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sequence of indeterminates ([Hi], p. 229). Gaeta then seems to be the ¬rst who indicated
a free resolution of Im (U ) in a comparatively general case ([Ga]); he considered an m — n
matrix U , m ¤ n, the entries of which are homogeneous polynomials over a ¬eld such
that grade I(U ) = n ’ m + 1. The ¬rst general construction was given by Eagon and
Northcott in [EN.1]: The di¬erentiation di of their complex
n m
dr d0
0 ’’ G — Sr (F ) ’’ . . . G ’’ A ’’ 0

de¬ned by means of bases of G and F in an obvious manner, depends on the special
choice of the basis for F (in case i > 0) and G (i = 0), resp. Our presentation, i.e. D 0 (g),
is independent of the bases choosen for F and G, up to the de¬nition of ν0 . It goes back
to [BE.1].
There are numerous generalizations of the Eagon-Northcott complex in a similar
direction as we took in Subsection C, and we do not claim the following list to be complete.
From the complexes considered in [Bu.1], [BR.1], [BR.2], [Gv], [BE.4], one gets (minimal)
free resolutions of Coker g and Coker Sp (g), 1 ¤ p ¤ m, if grade I(g) = n ’ m + 1. With
the same assumption the complexes in [Le.1], [Le.2] yield (minimal) free resolutions
p p
of Coker g , cf. (2.19), those in [Wm] corresponding resolutions of Coker g — and

Sp (Coker g — ), 1 ¤ p ¤ n ’ m + 1. The last three papers and [BE.4] make use of divided
powers which are also applied to the construction in [BV] giving (minimal) free resolutions
of Sp (Coker g), 1 ¤ p ¤ n ’ m as in (2.16). It seems that the complexes Di (g) have ¬rst
been constructed by Kirby ([Ki]) in terms of bases.
References for (2.20) and (2.21) will be given in Section 9.
[BE.4] also covers the Gulliksen-Neg˚ complex for which we followed the original
treatment in [GN]. With [Po] containing a (minimal) free resolution of In’1 (U ) in case
U is an n — (n + 1)-matrix (and grade In’1 (U ) = 6), we ¬nish our list of “classical”
contributions to the problem of constructing free resolutions for determinantal rings.
After some attempts which were more or less e¬ective, Lascoux [Ls] was the ¬rst
who found a minimal free resolution of Rt = B[X]/It (X), 1 ¤ t ¤ m, over B[X] in case
B contains the ¬eld of rational numbers. This resolution has also been constructed in
di¬erent ways by Nielsen [Ni.1] and Roberts [Rb.1]. In [PW.1] Pragacz and Weyman
give “another approach to Lascoux™s resolution”. It is at present not known whether
a minimal free resolution of Z[X]/It (X) exists which remains minimal after tensoring
over Z with any ring B. Of course such a resolution exists in the maximal minor case
(see above). Akin, Buchsbaum and Weyman [ABW.1] gave a positive answer in the
submaximal minor case by an explicit construction, following an idea ¬rst applied in
[Bu.3]. For further discussion of the subject we refer the reader to [Ni.2], [Rb.2], [Rb.3].
3. Generically Perfect Ideals

The determinantal ring B[X]/It (X), where X is a matrix of indeterminates, may be
written as
B[X]/It (X) ∼ (Z[X]/It (X)) —Z B.
It arises from the corresponding object over the “generic” ring Z by extension of coef-
¬cients. In this section we want to study how the arithmetic properties of Z[X]/I, I
an ideal in Z[X], carry over to (Z[X]/I) —Z B under the (inevitable) assumption that
Z[X]/I is Z-¬‚at.
Another type of extension to be investigated below is the substitution of a sequence
of indeterminates X1 , . . . , Xn by an A-sequence x1 , . . . , xn , A a Z-algebra.

A. The Transfer of Perfection

(3.1) Proposition. Let M be a (not necessarily ¬nitely generated) Z-module. Then
M is ¬‚at if and only if it is torsionfree.
Proof: A ¬‚at module is always torsionfree. For the converse one uses that a ¬nitely
generated torsionfree Z-module is free. Thus M , being the direct limit of its ¬nitely
generated submodules, is the direct limit of ¬‚at Z-modules and therefore ¬‚at itself. ”
Throughout this section X will merely denote a ¬nite collection X1 , . . . , Xn of in-
determinates. The most important property which descends from a Z-¬‚at Z[X]-module
M to M —Z B, is perfection: grade M —Z B = grade M , and for a free Z[X]-resolution
F of length grade M the complex F —Z B is a free resolution over B[X]. This will be
shown in Theorem (3.3) below. (Note that every ¬nitely generated Z[X]-module has
¬nite projective dimension for obvious reasons and that every projective Z[X]-module is
free, cf. [Qu].)
Definition. A ¬nitely generated Z[X]-module M is called generically perfect (of
grade g) if it is perfect (of grade g) and faithfully ¬‚at as a Z-module. An ideal I is called
generically perfect, if Z[X]/I is generically perfect.
Before we state the main theorem on generically perfect modules, we want to indicate
how this de¬nition could be modi¬ed:
(3.2) Proposition. A ¬nitely generated Z[X]-module M is generically perfect of
grade g if and only if M is a perfect Z[X]-module of grade g, and for every prime number
p the (Z/Zp)[X]-module M —Z (Z/pZ) is perfect of grade g.
Proof: The implication “only if” is covered by Theorem (3.3) below, whereas for
the “if” part we only need to prove that M is torsionfree as a Z-module: M —(Z/Zp) = 0
for all prime numbers p by hypothesis. Assume that an associated prime P ‚ Z[X] of
M contains a prime number p. Since a perfect module is unmixed, P is a minimal
prime of M , so grade P = g, and P/Z[X]p is a minimal prime of M — (Z/Zp), thus
grade P/Z[X]p = g, too. This is a contradiction. ”
28 3. Generically Perfect Ideals

(3.3) Theorem. Let M be a ¬nitely generated Z[X]-module which is faithfully ¬‚at
over Z. Then the following properties are equivalent (all tensor products taken over Z):
(a) M is (generically) perfect of grade g.
(b) For every noetherian ring B, M — B is a perfect B[X]-module of grade g.
(c) For every prime number p, M — (Z/Zp) is perfect of grade g.
If M is generically perfect of grade g, then for a Z[X]-free resolution F of M of length
g the complex F — B is a B[X]-free resolution of M — B.
Proof: The implications (b) ’ (c) and (b) ’ (a) are trivial. Let us discuss the
implication (c) ’ (a). We have to show that MQ is perfect of grade g over Z[X]Q for
every maximal ideal Q of Z[X] for which MQ = 0. There would be no chance to utilize
(c) if Q would not contain a prime number p. However such a prime number p ∈ Q exists,
cf. [Bo.2], § 3, no. 4, Th´or`me 3, Corollaire 1 for example. To have a compact notation
let A = Z[X], S = A/Ap, P = Q/Ap. Since MQ = 0, (M/pM )P = (M/pM )Q = 0, so

g = gradeSP (M/pM )P = pdSP (M/pM )P .

This implies pdAQ MQ = g, since p is not a zero-divisor of MQ (by ¬‚atness over Z !). On
the other hand one has

grade M = gradeA M/pM ’ 1 = (gradeS M/pM + 1) ’ 1 = g,

using again that p is not a zero-divisor of M (and A) and M — (Z/Zp) = 0.
For the proof of the implication (a) ’ (b) we need a lemma.
(3.4) Lemma. Let A be a noetherian ring, and M a perfect A-module of grade g.
Then the sets of zero-divisors of M and Extg (M, A) coincide.

Proof: Since Supp Extg (M, A) ‚ Supp M in general, and

M = Extg (Extg (M, A), A)

here, M and Extg (M, A) have the same support, and a prime ideal P is associated to
M if and only if MP = 0 and depth AP = g. Then it is associated to Extg (M, A), too,
and vice versa. ”
Let M be generically perfect of grade g now, A = Z[X], S = B[X], and

F : 0 ’’ Gg ’’ Gg’1 ’’ · · · ’’ G1 ’’ G0

a free resolution of M . Over Z the modules Gj are ¬‚at, so

Hi (F — B) = TorZ (M, B) = 0

for all i ≥ 1. Hence F —B is a free resolution of the S-module M —B, and pd M —B ¤ g.
The crucial point is to show that grade M — B = g. By — we denote the functor
HomA (. . . , A), by ∨ the functor HomS (. . . , S). F — is a free resolution of Extg (M, A), a
A. The Transfer of Perfection

¬‚at Z-module by (3.4). As above we conclude that

F — — B = (F — B)∨

is a free resolution (of the S-module Extg (M — B, S)), and Exti (M — B, S) = 0 for
i = 1, . . . , g ’ 1. Since M — B = 0 is granted, we have

grade M —B ≥ g ≥ pd M —B,

as desired.
The last contention of the theorem has been proved already. ”
We shall apply (3.3) mainly to ¬nitely generated graded Z[X]-modules M , in par-
ticular cyclic ones. Then M is ¬‚at over Z if and only if it is free (and therefore faithfully
¬‚at), for M is a direct sum of ¬nitely generated Z-modules. Our main approach to the
investigation of determinantal rings starts with the construction of an explicit Z-basis
of the determinantal rings (over Z), and therefore (3.3) is ideally suited to reduce the
problem of proving perfection for the determinantal ideals to the case where the ring B
of coe¬cients is a ¬eld.
Theorem (3.3) also explains why the resolutions constructed in Section 2 look the
same regardless of B: A resolution over Z[X] turns into a resolution over B[X] upon
tensoring with B.
Often one encounters determinantal ideals of matrices whose entries cannot be re-
garded as a family of algebraically independent elements generating the ambient ring
over a ring of coe¬cients, for example when the ambient ring is local. In general these
ideals are anything but perfect. On the other hand we have seen that an ideal of max-
imal minors is perfect as soon as its grade is su¬ciently large: the generic resolution
specializes to an acyclic complex then, and gives a free resolution of the desired length.
This fact admits a far-reaching generalization:
(3.5) Theorem. Let A be a noetherian ring, and M a perfect A-module of grade g.
Let S be a noetherian A-algebra such that grade M — S ≥ g and M — S = 0. Then M — S
is perfect of grade g (and grade(Ann M )S = g). Furthermore F — S is a free resolution
of M — S for every free resolution F of M of length g.
Proof: Note that Ann(M — S) and (Ann M )S have the same radical (by Nakaya-
ma™s lemma). Let P be a prime ideal of S such that grade P < g. Then Q = A © P ⊃
Ann M , and
F — SP = (F — AQ ) — SP
is split-exact. Now the claim follows from the acyclicity lemma (16.16). ”
For a typical application of (3.5) we consider an S-sequence x1 , . . . , xn and the powers
AXi , and M = A/J k .
I of the ideal I generated by it. Let A = Z[X1 , . . . , Xn ], J =
Then M is generically perfect (by (16.19), say), and one concludes immediately that I k
is a perfect ideal (of grade n): Proposition (2.14) has been proved again, and perhaps in
a simpler fashion now.
In the situations of (3.3) and (3.5) not only perfection, but also a free resolution
is preserved under the extension considered. As a particular consequence, the canonical
module of the extension is obtained as the extension of the canonical module:
30 3. Generically Perfect Ideals

(3.6) Theorem. Let I be a generically perfect ideal in Z[X] and R = Z[X]/I.
(a) The canonical module ωR is generically perfect, too.
(b) For a Cohen-Macaulay ring B with canonical module ωB one has

ωR—Z B ∼ ωR —Z ωB .

(c) For every Cohen-Macaulay Z[X]-algebra S such that grade IS ≥ grade I one has

ωR—Z[X] S ∼ ωR —Z[X] ωS ,

provided S has a canonical module ωS .
Proof: Let g = grade I. Then ωR = Extg (R, Z[X]), hence part (a) is a by-
product of the proof of (3.3), and parts (b) and (c) are proved essentially in the same
way as Theorem (2.20). ”

B. The Substitution of Indeterminates by a Regular Sequence

In the following we shall have to work with associated graded rings and modules.
Let A be a ring, I ‚ A an ideal. The associated graded ring with respect to I is

I i /I i+1 ,
GrI A =

and for an A-module M the associated graded module is given by

I i M/I i+1 M ;
GrI M =

it carries the structure of a GrI A-module in a natural way. To each element x ∈ M we
associate its leading form x— ∈ GrI M by

x— = x mod I d+1 M x ∈ I d M \ I d+1 M,
if and
x— = 0 I i M.
if x ∈

For a submodule U ‚ M the form module U — ‚ GrI M is generated by the elements
x— , x ∈ U . Then obviously

GrI (M/U ) ∼ (GrI M )/U — .
(1) =

If M = A and J ‚ A is an ideal, the isomorphism (1) implies that

GrI A ∼ (GrI A)/J —
(1 ) =

where we let A = A/J, I = (I + J)/J.
B. The Substitution of Indeterminates by a Regular Sequence

Let A be a noetherian ring and x1 , . . . , xn an A-sequence. Then the associated
graded ring GrI A with respect to the ideal I generated by x1 , . . . , xn is a polynomial ring
over A/I, cf. [Re], the indeterminates being represented by the residue classes x— , . . . , x—
1 n
modulo I 2 of x1 , . . . , xn . Suppose that M is a generically perfect module over Z[X],
X = (X1 , . . . , Xn ). Then, by (3.3) the module

M = M —Z A/I

is a perfect (GrI A)-module. We would like to conclude that

M = M —Z[X] A


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