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