From (3.5) it is clear that we only need to know that grade(Ann M )A ≥ grade M . Since

grade(Ann M )(GrI A) ≥ grade M , this should hold if (Ann M )A and (Ann M )(GrI A)

can be related in a reasonable fashion.

The following example shows that M may not be perfect if one replaces indetermi-

nates by an A-sequence without further precautions. Let A = Z[U, V, W ]. Then x1 = U ,

x2 = V (1 ’ U ), x3 = W (1 ’ U ) is an A-sequence. However (Z[X1 , X2 , X3 ]/(X2 , X3 )) — A

(via the substitution Xi ’ xi ) is not a perfect A-module. Though x— , x— is a GrI A-

2 3

sequence, grade(x2 , x3 ) = 1 only, and (x2 , x3 ) is not even unmixed. The di¬culty arises

from the fact that

xk R = 0 in R = A/Ax2 .

3

k≥0

The usual way out is the assumption that I be contained in the Jacobson radical.

(3.7) Lemma. Let A be a noetherian ring, I, J ideals in A, x ∈ A.

(a) If x— is not a zero-divisor modulo J — , then (J + Ax)— = J — + (GrI A)x— .

(b) If furthermore I is contained in the Jacobson radical of A, then x is not a zero-divisor

modulo J.

Proof: The isomorphism (1 ) above readily reduces the problem to the case in

which J = 0. Then the ¬rst statement follows from the equation a— x— = (ax)— which

always holds if a— x— = 0, and the second statement is trivial: a— = 0 for all a ∈ A. ”

(3.8) Lemma. Let A be a noetherian ring, I, J ideals in A such that I is contained

in the Jacobson radical of A. Suppose that grade J — ≥ g. Then J — contains a (GrI A)-

sequence x— , . . . , x— , xi ∈ J. Therefore x1 , . . . , xg is an A-sequence, and, in particular,

1 g

grade J ≥ g.

Proof: A homogeneous ideal which is not composed of zero-divisors must contain

a form which is not a zero-divisor. The rest is induction on g, the inductive step relying

on the preceding lemma. ”

We return to the situation discussed above (M = M —Z[X] A, M = M —Z A/I). If

I is contained in the Jacobson radical, then by (3.8)

grade(Ann M) ≥ grade(Ann M )— .

In general (cf. (3.10),(c) below) grade(Ann M )— < grade(Ann M ), and the argument

breaks down. However, if M is a graded Z[X]-module, then Ann M is generated by

32 3. Generically Perfect Ideals

forms f1 , . . . , fs ∈ Z[X] of positive degree (otherwise M would not be Z-¬‚at!). The ideal

(Ann M )A is generated by the elements fi (x) ∈ I whereas (Ann M )(GrI A) is generated

by the fi (x— ). Since for a homogeneous polynomial f ∈ A[X]

(f (x))— = f (x— ) f (x— ) = 0,

or

we conclude that ((Ann M )A)— already contains (Ann M )(GrI A). So

grade M = grade(Ann M )(GrI A) ¤ grade((Ann M )A)—

¤ grade(Ann M )A ¤ grade(Ann M),

and quoting Theorem (3.5) we complete the proof of:

(3.9) Theorem. Let M be a generically perfect graded Z[X]-module of grade g. Let

n

A be a noetherian ring, x1 , . . . , xn an A-sequence such that I = i=1 Axi is contained

in the Jacobson radical of A. Then, via the substitution Xi ’ xi , the A-module M — A

is perfect of grade g.

(3.10) Remarks. (a) The hypothesis on I can be slightly weakened. Whether

M — A is perfect of grade g, can be decided from the localizations M — AQ , where Q runs

through the maximal ideals containing (Ann M )A. Therefore one may localize ¬rst, and

it su¬ces that I ‚ Q for these maximal ideals Q.

(b) The assumption that I be contained in the Jacobson radical can be replaced by

the hypothesis that A is graded and the xi are forms of positive degree. We leave the

necessary modi¬cations to the reader. (Any hypothesis covering both cases has a rather

arti¬cial ¬‚avour.)

(c) Let p ∈ Z, p = 0, ±1, A = Q[[U, V ]]/(U V ’ pV ), and x1 the residue class of U

in A. The module M = Z[X1 ]/(X1 ’ p) is generically perfect and x1 is contained in the

Jacobson radical. Nevertheless M — A is not perfect, an example demonstrating that the

assumption on M being graded is essential.

(d) Theorem (3.9) has obvious consequences for determinantal ideals. Whenever C

is a matrix whose entries form a regular sequence inside the Jacobson radical, then the

ideals It (C) are perfect. (It will be proved in (5.18) that the ideals It (X) are generically

perfect.) Guided by this example we want to indicate a second approach to the proof of

(3.9). Let X be an m — n matrix of indeterminates over Z. As we shall see in (5.9) there

is a sequence y1 , . . . , , ys , s = mn ’ (m ’ t + 1)(n ’ t + 1), of elements in I1 (X) such that

s

Rad It (X) + Z[X]yi = I1 (X).

i=1

If C is as above and • : Z[X] ’ A the substitution X ’ C, then

s

Rad It (C) + A•(yi ) = Rad I1 (C)

i=1

and it follows that

grade It (C) ≥ grade I1 (C) ’ s = (m ’ t + 1)(n ’ t + 1)

since I1 (C) is contained in the Jacobson radical (cf. [Ka], Theorem 127). ”

33

B. The Substitution of Indeterminates by a Regular Sequence

Besides M —Z[X] A and M —Z (A/I) = M —Z[X] (GrI A) there is a third module of

interest, namely GrI (M — A). This is a graded module over GrI A generated by its forms

of degree zero, and unless the same, possibly after a shift of the graduation, holds for M

itself, we cannot expect that GrI (M — A) ∼ M — (GrI A).

=

(3.11) Proposition. Let M be a graded Z-¬‚at (thus Z-free) Z[X]-module generated

by its forms of lowest degree. Then, with A, x1 , . . . , xn , and I as in (3.9) one has

GrI (M — A) ∼ M — (GrI A),

=

the tensor products taken over Z[X].

Proof: After a shift we may assume that M is generated by its forms of degree

zero. Consider a homogeneous representation

g f

Rp ’’ Rn ’’ Rm ’’ M ’’ 0

over R = Z[X] in which the elements of the canonical basis of R m are assigned the

degree 0, and those of the canonical basis e1 , . . . , en of Rn are assigned degrees µ1 , . . . , µn

such that µi = deg f (ei ), i = 1, . . . , n. Let d1 , . . . , dp the canonical basis of Rp , and δj =

deg g(dj ), j = 1, . . . , p. We may assume that none of f (e1 ), . . . , f (en ) or g(d1 ), . . . , g(dp ) is

divisible by any q ∈ Z, q = ±1, since M and Im f are Z-¬‚at. Therefore M = M —Z (A/I)

has the homogeneous representation

g f

Gp ’’ Gn ’’ Gm ’’ M ’’ 0,

(2) G = GrI A,

in which none of f(ei — 1) or g(dj — 1) is zero, i = 1, . . . , n, j = 1, . . . , p. After tensoring

the representation of M with A (over R) we obtain a zero-sequence

˜

˜

g f

Ap ’’ An ’’ Am ’’ M ’’ 0, M = M — A,

which is exact at Am . Since f (ei —1) = 0, we have f (ei —1) ∈ I µi Am , f(ei —1) ∈ I µi +1 Am .

/

Hence (f (ei — 1))— = f(ei — 1), and Im f ‚ (Im f )— , where — now denotes leading forms

in Am , of course. We have to prove that (Im f)— ‚ Im f , too (cf. the isomorphism (1)

above). The crucial argument will be that one can “lift” any relation of the f(e i — 1)

aj f (ej — 1) ∈ I u Am . If su¬ces to show that there

because of the exactness of (2). Let

are aj ∈ I u’µj such that

aj f(ej — 1) = aj f(ej — 1).

Suppose that deg a— = ±j and that there is a j with ±j < u ’ µj . Let

j

κ = max{ u ’ µj ’ ±j }

j

and

if u ’ µj ’ ±j = κ,

aj

bj =

0 else.

34 3. Generically Perfect Ideals

b— f (ej — 1) = 0 and deg(b— , . . . , b— ) = u ’ κ. Therefore there are homogeneous

Then 1 n

j

elements r1 , . . . , rp , deg rk = u ’ κ ’ δk or rk = 0, such that

(b— , . . . , b— ) = rk g(dk — 1).

1 n

Let s1 , . . . , sp ∈ A such that s— = rk , k = 1, . . . , p, and de¬ne

k

(a1 , . . . , an ) = (a1 , . . . , an ) ’ sk g(dk — 1).

For the j-th component dkj of g(dk ) one has

deg dkj = δk ’ µj .

d kj = 0 or

The image of dkj under the extension Z[X] ’ A is the j-th component dkj of g(dk — 1).

We claim that

sk dkj ≡ bj ≡ aj mod I u’µj ’κ+1 .

(3)

Since obviously

aj f(ej — 1) = aj f(ej — 1),

we are done, for an iterated application of the procedure leading from aj to aj will produce

aj f (ej — 1) as desired. The congruence (3) is likewise obvious, since

a representation

bj ≡ aj by de¬nition, and the residue class of sk dkj modulo I u’µj ’κ+1 is rk dkj , dkj being

the j-th component of g(dk — 1), as seen by arguing with degrees. ”

The hypothesis that M be generated by its forms of least degree is satis¬ed for cyclic

graded Z[X]-modules, so in particular for our standard example, the residue class rings

modulo determinantal ideals. A typical application of (3.11) will be given in (3.13).

C. The Transfer of Integrity and Related Properties

As pointed out already, one of the consequences of Theorem (3.3) for determinantal

ideals, say, is that their perfection can be proved by considering ¬elds only as the rings

of coe¬cients. A similar reduction works for the properties of being a radical ideal, a

prime ideal or even a prime ideal with a normal residue class ring.

(3.12) Proposition. Let J be an ideal of Z[X] such that Z[X]/J is faithfully ¬‚at

over Z. Suppose that (Z[X]/J) —Z K is reduced (a (normal) integral domain) whenever

K is a ¬eld. Then (Z[X]/J) —Z B is reduced (a (normal) domain) if B is noetherian

and reduced (a (normal) domain).

Proof: Let F denote the total ring of fractions of B. Then the embedding B ’ F

extends to an embedding

(Z[X]/J) — B ’ (Z[X]/J) — F.

If B is reduced (a domain) then F is a direct product of ¬nitely many ¬elds (a ¬eld),

so (Z[X]/J) — F is a direct product of reduced rings (a domain). It remains to consider

normality, for which we can use the properties of ¬‚at extensions, since A = (Z[X]/J) — B

is B-¬‚at. The normality of A follows from the normality of B and the normality of the

¬bers of the extension B ’ A which are given as (Z[X]/J) — (BQ /QBQ ), Q running

through the prime ideals of B. ”

35

C. The Transfer of Integrity and Related Properties

This proposition has a variant concerning extensions Z[X] ’ A as discussed in (3.9)

and (3.11).

(3.13) Proposition. Let J be a homogeneous ideal in Z[X] such that Z[X]/J is

Z-¬‚at and (Z[X]/J) —Z K is reduced (a (normal) domain) for all ¬elds K. Let A be a

noetherian ring with an A-sequence x1 , . . . , xn inside its Jacobson radical such that A/I,

I = n Axi , is reduced (a (normal) domain). Then, via the substitution Xi ’ xi ,

i=1

(Z[X]/J) — A is reduced (a (normal) domain).

Proof: Let G = GrI A. Then, by virtue of (3.11), one has (JA)— = JG. Proposition

(3.12) implies that G/JG is reduced (a (normal) domain) and this in turn forces A/JA

to have the property claimed (cf. [ZS], p. 250). ”

Again the hypotheses of (3.13) could be slightly weakened as indicated in (3.10),(a).

The following proposition will sometimes help to compute the grade or height of an

ideal.

(3.14) Proposition. Let J be an ideal in Z[X] such that R = Z[X]/J is faithfully

¬‚at. Suppose that I is an ideal in R such that grade I(R—Z K) ≥ k for all ¬elds K. Then

grade I(R —Z B) ≥ k for all noetherian rings B, and if grade I(R —Z K) = k throughout,

then always grade I(R —Z B) = k. Analogous statements hold for height.

Proof: Let P be a prime ideal of A = R —Z B, P ⊃ IA, Q = B © P , and

K = BQ /QBQ . Then

depth AP = depth BQ + depth AP —A K

since A is a ¬‚at B-algebra. AP —A K is a localization of R —Z K with respect to a prime

ideal containing I(R —Z K), hence

depth AP ≥ depth AP —A K ≥ k

as desired.

Suppose now that grade I(R —Z K) = k for all ¬elds K. Then we choose P as

follows. First we pick a minimal prime Q of B. Next we take a minimal prime ideal P

of I(R —Z K), K = BQ /QBQ , such that depth(R —Z K)P = k. Then the preimage P of

˜

P in A satis¬es P © B = Q, (R —Z K)P = AP —A K, so depth AP = k by the equation

˜

above.

In order to get the statements for height, one replaces depth by dimension. ”

We note a consequence which will be used several times below:

(3.15) Corollary. Suppose that (the image of ) x ∈ R is not a zero-divisor in

R —Z K for all ¬elds K. Then x is not a zero-divisor in R —Z B for every ring B.

In fact, if x is a zero-divisor in R —Z B for some commutative ring B, then there is

a ¬nitely generated Z-algebra B ‚ B such that x is a zero-divisor in R —Z B, and one

obtains a contradiction from (3.14).

36 3. Generically Perfect Ideals

D. The Bound for the Height of Specializations

Let M be a perfect Z[X]-module of grade g, A a noetherian Z[X]-algebra, P a

minimal prime ideal of Ann(M — A). Then, applying (3.5) to M — AP , we see that

grade P ¤ grade P AP ¤ g.

This inequality can be sharpened:

(3.16) Theorem. Let M be a perfect Z[X]-module of grade g, A a noetherian

Z[X]-algebra. Then ht P ¤ g for every minimal prime ideal P of Ann(M — A).

Proof: We may obviously assume that A is local and P its maximal ideal, and, com-

pleting if necessary, that A is a complete local ring. P is a minimal prime of (Ann M )A,

too. By the Cohen structure theorem A may be written S/I, where S is a regular local

ring. The extension Z[X] ’ A can be factored through S, and