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(2.6) Theorem. Let R = Rt (X), P be a prime ideal of R and Q = P в€© B. Then
is regular if and only if BQ is regular and P вЉѓ Itв€’1 (X)/It (X).
RP
Proof: The statement is obvious if t = 1. Suppose that t > 1.
Abbreviating Is = Is (X), we claim that RP cannot be regular if P вЉѓ I1 /It . Oth-
erwise we may assume that P is a minimal prime of I1 /It and B = BQ . Since B is an
integral domain, I1 /It = P . Therefore Q вЉ‚ I1 /It в€© B = 0, and B is a п¬Ѓeld, say K. Now
2
it suп¬ѓces to note that (K[X]/It )I1 /It = K[X]I1 / It K[X]I1 is not regular since It вЉ‚ I1 .
For the rest of the proof we may therefore assume that the residue class xmn of Xmn
is not contained in P . According to (2.4) we have an isomorphism

R[xв€’1 ] в€ј Rtв€’1 (Y )[Xm1 , . . . , Xmв€’1,n ][Xmn ].
в€’1
mn =

Let P be the contraction to Rtв€’1 (Y ) of the image of P R[xв€’1 ] under this isomorphism.
mn
Then P contains Itв€’1 /It if and only if P contains Itв€’2 (Y )/Itв€’1 (Y ). And RP is regular
if and only if Rtв€’1 (Y )P is regular. The inductive hypothesis now immediately yields the
Лњ
required result. вЂ”
13
B. The Perfection of Im (X) and Some Consequences

B. The Perfection of Im (X) and Some Consequences

From now on we shall restrict our attention to the ideal Im (X) in B[X], generated
by the maximal minors of X. It will be shown that Im (X) is a perfect ideal. Because
of (2.5) this means that pdB[X] Rm (X) = n в€’ m + 1. In Subsection C we shall prove
this equation by constructing a free resolution of Rm (X) over B[X]. On the other hand
there is a simple proof, which does not use a concrete free resolution. We formulate the
following theorem for arbitrary matrices U . Apart from giving a more general result,
this formulation is better adapted to the method of proof being used below.
(2.7) Theorem. Let A be a noetherian ring and U an m Г— n matrix, m в‰¤ n, with
entries in A. Suppose grade Im (U ) = n в€’ m + 1. Then Im (U ) is a perfect ideal.
Using (2.5) and (16.19) we obtain:
(2.8) Corollary. The ideal Im (X) is perfect. In particular Rm (X) is a Cohen-
Macaulay ring if this holds for B.
As we shall see in Section 3, it would be equally justiп¬Ѓed to call (2.7) a corollary of
(2.8). (2.7) will follow from Proposition (2.9). In the proof of (2.9) we will frequently
use arguments from 16.A, and we assume that the reader is familiar with the material of
that subsection.
(2.9) Proposition. Let A be a noetherian ring, F and G free A-modules of ranks
m and n, resp. Further, let f : F в†’ G be a homomorphism such that the ideal I m (f ) has
pв€’1
grade at least p в‰Ґ 1. Then f is injective, and, M denoting the cokernel of f , M is
p
M в‰¤ p.
torsionfree and pd
First we will derive (2.7) from (2.9). Let f : Am в†’ An given by U , and r = n в€’ m.
Denote by u1 , . . . , um the rows of U , and consider the map
r n
n
An ,
A в€’в†’ ОЅ(x) = x в€§ u1 в€§ В· В· В· в€§ um ,
ОЅ:

n
An в€ј A. Put M = Coker f . Then
Obviously Im ОЅ = Im (U ) by an identiп¬Ѓcation =
we have a presentation

rв€’1 r r
m n n
AвЉ— A в€’в†’ A в€’в†’ M в€’в†’ 0,
xвЉ—y в€’в†’f (x) в€§ y,

r r r
so ОЅ factors through M . Since rk M = rk Im ОЅ = 1, and M is torsionfree by (2.9),
r
we conclude pd Im ОЅ = pd M = r.
proof of (2.9): By induction on m. The proposition is trivial for m = 0 (Im (f ) =
A in this case). Let m > 0. Since p в‰Ґ 1, Im f has rank m, and f is injective for trivial
reasons. Furthermore there is nothing to prove if p = 1, and we can proceed by induction
14 2. Ideals of Maximal Minors

pв€’1
M в‰¤ p в€’ 1. Since MP is free
on p. By the inductive hypothesis with respect to p, pd
for all prime ideals P in A with depth AP < p, we get

pв€’1
M вЉ— AP в‰Ґ min(1, depth AP )
depth

pв€’1
for all prime ideals P . Consequently M is torsionfree.
Write F = F вЉ• A, F free of rank m в€’ 1, let f : F в†’ G be the restriction of f ,
and put M = Coker f . Since Imв€’1 (f ) вЉѓ Im (f ), the inductive hypothesis on m can be
applied to M . We claim that there exists an exact A-sequence

pв€’1 p p
0 в€’в†’ M в€’в†’ M в€’в†’ M в€’в†’ 0.

p
This immediately yields pd M в‰¤ p.
Let ПЂ : M в†’ M be the natural projection, and y a generator of Ker ПЂ. Then we
have canonical presentations
p
pв€’1 p p
ПЂ
Пѓ
M в€’в†’ M в€’в†’ M в€’в†’ 0,
pв€’2 pв€’1 pв€’1
M в€’в†’ M в€’в†’ M в€’в†’ 0,
x в€’в†’ x в€§ y.

The second of these presentations shows that the map Пѓ introduced in the п¬Ѓrst one
pв€’1 pв€’1 pв€’1 p
factors through M . Since M is torsionfree and rk M = rk Ker ПЂ, we obtain
p pв€’1
Ker ПЂ в€ј M , as desired. вЂ”
=
As a consequence of (2.8) one can answer questions about the ideals Im (X) which,
from a naive point of view, concern their prime [sic] properties.
(2.10) Theorem. If B is an integral domain, then Im (X) is a prime ideal.
Proof: One may assume B to be noetherian, for the general statement is easily
reduced to this case. Then we use induction on m. If m = 1, the theorem is obvious.
We assume that m > 1. Since Im (X) is perfect of grade n в€’ m + 1 and grade I1 (X) =
mn > n в€’ m + 1, the ideal I1 (X) is not contained in any associated prime ideal of Im (X),
cf. (16.17).
Denote by xij the residue class of Xij in R = Rm (X). Since R[xв€’1 ] is a domain
mn
by (2.4) and the inductive hypothesis, there is exactly one associated prime ideal P of
R such that xmn в€€ P . If P is the single associated prime ideal, then xmn is not a
/
zero-divisor in R, and R is a domain, too. Suppose there is a second associated prime
ideal Q = P . By what we have stated above and since xmn в€€ Q, there is some xij в€€ Q. /
Arguing inductively again, we get xij в€€ P . Now P R[xв€’1 ] = 0, but the image of xij in
mn
P R[xв€’1 ] is diп¬Ђerent from 0, cf. (2.4). Contradiction! вЂ”
mn
15
B. The Perfection of Im (X) and Some Consequences

(2.11) Theorem. If B is reduced (a normal domain), then Rm (X) is reduced (a
normal domain), too.
Proof: Suppose that B is a domain. Then R = Rm (X) is a domain by (2.10).
In order to show that B is reduced or normal resp. we apply criteria based on SerreвЂ™s
conditions.
The statements are obvious if m = 1. Let m > 1 and suppose that B is reduced
(a normal domain). Consider a prime ideal P in R such that depth RP = 0 (в‰¤ 1).
Then grade P = 0 (в‰¤ 1). Because of grade I1 (X) = mn > n в€’ m + 2 there is an
indeterminate Xij which has residue class xij not contained in P . Clearly we may
assume xij = xmn . Then by (2.4) and the inductive hypothesis R[xв€’1 ] is reduced (a
mn
normal domain). Consequently RP is reduced (a normal domain), too. вЂ”
(2.12) Remark. In Section 5 we shall prove that It (X) is a perfect ideal for every
t, 1 в‰¤ t в‰¤ m. Since grade I1 (X) > grade It (X)+1 if t > 1, the arguments in the proofs of
(2.10) and (2.11) demonstrate that (2.10) and (2.11) hold for arbitrary t (the case t = 1
being trivial).
(2.13) Remark. As to converse statements of (2.8) and (2.12), applying (2.4)
one easily deduces that B is reduced (a normal domain, a Cohen-Macaulay ring) if
B[X]/It (X) is reduced (a normal domain, a Cohen-Macaulay ring). вЂ”
So far we have used Corollary (2.8) only, and it seems adequate to discuss an appli-
cation of (2.7) which is independent of (2.8). Let y1 , . . . , yk be elements of a commutative
ring A, J the ideal generated by them, and Y the m Г— (m + k в€’ 1)-matrix
пЈ«y 0пЈ¶
В·В·В· В·В·В· В·В·В·
y2 y3 yk 0
1
.пЈ·
.. .. ..
пЈ¬ .пЈ·
. . .
пЈ¬0 y1 y2 .
пЈ·.
пЈ¬.
.. .. .. .. ..
пЈ­. пЈё
. . . . .
. 0
В·В·В· В·В·В· В·В·В·
0 0 y1 y2 y3 yk

For trivial reasons Im (Y ) вЉ‚ J m . We claim Im (Y ) = J m . It is of course enough to
prove this for the case in which A = Z[y1 , . . . , yk ], the yi being indeterminates. Arguing
inductively we conclude y1 J mв€’1 вЉ‚ Im (Y ) and Ay1 + Im (Y ) = Ay1 + J m . Next it follows
that Ay1 в€© Im (Y ) = y1 J mв€’1 = Ay1 в€© J m , and altogether this yields the desired equality.
Letting n = m + k в€’ 1 we have
n в€’ m + 1 = k,
and (2.7) implies that J m is perfect (of grade k) if grade J = k:
(2.14) Proposition. Let A be a noetherian ring, and y1 , . . . , yk an A-sequence.
Then all the ideals J m , m в‰Ґ 1, are perfect (of grade k).
The matrix Y above helps us to get more information on the rings Rm (X). Given
an m Г— n matrix X of indeterminates, we put k = n в€’ m + 1 and choose Y1 , . . . , Yk
as indeterminates over B. Let S = B[Y1 , . . . , Yk ]/Im (Y ). Then the substitution which
assigns each entry of X the corresponding entry of Y (formed from Y1 , . . . , Yk ), induces
surjections
П€ : B[X] в€’в†’ S П• : Rm (X) в€’в†’ S.
and
16 2. Ideals of Maximal Minors

The kernel of П€ is generated by the linear polynomials

j в€’ i < 0 or j в€’ i > k в€’ 1,
Xij ,
Xij в€’ Xiв€’1,jв€’1 , i = 2, . . . , m, 0 в‰¤ j в€’ i в‰¤ k в€’ 1,

and the ideal Im (X) . The residue classes of the polynomials listed generate the kernel
of П•. Their number is exactly

nm в€’ (n в€’ m + 1) = grade Ker П€ в€’ grade Im (X)
= grade Ker П•

by virtue of (16.18): both Ker П€ and Im (X) are perfect. Here we assume B to be
noetherian, of course. Since the generators of Ker П• are homogeneous (of degree 1), one
concludes easily that they form an Rm (X)-sequence (in any order). This fact makes it
possible to transfer information from Rm (X) to S and vice versa. After all, S can be
considered a well-understood B-algebra.
We use the connection between Rm (X) and S to compute the multiplicity of Rm (X)
in case B = K is a п¬Ѓeld. The graded K-algebra Rm (X) then has a well-deп¬Ѓned multi-
plicity (given by the multiplicity of its localization with respect to the irrelevant maximal
ideal). We refer the reader to [Na] for multiplicity theory.
(2.15) Proposition. Let B = K be a п¬Ѓeld, X an m Г— n matrix of indeterminates
and y the Rm (X)-sequence generating Ker П•, as speciп¬Ѓed above. Then the multiplicity of
Rm (X) is given by

n
e(Rm (X)) = О»(Rm (X)/yRm (X)) = .
mв€’1

Proof: Since the sequence y is a вЂњsuperп¬Ѓcial sequenceвЂќ (deп¬Ѓned to be a sequence
of superп¬Ѓcial elements in the same way as an A-sequence is a sequence of elements
not dividing zero), the multiplicities of Rm (X) and Rm (X)/yRm (X) coincide. The
multiplicity of the latter ring is just its length. вЂ”
One could further exploit the relationship between Rm (X) and S in order to deter-
mine the Gorenstein rings among the rings Rm (X). We shall do this in (2.21), based on
a diп¬Ђerent argument.

C. The Eagon-Northcott Complex

In the preceding subsection we have investigated the ideal Im (X) by considering X
as the matrix of a linear map f : F в†’ G. In this subsection it is better to start from the
dual map f в€— : Gв€— в†’ F в€— . To avoid notational complications we replace Gв€— and F в€— by G
and F and f в€— by a map g : G в†’ F . Instead the map f will be treated as the dual of g,
and the ideal Im (f ) of Subsection B is Im (g) below. While the perfection of Im (g) has
been proved already, cf. (2.8), we will construct a free resolution of the corresponding
residue class ring and some related modules. The approach taken in the following may
be rather abstract, but it is certainly very eп¬Ђective.
17
C. The Eagon-Northcott Complex

Let A be an arbitrary ring, and suppose that F and G are п¬Ѓnitely generated free
A-modules of rank m and n, resp. Since the natural homomorphism

Gв€— вЉ—A F в†’ HomA (G, F )

is an isomorphism in this situation, one may view every A-homomorphism g : G в†’ F an
element of Gв€— вЉ—F . The free module F is the degree 1 homogeneous part of the symmetric
algebra S(F ), so we can consider g even an element of

Gв€— вЉ— S(F ) в€ј HomS(F ) (G вЉ— S(F ), S(F )).
=

Viewed as an S(F )-linear form on

G = G вЉ— S(F ),

g gives rise to a Koszul complex (cf. [Bo.4], В§ 9)
n nв€’1
в€‚ в€‚ в€‚ в€‚
C(g) : 0 в€’в†’ G в€’в†’ G в€’в†’ . . . в€’в†’ G в€’в†’ S(F ) в€’в†’ 0,

i+1 i
G в€’в†’
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