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 $\operatorname{Ext}^j(N,S)=0, \,\, \forall j<r$.)

  • 4
    $\begingroup$ What does Exercise 2.4 say ... ? $\endgroup$
    – Matt
    Sep 23, 2012 at 19:59
  • $\begingroup$ 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 products $\endgroup$
    – john
    Sep 23, 2012 at 20:01
  • $\begingroup$ 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. $\endgroup$
    – john
    Sep 23, 2012 at 20:32
  • $\begingroup$ I think this question is better suited for www.mathoverflow.net. I would be pleased to answer your question there. $\endgroup$
    – Ralph
    Sep 25, 2012 at 8:19
  • $\begingroup$ I think this was a typo. I believe he meant exercise 21.4 instead, that is, for a regular local ring $R$ of dimension $d$, the functor $\operatorname{Ext}_R^d(-, R)$ is a dualizing functor on the category of $R$-modules of finite length. Here we should apply the graded case. $\endgroup$
    – user782932
    May 19, 2022 at 19:16

2 Answers 2


The isomorphism comes from the local duality theorem for graded modules. The reference from Eisenbud is obviously wrong, but Theorem A4.2 (from the same book) may help.

In fact, as one can see in the book of Brodmann and Sharp, Local Cohomology, Example 13.4.6, $$^*\operatorname{Hom}_k(H_{\mathfrak{m}}^i(N),k)\cong\ ^*\operatorname{Ext}_S^{r-i}(N,S(-r)),$$ where $\mathfrak{m}=(X_1,\dots,X_r)$ and $N$ a finitely generated graded $S$-module. Now take $i=0$ and get $$^*\operatorname{Hom}_k(H_{\mathfrak{m}}^0(N),k)\cong\ ^*\operatorname{Ext}_S^{r}(N,S(-r)).$$ But when $N$ has finite length, $H_{\mathfrak{m}}^0(N)=N$ and therefore $$^*\operatorname{Hom}_k(N,k)\cong\ ^*\operatorname{Ext}_S^{r}(N,S(-r)).$$


$\def\Ext{\operatorname{Ext}}$To prove your last statement, notice you can prove it by induction on the length of $N$, starting from the computation of $\Ext_S(k,S)$ which you can do trivially by using the Koszul resolution of $k$. (If you know about the Künneth formula for $\Ext$, this can be reduced to the computation of $\Ext_{k[X]}(k,k[X])$, with one variable, as $\Ext_{k[X_1,\dots,X_n]}(k,k[X_1,\dots,X_n])\cong\Ext_{k[X]}(k,k[X])^{\otimes n}$, but really, there is no need to do this, as the computation required is almost immediate)

If $N$ is of length $n>1$, and $0\subsetneq N'\subsetneq N$ is a non-zero proper submodule, we have a short exact sequence $0\to N'\to N\to N/N'\to 0$, and by induction we know that $\Ext^j(N',S)=\Ext^j(N/N',S)=0$ for all $j<r$, so the long exact sequence for Ext gives what the desired conclusion for $\Ext^j(N,S)$.

This is probably the official proof of the fact, I dunno.

To prove your first formula, notice that what we have just shown together with the fact that $\operatorname{gldim}S=r$ implies that $F=\Ext^r(\mathord-,S)$ is an exact functor on the category $\mathcal{A}$ of finite length graded modules, just as $G=\hom_k(\mathord-,k)=\hom_S(\mathord-,k)=\Ext^0_S(\mathord-,k)$ is.

There is a natural map $\Ext^0(M,k)\to \Ext^r(M,S)$ given by Yoneda product with the class of the $r$-extension given by the Koszul complex, which is an element of $\Ext^r(k,S)$. This map is an isomorphism when $M$ is simple, so general nonsense —the Five Lemma, in fact— implies that it is an isomorphism on all of $\mathcal A$.

  • $\begingroup$ I dropped the grading throughout, but you can put it back correctly :-) $\endgroup$ Feb 12, 2013 at 7:37

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