Dual of $End_A(M)$

Question

Let $A$ be a finitely generated $\mathbb C$-algebra and an integral domain. Assume also $A$ is Gorenstein. Let $M$ be a finitely generated torsion-free $A$ module. Is it true that $Hom_A(End_A(M), A)\cong End_A(M)$?

Answer 1

As Mohan mentioned in the comments, this is false if one does not assume $M$ is reflexive, but $M$ being reflexive is still not good enough. I'll comment on the local case, and remark that this can easily be extended to more generality, including, for instance, the standard graded case.

Proposition: Let $(A,\mathfrak{m},k)$ be a Gorenstein local ring with $\dim A \le 1$. If $\operatorname{Hom}_A(\operatorname{End}_A(\mathfrak{m}),A) \cong \operatorname{End}_A(\mathfrak{m})$, then $\mu_A(\mathfrak{m}) \le 2$.

Proof: Let $(-)^*=\operatorname{Hom}_A(-,A)$ denote the $A$-dual. Then, as $A$-modules, we have $\operatorname{End}_A(\mathfrak{m}) \cong \mathfrak{m}^*$ (see, for example, this answer). Furthermore, since $\dim A \le 1$, $\mathfrak{m}$ is maximal Cohen-Macaulay, and thus reflexive, since $A$ is Gorenstein. In particular, we have $\operatorname{End}_A(\mathfrak{m})^* \cong \mathfrak{m}^{**} \cong \mathfrak{m}$. We claim that if $\mathfrak{m} \cong \mathfrak{m}^*$, then $\mu_A(\mathfrak{m}) \le 2$. To see this, we have the natural exact sequence $0 \to \mathfrak{m} \to A \to k \to 0$. Applying $(-)^*$ to this sequence, we get an exact sequence of the form $$ 0 \to k^* \to A^* \to \mathfrak{m}^* \to \operatorname{Ext}^1_A(k,A) \to 0.$$ Either $\dim A=0$ or $\dim A=1$. If $\dim A=0$, then, as $A$ is Gorenstein, $\operatorname{Ext}^1_A(k,A)=0$, so so we have a surjection $A \to \mathfrak{m}$, implying $\mu_A(\mathfrak{m}) \le 1$. In instead, $\dim A=1$, then $k^*=0$ and $\operatorname{Ext}^1_A(k,A) \cong k$; of course $A^* \cong A$. In particular, $\mu_A(\mathfrak{m}) \le \mu_A(A)+\mu_A(k)=2$. As a remark, note that in the dimension $1$ case we actually have the stronger claim that the Hilbert-Samuel multiplicity $e(A)$ of $A$ is at most $2$; I'll leave this as an exercise.

To point out, some work has been done on understanding for which local rings every maximal Cohen-Macaulay module is self dual, rather than endomorphism modules; see this paper which also contains part of the above Proposition, though the argument is slightly different and a bit more general. This condition is extremely restrictive (for instance it forces $A$ to be a hypersurface and it is conjectured that $e(A) \le 2$ without the $\dim A \le 1$ hypothesis), and I would expect the same of the condition that $\operatorname{End}_A(M)$ be self dual for even every maximal Cohen-Macaulay $M$.