for all prime ideals Q in B[X] satisfying P ⊃ Q ⊃ Ir (X). This contradicts (14.9). ”

D. Comments and References

Special cases of Theorems (14.10), (14.12) have already been treated in [Ve.1] (r =

1 = m ’ 1) and [Br.2] (r = 1). As mentioned in 9.E, Theorem (3.5) in [AH] implies the

grade formula of (14.12) for n = m + 1 and I = I1 (X) (cf. (9.27),(a)). Our presentation

of the general case follows [Ve.3] where the main results are covered by Theorem (3.4).

15. Derivations and Rigidity

With the notations of the previous section we continue the investigation of „¦1 . R/B

1

More precisely we shall treat the R-dual of „¦R/B which is just the module of B-derivations

from R to R. The main result will be that („¦1 )— is an almost perfect B[X]-module

R/B

which is perfect if and only if m = n.

For obvious reasons this result makes it possible to describe the syzygetic behaviour

of („¦1 )— as we did for the generic module in Section 13. As a consequence one obtains

R/B

some results concerning the rigidity of R as a B-algebra in case B is a (perfect) ¬eld

(cf. Subsection C).

To have a shorter notation we put „¦ = „¦1 . For δ ∈ ∆(X), δ ≥ [1, . . . , r|1, . . . , r],

R/B

we write

I(x; δ) = I(X; δ)/Ir+1 (X),

and correspondingly

Ir (x) = Ir (X)/Ir+1 (X)

etc. as introduced in Section 5.

A. The Lower Bound for the Depth of the Module of Derivations

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

r a nonnegative integer. In 13.B we have de¬ned a homomorphism

r+1 r+1

G— ’’ F — G— .

F—

• = •f,r :

In case A = R, F = Rm , G = Rn , f = x, the cokernel of • represents the module „¦.

To investigate „¦— we shall de¬ne two more homomorphisms χ, ψ in the general situation

considered above, which are connected with • in a sequence

• χ ψ

F0 ’’ F1 ’’ F2 ’’ F3

(1)

where

r+1 r+1

G— ,

F—

F0 =

F1 = F — G — ,

F2 = F — F — • G — G — ,

r+1 r r r+1

— ——

G — )— — G — .

F3 = A • G — F • F —( F— •( F—

G)

185

A. The Lower Bound for the Depth of the Module of Derivations

Let

χ(y — z — ) = y — f — (z — ) + f (y) — z — ,

ψ(y — y — + z — z — ) = y — (y) ’ z — (z) + f (y) — y — ’ z — f — (z — )

+ y — •— (y — ) + •— (z) — z — ,

f,r f,r

for all y ∈ F , y — ∈ F — , z ∈ G, z — ∈ G— , •f,r being as in 13.B and z viewed an element

of G as well as an element of G—— via the canonical map G ’ G—— . (We adopt this

convention for analogous situations.)

(15.1) Proposition. (a) If rk f ¤ r, then (1) is a complex.

(b) If Im f is a free direct summand of G and rk f = r, then (1) is split exact.

r+1 r+1

—

G— . Then

Proof: (a) Let yI ∈ F , zJ ∈

—

χ —¦ •(yI — zJ )

r

—

f (yU )) yI\U — f — (zJ\V ) + f (yI\U ) — zJ\V

— —

= σ(U, I\U )σ(V, J\V ) zV (

U ∈S(r,I)

V ∈S(r,J)

r

f — (zV )) f — (zJ\V )

— —

σ(U, I\U )yI\U —

= σ(V, J\V ) yU (

U ∈S(r,I) V ∈S(r,J)

r

— —

f (yU )) f (yI\U ) — zJ\V

+ σ(V, J\V ) σ(U, I\U ) zV (

V ∈S(r,J) U ∈S(r,I)

σ(U, I\U ) yI\U — f — —¦ •f — ,r (zJ — yU )

—

=

U ∈S(r,I)

— —

σ(V, J\V ) f —¦ •f,r (yI — zV ) — zJ\V

+

V ∈S(r,J)

=0

(cf. 13.B). To prove ψ —¦ χ = 0 we take y ∈ F , z — ∈ G— . Then

ψ —¦ χ(y — z — ) = ψ(y — f — (z — ) + f (y) — z — )

= f — (z — )(y) ’ z — (f (y)) + f (y) — f — (z — ) ’ f (y) — f — (z — )

+ y — •— —¦ f — (z — ) + •— —¦ f (y) — z —

f,r f,r

= 0.

(b) The assumption on f guarantees that G = Im f • C, C being a submodule of G

isomorphic with Coker f , and that there are elements y1 , . . . , yr ∈ F whose images under

f form a basis of Im f . As one easily checks,

Ker χ = Ker f — C — ,

Ker ψ = (Ker f — Im f — ) • (Im f — C — ) • A(yi — yj + f (yi ) — f (yj )— ),

—

i,j

186 15. Derivations and Rigidity

and the modules on the right hand side of these equations are direct summands of F1

and F2 resp. Now let y ∈ Ker f , z — ∈ C — . Then

•(y1 § · · · § yr § y — f (y1 )— § · · · § f (yr )— § z — ) = y — z — ,

so Im • = Ker χ, and

χ(y — f (yi )— ) = y — yi ,

—

χ(yi — z — ) = f (yi ) — z — ,

χ(yi — f (yj )— ) = yi — yj + f (yi ) — f (yj )— ,

—

1 ¤ i, j ¤ r, so Im χ = Ker ψ. ”

(15.2) Corollary. In case A = R, F = Rm , G = Rn and f = x, the sequence

(1) is exact. Consequently it yields an explicit presentation of „¦ = Coker • as a second

syzygy.

Proof: The assumption of (15.1),(b) is ful¬lled if we localize at a prime ideal

P ⊃ Ir (x). In particular (Coker •)P and (Coker χ)P are RP -free, and (1) is split-exact

in depth 1. From (14.10) we know that Coker • = „¦ is a second syzygy, so Coker χ is

torsionfree. But then Im • = Ker χ, Im χ = Ker ψ. ”

As in the corollary let A = R, F = Rm , G = Rn and f = x. Then Ker •— = „¦— . We

will now prove that pdB[X] Coker χ— ¤ grade R + 2. This will imply that „¦— is almost

perfect. The method of proof is very similar to the one used in Section 14: We shall

consider a ¬ltration of Coker χ— , the quotients of which are direct sums of well-known

ideals generated by poset ideals.

Let y1 , . . . , ym and z1 , . . . , zn be the canonical bases of Rm and Rn . Put N =

Coker χ— and abbreviate wij = yi — zj mod Im χ— , i ∈ S(1, m), j ∈ S(1, n). Let

—

k, l ≥ 0.

Nkl = Rwij

i>k,j>l

Obviously Nrr ‚ N0r ‚ N0r + Nr0 ‚ N . Using notations as at the beginning of 14.A we

can state the following

(15.3) Proposition. (a) Nrr is a free submodule of N .

(b) AnnR N0r /Nrr = I(x; δr+1,r ). As an R(X; δr+1,r )-module N0r /Nrr is isomorphic to

the (n ’ r)-fold direct sum of I(x; δr+2,r , δr+1,r+1 ).

(c) AnnR (N0r + Nr0 )/N0r = I(x; δr,r+1 ). As an R(X; δr,r+1 )-module (N0r + Nr0 )/N0r is

isomorphic to the (m ’ r)-fold direct sum of I(x; δr+1,r+1 , δr,r+2 ).

(d) AnnR N/(N0r + Nr0 ) = I(x; δr+1,r+1 ). As an R(X; δr+1,r+1 )-module N/(N0r + Nr0 )

is isomorphic to I(x; δr+2,r , δr,r+2 ).

Proof: (a) Assume that aij wij = 0. Then in particular

i>r,j>r

— —

aij (yi — zj ) —¦ • (y{1,...,r,k} — z{1,...,r,l} ) = 0, r < k ¤ m, r < l ¤ n,

i>r,j>r

187

A. The Lower Bound for the Depth of the Module of Derivations

whence akl [1, . . . , r|1, . . . , r] = 0, and consequently akl = 0.

(b) Observe that

n

— — —

— yi ) = [i|v] yk — zv , i, k ∈ S(1, m),

χ (yk

v=1

(2) m

χ— (zj — zl ) =

— —

[u|j] yu — zl , j, l ∈ S(1, n).

u=1

r

∈ Nrr for all j ∈ S(1, n), r < l ¤ n, and consequently

Therefore i=1 [i|j] wil

[1, . . . , r|J] wil ∈ Nrr , 1 ¤ i ¤ r < l ¤ n, J ∈ S(r, n).

This proves I(x; δr+1,r ) ‚ AnnR N0r /Nrr . To show the rest of the assertion we proceed

like in the proof of (14.5). Let R = R(X; δr+1,r ) and consider the homomorphisms

gj : (Rr )— ’’ N0r /Nrr , gj (y — ) = wij r < j ¤ n, 1 ¤ i ¤ r

mod Nrr ,

i

(y 1 , . . . , yr being the canonical basis of Rr ), and

r’1

Rr ’’ R, 1 ¤ i ¤ r.

•: •(y {1,...,r}\i ) = [1, . . . , i, . . . , r + 1|1, . . . , r],

By (14.2) Ker • is generated by the elements r (’1)i+1 [i|k] y{1,...,r}\i , k ∈ S(1, n).

i=1

Since r [i|k]y— ∈ Ker gj , r < j ¤ n, • induces a surjective map

i

i=1

n’r

I(x; δr+2,r , δr+1,r+1 ) ’’ N0r /Nrr ,

1

which is injective, too, since for example the residue classes of wr,r+1 , . . . , wrn mod-

ulo Nrr are linearly independent over R(X; δr+1,r ) as is readily checked: Assume that

n n — —

j=r+1 bj wrj ∈ Nrr , bj ∈ R, and apply j=r+1 bj (yr — zj ) —¦ • to y{1,...,r+1} — z{1,...,r,k} ,

r < k ¤ n.

(c) Analogously with (b) we obtain that AnnR Nr0 /Nrr = I(x; δr,r+1 ) and that

Nr0 /Nrr is isomorphic to the (m ’ r)-fold direct sum of I(x; δr+1,r+1 , δr,r+2 ) as an

R(X; δr,r+1 )-module. Since (N0r +Nr0 )/N0r ∼ Nr0 /(N0r ©Nr0 ) and Nrr ‚ N0r ©Nr0 , we

=

get a surjection Nr0 /Nrr ’’ (N0r + Nr0 )/N0r . The residue classes of wr+1,r , . . . , wmr in

(N0r +Nr0 )/N0r being linearly independent over R(X; δr,r+1 ), this map must be injective,

too.

(d) The inclusion I(x; δr+1,r+1 ) ‚ AnnR N/(N0r +Nr0 ) is again an easy consequence

of (2). Next we put R = R(X; δr+1,r+1 ) and consider the homomorphisms

g : (Rr )— — Rr ’’ N/(N0r + Nr0 ),

g(y — — yj ) = wij mod N0r + Nr0 , i, j ∈ S(1, r)

i

188 15. Derivations and Rigidity

(y 1 , . . . , yr being the canonical basis of Rr ), and

r’1 r’1

r

(Rr )— ’’ R,

R—

•:

•(y {1,...,r}\i — y— i, j ∈ S(1, r).

{1,...,r}\j ) = [1, . . . , i, . . . , r + 1|1, . . . , j, . . . , r + 1],

r

[i|k] y{1,...,r}\i — y —