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or less well known) homomorphism which will also play an essential role in the next two
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 )

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