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using the п¬Ѓrst inclusion of (a). So

aВµ в€€ R, a в€€ Вµ В· I(x; О¶j ).
a= aВµ Вµ + a,
Вµв€€ОЁ
Вµв€€ОЁ\{П„ }

Applying the inductive hypothesis we get the result for s = 1.
Let s > 1. From (12) and the inductive hypothesis on s we obtain

Qsв€’2 [I|H] + Qsв€’1 I(x; О¶j )
aП„ в€€
Iв‰ҐJ

(substituting J for K and {[I|H] : I в€€ S(r, m)} for ОЁ). We claim that

Qsв€’1 [I|H] + П„ В· Qsв€’1 I(x; О¶j ).
aП„ П„ в€€
(13)
I<J

To show this we must only look at terms of the form b[L|H]П„, b в€€ Qsв€’2 , in which [L|H]
and П„ = [J|H] are incomparable. According to (9.1) every standard monomial in the
standard representation of [L|1, . . . , r][J|1, . . . , r] is a product [U |1, . . . , r][V |1, . . . , r],
U, V в€€ S(r, m), U в‰¤ V . So [L|H]П„ has a standard representation whose monomials are
of the form [U |H][V |H], U, V as above. Thus we get

Qsв€’1 [I|H].
b[L|H]П„ в€€
I<J

So (13) holds. With that we obtain a representation

aВµ в€€ Qsв€’1 , a в€€ ВµQsв€’1 I(x; О¶j ),
a= aВµ Вµ + a,
Вµв€€ОЁ
Вµв€€ОЁ\{П„ }

and the proof of (11) can be п¬Ѓnished as for s = 1. вЂ”
197
C. Syzygetic Behaviour and Rigidity

C. Syzygetic Behaviour and Rigidity
We now investigate some homological properties of в„¦в€— as we did for the generic
modules in Section 13. Abbreviating
(= m + n в€’ 2r + 1)
s = grade Ir (x)
we state the following
(15.10) Theorem. (a) Let m = n. Then в„¦в€— is an (sв€’1)-th syzygy or, equivalently,
Exti (в„¦, R) = 0, 1 в‰¤ i в‰¤ s в€’ 3.
R
(b) в„¦в€— is an s-th syzygy in case m = n. Equivalently Exti (в„¦, R) = 0, 1 в‰¤ i в‰¤ s в€’ 2.
R

Proof: Since в„¦ = в„¦в€—в€— (cf. (14.10)) and в„¦в€— is free for all prime ideals P вЉ‚ R
P
such that depth RP < s, the assertions given in (a) and (b) are in fact both equivalent
to depth в„¦в€— в‰Ґ min(s в€’ 1, depth RP ) and depth в„¦в€— в‰Ґ min(s, depth RP ), resp., for all
P P
P в€€ Spec R (cf. (16.33)). From (15.7) we get depth в„¦в€— в‰Ґ depth RP в€’ 1 in case m = n
P
and depth в„¦в€— = depth RP in case m = n. This proves the theorem. вЂ”
P
To derive some supplementary results on the syzygetic behaviour of в„¦в€— we need the
map П•1 deп¬Ѓned in (14.8),(a), the cokernel of which coincides with the п¬Ѓrst syzygy M of
в„¦.
(15.11) Supplement to (15.10).
(a) Let m = n. Then в„¦в€— is not an s-th syzygy and consequently Extsв€’2 (в„¦, R) = 0.
R
(b) Let m < nв€’1. Then в„¦в€— is not an (s+1)-th syzygy and consequently Extsв€’1 (в„¦, R) = 0.
R
(c) Let m = n в€’ 1. Then:
(c1 ) Exti (в„¦, R) = 0, 1 в‰¤ i в‰¤ s, and Exts+1 (в„¦, R) = 0 in case r + 1 < m (and
R R
consequently в„¦в€— is at least an (s + 2)-th syzygy).
(c2 ) Exti (в„¦, R) = 0, 1 в‰¤ i в‰¤ s + 1 = 5, and Ext6 (в„¦, R) = 0 in case r + 1 = m (and
R R
therefore в„¦в€— is at least a seventh syzygy).
Proof: (a) Because of (15.8) depth в„¦в€— = s в€’ 1 for all minimal prime ideals P of
P
Ir (x), so в„¦в€— is not an s-th syzygy.
(b) If в„¦в€— were an (s + 1)-th syzygy, then it would be (s + 1)-torsionless (cf. 16.34)).
So M в€— were an s-th syzygy. We claim however that depth MP = depth RP в€’ 1 for all
в€—

prime ideals P вЉѓ Ir (x). To prove this, we п¬Ѓrst reduce to the case in which B = Z. We
may then argue as in the last part of the proof of (13.8).
в€—
Let B = Z. Clearly depth MP в‰Ґ depth RP в€’ 1 for all prime ideals P вЉѓ Ir (x) since
в„¦в€— is perfect. So it is enough to show that depth MIв€— (x) < depth RIr (x) . Localizing as
r
in the proof of (14.9) we can reduce to the case in which r = 1. Furthermore we may
replace Z by Q since Ir (x) в€© Z = 0. Then we need only to prove that M в€— is not perfect
(cf. (16.20)) or equivalently that Ext1 (M в€— , Qnв€’m ) = 0, Q being the ideal in R generated
R
by the entries of the п¬Ѓrst column of x. Consider the map

h : HomR (M в€— , R) в€’в†’ HomR (M в€— , R/Qnв€’m )
induced by the residue class projection R в†’ R/Qnв€’m . It is easy to see, that

[2|1]nв€’mв€’1 y1 вЉ— z1 в€’ [1|1][2|1]nв€’mв€’2y2 вЉ— z1
в€— в€—
198 15. Derivations and Rigidity

represents an element of HomR (M в€— , R/Qnв€’m ) which is not in Im h (y1 , . . . , ym and
z1 , . . . , zn denoting the canonical bases of Rm and Rn , resp.). So h is not surjective
and consequently Ext1 (M в€— , R/Qnв€’m ) = 0.
R
(c) In case r + 1 = m the kernel of the map П• (cf. Subsection A) is obviously
isomorphic with Im x. The assertion of (c2 ) follows therefore from (15.10) and (13.14),(b).
As to (c1 ) we consider the three R-modules Ker П•в€— (= M в€— ), Im П•в€— and Coker П•в€— .
1 1 1
The assertion is an easy consequence of the following claim:
(14) Ker П•в€— and Im П•в€— are perfect B[X]-modules whereas
1 1

depth(Coker П•в€— )P = depth RP в€’ 1
1

for all prime ideals P вЉѓ Ir (x).

We outline the proof of (14); the computational details are very similar to those
used up to now and may be left to the reader: Since Ker П•в€— and Im П•в€— are Z-free in
1 1
case B = Z, the proof of the perfection of Ker П•в€— can be reduced to this case (cf. (3.3)).
1
1 в€—
Then it is enough to show that ExtR (M , Q) = 0 where Q is the ideal in R generated
by the r-minors of the п¬Ѓrst r columns of x. The same way leads to the perfection
of Im П•в€— . (Observe that depth(Coker П•в€— )P в‰Ґ depth MP в€’ 2 for all prime ideals P in
в€—
1 1
R, so Coker П•в€— is R-torsionfree and therefore Z-п¬‚at in case B = Z.) It follows that
1
depth(Coker П•в€— )P в‰Ґ depth RP в€’ 1 for all prime ideals P in R. To get equality when
1
P вЉѓ Ir (x), one reduces to the case in which r = 1 and B = Q as one did in the proof of
(b). An easy computation yields Ext1 (Coker П•в€— , Q) = 0. вЂ”
R 1

(15.12) Remark. The proof shows that the assertions of (15.11) remain true if we
localize at some prime ideal containing Ir (x).вЂ”

Finally we shall derive some results concerning the rigidity of determinantal rings,
the base ring B presumed to be a п¬Ѓeld K from now. Some concepts and results in a more
general situation are needed.
Let S be a п¬Ѓnitely generated K-algebra, S = K[X1 , , . . . , , Xu ]/I, X1 , . . . , Xu being
indeterminates and I an ideal in K[X1 , . . . , Xu ]. вЂњTheвЂќ Auslander-Bridger dual of the
S-module I/I 2 (cf. 16.E) will be called an Auslander-module of S and is denoted by DS ;
up to projective direct summands it does not depend on the special presentation taken
for S.
An S-algebra T will be called a complete intersection over S if T is a factor ring of a
polynomial ring S[Y1 , , . . . , , Yv ] with respect to an ideal generated by an S[Y1 , , . . . , , Yv ]-
sequence.
We will not discuss here what it means that S is rigid. The reader may п¬Ѓnd detailed
information about this concept in the literature (cf. [Ar], [JВЁ], [KL] or [Sl] for instance).
a
The only fact we notice is that in case K is perfect and S is reduced, S is a rigid K-algebra
if and only if Ext1 (в„¦1 , S) = 0.
S S/K

Definition. Let S be as above and k a natural number. Assume S to be rigid. S
is k-rigid if the following condition holds: If T is a complete intersection over S whose
п¬‚at dimension as an S-module is at most k, then Tors (T, DS ) = 0 for all i > 0. If S is
i
k-rigid for all k, then S is very rigid .
199
C. Syzygetic Behaviour and Rigidity

(15.13) Proposition. Let S be as above.
(a) If S is k-rigid and a1 , . . . , aj an S-sequence, j в‰¤ k, then TorS (S/(a1 , . . . , aj )S, DS )
i
= 0 for all i > 0.
(b) If DS satisп¬Ѓes the condition (Sk ) (cf. 16.E), then S is k-rigid.
Proof: (a) Assume a1 , . . . , aj , 1 в‰¤ j в‰¤ k, to be an S-sequence. The S-module
S = S/(a1 , . . . , aj )S is a complete intersection over S which has п¬‚at dimension at most
k, so TorS (S, DS ) = 0 for all i > 0.
i

(b) Let T be a complete intersection over S, S = S[Y1 , . . . , Yv ], T = S/(f1 , . . . , fl )S,
where f1 , . . . , fl is an S-sequence. Assume that the п¬‚at dimension of T over S is at most
k. We claim that

depth(DS вЉ—S S)Q в‰Ґ l
(15) Лњ

for all Q в€€ Spec S containing f1 , . . . , fl . This implies

Лњ
S
TorS (T, DS ) = Tori (T, DS вЉ—S S) = 0
i

for all i > 0: Let l в‰Ґ 1 and
П•l П•2 П•1
F : 0 в€’в†’ Fl в€’в†’ В· В· В· в€’в†’ F1 в€’в†’ F0

be the Koszul-complex over S derived from f1 , , . . . , , fl . F is an S-free resolution of T .
To see that F вЉ—S (DS вЉ—S S) remains exact, we use Theorem (16.15), the numbers ri
Лњ
and the ideals Ji being deп¬Ѓned as there. It is well known that Rad Ji = Rad l Sfj ,j=1
i = 1, . . . , l. So the exactness we want to prove follows immediately from (15).
To verify (15) let Q be the image of Q in T and P the preimage of Q in S with
respect to the canonical maps. From the п¬‚atness of the extension SP в†’ S Q we obtain
Лњ

depth(DS вЉ—S S)Q = depth(DS )P + depth S Q в€’ depth SP .
Лњ Лњ

Since depth(DS )P в‰Ґ min(k, depth SP ) by assumption, we are done in case k в‰Ґ depth SP .
Let k < depth SP and consider elements a1 , . . . , aj в€€ P which form a maximal SP -
sequence. Denote by K = K(a1 , . . . , aj ) the Koszul-complex (over S) derived from
a1 , . . . , aj . Since the п¬‚at dimension of T over S is at most k, we get

Hi (K вЉ—S SP вЉ—SP TQ ) = TorSP (SP /(a1 , . . . , aj )SP , TQ )
i
= TorS (S/(a1 , . . . , aj )S, TQ ) вЉ—S SP
i

= TorS (S/(a1 , . . . , aj )S, T ) вЉ—T TQ
i
=0

for i > k. Thus the ideal (a1 , . . . , aj )TQ has grade at least j в€’ k (cf. Theorem (16.15) for
example). Consequently depth TQ в‰Ґ depth SP в€’ k. Since depth S Q в‰Ґ l + depth TQ , the
Лњ
proof of (15) is complete now. вЂ”
200 15. Derivations and Rigidity

(15.14) Corollary. Let S be Cohen-Macaulay. Then S is k-rigid if and only if D S
satisп¬Ѓes the condition (Sk ).
Proof: It is easy to see that in the case we consider, a п¬Ѓnitely generated S-module
M satisп¬Ѓes the condition (Sk ) if and only if TorS (S/(a1 , . . . , aj )S, M ) = 0 for every
i
S-sequence a1 , . . . , aj , 1 в‰¤ j в‰¤ k, and all i > 0. вЂ”
Now we specialize to the determinantal ring R = Rr+1 (X) with base ring B = K.
As above put s = grade Ir (x). From (15.14) and the syzygetic behaviour of в„¦в€— we derive:
(15.15) Theorem. Assume that K is a perfect п¬Ѓeld. Then R is rigid except for
the case in which r + 1 = m = n. Furthermore:
(a) If r + 1 < m = n then R is (s в€’ 4)-rigid but not (s в€’ 3)-rigid.
(b) If m < n в€’ 1 then R is (s в€’ 3)-rigid but not (s в€’ 2)-rigid.
(c) Let m = n в€’ 1.
(c1 ) If r + 1 < m then R is (s в€’ 1)-rigid but not s-rigid.
(c2 ) If r + 1 = m then R is very rigid.
Proof: First we observe that there is a commutative diagram

r+1 r+1 П•
m
(Rn )в€— Rm вЉ— (Rn )в€—
RвЉ— в€’в†’
Лњ
П•
2
Ir+1 (X)/Ir+1 (X)

where П• = П•x,r and
2
в€—
П•(yI вЉ— zJ ) = [I|J] mod Ir+1 (X) ,
y1 , . . . , ym and z1 , . . . , zn being the canonical bases of Rm and Rn resp. (cf. Section 14). It
2
is a well known fact that Ir+1 (X)/Ir+1 (X) and Im П• have the same rank as R-modules.
So their R-duals coincide, and consequently в„¦в€— is a third syzygy of an Auslander-module
DR of R. Clearly (DR )P is free for all prime ideals P вЉ‚ R such that depth RP < s.
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