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 $\text{depth } \text{Ext}^{n-t}\_R(M,\omega)=\dim \text{Ext}^{n-t}\_R(M,\omega)$?



For a Cohen-Macaulay module $M$ of dimension $t$ over a local CM ring of dimension $n$, is $\text{Ext}^{n-t}_R(M,\omega)$ Cohen-Macaulay?
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.


