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In the proof of Theorem 20.18 in Eisenbud Commutative Algebra, the following fact is stated:

If $S=k[X_1,\ldots,X_r]$ and $N$ is a finite length graded $S$-module, then

$$\operatorname{Ext}_S^r(N,S(-r)) \cong \operatorname{Hom}_k(N,k).$$

It says this follows from Exercise 2.4, but I don't see why...

(I think one needs the fact that if $N$ is of finite length, then $\operatorname{Ext}^j(N,S)=0, \,\, \forall j.)

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    What does Exercise 2.4 say ... ?2012-09-23
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    2.4 just asks to describe explicitly $Hom(Z/n,Z/m)$ and $Hom(k[x]/x^n),k[x]/x^m)$, and also the analogous statement for tensor products2012-09-23
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    Actually, I would also be interested in a proof of the last result I stated: I know in general $\min\{j | Ext^j(M,N)\ne 0\}=depth(Ann(M),N)$ and this easily implies it, but I was hoping for a more direct proof.2012-09-23
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    I think this question is better suited for www.mathoverflow.net. I would be pleased to answer your question there.2012-09-25

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