Introduction to Homological Methods in Commutative Rings

Table of contents :
Contents
Introduction
§0. Prerequisites
§1. Height
§2. Primary Decomposition
§3. Length
§4. The Principal Ideal Theorem (Hauptidealsatz).
§5. sxstems of Parameters.
§6. Polynomial Rings
§7. Power Series Rings.
§8. The Krull Intersection Theorem.
§9. An Application of the Intersection Theorem to Primary Decomposition.
§10. Cohen-Macaulay Rings
§11. R-sequences and Grade.
§12. Some Additional Properties of C - M rings.
§13. Regular Local Rings.
§14. Hilbert Rings and the Hilbert Nullstellensatz.
§15. Some additional remarks on maximal ideals in polynomial rings.
§16. Polynomial Rings over Dedekind Domains.
§17. The Homological Machine.
§18. Change of Rings and-Dimension of Polynomial Rings.
§19. More on Change of Rings.
§20. Homological Characterization of Regular Local Rings.
§21. A Homological Characterization of Grade.
§22. Localizing the Machine.
§23. Domains of Global Dimension \leq 1.
§24. Codimension.
§25. Regular Local Rings are UFD's.
§26. Lifting Finite Free Resolutions.
§27. Irreducible Ideals and Gorenstein Rings.
§28. Local Rings of Krull Dimension 0.
§29. A Homological Characterization of Gorenstein Rings and a conjecture of Bass.
Suggestions for Further Reading
References

Citation preview

INTRODUCTION TO HOMOLOGICAL METHODS IN COMMUTATIVE RINGS

by

Anthony V. Geramita and Charles Small

QUEEN'S PAPERS IN PURE AND APPLIED MATHEMATICS - NO. 43

QUEEN'S UNIVERSITY, KINGSTON, ONTARIO

1976

COPYRIGHT

@

1976

This book, or any parts thereof, may not be reproduced in any form without written permission from the authors.

Contents

----------------------------------------

3

0.

Prerequisites------------------------------------------

6

1.

Height-------------------------------------------------

10

2.

Primary Decomposition

---------------------------------

13

3.

Length ------------------------------------------------

33

4.

The Principal Ideal Theorem (Hauptidealsatz) -----------

36

5.

Systems of Parameters

---------------------------------

43

6.

Polynomial Rings

--------------------------------------

46

7.

Power Series Rings

------------------------------------

53

8.

The Krull Intersection Theorem-------------------------

5s

9.

An Application of the Intersection Theorem to Primary

Introduction

Decomposition -----------------------------------------

65

10.

Cohen-Macaulay Rings

----------------------------------

69

11.

R-sequences and Grade ---------------------------------

76

12.

Some additional properties of

13.

Regular local rings ----------------------------------- 110

14.

Hilbert rings and the Hilbert Nullstellensatz --------- 127

15.

Some additional remarks on maximal ideals in polynomial rings

C-M rings -------------- 100

------------------------------------------------ 140

16.

Polynomial rings over Dedekind domains -----------~---- 145

17.

The Homological Machine ----------------------~-------- 150 Exa~t S7quen~es -~---------------------------------ProJective Dimension-----~-----~-----------------~Projective Resolutions -----------------~---------~Injectives -------------------- ------------------ Ext(M,N) ------------------------------------------Long exact sequences for Ext, and the connection with projective dimension----------------------Global dimension-----------------------------------

- 1 -

150 152 158 164 167 177 183

contents (continued) ...

Page

18.

Change of rings and dimension of polynomial rings

191

19.

Mo~e on change of rings------------------------------

205

20.

Homological characterization of regular local rings --

215

21.

A homological characterization of Grade--------------

228

22.

Localizing the Machine-------------------------------

238

23.

Domains of global dimension

-------------------

248

24.

Codimension ------------------------------------------

252

25.

Regular local rings are UFD's

-----------------------

262

26.

Lifting Finite Free Resolutions----------------------

281

27.

Irreducible ideals and Gorenstein rings--------------

287

28.

Local rings of Krull dimension

305

29.

A Homological characterization of Gorenstein rings

30.

< 1

O

-------------------

and a conjecture of Bass-----------------------------

326

Suggestions for further reading----------------------

336

References

----------------------------------------------

- 2 -

350

--,

Introduction These notes reflect a graduate course we gave at Queen's University in 1974-75 .

A one term course in

Commutative Rings was a prerequisite so that, for example, such basic tools as localization, Nakayama's Lemma, and Noetherian rings, were available from the start.

(We list

exactly what we wish to assume in §0 , and thereafter the notes are almost completely self-contained.) Given these prerequisites, the course proceeded in three stages, as follows: 1) i.e.

A large chunk of non-homological. commut~ive algebra, essentially the theory of ideals in Noetherian rings.

For example, we do primary decomposition (§2)

, the principal

ideal theorem and systems of parameters (§§4-5) intersection theorem (§8)

, and the

This gives a feeling for the

subject, and puts us in a position to raise some significant questions whose solution can be given by homological methods. These questions chiefly concern three successively larger classes of local rings: regular, Gorenstein, Cohen-Macaulay. 2)

Introduce the homological machine:

This we do in §17

and §22 . This is spelled out in detail, since we want to keep the notes accessible to readers wi t.h no background in homological algebra. 3)

Use the Homological Machine: enough to demonstrate the

power of the methods, especially with respect to answering the problems r~ised earlie~.

For example, regular local rings

are introduced (non-homologically) in §13, and in §20 we give -

3 -

a homological characterization of them, due to Auslander and Buchsbaum, which resolves a problem of Krull, namely to show that if of

R

R

, then

R

is regular local and p

is regular.

p

is a prime ideal

As another example we use

the machinery to prove that regular local rings are U.F.D. 's (§24-§26).

These theorems on regular local rings were

historically the first successes of the homological invasion of the subject. As another example, Cohen-Macaulay rings and Gorenstein rings are introduced non-homologically (in §10-12 and §27 respectively), but as soon as the machinery is available the earlier definition can be re-interpreted, and this leads to a homological characterization of Gorenstein rings and h~lf of a homological characterization of Cohen-Macaulay rings (see §21 and §28-29).

(The missing half of the latter "theorem"

is a famous open problem usually referred to as Bass' conjecture.) With a subject as many-faceted as commutative rings there are always some topics that one cannot cover in a one-year course.

We reluctantly sacrificed the Hilbert-Samuel polynomial

and the Koszul complex as well as Tor, the derived functors of

©

These are important omissions and a serious student

in the subject would have to become acquainted with these ideas.

Apart from these omissions, to go further in the

subject is, to a large extent, to confront theorems in which the content, as well as the proof, is homological.

We have

been content to provide a sketchy bibliography of further -

4

-

readings (§30) to which the reader can repair to pursue the subject. It will be obvious, probably even to non-specialists, that we have borrowed freely from available sources with little effort to·, give due c·redi t for theorems or methods of proof; in particular, we have relied heavily, in places, bn the books of Kaplansky [18] and [2].

[1s}J and Atiyah-MacDonal~

In fact, much of what we have done in these notes is

already available somewhere in the literature always in a suitable form.

though not

What we feel has been lacking,

and what we have tried to provide, is a picture of the connections between pre-homological and homological commutative algebra that is accessible to readers -particularly graduate students ~with limited background in the subject. We have been helped by many people in preparing the course and these notes but we would like to single o~t our dee~ appreciation to: The late Professor M. Narita of the International Christian Univirsiiy in Tokyo whose lectures on local rings at Queen's University in 1~70-1971 were an important inspiration for o'ur course; the students in the course,whose impatience with nonsense occasionally.saved the day; Karen Lewis for a splendid typing job from an impossible manuscript;

Professor I. Kaplansky for letters clarifying the

proof of 26~3 ; Profess6r Max Boratynski for useful conversations concerning the material in §§27-28

and Professor

Mel Hochster, for communicating the lovely proof outlined on the last page of §11. - 5 ·-

u

.....

§0.

Prerequisites

Throughout these notes,

1

ring with

will be a commutative

R

This section lists the material we

shall assume.

-

If

0.1

R

+

R/I

I

is an ideal in

R

induces an inclusion-preserving bijection between

all (resp. all prime) ideals of prime) ideals of view

the natural map

,

R

R/I

and all (resp. all

which contain I

We occasionally

R/I-modules as R-modules killed by

I

, and con-

versely.

0.2

S

If

"""'-"""'

natural map / 1R

is a multiplicative set in +

s- 1 R.

p

if R

+

disjoint from

R

, the

induces an inclusion-preserving

bijection between all prime ideals of prime ideals of

R

is a,' prime ideal of

R

s- 1 R

S

and all

In particular,

, the natural map

induces an inclusion-preserving bijection between

RP

all prime ideals of contained in

p

RP

and all prime ideals of

R

•

The processes described in 0.1 and 0.2 commute:

0.3

~

if

p

is a prime ideal and

in

p

, there are canonical isomorphisms

(R/I)p/I

+ +

I

making

-

6 -

is an ideal contained

-

-.,.....

R

I\ R /IR

(R/I)p/I

p

p

commute.

RP

~

is an example of a quasi-local ring, i.e. a

commutative ring with exactly one maximal ideal. Equivalently, a commutative ring is quasi-local if and only if its non-units form an ideal.

0.5 ...-

The radical of an ideal

I

is by definition

Theorem: II= the intersection of all prime ideals containing

I

In

particular, the intersection of all the prime ideals of

R

0.6

is precisely the set of nilpotent elements.

The intersection of all the maximal ideals of

~

is called the Jacobson radical, Nakayama's Lemma:·· and

for all

Exercise:

, of

R

M is a finitely generated R-module

IM= M for some ideal

M = 0

0.7 .,..__,

if

J(R)

I

contained in

x e: J(R) l + yx

J(R)

generated.

is Noetherian if every ideal is finitely One

condition equivalent to this is the

"ascending chain condition": -

given ideals 7 -

then

is invertible

y e: R

R

R

Il c I2 ~ I3 ~ •••

, there is an integer

1 N = 1 N+l - •••

Another equivalent condition:

N

such that every

non-empty set of ideals contains maximal elements (with respect to inclusion). if

R

Theorem

is Noetherian, so is

(Basis theorem of Hilbert):

R[X]

Also, if

R

is

Noetherian, so is any homomorphic image or localization of

R

A local ring is, by definition, a Noetherian

quasi-local ring.

We abbreviate

m "to

with maximal ideal

"Risa local ring

"(R,m)

is a local ring".

A semilocal ring is a ring which has only finitely many maximal ideals. We will also refer occasionally to integral extensions

R c S

, and to basic facts about Dedekind

domains, discrete valuation rings (DVRs), and unique factorization domains (UFDs).

However these are not

serious prerequisites. We use basic properties of "base-change" or "extension of scalars":

if

R

+

S

is a homomorphism of

commutative rings, and we use it to view R-module, then to S-modules.

Rp~RM

M~ S~RM

S

as an

is a functor from R-modules

For example, we will identify

Mp

and

• We will need also a few basic facts about

localization, notably:

0.8

~

x

=0

If

x EM

~

1

=O

, where in

M

p

M is an R-module, then for every prime ideal

-

8 -

p

of

R

same statement with "maximal" in place of "prime").

(

It follows from 0.8, for example, that M

=0

f: M -+ N

for all

p

,

and that a map

is injective or surjective if and only if the

induced map all

=0

Mp f

. p' Mp

-+

Np

has the same property for

p

-

9 -

§1.

Height

If

1.1 ,...._

is a prime ideal in a ring

p

of length

n

going down from

R

, a chain

is by definition a

p

string

p = Po ~ P1 ~·. ·~ Pn

of proper inclusions of prime ideals.

Note that the

length of the chain is the number of inclusions, hence is one less than the number of prime ideals. admits such chains for all height and write ht p = n

if

p

nol'l: of length

--

ht p =

n

we say

definition

primes

; otherwise we write

~

admits such a chain of length

n

but

n + 1

sup ht p

, the

For example,

p

p

is of infinite

p

K-dim(R) , the Krull dimension of

1.2

If

sup

R, is by

taken over all

K-dim(Rp)

is just

we put .ht(I)

= min

ht p

'

by 0.2

1.3 .........,

the

'···For any ideal. I min

taken over all primes

Also we write is the

sup

dim(I)

p

= K-dim(R/I)

ht(p)

'

such that

p

.

dim(I)

'

by 0.1,

::>

I

of lengths of chains of. prime ideals

going up, and starting above

I

- 10 -

.

Clearly, by 0.1

,

dim(I) + ht(I) < K-dim(R) for any ideal

I

We shall see later (10.4) that this inequality can be strict, even when

Examples: "field". let R

=

V F

a)

is finite.

"Domain of dimension O" means

But there are non-domains of dimension O :

CF any field) and make

be an F-vector space

~

V b)

where

K-dim(R)

F

a ring by putting

xy

=

0

Any Dedekind domain (e.g. is a field) has dimension 1

x,y EV

for ~

, or

F[X]

every non-zero

prime ideal is maximal. c)

If

R

S

c

is an integral extension, the

"going-up" theorem of Cohen-Seidenberg implies that K-dirn(R) = K-dim(S).

(See Atiyah-MacDonald [2], Chapter

5. )

d)

If

F

(see 6.7 below).

is a field,

F[X 1 , ... ,Xn]

F[X 1 ,x 2 , ..• J

has Krull dimension n

(countably many

indeterminates) clearly has infinite Krull of course, it is not Noetherian. infinite Krull dimension exist:

dimension;

Noetherian rings of (See Nagata [22],

Example 1, page 203). We shall return to this example often.

We show

below (4.4) that ideals in

Noetherian

rings always have finite height, and so in particular local rings always have finite K-dimension. e) We shall see later (14.8, (3)) that even in -

11 -

a nice ring like

R[X 1 , ••• ,Xn]

where

R

if a DVR

,

distinct maximal ideals can have different heights. i

·'

(

- 12 -

§2.

Primary Decomposition

An ideal

~

R/q

q

is primary if every zero divisor in

is nilpotent.

q

XE

or

y

/q

E

xy E q

(Equivalently:

implies

.) . I

For example, prime id,als are primary. ( rP>

ideals of. the form

2.2

z

ar~ primary, p prime in

In a Noetherian ring, every ideal

I

I .

"""-"'""

In

contains

a power of its radical.

v'!

Proof:

x 1 , ... ,xm

is finitely generated, say by

Then for each

,

i

· lII'= { n q. = 1

p.1

some

·"

n~ 1

= np.1

(2.10), and hence

p

Hence

p

contains

contains a minimal

p.1

also •

.Corollaries:

1)

ideals .are equal: containing of

I

I

For any ideal {I

I

the following four

; the intersection of all primes

; the intersection of all associated primes

; the intersection of all isolated primes of (cf. 0. 5) •

I

2)

In a Noetherian ring there are only finitely

many minimal prime ideals (i.e., prime ideals of height 0) ,

.namely the isolated primes of

1

( 0)

Similarly,

there are only finitely many primes minimal over any ideal

I The corollaries are immediate from 2.26.

- 28 -

n I =

Let

I'\

q.

be a primary

P·1. = lq-:J.

'

1.

i=l

As we

decomposition, so that Ass(I) = {p 1 , •.. ,pn}

have already noted, the minimal elements of Ass(!) sometimes called the isolated primes of the minimal primes of Ass(I)

not minimal in

p.

.

(or simply I

Any associated prime of

is said to be embedded,

q. i

one sometimes calls radical

I )

I

are

and

isolated or embedded if its

is.

i

To illustrate the terminology, we return to the 2 I = ( X , XY )

example

1

(X) n (X 2 ,Y)

saw that

n

F [ X, Y ]

I

I

is the

(X,Y)

Y-axis

isolated component (X=Y=O)

I

, therefore, are

is isolated and

a fie 1 d ) .

(X) n (X,Y) 2

and

primary decompositions for primes of

(F

are both

The associated (X)

and

is embedded.

(X,Y)

, as is the

The embedded component

is just the origin.

decomposition of an ideal

CX)

Geometrically,

( x 2 = XY = O X = 0) (X)

\ve

I

In general, a primary corresponds to a

decomposition of the variety defined by

I as a union of

irreducible varieties defined by the isolated primes of I

; the embedded primes contribute certain distinguished

subvarieties of these isolated components. The preceding example shows that the embedded primary components in a primary decomposition are not uniquely determined by the ideal.

The reader who wants

more information on this point should study "syml.Jolic powers" of an ideal, in Atiyah-MacDonald [2] or

- 29 -

Zariski-Samuel [31]. §9.)

(See also Remark 2 at the end of

On the other hand, it can be shown that the isolated

primary components (which are the pictorially significant ones)~ uniquely determined by

(See p. 54 of

I

Atiyah-MacDonald [2] or theorem 8 on p. 211-212 of Zariski-Samuel [31]).

We don't pursue this here, as

we shall not need it. If rings and

f: R I

T

+

is a. homomorphism of Noetherian

is an ideal of

R

it is natural to

'

ask how the primary decompositions of

I

in

R

are

related to the primary decompositions of the ideal generated by

f(I)

happens when

S

in

T

We describe here what

is a multiplicative set in

is the canonical map

R +

sider the case where

f

s- 1 R

In

§6

R

and

f

we will con-

is the inclusion of

R

in

R[X] We need a few preliminaries, which we list in four steps (2.27-2.30). set in

R

'

f: R

let

for an ideal

I

generated by

f(I)

localization at

--2.27

S-lR

s =

'

0

for some then f-lJ

S-lR let

denote a multiplicative be the canonical map, and

I

denote the ideal of

I = S-lI

(Exercise: of the R-module

I

For any ideal

{xe:Rlsxe:I

p n

s

+

R

of

s

Let

s

f- 1

£

of

R

S}

=I

=J - 30 -

' If

I

.)

f- 1 I

'

S-lR

the

=

is p-primari and

For ani ideal

J

of

The proof of 2.27 is an easy exercise.

-

2.28

If

g: R

p -primary in

T

+

T

,

then

is any homomorphism, and f-lq

f

is

q

-1,p)- primary· . in.

l.S

R

. This is immediate from the definition.

2.29

If

""'-

and

q

then

is -·

s =0 f-l(q) = q

R where

p n S

v'q = P1 and p is prime since Finally, we saw in 2.27 that

'

and using this it is easy to check is primary.

q

f

With rtotation as in 2.29,

induces a

b ijection between all p-primary ideals in P-,:e.rimary ideals in

s- 1R

This is clear from 2.27-2.29.

-

Corollary:

R

and let. I

= ql

Let

S

n

Then

Pi

=

0

I = -ql

for

n ••• n

qn

pi = lq7 1

'

q.

decomposition of

1

and s P· ---qm P· = ~'

l

+

Note that

2

over

m

is Noetherian,

e: (y) n

with

we can assume

p

n=l

R

By

is local, and in any

case we can use Noetherianness to assume there are no primes between say

a1

p

Since

and

, is not in

imply that no prime of

p1 R

, some

a.l. '

Then our two assumptions contains -

38 -

(p 1 ,a 1 )

except the

maximal ideal

, so that p =

p

there is an integer all

i

2 < i < n

' k

a. = J.

with

a prime

q

Let

J

contradict the theorem for all i, 1 < i < n

-

R = R/q

in ~

-1--

p1

~

,

Ll_ Corollary; ht(I)

ht(p) > 2

If

is finite.

m

maximal ideal

I

£

( q 'a,... )

(q,a 1 )

for

contains

(q,al)

is minimal over

, hence

I

This means that

(a1 )

is minimal over

shows

q

-f'

a.kJ_

so any prime containing

p

there is

'

otherwise we Now

is the only prime containing

p

-

q ~ P1

p

'

(n-1)-generator

ht(pl) > n

n-1

But the maximal ideal

I

p

'

b. e: pl J.

'

be the

Hence

, and for

.:.. (pl,al)

J.

J

C

(0.5).

c. e: R

Since

P1

such that

'

P

there are

'

C

k

such that

b. + cial J.

(b2,···,bn )

ideal

k

/(pl,a;:>

; but

, contradicting 4.2

is an ideal in a Noetherian ring,

In particular, if

, then

is local with

R

d = K-dim(R)

m cannot be generated by fewer than

is finite, and d

elements.

To motivate the next result, notice that when

p

=.

I = (a 1 , ..• ,an)

minimal over

then

ht p

~

n

I d

then since

m

is the only prime of

, it is the only prime containing

lca 1 , .. .

,ad) = m

,

hence

(2.4).

-

45 -

(a 1 , •.. ,ad)

Ca 1 , ... ,ad) is

m-primary

§6.

Polynomial Rings Our aim in this section is to compare the ideal structure

in

R[X]

with that in

..

write

for the ideal

I

If

R

IR[X]

is an ideal of

I

of

R[X]

R

we

generated by

I

Observe that:

-

I

6. 1.

= {ta.xii a. c I l. l.

I ,

R/r[X] = R[X] I

for all

R(\

and

I

(or contracts to) R[X]

I

R[XJ

in

R ('\ J = I

if

its contraction

,

I= J

We say that an ideal

i}

lies over

If

J

P.. .

is prime in

R ('\ J

is prime in It is no

harder to verify this in the following more general form: Let I

f of

R

and

J

of

J

f(I)

generated by of

be any homomorphism of rings, and for ideals

T

R +

Then, if

is a prime ideal of

T

'

'

for the ideal of

f-l(J)

and call

J

-I

write

the contraction ---·- ·----~-----~-

is a prime ideal of R

contraction need not be maximal; if

R + R[X]

~~·

Let

R

Then

,

However when I

R

...

p

T

,

its contraction

The reader should verify this, and

find examples to show the following: if

not be prime.

T

f : R

prime ~oes imply

+

-

I

I

is prime,

T

-

46 -

I

need

is the inclusion

prime:

be any _commutative. rin;;, is_~rime ideal of

is maximal its

J

R[X]

p

a pr1.me ·-~d,:!?l ot_

lyin_g o_ver

p

There are infinitely maf!.Y.: other primes_pf

p

over

; they all contain

p

R[X]

lying

and there are no inclusions

among them. (Thus the picture over_ ~ivenyrime is

J •f Qf

.

~[X)

\h

ih R

course there may be complicated inclusions among primes

in

lying over different primes of

R[X]

Proof of 6.2.

p

Thus

(0 4 1)

R[X]

domain, and we are after the primes in

where R[X]

Let

.)

The first statement is clear from 6.1 .

the rest, we may factor out

(0)

R

F

be the quotient field of.

s is the multiplicative set contracts to

(0)

R

-

R

is a

lying over S-lR

R : F ::

{O}

For

A

prime in

if and only if it contains no

non-zero constants, and by

0.2

there is a bijective

inclusion-preserving correspondence between all such primes and all primes of

F[X]

But

F[X]

is a P~I.D. whose

prime structure is very well known: every non-zero prime is maximal and there are infinitely many such.

-

6. 3 •

Corollary.

Pi, p 2 , p 3

do not all contract to the same~rime __in

- 47 -

R

Ll,.-

Corollary.

and let

-

q ~

p

Let

be a prime of h~igh~

p

be a prime in

n < ht p < ht q < 2n+l

R[X]

n

lying over

in

p

1 + K-dim(R)

Consequently,

R

Then
ht p

p? P1 ~ ••• ~ Pm is a chain in R then ::> is a chain in R[X] Hence ~ Pm

ht q > ht p+l

and

chain in

R[X]

chain in

R

going down from

going down from

-

ht p < 2n

we see that

q

p

K-dim(R) = n

SEIDENBERG (28]

and

~

6•5•

I

in

Proof.

p

If

K-dim R[X] = m

--

; see

We need first two lemmas:

I

in

p

R,

is minimal over

R[X]

If

a prime

cohtained in

domain

R

However, Noetherian rings are better

is minimal over

contracting to

6.6.

2n+l

, there are domains

behaved, as we shall see in 6.7 .

--

contracts to a

p

and, using 6.3 to count,

'

ht q

and

or

n+l < m < 2n+l

Given any such that

On the other hand, a

-

Let R

p

R

p

q

satisfies

shows

R {'\ q

I~ q ~ p

=

and lying over

p

But the only prime

p

is

be a prime ideal of heig~t Then

-

ht(p)

=1

-

48 -

itself.

p

1

in a Noetherian

Proof.

is minimal over (a), for any

p

Hence

Thus ht(~) ~ 1

and clearly

ht(p)

>

(a)

in

by the Principal Ideal Theorem,

ht(p) = l

is the crucial case of:

6.6.

-

-

is minimal over the principal ideal

p

R[X]

Let

6. 7.

R

O# a E p

p

and let

Then

be ~rime of height be a EEime in

q # p

ht(p) = n

ht(q)

and

::

K-dim R[X] = 1 + K-dim(R)

1n a Noetherian ring -·----··--·R[X] 0E__over p n

n+l

In p~rt icular,

, and (by induction)

K-dim R[X 1 , ••• ,Xm] = m + K-dim(R)

Proof.

By induction on

minimal over

(0)

in this case

ht(q) < 1

(q ~

p)

smaller p

n

If n

'

by 6.4 >

0

ht(p)

>

ht(p) :: O then

ht(p) =

and then

n

Now let

of height

ht(pl)

'

If

n

but clearly

'

.

by 6.5

0

is

p

Also

ht(q) > 1

-

and assume the result for n

'

and, letting

p contains

then

P1 = R () p

a prime

, we have

< n-1 "" ht > n

p

ht = n

p

• I •

ht

•P1

ht < n-1

I

By inductive hypothesis, this forces Now go to

R = Rlp 1 :

- 49 -

= n

•p

~

-

ht p 1 = n-1 has height

1

and , but

in

p has height

contradiction shows that ht(q) = n+l

contradicts 6.6 .

R[X]

n

This

To show that

, it therefore suffices to show that any prime

properly contained in such a prime and let

has height

q

P1 =

R ('\ p

in which case in which case

ht(P) < n

l)

Then either

= p and ht(P) = n

p

.

Let

< n

; or

p

1

be

-

p

'

ht(pl) < n-1

-

by induction.

To complete this section we describe how a primary decomposition in

R

carries up to

We need two

R(X]

cute lemmas:

6.8. .,,,,._

Let

Then

f

some

O

a

T

is a ~

n . 1 , T[XJ f = . t 0a.X . J. 1= if. and onl_y~if there is

be a commutative ring, and 0-divisor in

c ET

with

T[X]

(In particular, if

cf= 0

0-divisor, then every coefficient of

is a

f

is

f

0-divisor;

The simplest possible example shows the converse is false: 2 + 3X

is not a

0-divisor in

Proof. '~" is trivial.

Conversely, if m

O

there is some g

with

~

ab .· m = a then le.ss than m

ab.

l.

,

and

•

g = t b,XJ j:O J

m minimal.

=

( Z/ ( 6)) (,f.l

with

for all

0
O 1 multiplication.

will denote

R

with the expected addition and

R[ [X 1 , ..• ,XmlJ

is defined inductively a;s

x~ O defines a homomorphism

·· .. R[ [Xi, ••• ,Xm-l]] [ [Xm l] . (j>:R[[X]]-+ R

-

7. l.

For any conunutative. rin_&· R

f = Ea.Xi

i)

i>O

~ · (j>(f)

= a0

ii)

If

..

R[[X]]

is invertible

is invertible. in- R

is a maximal ideal: of

M

is a maximal ideal of between .the set R[[X]]

in

'-.-.

1

.

R-

, and

Max CR[ [X]])

and the set

R[[X]] then

induces-a bijection

(.j)

of all maximal ide.als of

'Max(R)· of all maximal ideals of ·R

iii,)

R

is Noetherian ~ R[(XlJ 'is Noetherian. . .

iv)

R

is local ~ R[[X]]

v). · When

is iocal.

K-dimR[ [X]] = l + K..;dim R

R · · is local,

and hence by induction ·lO

-

-

Ci> 1)

b.

l.

-b 0

=m+

.J.

5.3 .. ·

i ··

is

f-l

inductively by

ii)

If

~

that

M is a maximal ideal of

R[(X]l, it follows from

is a proper ideal of

4>(M)

in some maximal ideal "" of

X

..

M = E1\.,X)

'

and since -M

~,X)

of

Conversely, if

m is a maximal ideal of

a maximal ideal of

It's clear that R

R

R[[X]]

'

then

for

m~

and

MH> (M)

is

R[[X]]_ generated

It is clear from this that is an ideal .of

.

M

is contained

is maximal, this implies

gerieral~ if .I

l.S

•l J. n n .I: b 0 .a. and g Then co I: b .f. E XR[ [X]] say . l 0 J. J.. . ' i=l .1 1 1= f.

l.

(

with

;

-

-

n

g

-· i=l

'I: b 0 .f. 1

1

= Xgl

•.

Now

gl '

· -

p .· (because

54 -

x;.

P)

,

and

n

E b 1 .f. = Xg 2 for suitable i=l 1 1 n j g = t h.f. for h.l. = 1: b .. x i=l 1 1 j;:::O J 1 g1 -

then Show

-

iv)

-

· v)

is trivial from ii) and iii) • For any ideal· I

R[[X]]

(verify).

of

v'(l,X)

if

then

But it's clear that

a Pi a •.•

~ Pct

(m,x>-a

ideals of

R[ [X]]

Thus ~

, then R[[XJ}

Ca 1 , ••. ,ad,X)

K-dim. R + l

K-dim J[[X]]

!

K-dim R + 1

is a chain of prime ideals in

mR[[X]]~ p 1 R[[X]]

argument shows .that

ll.

R ).

Cm ,X)-primary, so that . K-dim R[ [X]]

m

in

is the maximal ideal of

ihain of prime ideals iri

7,2.

I= Ca 1 , ••• ,ad)

m is the maximal ideal of

. by 5. 2 •

•

,x> = (/f ,X)

/(I

is a.system of

, and

R

= Clf,X) = Cm,X)

(where

R· ,

Hence if

parameters (5.3) in

is

Continue.

i ...

R([X]]

mR[ [X]]

and

which contain

'aPdR[[XJ]

R

is a

(Note that this (m ,X)

are the only prime

•

m .• )

Exercises. Use 7 .1. i and O. 6 to show that

ideal of

R[[X]]

X

is in every maximal

Use this in turn to show that if

($

• f ;g E R[ [X]] of

R[ [Xll

satisfy. cpf :. · cpg. , and

, then

f E: M. ~ g E M

M is a maximal ideal •.

. Then R[ [X] JM = ,U. Let M .be ~· maximal ideal of R([X]] , where m = cp(M) •· (Hint: consi.der the natural maps

·- 55 -

R[[X]]

R[[X]]M ---~ R [[X]] f m Use 7.1.i and 1) to show that

h

¢M

f 2 Ch)

Hence there is a map

f

is invertible whenever

making the diagram commute.

For a similar reason, there is a map

f

I

making

R

I .'\i R[ [ X ] ]

commute.

Extend

f

'

O

to be

an ideal in a Noetherian ring.

, when

(0)

I

is

In particular, it shows that

n

mn = CO) when · CR ,m) is local, a fact we shall use often n>O ·. in the sequel.. The proof will be easy after four little lemmas.

8.l.

be a multiplicative set in ·R

Let ·S

~

a finitely generated if

sM =

Proof.

=

•••

for some

0

If xn

=1

R-module.

Then

M be

and let

s- 1M =

O if and only

s -~ S

xl x1,•••,xn generate M then S-lM = 0 ~ l ·- 0 in S-lM ~ there are elements s.l. ~ s.. l.

with· s.x. = 0 l. l.

for each

i

l < i < n

=

The "if 11 part

is therefore clear, and for the "only if" one can take

E*ercise:

Use 8.1 to show that if

generated

R-module and

Pt AnnR(M) If

I

p

M i s • finitely

is a prime ideal, then

MP - o ~

•

is an ideal of

R

'

is clearly a multiplicative set.

S = 1 +I= {l+ala, I}· We will apply 8.1 to this

S . , after noting the following property:

- 58 -

S =1 + I

then

Jacobson radical of

s- 1R

8.2.

If

~

Proof. S-lR

is contained in the

It suffices to show that , for all

are of the form and

s- 1 I

a

S-lI b

Sl

S2

X (

s1s2+ab

1 + yx =

with Since

$.·J. E

But

x, y

a~I,bER

J.

I!)

, the

is in

S

, hence the fraction

is an ideal of

R

and

s 1 s 2 + ab

is

s- 1 R

invertible in

W..·

s

s. E: S ::: 1 + I

SlS2

numerator

is invertible in

y t S-lR(by 0.6)

-

'

l + yx

®

If

I

generated

R-module such that

(l+a)M = 0

Proof.

for some

Let

S

a

M

is a f ini t ~

IM= M , then

I

be the multiplicative set

hypothesis implies

(S- 1 I)(S- 1 M) =

by 8.2 and Nakayama's Lemma (0.6) .

s- 1M

l + I , hence

Then the conclusion

follows from 8.1 •

t:,..1!.;

Let

and let

I

be any ideal in the Noetherian ring

(Jl = f\ In n>O

R

I at = 07- .

Then

-

(The reader should pause to convince him(her)self that 8.4 is not a triviality.

It is often proved by

invoking the Artin-Rees lemma, which says that if

-

59 -

I, J,

are ideals in a Noetherian ring integer

r

R

, then there is an

In J ft K = In-r ( Ir J {\ K)

such that

See NAGATA (22], I, §3 .

n > r

for all

The simple approach

followed here is outlined in some exercises in ATIYAH•MACDONALD [2] (p. 43) .

For still another method

see §2.1 of KAPLANSKY (18].)

Proof of 8. 4. I Ol..

of

•

I

oi.

be a primary decomposition

q0

ot,cq. -

for all

1

01.~ q1n•••/1Qn =

; for then

-

I(Jt,C.

n ... ('\

,q 1

It suffices to show

1· < :.. i < n ''

Let

IQt..

The proof that OL C q. 1

is clear.

'

i

'

whereas for all

i

breaks into two cases: Case 1.

If

I

1=-

iinpl,ies that ~ If

Case 2.

.

by 2.2

8.5.

C q. -

'

l.

IC p. -

fii.l.

pi =

l.

Hence Ol

. n·

n.

1

'

then

-

1

C. - qi

for some

n.

1

1

Let

I

be an ideal

Then the following are

In = (0) 2) no element of the form n>O a~ I , is a zero-divisor in R

equivalent: 1 + a ,

n

R

I

IOl. C- q.1

C q.

(Krull Intersection Theorem).

in the Noetherian ring

.

by exercise 2.3(2)

= lq.l. ~ I

the fact that

'

1)

To illustrate, we display two important special cases of 8.5 before proving it:

- 60 -

-

8. 6.

Corollaries.

1)

m is the Jacobson radical of a

If

Noetherian ring (for example, the maximal ideal in

n

a local ring), then

mn = (0)

n>O

2)

n

is any proper ideal in a Noetherian domain,

I

If

In = ( 0)

n>O

Proof of 8.6. ( 0. 6)

1)

follows from 8.4 and Nakayama's lemma

(Alternatively, use 8.5, noting that a ( m

invertible for all For 2), if a= -1

l + a

; but

is a

1) ~2)

some

a f

say

0 ~ b

=

I

-ab

2) ===?1) :

' = a 2b = (fl

Let

0-divisor in a domain, then

.

is impossible if

1 + a

If

b(l+a) = 0

...

=

nn>O

= +anb

c

In

1 + I

(l+a)cJi

=

(0)

.,. b

'

then

nrn n>O Then

contains no

()i, = (0)

for some

a

is proper.

is a zero-divisor for 0

'

I

I OZ.

=at

by 8.4 ' so

-

that

is

, by the exercise of 0.6 .)

-1 =a~ I

Proof of 8 • 5 •

1 + a

e

I

'

by 8. 3

.

(l+a)Ot = (0)

0-divisors,

But if implies

, as desired. As an appiication of th~ Intersection Theorem, we

prove:

8.7.

Let

(R,m)

be local.

Then either

every principal prime ideal of

R

- 61 -

R

has height

is a domain, or 0

First we need a little lemma: Lemma.

If a prime ideal

p ~

, then

principal ideal

(x)

Proof of lemma~

Take

a 1 '- R

is properly contained ..i.~~

p

, and

y,

a2

a1

E.

(x)

l

Then

by 8.6(1) .

Remark.

If

R

p

Xe y : a

~

n

X

p

y =. a 1 x =

, and

p

n

Thus

n

for all

n

has a principal prime ideal

(x)n

p

(x)

properly contained in

n

b y th e l emma; b u t ·

n>O

Thus

for some

x ¢ p

since

, there is a prime p

Y -- a 1 X

Then

because

p

~

p

Continuing, we find

of height

X) n

n>O

, and in fact

Proof of 8.7.

n(

= (0)

(x) n

n>O

= (0)

is prime, and

R

is a domain.

This argument gives a pro~f that any principal

prime ideal in a Noetherian ring has height

< 1

independent of the Principal Ideal Theorem. 8.7 will be generalized to a result for prime ideals of height

n

, in 13.7 .

Another way to phrase 8.7 is

if

principal prime ideal in a local ring, then and only if,

x

n-generator

is a non-zero-divisor.

I = (x)

is

ht I = 1

For non-prime

principal ideals, this equivalence becomes slightly less simple:

- 62 -

a

if,

~·

Let

I

R.

be a principal ideal in a Noetherian ring

Consider the following three conditions: 1)

I

can be generated by a zero-divisor.

2)

I

consists entirely of zero-divisors.

3)

ht I= 0

Then 1) and 2) are equivalent, and 3) implies 2).

If

(O)

has no embedded primes then all three conditions are equivalent; but whenever is an

I

Proof.

has an embedded prime there

(0)

satisfying 2) but not

3)

•

The equivalence of 1) and 2) is trivial. If 2) is false, we can write

3) =;:>2)

a non-zero-divisor. prime of

minimal prime of

, so

R

x

Hence

x

lies outside every

ht(I) > 1

(In fact, therefore,

by the Principal Ideal Theorem, 4.2).

If

(0) · has no embedded primes, 2.20 shows that

every zero-divisor in

R lies in a minimal prime. ~

in this case we get 2) Finally, if use 2.22 to choose Then

X

have

ht I >

p X

is an embedded prime of

€ p

I

=

(x)

This completes the proof of 8.8

can be manufactured with ease. Krull dimension

(0)

> 1

'

x outside all minimal primes.

'

is a zero-di.visor, and, putting

a

Hence

3) .

We note that rings. in which (0)

of

with

(x)

lies outside every associated

x

, by 2.20 .

(0)

ht(I) = 1

Then

=

I

'

we

.

has embedded primes

In any Noetherian ring

choose prime ideals

- 63 -

pc:. q

·+

T

with

ht

p

is an ideal R

= T/I

~

l I

We will see (9.1 below) that there

of

, the image

T

of

such that q

Ass(!) = {p,q}

is an embedded prime of

(cf. 2.32).

- 64 -

In (0)

§ 9 ...

An Application of the Intersection Theorem to Primary

Decomposition. As another application of the Intersection Theorem, we will prove the following pretty res~li:

U·

Let

{p 1 , ... ,pn}

be a finite set of distinct prime

ideals, all of height

> 0

Theri there i~ an ideal

I

,

in~ Noetherian ring

of

R

As~(I) = {p1 , ••• ,pn}.

R with

We need, this time, only one new lemma:

-

9•2•

Let

be the canonical map, where

f :

prime ideal in the Noetherian ring q 1 f"\ ••• riqn

q's

with the qi

O -2)

9. 2 can also be used to show that if

p.1 = ./q.1

I -- q l'A I • • • , is a primary decomposition and p 1 is an

embedded prime of

I

, then there are infinitely many

pl -primary ideals

J

such that

a primary decomposition.

I = J (') q 2 () ••• (') q n

is

See p.231 of ZARISKI-SAMUEL

[31] ,

3)

The hypothesis in 9.1, that none of the primes be

m inimal, is clearly too strong: take , and

n = 1

.

R

to be a domain,

But the same

with

n > 1

shows that that hypothesis cannot be dropped entirely. It would be interesting to know what the precise necessary and sufficient condition is, on a set of primes for the existence of an

I

with

p1 , ... ,pn

Ass(!) = {p 1 , ••. ,pn}

- 68 -

---·----

,

1

§10.

Cohen-Macaulay Rings In the early part of this century the following

theorem was proved by F.S. Macaulay.

,

~

the complex

Ol = Cf 1 , .•. ,ft) has height then every associated prime of Ol has height t

numbers, and suppose

t

;

Note the strength of this theorem (compare with 4.7) and how much more than Krull's Principal Ideal Theorem

it gives for

R . : e.g. all the associated primes of a

principal ideal (not just the minimal ones) have height one.

There are no embedded components. This theorem points out, at first glance anyway,

what might be a difference between polynomi~l rings and arbitrary Noetherian rings.

We first give an example to

show that Macaulay's theorem does not hold in an arbitrary Noetherian ring.

~·

Example :

A Noetherian domain with a principal ideal

having embedded components~· Let subring,

A= k[X,Y]

'3

k

a field, and let

B = k[X 2 ,Y 2 ,XY,~]

Then

B be the

B .is a homomorphic

image of a polynomial ring in 4 variables and so is Noetherian; since it is a subring of

A it is a domain.

-

69 ·-

Claim 1.

A

Proof of claim 1. Y

and X

B

is integral over

It will be sufficient to show that

are each integral over f(t) = t 2 -

satisfies

g(t) = t 2 - Y 2 (

B[t]

B

X

and that is clear since

x 2 t B[t]

y

and

satisfies

This completes the proof of the

claim. Thus, in

A

K-dim A= K-dim B = 2

m r\ B is maximal in

then

m 1s maximal

and if

B

(These facts about

integral extensions can be found in ATIYAH-MACDONALD [2], chapter 5, for example). I= (X 3 )

Let

in

Principal Ideal Theorem,

Then, by Krull's

B

ht I= 1

minimal associated primes of

I

and further ~11 the

have height l (4.7).

shall show that there is a prime associated to

We

I

having

A

and

height 2 .

m = ( X , Y)

Let

m = mn B

~ I

m is a

since

in

having height 2 . I

It will be sufficient to show that every 0-divisor when considered in

x4 .x

f I x3 .x 2 € x4

First note that

x4 (m)

ht m = 2

then

is an associated prime of

Proof of claim 2. element of

A

is a maximal ideal in B

m

Claim 2.

~

=

-

, since I

70 -

and

B/I

Xi B Now x4 Y = x3 .(XY) EI

-

Thus all the elements of

are

'W\

0-divisors in

B/I

We shall now investigate this property of t[X 1 , .•. ,Xn]

that was illuminated by Macaulay's theorem.

Our approach will be ideal-theoretic, at first.

We shall

later (§21) re-introduce the ideas in a module setting and examine them from a homological point of view.

-

10.3.

l)

I

Let

be an ideal in the Noetherian ring

and let

Ass(I) = {p1,•••,Pt}

for all

1 < i¢j < t

2)

R

-

ht p.l. = ht P· J

If

, we say that

I

is called a Cohen-Macaulai ring i)

and

-

ii)

(0)

R

is height unmixed. (C-M) . if

is height unmixed

for every ideal ht I= n

I

I= Ca 1 , ... ,an)

where

is height unmixed.

(Cohen's name is attached to this definitiort because he showed that a class of rings, different from

C[X 1 , •.. ,Xn] ,

has the unmixedness property defined above). Our main goal will be to recover Macaulay's theorem in a more general setting.

As a guide to the reader we

note that some authors use semi-regular for what we have called

C-M

rings.

We begin the investigation of the unmixedness property by looking a bit more carefully at systems of parameters in local rings.

A first bbservation harkens back to a simple

- 71 -

formula we mentioned earlier (1.3) : in a ring of Krull dimenison

d

If

then

I

is an ideal

ht I+ dim I< d

We mentioned that the inequality might be strict.

The

following example shows this:

10.4. ..___

Let

k

be a field and

k[[u,v,w]]

ring in three variables over Clearly

R

and set

Cuv,uw) = Cu)/'\ Cv,w)

easily verifies that in

R = k[[u,v,w]J (uv,uw) If we let xi-+ x

k

denote the canonical map from

, so

K-dim k[[u,v,w]] = 3

k[[u,v,w]]

Cu) $- Cu, v> $

K-dim R > 2

the power series

to

Cu, v, w)

C7.1Cv))

in

R

, namely

2 •

K-dim k[[u]] = 1

a

prime chain

and we have factored out K-dim R < 2

Our description of the

(0) = (u)

is a prime of height

is

On the other hand, since

some non-zero divisors, we have by 4.6 that and hence equals

then one

R

f\

O in

(v,w) R and

(0)

, shows that R/(v,w)

~

'

ideal (v,w)

k[[u]]

and

Thus,

ht Cv,w) + dim(v,w) = 0 + 1 = 1 < 2

The next proposition shows how we can exercise some control over the dimension of ideals generated by subsets of a system of parameters.

-

72 -

--

10.5.

Let

u 1 , ••. , u.6

(R,m) E.

i) ii)

m ,

K-dim R = d

be local,

and let

< d

.6

dim I> d - .6 dim I= d subset of a system of parameters.

Proof.

Q

Let

R=

K-dim R = dim I= t

(R,m)

then

R/I

(say).

-

Let

-

u6 +1 , ... ,u-6+t

representatives of these elements in Clearly m-primary and so ii)

R

Cu 1 , .•. ,u 6 ,u.6+l'"""'u.&+t) = J .6

+ t > d

be

(5.2) hence

is

t > d -

.6

We use the notation of i) .

~:

If

~~

Let

Then

be a

u-6+1' ... ,u&+t

R and let

system of parameters for

is local and

dim I

=t =d -

u 1 , •.. ,u .6

u.6+l'"""'ud

then

.6

t +

.6

= d

and

so

, ••• , ud

be a system of parameter's.

generate an

m-primary ideal of the

K-dim R < d -

local ring

R

Thus,

dim I< d -

.6

By i) we always have

-0

,

i.e.

dim I> d -

.6

,

and

so we're done.

In view of this proposition we see that if u 1 , ..• ,u 6

live in the maximal ideal of a local ring

(R,m)

is a subset of a system I

of parameters and

2) ht(u 1 , ••. ,u 6 ) =

-

73 -

.6

then

ht(u 1 , ..• ,u 4 ) + dim(u 1 , .•• ,u~) = K-dim R In the example we gave (10.2) neither 1) nor 2) was satisfied.

{u 1 , .•• ,u,}

In the next example we show that if

is a subset of a system of parameters and

I= Cu 1 , ••• ,u 6

l) can hold without

~·

already seen that

{w,u+v}

Proof of claim.

, i.e.

2) • R = k[[u,v,w]]

Let

Exam;ele.

Claim.

ht I=~

then we need not have

)

(uv,uw) = 2

K-dim R

as before.

We have

is a system of parameters.

J =

It suffices to show that

(w,u+v)

is

m-primary and for that it's enough to show that mnc;,. J -2 = (u-2 ,v-2 ,w -2 ,uv,uw,vw) -- -- -for some n ( 2 • 7 ) • Now, m and since

UV: UW

=

0

these two generators of

-2 Obviously w and vw it's enough to show u-2 and V-2 are in

clearly in

ucu+v>, -2

V

E,

J

J

and equals

J

and so

-

Thus

so that

dim(w)

(v,w)

w

-l

u2

+UV:

u2

-2 m

are

J

and

are in J

But,

Similarly

-2 CJ m -

is a subset of a system of parameters and Since

Cw) K-dim R = d a

is not a

0-divisor.

by

4.6

we know

As a consequence

ht(a) = 1

(4.2) ,

dim(a) = K-dim(R/(a)) = d-1

Now,

of the Principal Ideal Theorem we know and by

Let

10.5 i i , we see that

a

can always be made part of

a system of parameters and all ingredients are present to assert

ht (a)+ dim(a) = K-dim R

-

75 -

§11.

R-sequences and Grade.

ll.:J:,·

An

R-sequence in

elements

a 1 , .•• ,a 4

~

R

such that

Ca 1 , ..• ,a~) -

i)

is an ordered collection of

R

and

R

Ca 1 , ... ,ai_ 1 ):ai = Ca 1 , ... ,ai_ 1 >

ii)

(where if

O:a 1 =0)

i=O

, 1

~

i
1

show that

J

~

J

gr(I,x) = 1

R then

is a non-zero-divisor in

is a non-zero-divisor we certainly

To complete the proof it is enough to

C Up

p£Ass(x) So, pick

f.,

a+ bx

y ( a+ bx )

a #. 0

(x) ~

(x ) Since

We want to find Now, since

a~ I

and

zero-divisors there is an Then

a~ I

a+ bx E J ,

a 1 (a+bx) € (x)

Yf

y

a+ bx I

and suppose

t

(x)

and

(x)

we must have

consists entirely of

a1 # O

such that

and we are done, unless

- 104 -

aa

1

a1

= O

£ (x)

In that case, aa 1

::

al = a 2 x

and

0

because

al

=

a 2 (a+bx)

and

we have

a 2x

a(a 2x)

is not a zero-divisor,

X

a 2 (a+bx) E. (x)

+ a 2 bx

= a 2a

aa 2

=

(aa 2 )x

=

0

and we are done, unless

an+l, (x)

and

0

Thus,

a 2 E. (x)

continue this process and eventually obtain and

=

Since.

We

an= an+lx

(a 1 ) ~ (a 2 )$ .. ~(an)~

; for if not we have

and all inclusions must be proper, contradicting the Noetherian The reason all the inclusions are proper is

assumption. that if and so X

(ak)

::

ak(l-Ax)

€ m

1

-

then

(ak+l)

AX

:::

ak+l

:::

(R,m)

But, since

0

is a unit of

R

and

Aak

1S

and hence

ak

::

ak+lx

local and

=

ak

0

= ak-1 =

= a 1 -- 0

and

show that

We leave that simple induction

an+la = 0

argument to the reader.

Lemma 3. suppose

Let

(R,m)

gr I< gr m

be local and let

I

be an ideal and

Then there is a prime

gr q = 1 + gr I

- 105 -

q

2 I

with

(of Lemma 3).

Proof

a maximal

R-sequence in

x e m with

X

r¥

x '¢ I

£

q E. Ass(L)

gr(I,x) = 1 + gr I

Since q

Proof of 12.5 --------

:

gr L

I/J

in

L

= gr

up

we note

as is easily seen R/J

and observe that

and so by 2.22,

= gr

have

I C

p~Ass(J)

L = Ca 1 , ... ,an,x)

Uq qEAss(L)

be

Thus, there is an

Since

by using Lemma 2 on the ideal

(I,x)

O\,•••,etn

J = (a 1 , ... ,an) CI

Set

I

\_}p p,Ass(J) Now,

Set

and let

gr J = gr I < gr m

and observe that

that

gr I = n

Let

(I,x)

small ht p .

to 11.16 the grade can only increase, and the small height is unaffected.

So, we assume that

gr m > small

(R,m)

ht m = n

We proceed by induction on n

=

O:

is a local ring and

If small

ht m

=

0

them

n

m

~

consists entirely of zero-divisors.

A~s(O) So

and so

gr m > 0

m

is a

contradiction. n > O!

Now assume

of small height< n

gr p

~

small ht p

for all primes

Since small ht m = n

- 106 -

p

there is a

p

prime ideal

with small

p ,

By the

-

~

q

p

gr m = 1 + (n-1) = n

p

gr q = gr

with

is the only prime containing

so

= n - 1

Since

we may invoke lemma 3 to assert the

existence of a prime

m

p

gr p < small ht p = n - 1

induction hypothesis, gr m > n > gr

ht

p

+ 1

But

q = m

and hence

and

This contradiction completes

the proof of the theorem.

If we now apply this result to

C - M

rings we

have a very interesting consequence.

Corollary 1.

EEime in

Proof.

Let then

R

If

R

be a

C - M

But, by 11.19,

p

If

is

ht p = small ht p

is prime in

p

ring.

R

gr p = ht p

then

in a

gr p C-M

the inequalities to be equalities.

~

small ht p

(R,m)

Note: This corollary

, and says, in this case, that

for any minimal prime

p

of

R

K-dim R = d

Let If

(R,m) p

be a local

The following corollary

C-M

is a prime ideal in

ht p + dim p = K-dim R

- 107 -

C - M

K-dim R = dim p

generalizes this to primes of height> 0

C~ollary 2.

ht p.

ring, forcing all

applies especially to the maximal ideal of a local ring

~

ring with

R

then

Proof. a

ht p = n

Let

C-M

gr p = n

, then

Thus there is an

ring.

since

R-sequence

is

R

a.,, ... ...

,a.

in

n

By 12.l ,

R = R/Ca 1 , •.. ,an) K-dim

C-M

is a

By the note before this

d - n

equal to

K-dim R = dim p = d - n

corollary we must have p

final word about terminology.

A

R

of

R

RM

such that

since

R

is a minimal prime of

ring

ring and by 10.5 it has

C-M

is

Nagata calls a

M

for any maximal ideal

, locally Macaulay and if, in addition, every

maximal ideal of Macaulay.

R has the same height he calls

In his Macaulay rings one has

R

ht p + dim p

constant for every prime ideal of the ring, while for our (non-local)

C-M

R = k[[X]]

example if

, k

a field, then

dimensional domain (7.1.v) and so is

R[Y]

by 11.21.

M1 -- (X,Y)

If

it is not a unit of

of

R[Y]

it is either l E. M2

'

(X)

or

(0)

ht M2 = 1

R[Y]

Thus, so is

ht M1 = 2

then

M2 f'"'\ R

XY -

and

1 E: R[Y]

If Thus Since

is prime in

M2 f"'i R = (X) M2 n R = (0) M2

R

and so

then

'

is maximal,

ht M2 + dim M2 '# ht M1 + dim Ml

terminology,

is a one

, so we can choose a maximal ideal

containing it.

a contradiction.

6.7 we have and so

R[Y]

R

C-M

Now, consider the element

M2

For

rings this may not be the case.

and by dim M2 = 0

In Nagata's

is locally Macaulay but not Macaulay.

- 108 -

---, j

Exercises. 1)

12.6 ......._..,.. .

p

$. q

be prime ideals of

between ht q 2)

Let

::

p

and

q

R R

be a

ring and let

such that there are no primes

Show that

gr q = 1 + gr P

and

1 + ht p

Find a Noetherian ring

R

and prime ideals

with no primes between them, where 3)

C-M

Same as 2) but now

gr q # l + gr P

ht q # 1 + ht p

- 109 -

p~ q

'

§13.

Regular Local Rings. If

(R,m)

obvious way)

m/m 2

is a local ring,

an

Rim-module, i.e.

is (in the

a vector-space.

As

such, it has a dimension; this number is called the V-dimension of

R

Using the Principal Ideal Theorem and V-dim R

Nakayama's lemma, we shall show that


..

I

=

>,.

=

u

R[X]

all in

Since

yu

'

p () R = 0 we have

-

degree in

p

4>

'

and

and

= cr

ycr

p

and since

we must have

y 2 f(X)

y ;. 0

y2g(X) ~ p

y 2 f (X) €

deg f(X) < deg ,:(X)

=

=

Since p

We can do this since

Claim.

Proof

m

=

'

be a

=

YT

=

y

and hence

¢

..(X) + f(X)µ(X)

Then

On the other hand

¢,(X)

yep

'

f(X) =

and

K[X]

both in

non-zero element of I

I

(X)q, (X)

0

= c

such that

act m

J(R) = 0

p + mR[X] "/. R[X]

(of claim).

m(X) € mR[X]

with

If not there is i(X) + m(X) = l

- 131 -

i(X) € p

and

By 14.5 , we know

that for some integer .6(X)

€ R[X]

Thus,

Now, by choice, -6

(X)

¢

a

-61 (X) (/ mR[X] -6 1 (X) = O

N

l/

m

-62(X)

€

mR[X]

and so

m

¢(X) = "6l(X) + &2(X) where

as

mR[X]

for some

N

and observe that

a

-61 (X) 'I-

Rewrite

,

For, if

0

, then the right-hand side of(*) is in

, whereas

¢

aN

m

a N = f(X)-6 1 (X) + f(X)-6 2 (X) + a Nm(X)

Thus,

= f(X)-6 1 (X) + rn(X) where

mOO € mR[X]

Write

-6l(X) = a.ix

and

0 ~ tl < .Q,2
1) in - € m hence 0 = aa. p + m for some m ' a contradiction since neither a nor

degree

That term is

(t)

aa (l

This

f m

p

is in

p

J.S

m

This completes the proof of the claim.

We can now finish the proof of the theorem. P

be a maximal ideal of

p + mR[X]

that

g(X)

Clearly

f/.

P r't R

g ( X) E P

~

p

containing the ideal and it remains only to show

P

Since have

P

R[X]

Let

P r'l R

= m

then

'2. m and

Now

f (X)

m ~

P

is maximal in

we

must

and if we also assume

g ( X) A ( X) + f ( X) µ ( X) = c

- 132 -

R

e.

P

But

c £ R

--, j

and so

C

E:

p

nR

= m

was chosen so that

c

m

This is a contradiction since

4

m

We obtain several corollaries to this proposition.

Corollary 1.

R

4:===? R[X 1 , ... ,Xn]

is a Hilbert ring

1.s

a Hilbert ring.

Pr'oof,

4'::::.'. We have already seen this ( Corollary to 14. 2).

~ : It is clearly sufficient to prove this for n = 1 A

R[X]

So, let

=

where

R[X];? p

We want to show that

p

=nm

prime ideal in

p

lS a

m

maximal in

-

m:> p

'

Let above

A

p

= p

p() R

Then the primes in

R[X]

R[X] = R/p[X] pR[X]

(and

"are" the primes in

living

we'll even call them by the same names.) Since

R

is a Hilbert ring we have

R/ [X] we have "p () Rip = (O)" . p then follows from 14.6 .

and in

Corollary 2.

If

F

is a field,

J(R/p) = (o) The result

F[X 1 , ... ,Xn]

Hilbert ring.

Corollary 3.

Z[X 1 , ..• ,Xn]

is a Hilbert ring.

- 133 -

is a

Corollary 4.

Let

R

be a Hilbert ring and

which is finitely generated as a rin~~~er S

a ring

S

Then

R

is a Hilbert ring.

Proof. over

is a homomorphic image of a polynomial ring

S

R

and so is Hilbert by virtue of

14r2

and Corollary 1 above.

--

14.7.

Let

R

maximal in

J(R) = W)

be a ring with

R[X]

with

M () R

=

Let

M

R

is a

then

(0)

be

field.

Proof.

M() R = (0)

Since

R and so

prime in

Choose least degree in we can find

m = (0)

R

(0)

we have that

is

is a domain.

f(X) M

to be of , then

Since

n > 1

m maximal in

R

J{R) = (0)

aim

with

If

we are done.

m # (0)

So, assume

at

Since

Mn

hence

ag(X) + h(X) = 1

R = (0)

We know that for

= R.(X)f(X)

and

t

and choose M and so

for some

(a,M)R[X] = R[X]

g(X) E".

sufficiently large,

a t g(X) = f(X)q(X) + r{X)

deg r(X) < deg f(X)

Now,

- 134 -

O #a£ m.

R[X]

'

a t h(X) = where

'

h(X) £ M

= a(f(X)q(X)+r(X)) + i(X)f(X) = ar(X) + f(X) (ag(X)+1(X))

f(X) €. M

Since

other hand, at

hence

= a.r(X)

Corollary 1.

a

t

-

ar(X)

and so

a

t

€

R/p

Let

R

hence

p

M (\ R

p

0

at E:- m

Let

R

be

is maximal in

R

R

So, let and that

By abusive notation

R/p = ~) and so, by 14.7,

Rip

is a field and

R

be a Noetherian Hilbert ring in ~

is a Noetherian Hilbert_ ring

ideals have height

Proof.

o.r(X) =

-·-·-·---

is prime in

which all maximal ideals have height R[X 1 , ... ,Xn]

-

M

n = 1

J(R/p) = (0)

Thus

is maximal in

Corollary 2.

t

Thus,

be a Hilbert ring ___a_!1_d__J..~!

We know that

Mn

a

, contradiction.

is Hilbert.

we have

on the

Thus

It is enough to prove this for

MnR=p

'

(a) Cm

then

Proof.

E- M

deg(a t -ar(X)) < deg f(X)

a€ m

and so

we have

n + 6

Use Corollary 1 and 6.7 .

- 135 -

Then

-1:.!l__w~ich __all __!I~~im~

'

~.

Examples. 2:_)

ideals in

If

F[X 1 , ... ,Xn]

In Z[X 1 , ... ,Xn]

2)

F

is a field, all maximal

have height

n

, all maximal ideals have height

n + 1

ll

The condition, in Corollary 2 above, that

Hilbert in order that all maximal ideals in the same height is, indeed, necessary.

R

be

R[X]

have

See the example

at the end of §12 . 4)

It is worth mentioning the following characterization

of Hilbert rings (although we shall not use it here) and we refer the reader to GOLDMAN [16] for proofs.

~'7

Hilbert ring M /JR

whenever

is maximal in

M is maximal in

R

is a

R[X]

R

We would like to say something about generators of maximal ideals in polynomial rings over liilbert rings. The next proposition is a step in that direction.

---

14.9.

Let

be a Hilbert ring and let ··----·-·-

R[X]

ideal in ;eolynomial

Proof.

R

If

p

f(X) E R[X]

= M () with

M

R

In

R/ [X] p

-

R

then there lS a manic -·---------------(p,f(X)) = M

By Corollary 1 to 14.7 we know that

maximal in

be a maximal

·-·-·-·

we can generate

by a single monic polynomial, since

R/

p

[X]

p

is

nM" is a principal

ideal domain.

A monic representative for this generator

will serve as

f(X)

- 136 -

l Corollary 1.

Let

R

be a Hilbert rin_g_ in whichdeve.E.Y

maximal ideal can be generated by every maximal ideal in

rn

R[X 1 ,. , ,Xn]

elements.

Then

can be_g_enerated by_

n + m elements. We sihgle out two special cases of specific interest.

Corollary 2.

Every maximal ideal in

a field, may be generated by maximal ideal of

n

r

rrxl'. ... ,x n J .

elemen"t:~ and eve..E_l

Z[X 1 , •.. ,XnJ. may be generated.by

n + 1

elements.

Remark.

By 14.8 , we know that maximal ideals in

F[X 1 , ... ,Xn]

and

respectively.

Z[X

1,... ,Xn]

have height

and

n + 1

By Krull's Principal Ideal Theorem we

know that we could not generate a height fewer than

n

k

k

ideal by

elements, so these results give the best

one could expect. We are now ready for>.·. a proof of Hilbert's famous Nullstellensat z.

14.10.

(Nullstellensatz-W~ak Version).

Let

be an -.·---

F

algebraically closed field and let

M C F[X 1 , ••. ,Xn]

maximal ideal.

a. 1 , ... ,a.n

There are elements .

such that

M

=

'.·

.·

.·

(X 1 -a. 1 ,.. ••.• ;X n --a n ) .. ·.·

.,. 137 -

of

F

be a

Proof.

We first note that

Since

M contains monic polynomials

is algebraically closed, each

F

of linear factors and since factors belongs to

F[X 1 , .•• ,Xn]

in

I

is clearly maximal

I= M

, and hence

1-1

This theorem then gives the between maximal ideals in when

F

the F

F[X 1 , •.. ,Xn]

is algebraically closed.

tioning that

1-1

F

l.

M ~ (X-a 1 , ... ,Xn-an) = I

But,

a 1 , ••• ,an £ F

for some

l.

1.s prime, one of these

M

Thus,

M

is a product

f.(X.)

correspondence and points of

Fn

It is, perhaps, worth men-

algebraically closed is really needed for

correspondence.

As an extreme case in point, if

is a finite field there are infinitely many prime

(hence maximal) ideals in

F[X]

of the corresponding fact for

(Recall Euclid's proof Z

).

We finish off this section with the full version of Hilbert's theorem:

:l,;_4..:,..,1~.

(Nullstellensat"Z,).

Let

f(a 1 , •.. ,an) = 0

= gk(a 1 , ... ,an) = 0 integer

Proof.

be an algebraically

be in

closed field and let Suppose

F

whenever

Then

f1

c

g 1 (a 1 , ..• ,an) = (g 1 , ... ,gk)

=

for some

R.

The conclusion of the theorem is that

and we prove the theorem in this form.

- 138 -

f

'=

/"(g1 , .... ,gk) ,

We first note that

To see this, define

f €: (X-a 1 , ... ,X -a) n n cp :

:: a

r [ x 1 , ... , xn J Then

n

cp

..... F

by

cj>/F

f(a 1 , ••• ,an) :: 0

f

P1

l

p

Q

l

/

P2

P/IP

~

Q

p 2f

K/IK

O

+

K

+

P

is).

is just

Since

ker g

K = 0

Corollary.

+

Q+ 0

+

K

is projective); so P

therefore

~

0

IQ

Hence

Q is finitely generated; Nakayama). Then

(Q

(because

Q = fP

is onto, we have

= ker f

splits

Q/IQ

~

P2

onto (because K

.)

gpl

/

g Since

1

/

/

/

Q/IQ

~

/

f

as

O

+

K

+

f

lS

Let

is exact, hence

is finitely generated P

(cf. 17.4)

+

Q + O is split, Hence

K = IK

, and

(Nakayama again).

Finitely generated projective_modules over

local rings are free.

Proof. P/mP

Given a finitely generated projective module is a finite-dimensional

dimension

~·

n

Rim-vector space, say of

Apply 17.5 with

Two R-modules

M and

equivalent (and we write some projective modules

N

M - N) P

P

and

Q = Rn

are said to be projectively if

Q

- 154 -

M fB P = N

(I)

Q for

Exercises.

±J

is an equivalence relation. M - N #

projective then

M

2) If

N

is

is projective.

The motivation for 17.6 and for what follows is the desire to study modules by "generators and relations". Given a module

, we approximate it by mapping a free

M

module onto it, then 1=ook at the kernel to see how good the approximation was. 1

0 + K +

Fl

l

M

0

+

Hn +

0 + K EB F2

Fl

EB

The kernel

1S

not unique: if

is exact, so lS ( j '0) + M+ 0 F2

17.7 shows that it

is unique up to projective equivalence:

--

+

~

K

17.7. 0 +

L

Proof.

f ("Schanu.el's Lerruna"). If 0 + K + p + M + 0 and g Q + M + 0 are exact, with p and Q projective,

-L

More 2reciseli,

Q XM p = {(x,y)

Put

obtaining

K 61 Q = L 83 p

~ QxPjfy

= gx}

0

,l, K 'II' 2

Q XM p

'11'1 Q-a.>

L--,.

.,.

p

l

L Q

,!,

~

M-1> 0

.i 0

where and

'If.

1

'If 2

Observe that

are induced by projections. are onto, and tha.t

ker

11' 1

- 155 -

=

K

and

ker n 2

'lr 1

-

L

Hence we have split exact sequences 0

+

K

+

Q xM P

Q

+

+

0

O

and

L

+

+

Q xM P

P

+

+

0

, so

that

-

17.8. -1

1) The 0-module is said to have proj~