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i+1
The kernel of • is generated by the elements i=1 (’1) {1,...,r}\l
r —
j+1
[u|j] y{1,...,r}\v — y {1,...,r}\j , k ∈ S(1, n), u ∈ S(1, m), l, v ∈ S(1, r)
and j=1 (’1)
r r
(cf. (14.2)). Since i=1 [i|k] y— — yl and j=1 [u|j] y— — y j are elements of Ker g, we get
i v
a surjection
I(x; δr+2,r , δr,r+2 ) ’’ N/(N0r + Nr0 ).
Obviously the residue class of wrr generates an R-free submodule in N/(N0r + Nr0 ),
whence the map must be bijective. ”
(15.4) Proposition. Choose notations as at the beginning of the section. Then „¦ —
is an almost perfect B[X]-module.
Proof: We consider the ¬ltration

Nrr ‚ N0r ‚ N0r + Nr0 ‚ N

preceding (15.3). Put g = grade R. Then pd Nrr = g since this module is R-free
and R is perfect. By (5.18) and (14.1) the rings R(X; δst ), B[X]/I(X; δs+1,t, δs,t+1 ),
r ¤ s ¤ m, r ¤ t ¤ n, are perfect, too, of grades g+(s’r)+(t’r) and g+(s’r)+(t’r)+1
resp., so

pd N0r /Nrr = pd(N0r + Nr0 )/N0r = g + 1,
pd N/(N0r + Nr0 ) = g + 2

by (15.3),(b)“(d). Consequently pd N ¤ g + 2 and therefore

pd Im χ— ¤ g + 1

since Im χ— is a ¬rst R-syzygy of N . This shows that Im χ— is almost perfect.
Consider the inclusion map Im χ— ’’ Ker •— = „¦— which is bijective if localized
at a prime ideal not containing Ir (x) (cf. (15.1),(b)). Since Im χ— is almost perfect, we
obtain
depth(Im χ— )P ≥ depth RP ’ 1
for all prime ideals P . Consequently Im χ— is re¬‚exive because (Im χ— )P is free when
P ⊃ Ir (x) (cf. (16.33)). Ker •— being torsionfree, we thus get Im χ— = Ker •— . ”
(15.5) Remarks. (a) It has just been demonstrated that Im χ— = Ker •— in case
f = x. Actually in this case the dual to (1) is exact everywhere as we shall see below.
(b) The system (2) of generators of Im χ— = „¦— is not minimal since the element
m n
— — —
u=1 yu — yu ’ v=1 zv — zv lies in Ker χ . This is nothing but the fact that the Euler-
m n
derivation i,j [i|j]yi — zj can be written as u=1 χ— (yu — yu ) and as v=1 χ— (zv — zv ).
— — —

On the other hand arbitrary m2 + n2 ’ 1 elements of (2) form a minimal system of
generators for „¦— .”
189
B. The Perfection of the Module of Derivations




B. The Perfection of the Module of Derivations

(15.4) leaves open when „¦— is perfect. To answer this question we need the ¬rst
syzygy of „¦— and some technical information about intersections of certain determinantal
ideals.
(15.6) Proposition. In case f = x, the dual to (1) is exact.
Proof: In view of (15.5),(a) it remains only to show that Im ψ — = Ker χ— . Since
the dual to (1) is a zero-sequence which is split-exact at all prime ideals not containing
Ir (x), it will be enough that grade(Ir (x), Coker ψ — ) ≥ 1. For Coker ψ — is torsionfree,
then, and consequently the canonical map Coker ψ — ’’ Im χ— is bijective.
Put C = Coker ψ — and let y1 , . . . , ym and z1 , . . . , zn the canonical bases of Rm and
Rn . We intend to apply (13.1). The ¬ltration of C needed for (13.1), is obtained as
follows: Let
⇐’
(i, j) (k, l) i < k or i = k, j > l,
and put
— —
mod Im ψ — .
Ryi — yj + Rzu — zv
Ckl =
u,v
(i,j) (k,l)

Let (k, l) ∈ S(1, r) — S(1, m). We claim:
(i) The module Ckl /Ck,l’1 is annihilated by I(x; [1, . . . , k, . . . , r, l|1, . . . , r]), m ≥ l > r,
and is free as an R(X; [1, . . . , k, . . . , r, l|1, . . . , r])-module;
(ii) the modules Ckr /Ckk and, if k > 1, Ck,k’1 /Ck+1,m are annihilated by the ideal
I(x; [1, . . . , k, . . . , r +1|1, . . . , r]), and are free as R(X; [1, . . . , k, . . . , r +1|1, . . . , r])-
modules;
(iii) C11 = C2m , and if k > 1, then the module Ckk /Ck,k’1 is annihilated by the
ideal I(x; [1, . . . , k ’ 1, . . . , r +1|1, . . . , r]) and is a free R(X; [1, . . . , k ’ 1, . . . , r +1|
1, . . . , r])-module;
(iv) Cr+1,m ∼ (Im x)m’r (Im x— )n .
=
Since [m’r+1, . . . , m|1, . . . , r] is not a zero-divisor modulo any of the ideals occuring
in (i)“(iii), the claim and (13.1) imply immediately that grade(Ir (x), C) ≥ 1.
To prove the annihilator assertions of (i)“(iii), we observe that Im ψ — contains the
following elements (cf. the de¬nition of ψ given in Subsection A):
m n
— — —
— yu ’ zv — z v ,
(3) ψ (1) = yu
u=1 v=1
m n
— — — —
— yj ) = — yj ’ [j|v] zi — zv ,
(4) ψ (zi [u|i] yu
u=1 v=1
i ∈ S(1, n), j ∈ S(1, m),

ψ — (yi — yI — zJ ) =
— — —
σ(u, I\u)[I\u|J] yi — yu ,
(5)
u∈I
i ∈ S(1, m), I ∈ S(r+1, m), J ∈ S(r, n).
190 15. Derivations and Rigidity

We abbreviate yij = yi — yj mod Im ψ — . From (4) one gets




k
[u|i] yul ∈ Ck+1,m , l ∈ S(1, m), i ∈ S(1, n),
u=1


so

[1, . . . , k|V ] ykl ∈ Ck+1,m , l ∈ S(1, m), V ∈ S(k, n),

(by elementary determinantal calculations), and from (5) we obtain

[I\l|J] ykl ∈ Ck,l’1 , r < l ∈ I ∈ S(r+1, l), J ∈ S(r, n).

This proves the ¬rst half of (i) and (ii). Clearly (3) implies C11 ‚ C2m . Assume that
k > 1. From (3) and (4) it follows that

k
yuu ∈ Ck+1,m ,
u=1
k’1
[u|i] yu,j’1 ∈ Ck,k’1 , i ∈ S(1, n), j ∈ S(1, k ’ 1).
u=1


The inclusions of the last line yield

[1, . . . , k ’ 1|V ] yjj ∈ Ck,k’1 , j ∈ S(1, k’1), V ∈ S(k’1, n),

so [1, . . . , k ’ 1|V ] ykk is an element of Ck,k’1 for all V ∈ S(k’1, n).
Now we turn to the second part of (i)“(iii). Since Im ψ — ‚ Ker χ— , a relation


buv zu — zv ∈ Im ψ —

aij yi — yj +
u,v
i,j


implies
m n
i ∈ S(1, m), v ∈ S(1, n)
(6) aij [j|v] + buv [i|u] = 0,
u=1
j=1


(cf. (2)). To prove the second assertion of (i)“(iii), resp., we therefore shall deduce that
a system of equations
n
1 ¤ i < k, v ∈ S(1, n),
buv [i|u] = 0,
u=1
l n
v ∈ S(1, n),
akj [j|v] + buv [k|u] = 0,
u=1
j=1
191
B. The Perfection of the Module of Derivations

yields

akl ∈ I(x; γ)
(7)

where
±
 [1, . . . , k, . . . , r, l|1, . . . , r] if l > r,

γ = [1, . . . , k, . . . , r + 1|1, . . . , r] or k < l ¤ r,
if l < k


[1, . . . , k ’ 1, . . . , r + 1|1, . . . , r] if 1 < l = k.

In any case, and provided v ∈ S(1, n) has been ¬xed, multiplying the i-th equation of the
system by (’1)i [1, . . . , i, . . . , k|1, . . . , k ’ 1], i = 1, . . . , k, and summing up leads to
l
[1, . . . , k ’ 1|1, . . . , k ’ 1] akj [j|v] ∈ I(x; [1, . . . , k, . . . , r + 1|1, . . . , r]),
j=1

whence
l
akj [j|v] ∈ I(x; [1, . . . , k, . . . , r + 1|1, . . . , r]), v ∈ S(1, n).
j=1

By the usual determinantal calculations we ¬nally obtain that

δakl ∈ I(x; γ),

where ±
γ if l > r,


 [1, . . . , k, . . . , l + 1|1, . . . , l] k < l ¤ r,
if
δ=
 [1, . . . , l|1, . . . , l] if l < k,



[2, . . . , k|1, . . . , k ’ 1] if 1 < l = k.
This implies (7).
As to assertion (iv), we consider the surjection
π
(Rm’r )— — Rm • (Rn )— — Rn ’’ Cr+1,m ,

which maps yi —yj , zu —zv to their images mod Im ψ — , i, j ∈ S(1, m), i > r, u, v ∈ S(1, n).
— —

So
— —
aij yi — yj + buv zu — zv ∈ Ker π
u,v
i,j
i>r

implies
n
i ∈ S(1, r), v ∈ S(1, n),
buv [i|u] = 0,
u=1
m n
r < i ¤ m, v ∈ S(1, n)
aij [j|v] + buv [i|u] = 0,
u=1
j=1
192 15. Derivations and Rigidity

(cf. (6)). Since the ¬rst r rows of x are linearly independent and rk x = r, this system of
equations is equivalent to

n
buv zu ∈ Ker x— ,

v ∈ S(1, n),
u=1
m
aij yj ∈ Ker x, r < i ¤ m.
j=1


Thus

buv zu — zv ∈ Ker(x— — 1).

aij yi — yj ∈ Ker(1 — x),
u,v
i,j
i>r

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