<<

. 39
( 47 .)



>>


Conversely this implies, that


buv zu — zv ∈ Im ψ —

aij yi — yj ,
u,v
i,j
i>r


since Ker(1 — x) = Im(1 — •x,r ) ‚ Im ψ — , Ker(x— — 1) = Im(•x,r — 1) ‚ Im ψ — (cf. (13.6)
and the de¬nition of ψ in Subsection A). ”
(15.7) Theorem. Choose notations as at the beginning of the section. Then „¦ — is
a perfect B[X]-module except for m = n in which case it is almost perfect.
Proof: We know already that „¦— is almost perfect in any case (cf. (15.4)). To
prove that it is perfect except for m = n we reduce to the case B = Z as usual: Let
R0 = Z[X]/Ir+1 (X). („¦1 0 /Z )— is faithfully ¬‚at over Z, so „¦— ∼ („¦1 0 /Z )— —Z B is a perfect
=R
R
B[X]-module if („¦1 0 /Z )— is a perfect Z[X]-module. In case („¦1 0 /Z )— is not perfect, we
R R
repeat the argument used in the last paragraph of the proof of (13.8), to obtain that „¦ —
is not perfect, too.
Let B = Z. In case m = n, R is Gorenstein (cf. (8.9)). Put P = Ir (x). Then
depth „¦P = 2 (cf. (14.9)). Thus „¦P is not a third syzygy. Consequently Ext1 P („¦— , RP )
R P
= 0. It follows that depth „¦— < depth RP , in particular that „¦— is not perfect in the
P
case just treated. Assume that m = n, say m < n, and let ω be the canonical module of
R. We must show that Ext1 („¦— , ω) = 0. Consider the exact sequence
R

ψ— χ—
— — —
F3 ’’ F2 ’’ F1

where „¦— ∼ Im χ— (cf. (15.6)). Let D = Im ψ — . Obviously Ext1 („¦— , ω) = 0 is equivalent
= R
to the fact that the induced map

HomR (F2 , ω) ’’ HomR (D, ω)

is surjective. Put s = n ’ m. As we know, ω can be represented by Qs , Q being
the ideal in R generated by the r-minors of arbitrary r columns of x. For technical
reasons we assume Q to be generated by the r-minors [I|2, . . . , r + 1], I ∈ S(r, m). Let
h ∈ HomR (D, Qs ). We have to ¬nd a g ∈ HomR (F2 , Qs ) such that g|D = h.

193
B. The Perfection of the Module of Derivations

Let D be the ψ — -image of
r+1 r r r+1
— —
G— ) — G ,
F —( F— G) • ( F—

the last two components of F3 (cf. Subsection A, F = Rm , G = Rn , of course).


Then F2 /D is a direct sum of m copies of Im x and n copies of Im x— (cf. (13.6)),


so Ext1 (F2 /D , Qs ) = 0 by (13.4). It follows that there exists a g ∈ HomR (F2 , Qs ) such
— —
R
that g|D = h|D . Thus we may assume that h|D = 0.
Put hij = h —¦ ψ — (zj — yi ), i ∈ S(1, m), j ∈ S(1, n) (y1 , . . . , ym and z1 , . . . , zn being


the canonical basis of Rm and Rn ). Since ψ — (1) generates a free direct summand of F2 , —

it su¬ces to ¬nd a g ∈ HomR (F2 , Qs ) satisfying g —¦ ψ — (zj — yi ) = hij for all i ∈ S(1, m),
— —

j ∈ S(1, n), and g|D = 0. This will be done by the following assertion:
(8) Assume that (1, 1) (i, j) (m, n) ( means lexicographically less or equal). Then
there exists a g ∈ HomR (F2 , Qs ) such that



g —¦ ψ — (zv — yu ) = huv ,

(u, v) (i, j),
g|D = 0.

To prove (8) we ¬rst observe the relations

I ∈ S(r+1, m), J ∈ S(r, n), k ∈ S(1, n),
(9) σ(u, I\u)[I\u|J] huk = 0,
u∈I

I ∈ S(r, m), J ∈ S(r+1, n), l ∈ S(1, m),
(10) σ(v, J\v)[I|J\v] hlv = 0,
v∈J

which come from h|D = 0 and from the obvious fact that

m
— — —
— yu ’ [l|k] yl — yI — zJ ,
σ(u, I\u)[I\u|J] zk I, J, k as in (7),
u∈I l=1
n
— —
σ(v, J\v)[I|J\v] zv — yl + [l|k] yI — zJ — zk , I, J, l as in (8),
v∈J k=1

are in Ker ψ — . We may further suppose that huv = 0 for all (u, v) (i, j). Let i > r.
Substituting {1, . . . , r, i} for I, {1, . . . , r} for J and j for k we obtain from (9) that
[1, . . . , r|1, . . . , r]hij = 0, so hij = 0. Using (10) we get analogously that hij = 0 in case
j > r.
One may therefore assume that i, j ¤ r. From (9), with I = {1, . . . , r + 1}, J =
{1, . . . , r}, k = j, it follows that

hij ∈ I(x; [1, . . . , i, . . . , r + 1|1, . . . , r]),

and from (10), putting I = {1, . . . , r}, J = {1, . . . , r + 1}, l = i, we derive

hij ∈ I(x; [1, . . . , r|1, . . . , j, . . . , r + 1]),
194 15. Derivations and Rigidity

so
hij ∈ I(x; [1, . . . , i, . . . , r + 1|1, . . . , r]) © I(x; [1, . . . , r|1, . . . , j, . . . , r + 1]) © Qs .
By Lemma (15.9),(b) below

hij = aI [I|1, . . . , j],
I∈S(j,m)

where
aI = aI;J [J|H],
J∈S(r,m)
{1,...,i}‚J

aI;J ∈ Qs’1 and H = [2, . . . , r + 1]. We put


bkl yk — yl ,
g=
k,l∈S(1,m)

bkl = (’1)i+ j σ(k, I\k)[I\k|1, . . . , j ’ 1] aI;J [l, J\i|H].
I∈S(j,m) J∈S(r,m)
k∈I {1,...,i}‚J

Then
m m
— — —
g—¦ψ — yu ) = g( — yu ) =
(zv [k|v] yk [k|v] bku
k=1 k=1
m
[k|v](’1)j’ 1 aI;J (’1)i’1 [u, J\i|H]
σ(k, I\k)[I\k|1, . . . , j ’ 1]
=
k=1 I∈S(j,m) J∈S(r,m)
k∈I {1,...,i}‚J

0 for u < i,
= j’1
σ(k, I\k)[k|v][I\k|1, . . . , j ’ 1] aI
k∈I (’1) for u = i
I∈S(j,m)


0 for (u, v) (i, j),
=
hij for (u, v) = (i, j).
Trivially g —¦ ψ — (yK — zL — zk ) = 0, K ∈ S(r, m), L ∈ S(r+1, n), k ∈ S(1, n). Furthermore


g —¦ ψ — (yl — yI — zJ )
— —


= σ(u, I\u)[I\u|J] blu
u∈I

σ(u, I\u)[I\u|J](’1)i+ j
= σ(l, K\l)[K\l|1, . . . , j’1] aK;L [u, L\i|H]
u∈I K∈S(j,m) L∈S(r,m)
l∈K {1,...,i}‚L

=± σ(l, K\l)[K\l|1, . . . , j’1] aK;L σ(u, I\u)[I\u|J][u, L\i|H]
u∈I
K∈S(j,m) L∈S(r,m)
l∈K {1,...,i}‚L

=0
for all l ∈ S(1, m), I ∈ S(r+1, m), J ∈ S(r, n), since the sum in parentheses is seen to be
zero when [u, L\i|H] is expanded with respect to the ¬rst row. ”
195
B. The Perfection of the Module of Derivations

(15.8) Remark. Let m = n. The generalization from Z to arbitrary B indicated
in the proof of (15.7) actually yields for all prime ideals P ⊃ Ir (x) that

depth „¦— = depth RP ’ 1.”
P


(15.9) Lemma. The ring B being arbitrary the following holds:
(a) For all I, J ∈ S(r, m) and K, L ∈ S(r, n)


I(x; [I|1, . . . , r]) · [I|K] ‚ R[U |K],
U <I

I(x; [1, . . . , r|L]) · [J|L] ‚ R[J|V ].
V <L

(Of course U < I and V < L mean [U |K] < [I|K] and [J|V ] < [J|L].)
(b) Let H = {2, . . . , r + 1} and Q be the ideal in R generated by the r-minors [I|H], I ∈
S(r, m). Then for all i ∈ S(1, r), j ∈ S(1, r+1), and s ≥ 1:
(b1 ) Qs © I(x; [1, . . . , i, . . . , r + 1|1, . . . , r]) = I∈S(r’i,m) Qs’1 [1, . . . , i, I|H]
(b2 ) Qs © I(x; [1, . . . , i, . . . , r + 1|1, . . . , r]) © I(x; [1, . . . , r|1, . . . , j, . . . , r + 1])
s’1
[1, . . . , i, I|H] · I(x; [1, . . . , r|1, . . . , j, . . . , r + 1]).
= I∈S(r’i,m) Q

Proof: (a) Of course it su¬ces to prove the ¬rst inclusion. Moreover we may
assume that K = {1, . . . , r} since I(x; [I|1, . . . , r]) is invariant under a permutation of the
columns of x. Let I = {a1 , . . . , ar }, ai < ai+1 for i = 1, . . . , r ’ 1, and δ a j-minor of the
¬rst aj ’1 rows of x. If δ and [I|1, . . . , r] are comparable then necessarily δ < [I|1, . . . , r],
and δ · [I|1, . . . , r] ∈ U <I R [U |1, . . . , r] for trivial reasons. If they are not, we observe
that every standard monomial in the standard representation of δ · [I|1, . . . , r] contains
a factor µ ¤ δ, [I|1, . . . , r]. Then µ = [b1 , . . . , br |1, . . . , r], bi ¤ ai for i = 1, . . . , r. Since
µ = δ there is an i such that bi < ai . So δ · [I|1, . . . , r] ∈ U <I R [U |1, . . . , r].
(b) We abbreviate ζj = [1, . . . , r|1, . . . , j, . . . , r + 1]. (b1 ), (b2 ) are immediate
consequences of the following assertion:
(11) Let K ∈ S(r, m) and Ψ be an ideal in the poset {[I|H] : I ∈ S(r, m)}. Then

Qs’1 µ © I(x; [K|1, . . . , j, . . . , r + 1]) ‚ Qs’1 [I|H] + Qs’1 µ · I(x; ζj ).
µ∈Ψ I≥K µ∈Ψ

(Here I ≥ K means [I|H] ≥ [K|H]; we adopt this convention for analogous situations.)
Substituting {1, . . . , i, . . . , r + 1]} for K, r + 1 for j and {[I|H] : I ∈ S(r, m)}
for Ψ, we obtain (b1 ) from (11). Similarly we get (b2 ): Put K = {1, . . . , r}, Ψ =
{[I|H] : {1, . . . , i} ‚ I} and observe (b1 ).
To prove (11) we use induction on |Ψ|. Take

aµ ∈ Qs’1 .
aµ µ ∈ I(x; [K|1, . . . , j, . . . , r + 1]),
a=
µ∈Ψ

If there is a maximal element „ of Ψ such that „ ≥ [K|H], then applying the inductive
hypothesis to a ’ a„ „ yields immediately that a is in the right hand ideal of (11). So we
196 15. Derivations and Rigidity

may assume that „ ≥ [K|H] for all maximal elements „ of Ψ. Let „ = [J|H] be such an
element. Then we get
a„ „ ∈ I(x; [J|1, . . . , j, . . . , r + 1])
because J ≥ K and µ ≥ [J|1, . . . , j, . . . , r + 1] for all µ ∈ Ψ\{„ }. It follows that

a„ ∈ I(x; [J|1, . . . , j, . . . , r + 1]) © Qs’1
(12)

since „ ≥ [J|1, . . . , j, . . . , r + 1]. Observing

I(x; [J|1, . . . , j, . . . , r + 1]) = I(x; [J|1, . . . , r]) + I(x; ζj ),

we obtain

a„ „ ∈ I(x; [J|1, . . . , r])„ + „ · I(x; ζj )

‚ µ · I(x; ζj )
R [I|H] +
I<J µ∈Ψ

<<

. 39
( 47 .)



>>