sections.

Let F, G be modules over an arbitrary ring A, f : F ’ G an A-homomorphism and

r, s, t integers such that 0 ¤ r ¤ min(s, t). Then

s t s’r t’r

—

G—

F— G ’’ F—

•f,r :

is de¬ned by

r

— — —

•f,r (yI — f (yU )) yI\U — zJ\V ,

zJ ) = σ(U, I\U )σ(V, J\V ) zV (

U ∈S(r,I)

V ∈S(r,J)

yI = yi1 § · · · § yis , yiσ ∈ F , and zJ = zj1 § · · · § zjt , zj„ ∈ G— . Clearly •f,0 is the identity

— — — —

map.

Here we are interested in the case in which s = r + 1, t = r. Then obviously

f —¦ •f,r = 0 if rk f ¤ r. If moreover Im f is a free direct summand of G and rk f = r,

then

r+1 r

•f,r f

G— ’’ F ’’ G

F—

is split-exact. Adopting the notations of the introduction, we can show:

166 13. Generic Modules

(13.6) Proposition. Suppose r < min(m, n). Then the sequence

r+1 r

•x,r x

m

(Rn )— ’’ Rm ’’ Rn

R—

(2)

and its dual are exact.

Proof: For r = 0 there is nothing to prove. Let r ≥ 1 and • = •x,r . It is clear

that (2) is split exact if localized at a prime ideal of R which does not contain the

ideal Ir (x). In particular (2) is exact in depth 0. So it su¬ces to show that Coker •

is a torsionfree R-module. By what we have just mentioned this will follow from the

inequality grade(Ir (x), Coker •) ≥ 1.

We argue as in the proof of (13.2): Let y1 , . . . , ym be the canonical basis of Rm ,

m’i

0 ¤ i ¤ m ’ r,

Fi = Ryj mod Im •,

j=1

Fm’r+1 = 0, and

γi = [1, . . . , r ’ 1, m ’ i|1, . . . , r], 0 ¤ i ¤ m ’ r.

Then for i = 1, . . . , m ’ r + 1:

(i) (Coker •)/Fi is annihilated by I(X; γi ).

(ii) Fi’1 /Fi is an R(X; γi )-free submodule of (Coker •)/Fi .

From these assertions and (13.1) we obtain

grade(Ir (x), Coker •) ≥ grade(Ir (x), R(X; γ0 ))

= grade Ir (x) ’ grade I(X; [1, . . . , r ’ 1, m|1, . . . , r])

= 1.

To prove (i) we observe that Im • is generated by the elements

I ∈ S(r+1, m), J ∈ S(r, n).

σ(i, I\i)[I\i|J] yi ,

i∈I

As to (ii) we assume that there is an equation

m’i

aym’i+1 = aj yj + y,

j=1

a, aj ∈ R, y ∈ Im •. Then

m’i

axm’i+1 = a j xj

j=1

where xk denotes the k-th row of x. Elementary determinantal calculation yields

aγi’1 ∈ I(X; γi’1 )/Ir+1 (X)

167

B. The Perfection of a Generic Module

and consequently a ∈ I(X; γi’1 )/Ir+1 (X) since γi’1 is not a zero-divisor mod I(X; γi’1 )

(cf. (5.11)). This proves (ii).

To demonstrate the exactness of the dual to (2) we replace x by x— . Then the dual

to (2) becomes the sequence

r+1 r

ψ

x

m n n

(Rm )—

’’ R ’’ R—

(3) R

(up to canonical isomorphisms) where ψ = (•x— ,r )— . Since (3) is exact in depth 0 and

Coker x is torsionfree (cf. (13.2),(a)), the exactness of (3) follows immediately. ”

(13.7) Remark. A routine argument shows that the cokernel of the map ψ in the

sequence (3) is torsionfree: Coker ψ is free in depth less than m+n’2r +1 and Coker x is

re¬‚exive (cf. (13.2),(a)). This fact will be used in the proof of the following theorem. ”

Of course it cannot be expected that Coker x is a perfect B[X]-module in general: If

r ≥ m and m < n then R = B[X] and Coker x fails to be perfect since it has projective

dimension 1 and rank ≥ 1. On the other side (2.16) says that Coker x is perfect in

case r ≥ n and m ≥ n as we have indicated already in the introduction. The following

theorem describes completely how the perfection of Coker x depends on the size of X.

(13.8) Theorem. With the notations of the introduction Coker x is a perfect B[X]-

module if and only if (i) r = 0 or (ii) r ≥ 1 and m ≥ n.

Proof: Obviously Coker x is perfect if r = 0. Taking into account what has just

been said we may further assume that 1 ¤ r < min(m, n).

We ¬rst consider the case in which B = Z. Since C = Coker x is almost perfect,

we have depth CP ≥ dim RP ’ 1 for all P ∈ Spec R. Perfection of C means depth CP =

dim RP for all P ∈ Spec R and hence is equivalent to Ext1 (C, ωR ) = 0 by the local duality

R

theorem, ωR being a canonical module of R (cf. [HK], 4.10 and 5.2). One has C — = Ker x— ,

—

so this module is a third (actually a fourth) syzygy by (13.2),(a). Furthermore C P and CP

are free RP -modules for all P ∈ Spec R such that P ⊃ Ir (x). Since grade Ir (x) = m + n ’

2r + 1 ≥ 3, C — is 3-torsionless (cf. (16.33)) and hence Ext1 (C —— , R) = Ext1 (C, R) = 0.

R R

In case m = n this already shows that C is perfect, R being a Gorenstein ring then

(cf. (8.9)).

In Section 9 we gave a representation of ωR as an ideal of R. From the exact sequence

0 ’ ωR ’ R ’ R/ωR ’ 0 one derives an exact sequence

h

0 ’’ HomR (C, ωR ) ’’ C — ’’ HomR (C, R/ωR ) ’’ Ext1 (C, ωR ) ’’ 0.

R

Thus the perfection of C is equivalent with the fact that h is surjective. Denote by

x1 , . . . , xn the columns of x. Then C — can be identi¬ed with the submodule of all

(a1 , . . . , an ) ∈ (Rn )— such that a1 x1 +· · ·+an xn = 0, and a homomorphism β : C ’ R/ωR

can be lifted to an element (b1 , . . . , bn ) ∈ (Rn )— with the property b1 x1 + · · · + bn xn ∈

ωR (Rm )— .

In case m > n the canonical module ωR is given by P m’n where P is the ideal in

R generated by the r-minors of the ¬rst r rows of x (cf. (9.20)). Lemma (13.9) below

says that there is an element ± ∈ C — which is congruent to (b1 , . . . , bn ) modulo ωR (Rn )— .

This means h(±) = β. Consequently h is surjective and C is perfect in this case.

168 13. Generic Modules

Let m < n and denote by M the prime ideal Ir (x). Clearly C is not perfect if

depth CM < depth RM . According to (2.4) we have an isomorphism

R[x’1 ] ∼ Rr (Y )[Xm1 , . . . , Xm’1,n ][Xmn ].

’1

(4) mn =

Put R = Rr (Y ), M = Ir’1 (Y ) and let C be the corresponding generic module. Denote

by y the matrix of residue classes modulo Ir (Y ) of the indeterminates Yij . Then the map

x — R[x’1 ] is represented by the matrix

mn

«

0

.·

¬ .

y . ·,

¬

(5) 0

0 ··· 01

so C — R[x’1 ] ∼ C — R[x’1 ]. Since RM ’ RM is a local and ¬‚at extension, the depth

mn = ˜

mn

inequality above is equivalent to depth C M < depth RM . We may therefore assume

˜ ˜

that r = 1. Furthermore we can replace the base ring Z by the ¬eld Q of rational

numbers because M © Z = {0}. But then it su¬ces to show that C is not a perfect

Q[X]-module (cf. (16.20)). In the case under consideration the canonical module ωR of

R is given by Qs , s = n ’ m, where Q is the ideal in R generated by the entries of the

¬rst column of x. Take b1 = xs’1 , b2 = · · · = bn = 0. Then (b1 , . . . , bn ) induces an

11

element of HomR (C, R/Q ). If there were a1 , . . . , an ∈ R such that a1 x1 + · · · + an xn = 0

s

and bj ’ aj ∈ Qs , one in particular would get xs’1 ∈ Qs + J, J being the ideal in R

11

2 n

generated by the components of x , . . . , x . This is obviously impossible. Therefore the

homomorphism h above is not surjective.

Now we treat the general case for B. Let m ≥ n. Since the cokernel of the map

(•x— ,r )— in (13.6) is Z-¬‚at (cf. (13.7)), the perfection of Coker x follows from the fact that

it is B-free and perfect in case B = Z (cf. (3.3)). It remains to prove that CB = C —Z B

is not perfect if 1 ¤ r < m < n. From (13.4) and the considerations above we obtain

that pdZ[X] C = grade Ir+1 (X) + 1. Let

F : 0 ’’ Ft ’’ · · · ’’ F0

be a Z[X]-free resolution of C of minimal length. Since C and the modules Fj are Z-¬‚at,

F —Z B is a B[X]-free resolution of CB . Let J be a prime ideal in B[X] containing Ir (X),

and I the preimage of J in Z[X]. Then (F —Z B) —B[X] B[X]J is a B[X]J -free resolution

of (CB )J . To see that it has minimal length, we consider the canonical isomorphism

(F —Z B) —B[X] B[X]J ∼ (F —Z[X] Z[X]I ) —Z[X]I B[X]J .

=

F —Z[X] Z[X]I has minimal length in view of the inequality depth CM < depth RM

above. Furthermore the extension Z[X]I ’’ B[X]J is local. Consequently pd(CB )J =

t = grade(CB )J + 1. ”

169

B. The Perfection of a Generic Module

(13.9) Lemma. Denote by P the ideal in R generated by the r-minors of the ¬rst

r rows of x. Let further s be a positive integer and b1 , . . . , bk ∈ R such that b1 x1 + · · · +

bk xk ∈ P s Rm , xj being the j-th column of x. Then there are elements a1 , . . . , ak ∈ R

such that a1 x1 + · · · + ak xk = 0 and bj ’ aj ∈ P s for j = 1, . . . , k.

Proof: Of course we may assume that r ≥ 1. Let Q be the ideal in R generated

by the r-minors of the ¬rst k ’ 1 columns of x. Taking linear combinations of b1 xi1 +

· · · + bk xik , 1 ¤ i ¤ m, with suitable minors of x as coe¬cients we get

bk δ ∈ P s + Q

where

[1, . . . , k ’ 1, r + 1|1, . . . , k] if k ¤ r,

δ=

[1, . . . , r ’ 1, k|1, . . . , r ’ 1, k] if k > r.

In case B is a ¬eld, δ is not a zero-divisor modulo P s +Q since P s +Q is (P +Q)-primary

and δ ∈ P + Q (cf. (9.18)). Furthermore R/(P s + Q) is Z-free in case B = Z. From

/

(3.14) we then obtain that δ is not a zero-divisor modulo P s + Q for arbitrary B.

Consequently bk ∈ P s + Q. Obviously this implies the assertion in case k ¤ r.

Assume that k > r. Let J ∈ S(r, k’1) and put J = J ∪ {k}. Then

σ(j, J \j)[I|J\j] xj = 0 for all I ∈ S(r, m).

j∈˜

J

A suitable linear combination of these determinantal relations of x1 , . . . , xk yields a re-

lation (a1 , . . . , ak ) such that bk ’ ak ∈ P s . Induction on k now completes the proof of

(13.9). ”

Since Coker x fails to be perfect in case 1 ¤ r ¤ m < n the cokernel of the map

ψ = (•x— ,r )— certainly cannot be perfect in this case. It requires a little more e¬ort to

see that Coker ψ is not perfect except for m = n.

(13.10) Proposition. The cokernel of the map ψ = (•x— ,r )— is perfect if and only

if m = n.

Proof: There is to prove something only in case m ≥ n. Assume ¬rst that B =

Z. If m = n then D = Coker ψ is almost perfect, so perfection of D is equivalent to

Ext1 (D, R) = 0 since R is a Gorenstein ring in that case. The vanishing of Ext1 (D, R)

R R

in any case follows from (13.6).

Suppose now that m > n. Imitating part of the proof of (13.8) we write M =

Ir (x) and prove that depth DM < depth RM . Again we use the isomorphism (4), put

R = Rr (Y ), M = Ir’1 (Y ) and D the cokernel of the map (•x— ,r’1 )— , y denoting the

matrix of residue classes modulo Ir (Y ) of the indeterminates Yij . Since x — R[x’1 ]

mn

’1 ∼ ’1

can be represented by the matrix (5), we obtain that D — R[xmn ] = D — R[xmn ] • F ,

F being a free R[x’1 ]-module. Consequently depth DM < depth RM is equivalent to

mn

depth DM < depth RM , so we may assume that r = 1. As in the proof of (13.8) we can

˜ ˜

replace the base ring Z by Q and have only to show that D is not a perfect Q[X]-module.

Since Ext1 (D, R) = 0 this is equivalent to the fact that the natural homomorphism

R

h

HomR (D, R) ’’ HomR (D, R/P m’n )