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1

A Course In Commutative Algebra
Robert B. Ash

Preface
This is a text for a basic course in commutative algebra, written in accordance with the
following objectives.
The course should be accessible to those who have studied algebra at the beginning
graduate level. For general algebraic background, see my online text “Abstract Algebra:
The Basic Graduate Year”, which can be downloaded from my web site
www.math.uiuc.edu/∼ r-ash
This text will be referred to as TBGY.
The idea is to help the student reach an advanced level as quickly and e¬ciently as
possible. In Chapter 1, the theory of primary decomposition is developed so as to apply to
modules as well as ideals. In Chapter 2, integral extensions are treated in detail, including
the lying over, going up and going down theorems. The proof of the going down theorem
does not require advanced ¬eld theory. Valuation rings are studied in Chapter 3, and
the characterization theorem for discrete valuation rings is proved. Chapter 4 discusses
completion, and covers the Artin-Rees lemma and the Krull intersection theorem. Chapter
5 begins with a brief digression into the calculus of ¬nite di¬erences, which clari¬es some
of the manipulations involving Hilbert and Hilbert-Samuel polynomials. The main result
is the dimension theorem for ¬nitely generated modules over Noetherian local rings. A
corollary is Krull™s principal ideal theorem. Some connections with algebraic geometry
are established via the study of a¬ne algebras. Chapter 6 introduces the fundamental
notions of depth, systems of parameters, and M -sequences. Chapter 7 develops enough
homological algebra to prove, under approprate hypotheses, that all maximal M -sequences
have the same length. The brief Chapter 8 develops enough theory to prove that a regular
local ring is an integral domain as well as a Cohen-Macaulay ring. After completing
the course, the student should be equipped to meet the Koszul complex, the Auslander-
Buchsbaum theorems, and further properties of Cohen-Macaulay rings in a more advanced
course.


Bibliography
Atiyah, M.F. and Macdonald, I.G., Introduction to Commtative Algebra, Addison-Wesley
1969
Balcerzyk, S. and Joze¬ak, T., Commutative Noetherian and Krull Rings, Wiley 1989
Balcerzyk, S. and Joze¬ak, T., Commutative Rings: Dimension, Multiplicity and Homo-
logical Methods, Wiley 1989
Eisenbud, D., Commutative Algebra with a view toward algebraic geometry, Springer-
Verlag 1995
Gopalakrishnan, N.S., Commutatilve Algebra, Oxonian Press (New Delhi) 1984
Kaplansky, I., Commutative Rings, Allyn and Bacon 1970
2

Kunz, E., Introduction to Commutative Algebra and Algebraic Geometry, Birkh¨user
a
1985
Matsumura, H., Commutatlive Ring Theory, Cambridge 1986
Raghavan, S., Singh, B., and Sridharan, S., Homological Methods in Commutative Alge-
bra, Oxford 1975
Serre, J-P., Local Albegra, Springer-Verlag 2000
Sharp, R.Y., Steps in Commutative Algebra, Cambridge 2000

c copyright 2003, by Robert B. Ash. Paper or electronic copies for noncommercial use
may be made freely without explicit permission of the author. All other rights are reserved.
Table of Contents

Chapter 0 Ring Theory Background
0.1 Prime Avoidance
0.2 Jacobson Radicals, Local Rings, and Other Miscellaneous Results
0.3 Nakayama™s Lemma


Chapter 1 Primary Decomposition and Associated Primes
1.1 Primary Submodules and Ideals
1.2 Primary Decomposition
1.3 Associated Primes
1.4 Associated Primes and Localization
1.5 The Support of a Module
1.6 Artinian Rings


Chapter 2 Integral Extensions
2.1 Integral Elements
2.2 Integrality and Localization
2.3 Going Down


Chapter 3 Valuation Rings
3.1 Extension Theorems
3.2 Properties of Valuation Rings
3.3 Discrete Valuation Rings



Chapter 4 Completion
4.1 Graded Rings and Modules
4.2 Completion of a Module
4.3 The Krull Intersection Theorem

1
2

Chapter 5 Dimension Theory
5.1 The Calculus of Finite Di¬erences
5.2 Hilbert and Hilbert-Samuel Polynomials
5.3 The Dimension Theorem
5.4 Consequences of the Dimension Theorem
5.5 Strengthening of Noether™s Normalization Lemma
5.6 Properties of A¬ne k-Algebras

Chapter 6 Depth
6.1 Systems of Parameters
6.2 Regular Sequences

Chapter 7 Homological Methods
7.1 Homological Dimension: Projective and Global
7.2 Injective Dimension
7.3 Tor and Dimension
7.4 Application

Chapter 8 Regular Local Rings
8.1 Basic De¬nitions and Examples

Exercises
Solutions
Chapter 0

Ring Theory Background

We collect here some useful results that might not be covered in a basic graduate algebra
course.



0.1 Prime Avoidance
Let P1 , P2 , . . . , Ps , s ≥ 2, be ideals in a ring R, with P1 and P2 not necessarily prime,
but P3 , . . . , Ps prime (if s ≥ 3). Let I be any ideal of R. The idea is that if we can avoid
the Pj individually, in other words, for each j we can ¬nd an element in I but not in Pj ,
then we can avoid all the Pj simultaneously, that is, we can ¬nd a single element in I that
is in none of the Pj . We will state and prove the contrapositive.


0.1.1 Prime Avoidance Lemma
With I and the Pi as above, if I ⊆ ∪s Pi , then for some i we have I ⊆ Pi .
i=1

Proof. Suppose the result is false. We may assume that I is not contained in the union
of any collection of s ’ 1 of the Pi ™s. (If so, we can simply replace s by s ’ 1.) Thus
for each i we can ¬nd an element ai ∈ I with ai ∈ P1 ∪ · · · ∪ Pi’1 ∪ Pi+1 ∪ · · · ∪ Ps . By
/
hypothesis, I is contained in the union of all the P ™s, so ai ∈ Pi . First assume s = 2, with
I ⊆ P1 and I ⊆ P2 . Then a1 ∈ P1 , a2 ∈ P1 , so a1 + a2 ∈ P1 . Similarly, a1 ∈ P2 , a2 ∈ P2 ,
/ / /
so a1 + a2 ∈ P2 . Thus a1 + a2 ∈ I ⊆ P1 ∪ P2 , contradicting a1 , a2 ∈ I. Note that P1
/ /
and P2 need not be prime for this argument to work. Now assume s > 2, and observe
that a1 a2 · · · as’1 ∈ P1 © · · · © Ps’1 , but as ∈ P1 ∪ · · · ∪ Ps’1 . Let a = (a1 · · · as’1 ) + as ,
/
which does not belong to P1 ∪ · · · ∪ Ps’1 , else as would belong to this set. Now for all
i = 1, . . . , s ’ 1 we have ai ∈ Ps , hence a1 · · · as’1 ∈ Ps because Ps is prime. But as ∈ Ps ,
/ /
so a cannot be in Ps . Thus a ∈ I and a ∈ P1 ∪ · · · ∪ Ps , contradicting the hypothesis. ™
/
It may appear that we only used the primeness of Ps , but after the preliminary reduc-
tion (see the beginning of the proof), it may very well happen that one of the other Pi ™s
now occupies the slot that previously housed Ps .

1
2 CHAPTER 0. RING THEORY BACKGROUND

0.2 Jacobson Radicals, Local Rings, and Other Mis-
cellaneous Results
0.2.1 Lemma
Let J(R) be the Jacobson radical of the ring R, that is, the intersection of all maximal
ideals of R. Then a ∈ J(R) i¬ 1 + ax is a unit for every x ∈ R.
Proof. Assume a ∈ J(R). If 1 + ax is not a unit, then it generates a proper ideal, hence
1 + ax belongs to some maximal ideal M. But then a ∈ M, hence ax ∈ M, and therefore
1 ∈ M, a contradiction. Conversely, if a fails to belong to a maximal ideal M, then
M + Ra = R. Thus for some b ∈ M and y ∈ R we have b + ay = 1. If x = ’y, then
1 + ax = b ∈ M, so 1 + ax cannot be a unit (else 1 ∈ M). ™


0.2.2 Lemma
Let M be a maximal ideal of the ring R. Then R is a local ring (a ring with a unique
maximal ideal, necessarily M) if and only if every element of 1 + M is a unit.
Proof. Suppose R is a local ring, and let a ∈ M. If 1 + a is not a unit, then it must
belong to M, which is the ideal of nonunits. But then 1 ∈ M, a contradiction. Conversely,
assume that every element of 1+M is a unit. We claim that M ⊆ J(R), hence M = J(R).
If a ∈ M, then ax ∈ M for every x ∈ R, so 1 + ax is a unit. By (0.2.1), a ∈ J(R), proving
the claim. If N is another maximal ideal, then M = J(R) ⊆ M © N . Thus M ⊆ N , and
since both ideals are maximal, they must be equal. Therefore R is a local ring. ™


0.2.3 Lemma
Let S be any subset of R, and let I be the ideal generated by S. Then I = R i¬ for every
maximal ideal M, there is an element x ∈ S \ M.
Proof. We have I ‚ R i¬ I, equivalently S, is contained in some maximal ideal M. In
other words, I ‚ R i¬ ∃M such that ∀x ∈ S we have x ∈ M. The contrapositive says
that I = R i¬ ∀M ∃x ∈ S such that x ∈ M. ™
/


0.2.4 Lemma
√ √
Let I and J be ideals of the ring R. Then I + J = R i¬ I+ J = R.
Proof. The “only if” part holds because any ideal is contained in its radical. Thus assume
that 1 = a + b with am ∈ I and bn ∈ J. Then

m+n i j
1 = (a + b)m+n = ab.
i
i+j=m+n


Now if i + j = m + n, then either i ≥ m or j ≥ n. Thus every term in the sum belongs
either to I or to J, hence to I + J. Consequently, 1 ∈ I + J. ™
0.3. NAKAYAMA™S LEMMA 3

0.3 Nakayama™s Lemma
First, we give an example of the determinant trick ; see (2.1.2) for another illustration.


0.3.1 Theorem
Let M be a ¬nitely generated R-module, and I an ideal of R such that IM = M . Then
there exists a ∈ I such that (1 + a)M = 0.
Proof. Let x1 , . . . , xn generate M . Since IM = M , we have equations of the form
n n
xi = j=1 aij xj , with aij ∈ I. The equations may be written as j=1 (δij ’ aij )xj = 0.
If In is the n by n identity matrix, we have (In ’ A)x = 0, where A = (aij ) and x is a
column vector whose coe¬cients are the xi . Premultiplying by the adjoint of (In ’ A),
we obtain ∆x = 0, where ∆ is the determinant of (In ’ A). Thus ∆xi = 0 for all i, hence
∆M = 0. But if we look at the determinant of In ’ A, we see that it is of the form 1 + a
for some element a ∈ I. ™
Here is a generalization of a familiar property of linear transformations on ¬nite-
dimensional vector spaces.


0.3.2 Theorem
If M is a ¬nitely generated R-module and f : M ’ M is a surjective homomorphism,
then f is an isomorphism.
Proof. We can make M into an R[X]-module via Xx = f (x), x ∈ M . (Thus X 2 x =
f (f (x)), etc.) Let I = (X); we claim that IM = M . For if m ∈ M , then by the
hypothesis that f is surjective, m = f (x) for some x ∈ M , and therefore Xx = f (x) = m.
But X ∈ I, so m ∈ IM . By (0.3.1), there exists g = g(X) ∈ I such that (1 + g)M = 0.
But by de¬nition of I, g must be of the form Xh(X) with h(X) ∈ R[X]. Thus (1+g)M =
[1 + Xh(X)]M = 0.
We can now prove that f is injective. Suppose that x ∈ M and f (x) = 0. Then

0 = [1 + Xh(X)]x = [1 + h(X)X]x = x + h(X)f (x) = x + 0 = x. ™

In (0.3.2), we cannot replace “surjective” by “injective”. For example, let f (x) = nx on
the integers. If n ≥ 2, then f is injective but not surjective.
The next result is usually referred to as Nakayama™s lemma. Sometimes, Akizuki and
Krull are given some credit, and as a result, a popular abbreviation for the lemma is
NAK.


0.3.3 NAK
(a) If M is a ¬nitely generated R-module, I an ideal of R contained in the Jacobson
radical J(R), and IM = M , then M = 0.
(b) If N is a submodule of the ¬nitely generated R-module M , I an ideal of R contained
in the Jacobson radical J(R), and M = N + IM , then M = N .
4 CHAPTER 0. RING THEORY BACKGROUND

Proof.
(a) By (0.3.1), (1 + a)M = 0 for some a ∈ I. Since I ⊆ J(R), 1 + a is a unit by (0.2.1).
Multiplying the equation (1 + a)M = 0 by the inverse of 1 + a, we get M = 0.
(b) By hypothesis, M/N = I(M/N ), and the result follows from (a). ™
Here is an application of NAK.

0.3.4 Proposition
Let R be a local ring with maximal ideal J. Let M be a ¬nitely generated R-module, and
let V = M/JM . Then
(i) V is a ¬nite-dimensional vector space over the residue ¬eld k = R/J.
(ii) If {x1 + JM, . . . , xn + JM } is a basis for V over k, then {x1 , . . . , xn } is a minimal
set of generators for M .
(iii) Any two minimal generating sets for M have the same cardinality.
Proof.
(i) Since J annihilates M/JM , V is a k-module, that is, a vector space over k. Since M
is ¬nitely generated over R, V is a ¬nite-dimensional vector space over k.
n
(ii) Let N = i=1 Rxi . Since the xi + JM generate V = M/JM , we have M = N + JM .
By NAK, M = N , so the xi generate M . If a proper subset of the xi were to generate
M , then the corresponding subset of the xi + JM would generate V , contradicting the
assumption that V is n-dimensional.
(iii) A generating set S for M with more than n elements determines a spanning set for
V , which must contain a basis with exactly n elements. By (ii), S cannot be minimal. ™


0.4 Localization
Let S be a subset of the ring R, and assume that S is multiplicative, in other words,
0 ∈ S, 1 ∈ S, and if a and b belong to S, so does ab. In the case of interest to us, S will
/
be the complement of a prime ideal. We would like to divide elements of R by elements
of S to form the localized ring S ’1 R, also called the ring of fractions of R by S. There
is no di¬culty when R is an integral domain, because in this case all division takes place
in the fraction ¬eld of R. We will sketch the general construction for arbitrary rings R.
For full details, see TBGY, Section 2.8.

0.4.1 Construction of the Localized Ring
If S is a multiplicative subset of the ring R, we de¬ne an equivalence relation on R — S
by (a, b) ∼ (c, d) i¬ for some s ∈ S we have s(ad ’ bc) = 0. If a ∈ R and b ∈ S, we de¬ne
the fraction a/b as the equivalence class of (a, b). We make the set of fractions into a ring
in a natural way. The sum of a/b and c/d is de¬ned as (ad + bc)/bd, and the product of
a/b and c/d is de¬ned as ac/bd. The additive identity is 0/1, which coincides with 0/s for
every s ∈ S. The additive inverse of a/b is ’(a/b) = (’a)/b. The multiplicative identity
is 1/1, which coincides with s/s for every s ∈ S. To summarize:
S ’1 R is a ring. If R is an integral domain, so is S ’1 R. If R is an integral domain and
S = R \ {0}, then S ’1 R is a ¬eld, the fraction ¬eld of R.
0.4. LOCALIZATION 5

There is a natural ring homomorphism h : R ’ S ’1 R given by h(a) = a/1. If S
has no zero-divisors, then h is a monomorphism, so R can be embedded in S ’1 R. In
particular, a ring R can be embedded in its full ring of fractions S ’1 R, where S consists
of all non-divisors of 0 in R. An integral domain can be embedded in its fraction ¬eld.
Our goal is to study the relation between prime ideals of R and prime ideals of S ’1 R.

0.4.2 Lemma
If X is any subset of R, de¬ne S ’1 X = {x/s : x ∈ X, s ∈ S}. If I is an ideal of R, then
S ’1 I is an ideal of S ’1 R. If J is another ideal of R, then
(i) S ’1 (I + J) = S ’1 I + S ’1 J;
(ii) S ’1 (IJ) = (S ’1 I)(S ’1 J);
(iii) S ’1 (I © J) = (S ’1 I) © (S ’1 J);
(iv) S ’1 I is a proper ideal i¬ S © I = ….
Proof. The de¬nitions of addition and multiplication in S ’1 R imply that S ’1 R is an
ideal, and that in (i), (ii) and (iii), the left side is contained in the right side. The reverse
inclusions in (i) and (ii) follow from

ab at + bs a b ab
+= , =.
s t st st st
To prove (iii), let a/s = b/t, where a ∈ I, b ∈ J, s, t ∈ S. There exists u ∈ S such that
u(at ’ bs) = 0. Then a/s = uat/ust = ubs/ust ∈ S ’1 (I © J).
Finally, if s ∈ S © I, then 1/1 = s/s ∈ S ’1 I, so S ’1 I = S ’1 R. Conversely, if
S ’1 I = S ’1 R, then 1/1 = a/s for some a ∈ I, s ∈ S. There exists t ∈ S such that
t(s ’ a) = 0, so at = st ∈ S © I. ™
Ideals in S ’1 R must be of a special form.

0.4.3 Lemma
Let h be the natural homomorphism from R to S ’1 R [see (0.4.1)]. If J is an ideal of
S ’1 R and I = h’1 (J), then I is an ideal of R and S ’1 I = J.
Proof. I is an ideal by the basic properties of preimages of sets. Let a/s ∈ S ’1 I, with
a ∈ I and s ∈ S. Then a/1 = h(a) ∈ J, so a/s = (a/1)(1/s) ∈ J. Conversely, let a/s ∈ J,
with a ∈ R, s ∈ S. Then h(a) = a/1 = (a/s)(s/1) ∈ J, so a ∈ I and a/s ∈ S ’1 I. ™
Prime ideals yield sharper results.

0.4.4 Lemma
If I is any ideal of R, then I ⊆ h’1 (S ’1 I). There will be equality if I is prime and disjoint
from S.
Proof. If a ∈ I, then h(a) = a/1 ∈ S ’1 I. Thus assume that I is prime and disjoint from
S, and let a ∈ h’1 (S ’1 I). Then h(a) = a/1 ∈ S ’1 I, so a/1 = b/s for some b ∈ I, s ∈ S.
There exists t ∈ S such that t(as ’ b) = 0. Thus ast = bt ∈ I, with st ∈ I because
/
S © I = …. Since I is prime, we have a ∈ I. ™
6 CHAPTER 0. RING THEORY BACKGROUND

0.4.5 Lemma
If I is a prime ideal of R disjoint from S, then S ’1 I is a prime ideal of S ’1 R.
Proof. By part (iv) of (0.4.2), S ’1 I is a proper ideal. Let (a/s)(b/t) = ab/st ∈ S ’1 I,
with a, b ∈ R, s, t ∈ S. Then ab/st = c/u for some c ∈ I, u ∈ S. There exists v ∈ S such
that v(abu ’ cst) = 0. Thus abuv = cstv ∈ I, and uv ∈ I because S © I = …. Since I is
/
prime, ab ∈ I, hence a ∈ I or b ∈ I. Therefore either a/s or b/t belongs to S ’1 I. ™
The sequence of lemmas can be assembled to give a precise conclusion.


0.4.6 Theorem
There is a one-to-one correspondence between prime ideals P of R that are disjoint from
S and prime ideals Q of S ’1 R, given by

P ’ S ’1 P and Q ’ h’1 (Q).

Proof. By (0.4.3), S ’1 (h’1 (Q)) = Q, and by (0.4.4), h’1 (S ’1 P ) = P . By (0.4.5), S ’1 P
is a prime ideal, and h’1 (Q) is a prime ideal by the basic properties of preimages of sets.
If h’1 (Q) meets S, then by (0.4.2) part (iv), Q = S ’1 (h’1 (Q)) = S ’1 R, a contradiction.
Thus the maps P ’ S ’1 P and Q ’ h’1 (Q) are inverses of each other, and the result
follows. ™


0.4.7 De¬nitions and Comments
If P is a prime ideal of R, then S = R \ P is a multiplicative set. In this case, we write
RP for S ’1 R, and call it the localization of R at P . We are going to show that RP is
a local ring, that is, a ring with a unique maximal ideal. First, we give some conditions
equivalent to the de¬nition of a local ring.


0.4.8 Proposition
For a ring R, the following conditions are equivalent.
(i) R is a local ring;
(ii) There is a proper ideal I of R that contains all nonunits of R;
(iii) The set of nonunits of R is an ideal.
Proof.
(i) implies (ii): If a is a nonunit, then (a) is a proper ideal, hence is contained in the
unique maximal ideal I.
(ii) implies (iii): If a and b are nonunits, so are a + b and ra. If not, then I contains a
unit, so I = R, contradicting the hypothesis.
(iii) implies (i): If I is the ideal of nonunits, then I is maximal, because any larger ideal J
would have to contain a unit, so J = R. If H is any proper ideal, then H cannot contain
a unit, so H ⊆ I. Therefore I is the unique maximal ideal. ™
0.4. LOCALIZATION 7

0.4.9 Theorem
RP is a local ring.
Proof. Let Q be a maximal ideal of RP . Then Q is prime, so by (0.4.6), Q = S ’1 I
for some prime ideal I of R that is disjoint from S = R \ P . In other words, I ⊆ P .
Consequently, Q = S ’1 I ⊆ S ’1 P . If S ’1 P = RP = S ’1 R, then by (0.4.2) part (iv), P
is not disjoint from S = R \ P , which is impossible. Therefore S ’1 P is a proper ideal
containing every maximal ideal, so it must be the unique maximal ideal. ™

0.4.10 Remark
It is convenient to write the ideal S ’1 I as IRP . There is no ambiguity, because the
product of an element of I and an arbitrary element of R belongs to I.

0.4.11 Localization of Modules
If M is an R-module and S a multiplicative subset of R, we can essentially repeat the
construction of (0.4.1) to form the localization of M by S, and thereby divide elements
of M by elements of S. If x, y ∈ M and s, t ∈ S, we call (x, s) and (y, t) equivalent if for
some u ∈ S, we have u(tx ’ sy) = 0. The equivalence class of (x, s) is denoted by x/s,
and addition is de¬ned by
xy tx + sy
+= .
s t st
If a/s ∈ S ’1 R and x/t ∈ S ’1 M , we de¬ne
ax ax
= .
st st
In this way, S ’1 M becomes an S ’1 R-module. Exactly as in (0.4.2), if M and N are
submodules of an R-module L, then

S ’1 (M + N ) = S ’1 M + S ’1 N and S ’1 (M © N ) = (S ’1 M ) © (S ’1 N ).
Chapter 1

Primary Decomposition and
Associated Primes

1.1 Primary Submodules and Ideals

1.1.1 De¬nitions and Comments

If N is a submodule of the R-module M , and a ∈ R, let »a : M/N ’ M/N be mul-
tiplication by a. We say that N is a primary submodule of M if N is proper and for
every a, »a is either injective or nilpotent. Injectivity means that for all x ∈ M , we have
ax ∈ N ’ x ∈ N . Nilpotence means that for some positive integer n, an M ⊆ N , that is,
an belongs to the annihilator of M/N , denoted by ann(M/N ). Equivalently, a belongs to
the radical of the annihilator of M/N , denoted by rM (N ).
Note that »a cannot be both injective and nilpotent. If so, nilpotence gives an M =
a(an’1 M ) ⊆ N , and injectivity gives an’1 M ⊆ N . Inductively, M ⊆ N , so M = N ,
contradicting the assumption that N is proper. Thus if N is a primary submodule of M ,
then rM (N ) is the set of all a ∈ R such that »a is not injective. Since rM (N ) is the radical
of an ideal, it is an ideal of R, and in fact it is a prime ideal. For if »a and »b fail to be
injective, so does »ab = »a —¦ »b . (Note that rM (N ) is proper because »1 is injective.) If
P = rM (N ), we say that N is P -primary.

If I is any ideal of R, then rR (I) = I, because ann(R/I) = I. (Note that a ∈
ann(R/I) i¬ aR ⊆ I i¬ a = a1 ∈ I.)
Specializing to M = R and replacing a by y, we de¬ne a primary ideal in a ring R
as a proper ideal Q such that if xy ∈ Q, then either x ∈ Q or y n ∈ Q for some n ≥ 1.
Equivalently, R/Q = 0 and every zero-divisor in R/Q is nilpotent.

A useful observation is that if P is a prime ideal, then P n = P for all n ≥ 1. (The
radical of P n is the intersection of all prime ideals containing P n , one of which is P . Thus
√ √
P n ⊆ P . Conversely, if x ∈ P , then xn ∈ P n , so x ∈ P n .)

1
2 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

1.1.2 Lemma

I is a maximal ideal M, then I is M-primary.
If

Proof. Suppose that ab ∈ I and b does not belong to I = M. Then by maximality of
M, it follows that M + Rb = R, so for some m ∈ M and r ∈ R we have m + rb = 1. Now

m ∈ M = I, hence mk ∈ I for some k ≥ 1. Thus 1 = 1k = (m + rb)k = mk + sb for
some s ∈ R. Multiply by a to get a = amk + sab ∈ I. ™


1.1.3 Corollary
If M is a maximal ideal, then Mn is M-primary for every n ≥ 1.

Proof. As we observed in (1.1.1), Mn = M, and the result follows from (1.1.2). ™



1.2 Primary Decomposition
1.2.1 De¬nitions and Comments
A primary decomposition of the submodule N of M is given by N = ©r Ni , where the
i=1
Ni are Pi -primary submodules. The decomposition is reduced if the Pi are distinct and
N cannot be expressed as the intersection of a proper subcollection of the Ni .
We can always extract a reduced primary decomposition from an unreduced one, by
discarding those Ni that contain ©j=i Nj and intersecting those Ni that are P -primary
for the same P . The following result justi¬es this process.


1.2.2 Lemma
If N1 , . . . , Nk are P -primary, then ©k Ni is P -primary.
i=1

Proof. We may assume that k = 2; an induction argument takes care of larger values.
Let N = N1 © N2 and rM (N1 ) = rM (N2 ) = P . Assume for the moment that rM (N ) = P .
If a ∈ R, x ∈ M, ax ∈ N , and a ∈ rM (N ), then since N1 and N2 are P -primary, we have
/
x ∈ N1 © N2 = N . It remains to show that rM (N ) = P . If a ∈ P , then there are positive
integers n1 and n2 such that an1 M ⊆ N1 and an2 M ⊆ N2 . Therefore an1 +n2 M ⊆ N , so
a ∈ rM (N ). Conversely, if a ∈ rM (N ) then a belongs to rM (Ni ) for i = 1, 2, and therefore
a ∈ P. ™
We now prepare to prove that every submodule of a Noetherian module has a primary
decomposition.


1.2.3 De¬nition
The proper submodule N of M is irreducible if N cannot be expressed as N1 © N2 with
N properly contained in the submodules Ni , i = 1, 2.
1.3. ASSOCIATED PRIMES 3

1.2.4 Proposition
If N is an irreducible submodule of the Noetherian module M , then N is primary.
Proof. If not, then for some a ∈ R, »a : M/N ’ M/N is neither injective nor nilpotent.
The chain ker »a ⊆ ker »2 ⊆ ker »3 ⊆ · · · terminates by the ascending chain condition, say
a a
at ker »a . Let • = »a ; then ker • = ker •2 and we claim that ker • © im • = 0. Suppose
i i

x ∈ ker • © im •, and let x = •(y). Then 0 = •(x) = •2 (y), so y ∈ ker •2 = ker •, so
x = •(y) = 0.
Now »a is not injective, so ker • = 0, and »a is not nilpotent, so »i can™t be 0 (because
a
ai M ⊆ N ). Consequently, im • = 0.
Let p : M ’ M/N be the canonical epimorphism, and set N1 = p’1 (ker •), N2 =
p’1 (im •). We will prove that N = N1 © N2 . If x ∈ N1 © N2 , then p(x) belongs to
both ker • and im •, so p(x) = 0, in other words, x ∈ N . Conversely, if x ∈ N , then
p(x) = 0 ∈ ker • © im •, so x ∈ N1 © N2 .
Finally, we will show that N is properly contained in both N1 and N2 , so N is reducible,
a contradiction. Choose a nonzero element y ∈ ker •. Since p is surjective, there exists
x ∈ M such that p(x) = y. Thus x ∈ p’1 (ker •) = N1 (because y = p(x) ∈ ker •), but
x ∈ N (because p(x) = y = 0). Similarly, N ‚ N2 (with 0 = y ∈ im •), and the result
/
follows. ™

1.2.5 Theorem
If N is a proper submodule of the Noetherian module M , then N has a primary decom-
position, hence a reduced primary decomposition.
Proof. We will show that N can be expressed as a ¬nite intersection of irreducible sub-
modules of M , so that (1.2.4) applies. Let S be the collection of all submodules of M
that cannot be expressed in this form. If S is nonempty, then S has a maximal element
N (because M is Noetherian). By de¬nition of S, N must be reducible, so we can write
N = N1 © N2 , N ‚ N1 , N ‚ N2 . By maximality of N , N1 and N2 can be expressed
as ¬nite intersections of irreducible submodules, hence so can N , contradicting N ∈ S.
Thus S is empty. ™


1.3 Associated Primes
1.3.1 De¬nitions and Comments
Let M be an R-module, and P a prime ideal of R. We say that P is an associated prime
of M (or that P is associated to M ) if P is the annihilator of some nonzero x ∈ M . The
set of associated primes of M is denoted by AP(M ). (The standard notation is Ass(M).
Please do not use this regrettable terminology.)
Here is a useful characterization of associated primes.

1.3.2 Proposition
The prime ideal P is associated to M if and only if there is an injective R-module homo-
morphism from R/P to M . Therefore if N is a submodule of M , then AP(N ) ⊆ AP(M ).
4 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

Proof. If P is the annihilator of x = 0, the desired homomorphism is given by r +P ’ rx.
Conversely, if an injective R-homomorphism from R/P to M exists, let x be the image of
1 + P , which is nonzero in R/P . By injectivity, x = 0. We will show that P = annR (x),
the set of elements r ∈ R such that rx = 0. If r ∈ P , then r + P = 0, so rx = 0, and
therefore r ∈ annR (x). If rx = 0, then by injectivity, r + P = 0, so r ∈ P . ™
Associated primes exist under wide conditions, and are sometimes unique.


1.3.3 Proposition
If M = 0, then AP(M ) is empty. The converse holds if R is a Noetherian ring.
Proof. There are no nonzero elements in the zero module, hence no associated primes.
Assuming that M = 0 and R is Noetherian, there is a maximal element I = annR x in
the collection of all annihilators of nonzero elements of M . The ideal I must be proper,
for if I = R, then x = 1x = 0, a contradiction. If we can show that I is prime, we have
I ∈ AP(M ), as desired. Let ab ∈ I with a ∈ I. Then abx = 0 but ax = 0, so b ∈ ann(ax).
/
But I = ann x ⊆ ann(ax), and the maximality of I gives I = ann(ax). Consequently,
b ∈ I. ™


1.3.4 Proposition
For any prime ideal P , AP(R/P ) = {P }.
Proof. By (1.3.2), P is an associated prime of R/P because there certainly is an R-
monomorphism from R/P to itself. If Q ∈ AP(R/P ), we must show that Q = P .
Suppose that Q = ann(r + P ) with r ∈ P . Then s ∈ Q i¬ sr ∈ P i¬ s ∈ P (because P is
/
prime). ™


1.3.5 Remark
Proposition 1.3.4 shows that the annihilator of any nonzero element of R/P is P .
The next result gives us considerable information about the elements that belong to
associated primes.


1.3.6 Theorem
Let z(M ) be the set of zero-divisors of M , that is, the set of all r ∈ R such that rx = 0
for some nonzero x ∈ M . Then ∪{P : P ∈ AP(M )} ⊆ z(M ), with equality if R is
Noetherian.
Proof. The inclusion follows from the de¬nition of associated prime; see (1.3.1). Thus
assume a ∈ z(M ), with ax = 0, x ∈ M, x = 0. Then Rx = 0, so by (1.3.3) [assuming R
Noetherian], Rx has an associated prime P = ann(bx). Since ax = 0 we have abx = 0, so
a ∈ P . But P ∈ AP(Rx) ⊆ AP(M ) by (1.3.2). Therefore a ∈ ∪{P : P ∈ AP(M )}. ™
Now we prove a companion result to (1.3.2).
1.3. ASSOCIATED PRIMES 5

1.3.7 Proposition
If N is a submodule of M , then AP(M ) ⊆ AP(N ) ∪ AP(M/N ).
Proof. Let P ∈ AP(M ), and let h : R/P ’ M be a monomorphism. Set H = h(R/P )
and L = H © N .
Case 1: L = 0. Then the map from H to M/N given by h(r + P ) ’ h(r + P ) + N is
a monomorphism. (If h(r + P ) belongs to N , it must belong to H © N = 0.) Thus H
is isomorphic to a submodule of M/N , so by de¬nition of H, there is a monomorphism
from R/P to M/N . Thus P ∈ AP(M/N ).
Case 2: L = 0. If L has a nonzero element x, then x must belong to both H and N , and
H is isomorphic to R/P via h. Thus x ∈ N and the annihilator of x coincides with the
annihilator of some nonzero element of R/P . By (1.3.5), ann x = P , so P ∈ AP(N ). ™

1.3.8 Corollary

AP( Mj = AP(Mj ).
j∈J j∈J

Proof. By (1.3.2), the right side is contained in the left side. The result follows from
(1.3.7) when the index set is ¬nite. For example,

AP(M1 • M2 • M3 ) ⊆ AP(M1 ) ∪ AP(M/M1 )
= AP(M1 ) ∪ AP(M2 • M3 )
⊆ AP(M1 ) ∪ AP(M2 ) ∪ AP(M3 ).

In general, if P is an associated prime of the direct sum, then there is a monomorphism
from R/P to •Mj . The image of the monomorphism is contained in the direct sum of
¬nitely many components, as R/P is generated as an R-module by the single element
1 + P . This takes us back to the ¬nite case. ™
We now establish the connection between associated primes and primary decomposi-
tion, and show that under wide conditions, there are only ¬nitely many associated primes.

1.3.9 Theorem
Let M be a nonzero ¬nitely generated module over the Noetherian ring R, so that by
(1.2.5), every proper submodule of M has a reduced primary decomposition. In particular,
the zero module can be expressed as ©r Ni , where Ni is Pi -primary. Then AP(M ) =
i=1
{P1 , . . . , Pr }, a ¬nite set.
Proof. Let P be an associated prime of M , so that P = ann(x), x = 0, x ∈ M . Renumber
the Ni so that x ∈ Ni for 1 ¤ i ¤ j and x ∈ Ni for j + 1 ¤ i ¤ r. Since Ni is Pi -primary,
/
we have Pi = rM (Ni ) (see (1.1.1)). Since Pi is ¬nitely generated, Pini M ⊆ Ni for some
ni ≥ 1. Therefore
j r

Pini )x
( Ni = (0)
i=1 i=1
6 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

so ©j Pini ⊆ ann(x) = P . (By our renumbering, there is a j rather than an r on the left
i=1
side of the inclusion.) Since P is prime, Pi ⊆ P for some i ¤ j. We claim that Pi = P ,
so that every associated prime must be one of the Pi . To verify this, let a ∈ P . Then
ax = 0 and x ∈ Ni , so »a is not injective and therefore must be nilpotent. Consequently,
/
a ∈ rM (Ni ) = Pi , as claimed.
Conversely, we show that each Pi is an associated prime. Without loss of generality, we
may take i = 1. Since the decomposition is reduced, N1 does not contain the intersection
of the other Ni ™s, so we can choose x ∈ N2 ©· · ·©Nr with x ∈ N1 . Now N1 is P1 -primary, so
/
as in the preceding paragraph, for some n ≥ 1 we have P1 x ⊆ N1 but P1 x ⊆ N1 . (Take
n’1
n

P1 x = Rx and recall that x ∈ N1 .) If we choose y ∈ P1 x \ N1 (hence y = 0), the proof
n’1
0
/
will be complete upon showing that P1 is the annihilator of y. We have P1 y ⊆ P1 x ⊆ N1
n

and x ∈ ©r Ni , so P1 x ⊆ ©r Ni . Thus P1 y ⊆ ©r Ni = (0), so P1 ⊆ ann y. On the
n
i=2 i=2 i=1
other hand, if a ∈ R and ay = 0, then ay ∈ N1 but y ∈ N1 , so »a : M/N1 ’ M/N1 is not
/
injective and is therefore nilpotent. Thus a ∈ rM (N1 ) = P1 . ™
We can now say something about uniqueness in primary decompositions.


1.3.10 First Uniqueness Theorem
Let M be a ¬nitely generated module over the Noetherian ring R. If N = ©r Ni is a
i=1
reduced primary decomposition of the submodule N , and Ni is Pi -primary, i = 1, . . . , r,
then (regarding M and R as ¬xed) the Pi are uniquely determined by N .
Proof. By the correspondence theorem, a reduced primary decomposition of (0) in M/N
is given by (0) = ©r Ni /N , and Ni /N is Pi -primary, 1 ¤ i ¤ r. By (1.3.9),
i=1


AP(M/N ) = {P1 , . . . , Pr }.

But [see (1.3.1)] the associated primes of M/N are determined by N . ™


1.3.11 Corollary
Let N be a submodule of M (¬nitely generated over the Noetherian ring R). Then N is
P -primary i¬ AP(M/N ) = {P }.
Proof. The “only if” part follows from the displayed equation above. Conversely, if P is
the only associated prime of M/N , then N coincides with a P -primary submodule N ,
and hence N (= N ) is P -primary. ™


1.3.12 De¬nitions and Comments
Let N = ©r Ni be a reduced primary decomposition, with associated primes P1 , . . . , Pr .
i=1
We say that Ni is an isolated (or minimal ) component if Pi is minimal, that is Pi does not
properly contain any Pj , j = i. Otherwise, Ni is an embedded component (see Exercise 5
for an example). Embedded components arise in algebraic geometry in situations where
one irreducible algebraic set is properly contained in another.
1.4. ASSOCIATED PRIMES AND LOCALIZATION 7

1.4 Associated Primes and Localization
To get more information about uniqueness in primary decompositions, we need to look at
associated primes in localized rings and modules. In this section, S will be a multiplicative
subset of the Noetherian ring R, RS the localization of R by S, and MS the localization
of the R-module M by S. Recall that P ’ PS = P RS is a bijection of C, the set of prime
ideals of R not meeting S, and the set of all prime ideals of RS .
The set of associated primes of the R-module M will be denoted by APR (M ). We
need a subscript to distinguish this set from APRS (MS ), the set of associated primes of
the RS -module MS .

1.4.1 Lemma
Let P be a prime ideal not meeting S. If P ∈ APR (M ), then PS = P RS ∈ APRS (MS ).
(By the above discussion, the map P ’ PS is the restriction of a bijection and therefore
must be injective.)
Proof. If P is the annihilator of the nonzero element x ∈ M , then PS is the annihilator
of the nonzero element x/1 ∈ MS . (By (1.3.6), no element of S can be a zero-divisor,
so x/1 is indeed nonzero.) For if a ∈ P and a/s ∈ PS , then (a/s)(x/1) = ax/s = 0.
Conversely, if (a/s)(x/1) = 0, then there exists t ∈ S such that tax = 0, and it follows
that a/s = at/st ∈ PS . ™

1.4.2 Lemma
The map of (1.4.1) is surjective, hence is a bijection of APR (M ) © C and APRS (MS ).
Proof. Let P be generated by a1 , . . . , an . Suppose that PS is the annihilator of the
nonzero element x/t ∈ MS . Then (ai /1)(x/t) = 0, 1 ¤ i ¤ n. For each i there exists
si ∈ S such that si ai x = 0. If s is the product of the si , then sai x = 0 for all i, hence
sax = 0 for all a ∈ P . Thus P ⊆ ann(sx). On the other hand, suppose b annihilates sx.
Then (b/1)(x/t) = bsx/st = 0, so b/1 ∈ PS , and consequently b/1 = b /s for some b ∈ P
and s ∈ S. This means that for some u ∈ S we have u(bs ’ b ) = 0. Now b , hence ub ,
belongs to P , and therefore so does ubs . But us ∈ P (because S © P = …). We conclude
/
that b ∈ P , so P = ann(sx). As in (1.4.1), s cannot be a zero-divisor, so sx = 0 and the
proof is complete. ™

1.4.3 Lemma
Let M be a ¬nitely generated module over the Noetherian ring R, and N a P -primary
submodule of M . Let P be any prime ideal of R, and set M = MP , N = NP . If
P ⊆ P , then N = M .
Proof. By (1.4.1) and (1.4.2), there is a bijection between APRP (M/N )P (which coin-
cides with APRP (M /N )) and the intersection APR (M/N ) © C, where C is the set of
prime ideals contained in P (in other words, not meeting S = R \ P ). By (1.3.11), there
is only one associated prime of M/N over R, namely P , which is not contained in P by
hypothesis. Thus APR (M/N ) © C is empty, so by (1.3.3), M /N = 0, and the result
follows. ™
8 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

At the beginning of the proof of (1.4.3), we have taken advantage of the isomorphism
between (M/N )P and M /N . The result comes from the exactness of the localization
functor. If this is unfamiliar, look ahead to the proof of (1.5.3), where the technique is
spelled out. See also TBGY, Section 8.5, Problem 5.

1.4.4 Lemma
In (1.4.3), if P ⊆ P , then N = f ’1 (N ), where f is the natural map from M to M .
Proof. As in (1.4.3), APR (M/N ) = {P }. Since P ⊆ P , we have R \ P ⊆ R \ P . By
(1.3.6), R \ P contains no zero-divisors of M/N , because all such zero-divisors belong to
P . Thus the natural map g : x ’ x/1 of M/N to (M/N )P ∼ M /N is injective. (If
=
x/1 = 0, then sx = 0 for some s ∈ S = R \ P , and since s is not a zero-divisor, we have
x = 0.)
If x ∈ N , then f (x) ∈ N by de¬nition of f , so assume x ∈ f ’1 (N ). Then f (x) ∈ N ,
so f (x) + N is 0 in M /N . By injectivity of g, x + N is 0 in M/N , in other words, x ∈ N ,
and the result follows. ™

1.4.5 Second Uniqueness Theorem
Let M be a ¬nitely generated module over the Noetherian ring R. Suppose that N =
©r Ni is a reduced primary decomposition of the submodule N , and Ni is Pi -primary,
i=1
i = 1, . . . , r. If Pi is minimal, then (regarding M and R as ¬xed) Ni is uniquely determined
by N .
Proof. Suppose that P1 is minimal, so that P1 ⊇ Pi , i > 1. By (1.4.3) with P =
Pi , P = P1 , we have (Ni )P1 = MP1 for i > 1. By (1.4.4) with P = P = P1 , we have
N1 = f ’1 [(N1 )P1 ], where f is the natural map from M to MP1 . Now
NP1 = (N1 )P1 © ©r (Ni )P1 = (N1 )P1 © MP1 = (N1 )P1 .
i=2

Thus N1 = f ’1 [(N1 )P1 ] = f ’1 (NP1 ) depends only on N and P1 , and since P1 depends
on the ¬xed ring R, it follows that N1 depends only on N . ™


1.5 The Support of a Module
The support of a module M is closely related to the set of associated primes of M . We
will need the following result in order to proceed.

1.5.1 Proposition
M is the zero module if and only if MP = 0 for every prime ideal P , if and only if MM = 0
for every maximal ideal M.
Proof. It su¬ces to show that if MM = 0 for all maximal ideals M, then M = 0.
Choose a nonzero element x ∈ M , and let I be the annihilator of x. Then 1 ∈ I (because
/
1x = x = 0), so I is a proper ideal and is therefore contained in a maximal ideal M. By
hypothesis, x/1 is 0 in MM , hence there exists a ∈ M (so a ∈ I) such that ax = 0. But
/ /
then by de¬nition of I we have a ∈ I, a contradiction. ™
1.5. THE SUPPORT OF A MODULE 9

1.5.2 De¬nitions and Comments
The support of an R-module M (notation Supp M ) is the set of prime ideals P of R such
that MP = 0. Thus Supp M = … i¬ MP = 0 for all prime ideals P . By (1.5.1), this is
equivalent to M = 0.
If I is any ideal of R, we de¬ne V (I) as the set of prime ideals containing I. In
algebraic geometry, the Zariski topology on Spec R has the sets V (I) as its closed sets.


1.5.3 Proposition
Supp R/I = V (I).
Proof. We apply the localization functor to the exact sequence 0 ’ I ’ R ’ R/I ’ 0 to
get the exact sequence 0 ’ IP ’ RP ’ (R/I)P ’ 0. Consequently, (R/I)P ∼ RP /IP .=
Thus P ∈ Supp R/I i¬ RP ⊃ IP i¬ IP is contained in a maximal ideal, necessarily P RP .
But this is equivalent to I ⊆ P . To see this, suppose a ∈ I, with a/1 ∈ IP ⊆ P RP . Then
a/1 = b/s for some b ∈ P, s ∈ P . There exists c ∈ P such that c(as ’ b) = 0. We have
/ /
cas = cb ∈ P , a prime ideal, and cs ∈ P . We conclude that a ∈ P . ™
/


1.5.4 Proposition
Let 0 ’ M ’ M ’ M ’ 0 be exact, hence 0 ’ MP ’ MP ’ MP ’ 0 is exact.
Then

Supp M = Supp M ∪ Supp M .

Proof. Let P belong to Supp M \ Supp M . Then MP = 0, so the map MP ’ MP is
injective as well as surjective, hence is an isomorphism. But MP = 0 by assumption, so
MP = 0, and therefore P ∈ Supp M . On the other hand, since MP is isomorphic to a
submodule of MP , it follows that Supp M ⊆ Supp M . If MP = 0, then MP = 0 (because
MP ’ MP is surjective). Thus Supp M ⊆ Supp M . ™

Supports and annihilators are connected by the following basic result.


1.5.5 Theorem
If M is a ¬nitely generated R-module, then Supp M = V (ann M ).
Proof. Let M = Rx1 + · · · + Rxn , so that MP = (Rx1 )P + · · · + (Rxn )P . Then Supp M =
∪n Supp Rxi , and by the ¬rst isomorphism theorem, Rxi ∼ R/ ann xi . By (1.5.3),
=
i=1
Supp Rxi = V (ann xi ). Therefore Supp M = ∪i=1 V (ann xi ) = V (ann M ). To justify
n

the last equality, note that if P ∈ V (ann xi ), then P ⊇ ann xi ⊇ ann M . Conversely, if
P ⊇ ann M = ©n ann xi , then P ⊇ ann xi for some i. ™
i=1


And now we connect associated primes and annihilators.
10 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

1.5.6 Proposition
If M is a ¬nitely generated module over the Noetherian ring R, then

P = ann M .
P ∈AP(M )


Proof. If M = 0, then by (1.3.3), AP(M ) = …, and the result to be proved is R = R.
Thus assume M = 0, so that (0) is a proper submodule. By (1.2.5) and (1.3.9), there is
a reduced primary decomposition (0) = ©r Ni , where for each i, Ni is Pi -primary and
i=1
AP(M ) =√ 1 , . . . , Pr }.
{P
If a ∈ ann M , then for some n ≥ 1 we have an M = 0. Thus for each i, »a : M/Ni ’
M/Ni is nilpotent [see (1.1.1)]. Consequently, a ∈ ©r rM (Ni ) = ©r Pi . Conversely, if
i=1 i=1
a belongs to this intersection, then √ all i there exists ni ≥ 1 such that ani M ⊆ Ni . If
for
n = max ni , then an M = 0, so a ∈ ann M . ™

1.5.7 Corollary
If R is a Noetherian ring, then the nilradical of R is the intersection of all associated
primes of R.

Proof. Take M = R in (1.5.6). Since ann R = 0, ann R is the nilradical. ™

And now, a connection between supports, associated primes and annihilators.

1.5.8 Proposition
Let M be a ¬nitely generated module over the Noetherian ring R, and let P be any prime
ideal of R. The following conditions are equivalent:
(1) P ∈ Supp M ;
(2) P ⊇ P for some P ∈ AP(M );
(3) P ⊇ ann M .
Proof. Conditions (1) and (3) are equivalent by (1.5.5). To prove that (1) implies (2),
let P ∈ Supp M . If P does not contain any associated prime of M , then P does not
contain the intersection of all associated primes (because P is prime). By (1.5.6), P does

not contain ann M , hence P cannot contain the smaller ideal ann M . This contradicts
(1.5.5). To prove that (2) implies (3), let Q be the intersection of all associated primes.

Then P ⊇ P ⊇ Q = [by (1.5.6)] ann M ⊇ ann M . ™

Here is the most important connection between supports and associated primes.

1.5.9 Theorem
Let M be a ¬nitely generated module over the Noetherian ring R. Then AP(M ) ⊆
Supp M , and the minimal elements of AP(M ) and Supp M are the same.
Proof. We have AP(M ) ⊆ Supp M by (2) implies (1) in (1.5.8), with P = P . If P is
minimal in Supp M , then by (1) implies (2) in (1.5.8), P contains some P ∈ AP(M ) ⊆
1.6. ARTINIAN RINGS 11

Supp M . By minimality, P = P . Thus P ∈ AP(M ), and in fact, P must be a minimal
associated prime. Otherwise, P ⊃ Q ∈ AP(M ) ⊆ Supp M , so that P is not minimal
in Supp M , a contradiction. Finally, let P be minimal among associated primes but not
minimal in Supp M . If P ⊃ Q ∈ Supp M , then by (1) implies (2) in (1.5.8), Q ⊇ P ∈
AP(M ). By minimality, P = P , contradicting P ⊃ Q ⊇ P . ™

Here is another way to show that there are only ¬nitely many associated primes.

1.5.10 Theorem
Let M be a nonzero ¬nitely generated module over the Noetherian ring R. Then there is
a chain of submodules 0 = M0 < M1 < · · · < Mn = M such that for each j = 1, . . . , n,
Mj /Mj’1 ∼ R/Pj , where the Pj are prime ideals of R. For any such chain, AP(M ) ⊆
=
{P1 , . . . , Pn }.
Proof. By (1.3.3), M has an associated prime P1 = ann x1 , with x1 a nonzero element
of M . Take M1 = Rx1 ∼ R/P1 (apply the ¬rst isomorphism theorem). If M = M1 ,
=
then the quotient module M/M1 is nonzero, hence [again by (1.3.3)] has an associated
prime P2 = ann(x2 + M1 ), x2 ∈ M1 . Let M2 = M1 + Rx2 . Now map R onto M2 /M1 by
/
r ’ rx2 + M1 . By the ¬rst isomorphism theorem, M2 /M1 ∼ R/P2 . Continue inductively
=
to produce the desired chain. (Since M is Noetherian, the process terminates in a ¬nite
number of steps.) For each j = 1, . . . , n, we have AP(Mj ) ⊆ AP(Mj’1 ) ∪ {Pj } by (1.3.4)
and (1.3.7). Another inductive argument shows that AP(M ) ⊆ {P1 , . . . , Pn }. ™

1.5.11 Proposition
In (1.5.10), each Pj belongs to Supp M . Thus (replacing AP(M ) by {P1 , . . . , Pn } in the
proof of (1.5.9)), the minimal elements of all three sets AP(M ), {P1 , . . . , Pn } and Supp M
are the same.
Proof. By (1.3.4) and (1.5.9), Pj ∈ Supp R/Pj , so by (1.5.10), Pj ∈ Supp Mj /Mj’1 . By
(1.5.4), Supp Mj /Mj’1 ⊆ Supp Mj , and ¬nally Supp Mj ⊆ Supp M because Mj ⊆ M . ™


1.6 Artinian Rings
1.6.1 De¬nitions and Comments
Recall that an R-module is Artinian if it satis¬es the descending chain condition on
submodules. If the ring R is Artinian as a module over itself, in other words, R satis¬es
the dcc on ideals, then R is said to be an Artinian ring. Note that Z is a Noetherian ring
that is not Artinian. Any ¬nite ring, for example Zn , is both Noetherian and Artinian,
and in fact we will prove later in the section that an Artinian ring must be Noetherian.
The theory of associated primes and supports will help us to analyze Artinian rings.

1.6.2 Lemma
If I is an ideal in the Artinian ring R, then R/I is an Artinian ring.
12 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

Proof. Since R/I is a quotient of an Artinian R-module, it is also an Artinian R-module.
In fact it is an R/I module via (r + I)(x + I) = rx + I, and the R-submodules are
identical to the R/I-submodules. Thus R/I is an Artinian R/I-module, in other words,
an Artinian ring. ™

1.6.3 Lemma
An Artinian integral domain is a ¬eld.
Proof. Let a be a nonzero element of the Artinian domain R. We must produce a
multiplicative inverse of a. The chain of ideals (a) ⊇ (a2 ) ⊇ (a3 ) ⊇ · · · stabilizes, so for
some t we have (at ) = (at+1 ). If at = bat+1 , then since R is a domain, ba = 1. ™

1.6.4 Proposition
If R is an Artinian ring, then every prime ideal of R is maximal. Therefore, the nilradical
N (R) coincides with the Jacobson radical J(R).
Proof. Let P be a prime ideal of R, so that R/I is an integral domain, Artinian by (1.6.2).
By (1.6.3), R/P is a ¬eld, hence P is maximal. ™

One gets the impression that the Artinian property puts strong constraints on a ring.
The following two results reinforce this conclusion.

1.6.5 Proposition
An Artinian ring has only ¬nitely many maximal ideals.
Proof. Let Σ be the collection of all ¬nite intersections of maximal ideals. Then Σ is
nonempty and has a minimal element I = M1 © · · · © Mr (by the Artinian property). If
M is any maximal ideal, then M ⊇ M © I ∈ Σ, so by minimality of I we have M © I = I.
But then M must contain one of the Mi (because M is prime), hence M = Mi (because
M and Mi are maximal). ™

1.6.6 Proposition
If R is Artinian, then the nilradical N (R) is nilpotent, hence by (1.6.4), the Jacobson
radical J(R) is nilpotent.
Proof. Let I = N (R). The chain I ⊇ I 2 ⊇ I 3 ⊇ · · · stabilizes, so for some i we have
I i = I i+1 = · · · = L. If L = 0 we are ¬nished, so assume L = 0. Let Σ be the collection of
all ideals K of R such that KL = 0. Then Σ is nonempty, since L (as well as R) belongs
to Σ. Let K0 be a minimal element of Σ, and choose a ∈ K0 such that aL = 0. Then
Ra ⊆ K0 (because K0 is an ideal), and RaL = aL = 0, hence Ra ∈ Σ. By minimality of
K0 we have Ra = K0 .
We will show that the principal ideal (a) = Ra coincides with aL. We have aL ⊆ Ra =
K0 , and (aL)L = aL2 = aL = 0, so aL ∈ Σ. By minimality of K0 we have aL = K0 = Ra.
From (a) = aL we get a = ab for some b ∈ L ⊆ N (R), so bn = 0 for some n ≥ 1.
Therefore a = ab = (ab)b = ab2 = · · · = abn = 0, contradicting our choice of a. Since the
1.6. ARTINIAN RINGS 13

assumption L = 0 has led to a contradiction, we must have L = 0. But L is a power of
the nilradical I, and the result follows. ™

We now prove a fundamental structure theorem for Artinian rings.

1.6.7 Theorem
Every Artinian ring R is isomorphic to a ¬nite direct product of Artinian local rings Ri .
Proof. By (1.6.5), R has only ¬nitely many maximal ideals M1 , . . . , Mr . The intersection
of the Mi is the Jacobson radical J(R), which is nilpotent by (1.6.6). By the Chinese
remainder theorem, the intersection of the Mi coincides with their product. Thus for
r r
some k ≥ 1 we have ( 1 Mi )k = 1 Mk = 0. Powers of the Mi still satisfy the
i
r
hypothesis of the Chinese remainder theorem, so the natural map from R to 1 R/Mk i
is an isomorphism. By (1.6.2), R/Mk is Artinian, and we must show that it is local. A
i
maximal ideal of R/Mk corresponds to a maximal ideal M of R with M ⊇ Mk , hence
i i
M ⊇ Mi (because M is prime). By maximality, M = Mi . Thus the unique maximal
ideal of R/Mk is Mi /Mk . ™
i i


1.6.8 Remarks
A ¬nite direct product of Artinian rings, in particular, a ¬nite direct product of ¬elds,
is Artinian. To see this, project a descending chain of ideals onto one of the coordinate
rings. At some point, all projections will stabilize, so the original chain will stabilize.
A sequence of exercises will establish the uniqueness of the Artinian local rings in the
decomposition (1.6.7).

It is a standard result that an R-module M has ¬nite length lR (M ) if and only if M
is both Artinian and Noetherian. We can relate this condition to associated primes and
supports.

1.6.9 Proposition
Let M be a ¬nitely generated module over the Noetherian ring R. The following conditions
are equivalent:
(1) lR (M ) < ∞;
(2) Every associated prime ideal of M is maximal;
(3) Every prime ideal in the support of M is maximal.
Proof.
(1) ’ (2): As in (1.5.10), there is a chain of submodules 0 = M0 < · · · < Mn = M ,
with Mi /Mi’1 ∼ R/Pi . Since Mi /Mi’1 is a submodule of a quotient M/Mi’1 of M , the
=
hypothesis (1) implies that R/Pi has ¬nite length for all i. Thus R/Pi is an Artinian
R-module, hence an Artinian R/Pi -module (note that Pi annihilates R/Pi ). In other
words, R/Pi is an Artinian ring. But Pi is prime, so R/Pi is an integral domain, hence
a ¬eld by (1.6.3). Therefore each Pi is a maximal ideal. Since every associated prime is
one of the Pi ™s [see (1.5.10)], the result follows.
14 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

(2) ’ (3): If P ∈ Supp M , then by (1.5.8), P contains some associated prime Q. By
hypothesis, Q is maximal, hence so is P .
(3) ’ (1): By (1.5.11) and the hypothesis (3), every Pi is maximal, so R/Pi is a ¬eld.
Consequently, lR (Mi /Mi’1 ) = lR (R/Pi ) = 1 for all i. But length is additive, that is, if
N is a submodule of M , then l(M ) = l(N ) + l(M/N ). Summing on i from 1 to n, we get
lR (M ) = n < ∞. ™

1.6.10 Corollary
Let M be ¬nitely generated over the Noetherian ring R. If lR (M ) < ∞, then AP(M ) =
Supp M .
Proof. By (1.5.9), AP(M ) ⊆ Supp M , so let P ∈ Supp M . By (1.5.8), P ⊇ P for some
P ∈ AP(M ). By (1.6.9), P and P are both maximal, so P = P ∈ AP(M ). ™

We can now characterize Artinian rings in several ways.

1.6.11 Theorem
Let R be a Noetherian ring. The following conditions are equivalent:
(1) R is Artinian;
(2) Every prime ideal of R is maximal;
(3) Every associated prime ideal of R is maximal.
Proof. (1) implies (2) by (1.6.4), and (2) implies (3) is immediate. To prove that (3)
implies (1), note that by (1.6.9), lR (R) < ∞, hence R is Artinian. ™

1.6.12 Theorem
The ring R is Artinian if and only if lR (R) < ∞.
Proof. The “if” part follows because any module of ¬nite length is Artinian and Noethe-
rian. Thus assume R Artinian. As in (1.6.7), the zero ideal is a ¬nite product M1 · · · Mk
of not necessarily distinct maximal ideals. Now consider the chain
R = M0 ⊇ M1 ⊇ M1 M2 ⊇ · · · ⊇ M1 · · · Mk’1 ⊇ M1 · · · Mk = 0.
Since any submodule or quotient module of an Artinian module is Artinian, it follows
that Ti = M1 · · · Mi’1 /M1 · · · Mi is an Artinian R-module, hence an Artinian R/Mi -
module. (Note that Mi annihilates M1 · · · Mi’1 /M1 · · · Mi .) Thus Ti is a vector space
over the ¬eld R/Mi , and this vector space is ¬nite-dimensional by the descending chain
condition. Thus Ti has ¬nite length as an R/Mi -module, hence as an R-module. By
additivity of length [as in (3) implies (1) in (1.6.9)], we conclude that lR (R) < ∞. ™

1.6.13 Theorem
The ring R is Artinian if and only if R is Noetherian and every prime ideal of R is
maximal.
Proof. The “if” part follows from (1.6.11). If R is Artinian, then lR (R) < ∞ by (1.6.12),
hence R is Noetherian. By (1.6.4) or (1.6.11), every prime ideal of R is maximal. ™
1.6. ARTINIAN RINGS 15

1.6.14 Corollary
Let M be ¬nitely generated over the Artinian ring R. Then lR (M ) < ∞.
Proof. By (1.6.13), R is Noetherian, hence the module M is both Artinian and Noetherian.
Consequently, M has ¬nite length. ™
Chapter 2

Integral Extensions

2.1 Integral Elements
2.1.1 De¬nitions and Comments
Let R be a subring of the ring S, and let ± ∈ S. We say that ± is integral over R if ±
is a root of a monic polynomial with coe¬cients in R. If R is a ¬eld and S an extension
¬eld of R, then ± is integral over R i¬ ± is algebraic over R, so we are generalizing a
familiar notion. If ± is a complex number that is integral over Z, then ± is said to be an

algebraic integer For example, if d is any integer, then d is an algebraic integer, because
it is a root of x2 ’ d. Notice that 2/3 is a root of the polynomial f (x) = 3x ’ 2, but f
is not monic, so we cannot conclude that 2/3 is an algebraic integer. In a ¬rst course in
algebraic number theory, one proves that a rational number that is an algebraic integer
must belong to Z, so 2/3 is not an algebraic integer.
There are several conditions equivalent to integrality of ± over R, and a key step is
the following result, sometimes called the determinant trick.

2.1.2 Lemma
Let R, S and ± be as above, and recall that a module is faithful if its annihilator is 0. Let
M be a ¬nitely generated R-module that is faithful as an R[±]-module. Let I be an ideal
of R such that ±M ⊆ IM . Then ± is a root of a monic polynomial with coe¬cients in I.
Proof. let x1 , . . . , xn generate M over R. Then ±xi ∈ IM , so we may write ±xi =
n
j=1 cij xj with cij ∈ I. Thus

n
(δij ± ’ cij )xj = 0, 1 ¤ i ¤ n.
j=1

In matrix form, we have Ax = 0, where A is a matrix with entries ± ’ cii on the main
diagonal, and ’cij elsewhere. Multiplying on the left by the adjoint matrix, we get
∆xi = 0 for all i, where ∆ is the determinant of A. But then ∆ annihilates all of M , so
∆ = 0. Expanding the determinant yields the desired monic polynomial. ™

1
2 CHAPTER 2. INTEGRAL EXTENSIONS

2.1.3 Remark
If ±M ⊆ IM , then in particular, ± stabilizes M , in other words, ±M ⊆ M .


2.1.4 Theorem
Let R be a subring of S, with ± ∈ S. The following conditions are equivalent:
(1) ± is integral over R;
(2) R[±] is a ¬nitely generated R-module;
(3) R[±] is contained in a subring R of S that is a ¬nitely generated R-module;
(4) There is a faithful R[±]-module M that is ¬nitely generated as an R-module.
Proof.
(1) implies (2): If ± is a root of a monic polynomial over R of degree n, then ±n and all
higher powers of ± can be expressed as linear combinations of lower powers of ±. Thus
1, ±, ±2 , . . . , ±n’1 generate R[±] over R.
(2) implies (3): Take R = R[±].
(3) implies (4): Take M = R . If y ∈ R[±] and yM = 0, then y = y1 = 0.
(4) implies (1): Apply (2.1.2) with I = R. ™

We are going to prove a transitivity property for integral extensions, and the following
result will be helpful.


2.1.5 Lemma
Let R be a subring of S, with ±1 , . . . , ±n ∈ S. If ±1 is integral over R, ±2 is integral
over R[±1 ], . . . , and ±n is integral over R[±1 , . . . , ±n’1 ], then R[±1 , . . . , ±n ] is a ¬nitely
generated R-module.
Proof. The n = 1 case follows from (2.1.4), part (2). Going from n ’ 1 to n amounts
to proving that if A, B and C are rings, with C a ¬nitely generated B-module and B a
¬nitely generated A-module, then C is a ¬nitely generated A-module. This follows by a
brief computation:
r s r s
Ayj xk . ™
C= Byj , B = Axk , so C =
j=1 j=1 k=1
k=1



2.1.6 Transitivity of Integral Extensions
Let A, B and C be subrings of R. If C is integral over B, that is, every element of C is
integral over B, and B is integral over A, then C is integral over A.
Proof. Let x ∈ C, with xn + bn’1 xn’1 + · · · + b1 x + b0 = 0. Then x is integral over
A[b0 , . . . , bn’1 ]. Each bi is integral over A, hence over A[b0 , . . . , bi’1 ]. By (2.1.5),
A[b0 , . . . , bn’1 , x] is a ¬nitely generated A-module. By (2.1.4), part (3), x is integral
over A. ™
2.1. INTEGRAL ELEMENTS 3

2.1.7 De¬nitions and Comments
If R is a subring of S, the integral closure of R in S is the set Rc of elements of S that
are integral over R. Note that R ⊆ Rc because each a ∈ R is a root of x ’ a. We say that
R is integrally closed in S if Rc = R. If we simply say that R is integrally closed without
reference to S, we assume that R is an integral domain with fraction ¬eld K, and R is
integrally closed in K.
If the elements x and y of S are integral over R, then just as in the proof of (2.1.6), it
follows from (2.1.5) that R[x, y] is a ¬nitely generated R-module. Since x + y, x ’ y and
xy belong to this module, they are integral over R by (2.1.4), part (3). The important
conclusion is that
Rc is a subring of S containing R.
If we take the integral closure of the integral closure, we get nothing new.

2.1.8 Proposition
The integral closure Rc of R in S is integrally closed in S.
Proof. By de¬nition, Rc ⊆ (Rc )c . Thus let x ∈ (Rc )c , so that x is integral over Rc . As in
the proof of (2.1.6), x is integral over R. Thus x ∈ Rc . ™
We can identify a large class of integrally closed rings.

2.1.9 Proposition
If R is a UFD, then R is integrally closed.
Proof. Let x belong to the fraction ¬eld K of R. Write x = a/b where a, b ∈ R and a and
b are relatively prime. If x is integral over R, there is an equation of the form

(a/b)n + an’1 (a/b)n’1 + · · · + a1 (a/b) + a0 = 0

with ai ∈ R. Multiplying by bn , we have an + bc = 0, with c ∈ R. Thus b divides an ,
which cannot happen for relatively prime a and b unless b has no prime factors at all, in
other words, b is a unit. But then x = ab’1 ∈ R. ™
A domain that is an integral extension of a ¬eld must be a ¬eld, as the next result
shows.

2.1.10 Proposition
Let R be a subring of the integral domain S, with S integral over R. Then R is a ¬eld if
and only if S is a ¬eld.
Proof. Assume that S is a ¬eld, and let a be a nonzero element of R. Since a’1 ∈ S,
there is an equation of the form

(a’1 )n + cn’1 (a’1 )n’1 + · · · + c1 a’1 + c0 = 0

with ci ∈ R. Multiply the equation by an’1 to get

a’1 = ’(cn’1 + · · · + c1 an’2 + c0 an’1 ) ∈ R.
4 CHAPTER 2. INTEGRAL EXTENSIONS

Now assume that R is a ¬eld, and let b be a nonzero element of S. By (2.1.4) part (2),
R[b] is a ¬nite-dimensional vector space over R. Let f be the R-linear transformation on
this vector space given by multiplication by b, in other words, f (z) = bz, z ∈ R[b]. Since
R[b] is a subring of S, it is an integral domain. Thus if bz = 0 (with b = 0 by choice of b),
we have z = 0 and f is injective. But any linear transformation on a ¬nite-dimensional
vector space is injective i¬ it is surjective. Therefore if b ∈ S and b = 0, there is an
element c ∈ R[b] ⊆ S such that bc = 1. Consequently, S is a ¬eld. ™


2.1.11 Preview
Let S be integral over the subring R. We will analyze in great detail the relation between
prime ideals of R and those of S. Suppose that Q is a prime ideal of S, and let P = Q © R.
(We say that Q lies over P .) Then P is a prime ideal of R, because it is the preimage
of Q under the inclusion map from R into S. The map a + P ’ a + Q is a well-de¬ned
injection of R/P into S/Q, because P = Q © R. Thus we can regard R/P as a subring of
S/Q. Moreover, S/Q is integral over R/P . To see this, let b + Q ∈ S/Q. Then b satis¬es
an equation of the form

xn + an’1 xn’1 + · · · + a1 x + a0 = 0

with ai ∈ R. But b + Q satis¬es the same equation with ai replaced by ai + P for all i,
proving integrality of S/Q over R/P . We can now invoke (2.1.10) to prove the following
result.


2.1.12 Proposition
Let S be integral over the subring R, and let Q be a prime ideal of S, lying over the prime
ideal P = Q © R of R. Then P is a maximal ideal of R if and only if Q is a maximal ideal
of S.
Proof. By (2.1.10), R/P is a ¬eld i¬ S/Q is a ¬eld. ™


2.1.13 Remarks
Some results discussed in (2.1.11) work for arbitrary ideals, not necessarily prime. If R
is a subring of S and J is an ideal of S, then I = J © R is an ideal of R. As in (2.1.11),
R/I can be regarded as a subring of S/J, and if S is integral over R, then S/J is integral
over R/I. Similarly, if S is integral over R and T is a multiplicative subset of R, then
ST is integral over RT . To prove this, let ±/t ∈ ST , with ± ∈ S, t ∈ T . Then there is an
equation of the form ±n + cn’1 ±n’1 + · · · + c1 ± + c0 = 0, with ci ∈ R. Thus

± cn’1 ± n’1 c1 ± c0
+ · · · + ( n’1 ) + n = 0
( )n + ( )( )
t t t t t t

with cn’j /tj ∈ RT .
2.2. INTEGRALITY AND LOCALIZATION 5

2.2 Integrality and Localization
Results that hold for maximal ideals can sometimes be extended to prime ideals by the
technique of localization. A good illustration follows.

2.2.1 Proposition
Let S be integral over the subring R, and let P1 and P2 be prime ideals of S that lie over
the prime ideal P of R, that is, P1 © R = P2 © R = P . If P1 ⊆ P2 , then P1 = P2 .
Proof. If P is maximal, then by (2.1.12), so are P1 and P2 , and the result follows. In the
general case, we localize with respect to P . Let T = R\P , a multiplicative subset of R ⊆ S.
The prime ideals Pi , i = 1, 2, do not meet T , because if x ∈ T © Pi , then x ∈ R © Pi = P ,
contradicting the de¬nition of T . By the basic correspondence between prime ideals in a
ring and prime ideals in its localization, it su¬ces to show that P1 ST = P2 ST . We claim
that

P RT ⊆ (P1 ST ) © RT ‚ RT .

The ¬rst inclusion holds because P ⊆ P1 and RT ⊆ ST . The second inclusion is proper,
for otherwise RT ⊆ P1 ST and therefore 1 ∈ P1 ST , contradicting the fact that P1 ST is a
prime ideal.
But P RT is a maximal ideal of RT , so by the above claim,

(P1 ST ) © RT = P RT , and similarly (P2 ST ) © RT = P RT .

Thus P1 ST and P2 ST lie over P RT . By (2.1.13), ST is integral over RT . As at the
beginning of the proof, P1 ST and P2 ST are maximal by (2.1.12), hence P1 ST = P2 ST . ™
If S/R is an integral extension, then prime ideals of R can be lifted to prime ideals of
S, as the next result demonstrates. Theorem 2.2.2 is also a good example of localization
technique.

2.2.2 Lying Over Theorem
If S is integral over R and P is a prime ideal of R, there is a prime ideal Q of S such that
Q © R = P.
Proof. First assume that R is a local ring with unique maximal ideal P . If Q is any
maximal ideal of S, then Q © R is maximal by (2.1.12), so Q © R must be P . In general,
let T be the multiplicative set R \ P . We have the following commutative diagram.
R ’’’ S
’’
¦ ¦
¦ ¦g
f

R T ’ ’ ’ ST
’’
The horizontal maps are inclusions, and the vertical maps are canonical (f (r) = r/1 and
g(s) = s/1). Recall that ST is integral over RT by (2.1.13). If Q is any maximal ideal
of ST , then as at the beginning of the proof, Q © RT must be the unique maximal ideal
6 CHAPTER 2. INTEGRAL EXTENSIONS

of RT , namely P RT . By commutativity of the diagram, f ’1 (Q © RT ) = g ’1 (Q ) © R.
(Note that if r ∈ R, then f (r) ∈ Q © RT i¬ g(r) ∈ Q .) If Q = g ’1 (Q ), we have
f ’1 (P RT ) = Q©R. By the basic localization correspondence [cf.(2.2.1)], f ’1 (P RT ) = P ,
and the result follows. ™


2.2.3 Going Up Theorem
Let S be integral over R, and suppose we have a chain of prime ideals P1 ⊆ · · · ⊆ Pn
of R, and a chain of prime ideals Q1 ⊆ · · · ⊆ Qm of S, where m < n. If Qi lies
over Pi for i = 1, . . . , m, then there are prime ideals Qm+1 , . . . , Qn of S such that
Qm ⊆ Qm+1 ⊆ · · · ⊆ Qn and Qi lies over Pi for every i = 1, . . . , n.
Proof. By induction, it su¬ces to consider the case n = 2, m = 1. Thus assume P1 ⊆ P2
and Q1 © R = P1 . By (2.1.11), S/Q1 is integral over R/P1 . Since P2 /P1 is a prime ideal
of R/P1 , we may apply the lying over theorem (2.2.2) to produce a prime ideal Q2 /Q1 of
S/Q1 such that

(Q2 /Q1 ) © R/P1 = P2 /P1 ,

where Q2 is a prime ideal of S and Q1 ⊆ Q2 . We claim that Q2 © R = P2 , which gives the
desired extension of the Q-chain. To verify this, let x2 ∈ Q2 © R. By (2.1.11), we have
an embedding of R/P1 into S/Q1 , so x2 + P1 = x2 + Q1 ∈ (Q2 /Q1 ) © R/P1 = P2 /P1 .
Thus x2 + P1 = y2 + P1 for some y2 ∈ P2 , so x2 ’ y2 ∈ P1 ⊆ P2 . Consequently, x2 ∈ P2 .
Conversely, if x2 ∈ P2 then x2 + P1 ∈ Q2 /Q1 , hence x2 + P1 = y2 + Q1 for some y2 ∈ Q2 .
But as above, x2 + P1 = x2 + Q1 , so x2 ’ y2 ∈ Q1 , and therefore x2 ∈ Q2 . ™
It is a standard result of ¬eld theory that an embedding of a ¬eld F in an algebraically
closed ¬eld can be extended to an algebraic extension of F . There is an analogous result
for ring extensions.


2.2.4 Theorem
Let S be integral over R, and let f be a ring homomorphism from R into an algebraically
closed ¬eld C. Then f can be extended to a ring homomorphism g : S ’ C.
Proof. Let P be the kernel of f . Since f maps into a ¬eld, P is a prime ideal of R. By
(2.2.2), there is a prime ideal Q of S such that Q © R = P . By the factor theorem, f
induces an injective ring homomorphism f : R/P ’ C, which extends in the natural way
to the fraction ¬eld K of R/P . Let L be the fraction ¬eld of S/Q. By (2.1.11), S/Q is
integral over R/P , hence L is an algebraic extension of K. Since C is algebraically closed,
f extends to a monomorphism g : L ’ C. If p : S ’ S/Q is the canonical epimorphism
and g = g —¦ p, then g is the desired extension of f , because g extends f and f —¦ p|R = f .

In the next section, we will prove the companion result to (2.2.3), the going down
theorem. There will be extra hypotheses, including the assumption that R is integrally
closed. So it will be useful to get some practice with the idea of integral closure.
2.3. GOING DOWN 7

2.2.5 Lemma
Let R be a subring of S, and denote by R the integral closure of R in S. If T is a
multiplicative subset of R, then (R)T is the integral closure of RT in ST .
Proof. Since R is integral over R, it follows from (2.1.13) that (R)T is integral over RT .
If ±/t ∈ ST (± ∈ S, t ∈ T ) and ±/t is integral over RT , we must show that ±/t ∈ (R)T .
There is an equation of the form
± a1 ± an
( )n + ( )( )n’1 + · · · + =0
t t1 t tn
n
with ai ∈ R and ti , t ∈ T . Let t0 = i=1 ti , and multiply the equation by (tt0 )n to
conclude that t0 ± is integral over R. Therefore t0 ± ∈ R, so ±/t = t0 ±/t0 t ∈ (R)T . ™


2.2.6 Corollary
If T is a multiplicative subset of the integrally closed domain R, then RT is integrally
closed.
Proof. Apply (2.2.5) with R = R and S = K, the fraction ¬eld of R (and of RT ). Then
RT is the integral closure of RT in ST . But ST = K, so RT is integrally closed. ™
Additional results on localization and integral closure will be developed in the exercises.
The following result will be useful. (The same result was proved in (1.5.1), but a slightly

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