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