If $A$ is a Noetherian ring and $M$, $N$ are finitely generated modules over $A$, it is easy to see that $\mbox{Ext}_{A}(M,N)$ is finitely generated by taking a finitely generated projective resolution of $M$. But if I take an injective resolution of $N$ instead, it is not at all clear to me why $\mbox{Ext} _{A}(M,N)$ should be finitely generated. It is my understanding that injective hulls are in general not finitely generated. Does this mean that if I take $\mbox{Hom}(M,-)$ of the injective resolution and compute its cohomology I magically get a finitely generated module?

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    $\begingroup$ Yes. It does mean that. $\endgroup$ – Simon Wadsley Jun 7 '10 at 15:13
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    $\begingroup$ You'll get the $\mathrm{Ext}_A^k(M,N)$ which will be finitely generated; whether that is magical or not isn't really a mathematical question. $\endgroup$ – Robin Chapman Jun 7 '10 at 15:14
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    $\begingroup$ You get finitely generated abelian groups from taking the homology of the singular chain complex of a CW-complex with finitely many cells. This chain complex is likely to be even more monstrously huge :-) $\endgroup$ – Robin Chapman Jun 7 '10 at 15:57
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    $\begingroup$ Heck, a slight variant of the same example, if you have a projective variety $X$ over $\mathbb{C}$ and a coherent sheaf $F$ on $X$, then the cohomology $H^i(X, F)$ are finitely dimensional vector spaces! They are also computed using an injective resolution of $F$. $\endgroup$ – Karl Schwede Jun 7 '10 at 19:20
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    $\begingroup$ I think the question has a lot of potential. As Robin and Karl pointed out, there are a lot of examples of complexes of "big" objects with "small" cohomology. It would be nice if someone can give some underlying reasons. $\endgroup$ – Hailong Dao Jun 7 '10 at 19:39

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