Rt is also a principal ideal.
First of all, observe that for any maximal ideal P of the ring T we have
T p  R pnR and that Pl'tR is not the maximal ideal of R (because t rp PnR).
In view of Corollary 2.1.6 we see that T p is a regular local ring of dimension
less than dim R. By the induction hypothesis T p is a unique factorization
domain. Let I = (u) Tn(v) T; according to Lemma 2.2.1, IT p = (u) Tpn
n(v) T p is a principal ideal, for any maximal ideal P in T. Lemma 2.2.3
shows that I is projective. Consider the ideal I nR. It is a finitely generated
module over a regular local ring and therefore admits a finite resolution
consisting of free modules ovet: R (see Theorems A13.7 and 2.1.4)
o  FlI -+ II. -)- F 1  F,o -+ InR  o.
(3)
Since the extension R c T is flat (see Theorem 1.4.17 in [B]), tensor multi-
plication of (3) by T yields a finite free resolution of the module (InR)@ R
@R T = (InR) T = lover T. Since I is projective, the resolution splits
and leads to an isonl0rphism of the form lEe TP  Til for some positive
integers p and q. Any localization of I with respect to a maximal ideal is
a principal ideal; hence q = p + 1. Thus Lemma 2.2.4 applies, showing
that I is a principal ideal. This ends the proof of the theorem. 0
Exercises
1. Prove that R is a unique factorization domain if and only if every prime ideal of
R of height 1 is principal.
2. Prove that a local ring R is a unique factorization domain if and only jf for any
ideal I of R generated by two elements we have pdR(I)  1.
2.3 STRUCTURE OF COMPLETE J.JOCAL RINGS
The basic tool used in deriving structure theoren1s is the concept of a ring
of representatives and the theorem on its existence in the case of complete
rings. Suppose that the residue class field Rim has characteristic p; a Noether-
ian subring W is called a ring of representatives of R if and only if the
embedding W c.... R induces an isomorphism WlpW  Rim and W is
complete in the pW-adic topology. The first part of this section contains
a proof of the existence of a ring of representatives for any ring which is
complete in the m-adic topology. The general construction is rather long
and involved; ho,vever, if p = 0 then W is a field and its existence is an
easy consequence of the Hensellemmal

86
Regular Local Rings
[Ch.
The next part of this section opens with theorems concerning arbitrary
complete rings. We approach to them with regular rings "from outside",
i.e. as factor rings of regular rings as well as "from inside" by constructing
a regular subring of the same dimension. Restricting attention to regular'
rings Rand \vriting dimR = n, we classify them as follows (Cohen's
classification) :
1 0 if char( R) = char(Rlnt) then R = Rim [[Xl' ..., XII]];
2° if char(R) = 0, char(Rlm) = p#:-O then
(a) R = W[[X 1 , ..., XII-I]] whenever p  m 2 (the un ramified case),
(b) R = W[[X 1 , ..., Xn]]/(u) whenever p em 2 (the ramified case),
where W is a discrete valuation ring (in both cases).
In the concluding part \ve show that every ramified conlplete regular
ring is a special type extension of an unramified regular ring (an Eisenstein
extension).
We recall that (R, m) always denotes a local ring R \vith Inaximal ideal
tn and K = .Rlm is the residue class field.
1. Lifting Theorems
Lemma 2.3.1 (Hensel)
Let (R, m) be a complete local ring and let FE R[X] be a monic polynomial
of degree n. Suppose there exist monic polynomials g, h E K[X] satisfying
the following conditions:
- -.
(i) F = gll, \vhere F denotes the image of F under the homomorphism
R[X]  K[X];
(ii) g and h are relatively priJne.
Then there exist monic polynomials G, HE R[X] such that:
1) F = GH,
-- -
2) G = g, H = 11,
3) degG = degg, degH = degh.
Proof
We construct a sequence of monic polynon1ials G s , Hs E R[X], s = 1, 2, ...,
- -
fulfilling the conditions: G s = g, Hs = h, degG,t: = degg, degH s = deg H
and G s -G S + 1 EtnSR[X], Hs-Hs+l em"R[X], GsHs-FEmSR[X]. The co-
efficients of those polynomials form Cauchy sequences. By the complete-
ness of R, we may define G = limGs, H = limBs in the topology defined
by mR[X]. Then clearly G and H satisfy statements 1), 2), 3).
We construct Os and H" by induction on s. For s = 1 we take arbitrary
monic polynomials G 1 ', Hl E R[X] such that 6 1 = g, Ht = hand degG 1
= degg, degH l = degh. Then F = G 1 H i , whence
G 1 H 1 -FE mR[X].

II]
Structure of Complete Local Rings
87
Let s  1 and assume that we have already defined polynomials G 1 J ...
..., Go HI, ..., Hs with properties as stated. Let {m,} be a system of
generators of the ideal m S . We have GsHs-F = 2: m,D 'J D, E R[X], and
we may assume degD i < n since G s , Hs and F are monic. There exist
polynomials U iJ V, E R[X] such that
i5 = U, g+ V, h, degU, < degg, degV , < degh. (4)
To see this, note that (as g, h are relatively prime) there exist polynomials
U, v E K[X] such that 1 = ug+vh; thus D j = ( D ju) g+ ( Di v) hand Dt f)
= cg+Vj for some v, E K[X] of degree less than degg. Hence D j = (D ,u+
+ch) g+v,h. Writing u, = Diu+clz we have- deguj < degh, and choosing
polynomials U" Vi (of degree less than degg, degh, respectively) so that
- -
U, = U" Vi = V" we obtain representation (4), as desired. Now we define
G S + 1 = G s - 2: m i V j, HS+1 = Hs- 2:m , U , . Evidently, G S + 1 and HS+l
are monic polynomials of degree equal to degg, degh, respectively and
\ve have G S + 1 = G s = g, HS +l = Hs = h. It follows directly from the
above definition that G S + 1 -G s and H.s+l -Hs are in mSR[X].
Moreover,
G s + 1 Ha+l-F = (G s - 2:m, Vi) (Hs- 2:m , U,)-F
= 2: m,(D , - G s Uj-H s V,) + 2: m,mJ U, 'V.J Em s + 1 R[X],
in view of (4). 0
Corollary 2.3.2
Let (R, m) be a complete local ring, let FE R[X] be a monic polynomial
and suppose that F has a simple root a E K. Then F has a simple root
a E R such that a = a.
Proof
By assumption, F = (X -a) h, h(a) :/= o. According to Lemma 2.3.1,
F = (X-a) H, where (X-a) = X-a, H = h. Thus a = OG and H(a)
= h(a) :j:. 0, whence H(a) :j:. O. 0
Corollary 2.3.3
Let (R, m) be a complete local ring, char K = O. Then there exists a field
of representatives of R (more precisely, a ring of representatives, which
is a field), i.e. a subring L c: R such that the natural homomorphism
l{J: R  K defines an isomorphism between Land K.
Proof
According to assumption, R contains the field of rationals Q, because
Zf\m = O. Hence the family of all fields contained in R is non-empty.

88
Regular Local Rings
[Ch.
By the Kuratowski-Zorn lemma there exists a maximal field L contained
in R. It must be shown that rp(L) = K.
Suppose there exists an element a E K, a  q;(L), and consider two
cases:
1 0 <X is transcendental over q;{L). Choose a e R such that a = a. Then
F{a) m, for any non-zero polynomial FeL[X], for otherwise a would
be a root of a polynomial with coeffici"ents in lp{L). Therefore the field
generated by L and a is contained in R, which in view of a  L contradicts
the maximality of L.
2 0 rx is algebraic over cp{L), i.e. there is a non-zero polynomial h E lp{L) [X]
such that h{a) = O. If h is the minimal poIynon1ial of a, then rx is a simple
root (since charcp{L) = 0). Take a polynomial Fe L[X] such that F = h.
By Corollary 2.3.2 F has a simple root a E R such that a = eX. Then a  11t,
since rx ::/: 0; moreover, G{a)  m for any polynomial G E L[X] of degree
less than degF = degh. Further reasoning is as in case 10: R contains
the field generated by L and a, contrary to the maximality of L, since
a = rx  rp(L). This ends the proof. D
The last corollary is a special case of a more general theorem.
Theorem 2.3.4
Let (R, m) be a complete local ring, char K = p. Then there exists a ring
of representatives of R, i.e. a subring W c R such that:
(i) the embedding W  R induces an isomorphism W/pW  K,
(ii) W is complete in the p W-adic topology,
(Hi) W is Noetherian.
Before passing to the proof we point out certain in1mediatc COllse-
quences of this theorem.
Corollary 2.3.5
W is a local ring of dimension  1 with maximal ideal p W.
Proof
If x e W\JJW, then the residue class x e W/pW is an invertible element.
Let xy = 1, yE W; then t = l-xYEpW and (l-t)-l = I-t-t+t 2 + ...
e W, by the completeness of W. Hence xy = 1 - t is invertible, and
so is x. Finally, dinl W = ht(pW)  1, by Krull's theorem (Theorem 1.2.10).
o
CoroUarly 2.3.6
W is a field if and only if char R = char K.
Proof
Let p = char K; W is a field if and only if p = 0 in W, by Corollary 2.3.5;
hence the assertion. D

II]
Structure of Complete Local Rings
89
Recall that the characteristic of a local ring is either zero or a power of
a prime number, and is just a prime number whenever the ring has no zero
divisors.
CoroUarly 2.3.7
If R is a complete local dOlnain (e.g. a regular ring) and W is a ring of
representatives of R, then
(i) W is a field whenever the characteristic of Rand K are equal,
(ii) W is a discrete valuation ring if these characteristics are distinct
(i.e. when char R = 0, char K = p "# 0).
Proof
The first statement follows from Corollary 2.3.6. In the case where p
= charK '# 0 and charR = 0, we have pW '# 0 and by Corollary 2.3.5,
dimR = 1, because W has no zero divisors (W is a subring of R). Accord-
ing to Theorem 3.6.16 in [B], Wis a discrete valuation ring. D
Under the hypothesis of Theorem 2.3.4, if char K = 0, the assertion
of the theorem follows from Corollary 2.3.3. Thus, henceforth, we assume
that char K = p "¥= o.
The construction of W in this case will be preceded by the definition
and basic properties of a p-basis of a :field.
To begin with, observe that the operation of raising to power pn is
a homomorphism of K whose image KP" is a subfield of K.
Definition 2.3.8
A p-basis of K is defined to be any subset B c K satisfying the foIlo\ving
conditions :
(i) K = KP(B), i.e. K is the field generated by B over KP;
(ii) the set B is p-independent, i.e. for any pairwise distinct elements
b 1 , ..., b r E B we have [KP(b 1 , ..., br):KP] = pro
In other words, the set B consisting of all elements of the form bl ... b,
with b 1, ..., b, distinct elements of B and with 0  (X, < p, r = 1, 2, ...,
constitutes a basis of the extension KP c K.
Jemma 2.3.9
Every field K of characteristic p =F 0 has a p-basis.
Proof
Consider the family of p-independent subsets of K: f!JJ = {B c: K; B is
linearly independent over KP}. 91 is non-empty, since B = 0 E 01 (then
B = {I }). f!JJ satisfies the conditions of the Kuratowski-Zorn :lemma, and
so it has a maximal element B E 01. It suffices to show that K = K'(B).

90
Regular Local Rings
[Ch.
Assume, on the contrary, that there exists t E K, t  K'(B). We will
prove that Bu {t} E 01, contradicting the maximality of B. It is enough
to show that the system {I, t, ..., t P - 1 } is linearly independent over KP(B),
i.e. the minimal polynomial of t over K'(B) is of degree p. Of course, t is
a root of the polynomial XP - t P , which therefore lDUSt be divisible over
K'(B) by the minimal polynomial of t, and so the latter is of the forIn
(X-t)r, O<rp. Since (X-t)r=X'"-rtX r - 1 +...eKP(B)[X] and
t  KP(B), r is not invertible in KP(B), and consequently,. = p. 0
Lemma 2.3.10
If B is a p-basis of a field K, then the following t\vo conditions are fulfilled
for any positive integer s:
(i) K = KP'(B),
(ii) if b I , . II , b r are distinct then [KP'(b 1 , ..., b r ): KP] = prs.
Proof
Consider the sequence of fields K =:) KP =:) KP2 ::> ... =:) Kp fJ . It is easy to
see that Bpi is a p-basis of Kp l ; i.e., B pi is a basis of the extension Kpl+l c: KPf,
i = 0, 1, ..., s - 1. The product of these bases is a basis of Kover KP'.
Hence the set
{bl1+...+OCllP'-l ... bP1+...+a.rspS-l; b i E B, 0  (X'I < p}
= {b1 ... br: b , e B, 0  (X, < pS}
is a basis for the extension KP s c K. This proves the lemma.
o
In the construction of W we shall also need
Lemma 2.3.11
Let (R, m) be a local ring and suppose that pEnt. Then
(i) if a-b em then a""-b pn Emn+l.
In particular,
(ii) if mn+l = 0 then for every s  n there exists an injective multi..
plicative mapping a: K  R, a(a+m) = aPse
Proof
Let c = a-b em; then
apn_b Pn = L (pn)btc'.
k+J=pn J
Jt:-O
If jn = P1', (j',p) = 1, 0 <.i  p", then
pn-t (.) and (n)CJEmn-t+Jc:mn+l,
because j - t  1. Hence we get (i). The mapping C1 defined in (ii) is an
injection, since the equality a P ' = b pa implies (a-b)P' Em, \vhence a-b Em.
o

II]
Structure of Complete Local Rings
91
Construction of W
We shall define an inverse system of rings {" T,.}, W,. c: Rlmn, which
is a subsystem of the standard system {Rlm n , f{J,,}, Pn: Rlm n  Rlmn-l,
<JJn(r+m n ) = r+mn-l. Since R is complete, R = Iim {Rim"}. We define
-
W=lim{W n }.
-
Construction of W n
According to Lemma 2.3.11, there exists, for every n, the injective multi-
plicative mapping 0',,: K  Rim", (1,.(a+m) = (a+m n )p2". Writing An
= (1,.(K.) we obtain the following commutative diagram:
Rim'"
J
--.. K
J
K 
0',.
... A"....
V 2.
.. L\.p
in which the upper horizontal map is the natural epimorphism and the
composition of the lower maps is raising to the power p2rt. This operation is
one-to-one and hence it defines a one-to-one correspondence An <H KP2".
Let B* be a p-basis of K = Rim and let B be a subset of R of represen-
tatives of B*, i.e. B* = {b+m; b eB}. Further let Bn = {b+m n ; b eB}
for n = 1, 2, ...; in particular, B 1 = B*.
Let 8n be the subset of Rim" obtained by polynomial operations on the
elements of Bn, the polynomials having coefficients from An and being
of degree less than p2n in each indeterminate. That is to say,
S" = { 2:: (a", + nt")p2n (b 1 + m")"" ... (b k + m ")a:k ;
OCXI<p2n
a", E R, b , E B distinct}.
Since K = KP2n(B*), in view of Lemma 2.3.10, there is a one-to-one corre-
spondence between 8,. and K induced by the natural homomorphism
Rlm n --) K. The set 8n arises by the "lifting" of K to Rlm n by means of
the p-basis B*; in particular, S 1 = K. We obtain the commutative diagram
R/m ..,..-
J
Sn ·
J
 K
.- K
J
An 
2"
 KP

92
Regular Local Rings
[Ch.
We now define . as
n-t
W n = Sn+pSn+ ... +pn-1S n = {2:: C ,p i; C, ES,,}.
1=0
Properties of W,.
Lemma 2.3.12
w,. is a sub ring of Rim".
Proof
We must establish the inclusions
. ( )} -1 5
pJ W n + W n ) c p W n , P (W n · W n ) c: p W" ( )
for 0  j  n. We prove this by induction, starting withj = n and ending
with j = O. The case of j = n is obvious, since p" = 0 in Rltu n . Assume
that the inclusions in (5) are true for j = i + 1. We first show that
pi(Sn+8n) c: pi" p'(Sn. 8n) c p'W,.. (6)
Let L all H, L CH H e Sri, where aH, CH E An and H stands for a mono-
mial of products of elements of Bn. We have aH = di, Cs = eJ" where
q = p2n and dH, eH E Rim". It can be assumed that dR, eH E Sn; indeed,
if dB = d+m n then there exists d' +m" e S,. assigned to d+m in the corre..
spondence SrI ++ K; thus 'd-d' e tn and so au = (d+m Jl )' = (d' +ttt")',
by Lemma 2.3.11.
We have
pi (2::aH H + 2:: CH H) = pi (2: (d,t + el,) H)
q-t
= pi (2:: (dl{ + eH)IlH) _pi L (2: (k) dZ-ke) H.
kat
(7)
The first summand obviously belongs to piS n . As regards the second one,
note that -1 e S" c: W n , provided p :F 2; if p = 2 then -1 = 1 +2+ ...
... + 2,.-1 e fV;. since pn = 0 in Rlm n . From the induction hypothesis we get
i+1 (W: + + w: ) i+1 W. i+l (W: W. ) i+1 JXf
P n. · · n c: P n , P n ... n C P "Y n ,
for an arbitrary finite number of copies of W II . Hence in vie\v of the fact
that p (:) for 1  k  q - 1, we see that the second summand in (7)
belongs to p'+l W n . Consequently
pi(Sn+S,,) c: piSII+pi+1 W n = p'(Sn+PS,.+ ...) = piW n .
Before passing to the proof of the second inclusion in (6) let us remark
that p'(Sn+Sn) c p'W n yields p'(S,,+ ... +Sn) c: p'W" for any finite
number of summands; e.g. for three summands we have

II]
Structure of Complete Local Rings
93
pi(S" + S,. + S,,) = pi(Sn + S,,) + piS,. c: pi w,. + piS"
c pi(SII + S,.) +pi+l W n c: pi w,. + pi+l W n
c: piS" + pi+l w,. + p+l W" c: piS n + pi+1 W" C pi w,.;
we again used here the induction hypothesis for i+ 1.
In order to prove the inclusion pi(S". Sn) c: pi W,. observe that
(L: aGG) (L: CH H) = L: aG CH GH; hence by the preceding remark, it is
enough to show that aGcHGH E S,.. Thus let aG = (a+m,,)q, CH
= (c+m")q, G = n (bj+mnyxJ, H = n (bj+m,,)vJ. Then
j J
aGc1IGH= (d+ntn)qn(bJ+mnY'JeS n , .
j
where (Xj+YJ = l"jq+ j, 0  j < q. Thus (6) is proved.
Now \ve show that
i f i i
P (W n + W") c: p Wn, P (W". W n ) c p W n ,
and the induction will be complete. Using (6) and the induction hypothesis
we obtain the following inclusions for the sum:
pi(W n + w,.) c: pi(8 n + 8") + ]J'+l(W I1 + W n ) c: pi w,. +pi+l Wit
c: p i S"+pi+l W n +p'+l W" C p i Sn+pi+l W" = piW n
and for the product:
pi(W". W") c: pieS,. · 8") + pi+l JY,. c:.pi W n + pi+1 W n c: piw,.. 0
Lemma 2.3.13
(i) Wnnttt/tn'. = P W,,;
(ii) if CPn: R/mn -+ R/m"-l denotes the canonical homomorphism then
qJ,.( I) = W,._ 1 ;
(iii) let /e n = max {k; pk ,m"} = max {k; pk W n :F O}; the sequence
k'i is non-decreasing and Annwn(pi) = pkn-i+l W n for 0  i  k,,+ 1.
Proof
(i) Clearly pW II c W"nm/m". As for the opposite inclusion, it has to
be sho,vn that if x e W n V W n then x  m/mn, i.e. the image of x under
the natural homomorphism R/m" --+ K is different from zero. We may
assume that x e 8,,; the statement is then obvious because that mapping
establishes a one-to-one correspondence between 8" and K.
(ii) It follows from the definition of W n that the inclusion rpn(W n ) c: W n - t
is true, provided that rp,.(A,,) c: An-l and qJ,,(B,.) c: B"-t. The latter inclu"
sion is obvious and the fortner one also holds:
tpn(a+ tn n )p2n = (a+ tnU-l)p2n = (a P2 + 11t,,-1)p%(n-l) E A,I-i'
For the proof of the opposite inclusion we again apply induction fronI
i = n -1 to i = 0 to show that pi W'.-l c: piP,I(W n ), For i = n -1 this is
true because p" -1 = 0 in R/tn,,-t.

94
Regular Local Rings
[Ch.
Recall that we have the following commutative diagranl:
RJnt n (/In R/nln -1  K
,.
J I
s ..c )IiJ K
n
Thus for every x E W Il - 1 there exists Y E Sit havin.g in K the saine ilnage
as x. Also 9'n(Y), has the same image ill K. Hence X-9'n(Y) E W,,_lnm/m U - 1
= p W,,-I, in view of (i). Consequently x = tp,.(y) + pz, piX = pi9'n(Y) +
+p'+IZ for some Z E W,.-I. By the induction hypothesis pi+lZ epi+l9'n(Wn)
so that pi Z E pilpn(W II ).
(iii) Once more we employ descending induction from i = k n + 1 (the
equality is then obvious) to i = 1.
If pt = 0 then  = pkn-irJ, rJ E w,., by the induction hypothesis. Hence
pn'fJ = 0, and since pn i= 0 in R/m fJ , we have rJ em. According to (i),
'fJ = p for a certain a E W n . Thus, eventually, E e pkn-l+ 1 W n . D
Proof of Theorem 2.3.4
By Theorem 2.6.12 in [B] (exactness of the inverse limit functor) and by
the completeness of R we have lim{W n }  Iim {R/m"} = R. We define
- ..--
W = lim{W,,}.
...-
(i) At first we show that the embedding W c: R induces an isomorphis111
W/mn W  K. Indeed, WI = K atld for every A E K, there is  = <II>
E W such that 1 = A by Lemma 2.3.13 (Hi).
Since p W c: mn W, it relnains to show that mn W c; p w.
Let <,,) E Wn1n; we will construct a sequence {1]n} such that
(a) rJn E U/',,,
(b) 9'11 ( 1] n) = 'fJ n - 1 ,
(c) E,. = P1Jn.
Then <II> = p('fJ,,) EpW.
Notice that II epW n by Lemma 2.3.13. Consider two cases:
1 0 char R = pk i:- O. Then pk-l  m" for large n; in other words, there
exists N such that k" = k-l (in the notation of Lemma 2.3.13) for n  N.
We define rJN to be any element such that PrJN = EN and 'YJi, i  N, to
be the image of 'fJN under the natural homomorphism R/mN --+ R/mi.
Assume we have already defined rJl, ... 'Y/n-l satisfying conditions
(a)-(c). Let 'Y/ be any element in W n for ,vhich  = P'YJ. Then the difference
'Y/n- 1 - ffJ,,('YJ) is annihilated by p and so, in view of Lemma 2.3.13, there
exists y E W n - 1 such that 1]"-1 - f/Jn(1J) = pk--l9'n(Y)' 'Y e W n . Putting 17"
= 'fJ + pk-ly, we see without difficulty that 'Y/ l' ... , '1711 fulfil conditions
(a)-(c).

II]
Structure of Complete Local Rings
95
2 0 char R = O. In this case we can find, for any i, an integer n(i) > i
such that pi rp ttt"(i) and kn(l)  i. We may assume that the sequence {n(i)}
is non-decreasing. Further, for each i, there exists 'Y} such that "(i) = P17,
1] E Wn(i). Define 'Y/I as the image of 1]; under the natural homomorphism
Rfmn(i) -+ R/m. Then clearly 1]i e Wi and i = P'YJi. Moreover, p(tp('Y/;+1)-
-'Y}) = 0, ,vhere rp: R/mn(i+l)  R/m"(I). By Lemma 2.3.13,
<P('YJ+1)-'Y} Epkn(I)Wn(1) C pi JVn(i) c: nti/mn(i),
so that CP('Y}i+ 1) = 'YJi.
(ii) Let us remark that, in view of Theorem 2.6.11 in [B], W can be
identified with the set of ll limits lim {x,,}, X n e R, where (x n + 11t 1l ) e W
= Iim {W n }. Now, the completion of W in the pW-adic topology can be
.....-
viewed as a subring of R, since R is complete in the m-adic topology, and
the topology induced on W is precisely the p W-adic topology. All that
A A
must be shown is that W = W. Let c e W, i.e. c = lim {cn}, C n E W,
Cn+l-Cn EpnW. By the preceding remark, ell = Iim{x n i}, where (X"i+m")
e W = lim {W n }. It is readily verified that c = lim {xnr,} and (xnn+m") E W,
4-
i.e. C E W.
(iii) For every non-zero ideal I there exists 11 such that I c: pn W,
1. q: pn+l W. Thus for some x E I we have x = pny, YEP W. The argument
used in the proof of Corollary 2.3.5 now applies and shows that y is inver-
tible in W. Hence pn e I and so, eventually, I = p"W. D
2. Cohen's theorems
Let Xl' ..., x" be any elements in m, the maximal ideal of a local ring R,
and let UI' be a ring of representatives of R. The homomorphisnl cp:
W[X 1 , ..., Xn] -)0 R defined by tp(X.) = Xi, i = 1, ..., n, cpl W = Id, in-
duces a homomorphism q;: W[[X 1 , ..., Xn]] -)0 R, (Xi) = Xh I W = Id,
because the power series ring is the completion, in the (Xl' ..., Xn)-adic
topology, of the ring of polynomials. If moreover R is itself complete in
the m-adic topology, we get a homomorphism : W[[X 1 ,..., XII]] -)0 R.
The theorems that follow state certain properties of this homomorphisnl,
depending on the choice of Xl' ..., XII'
Theorem 2.3.14
Let (R, m) be a complete local ring and Waring of representatives of R.
If Xl' ..., X n is a set of generators of m then the induced homomorphism
fp: W [[Xl' ..., XII]] .-)0 R is surjective.
Proof
Take an element x in R. By the definition of a ring of representatives there
exists }V E W such that x = w+ 2:rIX" I"i E R. Applying the same argu-
ment to each ti we obtain a representation rl = Wi+ 2: r'Jxi, so that

96
Regular Local Rings
[Ch.
x = W + 2: }Vi Xi (modm 2). Continuing, we arrive at the conclusion that
x = W+LWiXI+LWijXIXJ+ ..., by the completeness of R in the m-adic
topology. Thus  is a surjection. 0
Corollary 2.3.15
Every complete local domain is a homomorphic image of a complete
regular ring.
Proof
According to Corollary 2.3.7, W is either a field or a discrete valuation
ring. Therefore W [[XI' ..., Xn]] is a complete regular ring, and by Theoreol
2.3.14 R is a homomorphic image of that ring. 0
Remark 2.3.16
Corollary 2.3.15 is valid for an arbitrary local ring (not only a domain).
In the general case this requires a proof of the fact that the ring of repre-
sentatives is an image of a discrete valuation ring; we shall not pursue this
subject.
For the proof of the next theorem we state the following lemma.
Lemma 2.3.17
Let (T, n) be a local ring complete in the n-adic topology and let N be
a T-module such that the 1t-adic topology on N is Hausdorff. If NlttN is
a finitely generated T-module then so is N.
Proof
We shaH prove that whenever the residue classes of YI, ..., Ys EN generate
N InN then the elements Yl' ..., y" generate N.
s
Let A = L TYi c: N and let c be any element of N. We construct
/=1
a sequence {at}, ak E A, such that
s
1) ak = L fXklYb fXki E nk-l,
i=l
k
2) c- L oJ En"N.
J=1
We proceed by induction on k. If k = 1 then by assumption there are
elements all e T, for which c- LCXI1Yi ettN; we set at == LallY" Let
k > 1 and suppose we have already defined elements a1' . to , ak satisfying
conditions 1), 2). Then
k
c- LOJ= La,h" IX/Enk, b,EN.
J=1

II]
Structure of Complete Local Rings
97
The argument which we used in case of k =:: 1 no\\' applies to each of the
hi's showing that there exist elements d i E A such that hi - d i E "N. Define
ak+ 1 = L (X,i d ,. Since i E nk, equality 1) holds with suitable (Xk+ 1.i. Con-
dition 2) is also satisfied:
Ic+l
c- 2: a j = L rx,b,- L (X,d, Enk(nN) = nt+1N.
)=1
By the completeness of T, the following elements are well defined:
a
a.t = L fXti e T, a* = L r y, E A. Moreover, we have
k 1=1
s k
c"_. a* = lim (c - 2:: (2:: (XII') y,)
k 1=1 n=1
k s k
= lim (c- 2:: (2:: (XnIY')) = lim(c- 2: an) = 0,
k n=1 1=1 k 11=1
in view of condition 2) and the assumption that the n-adic topology is
Hausdorff on N. Thus c = a* E A and so N = A. Consequently, N is
finitely generated over T. D
Theorem 2.3.18
Let (R, m) be a complete local domain and let W be a ring of representa-
tives of R. Suppose u 1, ..., Un is a set of parameters of R such that u 1 = P
when charRi:-charK=p. Let T denote W[[X 1 ,...,X n JJ when charR
= char K, and W[[X 2 , ..., Xn]] when char R i= char K. Then
(i) the homomorphism tp: T -+ R corresponding to the system u l' ... , Un
(i.e. given by q,(X t ) = U" I W = Id) is injective,
(ii) .R is a finitely generated T-module.
Proof
Notice that every complete ring R has a set of parameters with the property
described in the theorem, since p is not a zero divisor in R and we can
find a set of parameters containing p; see Corollary 1.4.3.
We start with the proof of (ii).
(ii) We have W/p W  K, in virtue of Theorem 2.3.4. Therefore, in
either case, the homomorphism tp induces an isomorphism of the residue
class fields. Let n be the maximal ideal of T. By the definition of  we have
nR = (Ul, ..., u,,), and, since li1, ..., U'I is a set of parameters of R, it
follows that m ::> nR :::> m k for some k > O. Hence R/nR is a homomorphic
image of R/rtt k . To show that R/ttR is a finitely generated T-module it
suffices to prove the same for R/mk. This will be done by induction on k.

98
Regular Local Rings
[Chi
For k = 1, Tin  RIm and lIenee the assertion. The inductive step is
achieved with the help of the exact sequence
o -+ mklmk+l -+ Rlm"+l -+ Rlm t -+ 0;
tn k /mk+ 1 is a finitely generated RIm-module, hence is also finitely gener-
ated over T. Thus if Rlm k is finitely generated over T, then so is R/mk+l.
Now, applying Lemma 2.3.17, we conclude that R is a finitely generated
T-module.
(i) It follows directly from the definition of T that dim T = ditn R. Let
T' = Im; since by (ii) R is a finitely generated T'-module, we get dimT'
= dimR, by Example 1.2.2. Thus dim T = dim T'. Supposing Kerq, -:/= 0,
we get dim T' < dim T, since T' is a domain, as a subring of R. We arrived
at a contradiction, which shows that Ker = 0, so that T  T' c R. D
Corollary 2.3.19
Every complete local domain R contains a regular subritIg T of the same
dimension as R such that R is a finitely generated T-module.
As a consequence of the two preceding theorems we obtain the main
result of Cohen.
Theorem 2.3.20 (Cohen)
Let (R, m) be a complete regular local ring, dim R = n, and let W be a ring
of representatives of R. Then:
(i) if char R = char K, then W is a field and R  W[[X 1 ,..., ,]];
(ii) if char R = 0, char K = P -:/= 0, then W is a discrete valuation ring
and
(a) if p m2, then R  W[[X 2 , ...,X lI ]] (the unramified case),
(b) if p em 2 , then there exists an element U E W[[X 1 ..., XII]]' U  9)(2
(9Jl denoting the maximal ideal of W[[X 1 , ..., XII]])' such that
R  W[[X 1 , ..., Xn]]/(u) (the ramified case).
Proof
In cases (i) and (ii) (a) there ex.ists a regular set of paralneters Ul, ..., Un,
with Ul = P in case (ii) (a). Since R has no zeo divisors (see Theorem 1.4.15),
the desired isomorphism is established by the homomorphism determined
by u l' ... , Uti, in accordance with Theorems 2.3.14 and 2.3.18.
(ii) (b) Let Ul, ..., 1,1" be any regular set of parameters of R. By Theorcl11
2.3.14, the induced homomorphism
1\
q;: W[[X 1 , ..., Xn]] = T -+ R
is a surjection. Since dim W = 1, then dim W[[X 1 , ..., Xn]] = n+ 1 and
so p = Ker is a prime ideal of height 1.
The condition p em 2 implies PEL TXiXJ+P in T. Thus there exists
U E P such that u = P- L tijXiX J . If IDl denotes the maximal ideal of T

II]
Structure of Complete Local Rings
99
then clearly p  rol 2 ; thus also u  9)12. Consequently, u is an element of
a minimal set of generators of ID1, and so T/(u) T is a regular ring. Since
a regular ring has no zero divisors, (u) is a prime ideal of T. From the
relations (u) c: P and ht (P) = 1 we obtain P = (u). 0
In the concluding part of this section we will give one more charac-
terization of ramified complete regular rings.
To this purpose we adopt the following definition based on the Eisen-
stein irreducibility criterion, wen known in nunlber theory.
Definition 2.3.21
Let (R,m) be a local ring. A monic polynomial 1= X"+ a1 X"-1+ ...
... +a n E R[X] such that a, Em, an  nt 2 will be called an Eisenstein poly-
nomial over R.
An extension of type R c: R[x]/(f)R[X] with an Eisenstein polynomial
1 will be called an Eisenstein extension.
As in the classical case (see [N], p. 128) one can prove that if R is a
unique factorization domain then an Eisenstein polynomial is irreducible
over R as well as over the field of fractions of R.
Theorem 2.3.22
(i) Every ramified complete regular local ring (R, m) is an Eisenstein
extension of an unramified complete regular local ring (T, n).
(ii) Any Eisenstein extension of a regular local ring is also a regular
local ring.
Proof
(i) According to Theorem 1.4.23 there exists a regular set of parameters
21.1' ..., Un of R such that p, U2, ..., Un is a set of parameters of R. By The-
orem 2.3.18, R contains a regular ring T isomorphic to W[[X 2 ,..., X n ]],
with the maximal ideal n generated by p, U2, ..., Un. We have the following
commutative diagram:
I'J
/"ttJ
TIn ... RIm
 R/nR /
We also know that R is a finitely generated T-module. Since p, U2, ..., U"
is a set of parameters of R, UI E nR for some k. Let 8 be the least integer
with this property. Then RlnR is generated over T/n by the residue classes
containing the powers 1, U 1, ..., U f-l. By Lelnma 2.3.17 these elements
generate Rover T, i.e. R = T[Ul]. Consequently there is a monic poly-
nomial fe T[X] of degree 8,1= X S +t 1 X S - 1 + ... +t s - 1 X+t s , such that
Ul is a root of f. We shan prove that f is an Eisenstein polynomial over T.

100
Regular Local Rings
[Ch.
Assume that some of the ti'S do not belong to tt and let h be the greatest
i for which Ii ,p n; then t h + I' ..., t:J E tt. From the equality !(Ut) = 0 we get
u + I 1 U-l + o.. + t h ul- h = t4- h (u + ... + t h ) E nR.
Since t h  nand Ul E m, the expression in the brackets is an invertible
element in R and thus 14- 1 ' E nR, contrary to the definition of s.
Now we show that I fj n 2 . Suppose not. Then !(Ut) = 0 implies
 em(nR) c: mp+(u2, ..., u,,)R = (pul)R+(U2, ..., u,,)Rand so uI(u1- 1 -
-ap) E (U2, ..., un)R for some a E R. The ideal (U2, ..., ulf)R is prime
(for it is generated by a subset of a regular set of parameters) and does
not contain Ut. Thus u-t E (p, U2, ..., u,,) R = nR, again contrary to the
definition of s.
To finish the proof of assertion (i) it remains to show that R  T[u 1]
 T[X]/(f) T[X]. Notice that if g(u J) = 0, g E T[X], g :F 0, then degg  s.
This is a consequence of the fact that f is irreducible 110t only over T but
also over the field of fractions of T; and this, in turn, follows froln the
regularity of the ring T, ,vhich therefore is a unique factorization domain
(Theorem 2.2.5). Applying the Euclid algorithm to the polynomials g
and f in T[X] we conclude that g is a multiple of f; and this yields the
required isomorphism.
(ii) Let (R, m) be a local ring and f = XS+Ct xs - t + ... +c s E R[X]
be an Eisenstein polynomial over R. We first show that R = R[X]/(j)R[X]
is a local ring with the maximal ideal fit = m+Rx+ ... +R X S-l, where
x = X + (f). Clearly fit is a maximal ideal. Suppose 9R is any maximal
N IV
ideal of R. Since R c: R is an integral extension, 9J1 nR is a maximal ideal
of R and so 9J1 nR = In. In view of C:J Em we have C s = -r - Cl X S - 1 _ ...
... - Cs-I X E IDl. Hence either x E 9R or r- l + Cl x'-2 + ... + C S -l E IDt
Repeating the argument with other c,'s in place of C s we arrive at the con-
clusion that x E 9Jl. Thus tit c: IDl. By the maximality of m this shows that
fit = 9R and hence (R, tit) is a local ring.
Now assume that (R, m) is regular. Since C s Em"m2, there exists
a regular system of parameters Xl'.." XII containing C.; say, X n = c..
We shall prove that tit = (Xl' ..., X II -l, x)R. Evidently, Xl'...' X,,-l ,
x Em. On the other hand, in view of f{x) = 0 we have C s E (x).R, and so
fit C (Xl' .o., X n -l, x)R. Since dimR = dimR = n because R c: R is an
integral extension, R is a regular local ring. 0
NOTES AND REFERENCES
Theorem 2.1.5 was proved by Auslander and Buchsbaum in [1], and, independently,
by Serre in [38]. Theorem 2.2.5 is taken from [2].
Theorem 2.3.20 on the structure of complete regular local rings is taken from Cohen
[7]. The proof we have presented, and the construction of the ring of representatives, are
based on the Nagata's book [R]. The Hensel lemma (Lemma 2.3.1) for complete valu-
ation rings was first published in [11].

Chapter III
Cohen-Macaulay Rings
In Chapter I we proved the classical Macaulay theorem (Theorem 1.3.7)
which asserts that in the polynomial ring K[X 1 , ..., Xn] over a field K the
heights of all the prime ideals associated with an ideal of height rand
generated by r elements are the same and are also equal to r. This theorem
has a very clear geometrical interpretation. To an ideal I corresponds an
algebraic set V = V(I) in the affine space Kn. Its irreducible components
Vi' ..., V s correspond to the associated ideals of I. If I is of height r, then
the dimension of the set V is equal to n - r. By Krull's theorem (Theorelll
1.2.10), the ideal I cannot be generated by less than r elements, which
means geometrically that the set V cannot be expressed as an intersection
of less than r hypersurfaces. If, however, the set V is expressible as an
intersection of r hypersurfaces, then from Macaulay's theorem we kno\v
that all the irreducible components V 1 , ..., V 3 of V have the same dimension
as V. This is in accordance with geometrical intuition suggesting that if
a hypersurface is "in general position" with respect to a given algebraic
variety, then it cuts out in it a variety, whose irreducible components are
all of dimension less by 1. We can say, not very precisely, that algebraic
varietes described by a "proper" number of equations are "unmixed", i.e.
all their irreducible components are of the same dimension. This shnple
geometrical fact has served as a point of departure for developing the theory
of an interesting class of commutative rings, called the Cohen-Macaulay
rings.
In this chapter we present the basic elements of this theory.
The name of I. S. Cohen appears here side by side, with that of F. S.
Macaulay because Cohen proved in the late 19408 that the property de-
scribed by Macaulay for polynomials rings, is also valid for regular local
rings. In the 1950s investigations were carried out on those commutative
rings which satisfy the Cohen-Macaulay theorem. This domain is still
developing and is an object of intense study.
In the meantime it turned out that Cohen-Macaulay rings can be
characterized in various ways, that they have n1any natural properties,
and that the principal tool for their study is a concept which expres'"
an algebraic way the "general" position of a hypersurface with rp
a given variety, ensuring that the intersection is unmixed. We are

102
Cohen-Macaulay Rings
[Ch.
here about regular sequences. This simple concept, which we have already
met in the preceding chapter, is analysed in detail in Section 3.1 and leads
to a very important invariant of an ideal I in a Noetherian ring, namely
to its depth; this is the length of a maximal regular sequence contained
in the ideal I. Furthermore, Section 3.1 contains an effective character-
ization of depth in terms of the functors Ext.
In Section 3.2 we present the relations between depth (I) and the pre-
viously introduced invariants of the ideal I: the height ht(l) and the mini-
mal number of generators v (1). It turns out that we always have depth(l)
 ht(l)  v (1). The equalities hold if and only if I can be generated by
a regular sequence (Theorem 3.2.3). Section 3.2 contains also the proof
of the theorem which describes regular sequences in a local ring in terms
of the graded ring associated with an ideal (Theorem 3.2.9).
In Section 3.3 we give various characterizations of Cohen-Macaulay
rings. It turns out, for example, that in a Noetherian ring the Cohen-Ma-
caulay theorem is valid if and only if depth is equal to height for any or
only for any maximal ideal. The characterizations of local Cohen-Macaulay
rings include Theorem 3.3.6 which describes them in terms of multiplicities
eo(R) for ideals Q generated by sets of parameters (for multiplicities see
Chapter I).
In Section 3.4 we deal with the basic properties of Cohen-Macaulay
rings. This class is closed under localization and under factoring by ideals
generated by regular sequences; also the formal power series ring R [[X]]
is a Cohen-Macaulay ring if and only if the ring R is Cohen-Macaulay.
Of particular interest is the property stating that in a Cohen-Macaulay
ring all saturated chains of prime ideals between two fixed prime ideals
have the same length.
In Section 3.5 we prove that R is a Cohen-Macaulay ring if and only
if the polynomial ring R[X] is Cohen-Macaulay. As a corollary we derive
once again the classical Macaulay theorem.
In Section 3.6 we deal with the problem for which ideals I of a Cohen-
Macaulay ring R the factor ring R/I is again Cohen-Macaulay. Under
the assumption of the finiteness of the projective dimension of RjI these
are precisely those ideals for which depth (1) = pd R (Rj I) (the so-called perfect
ideals, also already studied by Macaulay in the case of a polynomial
ring) and the ideal I is unmixed (Theorem 3.6.11). We obtain these results
using the Auslander-Buchsbaum theorem (Theorem 3.6.6), which estab-
lishes a relationship between depth and projective dimension of a module
over a local ring.
In Section 3.7 we concentrate on ideals of depth 2 in a local ring froDl
the point of view of their property of being perfect. Such an ideal, of finite
projective dimension, is perfect if and only if it is generated by all the
(n-I) x (n-l) minors of an n x (n-l) matrix with entries in a given local
ring (the Hilbert-Burch theorem).

III]
Regular Sequences and the Depth of a Module
103
3.1 REGULAR SEQUENCES AND THE DEPTH OF A MODULI:
The notion of a regular sequence, which played a fundamental role in the
description of regular rings, will now be extended to arbitrary modules.
Definition 3.1.1
A sequence of elements Xl' ..., X" of a ring R is caned a regular sequence
on an R-module M (or an M-sequence) if:
(i) (Xl' ..., X p ) M =F M,
(ii) the element Xi is not a zero-divisor 011 the l110dule M / (x l' ..., Xi - l)M
for i = 1, ...,]} (for i = 1 we set (Xl' ..., Xi-l) M = 0).
Remark 3.1.2
Condition (i) in Definition 3.1.1 is technical; it el1sures that the modules
M/(Xl, ..., x;)M are non-zero, and it makes sense to talk about non..
zero-divisors on them. Observe also that if Xl' ..., x p is a regular sequence
on the lnodule M then none of the elements Xi is either zero or an invertible
element of R.
In the sequel, we shaH also use the term regular sequence when it is
clear from the context which module is n1eant.
An example of a regular sequence on the R-module R, in the case where
R = T[X 1 , ..., X,,], is provided by the sequence of indeterminates Xl' ..., Xn.
Any regular sequence of a ring enjoys many properties of indeterminates
(see, e.g. Theorem 3.2.9).
If Xl' ..., x p is a regular sequence on a Jnodule M, then in the sequence
of ideals
(Xl) c: (Xl' X2) c... c (Xl' ..., X,)
all inclusions are proper. Indeed, if we had X n E (Xl' ..., Xn-t) for SOlne II,
then X,. E 3(M/(x 1 , .f.' x,,-t)M). This shows that for a Noetherian ring
there exist maxitnal M-sequences.
OUf first aim is to prove that any t\\'o maxiInal M-sequences contained
in a given ideal have the same number of elements. An important step in
this direction is to express regularity in terins of the functors Ext.
Theorem 3.1.3
Let R be a Noetherian ring and I an ideal of R. Let a sequence Xl' ..., X JJ
of elements of I be a reguI,ar sequence on a finitely generated R-module M.
Then
.
Ext(R/I, M)  HomR(R/I, M/(Xl' ...,xp)M).
To prove the theorem we shall need the following lemma.

104
Cohen-Macaulay Rings
[Ch.
Lemma 3.1.4
A sequence X1, ..., x p is a regular sequence on a module M if and only
if for each n, 1  n < P, Xl' ..., X n is a regular sequence on M and X n + I' ...
... , x p is a regular sequence on the module M /(Xt, ..., x n ) M.
The proof of the lemma follows directly from Definition 3.1.1 if we
make use of the isomorphism
(xn+t, ..., x,) (M/(XI , ..., xn)M)  (Xl' ..., x,)¥/(XI, ..., x,,)M
for i > n.
Lemma 3.1.5
Let R be a Noetherian ring, I an ideal in R, and M a finitely generated
R-module. Then HomR(R/I, M) = 0 if and only if there exists an element
in I which is not a zero-divisor on M.
Proof
If x E I, then for every homomorphism f: R/I  M and for any u e R/I
,ve have 0 = f(xu) = xf(u). If, moreover, x; 3(M), then feu) = 0, and
hence f = o.
Assume now that every element of I is a zero-divisor on M, i.e. I c: 3 (M).
It follows from Corollaries 1.1.8 and 2.4.4 in [B] that I c Ann(a) for
some a EM. We thus have a non-zero homomorphism R/I  R/Ann(a)
c 0
Proof of Theorem 3.1.3
The proof is by induction on the length p of the regular sequence.
For p = 0 the assertion is obvious since
Ext(R/I, M)  HomR(R/I, M).
Suppose that p > 0 and write M. = M/(XI t ..., xi)M for i = 1, ..., p.
By Lemma 3.1.4, X2' ..., x" is an M I-sequence, whence by the inductive
hypothesis
Ext-l(R/I, Mt)  HomR(R/I, M 1 /(X2, ..., xp)M 1 )
 HomR(RII, M p ).
Consider now the short exact sequence 0  M :; M  M 1  0 and a part
of the long exact sequence of functors Ext induced by it:
Ext-l(R/I, M)  Ext-l(R/I, M 1 )  Ext(R/I, M)
Xl
 Extft(R/I, M). (1)
By the inductive hypothesis, Ext'l1- 1 (R/I, M)  HomR(R/I, M p - 1 ). There-
fore, by .Lemma 3.1.5 and the assumption x p ft 3(M,-t), we get
Ext-(R/I, M) = o. Furthermore, the homomorphism which consists in the

III]
Regular Sequences and the Depth of a Module
105
multiplication by Xl in sequence (1) is zero because Xl E I. Hence, finally,
Ext(R/I, M)  Ext-l(R/I, M 1 )  HomR(RjI, M p ). 0
In the following, in 3.1.6 - 3.1.10, R denotes a Noetherian ring, I an ideal
of R, and M a finitely generated R-module satisfying the condition 1M #: M.
Theorem 3.1.3 and Lemma 3.1.5 directly imply
Theorem 3.1.6
If the sequence Xl' ..., x p is a maximal regular sequence on the module
M contained in I (i.e. if it cannot be extended to a longer regular sequence
on M contained in 1), then
Ext(RII, M) :F 0, Extk(R/I, M) = 0 for 0  i < p.
Corollary 3.1.7
Any two maximal regular sequences on the lnodule M contained in I have
the same length, equal to the least non-negative integer p for which
Ext(R/I, M) #= O.
The first part of Corollary 3.1.7, asserting that any two M-sequences
contained in I have the same length, can also be established without re-
sorting to the functors Ext. We suggest that the reader should try to
work out a direct proof. We hope that this experience will bring into relief
the power and efficiency of homological methods.
Definition 3.1.8
The common value of lengths of all the maximal regular sequences on the
module M contained in I is called the depth of the ideal I with respect to M,
and it is denoted by depth(I; M). If M = R then we write simply depth (I).
As a consequence of the discussion preceding Theorem 3.1.3 we have
Corollary 3.1.9
The number depth (I; M) is finite.
In order that certain formu]ae below should apply without restriction
we adopt the convention that depth(R; M) = 00 (see e.g. Lemma
3.1.1).
Let x E I and 1* = I/(x), Ai = MjxM.
Lemma 3.1.10
If x fft 3(M), then depth(I*; M) = depth(I; M)-l.
Proof
Let E 1, ..., En be any elements of the ideal 1* and. let Xl' . "' X n be the
representatives of the corresponding residue classes E 1 , ..., n modulo the

106
Cohen-Macaulay Rings
[Ch.
ideal (x). The isonl0rphisn1 MI(l"'" I;p) M  Mj(x, XJ , ..., x,,)M and
direct computation readily imply that E p + t  3 (M I(1' ..., p) M) <=> Xp+l
rp 3(M/(x, Xl' ..., xp)M). This shows that 1' ..., p is a maximal M-se-
quence if and only if x, Xl' ..., x p is a Inaximal M-sequence. The assertion
of the lenlma follows. 0
To conclude this section, we shall investigate the behaviour of depth
in passing to a ring of fractions (other properties of the depth, which are
not needed at present are given in the exercises).
Lemma 3.1.11
Let R be a Noetherian ritlg, I an ideal of R, and M a finitely generated
R-module. Then:
(i) for any multiplicative subset S of R, we have
depth(I;M)  depth(IRs;Af s );
(ii) depth(I; M) = infdepth(IR ttt ; M m ), where tn ranges over all maximal
111
ideals of R.
In particular, for M = R we have depth(I) = infdepth(IR m ). (If 1Rm
m
= Ran, then depth(IR m ) = 00 according to the remark made after Corollary
3.1.9.)
Proof
(i) The required inequality follows immediately from Corollary 3.1.7
and from the commutativity of functors Ext and localization (Theorem
A.3.6).
(ii) In view of (i), it is enough to show that depth(I; M)  depth(IRm;M m)
for SOlne maxin1a1 ideal m. Assume conversely that depth(I; M) = p but
depth(IR m ; M m ) > p for each m. The isolnorphism (see Theorem A.3.6)
(Ext(R//, M)ut  Extk (Ranl/R uo M m )
m
and Corollary 3.1.7 yield (Extk(RII, M»m = 0 for every maximal ideal m.
Applying Theorem 1.4.22 of [B], we conclude that Extk(RII, M) = 0,
which contradicts the assumption that depth(I; M) = p and the charac..
terization of depth in terms of functors Ext (Corollary 3.1. 7). 0
Exercises
1. (a) Let Xl, X2 be a regular sequence of a ring R. Prove that Xl rt 3(R/(x,,»).
(b) Give an example of a regular sequence Xl, X2 such that X2 is a zero-divisor in R.
2. Prove that if a sequence Xl, X2 of a local ring is regular, then the sequence Xl, Xl
is also regular.
3. Using the preceding exercise, prove that in a local ring any permutation of a regular
sequence is also a regular sequence.

III]
Regular Ideals
107
4. Prove that in a Noetherian ring any regular sequence contained in an ideal] can
be extended to a maximal regular sequence contained in 1.
5. Let R = K[X, Y, Z], where K is a field. Show that X, Y(l - X), Z(l - X) is a regular
sequence in R, while the sequence Y(l- X), Z(l- X), X is not.
6. Prove that 1 C J implies depth(I)  depth (J).
7. Show that depth(l) = depth(radl) for any ideal] of a Noetherian ring.
8. Establish the equalities depth(IJ) = depth(If"'\J) = min( depth(l), depth(J») for
any ideals 1, J of a Noetherian ring.
9. Let PI, ..., P" be all the prime ideals associated with an ideal I of a Noetherian
ring. Prove that depth(I) = min depth(P,). [Make use of preceding exercises and of
1 is
the existence of a primary decomposition.]
10. Let R be a ring and M an R-module. Show that jf Xl, ..., XII is an M..sequence,
then
Xi f/ a(kf/(Xl, ..., Xi-I, Xl+l, ..., XII) AI) for i = 1, ..., II.
11. Under the assumptions of the preceding exercise, let Xl = yz for a fixed i. Prove
the following statements:
(a) jf Xl, .oo, X'-I, y, X'+l, ..., X n and Xl, ..., X'-l, Z, X'+l, ."'X II are M-sequences,
then so is the sequence Xl, ..., Xn.
(b) if Xl, ..., x" is an M-sequence and (XI,..., Xl-I, y, x,+ 1, ..., XII) M -:1= M,
then Xl, ..., X'-I, y, X'+I, ..., X" is an M..sequence.
12. Let R be a ring. Show that jf Xl' ..., X n is a regular sequence of R, then xU, ...
. .. , X: n is also a regular sequence of R for any sequence of positive integers SI, ..., SII'
Prove furthermore that the regularity of a sequence xI 1 , ..., xn for son1e 81, ..., Sn implies
the regularity of the sequence Xl, ..., Xn.
13. We say that a sequence Xl, ..., X n of elements of a ring R is a n1aximal regular
sequence on an R-module M (without reference to a specific ideal) if for any ideal I
satisfying the conditions: 1 0 1M rI= M, 2 0 (Xl' ..., XII) C I, we have the equality
11 = depth(I; M). Let R = K [[t]] [X], where K is a field. Prove that tX-l and I, X are
maximal regular sequences of R. Thus there exist maxin1al regular sequences of R with
different lengths.
14. Give an example of an ideal I of a ring R and of a multiplicative subset S of
R such that depth(l) < depth (IRs). Find an example such that, the difference
depth(IR s ) - depth(l) is arbitrarily large.
15. Let R be a unique factorization domain. Show that every non-zero ideal of R
is isomorphic either to R or to an ideal of depth at least 2.
16. In the ring K[X 1 , ..., X,,] of polynomials over a field K let the ideal I be generated
by monomials in Xl' ..., Xn. Prove that depth(l) is equal to the minimal number k such
that I c (X, 1 ' ..., X,,) for some ;1, ..., i k .
17. Let In, II be the ideal of K[X 1 , ..., Xn] generated by an the square-free monomials
of degree p in Xl, ..., X,., i.e.,
I", II = (X, 1 ' X ,2 , oo., X,,,), 1  i 1 < ... < ; p  /I.
Show that depth(I n ,,,) = n-p+ 1.
3.2 REGULAR IDEALS
We already know three numerical invariants of any ideall of a Noetherian
rJng:
(1) height-ht(I),

108
Cohen-Macaulay Rings
[Ch.
(2) the minimal number of generators-v(1),
(3) depth-depth(l).
We also know the relation between the first two: Krull's theorem (Theorem
1.2.10) implies that ht(I)  v(1). Now we shall relate to them the third
invariant-the depth of an ideal, introduced in the preceding section.
Theorem 3.,2.1
Any ideal i of a Noetherian ring satisfies the inequality depth(I)  ht(I).
Pl'oof
The proof is by induction on depth(I) which is a non-negative integer in
view of Corollary 3.1.9. The case depth(I) = 0 is trivial. If depth(l) > 0,
choose an element x in I which is not a zero-divisor, and denote 1* = I/(x).
By Lemma 3.1.10, we have depth(/*) = depth(I)-l < depth(1). Thus
it follows from the inductive hypothesis that depth(I*) < ht (1*). Since
x  3(R), Corollary 1.2.17 implies ht (/*) = ht (1) -1, and finally depth(I)
ro. 0
Corollary 3.2.2
Any ideal I of a Noetherian ring satisfies the inequalites depth(I)  ht(l)
 p(1).
The purpose of the present section is to investigate the structure of
those ideals for which we have equalities in the above corollary.
Theorem 3.2.3
Let / be an ideal of a Noetherian ring R. Then depth(/) = v(I) if and
only if I is generated by a regular sequence.
If R is a local ring, depth(l) = n and I = (Xl' ..., x,,), then Xl' ..., x"
is a regular sequence.
In proving the theorem we shall make use of the following lemma.
Lemma 3.2.4
Let PI' ..., P" be prime ideals of a ring R, let I be an ideal of R, and let
x be an element of R such that (I, x) q: PI U ... uP". Then there exists
an element y e / such that x+y  Ptu ... uP".
Proof
We may of course assume that none of the prime ideals P, is contained
in any of the ideals PJ, i #: j. Suppose that the ideals Pi are numbered so
that xeP 1 n... nP"xP'+lU... uP k , 0  r  k. If r = 0 we can
take y=O. We may thus assume r 1. Clearly, Iq: P 1 u... uP" for
otherwise we would have (I, x) c PIU ... uP" contrary to the assumption.
Therefore there exists an element Yo eI,yo P1U ... uP,. Choose an

III]
Regular Ideals
109
element A such that A  P 1 u ... uP" A E P,+ln ... nP k . This is possible,
for otherwise we would have P r + 1 n ... nPk C PIU ... uP" and conse..
quently (by Corollary 1.1.8 in [B]) Pi c: PJ for some distinct i,j, which
contradicts the initial assumption.
The element y = AYo has the required property. Indeed, ;'Yo  PI U ...
. .. uP" which, since x e P 1 r. ... nP" shows that x + y  P 1 U ... uP r.
On the other hand, x  P r + 1 U ... uP", and thus also X+ y  P r + 1 U ... UPk
since AYo E P,+ In ... nP". 0
Proof of Theorem 3.2.3
If 1 is generated by a regular sequence of length n, then depth(I) = nand
'P(I)  n. By Corollary 3.2.2, we have depth(I) = 'P(I) = n.
Assume now that depth(/) = 'P(I) = n; we shall prove by induction
on n that I is generated by a regular sequence of length n.
If n = 1, the theorem is obvious since if an element x E R is a zero-div-
isor then so are also all its non-zero multiples.
Let n > 1 and let I be generated by elements Xl' ..., x,.. Since 1 q: 3(R)
and 3(R) is a union of a finite number of prime ideals (Theorem 2.3.22
of [B]), by applying Lemma 3.2.4 to the element Xl and to the ideal
(x 2 , ..., XII)' we deduce that there exists an element y E (X2, ..., XII) such
that U I = Xl + y  3(R); obviously, 1 = (Ul, X2, X3, ..., XII)' Denote R*
= R/(Ul),I* = I/(Ul)' By Lemma 3.1.10' we have depth(/*) = n-l
= 'P(l*), and therefore we can apply the inductive hypothesis to the ideal 1*.
If u, ..., u: is a regular sequence generating the ideal 1* , where uf = u, +
+ (Ul), then, by Lemma 3.1.4, the sequence "1' "2, ..., Un is a regular se-
quence generating I.
To prove the second part of the theorem, we assume R to be a local ring.
By the inductive hypothesis, the sequence x, ..., x: is an R*-sequence,
that is to say, Ul, X2' ..., XII is an R-sequence. R being a local ring, any
permutation of a regular sequence is also a regular sequence (Exercise
3, Section 3.1), and hence X2' ..., XII' "1 is regular. It follows that "1
 3(R/(X2' ..., X,,)), which, in view of "1 = Xl +Y, Y E (X2, ..., X,.), yields
Xl  3(R/(X2, ..., X,.»). Finally, the sequence X2' ..., X,., Xl and hence the
sequence Xl' ..., X,. are regular. 0
Corollary 3.2.5
If Xl' ..., X,. is a regular sequence of elements of a Noetherian ring, then
ht(x 1 , ..., XII) = n.
Definition 3.2.6
An ideal generated by a regular sequence is called a regular ideal.
In order to make later results (see Section 3.3) applicable to the trivial
case, we adopt the convention that 'P(O) = 0, and we shall regard the zero
ideal as a regular ideal.

110
Cohen-Macaulay Rings
[Ch,
To state a useful property of regular ideaJs, we shall need one more
definition.
DefiDitiOD 3.2.7
An ideall is said to be depth-unmixed (height-unmixed) if all the associated
prime ideals of I have the same depth (height).
Lemma 3.2.8
A regular ideal of a Noetherian ring is depth-unmixed.
Proof
Let Xl' ..., X n be a regular sequence generating the ideal I and let P be
any prime ideal associated with I. By Theorem 2.3.22 of [B], each element
in P is a zero-divisor on the module R/ I, and therefore Xl' ..., XII is a maxi-
mal regular sequence contained in P, i.e. depth(P) = n. D
A regular ideal of a Noetherial1 ring may not be height-unmixed (see
Exercise 3). In Section 3.3 we shall deal in more detail with rings whose
regular ideals are all height-unmixed (see Theorem 3.3.5).
To conclude this section, we shall establish a theorem which character-
izes a regular ideal in a local ring in terms of the graded ring associated
with it (see Chapter I). This characterization will be a basis for the descrip-
tion of Cohen-Macaulay local rings by means of multiplicities (Theorem
3.3.6).
Let us recall that by the graded ring GrI(R) associated with an ideal I
of a ring R we mean the ring
RllEaIII 2 Ea ... IIIIIn+lffJ ...
with multiplication defined by
(r+IP+I)(s+If+ 1 ) = rs+IP+Q+l for ,. E IP, S E [f.
If I is generated by elements Xl' ..., X n , then GrI(R) is the algebra over
R/ I generated by the elements X, + 1 2 , i = 1, ..., n.
The theorem belo\v is a generalization of Theorem 1.4.11.
Theorem 3.2.9
Let R be a local ring and I an ideal of R. The following properties are
equivalent:
(i) the ideal I is generated by a regular sequence Xl' ..., X,. of R,
(ii) the ring GrI(R) is isomorphic to the polynomial ring
(RII) [Xl , ...,X,,] via the isomorphism mapping X, to xi+I2.
Proof
(ii) => (i). Let X l = x,+/2, and GrI(R) be the ring of polynomials
in X l' ..., X ,. over RII. First we shall prove the follwing fact: if
rXI +CP(X2' ..., x n ) = 0, (2)

III]
ReguJar Ideals
111
where rp is a form of degree s with coefficients in R, then r e 1'-1 and the
coefficients of rp belong to I.
Suppose that r  1&-1. Then there exists t < s-1 such that r e It but
r, p+l. By considering equality (2) in GrI(R), we deduce that (r+1 t + 1 ) X l
= O. Since X l is not a zero-divisor in GrI(R), we have r+ [t+ 1 = 0, which
contradicts the choice of t; thus r e Ill-I. Consider now the equality (2)
in [lIlls+l. As X l' ..., X II are algebraically independent over RjI, all the
coefficients of rp are in I.
We now turn to the proof of the implication (ii) => (i). First, we shall
show that Xl is not a zero-divisor of R. If rXI = 0, then using (2), we
conclude that r belongs to every power of I. R being a local ring, by virtue
of Krull's theorem (Theorem 2.5.5 in [B]) we get r = O.
Let R' = RI(Xl)' I' = II(xl)' x = Xi + (Xl), x ; = x + (1')2, i = 2, ...
. . . , n. In what follows, we shall employ induction on n, proving first that
Grl(R')  Grl(R)/(x l ).
We have the homomorphism 0': Grl(R) -+ Grl,(R') defined by the
formula
a(y 1- IS+ 1) = y' + (I')S+ 1 for y e IS,
where we write y' for the residue class of y in (I')' = (l"'+(xl»)/(x 1 ). Ob-
serve that O'(x,) = x  for i = 2, .0.' n and that 0' is an epimorphism. We
claim that Ker( 0') is generated by X l.
Let! be a form of degree k in n -1 indeterminates with coefficients in
R and let f denote the form over RII obtained from f by replacing the
coefficients of f by their residue classes modulo I. If
I" ) K ( ) tl 1"( -' -' ) (/ ' ) "+1
X2' ..., X n E er 0' , len J \X2, ..0' x" E .
Since ([')"+1 = (/"+1 + (Xl) )/(X l ), this means that there exists a form
g of degree k + 1 in n - 1 indeterminates with coefficients in I such that
f(X2, ..., x,,) = g(X2, ..., XII) + UX 1 , U e R.
Using the implication (2) applied to the form g- f, we see that all the
coefficients of the form f are in I, i.e. !(X 2, 00.' x lt ) = 0 in GrI(R). This
ends the proof of the fact that GrI,(R')  Grl(R)I(Xt) is the ring of poly-
nomials in n - 1 indeterminates x , .00' x . By the inductive hypothesis,
we infer that x , ..0' x  is a regular sequence of R', i.e. finally, the
sequence Xl' ..., x" is regular in R.
(i) => (ii). Let Xl'...' x" be a regular sequence of R and let
f = l:" «,(Xl' ..., X,,) e R[X 1 , ..., XII] be a form of degree s in Xl' ..., XII'
where <Xi(X 1 ,. .. ,XII) are distinct monomials in Xl' ..., XII of degree s.
In order to show that the homomorphismqJ: (RII) [Xl' ...,X n ] -+ Grl(R).
rp(X,) = x, + 1 2 is an isomorphism it is sufficient to prove that

112
Cohen-Macaulay Rings
[Ch.
f(x 1 , ..., X n ) = 0 implies r, E I for any form f and any s  1.
(3)
Indeed, if g = L (t, + 1) P,(X 1 , ..., X n ) E (Rf 1) [Xl' ..., X n ], where t i E R,
and Pi are all the monomials in Xl' ..., XII of degree s, and q;(g) = 0,
then L t,{Ji(Xl, ..., XII) E /S+l, i.e. L (I, + Ui) (J,(Xl, ..., x n ) = 0 for some
u, e I. The validity of the implication (3) yields t, + u, E I, that is, Ij + I = 0,
and finally g = O.
The method of the proof of (3) requires introducing the following
definition. Given two monomials in Xl' ... , XII:  = xr 1 ... XCn, {J = xt l ... xn,
we say that (X is a formal multiple of fJ if p,  q, for all i. Property (3)
follows directly from a slightly more general fact. 0
Lemma 3.2.10
Let Xl' ..., XII be a regular sequence of a ring R, I = (Xl' ..., XII)' Further,
let (Xl' . "' «I;, P be monomials in Xl' ..., X n such that {J is not a formal
multiple of any of the monomials «1' ..., «t. Then
rfJ = rl C'l1 + . II + r"!Xi for r, rj E R,
implies rEI.
Proof
We proceed by induction on the degree of the monomial fJ. If {J is of degree
0, i.e. {J = 1, then (Xl' ..., (Xt are of positive degree in view of the assumption
that (J is not a formal multiple of any of the monomials «,. Thus relation
(4) shows that reI.
Assume that fJ is of positive degree, fJ = X11 ... xi", L q, > O. Since
fJ is not a formal multiple of any of the monomials (X1t for every i, 1  i  k,
there exists s, such that qs, is smaller then the power in which x s , occurs in
(x,. Consider all the monomials (x, for which s = SI = SIt and rewrite
equality (4) by transposing these monomials to the left-hand side. Setting
fJ = x:i' we have the equality
(4)
(ry-txJx:. = 2:>,x::.
where the sum is over all i such that Si #: sand ai > qSJ' By Exercise 12,
Section 3.1, a sequence consisting of powers of the elements XI' ..., x,.
is also regular. Applying to it the result of Exercise 10, Section 3.1, we
deduce that rr - tx s can be expressed as a linear combination of the powers
x::' S, :f:: s. Furthermore, " is not a formal multiple of either X. or any of
the powers x::. The degree of " being less than that of fJ, we finally
conclude from the inductive hypothesis that reI. 0
Let us observe that in the second part of the proof of Theorem 3.2.9 no
use is made of the fact that the ring R is local.

III]
Characterizations of Cohen-Macaulay Rings
113
Exercises
1. Let K be a field, R = K(X, YJ/(X", XY), and let I be the image of the ideal (X, Y)
in R. Prove that depth(I) = 0, ht (I) = 1.
2. Prove that every minimal ideal associated with a regular ideal, generated by a regu-
lar sequence of length n, has height n.
3. Let R = K[X, Y, Zj/(XY, XZ), where K is a field. Show that the element x+ y
is not a zero-divisor of R. Find the prime ideals associated with the ideal I = (x + y),
and prove that I is height-mixed (in spite of the fact that, by Lemma 3.2.8, I is depth...
unmixed).
4. Prove that if (l and p are monomials with respect to elements of a regular sequence
of a ring R, then (l is a multiple of P (i.e., « = rp for some r E R) if and only if « is a for.
mal multiple of p.
5. Let R = K[X, Y, ZlI(X1.Y-Zy2), where K is a field. Consider the monomials
« = x 2 y, p = y2 in x, y (the residue classes of X, Y respectively). Show that « is a multiple
of p in R, but it is not a formal multiple of p.
3.3 CHARACTERIZATIONS OF COHEN-MACAULAY RINGS
Referring to the introduction to this chapter we define the following notion.
Definition 3.3.1
A Noetherian ring R is called a Cohen-Macaulay ring if any ideal! of R
generated by ht(1) elements (i.e., ht(I) = ",(1) is height-unmixed.
We note that in accordance with the convention adopted in Section 3.2
(,,(0) = 0) the zero ideal of a Cohen-Macaulay ring is height-unmixed,
that is, the only prime ideals associated with the zero ideal are the minimal
prime ideals of the ring.
By virtue of the properties of regular sequence stated ill Sections 3.1
and 3.2, we can give the first characterization of Cohen-Macaulay rings.
Theorem 3.3.2
Let R be a Noetherian ring. The following properties are equivalent:
(i) R is a Cohen-Macaulay ring,
(ii) for every maximal ideal tn of R, we have depth(m) = ht(m),
(Hi) for every prime ideal P of R, we have depth(P) = ht{P),
(iv) for every ideall of R, we have depth(I) = h(I).
In proving this theorem we shall need the following lemma.
Lemma 3.3.3
Let (R, m) be a local ring. Then, for every prime ideal P #: m, there exists
a prime ideal Q * P such that ht(Q)-depth(Q) ;;:: ht(P)-depth(P).
Proof
Let depth(P) = p and let Xl' ..., x p be a maximal regular sequence con-
tained in P. By Theorem 2.3.22, Corollary 1.1.8, and Theorem 2.3.15 in [B],

]14
Cohen-Macaulay Rings
[Chi
there exists y ri (XI' ..., x p ), such that yP C (Xl' ..., x,,). Let R = R/
j(x l' ..., x p ), and consider two cases.
1 0 tn C '3( R) ; then depth(m) = p, 111  P, and so we can take Q to
be nt.
2 0 m q: 3( R) , that is to say, in m there exist elements which are not
zero-divisors on the module R. We claim that there exists an element
Z Em such that z  3( R) and y  (Xl' ..., Xl" z). Indeed, if we had, y
E (Xl' ..., xP' z) for every z f/ 3(.R), then, since Zk  3 (R) for each k, we
co
would have Y En Z" in R. Hence, by Krull's intersection theorem, y
k=1
E (Xl' ..., x p ), contrary to the assumption.
Thus depth (P, z) = p + 1 because Xl' ..., xl" Z is a regular sequence
of lengthp + 1, yep, z) c: (Xt, ... xl" z), andy, (Xl' ..., Xl" z). By CO'rollary
1.1.8 of [B], it follows from the inclusion (P, z) c: 3(R/(Xl, ..., xp, z),
that there exists a prime ideal Q associated with (Xl"'.' xp, z), which
contains (P, z). Of course, depth(Q) = p+ 1, ht(Q)  ht(P) + 1 and Q * P,
\vhence eventually ht(Q)-depth(Q)  ht(P)-depth(P). 0
The proof of the above lemma immediately yields
Corollary 3.3.4
If (R, m) is a local ring and if an ideal I of R fulfils the condition depth(I)
< depth (m), then there exists a prime ideal Q, Q * 1, such that depth(Q)
= depth(I) + 1.
Proof of Theorem 3.3.2
(i) => (ii). Let m be a maximal ideal of height d. If d = 0, then depth(m)
= 0 by Theorem 3.2.1. So assume that d = ht(m) > O. By the proof of
Theorem 1.2.16 it follows, that there exists a sequence Xl' ..., Xd of el-
ements of m such that ht (Xl' ..., Xi) = i, i = 1, ..., d. We shall show that
Xl' ..., Xd is a regular sequence of R. Note that, by the assumption, aU
the ideals (Xl"'" x,) are height-unmixed. Therefore, if X,+ I were a
zero-divisor modulo (Xl' ..., X,), then the whole ideal (Xl' ..., X,+ 1) would
be contained in one of the prime ideals associated with (x I' ..., x,), and
hence in a prime ideal of height i. We know, however, that ht(xl' ..., X'+l)
= i + 1. Accordingly XI' ..., Xli is a regular sequence of R contained in
m, i.e. depth(m)  d = ht(m). From Theorem 3.2.1 we obtain the equality
depth(m) = ht(m). (Observe that the same argument establishes the
implication (i) => (Hi).)
(ii) => (iii). First let us assume R to be a local ring, and let P be an arbitrary
prime ideal. Applying Lemma 3.3.3 repeatedly, we get an increasing chain
of prime ideals P = Po c P 1 C ... which must reach the maximal ideal
m after a finite number of steps since the ring R is Noetherian. Conse-

III]
Characterizations of Cohen-lVIacaulay Rings
115
quently, 0  ht{P)-depth(P)  ht{m)-depth(m) = 0, and finally ht{P)
= depth(P).
Now we pass to an arbitrary Noetherian ring R. If P is any prime
ideal, then, by Lemma 3.1.11, there exists a maximal ideal nt :::> P such
that depth(P) = depth (PRm), Since ht(m) = ht(mR m ) and depth (tn)
= depth(mR m ) (also by LeJnma 3.1.11), Rut has property (ii). Frolu the
implication (ii) => (iii), already proved for local rings, we infer that ht (P Rm)
= depth(PR m ). Thus ht{P) = depth(P), as height remains ullchanged
under localization.
(Hi) => (iv). Observe first that for any ideal I, there exists a prime ideal
P :::> I which has the same depth as I. Indeed, let depth(J) = p and let
Xl' ..., x p be a maximal regular sequence in I. Accordingly I c: 3{R/
I(Xt, ...,x p » and, by Corollary 1.1.8 in [B], the ideal I is cOl1tained ill
a certain pri.tne ideal P associated \vith (Xl' ..., X,), As P consists only
of zero-divisors on the module R/(Xl,"" x p ), we have depth(P) = p
= depth(l).
III view of this, depth(P) = depth(I)  ht{l)  ht(P). Therefore the
assumption depth(P) = ht(P) implies depth(l) = ht{I).
(iv) => (i). Let I be an ideal of height p and let 'V(/) = p. Since, by the
assumption, depth(I) = p, 1 is a regular ideal by Theorem 3.2.3. Applying
Lemma 3.2.8, we deduce that the ideal I is height-unmixed, once more
by (iv). 0
Notice that from Theorem 3.3.2 \ve infer in particular that a local ring
is Cohen-Macaulay if and only if the equality ht(rn) = depth(ln) holds
for the unique maximal ideal m.
As an application of the theorem proved above, we give the first
examples (except for polynomial rings over a field mentioned above) of
Cohen-Macaulay rings:
(1) Artin rings (see Section 2.8 of [BD; in this case depth(m) = ht(m) = 0
for every maximal ideal.
(2) Noetherian domains of dimension 1, e.g. Dedekind domains (see
Chapter III of [B]); then, for every maximal ideal, depth(ttt) = ht{m) = 1.
(3) Normal Noetherian domains of dimension 2. Let ttt be a maximal
ideal. Ifht(m) = 1, then clearJy depth(m) = ht(m). Suppose that ht(n1) = 2
and let x Em, x i= O. By Theorem 1.2.19, aU the prime ideals associated
with (x) have height 1, and so
m  U Pi ·
P ,eABs(R/(x»
Thus there exists Y Em, y  U Pi, and x, y is a regular sequence contained
in tn.
(4) Regular local rings since, by Theorem 1.4.19, the maxitnal ideal is
regular.

116
Cohen-Macaulay Rings
[Ch.
Further examples will be provided in subsequent sections. Here we
shan present another characterization of Cohen-Macaulay rings.
Theorem 3.3.5
Let R be a Noetherian ring. The foJIo\ving properties are equivalent:
(i) R is a Cohell-Macaulay ring,
(ii) every regular ideal of R is height-unmixed,
(iii) all the ideals I of .R which satisfy the condition ht (1) = p(1) are
a regular ,
(iv) for any prime ideal P of R of height  1 there exists a set of par-
ameters of the ring R p which is a regular sequence.
Further if R is a local ring, then the properties given above are equiv-
alent to each of the following:
(v) there exists a set of parameters of R, ,,,hich is a regular sequence,
(vi) every set of parameters of R is a regular sequence.
Proof
(i) => (ii). Let I be an arbitrary regular ideal of R. Then depth(l) = ht (I)
= ')1(1) by Theorem 3.2.3. From (i) we therefore conclude that I is height-
unmixed.
(ii) => (Hi). Let I be an ideal which satisfies the condition ht (I) = 'IJ(I).
Suppose that depth(I) = k, and let XI' ..., x" be a maximal regular se-
quence contained in I. Accordingly I is contained in some prime ideal P
associated with the regular ideal (XI' ..., x,,). By virtue of (ii), every regular
ideal is height-unmixed, i.e., ht (P) = ht (x l' ..., Xk) = k. Hence ht (I)
 ht(P) = k = depth(I), and finally depth(I) = ht(I) = v(1). In view
of Theorem 3.2.3, the ideal I is regular.
(iii) => (iv). Let P be a prime ideal of height n  1. Theorem 1.2.16
shows that there exists a sequence of elements ai' ..., all e P such that
ht (at, ..., a,.) = n. From (iii) we deduce that (al"'" a,.) is a regular
ideal. This means that there is a regular sequence b 1 , ..., b,. which generates
the ideal (at, ..., an)' The images of the elements of the sequence b 1 , ..., b n
form of course a set of parameters of the ring R p .
(iv) => (i). Let P be a prime ideal of R of height n ;;:: 1. By (iv), there
exists a set of parameters of the ring R p forming a regular sequence. There-
fore n  depth(PR p )  ht(PR p ) = n, i.e. depth(PR p ) = ht(PR p ). Obvi-
ously, this equality is also valid for prime ideals of height zero. Consequently
R p is a Cohen-Macaulay ring for any prime ideal P of R by Theorem 3.3.2.
This proves that R is a Cohen-Macaulay ring in view of Property 3, Section
3.4, whose proof is independent of the theorem we are proving.
Assume now that R is a local ring and m is its only maximal ideal.
(iii) => (vi). If d = dimR = ht(m) and Xl' ..., XII is an arbitrary set
of parameters of R, then ht(Xt, ..., x,,) = d. It follows from (iii) that the

III]
Characterizations of Cohen-Macaulay Rings
117
ideal (Xt, ..., XII) is regular. Using Theorem 3.2.3, we conclude that Xl' ...
. . . , XII is a regular sequence.
(vi) => (v). Obvious.
(v) => (i). It follows from the assumption that depth(m) = I1t (m),
whence R is a Cohen-Macaulay ring by Theorem 3.3.2. 0
The last theorem of this section will be a characterization of local
Cohen-Macaulay rings in terms of multiplicities. We recall that if d = dhnR
and Q is an ideal generated by a set of parameters of R, then the function
n.H I(RfQ") is a polynomial function, for large n, of degree d (see Theorem
1.5.5). The coefficient of the highest power in the polynomial has the form
eQ(R)fd!, where eQ(R) is a non-negative integer called the multiplicity of
the ideal Q. We always have (see Example 1.5.17) the inequality eQ(R)
 l(RIQ). The equality characterizes Cohen-Macaulay rings; more pre-
cisely, the following theorem is valid:
Theorem 3.3.6
Let R be a local ring. The following properties. are equivalent:
(i) R is a Cohen-Macaulay ring,
(ii) there exists an ideal Q generated by a set of parameters of R such
that eQ(R) = I(RfQ),
(iii) for any ideal Q generated by a set of parameters of R, \ve have
eQ(R) = I(RfQ).
Pl'oof
(i) => (iii). The multiplicity of the ideal Q can also be expressed in
terms of the graded ring Gro(R) = Qn/Qn+l associated \vith Q. For each
11, we have an exact sequence
o -+ Qn/QIt+l  R/Q"+l  R/Qn -+ 0,
and consequently I(Q"/Qn+l) = /(R/Qn+)-I(R/Qn). Thus it follows from
Section 1.5 that the function n f-+ I(Qn /Q" + 1) is also a polynoDlial function,
for large n, of degree d- ]. Moreover, the coefficient of the highest power
If-I is the number eQ(R)/(d-l) !.
In order to establish the implication (i) => (Hi), consider an arbitrary
ideal Q generated by a set of parameters Xl' ..., Xci. Theorem 3.3.5 shows
that Xl' ..., Xci is a regular sequence. Hence by Theorem 3.2.9 we get an
isomorphism GrQ(R)  (R/Q) [Xl , ..., Xci]. This implies directly I(Qn /Qn+l)
= I(R/Q) (n d l1) = [/(R/Q)/(d-l)!] n d - 1 + ..., i.e. eo(R) = /(R/Q).
(iii) => (ii). Th is is obvious.
(ii) => (i). Let Xl' ..., XII be a set of parameters' of the ring R and let
Q = (Xl' ..., Xd)' We want to show that Xl' ..., Xci is a regular sequence.
To this end, just as in Theorem 3.2.9, we define a homomorphism
q;: (R/Q) [Xl' ..., X,,] -+ Gr Q(R), cp(X i ) = Xi + Q2 .

118
Cohen-Macaulay Rings
[ChI
Further we put A = (R/Q) [Xl' ..., X 4 ], I = A(X 1 , ..., "Y d ), J = Ker(tp).
We shall prove that if I #= 0, then ea(R) < I(R/Q). Froln the renlarks
made in the course of proving the impJication (i) => (iii) it follows that
n
I(R/Q") = L I(Qi-t /Qi). Moreover, Gra(R)  All. Thus I(RfQ")
1=1
= I(A/(]n+J») for each n.
If J =1= 0, then I contains a form f =1= 0 which is, say, of degree p, and
hence, all the products off with luonomials of degree less than n - p. These
products generate an R/Q-module, whose length is equl at least to the nwn-
ber of all such different products, i.e. at least to (n-pd-l). Consequently
I(RIQn) = I(AIW+J)):::;; I(Aw)-(n-p"dd-l)
= I(RIQ)(n+-l) _ (n-Pdd-l)
= (I(R/Q)/d!)n d + ... - (n fd! + ...)
= [(l(RfQ) -1 )/d!]n d + ...
for all n > p. From the definition of multiplicity we get eQ(R) < I(R/Q).
The above reasoning also shows that if e{J(R) = I(RfQ), then rp lllust
be an isomorphism. By Theorem 3.2.9, the sequence Xt, ..., Xd is regular
and R is a Cohen-Macaulay ring in vie\v of Theorem 3.3.5. 0
Exercises
1. Prove that any regular sequence of a local ring may be extended to a set of par-
ameters of that ring.
2. Prove that R is a Cohen-Macaulay ring if and only if no regular ideal of R has
embedded colnponents.
3. Let R = K[X, Y, ZJ/(XY, XZ), where K is a field, and let I = (z, x+Y), where
x, y, z denote the residue classes of the indeterminates X, Y, Z in R. Show that ht(l) = 2,
depth(I) = 1. Deduce frotn this that R is not a Cohen-Macaulay ring.
4. Prove that sup(ht(P).-depth(P») = sup(ht(m)-depth(m») for any Noetherian
ring R, where the supremum on the left relates to all prime ideals, and that on the right
to an maximal ideals.
5. Show that the ring K[X, YJ/(X 2 , XY), where K denotes a field, is not a Cohen-
Macaulay ring.
6. Prove that a local ring (R, nt) is Cohen-Macaulay if and only if for every set of
parameters Xl, .oo, Xd of R the ideal m is not associated with the ideal (Xl, ..., X,-I)'
3.4 BASIC PROPERTIES OF COHEN-MACAULAY RINGS
Property 1
If R is a Cohen-Macaulay ring, then the ring of fractions Rs with respect
to any multiplicative subset S of R is also a Cohen-Macaulay ring.
Proof
Every prime ideal of .Rs has the forl11 PRs, where P is a prime ideal of R,
and Pf\S ¥: 0 (see Corollary 1.4.8 in [8]). By Lemma 3.1.11 we have

III]
Basic Properties of Cohen-Macaulay Rings
119
depth(P)  depth (PRs), and hence depth(P)  depth(PR s )  ht(PRs}
= ht(P). Therefore depth(PR s ) = ht(PRs) and Rs is a Cohen-Macaulay
ri. 0
Property 2
A ring R is Cohen-Macaulay if and only if, for any maximal ideal m of R
the ring Rm is Cohen-Macaulay.
Proof
The necessity of the condition foIlows from Property 1; we shall prove
the sufficiency. Let m be a maximal ideal of R. Observe that depth(m)
= depth(mR m ) by Lemma 3.1.11. Furthermore, obviously ht(m) = ht(mR m ).
Thus if Rm is a Cohen-Macaulay ring, then depth(m) = ht(m), and R
is a Cohen-Macaulay ring. 0
As an immediate consequence of Properties I and 2 we obtain
Property 3
A ring R is Cohen-Macaulay if and only if, for every prime ideal P of R,
the ring R p is Cohen-Macaulay.
Property 4
If R is a Cohen-Macaulay ring and x  3(R), then R/(x) is also a Cohen-
Macaulay ring.
Proof
Any maximal ideal of R/(x) is of the form m/(x) for some maximal ideal
m of R. Since x 3(R),i t follows from Corollary 1.2.17 that ht(m/(x»
= ht(m)-l. On the other hand, depth(m/(x» = depth(m)-1 by Lemma
3.1.10. Hence the equality of the height and depth ofm implies their equality
for m/(x), and so Rf(x) is Cohen-Macaulay. 0
Property 5
Let x be an element of the Jacobson radical of a ring R and let x  3(R).
If R/(x) is a Cohen-Macaulay ring, then R is also a Cohen-Macaulay
rIng.
Proof
Let m be a Dlaximal ideal of R. As x is in the Jacobson radical, x belongs
to each maximal ideal of R; in particular, x E m. Just as in the proof of
Property 4 one has ht(m) = ht{mf(x»)+I, depth(m) = depth (m/(x»+ 1,
and consequently R is a Cohen-Macaulay ring if R/(x) has this property. 0
Properties 4 and 5 immediately yield

120
Cohen-Macaulay Rings
[ChI
Property 6
The ring R [[X]] of formal power series with coefficients in R is a Cohen-
Macaulay ring if and only if R is a Cohen-Macaulay ring.
The next property concerns the completion of a local ring in the m-adic
topology. o prove it we shall need the following
Lemma 3.4.1
Let R be a Noetherian ring, I an ideal of R such that the /-adic topology
on R is Hausdorff, and M a finitely generated R-module. Then depth(/; M)
. A 1\ A "
= depth(/R; M), where Rand M denote the completiotlS of Rand M,
respectively, in the I-adic topology.
Proof
A
To begin with we observe that 3(M) = 3(M)nR. Indeed, it is enough to
"
show that if x  3(M), then x '3(M), but this follows from the fact that
tensor multiplication by R, preserves the n10nomorphism 0 --. M  M,
1\
since R is a flat R-algebra (see Corollary 2.6.20 in [B]).
Let depth(/; M) = p and let Xl' ..., X" be a maximal regular sequence
on M contained in I. It follows that X'+l  3(M/(Xl, ..., x,)M) implies
A A ______ A A
Xi+ 1 e 3(M /(X, ..., X,)M) because M /(Xl, ..., x,)M  M/(Xl, ..., xj)M by
A
Corollary 2.6.21 in [B]. Thus Xl' ..., x" is a regular sequence on M con-
 1\ A
tained in IR, i.e. depth(IR; M)  p.
A A
On the other hand, if I c 3(N), then IR c 3(N) for every finitely gener-
ated R-module N. Indeed, 3(N) = P l u ... uP., where Pi = Ann(Yi),
y, eN, y, :p 0, are prime ideals (Corollary 2.4.4 in [B]). Consequently
1\ A 1\
Iy, = 0 for some ;, and hence (IR) Yi = 0, i.e. IR c 3 (N).
Applying the above implication to N = M / (x 1, ..., X,,) M, we obtain
1\ 1\
the required equality p = depth(IR; M). 0
Corollary 3.4.2
1\ A
If (R, m) is a local ring, then depth(m) = depth(mR), where R is the com-
pletion of R in the m-adic topology.
Property 7
A local ring (R, m) is a Cohen-Macaulay ring if and only if its completion
1\ A
(R, mR) in the m-adic topology is a Cohen-Macaulay ring.
Proof
A
According to Corollary 3.4.2, we have depth(m) = depth (mR). On the
A
other hand, dimR = dimR by Corollary 1.5.10. This shows that ht(m)
A 1\
= depth(m) if and only if ht(mR) = depth(mR), and therefore, by The-
orem 3.3.2, we obtain the required equivalence. 0

III]
Basic Properties of Cohen-Macaulay Rings
121
In order to formulate the next property, let us recall that a chain of
prime ideals Po c PIC ... c: P 11 is said to be saturated if there is no
prime ideal between Pi and P i + 1 for i = 0, I, ..., n -I. Using this notion,
we can say that the height ht (P) of a prime ideal P is equal to the least
upper bound of the lengths of saturated chains of prime ideals en.ding
with P. In an analogous way we define the number ht' (P) as the greatest
lower bound of the lengths of saturated chains of prime ideals connecting
P with minimal prime ideals. Clearly, ht'(P)  ht(P), whence for the
ideals of a Noetherian ring both these numbers are finite (see Corollary
1.2.11). However, they need not be equal as can be seen from a simple
example.
Example 3.4.3
Let K be a field, let R = K[X, Y, Z]f(XY, XZ) = K[x, y, z], where x, y, z
denote the residue classes of X, Y, Z, respectively, and let P = (x, y, z).
One can readily prove that ht(P) = 2. Indeed, the chain (x) c (x, y) c P
is a chain of prime ideals of R, of length 2; on the other hand, dimR
< dimK[X, Y,Z] = 3. It can easily be checked that the chain (y, z) c P
of length 1 is also a saturated chain of prime ideals of R, and hence ht' (P)
= 1. Indeed, it is evident that there are no proper prime ideals between
P and (y, z) because R/(y, z)  K[X] and dimK[X]:= 1. Suppose that
there exists a prime ideal Q of K[X, Y, Z] between (Y, Z) and (XY, XZ).
Since (XY, XZ) = (Y, Z)n(X) and (Y, Z), (X) are both prime ideals, two
cases are possible:
1 0 either (X) c Q, and we arrive at the contradiction (X) c Q c (Y, Z);
2 0 or (Y, Z) c Q, which yields the equality Q = (Y, Z).
The situation above cannot arise in a Cohen-Macaulay ring.
Property 8
For any prime ideal P of a Cohen-Macaulay ring, ht'(P) = ht(P), that
is to say, all the saturated chains of prime ideals ending with a fixed prime
ideal are of equal length.
This property follows directly from Theorem 3.3.2 and from
Lemma 3.4.4
Any prime .ideal P of a Noetherian ring, satisfies the inequality depth(P)
 ht' (P).
Proof
Suppose that depth(P) > ht'(P). We may assume R to be a local ring
and P to be its maximal ideal, since in passing to the localization (i.e. to
the ring R p ) ht'(P) remains unchanged while the depth may only increase
(IJemnaa 3.].11).

122
Cohen-Macaulay Rings
[ChI
Set ht'(P) = n; we carry out the proof by induction on n. For n = 0
the assertion is trivial. Let n > 0 and let Po c ... C Pn-1 c Pn = P
be a saturated chain of prime ideals ending with P. Thus ht' (P It- 1) = n-l
and there are no pritne ideals between P and P,. _ l' According to the indue...
tive hypothesis, depth(P"_I)  n-l. Since depth(P) > ht'(P) = n
> depth(P n _ 1), we conclude applyin:g Corollary 3.3.4 to the ideal Pn- h
that there exists a prime ideal Q, Q ::> Pit-I' satisfying the equality depth(Q)
= depth(P.._ 1 )+I. Of course, Q:/= P,.-I; also Q #= P since depth(Q)  n
and depth(P) > n. In view of this, Q is strictly between P and Pit-I. The
resulting contradiction completes the proof. 0
Property 8 yields at once
Property 9
In a Cohen-Macaulay ring, all the saturated chains of prime ideals between
two fixed prime ideals have the same length, i.e. a Cohen-Macaulay ring
is catenary in the terminology of Section 1.2 (see the definition preceding
Exercise 6).
Proof
Let P and Q be prime ideals, P c Q. If Po c... c P" = P is any saturated
chain of prime ideals ending with P and if
P = Pkc ...C P k + r = Q (5)
is any saturated chain of prime ideals between P and Q, then the composite
chain Po c... c P"+r = Q is a saturated chain of prime ideals ending
with Q. Property 8 shows that k+l. = ht(Q), k = ht(P), i.e. the length
of the chain (5) is equal to r = ht(Q)-ht(P) and depends only on P and Q.
o
From the above proof we see that any ring with Property 8 also has
Property 9. The converse, however, is not true, a counterexample being
the ring in Example 3.4.3. Indeed, homomorphic images inherit the Prop..
erty 9; hence, by Corollary 3.4.6, the ring K[X, Y, ZJ/(XY, XZ) has this
property. We know, however, from Example 3.4.3 that the ring does not
have Property 8.
The example given in 1.2.21 shows that there exist also rings which
do not have Property 9.
Property 10
If (R, m) is a local Cohen-Macaulay ring, then for any ideal I of R we
have equality
ht(I)+dim(R/I) = dimR.

III]
Basic Properties of Cohen-Macaulay Rings
123
Proof
Let P be an arbitrary prime ideal of R. By Properties 8 and 9, any satu-
rated chain of prime ideals descending from m and going through P has
length dim R. Accordingly every saturated chain descending from m and
ending with P is of length dim R - ht (P). If P ranges over all the prime
ideals containing I, the number is largest when ht (P) is smallest, i.e. when
ht(P) = ht(I). Thus dimR/I = dimR-ht(P) = dimR-ht(J). 0
Property 11
A Noetherian ring R is a Cohen-Macaulay ring if and only if the poly-
non1ial ring R[X] is a Cohen-Macaulay ring.
In the proof we shall use a simple lemma.
Lemma 3.4.5
Let R be a a commutative ring. Then any maximal ideal ID1 of the poly-
nomial ring R[X] contains an element which is not a zero-divisor.
Proof
Assume the contrary. Then X  9J1 since X is not a zero-divisor. Therefore
9R+XR[X] = R[X], i.e. 1 = a+bX, a E 9)1, b E R[X]. But a = I-bX is
not a zero..divisor in R [X], and the resulting contradiction ends the proof.
o
Proof of Property 11
By Theorem 3.3.2, it is sufficient" to establish the equality of the height and
depth for any maximal ideal m of R[X]. Let P = 9)1 nR; by the assump-
tion, R is a Cohen-Macaulay ring, hence ht (P) = depth (P). Clearly
PR[X] is properly contained in the prime ideal PR[X]+XR[X], hence
9J1 =1= PR[XJ. By Lemma 1.3.3, we conclude that ht(9J1) = ht(P) + 1.
Let x 1, ...  Xs be a maximal regular sequence contained in P. Since
R[X]/R[X] (Xl' ..., Xs)  R/(XI, ..., x,)[X], by Lemma 3.4.5 the maximal
ideal Wl/(XI'.'" xs) contains an element which is not a zero-divisor.
Consequently depth (9J1)  s + 1 = ht (9)1), and an application of Theorenl
3.2.1 gives the required equality.
The opposite implication follows immediately frOln Property 4. D
As a simple consequence of Property 11, \ve obtain once Dl0re, although
in a slightly different form and independently of the proof in Chapter I,
Macaulay's original theorem ('fheorem 1.3.7).
orollary 3.4.6
1"he poJynomial ring K[X 1 , ..., X,,] over a field K is a Cohen-Macaulay
ring.

124
Cohen-Macaulay Rings
[Ch.
Exercises
1. Let I be an ideal of a Cohen-Macaulay ring R. Prove that if I is generated by
n elements and ht(l) = n, then R/ I is also a Cohen-Macaulay ring.
2. Prove that if Rand T are Cohen-Macaulay rings, then so is their product R x T.
In the following exercises (R, m) denotes a local ring and M denotes a finitely gener-
ated R-module.
3. Show that if Q e Ass(M/ xM), x  3(M), then there exists P E Ass(M) such that
PQ.
4. Let P e Ass(M). Prove that
depth(m; M)  dim(R/P).
[Apply induction on depth (m; M), using the results of Lemma 3.1.10 and Exercise 3.]
It follows from the result of Exercise 4 and the definition of the dimension of a mo-
dule (Definition 1.5.11) that depth (m; M)  dimM. M is called a Cohen-Macaulay
module irthe above inequality becomes an equality.
5. Show that if M is a Cohen-Macaulay module then dimM = dim(R/P) for any
P e Ass(M), i.e. M has no embedded components.
1\
6. Prove that M is a Cohen-Macaulay R-module if and only if M is a Cohen-Ma-
A A A
caulay R-module, where R, M denote the completions of R, M in the m-adic topology.
7. Let x 'f: 3(M). Prove that Mis a Cohen-Macaulay R-modute if and only if M/xM
is a Cohen-Macaulay R/(x)-module. [Apply Corollary 3.1. 10 and the result of Exercise
12, Section 1.5.]
3.5 PERFECT IDEALS
No\v we shall concentrate on describing those ideals of a Cohen-Macaulay
ring for which the corresponding factor rings are also Cohen-Macaulay.
Under the assumption of the finiteness of projective dimension, they
are precisely perfect ideals which were already studied by Macaulay in
the early years of the present century.
For the proof of the principal result in this section (the Auslander-Buchs-
baum theorem (Theorem 3.5.6) some auxiliary facts are required. The
first is contained in. McCoy's theorem, which is interesting in itself, and
which will also be used in Section 3.6.
Theorem 3.5.1 (McCoy)
Let R be a ring. A system of n homogeneous linear equations in p unknowns
p
LaljxJ=O. i= I.....n. a'JER. (6)
j=l
has a nontrivial solution in R if and only if either p > n or p  n and there
exists a non-zero element in R annihilating all the p xp minors of the
matrix [alj].
We shall formulate the problem of the existence of solutions of system
(6) in other terms. Let E be a. free R-module with basis el, .oo, e,J and let
F be a free R-module with basis /1, ... ,In. The system (6) has a solution

III]
Perfect Ideals
125
p n
[Xl' ..., x p ] if and only if I: x/v} = 0, where .Vj = I: akJJk E F. Let
j=1 k=1
cp: E -}> F be the homomorphism defined by rp(ej) = 'VJ,j = 1, .11 ,p.
Then (6) has a nontrivial solution if and only if Ker(rp) ¥= O.
To decide when the kernel of rp is non-zero, we shall use the following
Ielnma:
Lemma 3.5.2
The monomorphism rp: E -}> F of free modules over a ring R induces
a monomorphism !\ t rp: !\. t E -}> !\. t F of t-th exterior powers for any t > O.
Proof
We recall that!\.t E = (@tE)/N, where (j!)tE stands for the t-th tensor
power of E and N is the submodule generated by the elements of the fornl
Ul @ ... (g)u t , U, = Uj for some i,j, i =F .i.
For free modules there is another description of the module N, namely
a description in terms of alternating mappings. The symmetric group St
acts on @tE via 0'-1 (Ul (g) ... @U t ) = U(I) @ ... @U(t), 0' E St. This action
enables us to define an antisymmetrization operator
OCE: fi)' E --t @'E, OCE = L sgn(O') 0'.
oeS,
It is easily seen that N c Ker(cxp); indeed, if u is a generator of N, then
tl = 7:U for some transposition 7: E St. Then cxs(u) = I: a(u - 1:u) = 0,
a
where (] runs through all even permutations of the group St. Moreover,
it turns out that if E is a free module then N = Ker(cxB).
In order to prove this equality, let us consider a basis e 1, ..., e p of E,
and denote by L the set {I, 2, ..., p}. It is known that @tE is a free module
with basis {ell @ II. @e,,}, where [iI' ..., it] ranges over the set' L t = Lx
x... xL. Put ei = el t @ ... @ei, for i= [i 1 ,...,i t ]. We let d denote
the subset of Lt consisting of those i for which at least two components
are identical. The renlaining part L"".9I can be split into equivalence
classes of the relation I'V : i I'V j  if there exists (1 E S, such that i = (1U).
Let P4 be a subset of Lt"'-d consisting of representatives (one from each
class) of all these equivalence classes. Then the above-mentioned basis
of @tE may be obtained by taking:
1 0 the elements e, corresponding to the sequences i E dB,
2 0 the elements eo(i) corresponding to the sequences i E dI and to all
the non-identity permutations 0' E S, ,
3 0 the elements ei corresponding to the sequences i e d.
Fronl the fundamental properties of bases, it foJlows that we shall
also obtain a basis @'E by taking:

126
Cohen-Macaulay Rings
[Ch.
1) the elements e, corresponding to the sequences i E fJJ,
2) the elements sgn( cr)ea(i) - ej correspondin.g to the sequences i E f!A
and to all the non-identity permutations (] E Sh
3) the elements e, corresponding to the sequences i E d.
The elements of type 2) and 3) of that basis, belong to the submodule
N (as regards 2) see Exercise 1). Thus to show that N = Ker«(XB)' it is sufi1-
cient to prove that if u = L aiei E Ker(llE)' then OJ = O. However, if
ie91
fXE(U) =  a,sgn(a)e(1(1) = 0,
iei':tes r
then aj = 0 because the elements ea(1) are linearly independent.
Now the proof of the lemma no longer presents difficulties. The modules
E and F being free, we have monomorphism
E(g)E rpfg)l, Ffi)E and FE l@rp , F@F,
and hence also a monomorphism E@E - FF. Repeating these argu-
ments, vve arrive at the conclusion that (i!)tqJ: @tE  fi!)t F is a mono-
morphism, too. Moreover, the operators ex commute with @trp, i.e. we have
a commutative diagram
@'E
 (l)'F
Ot,!
0(1'
0 t E
:. (8)tF
Since the horizontal mappings are monomorphisms, we may regard (8)'E
as a submodule of @'F, which shows that Ker(ClE) = Ker(llF)n(@tE).
Therefore the induced mapping
1\' E = @tE/Ker(tXE)  @tFIKer(fXp) = 1\' F,
which is equal to 1\ I rp, is also a monomorphism.
D
Proof of Theorem 3.5.1
Returning to the previous interpretatiol1 of the problem of existence of
solutions of the system (6) in terms of the mapping cp: E -+ F, we have to
investigate \Vl1en Ker(qJ) =F o. Clearly, Ker(<p) i= 0 if and only if the ele-
ments Vt, ..., '0, are linearly dependent. The latter property, however,
is equivalent to the linear dependence of the element VIA ... A v, in !\p F.
Indeed, if Vt, ..., v p are linearly dependent and L b/vj = 0, b k #- 0, then
b.(vt A ... A V,,) = VIA ... Abk'iJtA ... AV p = VIA ... A (LbJvJ) A ... A
A VII = o. On the other hand, if VI' ..., v II are linearly independent, then

III]
Perfect Ideals
127
the mapping ({J is a nl0nomorphism, and hence so is !\.P cp by Lemma 7.5.2,
which means that V1A ... "VII is linearly independent.
of course, the element VIA ... "v p is Jinearly dependent if p > n.
If p  n, then, expressing VI" ... "v p in terms of the basis {fit A ... AJi,,},
1  i 1 < .., < i"  It, of the module!\. fJ F, we obtain as coefficients all
the p x p minors of the matrix [a'l]' Thus the element VI" ... A v p is linearly
dependent if and only if there exists a non-zero element in R annihilating
aU the p x p minors of the matrix [ail]' D
Corollary 3.5.3
Let Ebeafree nlodule ofrankp and Fa free module of rank n over a Noeth-
erian ring R. If there exists a monomorphism qJ: E  F, then p  nand
depth(I)  1, where I is the ideal generated by aU the p xp minors of the
matrix of the mapping rp.
Proof
If p > n, then!\.P qJ = 0 since !\.P F = O. On the other hand,!\.P E =F 0,
and hence !\.P cp is not a monomorphism, contrary to Lemma 3.5.2. Thus
p  n, and the ideal I is defined. If we had depth(l) = 0, then I would be
contained in one of the prime ideals associated with the zero ideal (Theorem
2.3.22 of [B]), and therefore every p xp minor of the matrix of rp would
be annihilated by a fixed non-zero element of R. By Theorem 3.5,1, this
means that Ker(qJ) ¥= 0, contradicting the assumption. D
The next two lemmas will also be useful in proving Theorem 3.5.6.
Lemma 3.5.4
Let R be a Noetherian ring and M a finitely generated R-module. If
x  3(R), x  3(M), and x belongs to the Jacobson radical J(R), then:
(i) the module M is free over R if M/xM is free over R/(x),
(i') the module Mis projective over R jf M/xM is projective over R/(x),
(ii) pdR(M) = 1 implies pdRj(x)(M/xM) = 1.
Proof
(i) We assume M/xM to be free over R/(x). Choose elements Ul, ,.., Un
in M, such that their images UI, ..., Un in M/xM form a basis over R/(x).
Let M' be the submodule generated by "1' .,., tin and consider an exact
sequence of R-modules
o  L  E .:: M'  0,
(7)
where E is a free module with basis el, ..., en, and n(ei) = Ui. Since by the
assumption M' +xM = M, we have M/M' = x(M/M'). From the Naka-
yama lemma (because x E J(R» we deduce that M' = M. Furthermore,
the sequence L/xL  E/xE  M/xM  0 is exact because N/xN

128
Cohen-Macaulay Rings
[Ch.
= NQ!)R/(x) for any R-moduIe N and because of the right exactness of the
R
tensor product functor. It follows from the definition of homomorpisbm
n that the mapping ElxE  l/xM is an isomorphism, which gives L c xE.
We shall show that L = xL. Indeed, if y = xe E L, then xn(e) = 0,
because of the exactness of sequence (7). Since x  3(M), we find nee) = 0,
i.e. eEL. Thus Y E xL. From the Nakayama lemma we get L = 0, and
consequently M  E is a free module over R.
(i') Suppose now that M/xM is projective over R/(x). Consider an
exact sequence 0 -+ L -+ F -+ M -+ 0, F being free over R. This sequence
induces an exact sequence 0 -+ L/xL  F/xF --+- MlxM -+ O. Indeed,
in view of an isomorphism N/xN  N@R/(x) for any R-module Nand
R
the right exactness of the tensor product functor the sequence L/xL
-+ F/xF --+ M/xM -+ 0 is exact. To prove that L/xL --.. F/xF is a mono...
morphism we apply the standard technique of diagram chasing to the
diagram
0 0 0
t 1 1
L/xL ... F/xF  M/xJ
t 1 i
 L ,. F  M
t x Ix Ix
)1Ir L - F  M
t i i
0 0 0
 o
o
 0
o
)a 0
whose rows are obviously exact and the columns are exact because x f: 3(R)
and x; a(M). Since MlxM is projective over Rf(x), the sequence splits,
and F/xF  LjxLEfJM/xM. Hence (LEaM)/x(L(j)M) is free, and also
LEa M is free by (i), i.e. M is projective.
(ii) Let 0 -+ F 1 -+ Fo  M -+ 0 be a projective resolution of Mover R.
By the arguments given in (i') above, the sequence 0  F 1 /xF 1 --.. Fo/xFo
-+ M/xM -+ 0 is exact. Thus the projectivity of F 1 jxF 1 , Fo/xFo over
R/(x) imply that pdR(x)(MfxM)  1. By (i') we obtain the equality. 0
Lemma 3.5.5
Let
O-.N-+FjVf-+O
(8)

III]
Perfect Ideals
129
be an exact sequence of non-zero, finitely generated modules over a local
ring (R, m). If depth(m; N) < depth(m; F), then depth(m; N) > 0 and
depth(m; M) = depth(nt; N)-I.
Proof
Put K = RIm, depth(m; N) = n, depth(m; F) = p and dcpth(m; .M) = In.
We shall make use of the characterization of depth (Corollary 3.1.7) in
terms of functors Ext. The sequence (8) induces, for each i > 0, an exact
sequence
Ext i - 1 (K, F) -+ Ext ' - 1 (K, M) -+ Exti(K, N) -+ Exti(K, F). (9)
Since p > 0, we have ExtO(K, F) = 0, and therefore ExtO(K, N) = 0,
which shows that n > O. Using the exactness of (9) for i  n and the
assumption n < p, we conclude that Ext'(K, N)  Ext'-l(K, M), and
m = n-l. 0
'Theorem 3.5.6 (Auslander-Buchsbaum)
Let (R, m) be a local ring and M a non-zero finitely generated R-module.
If pdR(M) < ex) then
pdR(M)+depth(nt; M) = depth(m).
Proof
We proceed by induction on s = pdR(M). Set depth(nt) = p, depth(m;M)
= m. If 8 = 0, then M is free, clearly m = p. Let s = 1 and consider an
exact sequence 0  Ft -+ Fo --.. M --.. 0, where F 1 and Fo are free modules
and the homomorphism Fo --.. M is a minimal epimorphism (see Lemma
A.5.3). It follows that the matrix of the mappitlg F L -+ Fo with respect to
any bases has entries in m. This, together with Corollary 3.5.3, gives p  1.
We shall prove by induction on m that m + 1 = p. If m = 0, then ExtO(K, M)
=1= O. But, we have the exact sequence 0 --.. ExtO(K, M) -+ Ext 1 (K, F t ),
which implies that also Ext 1 (K, F 1 ) :p 0, i.e. p = 1. Assume that m > 0
and let x  3(M). By Lemma 3.5.4, we have pdR/(x)(MlxM) = pdR(M) = 1,
while depth(m/(x); M/xM) = m-l, depth(m/(x») = p-l in view of
Lemma 3.1.10. By the inductive hypothesis, (m-l)+ 1 = p-l, i.e. m+ 1 = p.
We assume that s > 1 and proceed inductively on s. We have an exact
sequence 0 -+ N -+ F --.. M -.. 0, where F is free. Theorem A.4.t shows
that pdR(N) = s-1 > O. Applying the inductive hypothesis to N, we
obtain the equality 8-1 +depth(m; N) = p. Observe that depth(m; N) < p,
whence, using Lemma 3.5.5, we conclude that m = depth(m; N)-l, and
eventually m+s = p. 0
Corollary 3.5.7
Let R be a Noetherian ring and M a finitely generated R-module. Then,
for any prime ideal P associated with M, we have the inequality depth(P)
 pdR(M).

130
Cohen-Macaulay Rings
[Ch.
Proof
If P E Ass(M), then PR p E Ass(M p ) by Lemma 2.4.14 in [B]. By Theorel11
3.5.6, we get depth(PR p ) = pdR(M p ) because depth(PR p ; M p ) = 0 (we
may assume of course that pdR(M) < (0). In view of the known inequal-
ities depth(P)  depth(PR p ) and pdRp(M p )  pdR(M), the assertion
follows. 0
If J = Ann(M), then clearly every prime ideal P associated with M
contains J. Hence depth(J)  depth(P)  pdR(M). This leads to the
following definition.
Definition 3.5.8
Let M be a finitely generated module over a Noetherian ring R and let
Jbe its annihilator. We say that M is a perfect module if depth(J) = pdR(M).
In the case where M = RII is a cyclic module, the ideal I is said to be
perfect if R/I is a perfect R-module, i.e. if depth(I) = pdR(R/I).
CoroJlary 3.5.7 yields a simple coroJlary, which we shall state only for
cyclic modules.
Corollary 3.5.9
Any perfect ideal is depth-unmixed.
The simplest example of a perfect ideal is provided by a regular ideal.
Theorem 3.5.10
Every regular ideal of a Noetherian ring is perfect.
Proof
By Theorem A.6.2, from the Koszul complex associated with a regular
sequence of length 11 we get a free resolution of length n - 1 of the ideal
I generated by that sequence. Hence and from Theorem 3.2.3 we deduce
that n = depth(l)  pdR(R/I)  n, and therefore I is a perfect ideal. In
the case of a local ring, the assertion of the theorem follows immediately
from Lemma 2.1.3.
The importance of perfect ideals lies in their connection with Cohen-
Macaulay rings, as specified in the theorem below. Notice that in a Cohen-
Macaulay ring the notion of being depth-unmixed and of being height-
unmixed coincide in view of Theore1l1 3.3.2.
1nbeoremm 3.5.11
Let R be a Cohen-Macaulay rin.g and I an ideal of R. Assume
that pdR(RII) < 00. Then the foJlowing properties are equivalent:
(i) R/! is a Cohen-Macaulay ring and I is unmixed,
(ii) I is a perfect ideal.

III]
Perfect Ideals
131
If R is a local ring, then the assumption of the unmixedness of I made
in (i) may be omitted, i.e. R/I is a Cohen-Macaulay ring if and only if I
is perfect.
Proof
We first assume (R, m) to be a local ring. Using Theorems 3.5.6, 3.3.2
and Property 10 in Section 3.4, of Cohen-Macaulay rings, we obtain the
equalities
ht(m/J)-depth(m/I) = ht(ln)-ht(I)-depth(mlI)
= depth(m)-depth(m; R/I)-ht(I) = pd R (R/1)-depth(I).
Again by Theorem 3.3.2, R/I is a Cohen-Macaulay ring if and only
if the expression on the left-hand side of the above equality is zero, and
hence if and only if I is perfect.
Let R be an arbitrary Cohen-Macaulay ring and I a perfect ideal. For
any maximal ideal m containing 1, we have the inequalities
depth (I)  depth(IR m )  pdRm{Rm/IRm)  pdR(R/I),
hence, by the assumption, we deduce that the ideal 1Rm is perfect. Applying
the already proved local version of the theorem, we find that Rm/IRm
 (R/l)m is a Cohen-Macaulay ring, and hence so is R/I by Property 2 in
Section 3.4. The unmixedness of I follows from Corollary 3.5.9.
To prove the opposite implication, assume that pdR(R/I) < 00. Thus
by Theorem A.4.5, there exists a prime ideal P containing I, such that
pdR(R/I) = pdRp(Rp/IR p ). Moreover, the two conditions of the theorem
being equivalent in the local case, we have pdRp(Rp/IR p ) = depth(IR p ).
The ideal P contains one of the prime ideals associated ,vith I, say Q, and
therefore, by the assumption and the fact that both Rand R p are Cohen-
Macaulay rings, we get a sequence of equalities
depth(IR p ) = ht(IR p ) = ht(QR p ) = ht(Q) = ht(/) = depth(I),
which proves that I is perfect. 0
In the global case, the assumption in condition (i) that I is unmixed
is essential, as is shown by the following example.
Example 3.5.12
Let A = K[X o , Xl' ..., Xn] be the ring of polynomials in indeterminates
Xo, ..., X n over a field K, n > 1. Denote by S the following multiplicative
subset of A: S = A" {(XO)u(X I , ..., X n )}. We take R = As, I = R(X O X 1 ,
... , Xo,). R is Cohen-Macaulay by Properties 1 and 11 of Section 3.4.
Furthermore, the only maximal ideals of R are the ideals corresponding to
(X o ) and (Xl' ..., ,) in A, and they have I as their intersection, whence
R/I is also Cohen-Macaulay as an Artin ring. On the other hand, however,
depth(I) = 1 < pdR(R/I) < 00 (the simple proof is left to the reader)

132
Cohen-Macaulay Rings
[Ch.
and consequently the ideal I is not perfect. The reason for this is that I is
mixed, namely
htR(X o ) = 1., htR(X t , ..., X n ) = n > 1.
Exercises
1. Let E be an arbitrary module and @ t E the t-th tensor power of E. Let N be the
submodule of t E generated by the elements of the form Ul  ... Uh U, = UJ for
some;, j, i :F j. Prove that, for every element u e @ t E and for every permutation G E St,
the element u-sgn(O') G(II) belongs to N. [Establish the assertion at first for transposi-
tions and then by induction 011 the length of the expression of 0' as a product of trans-
positions.]
2. Let M be a module generated by n element and Jet E be a free submodule of M
of rank n. Show that M is free.
3. Prove that, under the assumptions of Lemma 3.5.4, we have pdR(M)
= pdR/(;JC)(M/xM).
4. Prove that if, under the assumptions of Theorem 3.5.11, the ring R/I is Cohen-
Macaulay and there exists a prime ideal containing all the associated prime ideals of 1,
then I is unmixed.
5. Let Xl' ..., XII be a regular sequence of a local ring and let u, = Xl ... X'-l X,+ 1...
... X n , i = 1, ..., fl. Show that the ideal (U1' ..., u.) is perfect and has depth 2.
6. Prove that any power of a regular ideal in a local ring is a perfect ideal.
7. Let R c T be a finitely generated extension of local rings, and let R be a regular
local ring. Show that T is a Cohen-Macaulay ring if and only if T is a free R-n10dule.
3.6 STRUCTURE OF PERFECT IDEALS OF DEPTH 2
In a local ring, there are no non-zero perfect ideals of depth 0, because no
module of the form RjI, I =1= 0, is free. The only perfect ideals of depth 1
are the principal ideals generated by a non-zero-divisor, since they are the
only ideals that are free R-modules.
The aim of the present section is to give the following description of
perfect ideals of depth 2.
Theorem 3.6.1 (Hilbert, Burch)
Let (R, m) be a local ring and I an ideal of R of depth 2 and of finite pro-
jective dimension. Then I is a perfect ideal if and only if there exists a posi..
tive integer n and an n x (n -1) matrix Cover R such that I is generated
by all the (n-l) x (n-l) minors of C.
Proof
Suppose that I is a perfect ideal of depth 2. Since every finitely generated,
projective module over a local ring is free (Theorem A.5.4), there exists
a resolution of the form
tp f
o -+ R S -+ R" -+ R -+ R/I -+ 0>
(10)

III]
Structure of Perfect Ideals of Depth 2
133
where Rl' denotes a free module of rank p. We shall first show that s = n - 1.
Since depth(I) > 0, there exists a prime ideal P of R such that I cf: P, and
thus (R/I)p = O. By localizing sequence (10) with respect to P, we obtain
an exact sequence of free modules over R p : 0  R -+ RJ -+ R p -+ O.
Since the rank of a free module is well determined we get s = n - 1.
Fix a basis h 1 , ..., h n - 1 in R,,-l and el, ..., en in R". Let vJ = qy(h j )
11
= L aijei,j = 1 , ..., n-l,f(ei) = bj, i = 1, ..., n. Accordingly, lis gener-
i = 1
ated by b 1 , ..., b". Further, let A = [aij] and write M i for the minor of
A obtained by deleting the i-th row. Set A, = (-l)'M,. We shall prove
that there exists an element x E R such that hi = xA, for all i, 1  i  n.
Then the ideal I will be generated by the maximal minors of the matrix
C obtained from A by multiplying one column by x.
We show first that b,Aj = bJA,. Indeed, from the above definition and
from the exactness of sequence (10), it follows that L aljb, = O. Applying
Cramer's rule for solving systems of linear equations, we arrive at the
required equalities.
By the assumption, there exists in I an element t which is not a zero-
divisor, t = Lt,b l ; consider at the same time the element u = LtiA;.
We have ubJ = L t,bJA, = L tib,A) = tAl. To establish our claim, it is
sufficient to show the existence of x E R such that t = ux, and to prove
that u is not a zero-divisor of R. Then ubJ = tAl = uxA} will give b} = xA}.
If yu = 0, then yA i = 0, as t is not a zero-divisor. On the other hand,
applying Lemma 3.5.2 to the mapping rp, we conclude that the element
VI A ... A V..-l is linearly independent. Since
VIA ... AV II -l = LM t e 1 A ... Aet-l A et+lA ... A ell'
it follows that y( V I A ... A V.. _ 1) = 0, thus J' = O. Therefore u is not a
zero-divisor.
In order to find an element x with the required property, consider the
factor ring R = R/(u). We shall show that t :F 0 in R implies a contra-
diction, i.e. we must have t = ux for some x E R. Denote by q; : R,, -1 -.. R"
the homomorphism induced by rp, and by V, the corresponding resid.ue
classes of the elements V, in R". Since tA, = 0, applying the above argu-
ments to ({J , we get t(V l/\ ... A V..-l) = o. If t '# 0, then the element Vl A ...
... /\ V,,-l is linearly dependent, and consequently, by Lemma 3.5.2, we
see that qy is not a monomorphism. Hence LCjf)j = 0 and cJ =1= 0 for at
least one j. The resulting relation L C,V, = uz yields uz E Ker(f), and
therefore Z E Ker(f) = Im(rp), since u is not a zero-divisor. If z = Ld,Vh
then, since 'VI, ..., V II _ 1 are linearly independent, we get c} = d J u, and
cJ = 0 for all j, 1  j  n - 1, contradicting the previous observations.

134
Cohen-Macaulay Rings
[Cb.
We shall now pass to the proof of the converse implication. Assume
we are given an n x (n-I) matrix C = [CiJ], C;j E R, such that the ideal
1= (M 1 , ..., M,.) has depth 2, where M, denotes the (n-I)x (n-I)
minor of C obtained by deleting the i-th row. We define a sequence of
mappings
o -+ R n - 1  R n  R (11)
in the following manner:
f( e,) = (- 1)' M, , i = 1, ..., n,
q;(h J ) = EC'Je" j = 1, ..., n-l;
{h J }, {e,} denote the canonical bases of R,,-l and Rn respectively. Applying
Laplace expansion to the matrix C, we find that (11) is a complex. We
shall now show that it is exact.
1) Ker(9') = 0; from Theorem 3.5.1 and the discussion following it,
we see that if Ker(9') ¥: 0, then depth(I) = 0, contrary to the assumption.
2) Ker(f) c: Im(9'); let t = L(-I)'t,M, be a non-zero-divisor in I,
and put e = L:tje, eRn. Sincef(e) = I, we have Im(q;)()Re = O. Therefore
the sequence
o  R"  Rn -+ Rnl(Im(cp)+Re)  0
is exact, where rlRn-1 = 9', r(h n ) = e. Writing C' for the matrix of the
homomorphism r with respect to the canonical bases we conclude by
Corollary 3.5.3 that y = detC' is not a zero-divisor. Furthermore, yell
n
= L fslr(h , ) for all s, where rsf denotes the cofactor of c;, in C'. Thus y
1=1
annihilates Coker(y), i.e.yR" c: Im(q;)+Re. Ifu e Ker(D, thenyu E Im(cp) +
+Re, and yu E Im(q;) because t is not a zero-divisor; let
n-l
yu = I>J9'(h J ).
J=-1
(12)
Consider the factor ring R = R/(y) and note that r = (I, y)/(y) ¥: 0
since depth(I) > depth(y) = 1. Moreover, the induced mapping cp : R,, -1
 Rn is a monomorphism; otherwise, arguing as in 1) we arrive at the
conclusion that depth(I) = 0, i.e. that I is contained in one of the prime
ideals associated with (y). But this is impossible because depth(y) = 1
and (y) is depth-unmixed (Lemma 3.2.8) while depth(I) = 2.
Since LOj h j e Ker( cp) in view of (12), aJ = ajy for 1  j  n-l. But
y is not a zero-divisor; hence finally u = Lajcp(h j ) E Im(q;).
It follows from the above considerations that the sequence (11) is
a free resolution of Rll. Thus 2 = depth(I)  pdR(R/I)  2, and so the
ideal I is perfect. 0

III]
Notes and References
135
TheorenlS 3.6.1 and 3.5.11 irnnlediately yield
Corollary 3.6.2
If in a local Cohen-Macaulay ring R an ideal I has depth 2 and finite pro-
jective dimension, then R/I is a Cohen-Macaulay ring if and only if I is
generated by the (n-I) x (n-l) nlinors of a certain n x (n-I) matrix
over R.
Exercises
1. Let X be an Il x p matrix with entries in a ring R, p  11, and let Ip(X) denote the
ideal generated by all the p x p minors of the matrix X. Prove that if x 11 is invertible
n R, then there exist invertible matrices C and D such that
CXD = ( Xl1 I  )
. 0 I X' '
where X' is an (It-I) x (p-l) matrix over R. Moreover prove that 1,,(X) = /"-1 (X').
2. Let (R, m) be a local ring and J an tn-primary ideal. Show that if J' is an ideal
in the polynomial ring R[Y] such that J' c 1nR[y] and (J', Y) = (JR[Y], Y) then mR[YJ
is one of the minimal prime ideals that contain J'.
3 (Macaulay). Let X be an n xp matrix with entries in a Noetherian ring R, p  n,
and let I,,(X) denote the ideal generated by the p x p minors of X. Prove that any minimal
prime ideal containing Ip(X) has height at most n - p + 1. [Proceed by induction on p,
making use of Exercise 1. Reduce the probleln to the local case. Replace the n1atrix
X with the matrix X', where Xl = XII + Y, x) = XI) for (i,j) i:- (1,1), and Yis an inde)
terminate. Apply Exercise 2.]
4. Let K[X'J], i = 1, ..., ", j = 1, ..., 11-1, be the ring of polynomials in 11(11-1)
indeterminates XI} over a field K. Denote by X = [Xlj] the nlatrix consisting of these
indeterminates. Show that depth( In-l (X)) = 2.
5. Let K be a field and let R = K[X t , X 2 , X 3 ) or K[[X 1 , X 2 J X 3 ]] and 1 = (Xl X J ,
X 1 X 3 ,X 2 X 3 ). Find a minimal resolution of RII over R (see Exercise 7, Section A.S 0
Appendix) and prove that the ideal I is perfect.
6. Let 1 be an ideal of a ring R, F a free resolution of the module R/I over R, and
E a complex of the form R  R for some x E R, i.e. let Eo = £1 = R, E, = 0 for
j > 1, and let d 1 be the multiplication by x. Under what condition on x is the tensor
product E(j!Jft' a free resolution of RI(I, x)?
7. Construct a minimal free resolution of the module K[X 1 , X z , X 3 ]/I over the ring
K[X 1 , X: h X 3 ] where I = (X 1 X Z , X 1 X 3 , X 2 X 3 , Xl +X 2 +%3).
NOTES AND REFERENCES
The notion of a regular sequence was implicit in the study of perfect ideals in a poly-
nomial ring by Macaulay [0]. Yet it was only in the middle fifties that, together with
the development of homological methods, the actual role of regular sequences in the
theory of rings was revealed. We should mention the pioneering papers by Auslander
and Buchsbaum [1], Rees [30] and Serre [381 in which also the general concept of the
depth of an ideal with respect to a module emerged. Regular ideals in arbitrary Noeth-
erian rings were studied by Rees (who caned them general ideals) in a series of papers

136
Cohen-Macaulay Rings
[ChI
[29], [30], [31]. Theorem 3.2.9 was proved for the first time in [30]. The proof given
in the text is based on [L].
The property adopted by us as a definition of a Cohen-Macaulay ring was proved
by Macaulay in [0] for the polynomial rings K[X I , ..., Xn] over a field K, and by Cohen
in [7] for regular local rings. Most of the results presented in Sections 3.3 and 3.4 are due
to Northcott and Rees, and are contained in papers [24], [26], [27]. The authors'
approach frequently differs from ours, e.g. Northcott introduced in [24] the class of
semi-regular local rings (defined by the property in Exercise 6, Section 3.3), and charac-
terized it in terms of multiplicity by the equivalent conditions 3.3.6 (i) and 3.3.6 (ii),
proving thereby that it is the class of Cohen-Macaulay rings. Theorem 3.3.6, and many
other properties of Cohen-Macaulay rings (Property 11 among them) are also to be
found in Nagata [22].
Perfect ideals in a polynomial ring over a field were studied first by Macaulay ill
[0] and then by Grabner in [10]. The equivalence of the different definitions \vhich they
give is demonstrated in [20]. The definition in case of arbitrary rings which we give in
this chapter is due to Rees [30]. The proof of the implication (ii) => (i) in Theorem
3.5.11 and the assertion in the case of local rings for regular local rings can also be found
there. The proof of the converse implication and Example 3.5.12 are taken fron1 [14].
The theorem of Auslander and Buchsbaum (Theorem 3.5.6) was published first in [1].
Theorem 3.6.1 in the form presented here is due to Burch [4]. The form of the res-
olution (11), for a determinantal ideal of the type under consideration, in a polynomial
ring over a field was known to Hilbert [12].

Chapter IV
Gorenstein Rings
Between the class of regular local rings an.d that of Cohen-Macaulay
rings, there is yet another class which has been intensively studied in recent
years because of its ubiquity and the various characterizations of rings
belonging to it. We mean here Gorenstein rings, named so after D. Goren-
stein who, at the beginning of the 1950s, published papers concerned with
certain algebraic curves. But it is H. Bass who deserves the credit for the
proper discovery and formulating basic properties of this class of rings
(see [3]).
According to the spirit of the present book, we adopt here a purely
ring theoretic definition with the aim of proving, via homological character-
izations, the basic properties and of providing some applications. The
starting point is an analysis of the presentation of an ideal as an
intersection of irreducible ideals. Emmy Noether proved that any two
irredundant decompositions of this kind have the same nUluber of com-
ponents. Moreover, in a Cohen-Macaulay ring, all ideals generated by
sets of parameters have irredundant decompositions of the same length,
which leads to the notion of the type of a Cohen-Macaulay ring (Section
4.1). We define Oorenstein rings to be the rings of type equal to 1. It turns
out that these are precisely those rings which, when regarded as modules
over themselves, have finite injective dimension. This is the main homo-
logical characterization. We also give alternative ones in Section 4.3, whereas
Section 4.2 is entirely devoted to zero-dimensional rings.
In Section 4.4 we examine when factor rings of Gorenstein rings are
again Gorenstein rings. We present a characterization on which A. Oro-
thendieck based his definition of Gorenstein rings. This theorem, as well
as the methods developed in Section 4.4 form a basis of the duality the-
orems in the theory of local cohomology.
4.1 THE TYPE OF J.JOCAL COHEN-MACAULAY RINGS
We recall the definition of an irreducible ideal. This is a proper ideal which
cannot be expressed as an intersection of t,vo ideals properly containing
it. We know that in a Noetherian ring, every proper ideal is an inter-
section of a finite number of irreducibIe ideals (see Lemma 2.3.9 of [B]).

138
Gorenstein Rings
[Ch.
We proceed to analyse more precisely the various expressions of an ideal
as an intersection of irreducible ideals.
Lemma 4.1.1
Let (R,111) be a local ring and Q an m-primary ideal. The following prop-
erties are equivalent:
(i) Q is irreducible,
(ii) l(Q: m)/Q) = 1, i.e., (Q: tn)/Q  RIm,
(Hi) the ideal Q: tn is the only minimal ideal among aU the ideals prop-
erly containing Q.
Proof
(i) => (ii) Since the ideal Q is m-pritnary, by Corollary 2.4.4 of [B].
there exists p such that nt P c Q. Let p be the minimal number with this
property, that is, m,-I ct: Q. Then there exists x  Q such that A'11t c Q.
Thus Q:m  Q, and 1(Q:m)/Q)  1.
Observe that (Q:m)/Q is a linear space over the field R/nt. If 1(Q:m)/
/Q) > 1, then there exist two non-zero subspaces V 1 , V 2 in (Q :tn)IQ
whose intersection is zero. Denote by Ql, Q2 the ideals of R deter.mined
by VI' V 2 . Then Q = Ql()Q2 and Ql  Q, Q2  Q, contrary to the assump-
tion that Q is irreducible. Hence l(Q:m)/Q) = I.
(ii) => (Hi) Let Q' be a minimal ideal of R proper1y containing Q.
Since Q c mQ' +Q c: Q', we must have Q = mQ' +Q by the Nakayama
lemma (because Q =F Q'). Hence we get the inclusion mQ' c Q, i.e.
Q' c: Q:m. Since Q =F Q' and 1(Q:tn)/Q) = 1, then Q' = Q:nt.
(Hi) => (i) If Q = Ql()Q2 and Q ( ¥= Q, Q2 :/= Q, then Ql ::> Q :m, Q2
:::> Q:m by the assumption. Accordingly Q = QlnQz :J Q:m, a contra..
diction. 0
We recall that the presentation of an ideal I as an intersection 1 = /1 n ...
. .. nIp is said to be irredundant if l,,:p n IJ for every k.
J=f-k
Theorem 4.1.2
Let (R, nt) be a local ring and Q an m-primary ideal. The following prop-
erties are equivalent:
(i) there exists an irredundant expression of length s, Q = Q 1 () ... ()Q,u
,,,here the Q, are irreducible ideals,
(ii) l«Q:m)IQ) = s.
Proof
We define a mapping of linear spaces over R/nt:
cp: (Q:m)/Q -+ (Ql:nl)/Ql ... EB(Qs:nt)/Qs,
91(X+Q) = (X+Ql' ..., x+Qs)'

IV]
'fype of Local Cohen-Macaulay Rings
139
In order to establish the equivalence of (i) and (ii), it is enough to show
that lfJ is an isomorphism. In fact, the ideals Q" are m-primary, and by
Lemma 4.1.1, the spaces (Qk:m)/Q" have dimension I, hence 1«Q:m)/Q) = s
which proves implication (i) => (ii). To get (ii) => (i) we note that Q has
at least one irredundant presentation of the form
Q = Qn ... nQ;
where Q are irreducible. Applying the impJication (i) => (ii) to this pres-
entation gives s = t.
Since Q = Q 1 n ... rlQ,SJ lfJ is an injection. To show that this is also
a surjection, it is sufficient to prove that some non-zero element of the sitnple
module (Qk:m)/Q" lies in the image of rp.
To simplify the notation, take k = 1 and observe that we on1y need
to prove
(Ql:m)nQ2n ... nQs  Q;
(1)
indeed if x belongs to the illtersection on the left-hand side, but does not
belong to Q, then xrtQl,xEQ:m c Ql:m and (X+Ql'O, ...,0)
= (X+Ql' ..., x+Qs) = q>(x+Q).
To prove* (1) asume s > I; observe that Ql +Q2 r1 ... nQs * Ql
by the irredundancy of the decomposition of Q. Since R/Q is an Artin ring
every non-zero ideal of R/Q contains a minimal non-zero ideal; therefore
Ql +Q2 n ... f1Qs ::> Ql:m by Lemma 4.1.1(iii). Intersecting both sides of the
last inclusion with Ql :m, we deduce Ql + (Ql :m)nQ2() ... f1Qs = Ql: m
:::> Ql; hence (Ql :m)nQ2 ... nQ * Q. 0
Corollary (and Definition) 4.1.3
Any two irredundant presentations of an m-primary ideal Q in the fornl
of an intersection of irreducible ideals have the same length. This length
is called the type of the ideal Q, and is denoted r(Q).
It turns out that in Cohen-Macaulay rings a certain class of m-primary
ideals have the same type, which therefore becomes an invariant of a ring
itself. More precisely, we have
Theorem (and Definition) 4.1.4
If (R, m) is a local Cohen-Macaulay ring of dimension d, and if Q, Q'
are both ideals generated by sets of parameters of R, then r(Q) = r(Q').
This number is equal to
dimRlmExt(Rm, R).
We call it the type r(R) of the ring R.
* This argument was suggested to us by Dr. D. Kirby.

140
Gorenstein Rings
[Ch.
Proof
Let an ideal Q be generated by a set of parameters Xl' ..., Xd. By Theorenl
4.1.2, we have r(Q) = 1(Q:m)IQ). Remark, however, that (Q:m)/Q
 HomR(R/m, R/Q). The isomorphism is defined by associating with
a homomorphism f: RIm  RIQ the element 1(1 +m) E (Q:m)IQ. FroDl
Theorem 3.3.5, we know that every set of parameters of a local Cohen-
Macaulay ring is a regular sequence. Applying this statement and Theorem
3.1.3 to the sequence Xl' ..., Xd, we obtain an isomorphism HomR(R/m,
R/(Xl, ..., Xci»  Ext(Rlm, R). It follows that r(Q) does not depend
on a specific ideal Q; at the same time we obtain the announced inter-
pretation of the type of a ring. 0
The Cohen-Macaulay rings of type 1 form a class of particular interest,
and the present chapter is devoted to them. Here is their first character-
ization.
Theorem 4.1.5
The following properties are equivalent for a local ring (R, m):
(i) the ring R is Cohen-Macaulay and there exists a set of parameters
of R generating an irreducible ideal, i.e. R is a Cohen-Macaulay ring of
type 1,
(ii) every set of parameters of the ring R generates an irreducible ideal.
Proof
The implication (i)  (ii) is obvious by Theorem 4.1.4.
To prove the converse implication, put d = dimR. If d = 0, then R
is a Cohen-Macaulay ring, and both conditions mean simply that the zero
ideal is irreducible.
Now let d > 0 and let Xl' ..., Xli denote a set of parameters of R.
Further, set Q, = (x, ..., xj), i = 1, 2, ... Since x,..., x also form
a set of parameters of R, all the ideals Qi are irreducible by the assumption.
Our principal aim is to show that R is a Cohen-Macaulay ring, i.e. that
depth(m) = d. First we shall prove the inequality depth(m) > 0..
Note that Q,+ 1 $ Q, for each i. Indeed, x rp Q,+ 1; otherwise we would
have an equality of the form
d
x , -  a X f+1
d -  J J ,
J-I
and hence
d-I
x(1-a"x,,) = L:a J xJ+1.
J=1
But the factor 1 - ad Xd is a unit in R, so we would obtain xj e (x+ 1 , ...
. .. , x:!: f) c (x, ..., X-l) and finally Q, = (x, .... Xj-l). This, however,

[V]
Type of Local Cohen-Macaulay Rings
141
is impossible because ht (m) = d, and the Q, are m-primary ideals (see
Theorem 1.4.1).
All the ideals Q, are irreducible, and therefore, applying Lemma 4.1.1,
we conclude that Q'+l:m is the only minimal ideal among those which
properly contain Q,+ 1. Thus the inclusion Q,+ 1  Q" proved above,
to
implies Q'+l: m c Q,. Since, by Krull intersection theorem, n Q, = 0,
1=1
we get
00 co 00
O:m = (n Q,):m = n(Q,:m)c: n Q, = 0,
1=1 ;=1 l=al
and O:m = O. But this means that depth(m) > 0 by Theorems 2.3.22 and
2.3.15 in [B].
Accordingly there exists x E m"'3(R). Setting R' = Rf(x), we obtain
dimR' = d-l (in view of Corollary 1.4.3). The next part of the proof
proceeds by induction on d. We assume the implication (ii) => (i) to be
valid for R' and consider any set of parameters Y, ..., Y.f-1 of the ring
R', where Y; = YI+(X). The sequence Yl, ..., Y4-1, X forms a set of par-
ameters of R since the ideal generated by these elements is m-primary and
d = dimR. By the assumption, the ideal (Yl, ..., Y4-1, x) is irreducible,
and so is the ideal (y, ..., Y-l) in R'.
We have proved that every set of parameters of R' generates an irre-
ducible ideal. Using the inductive hypothesis, we conclude that R' is a
Cohen-Macaulay ring. As x is a non-zero-divisor in R, we finally have
that R is a Cohen-Macaulay ring, by Property 5 in Section 3.4. 0
The assumption, in the condition (i), that R is a Cohen-Macaulay
ring is essential, as the following example indicates.
Example 4.1.6
Let K be a field and let R = K[[X, Y]]/(X 2 , XY). We shall prove that the
element y, the residue class of Y in R, forms a set of parameters of R, and
generates an irreducible ideal, while the ideal (y2) is reducible. The reason
for this is, of course, the fact that R is not Cohen-Macaulay (see Exercise 5,
Section 3.3). Since Rf(y) = K[[x]]f(X2), Y is a set of parameters and the
space «y):m)/(y) has dimension 1, which shows that (y) is irreducible by
Lemma 4.1.1. On the other hand, Rf(y2)  K[[X, y]]f(X, Y)2, whence
«y2):m)f(y2)  (X, Y)2:(X, y»)f(X, y)2.
The last space has dimension 2, and so, in view of Theorem 4.1.2, the ideal
(y2) is not irreducible.
Definition 4.1.7
A local ring is called a Gorenstein ring when it has the equivalent properties
stated in Theorem 4.1.5.
As an immediate corollary to the proof of Theorem 4.1.5, we have

142
Goren stein Rings
[Ch.
Coronary 4.1.8
If R is a loea] Gorenstein ring and x E R is not a zero-divisor of R, then
R/(x) is also a Gorenstein ring.
Proof
In fact, in proving the implication (ii)  (i), we established that if in R
every set of parameters generates an irreducible ideal, then the same is true
for the ring R/(x). D
Using the definitions only, we may provide the first examples of Gorenstein
rings, namely regular local rings. Such rings are obviously Cohen-Macaulay
\ith the maximal ideal irreducible and generated by a set of parameters.
Gorenstein rings form, however, a class larger than that of regular local
rings. They also admit a very efficient, homological characterization. To
these, and other problems concerning Gorenstein rings \ve devote the
subsequent sections, beginning with the characterizations and properties in
the zero-dimensional case.
Exercises
1. Let R == K [(XI' ,.., X.., Yl' . ,., YnJJ/(X. YJ-X J Y,), i, j = 1, ..., n, where K
is a field. Prove that:
(a) R is a domain.
(b) dimR = n+l,
(c) R is a Cohen-Macaulay ring,
(d) the type r(R) is n-l.
2. Let K[[XIJ]], i = 1, ,.., n, j = 1, .,., n- 1, be the ring of formal power series in
n(n -1) variables XI} over a field K. Denote by X = [Xu] the matrix consisting of the
variables, and by I(X) the ideal generated by all the n -1 minors of X. Show that the
type of the factor ring K[[XI)]]/I(X) is n-l.
4.2. ZERO-DIMENSIONAL GORENSTEIN RINGS
We recall that a zero-dimensional local ring is Gorenstein if and only if the
zero ideal is irreducible.
Theorem 4.2.1
Let (R, m) be a local ring. The following properties are equivalent:
(i) the ring R is a zero-dimensional Gorenstein ring,
(ii) the ring R is an injective R-module,
(iii) dimR = 0 and the mapping IH> 0:/ between the ideals of R,
sends finite intersections of ideals to their sums (the converse is always
true),
(iv) 0:(0:1) = I for any ideal Iof R.

IV]
Zero-dimensional Gorenstein Rings
143
Proof
(i) => (ii) For the sake of brevity let us denote A* = HomR(A, R),
where A is an R-module. We shall first show that I(A*)  leA) for every
R-module A of finite length. We employ induction on the length of A.
If leA) = J, then A  RIm, and since O:m  HomR(R/m, R) = (R/m)*,
we obtain i«Rlm)*) = 1 from the assumption and Lemma 4.1.1.
Suppose that I(A*)  I(A) for all R-modules of length < n, and con-
sider an R-module C of length n. There exists an exact sequence
o  B  C  D -. 0, (2)
in which the modules Band D have length < n. Thus 1(!J*)  I(B), I(D*)
 I(D). By left exactness of the functor HomR(-, R), the sequence (2)
induces an exact sequence 0 -+ D* -+ C* -. B*, from which, by Theorem
1.3.7 in [B] we get
l(C*)  /(B*)+I(D*)  I(B)+/(D) = I(C).
Consider now the exact sequence
o  (RIm).  R* -+ m* -+ Extj(Rlm, R)  0
induced by the short exact sequence 0 -+ m -. R -+ RIm -. O. In view of
/(m*)  l(m), using the additivity of the function I, we find
I(Exti(Rlm, R») = 1«R/t1t)*)+I(m*)-I(R*)
= I(Rlm)+I(m*)-/(R) = I(m*)-l(m)  0,
and consequently Extj,(R/m, R) = O. From this, by induction on the
length of the module A we deduce that ExtA(A, R) = 0 for every finitely
generated R-module A. This is equivalent to the ring R being injective
(see Theorem A.4.3).
(ii)  (iii) Consider two ideals 11' 1 2 of R, and the exact sequence of
R-modules
o  R/(/ 1 nI 2 )  R/l1$R/I2  RI(/ 1 +1 2 )  0,
where rp(x+I 1 nI 2 ) = (x+lt, -X+/2)' 1p(Xl +1 1 , X2 +1 2 ) = (Xl + X2)+
+ (II + 1 2 ), Since R is injective, the induced sequence
'P.
o -+ HomR(RI(/ t +/2)' R)  Hom R (RlI 1 , R)HomR(RI12' R)
tp*
-+ HomR(R/(/ 1 nI 2 ), R)  0
is also exact. Since HomR(R/l, R)  O:/for an arbitrary ideal I, the above
sequence can be rewritten in the form
'P* ,.
o -. 0:(1 1 +1 2 ) -+ (0:l t )E9(0:12) -+ 0:(/ 1 nI 2 ) -+ 0,
where V'*(x) = (x, x), f{J*(x, y) = x- y. On the other hand, we also have
the exact sequence
(I {J
o  (0:/ 1 )n(0:1 2 ) -+ (0: 1 1 )e (0: 1 2 )  (0:1 1 )+(0:1 2 )  0,

144
Gorenstein Rings
[Ch.
where oc(x) = (x, x), P(x, y) = x - y. One has trivially 0 :(/ 1 + 1 2 ) = (0: 1 1 )f1
(1(0:1 2 ), and hence also 0:(/ 1 n/ 2) = (0:/ 1 )+ (0:1 2 ),
It remains to prove that dimR = O. Suppose on the contrary that
dimR > 0, and note that, under this assumption, there exists a prime
ideal P :f.: m satisfying the condition HomR(RIP, R) ::f: O. Indeed, the set
Ass(R) of associated prime ideals of the zero ideal is non-empty (see
Example 2.4.3 in [B]). If P E Ass(R), then there exists an injection RIP -+ R.
When P :f.: m, our statement is proved; when P = m, then from the assump-
tion dooR> 0, it follows that there exists a prime ideal Q * m. Thus we
have a non-zero homomorphism R/Q -+ Rim -+- R.
Next we show that the injectivity of R yields HomR(R/P, R) = 0 for
any prime ideal P :f.: m. This contradiction wilt prove that dimR = O.
Since P  m, there exists x , P, X Em. It follows, from the injectivity of
R, that every diagram
 RIP
! //..//
R
x
.. RIP
o
may be completed to a commutative one. This means that xHom(RIP, R)
= Hom(RIP, R), and, by the Nakayama lemma, Hom(R/P, R) = O.
(iii) => (i) Assume that the zero ideal is reducible, i.e., there exist
non-zero ideals 1 1 , 1 2 having 0 as their intersection. On the one hand,
0:(1 1 nI 2 ) = R, on the other, (from (iii», we get 0:(11n/2) = (0:/ 1 )+
+ (0: 1 2 ) c m +m = m. This contradiction establishes the irreducibility
of the zero ideal.
(ii) + (iii) => (iv) First consider 1 = (x), a principal ideal. If y e (0 :(0: I) ),
then the diagram
· RI(O:!)'
%
o
... R
/'
./
,,;'
./
/
R/
.y
can be completed to a commutative one, R being injective. This shows that
there exists Z E R such that y = xz, whence Y E I.
To prove (iv) for any ideal, we proceed by induction on the number
of generators of an ideal. When J = 1+ (x), and the equality (iv) is valid
for I, then, applying the property (iii) once more, we obtain
0:(0:1) = (0:0: (1+ (x»)) = 0:[(0:1)(") (O:(x»)]
= (O:(O:1))+(O:(O:(x))) = 1+(x) = J.

IV]
Zero-oonensional Gorenstein Rings
145
(iv)  (iii) The inclusion 0:(/1n/2) ::> (0:/ 1 )+(0:1 2 ) always holds.
Applying to both sides the mapping 1 H> 0: 1 and using formula (iv), we get
1 1 n1 2 c 0:«0:/ 1 )+(0:/ 2 ») = [0:(0:/ 1 )](")[0:(0:1 2 )] = 1 1 n1 2 .
The first inclusion is actually an equality, and the required equality follows
by applying (iv) again.
In order to show that dimR = 0, consider the decreasing sequence
m :) m 2 :) ... It induces an increasing sequence of annihilators O:m
c 0:m 2 c ... Since the ring R is Noetherian, O:m" = 0:m s + 1 for some s.
Using (iv) , we obtain the equality m' = m S + 1 , which, by the Nakayama
lemma, implies m' = o. Thus m is the only prime ideal of R, and dimR = O.
o
Corollary 4.2.2
Let R be a zero-dimensional Gorenstein ring and Q an ideal of R. The
following properties are equivalent:
(i) Q is in irreducible ideal,
(ii) 0: Q is a principal ideal,
(Hi) Q = O:(x) for some x e R, x :f.: 0,
(iv) O:Q  R/Q, or equivalently HomR(R/Q, R)  R/Q.
Proof
(i)  (ii) We assume Q to be irreducible and let O:Q = (t 1 , ..., t,,).
We shall show that if Q, = O:(t,), i = 1, ..., p, then Q = Qln ... nQ,.
Obviously, Q c Ql() ... nQ,,; on the other hand, if A E Qln ... nQ"
then At, = 0 for all i, i.e. A E O:(O:Q) = Q (the last equality by Theorem
4.2.1). It follows, from the irreducibility of Q, that Q = Q" = 0: (t,,) for
some k. Using again Theorem 4.2.1, we see from (iv) that O:Q = (t,,) ,
thus O:Q is a principal ideal.
To prove the implication (ii) => (i), we assume that 0: Q = (t) and
Q = Ql n Q2 for some ideals Ql, Q2. Since (t) = O:Q . (0:Ql)+(0:Q2)
by Theorem 4.2.1 (iii), there exist elements tiE 0: Q 1 , t 2 EO: Q2 such
that t = t 1 + t 2 . At the same time t 1 = ,1,1 t, t 2 = A2 t for some AI' A2 E R.
Therefore 1- Al - A2 EO :(t) = Q, and since Q c m, either ,1,1 or ,1,2 must
be invertible. Accordingly either (t) = O:Ql, or (t) = O:Q2. But this
means, by Theorem 4.2.1, that either Q = Ql or Q = Q2.
The property (ii) is equivalent to (iii) in view of Theorem 4.2.1 (iv).
(ii)  (iv) If 0 : Q = (x) is a principal ideal, then the kernel of the
homomorphism R -+ O:Q which maps 1 to x, is equal to O:(x). But
O:(x) = Q, and O:Q  R/Q.
(iv)  (ii) If 0: Q  RIQ, then clearly 0: Q is a principal ideal. 0
As an immediate consequence of Theorem 4.2.1 and Corollary 4.2.2
we obtain

146
Gorenstein Rings
[Ch.
Corollary 4.2.3
Let (R, m) be a local ring and Q an irreducible m-primary ideal. Denote
by {!jJ the family of all ideals of R which contain Q. Then:
(i) the mapping I  Q: I sends finite intersections of ideals of f} into
their sums.
(ii) Q:(Q:I) = [for any IE,
(iii) an ideal I of &J is irreducible if and only jf I = Q: (x) for some x E R.
Exercises
1. Prove that if dirnR = 0, then I(A*)  /«Rlm)*) /(A) for every finitely generated
R-module A.
2. Prove that a zero-dimensional local ring is Gorenstein jf and only if /(M) = I(M *)
for every finitely generated R-n10dule M.
3. Prove that a zero-dimensional local ring R is a Gorenstein ring if and only if
(l)+/(O:l) = l(R) for every ideal 1 of R.
4. Let (R, m) be a Gorenstein ring and let XI, ..., Xd be a maximal regular sequence
of R contained in Ut. Prove that Q is an irreducible m-primary ideal precisely when there
exists a positive integer s and an element x e R such that Q = (x1, ..., x) : (x).
s. Give examples of zero-dimensional Gorenstein rings by making use of Exercise
4 for the formal power series ring K [(Xl' ..., X,]].
6. Let dimR = 0 and i  1. Prove that Ext(R/m, R) = 0 implies Ext+l(Rlm, R)
= o.
7. Prove that every zero-dimensional local ring is a homomorphic image of a zero-
dimensional Gorensteil1 ring. [Apply a generalization of Corollary 2.3.15, see Remark
2.3.16.]
4.3 HOMOLOGICAL CHARACTERIZATION AND PROPERTIES
OF ARBITRARY GORENSTEIN RINGS
In this section we present characterization of local Gorenstein rings of
arbitrary dimension. This is a continuation and development of the results
derived in Section 4.2.
We shall prove the fundamental Theorem 4.3.5 by induction on dimen.
sion using the connection which exists between the depth of a maximal
ideal and the injective dimension of a finitely generated R-module (Corollary
4.3.2), and a relation between the injective dien1ensions of the modules A
and A/xA, where x is a non-zero dvisor on A (Lemma 4.3.4). We shall
prove these auxiliary facts before formulating the main theorem.
Lemma 4.3.1
Let (R, m) be a local ring and A a finitely generated R-module such that
its injective dimension, id R A, is :finite. Then for any finitely generated
R-module M
idR(A) = depth(m; M)+q(M),

IV]
Homological Characterization of Gorenstein Rings
147
where q(M) is th largest non-negative integer t such that Extk(M, A) #= 0
(idR(A) < 00 implies that q(M) < (0).
Proof
We proceed by induction on s = depth(nt; M). If s = 0, then there exists
an exact sequence
o -+ RIm -+ M -+ N -+ O. (3)
If n = idR(A), then Ext(Rlt1t, A) i= 0 by Theorem A.S.7. On the other
hand, the sequence (3) induces an exact sequence
Ext(M, A) -+ Ext(Rlnt, A) --. 0,
since Ext+l(N, A) = O. Therefore Ext(M, A) i= 0, and q(M) = n = idR(A).
Let s = depth(m; M) > 0, choose x em"'3(M), and set M' = MlxM.
Since depth(m; M') = s-1 (see Lemma 3.1.10), it is sufficient to establish
the equality q(M') = q(M) + ]. We shall derive this by using the sequence
x
o -+ M --+ M  M'  0 and the induced homology sequence Ext(M, A)
-+ EX+l(M', A)  O. Indeed, ExtR(M, A) =F 0 implies Ext+l(M', A) i= O.
Since higher functors Ext vanish this yields the desired equality. 0
Applying Lemma 4.3.1 to R, we obtain
Corollary 4.3.2
Under the assumptions of the previous lemlna, idR(A) = depth(ln) for any
finitely generated R-module A of finite injective dimension.
The next lemma will enable us to prove Theorem 4.3.5 inductively.
Lemma 4.3.3
Let A be a module over the ring R and let x be an element of the ideal m
such that x  3(R), x -t 3(A). If R' = RI(x), A' = AlxA, then for an arbit-
rary R'.module B
Ext+l(B, A)  Ext,(B, A') for n == 0, t, .11
Proof
We use induction on n. First, let 11 = 0 and consider the exact sequence
x I
o  A  A -+ A' -+ 0 and the associated long exact sequence
o  HomR(B, A)  HomR(B, A) -+ HomR(B, A')
-+ Exti(B, A)  Extj(B, A).
Because x annihilates B, HomR(B, A) = 0 and HomR(B, A')  Extj(B, A).
On the other hand, clearly HomR(B, A')  HomR,(B, A'), and finally
ExtA(B, A)  HomR,(B, A') = Ext(B, A'), this isomorphism being natu-
ral with respect to B.

148
Gorenstein Rings
[Ch.
Before we go to the case n = 1, let us note that Extk(B, A) = 0 if B
is free over R' and p  2. In fact, in this case there exists a free module
x
E and an exact sequence of R-modules 0  E  E  B  O. Thus the
sequence Extk-1(E, A)  Ext(B, A)  Ext(E, A) is exact and we get
Ext};(B, A) = 0 whenever p-l  I.
We are now in a position to discuss the case n = I. Consider an exact
sequence of R'-modules
OLFB 
where F is free over R ' . Using the equality proved above, we obtain the
fol1owing commutative diagram with exact rows:
HomR,(F, A')
. HomR'(L, A')
r
.... Exth,(B, A')
 0
r
Ext1 (F, A)
 Ext1(L, A)
. Ext (B, A)
.- 0
the vertical mappings being the isomorphisms which we mentioned in the
case n = O. Accordingly Ext(B, A)  Exti/(B, A').
Finally, let n  2. Applying the sequence (4) once more, we get iso-
morphisms
Exttl(B, A')  Ext,(L, A') for p  1,
and
Ext+2(B, A)  Ext+l(L, A) for ]);;:: 1.
We obtain the last isomorphism from the previously proved equalities
Ext(F, A) = 0 for r  2. By the inductive hypothesis, the right-hand
sides of the above formulae are isomorphic, which implies the same for
the left-hand sides and the lemma follows. D
With the aid of Theorem A.517 and Lemma 4.3.3 we shall prove the
last lemma we need.
Lemma 4.3.4
Let (R, m) be a local ring, A a finitely generated R-module, and x an element
of m, x f 3(R), x, 3(A). Denote further R' = R/(x), A' = A/xA. Then
idR,(A') = id R (A)-I.
Proof
It follows from Lemma 4.3.3, that idR,(A')  id R (A)-I. If idR(A) = 00,
then, by Theorem A.5.7, for each number k there exists p ;>.: k such that
Ext(R/m, A) "# O. Put m' = m/(x). Applying Lemma 4.3.3 and taking

IV]
Homological Characterization of Gorenstein Rings
149
into consideration the isomorphisln RIm  R'/m', we conclude that
Exth-;l(R'/m', A') i= 0, too, which means idR,(A') = 00.
Suppose that idR(A) = 8 < 00, so we have Ex(Rlm, A) i= O. Using
again Lemma 4.3.3, we get Ext-;l(R' 1m', A') :/: 0, and consequently
idR,(A')  8-1 = id R (A)-l. 0
Theorem 4.3.5
Let (R, m) be a local ring of dimension d. The foIlowing properties are
equivalent:
(i) R is a Gorenstein ring,
(ii) id R (R) < 00,
(iii) idR(R) = d,
. f { o for i :f= d,
(IV) ExtR(Rlm, R) = R/ .c · - d
m lor , - ,
(v) R is a Cohen-Macaulay ring and Ext(Rlm, R)  Rim.
Proof
We shall use induction on d.
If d = 0, then the equivalence of all the conditions except (ii) follows
from Theorem 4.2.1. The implication (Hi) => (ii) is trivial, so there remains
the converse implication (ii) => (iii). By Lemma 4.3.2, we have idR(R)
= depth (m). But depth(m)  dimR = 0, which gives idR(R) = O.
In the sequel, we assume that d = dimR > 0, and put R' = RI(x),
where x em is a non-zero-divisor in R.
(i) => (ii) Since R is a Cohen-Macaulay ring, then depth(m) = d > 0, and
consequently there exists x em"'-3(R). In view of Corollary 4.1.8, we see that
R' is also a Gorenstein ring of dimension d-l. By the inductive hypothesis
idR,(R') < 00, which together with Lemma 4.3.4 yields idR(R) = idR,(R') + 1.
(ii) => (Hi) If idR(R) = 0, then by the inductive hypothesis, d = dimR
= 0, contrary to the assumption. Therefore, in accordance witll Corollary
4.3.2, one has depth(m) = idR(R) > O. Let x em"'-3(R); from Lemma 4.3.4
we deduce that idR,(R') < 00, and hence idR,(R') = d-l by the inductive
hypothesis. Applying Lemma 4.3.4 again, we obtain idR(R) = d.
(iii) => (iv) By Corollary 4.3.2 and (Hi) we have depth(m) = idR(R)
= d > 0 so there exists X E m"3(R). From Lemma 4.3.4 it follows that
idR,(R') = d -1.
By the inductive hypothesis, we get
Ext,(R' 1m', R) = {, 1m'
for
for
i#d-l,
i = d-l.
But R' Int'  Rim, and therefore the desired result follows by Lemma
4.3.3.
(iv) => (v) From the assumption and the characterization of depth in

150
Gorenstein Rings
[ChI
Corollary 3.1.7, it follows that depth(m) = d. Thus R is a Cohen-Macaulay
ring by Theorem 3.3.2.
(v) => (i) Let Xl' ..., XII be any set of parameters of the ring R. Since
R is a Cohen-Macaulay ring, Theorem 3.3.5 shows that the sequence
Xl' ..., Xtl is regular. Repeated use of Lemma 4.3.3 and the assumption
yields
RIm  Ext(Rlm, R)  HomR(Rlm, Rj(Xl' ..., Xci»
 «Xl' 01" x d ):m)/(Xl, 01" xiJ.
This Ineans, by Lemma 4.1.1, that the ideal (Xl' ..., X4) is irreducible,
and hence, finally, that R is a Gorenstein ring. 0
We state the basic properties of local Gorenstein rings which follo,v frotn
the theorem above.
The first one complements Corollary 4.1.8.
Corollary 4.3.6
If x E m'3(R) and RI(x) is Gorenstein, then so is R.
The assertion follows, for example, from the part (iii) of the theorem
and Lemma 4.3.4.
Corollary 4.3.7
The format power series ring R [[X]] is Gorenstein if al\d only if R is Go..
renstein.
Corollary 4.3.8
If R is Gorenstein, then any localization R p with respect to a prin1e ideal
P is also Gorenstein.
Proof
By Theorem A.4.5, we have idRp(R p )  idR(R). Thus the corollary follows
from the part (ii) of the theorem. 0
Corollary 4.3.9
A local ring (R,1n) is Gorenstein if and only if its completion (R, tnR)
is Gorenstein.
Proof
We shall show that for a finitely generated R-module M,
i "" . "
Extn(M, R)  Extk(M, R)&JRR
for all i.
,..
Let F be a free resolution of Mover R. Since .R is flat over R (Corollary
" A A
2.6.22 in [B]), F(8)RR is a free resolution of MQ?)RR = Mover R. More-
A A A
over, HomR(}'@RR, R)  HomR(F, R)fi!)RR, whence
. A A ,.. A '"
Ext(M, R) = H'(HomR(F@RR, R) = li'(HomR(F, R)&;JRR)
A A
 Hi (HomR(F, R»)@RR  Extk(M, R)@RR.

IV]
Gorenstein Ideals
IS1
The last but one isomorphism follows from the exactness of the rune-
A
tor-@RR.
We apply the above formula to M = Rim. Because of the isomorphism
A A A
RlmR  Rim (see Corollary 2.6.21 in [B]) and dimR = dimR our asser-
tion follows by Theorem 4.3.5 (iv).
Exercises
1. Let R be a Noetherian ring and A a module over R. Show that idR(A)  "if and
only if Ext(RI P, A) = 0 for each s > n and for all prime ideals P of R.
2. Let P be a prime ideal of a Noetherian ring R, x; P and x e J(R). Prove that
for a finitely generated R-module A, if ExtL(R/P, A) :;: 0, then Ext+t(R/(P, x), A) ,& o.
3. Prove Lemma 4.3.4 for an arbitrary Noetherian ring R, under the assumption
that x belongs to the Jacobson radical J(R) of R. [Use Exercise 2.]
4. Show that for any local ring (R, m), there exists a non-negative integer s such
that Ext(Rlm, R) :f:: o.
S. Prove that a d-dimensional local ring (R, m) is Gorenstein if and only
if Ext1(R/m, R) = 0 for i :/: d.
6. Let P = Po c ... C Ps-t C P, = m be a saturated chain of prime ideals of
a local ring (R, m). Show that if Ext1+I(R/m, R) = 0, then Extp(Rp/PRp, R p ) = O.
7. Prove that if (R, m) is a d-dimensionallocal ring, then Ext(Rlm, R) :;: O.
8. Making use of Exercise 7 prove that the following conditions are equivalent:
(i) idR(R) < 00,
(ii) idR(R) = dim R .
9. Prove that (R, tn) is a Gorenstein ring if and only if
, ( { 0 for i < dimR,
Ext R Rim, R) =
RIm for i == dimR.
10. Prove that (R, m) is a Gorenstein ring if there exists ; > dimR such that
Ext1(Rlm, R) = O.
11. Prove that if in a local ring (R, m) the equality v(m) = 1 + depth(m) holds
(v(m) = the minimal number of generators of m), then R is a Gorenstein ring.
12. Prove that, for a local ring (R, m), if idR(m) < 00, then R is a regular ring.
13. Let R be a local Gorenstein ring and A finitely generated R-module. Prove that
pdR(A) < 00 if and only if idR(A) < 00.
4.4 GORENSTEIN IDEALS
We shall be concerned in this section with the description of those ideals
I in a Gorenstein ring R, for which the factor ring RII is also Gorenstein.
These ideals will be called Gorenstein ideals. We have solved similar prob..
lelD for Cohen-Macaulay rings in Chapter III. We already know some
examples of Gorenstein ideals because, by Corollary 4.1.8, regular ideals
are Gorenstein.
We shall provide a characterization of Gorenstein ideals of depth 1 and
2 as corollaries of the main Theorem 4.4.4. Another interesting corollary
is Theorem 4.4.7, due to M. P. Murthy.

152
Gorenstein Rings
[ChI
We begin the proof of the fundamental theorem with two lemmas.
The first one is a multidimensional version of Corollary 4.2.2.
Lemma 4.4.1
Let (R, m) be a Gorenstein ring of dimension d and let Q be an m-primary
ideal. Then Q is irreducible if and only if Ext(R/Q, R)  R/Q.
Proof
The ring R is a Cohen-Macaulay ring, hence depth (Q) = ht(Q) = d.
Therefore there exists a regular sequence Xl' ..., x" contained in Q. By
Theorem 4.3.3, we have
Ext(R/Q, R)  HomRI(Xl'....Xd) (R/Q, R/(x 1 , ..., x d »).
Furthermore, it follows from Coronary 4.1.8 that Rj(Xl' ..., XII) is a Goren-
stein ring of dimension zero. Applying Corollary 4.2.2 to the ring
R/(x 1, ..., x,,), we get the assertion. 0
An immediate consequence of the lemma is
Corollary 4.4.2
Under the assumptions of Lemma 4.4.1, the ring R/Q is Gorenstein if and
only if Ext(R/Q, R)  R/Q.
In the following, we fix a local Gorenstein ring (R, m) of dimension d. If
1 is an ideal. of R such that dimR/I = n, then by Property 10, Section 3.5,
we have depth(l) = d-n. We set T = R/l and D(T) = Ext-n(T, R).
The number n depends, of course, on T and hence also on I.
Lemma 4.4.3
Let the notation be as above. We assume that T is a Cohen-Macaulay ring
and that pdR(T) < 00. Then:
(i) if an element y e m is a non-zero-divisor on the R-module T, then
it is also a non-zero-divisor on D(T) and fJ(T)/yD(T)  D(T/yT),
(ii) D(T)  T if and only if D(T/yT)  T/yT for some Y E m
which is a non-zero-divisor on the R-module T.
Proof
(i) Since T is a Cohen-Macaulay ring and pdR(T) < co, Theore111
3.5.11 applies, and shows that pdR(T) = depth(I) = d-n. Therefore
Ext-n+l(T, R) = O. Moreover, under the assumption that y E m"'3(T), we
have dim(T/yT)=dimT-l=n-l. Since T/yTR/(I,y), we have
depth(I, y) = d-n+ 1 by Property 10, Section 3.5. Hence
Ext-n(R/(I, y), R) = Ext-n(T/yT, R) = o.
As these modules vanish, there results an exact sequence
y
o -+ Ext-Jl(T, R)  Ext-IJ(T, R)  Ext-"+l(TlyT, R)  0 (5)

IV]
Gorenstein Ideals
153
induced by the sequence 0 -+ T  T  T/yT -+ O. Because we have
Ext"+l(T/yT, R) = (J(T/yT),
the exactness of the sequence (5) implies the existence of the desired
isomorphism and shows that y t$ a (D(T».
(ii) By Lemma 3.5.4, T-module Q(T) is free if and only if the T/YT-mod-
ule !J(T)/y!J(T) is free. Furthermore, both these modules are of the same
rank provided that they are free over the respective rings. Applying the
isomorphism (i) we obtain the desired equivalence. 0
Theorem 4.4.4
Let R be a local Gorenstein ring and I an ideal in R of depth p and of
finite projective dimension. Then I is a Gorenstein ideal (i.e. R/I is Goren-
stein ring) if and only if:
(i) R/I is a Cohen-Macaulay ring, and
(ii) Ext(RII, R)  R/I.
Proof
We preserve the previous notation T = R/I, d = dimR, n = dim T. Thus
p = d-n and Q(T) = Extk(T, R).
It is enough to show that, under the assumption that T is a Cohen-
Macaulay ring, the ring T is a Gorenstein ring precisely when Q(T)  T.
We shall prove this equivalence by induction on n. If n = 0, then I is
an m-primary ideal, whence p = d, and the desired conclusion is just the
content of Corollary 4.4.2.
Assume that n > O. T is a Cohen-Macaulay ring of dimension > 0, thus
there exists y ,m "3(T). Corollaries 4.1.8, 4.3.6, the inductive hypothesis,
and Lemma 4.4.3 yield the equivalences: T is a Gorenstein ring  TlyT
is a Gorenstein ring  !J(T/yT)  T/yT <=> fJ(T)  T. 0
Corollary 4.4.5
Under the assumptions of Theorem 4.4.4, the ideal I is a Gorenstein ideal
if and only if the following conditions are satisfied:
(i) a minimal free R-resolution F of the module R/I has the form
d"
F: 0  F"  Fp-l  ...  PI -+ Fo,
where Fo  Fp  R,
(ii) the dual complex F* = HomR(F, R) is a minimal free resolution
of Coker(d:)  RII.
Proof
We shall show the above conditions (i), (ii) to be equivalent to the prop-
erties (i), (ii) in Theorem 4.4.4.
Suppose that T = R/I is a Cohen-Macaulay ring and Extl(T, R)  T.
According to Theorem 3.5.11 we have pdR(T) = depth(/) c:: p, and hence

154
Gorenstein Rings
[Clt.
T has a minimal free resolution F of length p in which, of course, Fo = R
(Section A.S, Exercise 7).
Let us consider the dual complex F*  HomR(F, R). It follows immedi-
ately from the definition that Ext,k(T, R) = H'(F*), whence the equality
depth(I) = p and the condition (ii) in Theorem 4.4.4 imply
{ 0 when
H'(F*) = T when
i < p,
i = p.
Therefore the sequence
d*
0-+ F3 -+ Ft -+ ... -+ F't-t -: F3 -+ T-+ 0
is exact and F* is a minimal resolution, because it is the dual of a minimal-
one. Moreover, the isomorphism Fi/Im(d;)  T and the inclusion
Im(d:) c:: mF: make the composition T  F:/Im(d;) -+ F:/mF: an epi-
morphism, and so F; /mF: is a space of dimension lover RIm. This shows,
by Lemma A.S.I, that F,  R.
On the other hand, if the properties (i), (ii) in Corollary 4.4.5 are satis-
fied, then pdR(T) = p by Corollary 3.5.7. From this and from Theorem
3.5.11 we conclude that T is a Cohen-Macaulay ring. Moreover,
Ext.&(T, R) = Coker(d:)  T. 0
Using the last corollary and the results of Section 3.6, we can provide
the description of Gorenstein ideals of depth I and 2. .
Theorem 4.4.6
Let R be a local Gorenstein ring and I an ideal of depth 1 or 2, and of
finite projective dimension. Then I is a Gorenstein ideal if and only if it
is regular.
Proof
Every regular ideal is obviously a Gorenstein ideal. To prove the converse
assume at first that depth(I) = 1. By Corollary 4.4.5, the ideal I is a prin-
cipal ideal generated by a regular element.
Now let depth(I) = 2. Again, in view of Corollary 4.4.5, the R-module
R/I has a resolution of the form 0 -+ R -+ RS -+ R -» R/I -» o. It follows
from the proof of Theorem 3.6.1, that s = 2, and consequently the ideal
I can be generated by two elements. Applying Theorem 3.2.3, we deduce
that I is regular. 0
Theorem 4.4.7 (Murthy)
Let R be a local Gorenstein ring and I eRa prime ideal of :finite projective
dimension. If RII is a Cohen-Macaulay, unique factorization domain,
then RII is a Gorenstein ring.

IV]
Notes and References
155
Proof
We retain the previous notation, T = Ril. By Theorem 4.4.4, it is sufficient
to show that Q(T)  T. If follows, from Lemma 4.4.3, that every T-regular
sequence is also .Q(1)-regular. Hence we conclude that .Q(T) is a torsion-
free R-module. Denote by L the field of fractions of the ring T.
We shall show that Q(1)@TL  L. The module Q(T)@TL is iso-
morphic to the localization Q(1)1 of the R-module Q(1). From Theorem
A.3.6 we get
Q(T)I = (Ext(RII, R))I  Ext,«R/I)I' R 1 )
 Extk,(RIIIRI, R 1 ).
Applying Theorem 4.4.4 to the ring Rl and to the ideal IR 1 (or by direct
computation using the regularity of R 1 ) we obtain {J(T)1  RI/lRr  L.
It immediately follows, from the formula obtained, that Q(1) is iso-
morphic to some ideal J of the ring T. By Exercise 15, Section 3.2, we may
assume that the depth of J is at least 2. Let x, y E J constitute a J-regular
sequence and t E T. We then have xt E J and y(xt+xJ) = 0 in Jlxl. We
deduce from the I-regularity of the sequence x, y that xt E xl, and therefore
tEl. Thus T = J  Q(T), and the proof is complete. 0
NOTES AND REFERENCES
The rings considered in this chapter were named after D. Gorenstein. His paper [9]
is concerned with plane algebraic curves and it would be difficult to recognize his results
in our presentation. It is H. Bass who should be given full credit for developing the
theory of Gorenstein rings. His basic paper [3] contains the various characterizations,
including that as rings of finite injective dimension; also presents an historical outline
of the subject. Our approach is based partly on lecture notes from Greco's seminar [E],
and presents in an elementary way the most important facts of the theory without use of
structure theorems of injective modules and duaJity theorems. Another approach based
on these topics can be found in [W].
Theorem 4.1.4 which allowed us to define the type of a local Cohen-Macaulay ring
comes from Northcott [25]. Local zero-dimensional rings whose zero ideal is irreducible
(i.e. zero-dimensional Gorenstein rings) were already studied in 1934 by Grabner [10].
He proved there the implications (i) => (iii) and (i) => (iv) of Theorem 4.2.1, and Corollary
4.2.2. Theorem 4.4.7 comes from Murthy's paper (21].
FinaJly, we mention a recent result of P. Roberts [33] which weakens condition (iv)
of Theorem 4.3.5. Roberts proved that (R, m) is a Gorenstein ring if and only if
Ext(Rlnt, R)  Rlnt where d = dimR.

Appendix
Homological Background
The purpose of this Appendix is to present in a concise manner those
concepts and theorems of homological algebra which are used in this book.
We shall be mainJy concerned with the description of injective, projective
and global dimension in terms of the functors Ext R and Tor:. For the
readers convenience, we also provide an outline of the construction of
derived functors, of which the construction of the functors Ext and Tor
is a particular case. For the proofs of the theorems which are not proved
here, the reader is referred to [S] and [U].
A.l PROJECTIVE AND INJECTIVE DIMENSION
It is known that a homomorphism of a free module F to an arbitrary mod-
ule is determined uniquely by its values on a basis of F. Consequently
every diagram of the form
F
1
M
.. N .-
 o
can be completed to a commutative diagram
.,F
/////'/ !

M .- N . 0
for any epimorphislu M  N.
This last property is also valid for direct summands of ftee modules.
It turns out that, conversely, every module possessing this completing
property for diagrams is a direct summand of a free module. Such modules
are called projective. Notice that a module F is projective precisely when
the functor HomR(F, -) is exact (the functor HomR(M, -) is left exact
for an arbitrary module M).

158
Homological Background
With the aid of projective modules we can associate with any module
an invariant called projective dimension. In general not every R-module
is projective (this depends on the ring R), but for any R-module M, there
exists an epimorphism Fo -)- M of a projective (even free) R-module Fo.
Iterating this procedure, we construct an exact sequence
... -+ F 2  F 1 -+ Fo -+ M -+ 0,
where the Fn are projective modules.
This leads to the following definition.
Definition A.I.1
A projective resultion of an R-module M is a sequence
... -)- Fn -+ Fn-l -. ... -+ F 1 -+ Fo -+ 0
of projective R-modules and their homomorphisms, together with a hOlno-
morphism e: Fo -+ M, such that the sequence
B
...  Fn ... F,'-1 -+ ... -+ Fl -+ Fo --+ M -+ 0
(1)
is exact.
Occasionally we refer simply to the sequence (1) as a projective resol-
ution of M.
Therefore every module has a projective resolution. In general there
are many of them because already at the first step of the construction \ve
can freely choose the module Fo and the epimorphism Fo -)- M.
Definition A.l.2
H an R-module M has a projective resolution of the form
o -)- Fk -)- ... -+ Fo  M -+ 0,
(2)
where Fk #- 0 and there is no shorter projective resolution, then the ntunber
k is called the projective dimension of the module M, and it is denoted
pdR(M) (or pd(M) if it is clear which ring R is meant). If the module M
has no finite resolution of the form (2), then we put pdR(M) = 00; con-
ventionally we define pd(O) = -1.
As a rule, it is not easy to compute the projective dimension of a module
using only the above definition. In Section A.3 below we shall give an
outline of the theory of the homological functors Torn and Ext" by means
of which one can prove many properties concerning projective dimension
and its relations to other invariants of modules. A sketch of the theory of
derived functors, of which the above-mentioned sequences of functors Ext,
Tor are particular cases, will be discussed in Section A.2.
Before we take up the task, let us look at the concept dual to projec-
tivity, which leads to yet another invariant of a module.

Projective and Injective Dimension
159
A module E is said to be injective if every diagram of the form
N
"" M
!
E
can be completed to a commutative diagram
N -. M
1 //////
E
for any monomorphism N -+ M. In other words, the module E is injective
precisely when the functor Hom(-, E) is exact (the functor HomR(-, M)
is left exact for every module M).
It turns out that the range of the mOllomorphisms N -)- M in the above
definition may be restricted. We shall formulate this fact as a theorem,
whose proof the reader can find in [U], p. 31.
Theorem A.I.3
An R-module E is injective if and only if for any ideal I of R, every homo-
morphism I -+ E can be extended to a homomorphism R -)- E.
One can prove that every R-module can be embedded in an injective
module (see [U], p. 34 or [8], p. 70) although this is much more difficult
than in the case of the analogous property of projective modules. There-
fore, an arbitrary module has an injective resolution in the sense of the
following definition:
Definition A.l.4
An injective resolution of an R..modulc M is a sequence
0-+ EO -+ El-+... -+ En -). ER+l-+ ...
of injective R-modules and their homomorphisms, together with a homo-
morphism 'f}: M  EO, such that the sequence
'1
o - M  EO - £1  ... -+ E II -+ En+l  ...
j s exact.
DefinitioD A.l.5
If an R-module M has an injective resolution of the form
o -)- M  EO  ... -+ E k -+ 0)
(3)

160
Homological Background
where E k :p 0 and there is 110 shorter injective resolution, then the number
k is called the injective dimension of the module M, and it is denoted
idR(M) (or id(M) if it is clear which ring R is meant). If the module M has
no finite resolution of the form (3), then we put idR(M) = 00; furthern10re
we define id(O) = - 1.
In concluding this section, we recall that an R-module F is flat if a11d
only if the functor-(8) R F is exact. Every projective module is flat.
A.2 DERIVED FUNCTORS
Let R be a ring. We begin with the general properties of hon101ogy.
By a complex of R-modules one means a. sequence consisting of R-tl1od-
ules and their homomorphisms
X: ... -.. X n - 1 .!!.1I-1  X,. d n .. x n + 1 d"+ 1 + X'.+ 2 -+ ...
such that d n d,,-1 = 0 for all n. The module x n is called the component
of degree n, and the homomorphisms d'i are called differentials of the com-
plex X. A complex is said to be projective (free, injective) when all its
components are projective (free, injective) modules. A complex X is finite
if X" :p 0 for a finite number of indices n.
Sometimes we use subscripts instead of superscripts setting I = x-n,
d n = d-n:X" -+ X n - 1 . Therefore a projective resolution of a module is
a projective complex.
By a homomorphism of complexes f: X -+ Y one means a sequence
f = ifn} of homomorphisms of components fn: XII  yn such that the
diagrams
x
I"
.... yn
d fl
apt
xn+l
fit + t
 yn+ t
are commutative for all n, where the an denote differentials of the complex Y.
Complexes of R-modules form, in an obvious way, a category.
The module zn(X) = Ker(d ll ) is said to be the module of n-cycles of
the complex X, and the module BII(X) = Im(d"-l)-the module of n-bound-
aries of the complex X.
The condition d"d n - 1 = 0 implies B"(X) c: Z"(X), and the factor
module
H"(X) = zn(X)jBn(X)
is called the n-tb hOD10logy module of tbe complex X.

Derived Functors
161
A homomorphism of complexes f: X -+ Y induces homomorphisms
of homology modules Hn(f): HIt(X) --.. HIt(Y) defined by the formula
H1I(/) (z+BIt(X» = fn(z)+B"(Y) for Z E Z"(X);
H1I thus becomes a covariant functor.
A sequence of complexes and their homolnorphism 0  X  Y  W
In g"
-)- 0 is called exact if for each n, the sequence 0  X"  y"  w n  0
is exact.
Theorem A.2.1
An exact sequence of complexes
f g
o-+x- Y--..W-+O
induces an exact sequence of homology modules
... --.. Hn-l(W) Hn(x). Hn(fl.  n"(y) JI1f(g) + Hn(w)
d"
 H" + 1 (X) -+ ...,
where the homomorphism n, termed the connecting homomorphism, is
defined in the following manner: if H' E Z"(W), }v = f'l(y) for Y E yn, then
there exists an element X EZ"+l(X), such that f" + 1 (X) = an(y); the formula
n(w+Bn(w)) = x+Bn+l(X)
is a correct definition of the homomorphism n (see [8], p. 60).
Theorem A.2.2
A commutative diagram of complexes, with exact rows
o wq )II X .. Y  w
! ! !
0  r . Y l ... W'
induces commutative diagrams
6"
H"(W)  Hn+l(x)
! f5,n !
H"(W')  Hn+ (X')
for every n (see [S], p. 60).
. 0
v. o
If f: N  N' is a hOlnomorphism of R-modules and if E, E', together
with homomorphism 17: N --.. EO, r/,: N' 4> E'o are injective resolutions

162
Homological Background
of the modules N, N', then a homomorphism of complexes I: E  E'
is called an injective resolution of the homomorphism f provided that the
diagram
'I'J
. EO
.1 10
N
11
1J'
. E 'O
N'
is commutative. A projective resolution of a homomorphism is defined
analogously.
Every homomorphism of modules has an injective and projective resol-
ution. In the case described above we construct a commurative diagram
11  Eo I E i ]IIr ... )I Ert.- i  E"  ...
0., p N
f! fJ' f O ! f1!  f"- i  f"
O .L " N' .. £'0 ,. E' 1  ... . E''!-l  E'"  .. .
using the exactness of the upper row and the injectivity of the modules E'n.
Theorem A.2.3
For any exact sequence of modules
f g
o -+ N' --)- N  N" --)- 0,
there exists an exact sequence of complexes
o  E' .!; E  E" -+ 0
such that E', E, E" are injective resolutions of the modules N', N, Nil
respectively, and f, g are injective resolutions of the homomorphisms f, g
(see [8], pp. 80-84).
An analogous fact is valid for projective resolutions.
We can now start with a presentation of the scheme of construction
for derived functors.
Let Ube an additive covariant functor defined on the category of R-mod-
ules with values in the same category. The functor HomR(M, -) serves
as an example, \vhere M is a fixed R-module.
We define the sequence {R"U}no of right derived functors of the
functor U as follows.
For an R-module N, we choose an injective resolution (E, d) and put
(RnU)(N) = H"(U(E)), 11 = 0, 1, ...,
where U(E) is the complex whose components are the Inodules U(E'l),
and whose differentials are the homomorphisms U(d n ).

Derived Functors
163
For a homomorphisn1 of R-modules f: N --+ N', we choose an injective
resolution I: E --.. E' and define
(RlI U)(f) = Hn (U(f) ), n = 0, 1 , . to ,
where U(f) = {U(fll)}.
One can prove that the modules (R"U)(N) and the homomorphisms
(RnU)(f) do not depend (up to isomorphism) on the choice of the resol-
utions used in their construction (see [8], pp. 101-102), and that they form
a sequence of additive covariant functors.
An exact sequence of R-modules 0 -+ N' --. N --. N" -+ 0 has, by
Theorem A.2.3, an injective resolution 0 --. E' --. E -+ E" -)- O. By applying
the functor U to this resolution we obtain, on account of the injectivity
of the complex E', an exact sequence of complexes
o --. U(E')  U(E) -+ U(E") -. 0
which, by Theorem A.2.1, induces an exact sequence
o -+ ROU(N') -+ ...  R"U(N') -+ RnU(N)
.,n
-)- R"U(N") -+ Rn+l U(N') .-. ...
d" is the respective connecting homomorphism.
Moreover, a commutative diagram of modules, with exact rows,
 M"
 O
o
 M'
- M

 N

o

. N'
... N' ,
 0
has an appropriate injective resolution, from which, by applying the functor
U, we obtain a similar diagram. That diagram induces, by Theorem A.2.2,
a commutative diagram
R"U( M")
!
it
 Rn+IV(M')
J
()FI
R"U(N")
,. R n + 1 U(N')
where n , <5 n denote the respective connecting homomorphisms.
The right derived functors of a contravariant functor and left derived
functors are defined analogously. If U is an (additive) covariant functor,
then the left derived functors {L" U}n>O are defined as above, but ,vith
the aid of projective resolutions. These functors have properties analogous
to those of right derived functors, formulated above.
In the sequel, we shall use the right derived functors of the covariant
functor of the form Homn(M. -t constructed by means of injective resol-

164
Homological Background
utions, and the left derived functors of the contravariant functor of the
form HomR(-, M), and of the covariant functors of the form
M(g)R-, -QSJRM,
constructed by means of projective resolutions.
A.3 FUNCTORS Ext" AND Tor"
Definition A.3.t
Let M, N be R-modules, F a projective resolution of M, and E an injective
resolution of N. Then we have isomorphisms
(RnHomR(-, N)(M) = Hn(HomR(F,N»)  HtI(HomR(F, E)
 HfJ(HomR(M, E» = (RIIHomR(M, -)(N)
(see [8], p. 133; the complex HomR(F, E) is described in [8], pp. 94-97).
This module is denoted by Ext(M, N) and is called the n-th extension
module of the modules M, N, n = 0, 1, ...
Ext1t is a functor in two variables, contravariant in the first variable,
covariant in the second. In the sequel, if there is no confusion about the
ring R, we write Extra instead of Ext.
The basic properties of the extension functors Ext" are stated in the
following theorem.
Theorem A.3.2
Let M, N be R-modules. Then:
(i) we have isomorphisms of functors
ExtO(M, -)  Hom(M, -), ExtO(-, N)  Hom(-, N),
(ii) the module M is projective if and only if Ext! (M, -) = 0,
(iii) the module N is injective if and only if Ext 1 (-, N) = 0,
(iv) an exact sequence of R-modules
o -+ M' --)- M -+ M" -+ 0
induces, for every N, an exact sequence of R-modules
0-+ Hom(M", N) -+ Hom(M, N) -+ Hom(M', N)
dO
-+ Ext 1 (M", N)  ... 4> Extll(M", N) -+ Extn(M, N)
d"
.-. Extrr(M', N) -+ Ext n + 1 (M", N) -+ ...,
(v) an exact sequence of R-modules
o -+ N' --.. N -+ N" --. 0

Functors Ext n and Torn
165
induces an exact sequence of R-modules
o -+ Hom(M, N ' ) -+ Hom(M, N) -+ Hom(M, N")
o
-+ Ext 1 (M, N') -+ ... -+ Extll(M, N') -+ Extn(M, N)
d l
-+ Ext"(M; N")  Ext"+ 1 (M, N') -+ ... ,
(vi) a commutative diagram of modules, with exact rows
0  M'  M  M" . 0
f  g
0  M'  M ... M"  0
induces a commutative diagram
«5"
Extn(M', N) -- :. EKt ft + 1 (M", N)
Ex.tn(f, 1)1 i Ex.t" + l(g, 1)
5' Ext" + 1 ( M il, N)
Extn(M', N) ..
where n, n are the respective connecting homomorphisms,
(vii) a commutative diagram of modules, with exact rows
 N

 0
o
 N'
l
· N"

o
.- N '
 N
 Nil
 0
induces a commutative diagram
<5"
Ext"(M, N")  Ext,H- J(M, N')
Ext"(l,l>! 1 Ext"+l(I,g)

Extft(M, N")  Ext ft + 1 (M, N ')
where ", (511 denote the respective connecting homomorphisms.
Similar facts are valid for the derived functors of tensor product.
Definition A.3.3
Let M, N be R-modules and F, F' projective resolutions of M, N. Then we
have isomorphisms

166
Homological BackgroW1d
(L,I( - (6)N) )(M) = H,,(F(8JN)  H,,(F@F')  H,,(M@F')
= (Ln(M@-»)(N)
(see [S], p. 127; the complex F@F' is described in [8], pp. 94-97). This
module i denoted by Tor:(M, N) and is called the n-th torsion product of
the modules M, N, n = 0, 1, ...
Tor: is a functor in two variables, covariant in each of them. We write
Torn instead of Tor: if there is no confusion about the ring R.
The basic properties of torsion products are contained in the following
theorem.
Theorem A.3.4
Let M, N be R-modules. Then
(i) we have isomorphisms of functors Toro(M, -)  M(i9 -,
Toro(-, N)  - @N,
(ii) Tor,.(M, N)  Torn(N, M),
(iii) the module M is flat if and only if TorI (M, -) = 0; in particular,
if either M or N is a projective module, then Torn(M, -) = 0 for n > 0,
(iv) an exact sequence of R-modules
o  M'  M .-. M" -+ 0
induces an exact sequence of R-modules
... -+ Torn(M', N) -+ Tor,,(M, N)
-+ Tor,.(M", N)  Tor n - 1 (M', N) -+ ...
-+ Torl(M", N)  M'@N -+ M@N -+ M"N -+ 0,
(v) a commutative diagram of modules, with exact rows
0  M'  M  M"  0
'l  l'
0  M' - M  M"  0
inducesa commutative diagram
6,. Torn_.(',.N)
Tor,,(M", N) 
Tor.(g. 1) ! 8" . ! TOf/h.l(f, 1)
" -
Tor,,(M", N)  Tor n -l(M', 'N)
where ", n denote the respective connecting homomorphisms.
As an immediate consequence of Theorems A.3.2 and A.3.4 we obtain

Functors Ext" and Torn
167
Corollary A.3.5
If in an exact sequence of modules
o --» M' -+ F...... M -+ 0
the module F is projective, then the connecting homomorphisrl1s
n+l: TOf n +l(M, N) -+ Torll(M', N),
d n : Extll(M/, N) -)- Ext n + 1 (M, N)
are isomorphisms for n > 0 and for any module N.
If in an exact sequence of modules
o -+ N' -+ E --+- N--+-O
the module E is injective, then the C01111ecting homoJnorphisms
n: Ext"(M, N) 4> Ext ll + 1 (M, N')
are isomorphislns for n > 0 and for any Inodule M.
The above corollary is known as the reduction theorem.
The behaviour of the functors Ext and Tor: under locaJization of the
ring R is described in the following theorem.
Theorem A.3.6
Let S be a multiplicative subset of a ring R.
(i) If M, N are R-modules, then for n  0,
. (Tor:(M, N»)s  Tor::S(M s , N s ),
(ii) If R is a Noetherian ring, M, N are R-modules and M is finitely
generated, then for n  0,
(Ext(M, N»s  Ext1ts(Ms, N s ).
In the proof of the theorem one uses the fact that the localization
functor commutes with the tensor product and the homology functors
([8], p. 171).
Theorem A.3.7
If R is a Noetherian ring and if M, N are finitely generated R-modules,
then:
(i) the module M has a projective resolution whose components are
finitely generated modules,
(ii) the functors Tor:(M, N) and Ext(M, N) are finitely generated
R-modu1es.
Proof
(i) Since R is a Noetherian ring, the kernel of an epimorphism Fo  M
is finitely generated provided that Eo is a finitely generated R-module. It
is therefore possible to construct the desired resolution.

168
Homological Background
(ii) Consider a projective resolution F of the module M, whose compo..
nents are finitely generated R-modules. Then F(g)RN and HomR(F, N)
are complexes with finitely generated components. The assertion (ii) follovls
by the formulae
Tor:(M, N) = Iln(Fr&JRN), Ext(M, lV) = Hn(HomR(F, N)). 0
A.4 INJECTIVE, PROJECTIVE AND GLOBAL DIMENSION IN
TERMS OF THE FUNCTORS Ext AND Tor::
We begin with the characterization (promised in Section A.I) of projective
and injective dimension in terms of the functors Ext R .
Theorem A.4.1
For any R-module M, the following conditions are equivalent:
(i) pdR(M)  n,
(ii) Ext n + 1 (M, -) = 0,
(Hi) if a sequence 0  K -+ F,.-1  ...  Fo 4- M  0 is exact and
the modules Fo, ..., Fn-l are projective, then the module K is projective.
Hence we obtain an equality
pdR(M) = sup{n; Ext(M, -) ¥= O}.
Proof
The implication (i) => (ii) follows directly from the definition. In order
to establish the implication (ii) => (iii), observe that Ext n + 1 (M,-)
 Ext 1 (K, -) by the reduction theorem (Corollary A.3.5). Thus (ii) and
Theorem A.3.2 (ii) together imply that K is projective. The implication
(iii) => (i) is trivial. 0
From Theorem A,4.I we get an interesting property of the functors
Ext: Extn(M, -) = 0 implies Exti(M, -) = 0 for i > 11.
One similarly proves the theorem characterizing injective dimension.
Theorem A.4.2
For any R-module M, the following conditions are equivalent:
(i) idR(N)  n,
(ii) Extn+ 1 (-, N) = 0,
(iii) if a sequence 0  N  EO  ,..  En-l  L  0 is exact and the
modules EO, ..., En-l are injective, then the module L is injective,
Hence we get an equality
idR(N) = sup {n; Ext ft (-, N) :f:. O}.
Theorem A.l,3 enables us to give the following description of injective
dimension, which has no analogue for projective dimension.

Injection, Projection and Global Dimension in Terms of Ext and Tor: 169
Theorem A.4.3
Let N be an R-n10dule. Then:
(i) N is an injective R-module if and only if, for any ideal I c: R,
Extj(R/I, N) = 0,
(ii) idR(N) = sup {i; Extk(R/I, N) # 0 for SOlne id.eal I c R}.
Proof
(i) If I c: R is an ideal, then the exact sequence 0  I ..: R  R/I  0
induces, in view of Theorem A.3.2, an exact sequence Hom(R, N)
co.
-+ Hom(I, N) -+ Ext 1 (R/I, N)  Ext 1 (R, N) = O. By Theorem A.l.3, the
module N is injective if and only if 00* is an epimorphism, which is equiv-
alent to the condition Ext 1 (R/I, N) = O.
The property (ii) follows from (i) by applying the reduction theorem
(Corollary A.3.5). 0
The following fact can be proved in a way similar to the proofs of
Theorems A.4.1 and A.4.2.
}"'heorem A.4.4
For any R"lnodule M, the following conditions are equivalent:
(i) Tor ll + 1 (M, -) = 0,
(ii) if a sequen.ce 0  K -+ Fn-l  ...  Fo  M -+ 0 is exact and the
modules Fo, ..., F,.- 1 are projective, then the module K is flat.
The relations between the dimension of a module and that of its local-
ization are given for modules over Noetherian rings.
Theorem A.4.5
Let R be a Noetherian ring and M an R..module. Then:
(i) idR(M) = sup {idRp(M p )} = sup{idRm(M m )},
P m
(ii) if M is a finitely generated R-module, then
pdR(M) = sup {pdR(Mp)} = sup{pdRm(M m )},
P m
where, in the above formulae, P ranges over the set of prime ideals of R,
and m over the set of maximal ideals.
Proof
(i) We shall first prove that if E is an injective R-module, then E p is an
injective Rp-module for any prime ideal P c: R. Indeed, every ideal of
R p has the form IR p for some ideal I c: R. It follows, from the exactness
of the localization functor (cf. Corollary 1.4.16 in [B]), that Rp/IR p  (R/I)p.

170
Homological Background
Therefore, by Theorems A.3.6 and A.4.3, we get
ExtAp(Rp/1R p , E p )  (ExtA(RII, E»p = O.
Theorem A.4.3 thus shows that E p is an injective Rp-module.
Accordingly, by the exactness of the localization functor, an injective
resolution E of the module M determines an injective resolution E p of the
Rp-module M p. Hence
idR(M)  sup {idlp(Mp)}  sup {idRm(M m )}.
P 1n
If Ext(L, oM) #- 0 for some cyclic R-module L  R/I, then by The...
orem 1.4.22 in [B], there exists a maximal ideal tn such that (Exti(L, M))tn
:/:. o. Appealing to Theorem A.3.6, we get
ExttIt(LIl" M m ) ¥:: o.
By Theorem A.4.2 we deduce that sup {idRtn(M m )}  idR(M), which com-
m
pletes the proof.
One proves (ii) in a similar manner, the part corresponding to the
first part of the proof of (i) being simpler because F p is obviously a pro..
jective resolution of the Rp-module M p when F is a projective resolution
of the module M. 0
To conclude this section, we introduce yet another hOlnological invari..
ant.
By Theorems A.4.1 and A.4.2,
sup {PdR(M)} = sup{n; Ext ::f. O} = sup{idR(M)}, (4)
M M
where M ranges over all R-modules. This leads to the following definition.
Definition A.4.6
The comnlon value occurring in the forn1ula (4) is called the global dimen..
sion of the ring R and is denoted by gl.dimR.
Theorenl A.4.7
For any ring R,
gl. dim R = sup {pdR(R/I)},
I
where I runs through all the ideals of R.
Proof
By Theorems A.4.3 and A.4.1, \ve have
gl.dimR = sup {idR(N)} = sup{n; Ext(R/I,N) :FO}
N f,N
= sup{pdR(R/I)}.
I
o
Theorelns A.4.5, A.4.7 and 2.1.11 froln [B] immediately yield

Injective, Projective and Global Dimension over Local Rings 171
Deorem A.4.8
If R is a Noetherian ring, then
gl. dim R = sup {gl. dim R p } = sup {gI. dim Rnt } .
P tn
A.5 INJECTIVE, PROJECTIVE AND GLOBAL DIMENSION OVER
LOCAL RINGS
Theorems A.4.5 and A.4.8 show that the problem of determi11ing the
injective and projective dimension of a finitely generated module, and the
global dimension of a ring call be reduced to the case in which the ring is
local. Before we begin a more detailed study of that case, we shall present
some special properties of modules over a local ring.
Lemma A.S.t
Let (R, m) be a local ring and M a finitely generated R-module. Let Xl' ...
..., X n eM. Then:
(i) the elements Xl' ..., X n generate the module M if and only if their
residue classes x I' ..., X" generate the module M /mM.
The following conditions are equivalent:
(ii) the elements x I, ..., X n generate the module M and no proper
su bset of the set x I, ... , x" generates M,
(iii) the residue classes X l' ..., x n constitute a basis of the RIm-module
M/mM.
Proof
(i) The implication => is trivial. In order to prove the opposite impli-
cation, let us suppose that the elements X l' ..., X n generate the RIm-module
M/mM, and denote by N the submodule generated by Xl' ..., Xn. We then
have N+mM = M, and consequently m(M/N) = (N,-mM)/N = MIN.
It follows, from the Nakayama lemma, that N = M.
(ii) => (iii) In view of (i), the condition (ii) is satisfied if and only if
the elements X l' ..., X n generate M/mM and no proper subset of the set
X l' ..., x,. generates M ImM, i.e. if X l' ..., X n form a basis of M /mM over
R. D
Definition A.5.2
Let (R, m) be a local ring. A finite set of generators of an R-module M,
satisfying the equivalent conditions (ii), (iii) in Lemma A.5.1 is called
a minimal set of generators of the module M.
Obviously, from every set of generators, one can select a minimal one.

172
Homological Background
Lemma A.5.3
Let (R, m) be a local ring and M a finitely generated R-module. Then there
exists a free, finitely generated R-module .F, and an epimorphism h: F -+ M
such that the induced mapping h: FlmF  MlmM is an isomorphism.
Proof
Let Xl' ..., X n be a minimal set of generators of M. We take Fto be a free
module of rank n, with basis e 1, ..., ell. We define a homomorphism It by
the conditions h(e,) = Xh i = 1, ..., n. According to Lemma A.S.I (iii),
the homomorphism h sends a basis of FlmP onto a basis of M ImM, and
hence it is an isomorphism. 0
An epimorphism g: N  M which induces an isomorphism g :
N/mN --. MlmM is said to be a minim al epimorphism.
Theorem A.5.4
If (R, m) is a local ring and M a finitely generated R-module, then the
following conditions are equivalent:
(i) M is a free module,
(ii) M is a projective module,
(iii) M is a flat module,
(iv) Torf(M, R/m) = O.
Proof
The implications (i) => (ii) => (iii) are obvious, while the implication
(iii) => (iv) follows from Theorem A.3.4 (ii). To prove that (iv) implies (i),
consider an exact sequence 0  N  F  M  0, where F is a free module
and h is a minimal epimorphism. This sequence induces (in view of The-
orem A.3.4 (iv» an exact sequence
Torf(M, Rim) --. N(g;R/m --.. F (g; Rim h@J +> M@R/m -+ O.
Since h(6)l coincides with h: FlmF --.. M/mM, h@l is an isomorphism.
By the assumption Tor(M, R/m) = 0, we conclude that N/mN  N(t)
@R/m = O. The Nakayama lemma shows that N = 0, and M  F is
a free module. 0
It is easily seen that the above statements are valid for quasi-local
rings.
Corollary A.5.5
If (R, m) is a local ring and M is a finitely generated R-module, then
pd(M) = sup{n; Tor:(M, Rim)  OJ.

Injective, Projective and Global Dimension over Local Rings 173
Proof
It follows, from the definition of the functors Tor, that if the right-hand
side of the above equality were not finite, then the module M would have
no finite projective resolution; the equality is then obvious.
We assume the number is finite and equal to k, i.e. Torf(M, RIm) i= 0,
Tor:(M, Rltn) = 0 for n > k. Let 0  K  F"-l  ... --. Fo --. M  0
be an exact sequence, where Fo, ..., F"-l are free modules of finite rank.
By the reduction theorem (Corollary A.S.4), we have
Torf{K, Rim)  Tor:+l(M, Rim) = 0,
and so K (as a submodule of a finitely generated module over a Noetherian
ring) is a free module in view on Theorem A.5.4. Thus pdR(M)  k, and,
since Torf(M, Rim) i= 0, pdR(M) = k. 0
Corollary A.5.6
If (R, m) is a local ring, then
gl. dimR = pdR(Rlm) = sup {n; Tor:(Rlnt, Rim) #: O}.
Proof
We may assume that pdR(Rlm) is finite, pdR(Rlm) = k, say. By Theorem
A.4.7, it is sufficient to show that pdR(M)  k for any finitely generated
R-module M. Let F ' be a free resolution of length K of the module RIm.
Consider an exact sequence
o  K  F"-l -+ ...  Fo --. M --.. 0,
where Fo, ..., F k - 1 are free modules of finite rank. Coronary A.3.5 shows
that
Torf(K, RIm) = Torf"+l(M, Rim) = H"+1(M@F') = 0,
whence K is a fee module by Theorem A.5.4, which gives pdR(M)  k. 0
The itijective dimension of a module also admits a characterization
similar to that described in Corollary A.5.S for projective dimension.
Theorem A.5.7
If (R, m) is a local ring and N is a finitely generated R-module, then
idR(N) = sup{n; Ext(Rlm,N)  O}.
Proof
As in the proof of Corollary A.5.5, we may restrict ourselves to the cas.e
where the right-hand side of the above formula is finite; let us denote it
by k, i.e. Ext(Rlm, N)  0, Ext(Rlm, N) = 0 for n > k. The condition
Bxt(Rlm, N)  0 implies idR(N)  k. We shall prove the assertion of the
theorem by assuming that idR(N) > k and deriving a contradiction. By
Theorem A.4.3, the family d of all ideals of the ring R, such that

174
Homological Background
Extk{R/I, N) ::j; 0 for some i > k, is non-empty (the number i depends
on the ideal I). As R is Noetherian, the family .91 has a maximal element;
let it be I. By the assumption, I :f: m, hence there exists an element x E 11t,
X ;. I. Consider the exact sequence
x
o --. R/J  R/I  R/(I, x)  0,
where J = I:(x) ::) I, and that part of the induced exact sequence of the
functors Ext
Extk(R/(I, x), N)  Ext(R/I, N)  Extk(RIJ,N)
 Extk+ 1 (R/(I, x), N), (5)
for i > k such that Exti(R/I, N) i= O. Since (I, x)  I, we have, by the
definition of .91, Extk(R/(I, x), N) = Extk+1(R/(I, x), N) = 0, and there-
fore Ext(R/J, N) =F O. Noting that J::> I, we conclude that J = I, as
I is maxima] in .91. It follows, from the exactness of (5), that xExt(R/I, N)
= Extk(RII, N). By Theorem A.3.7, the module Extk(R/I, N) is finitely
generated, and hence the Nakayama lemma yields Extk(R/I, N) = O.
But this contradicts the choice of i, and we must have idR(N) = k. 0
Remark A.S.8
An analysis of the above proof shows that in fact, we have established that
Extk(R/m, N) = 0 for some n forces idR(N) < 00. Similarly, it follows,
from the proof of Corollary A.5.S, that Tor:(M, Rim) = 0 for some n,
forces pdR(.M) < 00.
Exercises
In the exercises below, (R,1tt) denotes a local ring.
1. Prove that if h : F -+ M is a minimal epimorphism of R-modules and if M is finitely
generated, then the homomorphism h*: Hom(M, RIm) -+ Hom(F, RIm) induced by
h is an isomorphism.
2. Using the preceding exercise, sho\v that a finitely generated R-module M is free
if and only if ExtA(M, RIm) = O.
3. Prove that for a finitely generated R..module M, one has
pd(M) = sup{ll; Extjl(M, Rltn) #= O}.
4. Prove that Extjl(M, RIm) = 0 for some n, implies pdR(M) < 00.
5. Show that gl.dimR = sup {n; Extft(Rlm, Rlttt) :/: O}.
6. Prove that gl.dimR = idR(Rlm).
7. Let M be a finitely generated R-module. Prove that there exists a free resolution
F = (F n , d n ) of the module M, with finitely generated components F" satisfying the con-
dition Im(d cl + 1 ) c mF n for n = 0, 1, ... Such a resolution is called a minimal resolution.
[Apply Lemma A.S.3.]
8. Prove that any two minimal resolutions of a finitely generated module are iso-
morphic.
In the following exercises, by a graded algebra A we mean a K-algebra eAn with
grading over a field K, satisfying the condition Ao = K.

KoszuI Complex
175
9. Denote by I the ideal generated by homogeneous elements of positive degree
in A. Show that if M is a graded A-module then M = 1M implies M = o. (This is the
analogue of the Nakayama lemma for graded algebras.)
10. Using Exercise 9, formulate and prove the analogues of Lemmas A.5.1 and
A.5.3 for graded algebras.
11. Prove the existence of minimal resolutions of graded modules over algebras,
and prove the analogue of Exercise 8 for them.
A.6 THE KOSZUL COMPLEX
In defining a Koszul complex, we shall need the concept of exterior powers
/\ P(M), and of the exterior algebra /\ (M) of an R-module M. The basic
definitions and properties connected with these concepts can be found in
[N], p. 424.
Here we only recall that for p > 0, the R-module !\ '(M) may be de-
fined as the factor of the tensor product @"M of the module M, by the
submoduJe generated by elements of the form U1@ ... @u p , where u, = uJ
for some i :/:. j. We have the canonical p-Iinear mapping
cop: Mx ... xM --"/\'(M),
p
which is a universal alternating mapping, that is, for any p-linear alterna-
ting mapping f: M x ... x M --. N, there exists a unique homomorphism
g:!\ '(M)  N such that f = gro,. The element ro,(Ut, ..., up) is written
Ul /\ ... "u,. We adopt the convention that /\ O(M) = R.
00
The direct sum /\ (M) = Ef) /\P(M) can be endowed with a structure
p=o
of a graded R-algebra in such a way that /\ (M) is generated as an R-algebra
by M = 1\1(M), and an element Ul A... /\ Up is the product of elements
U1, ..., u, belonging to M. In the sequel, we denote the multiplication in
!\ (M) by the symbol 1\.
The algebra !\ (M) is not, in general, commutative but for homogeneous
elements x E /\"(M), Y E /\ q(M), we have the formula
XAY = (-l)pqy/\x.
Let tp e Hom(M, R); ,ve define the complex K(rp) by setting Kp(cp)
= !\P(M) for p  0, K,(cp) = 0 for p < 0; the differentials d,: !\P(M)
-+ !\P-l(M) are given by the formula
d p (Ul A ... A Up)
p
= ..2; (-l)J-ltp(UJ)Ul A .., AUJ-I AU J+I A ... AU p '
i=!
(6)
A simple calculation shows that d'-1d p = 0 for all p, and therefore K(q;)
is a complex of R-modu1es.

176
Homological Background
It follows, from formula (6), that for any x E ,!\P(M), y E 1\ '-(M),
we have
d p + ll (x/\y) = dp(x)Ay+(-l)PxAdq(y).
(7)
Definition A.6.1
The complex K(cp) described above is called the Koszol complex of the
homomorphism rp: M --.. R. If x denotes a sequence of elements Xl' ..., X n
of the ring R and if F is a free module of rank n with basis e 1, ..., en, then
the Koszul complex K(rp) of a homomorphism cp: F -+ R for which rp(ei)
= X" i = 1, ..., n, is also denoted by K(x).
If N is an R-module, then we write K(rp; N), K(x; N) for the complexes
K(rp)(8)RN, K(X)(8)RN, respectively, an call them the KoszuI complex of
the homomorphism rp (of the sequence x) with coefficients in the module N.
Observe that the differentials of the complex K(cp; N) have the form d p (8)l,
the d p being the differentials of the complex K(rp).
We recall that if Fis a free module of rank n, then !\P(F) = 0 for p > n;
for 1  p  n, the module !\ P(P) is a free module of rank (;), and if
e 1, ..., ell constitute a basis of the module P, then a basis of the module
!\ P(F) consists of elements of the form e'l " ... A e' ll for. all the sequences
of positive integers subject to 1  i 1 < ... i p  n.
Thus the Koszul complex K(x) is a finite complexes of free modules.
The principal properties of Koszul complex are stated in the following
theorem:
Theorem A.6.2
Let x denote a sequence of elements Xl' ..., XII of a ring R and let N be an
R-module. Then:
(i) HoC K(x; N» = N/(Xl' "" XII) N,
(ii) the homologymodulesH,(K(x;N») of the Koszulcomplex K(x;N)
are annihilated by the ideal (Xl' ..., XII) + Ann(N), p = 0, ..., n,
(Hi) if the sequence x is a regular sequence of the ring R (see Definition
1.4.16), then the Koszul complex K(x) is a free resolution of the module
R/(XI, "" XII)'
Proof
The equality (i) follows immediately from the definition of a differential
of the complex K(x; N).
(ii) Let an element z be a p-cycle of the complex K(x; N). From the
property (7) of the differential of K(x), we conclude that (d p + 1 @1) (eJAz)
= xJz-eJA (d p Q5Jl)(z) = xJz, and hence the element xJz is a boundary
in the complex K(x; N) for j = 1, ..., n. This means that the homology

Koszul Complex
177
class of the cycle z is annihilated by the whole ideal (Xl' ..., x n ). Of course,
this class is also annihilated by Ann(N).
(iii) We use induction on n. When n = 1, then K(x) is a complex of
d 1
the form 0  Ret -+ R --. 0, d t (e1) = Xl' and consequently Ho(K(x»)
= Ker(d l ) = O:(Xt) = 0 because, by the assumption, Xl is not a zero..
divisor.
We assume the assertion to be valid for the numbers < n, and observe
that F = ERen' where E = Rel ... Ren-l. The pomplex K = K(x)
contains the subcomplex X = K(Xl' ..., X n -l); this follows from the formula
/\P(F) = 1\ P(E)(j) /\ p-1 (E)/f8J Re n and from the definition of differ..
entiaIs of a Koszul complex. Let Y be the factor complex K!X, then one
has an exact sequence of complexes
O-+X-+K-+ Y--.O
(8)
We note that Y is isomorphic to the complex arising from X by shifting
the grading of the components by 1, i.e. Y"  X,-l. In view of this,
H,,( Y)  H"-l (X). Since, by the inductive hypothesis, Hp(X) = 0 for p > 0,
it follows that Hp(Y) = 0 for p > 1. The exact sequence, induced by (8),
Hp(X) -+ Hp(K) --. Hp(Y) yields H,(K) = 0 whenever p > 1.
We have, moreover, the exact sequence
"
0-+ Ht(K) -+ Ht(Y) -+ Ho(X) ,
whence Ht(K) = Ker(<5). A direct computation shows that the composi-
tion of the isomorphism Ho(X)  Ht(Y) with the connecting homo-
morphism <5 is the mapping consisting in multiplying by Xa: Ho(X) -+ Ho(X).
By (i), we therefore deduce that
Ht(K)  Ker(R!(xI, ..., X n -l):: R/(Xt, ..., Xn-l»)
= (Xl' ..., x n - 1 ) : (XII)!(Xt , ..., Xn-t),
and hence Ht (K) = 0 according to the assumption that the sequence
Xl ..., X n is regular. 0
Corollary A.6.3
If Xt, ..., X n is a regular sequence of a ring R, then pdR(R!(xI, ..., x n ») = n.
Proof
It follows, from the conditions (i), (iii) in Theorem A.6.2, that the dimen-
sion is not greater then n. The formula
Torn (R/(xt, ..., x n ), R/(Xt, ..., x n »)
= Hn(K(x)@RR/(Xt, ..., x n ») = R/(Xt, ..., x n )  0
completes the proof.
o

178
Homological Background
:Xercises
1. Prove tbat if x= (Xh""X.), then Hn(K(x;N)) = O:(Xh""X n ).
2. Let 1 be an ideal of a Noetherian ring R, and 1et N be a finitely generated
R-module. Write lp: F-+R for a homomorphism of a free module F of rank 11, such
that Imtp == I. Prove that
depth(I; N)+ suP{j; HJ(K(p)f$9N) =#= O} = 11,
[Use Lemma 3.1.10 and apply induction on depth(l; N).]
3. Let x be a sequence Xt, ..., X n of a loca' ring R. Prove that x is regular if
and only it Ht(K(x») = o. From this, in wiew of Theorem (A.6.2), it fol1o\vs that
for local rings, 111(K(x») = 0 implies H»K(%») = 0 for p  1.

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Arne,.. Math. Soc., 62(1947), 1-52.

Index of Theorems and
Definitions quoted from
Commutative Noetherian and
Krull Rings
Theorem 1.1.7
Every ring R satisfies the following conditions:
(i) if 11, ..., Is c: R are ideals, peR is a prime ideal and 1 1 ... /. c P (or in par-
ticular 1 1 (1 ... nIs C P), then I" C P for some k, 1  k  s,
(ii) if s  2, and /, PI, ..., P, C R are ideals, where P 3 , ..., p. are prime and
1 C P 1 u ... uPs, then I C Pic for some k, 1  k  s,
(Hi) if J $ I c R are ideals, Pl. ..., Ps c: R are prime ideals and 1""-1 C Pt ... uP.,
then I c P" for some k, 1  k  s.
Corollary 1.1.8
If I c R is an ideal, P 1, II., P, are prime ideals, and I C PI U ... uP., then 1 c P" for
some k, 1  k  s.
Lemma 1.3.2
If an R-module M is generated by n elements, I c: R is an ideal, r e R, and the condition
rM c: 1M is satisfied, then there exist elements ao, ..., aR-l e I such that
,ft+ aa _ 1 ,R-l + ... +ao e Ann(M).
In particular, if 1M = M, then there exists an element a e I such that 1 + a e Ann(M).
Lemma 1.3.3 (Nakayama)
If M is a finitely generated R-module, 1 c R is an ideal contained in the Jacobson radical
J(R) and 1M = M, then M = O.
Theorem 1.3.7
If a sequence of R-modules
o -+ M' -+ M -+ Mil -+ 0
is exact, then 1(M) = I(M')+I(M").
Theorem 1.4.7
Let S be a multiplicative subset of a ring R. Let I c: R, 1 c: Rs be ideals and write w:
R -+ Rs for the canonical mapping given by the formula w(r) = r/1, r e R. Then

184
Index of Theorems
(i) w- 1 (IR..,) ;:) I, w- 1 (J)R s = J,
(iv) rad(I)R s = rad(IR s ), w- 1 (rad(J» = rad(w- 1 (J»).
Corollary 1.4.8
The correspondence P  P Rs between the set of those prime ideals of a ring R which
satisfy the condition Pr.S = 0 and the set of all prime ideals of the ring Rs is one-to-one
and preserves the relation of inclusion. The formula P' ... w- 1 (PI) defines the inverse
mapping.
Corollary t .4.17
If S is a multiplicative subset of a ring R, then Rs is a flat R-module (i.e. the functor
- Q?)R Rs is exact).
Corollary 1.4.20
If meR is a maxhnal ideal, then there exists an isonlorphism of R m -modules (ntRm)'J j
j(mR In )ft+l  m ft /tn,,+I, " = 0, 1, ...
Theorem 1.4.22
If M is an R-module, and Mtn = 0 for every n1aximal ideal1tt c R, then tv! = 0..
Definition 1.5.6
Let I be an ideal of a ring R. The descending sequence of ideals
R = [0 ::) [I ::) 1 2  ... => I" => 1"+1 ::) ...
determines a graded ring
00
Grl(R) = Ef) 1"//"+1
p=o
with multip1ication defined by the formula
(r+l"+I)(8+14+1) = rs+["+1+1 for reI", 8 elf.
We call this ring the graded ring associated with the ideal 1.
Theorem 1.5.9
ex>
Let R = E9 R" be a graded ring. Then
p=O
(ii) a minimal prime ideal of a homogeneous ideal is also homogenous,
00
(iv) a homogenous ideal [ is maximal if and only if it has the form I = m Ef) e R"
p-=1
where m is a maximal ideal of the ring Ro.
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.
Corollary 2.1.8
A finitely generated module over a Noetherian ring is a Noetherian module.

Index of Theorems
185
Corollary 2.1.12
H R is a Noetherian ring and P is a priJne ideal of R, then the localization R p is also
Noetherian.
Theorem 2.1.15 (Cohen)
R is a Noetherian ring if and only if each prime ideal of R is finitely generated.
Tbeorem 2.2.1 (Hilbert Basis Theorem)
If R is a Noetherian ring, then the polynomial ring R [X) is also a Noetherian ring.
Corollary 2.2.2
If R is a Noetherian ring, then the polynomial ring R[X 1 , ..., X,,] is also Noetherian.
Lemma 2.3.5
Let R be a Noetherian ring, 111 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" c Q c: 111 for some k > O.
Theorem 2.3.15
Let I = Q 1 () ... ()Qs be an irredundant primary decolnposition of an ideal 1 in a
Noetherian ring R, and let P, = rad(Q,), 1  i  s. A prime ideal P of R is equal to
some P, if and only if there exists an element x E R such that I:(x) = P.
Corollary 2.3.18
An ideal 1 of a Noetherian ring is primary jf and only if the set Ass R (R/1) consists of
one element.
Lemma 2.3.20
In a Noetherian ring, the set of isolated prime ideals of an ideal 1 is identical with the
set all ideals minimal among the prime ideals containing 1.
Theorem 2.3.22
Let 1 be an arbitrary ideal of a Noetherian ring R and let PI' ..., P, denote all the prime
ideals associated with I. Then the set of all zero-divisors on the R-module R/I is equal
to the union Ptu ... uPs, i.e., 3(R/I) = Ptu ... uPs.
Corollary 2.3.23
In a Noetherian ring, each element of a minimal prime ideal is a zero-divisor.
Lemma 2.4.3
Let R be a Noetherian ring an M and R-module. Then Ass(M) = 0 if and only if
M= O.
Corollary 2.4.4
The set of zero-divisors on any module over a Noetherian ring is the union of all associated
prime ideals of the module.

186
Index of Theorems
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
o = Mo c: M 1 c: .o. c: M n = M
such that MJI M J - 1  R/ Ph where the p} are prime ideals in R, 1  j  n,
(ii) for any such chain we have Ass(M) c: {PI' ..., p,.}; hence the set Ass(M)
is finite.
Lemma 2.4.14
Let R be a Noetherian ring, S a multiplicative subset of R, and D a subset od Spec(R)
consisting of all prime ideals disjoint from S. Then, for any R-module M, the mapping
P  P Rs is a bijection between Ass R (M) n nand Assns(M s ).
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 tat M p  o.
1nheorem 2.4.21
(iv) If M, N are finitely generated R-modules, then SUPP(MRN) = Supp(M)
(iSUpp(N).
Theorem 2.4.22
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).
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 k such that
InBnC = In-"(I"BnC)
for any n  k.
Lemma 2.5.2
(i) If R is a Noetherian ring, then the ring
T = R+IX+12X2+ ... +1"X"+ ...
is also Noetherian.
(ii) If M is a finitely generated R-module, then
E = M+IMX+12MX 2 + ... +InMx n + ...
is a finitely generated T-module.
Corollary 2.5.5
Let R be a Noetherian ring, 1 an ideal of R, and M a finitely generated R-module. H
I is conta.ined in the Jacobson radical J(R) of the ring R, then n IftM = O. In par-
n>O
ticular if R is a local ring, then for any proper ideall of R we have n 1"M = O.
n__O

Index of Theorems
187
Example 2.6.10
The completion of the polynomial ring K[X 1 , ..., X A ] in the topology determined by the
powers In, where I = (Xl, ..., Xt), is the formal power series ring K[X 1 , ..., X].
Theorem 2.6.11
"
Themappingcx: Iim{M/M,,} -+ M is an isomorphism of R-modules, where ex (x,,+M n <
,-
= {Xn} + Co and Co denotes the R-submodule of all sequences of M which converge to
zero.
Theorem 2.6.12
H every homomorphism ex of an inverse system {L:', ex} is an epimorphism, then the
exact sequence of inverse systems
o -+ {L, a} -+ {Ln, ex n } -+ {L/, ex/} -+ 0
induces an exact sequence of their respective limits
o -+ Jim {L} -+ lim {La} -+ Jim {L'} -+ O.
- -- --
Corollary 2.6.20
If R is a Noetherian ring, then the completion R of R in the I-adic topology is a flat
R-algebra.
oroUary 2.6.21
(iii) Let M be a finitely generated module over a Noetherian ring R with the I-adic
topology. Then
" "
M/lnM  M/lnM for each 1'1 > o.
Corollary 2.7.9
A homomorphic image of an Artin ring is also an Artin ring.
Theorem 2.7.11
Let R be a Neotherian 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).
Theoretn 2.7.12
A ring R is an Artin ring if and only if the following two conditions are satisfied:
(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 nil-
potent.
Corollary 2.7.13
If R is an Artin ring and M is a finitely generated R-n10dule, then M is of finite length,
and Ass(M) = Supp(M).
Corollary 2.7.14
If R is a Noetherian ring, then a finitely generated R-module A1 is of finite Jength jf and
only if R/ Aml(M) is an Artin ring.
Theorem 3.1.6
Let R c T be a ring extension. For every element t E 1', the following conditions are
equivalent:

188
Index of Tbeorems
(i) the element t is integral over R,
(ii) the ring R[t] is finitely generated as an R-Inodule,
(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"'lnodule M c: T satisfying the following con...
ditions :
(a) tMC M,
(b) if u e R[t] and 11M = 0 then II = o.
Corollary 3.1.9
If R c: Rl' Rl C T are integral ring extensions, then R c T is also an integral extension.
Theorem 3.1.12
If a ring extension R c T is integral. then a prime ideal Q of T is maximal if and only
if the ideal Q nR of R is tnaximal.
Theorem 3.1.13
If a ring extension R c T is integral, and if Q c Q.t at"e prhne ideals of T and Qr-R
= Ql nR , then Q = QI.
Theorem 3.1.16
If a ring extension R c T is integral, and if P is a prime ideal of R, then there existi
a prime ideal Q of Tsuch that Qf1R = P (i.e., the extensionj: R C Tinduces a surjection
Spec(j): Spec (T)  Spec (R».
Theorem 3.1.17 ("Going up")
If a ring extension R C T is integral, and if
PIC ... C P nl c ... C Pn
is a chain of prime ideals of R, and if further
Q. C ... c Qm
is a chain of prime ideals of T, 0  '" < n, such that
Qtt1R=Pl, ..., Q".nR = Pm,
then there exists an extension of the chain (*) to a chain of prime ideals of T of the form
(*)
Q l c: ... c Qm c ... c Qn
such that
Q,f1R == Ph ; = t, ..., II.
Theorem 3.1.4 ("Going down")
Let R C T be an integral ring extension. If R is a normal domain no elen1ent of which
is a zero-divisor in T, if
p. C... C Pm c: ... C Pn
is a chain of prime ideals of R, and if, further
QIIC c ... C Qn
is a chain of pl.ime ideals of T, 1 < III  II, such that
Qmf1R == Pm, ..., QnnR::r.: Pn,
(*.)

Index of 'fheorems
189
then there exists such an extension of chain (**) to a chain of pritne ideals of T of the
form
Q 1 C ... C Q", C ... c Q,.
that
Q,nR=P" i= 1,...,11.
Tbcorem 3.4.3
If R is a Dedekind dOJnain then the ideal class group Cl(R) is zero if and only if R is
a unique factorization domain.
Theorem 3.6.16
Let R be a local (and hence Noetherian) d0l11ain with the n13ximal ideal 111 i= O. The
following properties are equivalent:
(i) R is discrete valuation ring,
(iv) 11t is a principal ideal.

Index
Algebra
affine 3
exterior 125, 175
of polynomial functions 2
Auslander-Buchsbaum theorem 84
Hensel lemma 86
Hilbert
-Burch theorem 132
Nullstellensatz 7
-Samuel polynomial 50
theorem 48
homomorphism
cOlU1ecting 161
of complexes 160
p-basis of a field 89
Cohen theorem 98
complex
fini tc 160
free 160
injective 160
Koszul 176
of modules 160
projective 160
Depth of an ideal 105
dimension
global 170
injective 160
of an algebra 4
of a variety 3
projective 158
Eisenstein
extension 99
polynomial 99
Euler-Poincare characteristic 55, 61
exact sequence of complexes 161
exterior
algebra 125, 175
powers 125, 175
Functors
derived 162
of extensions 164
of torsion product 166
Group of cycles on a variety 70
Height
of a module 72
of an ideal 13
Ideal
Gorensteil1 151
perfect 130
regular 109
unmixed 110
Krull
dimension 12, 53
theorem 13
Leading form 46
length of a chain of prime ideals 12
Macaulay theorem 28
McCoy theorem 124
minimal epimorphism 172
module
Cohen-Macaulay 124
flat 160
injective 159
n-th homology 160
of n-boundaries 160
of n-cycles 160
perfect 130
projective 157
multiplicity
intersection 61
of a module 54
of an intersection of subvarieties 58, 69
Norma1ization theoreJn 5, 10

192
Index
Point
regular 39
singular 45
polynomial function 2
proper intersection
of cycles 70
of subvarieties 69
ring
catenary 25
Cohen-Macaulay 113
Gorenstein 141
of representatives 85
regular local 39
Saturated chain 9, 121
. set
algebraic 1
irreducible algebraic 2
minhnal set of generators 171
of parameters 35
l"egular set of parameters 39
Tensor product of algebras 11
transcendence degree 3
type
of an ideal 139
of a ring 139
Variety 2
Reduction theorem 167
regular
ideal 109
local ring 39
point 39
sequence 40, 103
set of parameters 39
subvariety 39
resolution
itijective 159
of a homomorphism J 62
projective 158

193
Mathematics and its Applications
Series Editor: G. M. IJELL, Professor of Mathetnatics, King's College Loudon
(KQC), University of London
B R Topology
rown t .
Burghes, D. N. & Borrie, rvt. Modelling with Differential Equations
Burghes, D. N. & Downs, A. M. Modern Introduction to Classical Mechanics and Control
Burghes, D. N. & Graham, A. Introduction to Control Theory, including Optimal Control
Burghes, D. N. t Huntley, I. & McDonald, J. Applying Mathematics
Burghes t D. N. & Wood, A. D. Mathematical Models in the Social, Management and Life
Sciences
Green's Functions and Transfer Functions Handbook
Fourier Methods: Applications in Mathematics, Engineering and Science
Complex Domain Anal)'sis
Textbook of Dynamics, 2nd Edition
Vector and Tensor Methods
Computability and Logic
Shape Tbeor': Cutegorical Methods of Approximation
Introduction to Water Waves
Learning the Art of Mathematicall\1odelling
Linear Models in Biology
Mathematical Methods for Mathematicians, Physical Scientists and
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Eason, G. t Coles, C. w. & Gettinby, G. Mathematics and Statistics for the Bio-sciences
El Jai, A. & Pritchard, A. J. Sensors and Controls in the Analysis of Distributed Systems
Exton t 1-1. Multiple Hypergeometric Functions and Applications
Faux, I. D. & Pratt, M. J. Computational Geometry for Design and l\'lanufacture
FirbYt P. A. & Gardiner, C. F. Surface Topology
Gardiner t C. F. Modem Algebra
Gardiner, C. F. Algebraic Structures: with Applications
Gasson t P. C. Geometry of Spatial Forms
Goodbody, A. M. Cartesian Tensors
Goult t R. J. Applied Lin.ear Algebra
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Grahaln, A. Nonnegative Matrices and Applicable Topic. in Linear Algebra
Griffel t D. H. Applied Functional Analysis
Griffel, D. H. Linear Algebra
Guest, P. B. The Laplace Transform and Applications
Hanyga, A. Mathematical Theory of Non-linear Elasticity
Harris, D. J. Mathematics for Business, Management and Economics
Hart, D. & Croft, A. Modelling with ProjectUes
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Hunter, S. C. Mechanics of Continuous Media, 2nd (Revised) Edition
Huntley, I. & Johnson, U.. M. Linear and Nonlinear Differential Equations
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ChorIton, F.
Cohen, D. E.
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Cross t M. & Moscardini, A. o.
Cullen, M. R.
Dunning- Davies, J.
Jones, R. H. & Steele, N. C.
Jordan t D.

194
Kel1y, J. C. Abstract Algebra
Kim, K. H. &: Roush. F. W. Applied Abstract Algebra
Kim, K. H. &: Roush, F. W. Team Theory
Kosinski, W. Field Singularities and Wa1'e Analysis in Continuum Mechanics
Krishnamurthy, V. Combinatorics: Theory and Applications
Lindfield, G. & Penny, J. E. T. Microcomputers in Numerical Analysis
Livesley, K. Engineering Mathenlatics
Lord, EI AI & Wilson, C. B. The Matbeu1atical Description of Shape and Form
Malik, MI, Riznichenko, G. Y. &: Rubin, AI B. Biological Electron Transport Processes and
their Computer Simulation
Massey, B. S. Measures in Science and Engineering
Meek, B. L. & Fairthorne, S. Using Computers
Menell, A. & Bann, M. Mathematics for the Biosciences
Mikolas, M. Real Functions and Orthogonal Series
Moore, R. Computational Functional Analysis
Murphy, J. A., Ridout, D. &: McShane, B. Numerical Analysis, Algorithms and Computation
Nonweiler, TI R. F. Computational Mathematics: An Introduction to Numerical
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Page, S. G. Matltenlatics: A Second Start
Prior, D. & Moscardini, A. O. Model Formulation Analysis
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Wheeler, R. F.
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Willmore, T. J. & Hitchin, N.
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195
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Editor: B. W. CONOLLY, Professor of Mathematics (Operational Research),
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Beaumont, G. P. Introductory Applied Probability
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Conolly, B. W. Lecture Notes in Queueing SysteolS
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Stanislaw Balcerzyk has been 8 Professor at the
Institute of Mathematics, Polish Academy of Sciences
since 1967. He was previously Associate Professor at the
same Institute from 1962-67. He was awarded the
degrees of Ph. D. (1959) and Habilitation (1962), both in
Algebra from the Institute of Mathematics, Polish Acad-
emy of Sciences. He has written, with Tadeusz J6zefiak,
this book, and Commutative Noetherian and Krull Rings
(Ellis Horwood Limited, due 1989).
TadeU8Z J6zefiak has been a Professor at the Ititute of
Mathematics, Polish Academy of Sciences since 1 85. H
was previously Associate Professor at the sarna In':'
from 1975-85. He was awarded the degrees ,.'
(1969) and Habilitation (1974), in Algebra f.
Institute of Mathematics, Polish Academy flf Sci)\

.  EN' T · a a D RULL.
S. BALCERZYK and T JOZEFIAK Institute of Mathematics, Polish Academy of Sciences,
Torun, Poland
Translation Editor: D. KIRBY, Faculty of Mathematical Studies, University of Southampton
The fundamental concepts of commutative ring theory are illustrated in this book by material
on algebraic geometry and algebraic number theory; the notions of a normal domain and
of integral extensions are emphasised The book commences with a description of how the
classical Fermat's problem is related to the property of unique factorization, and shows that
normal domains are the most natural class of rings suitable for the generalisation of classical
a rith metic.
TOPOLOGY: A Geometric Account of General Topology, Homotopy Types and the
u. .. It. ro. . · ·
R. BROWN, School of Mathematics, University College of North Wales, Bangor
This thorough and geometric treatment of general topology up to the level of identification
spaces, finite cell complexes, with material on function spaces, covers a range of topics
known to specialists, that are not easily available, for example, the construction of homotopy
equivalences.
1 . a N a L Y S, ALGORIT  S AND OM · T A 10
J. MURPHY, Head of Department of Computational Physics, Sowerby Research Centre,
British Aerospace pic, Bristol; D. RIDOUT, Department of Computer Science and
Mathematics, University of Aston, Birmingham; and BRIGID McSHANE, Department of
Mathematics, Cheadle Hulme School, Cheadle, Cheshire
This book takes an algorithmic pathway to the teaching of numerical analysis and
computation which are important for degree courses in mathematics, engineering and
science. It simplifies and minimises the number of advanced mathematical concepts to make
the text more acceptable to the reader. A knowledge of calculus is assumed.
· · . Sa: , · AN , OM a..:
G. P. BEAUMONT, Department of Statistics and Computer Science, Royal Holloway and
Bedford New College, Egham
"the book is clear and informative and will justify its positions on my shelves" - T. Sparks, National
Institute of Agricultural Botany, Cambridge, in The Statistician
· . . .. a. PLIE' PROS a : ILl
G. P. BEAUMONT, Department of Statistics and Computer Science, Royal Holloway and
Bedford New College, Egham
uprovides the attentive reader (who tries the problems) with a thorough grounding in the construction
and analysis of probablistic models as well as introducing them to a number of important ideas and
results" - N. L. Lawrie, University of Strathclyde, in European Journal of OperatIonal Research
a P · LIC a' . MOD LL G L a RN AND EA NG, aTE · TIC
Editors. W BLUM, Gesamthochschule Kassel, University of Kassel, Federal Republic of
Germany; J. S. BERRY, Department of Mathematics and Statistics, Plymouth Polytechnic;
R. BIEHLER; I. D. HUNTLEY, Head of Department of Mathematics, Sheffield City Polytechnic;
G. KAISER-MESSMER, Department of Mathematics, University of Kassel, Federal Republic
of Germany; and L. PROFKE, Professor of Mathematics Education, University of Giessen,
Federal Republic of Germany
This book presents the appropriate recent research and development studies in the teaching
and learning of mathematics in connection with real-life situations, e.g. theoretical concepts
empirical research concrete applications, and classroom examples for all levels from lower
secondary up to university It provides an international forum for the discussion and
exchange of experiences with applications and modelling, considering in particular the
social and technological developments and changes with respect to mathematics teaching.
published by
ELLIS HORWOOD LIMITED
Publishers. Chichester
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