By S. T Hu
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Extra resources for Cohomology theory
6 an independent proof is useful, however. (b) ) (a): Using induction again we may assume that F 0 : 0 ! Fs ! Fs;1 ! F2 ! F1 ! 0 is acyclic. We set Mi = Coker 'i+1 for i = 1 . . s, and show by descending induction that depth(Mi) minfi depth R g for all 2 Spec R and i = 1 . . s. As Ms = Fs , this is trivial for i = s. Let i < s and consider the exact sequence 0 ;! Mi+1 ;! Fi ;! Mi ;! 9. If depth R i, then Iri ('i+1) 6 . +1 . 4. Some linear algebra other hand rank Mi+1 = rank 'i+1 = ri+1, and therefore It('i+1) = 0 for t > ri+1.
S. Furthermore, It ('i) = 0 for all i = 1 . . s and t > ri, if one of these conditions holds. If Ass M, then (a) and (b) are equivalent to (c) F. M is acyclic. Proof. We may suppose that R = R . (a) (b): If F. is split acyclic, then F. R= is a (split) acyclic complex of vector spaces over R= so we can refer to elementary linear ; 6 p p p p p 2 p ) p p algebra. (b) ) (a): We again use induction, and may assume that Coker '2 is a free R -module of rank r1 . 8, Im '1 contains a free direct summand U of F0 of rank r1 .
Proof. We may assume that R = R . Then each of (b) and (c) is equivalent to the split exactness of the sequence 0 ! Im ' ! F0 ! M ! 0. If (a) holds, then, with respect to suitable bases of F1 and F0 , the matrix of ' has the form idt 0 0 0 where idt is the t t identity matrix. This implies (b). The converse is seen similarly. Let M be a nite module over a Noetherian ring R . Then M is a projective module (of rank r) if and only if M is a free R -module (of rank r) for all 2 Spec R . 10. Let R be a Noetherian ring, and M a nite R-module ' with a nite free presentation F1 ;!
Cohomology theory by S. T Hu