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From (15.7) and (15.8) we therefore obtain that DR satis¬es the condition (Ss’3 ) in
case m < n and (Ss’4 ) in case r + 1 < m = n. On the other hand DR satis¬es (St ) for
t ¤ s if and only if it is a t-th syzygy (cf. (16.33)). So (Ss’2 ) and (Ss’3 ) do not hold for
DR in case m < n ’ 1 and r + 1 < m = n resp. (cf. (15.11)). Since R is Cohen-Macaulay
in any case, (a) and (b) follow immediately from (15.14).
Let m = n ’ 1. Consider the map •1 we have de¬ned in (14.8),(a). If r + 1 =
m, we obtain from (9.18) that Im • ∼ Ir+1 (X)/Ir+1 (X)
(2) ∼
= Im •, so Coker •— is an
= 1
Auslander-module of R. Moreover Coker •— is isomorphic with Coker x— (in the case
1
under consideration) which is a perfect module in view of (13.8). Thus Coker •— satis¬es 1
(Sk ) for all k, and (c2 ) holds because of (15.14). In case r+1 < m the module DR satis¬es
(Ss’1 ) since „¦— is an (s + 2)-th sygyzy (cf. (15.11),(c1 )), so R is (s ’ 1)-rigid. If it were
s-rigid, DR would be an s-th syzygy. Then Im •— , which is isomorphic with Coker •— in
1
any case, would be an (s + 1)-th syzygy and thus even (s + 1)-torsionless (cf. (16.34)).
This implies Exts’1 (Ker •, R) = 0 since (Im •— )— = Ker •. But Exts’1 (Ker •, R) =
1
R R
s+1
ExtR („¦, R) = 0 (cf. (15.11),(c1 )). ”
201
D. Comments and References




D. Comments and References

The content of the ¬rst two subsections is taken from [Ve.4]. The history of the
results given in Subsection C is a little more complicated.
(15.10) is due to Svanes ([Sv.3], 6.8.1). The vanishing of Ext1 („¦, R) and, as a
R
consequence, the rigidity of R in case B is a perfect ¬eld was independently shown by
J¨hner (cf. [J¨], 7.6)). The methods used by Svanes are far from being elementary (and
a a
yield a more general result than (15.10)), in contrast to J¨hner™s proof which is quite
a
adjusted to the special (determinantal) situation and works by methods very similar to
those we developped in the ¬rst two subsections (cf. [J¨], (7.1)“(7.5)). We mention that
a
the rings R have been suspected to be rigid for a long time. The special case in which
r = 1, m = 2 < n, has been treated already in [GK]. The statements of (15.10) are
sharpened by (15.11) which result can also be found in [Ve.4].
In [Bw], (4.5.4) and (5.1.1) Buchweitz has introduced the concepts of an Auslander-
module and k-rigidity, resp. Part of (15.13) is contained in [Bw], (5.1.3), and (15.15) is
a generalization of [Bw], (5.2.1). Theorem (15.15) is also suggested by Buchweitz who
has given a somewhat weaker version based on the Theorem of Svanes mentioned above
(cf. [Bw], (5.3)).
16. Appendix


In the appendix we discuss topics in commutative algebra for which we cannot
adequately refer to a standard text book.

A. Determinants and Modules. Rank

Let A be an arbitrary commutative ring. With every homomorphism f : F ’ G of
¬nitely generated free A-modules F and G we associate its determinantal ideals Ik (f ) in
the following manner: With respect to bases of F and G resp., f is given by a matrix
U , and we simply put Ik (f ) = Ik (U ). This de¬nition makes sense since Ik (U ) obviously
depends on f only and not on the bases chosen. It is equally obvious that Ik (f ) is an
invariant of the submodule Im f of G. It is a little more surprising that Ik (f ) is completely
determined by the isomorphism class of Coker f :
(16.1) Proposition. Let A be a commutative ring, M an A-module with ¬nite free
presentations

˜ ˜
f g
f g
F ’’ G ’’ M ’’ 0 F ’’ G ’’ M ’’ 0
and

Let n = rk G, n = rk G. Then In’k (f ) = In’k (f ) for all k ≥ 0.
˜
Proof: Let e1 , . . . , en and e1 , . . . , e˜ be bases of G and G resp. Then one has
n
equations
n
aij ∈ A, i = 1, . . . , n.
g(ei ) = aij g(ej ),
j=1

Therefore M has a presentation

˜
˜
g
˜h
F • An ’’ G • G ’’ M ’’ 0, g(x, x) = g(x) ’ g(x),

for which h has the following matrix relative to a matrix U = (uij ) of f :
« 
0 ··· ··· 0
. .·
¬ . .·
¬ . .·
U
¬
¬ 0·
0 ··· ···
¬ ·
¬ ·
H=¬ ·
0 ·.
···
1 0
¬
¬ .·
.. .. .·
¬ . .
0 .·
¬ aij . .. ..
¬ ·
. . .
. 0

0 ··· 0 1
203
A. Determinants and Modules. Rank

Evidently In+n’k (h) = In+n’k (H) = In’k (U ), and, as stated above, this ideal is
˜ ˜
determined by Ker g. By symmetry In+˜’k (h) = I˜’k (f ) as well. ”
n n
With the notation of the preceding proposition, the ideal In’k (f ) is also called the
k-th Fitting invariant of M . It is not di¬cult to see that the zeroth Fitting invariant
annihilates M . Let e1 , . . . , en be a basis of the free A-module G as above, and let
n
uij ∈ A,
yi = uij ej ,
j=1

be a system of generators of Ker g. We take minors with respect to U = (u ij ). By
Laplace expansion
n
(’1)i+1 [a1 , . . . , ai , . . . , an |1, . . . , j, . . . , n]yai = [a1 , . . . , an |1, . . . , n]ej .
i=1

So all the n-minors of U annihilate M . In general Ann M = In (f ), but the radicals of
Ann M and In (f ) conincide:
(16.2) Proposition. Let A, f, M, n be as in the preceding proposition. Then
In (f ) ‚ Ann M and Rad In (f ) = Rad Ann M.

The inclusion has been proved already. It remains to show that a prime ideal P
does not contain In (f ) if MP = 0. This may be considered a special case of Proposition
(16.3) below.
The ideals Ik (f ) control the minimal number of local generators of M in the same
way as they control the vector space dimension of M when A is a ¬eld. For a prime ideal
P of A we denote by µ(MP ) the minimal number of generators of the AP -module MP ;
by virtue of Nakayama™s lemma:
µ(MP ) = dimK M — K, K = AP /P AP .
(16.3) Proposition. Let A be a commutative ring and M an A-module with ¬nite
f
free presentation F ’’ G ’’ M ’’ 0. Let n = rk G and P a prime ideal of A. Then
the following statements are equivalent:
(a) Ik (f ) ‚ P ,
(b) (Im f )P contains a (free) direct summand of GP of rank ≥ k,
(c) µ(MP ) ¤ n ’ k.
Proof: We may assume that A is local with maximal ideal P . Let denote residue
classes mod P . The presentation of M induces a presentation
f
F ’’ G ’’ M ’’ 0.
Since Ik (f ) = Ik (f ) and because of Nakayama™s lemma one can replace A, P, M, f by
A, P , M, f without a¬ecting the validity of (a), (b), or (c). Now we are dealing with
¬nite-dimensional vector spaces over the ¬eld A, for which the equivalence of (a), (b),
and (c) is trivial. ”
For later application we note a consequence of (16.3):
204 16. Appendix

(16.4) Proposition. Let A be a commutative ring,

fn f1
F : 0 ’’ Fn ’’ Fn’1 ’’ · · · ’’ F1 ’’ F0

n
be a complex of ¬nitely generated free A-modules, and rk = i=k (’1)i’k rk Fi . Let P
be a prime ideal. Then F — AP is split-exact if and only if Irk (fk ) ‚ P for k = 1, . . . , n.
Proof: We may certainly suppose that A = AP . Let ¬rst F be split-exact. Then,
by a trivial induction, Im fk is a free direct summand of Fk’1 , rk Im fk = rk , whence
Irk (fk ) ‚ P by virtue of (16.3). For the converse one applies induction, too: One may
assume that Coker f2 is free of rank ¤ r1 . On the other hand Im f1 contains a free direct
summand of F0 whose rank is ≥ r1 . Therefore the natural surjektion Coker f2 ’ Im f1
is an isomorphism, and, ¬nally, Coker f1 is free of rank r0 . ”
Usually the rank of a module M over an integral domain with ¬eld of fractions L is
the dimension of the L-vector space M — L. We extend this notion to all rings without
attempting to assign a rank to every module.
Definition. Let A be an arbitrary commutative ring, Q its total ring of fractions.
An A-module M has rank r, abbreviated rk M = r, if M —Q is a free Q-module of rank r.
Over a noetherian ring the rank of a module can be computed in several ways:
(16.5) Proposition. Let A be a noetherian ring, and M a ¬nitely generated A-
f
module with a ¬nite free presentation F ’’ G ’’ M ’’ 0. Let n = rk G. Then the
following are equivalent:
(a) M has rank r.
(b) M has a free submodule N of rank r such that M/N is a torsion module.
(c) For all P ∈ Ass A the AP -module MP is free of rank r.
(d) In’r (f ) contains an element which is not a zero-divisor of A, and Ik (f ) = 0 for all
k > n ’ r.
Proof: (a) ’ (b): Let x1 , . . . , xr be a basis of M — Q. Multiplying with a suitable
element of A which is not a zero-divisor, we obtain elements y1 , . . . , yr (which are images
of elements) in M . Now take N = Ayi .
(b) ’ (c): This is as trivial as the implication (b) ’ (a).
(c) ’ (d): Replacing A by a localization AP , P ∈ Ass A, it is enough to show that
In’r (f ) = A and Ik (f ) = 0 for k > n ’ r if M is free of rank r over the ring A. One
replaces the given presentation by 0 ’’ M ’’ M ’’ 0 and applies (16.1) above.
(d) ’ (a) Replacing A by its total ring of fractions Q, one may assume that
In’r (f ) = A and Ik (f ) = 0 for k > n ’ r. Let P be a prime ideal of A. By virtue
of (16.3) (Im f )P contains a free direct summand of GP . After a suitable choice of basis
for GP it is evident that MP is free of rank r. Therefore M is a projective A-module of
constant local rank r. The ring A under consideration has only ¬nitely many maximal
ideals, whence M is free. (The last conclusion can be proved in the following manner:
Let P1 , . . . , Pu be the maximal ideals of A. There are elements ai ∈ P1 © · · · © Pi © · · · © Pu
such that ai ∈ Pi . Now one chooses elements gij ∈ M , i = 1, . . . , u, j = 1, . . . , r, such
that gi1 , . . . , gir are mapped to a basis of M/Pi M . Obviously ai gi1 , . . . , ai gir then
is a basis of M .) ”
205
A. Determinants and Modules. Rank

The reader may ¬nd out which of the implications between (a), (b), (c), and (d) in
(16.5) remain valid if one drops the hypothesis “noetherian.”
Over a noetherian ring a projective module has a rank if and only if the rank of its
localizations is constant. For ¬nitely generated A-modules M , N with ranks the modules
HomA (M, N ), M — N , k M , Sj (M ) have ranks, too, and these are computed as for free
modules M , N : after all, their construction commutes with localization.
Rank is an additive function along sequences which are exact in depth 0:
(16.6) Proposition. Let A be noetherian ring, M, M1 , M2 ¬nitely generated A-
modules for which there is a sequence of homomorphisms

f
C : 0 ’’ M1 ’’ M ’’ M2 ’’ 0

such that C — AP is exact for all prime ideals P ∈ Ass A. If two of M, M1 , M2 have a
rank, then the third one has a rank, too, and

rk M = rk M1 + rk M2 .

Proof: One may directly assume that A is local of depth 0 and C is exact. Only
the case in which M1 and M are supposed to be free, is nontrivial. Then pd M2 ¤ 1 and,
since depth A = 0, even pd M2 = 0 by the Auslander-Buchsbaum formula, or one proves
this afresh: Let P be the maximal ideal. M2 is a free A-module if and only if

f
C — A/P : 0 ’’ M1 /P M1 ’’ M/P M ’’ M2 /P M2 ’’ 0

is exact. If it is not exact, there is an element x ∈ M1 \ P M1 such that f (x) ∈ P M .
Because of depth A = 0 and the injectivity of f , the element x is annihilated by a nonzero
element of A. On the other hand it belongs to a basis of M1 . Contradiction. ”
(16.7) Corollary. Let A be a noetherian ring, and M an A-module with a ¬nite
free resolution
0 ’’ Fn ’’ · · · ’’ F1 ’’ F0 .
n i
Then rk M = i=0 (’1) rk Fi .
(16.8) Corollary. Let A be a noetherian ring, and I an ideal of A. Then I has a
rank if and only if I = 0 (and rk I = 0) or I contains an element not dividing 0 (and
rk I = 1).
Proof: Consider the exact sequence 0 ’’ I ’’ A ’’ A/I ’’ 0 and apply
(16.6). ”
The following corollary is absolutely trivial now. We list it because it contains an
argument often e¬ective:
(16.9) Corollary. Let A be a noetherian ring, f : M ’ N a surjective homomor-
phism of A-modules M, N such that rk M = rk N and M is torsionfree. Then f is an
isomorphism.
We conclude this subsection with a description of the free locus of a module with
rank.
206 16. Appendix

(16.10) Proposition. Let A be a noetherian ring, M a ¬nitely generated A-module

<<

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