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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 —

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