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Let $(R,\mathfrak m)$ be a local Cohen-Macaulay ring of dimension $n$ with a canonical module $\omega$. Let $M$ be a finitely generated $R$-module with $\text{depth } M=\dim M=t$. Using Bruns&Herzog's book, Cohen-Macaulay rings, Corollary 3.5.11 (a consequence of Grothendieck local-duality), I can see that $\text{Ext}^i_R(M,\omega)\ne 0$ if and only if $i=n-t$ and $\dim \text{Ext}^{n-t}_R(M,\omega)\le n-(n-t)=t.$

My question is: Is the module $\text{Ext}^{n-t}_R(M,\omega)$ Cohen-Macaulay? i.e., is it true that $\text{depth } \text{Ext}^{n-t}_R(M,\omega)=\dim \text{Ext}^{n-t}_R(M,\omega)$?

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    $\begingroup$ Yes. The reference I know is in french, sorry: Bourbaki's Commutative algebra X, §9, no. 1, Corollaire of Proposition 3. $\endgroup$
    – abx
    Apr 29 at 10:35
  • $\begingroup$ @abx: I see, thank you. I do have an English version of Bourbaki's Commutative Algebra, but it only goes up to Chapter VII and has 642 pages. Would it be possible for you to tell the page number of the Corollary you are referring to? Also, if it would not be too much trouble, can you please provide the argument in the proof as an answer? I would be happy to accept your answer then ... Thanks $\endgroup$
    – strat
    Apr 29 at 11:01
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    $\begingroup$ This result you are looking for appears earlier in Burns and Herzog's "Cohen-Macaulay Rings" as the implication (a)$\Rightarrow$(c)(i) in Theorem 3.3.10. $\endgroup$ Apr 29 at 13:05
  • $\begingroup$ @Uriya First: you are right, thank you very much. $\endgroup$
    – strat
    Apr 29 at 13:26
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At the request of the OP, I write down Bourbaki's proof in Commutative algebra X, §9, no. 1, Corollaire of Proposition 3.

The proof is by induction on $t=\dim(M)$. If $t=0$, $ \operatorname{Ext}^{n}_{R}(M,\omega ) $ has dimension 0, hence is Cohen-Macaulay. If $t>0$, we choose an element $x$ of $\mathfrak{m}$ such that $\times\, x$ is injective on $M$. Then $\operatorname{depth} M/xM=\dim M/xM=t-1$, hence $\operatorname{Ext}^{i}_{R}(M/xM,\omega )=0 $ for $i\neq t-1$, and we have an exact sequence $$ 0\rightarrow \operatorname{Ext}^{n-t}_{R}(M,\omega ) \xrightarrow{\ \times \,x\ } \operatorname{Ext}^{n-t}_{R}(M,\omega ) \rightarrow \operatorname{Ext}^{n-t+1}_{R}(M/xM,\omega ) \rightarrow 0$$ by the induction hypothesis the right hand term has $\operatorname{depth} = \dim =t-1$; it follows that $\operatorname{Ext}^{n-t}_{R}(M,\omega ) $ has $\operatorname{depth} = \dim =t$, hence is Cohen-Macaulay.

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I think that another answer using derived categories can help explain why this result is actually trivial (and you do not even need to assume that the ring is Cohen-Macaulay).

In general, a finitely generated module $M$ over a noetherian local ring $R$ which has a dualizing complex $D$ is Cohen-Macaulay if and only if $RHom_R(M,D)$ has non-zero cohomology only in a single degree (as said by the OP, this follows from local duality).

Now, what you are asking can be stated as: is (a shift of) $RHom_R(M,D)$ Cohen-Macaulay?

Well, of course it is, because to check this, we need to compute its dual, but its dual is just $RHom_R(RHom_R(M,D),D)) = M$, so obviously it is concentrated in a single degree.

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