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 $Hom_A(End_A(M), A)\cong End_A(M)$?

  • $\begingroup$ What's the intuition behind this you suggest? What makes you suspect this? $\endgroup$ Dec 14, 2019 at 20:11
  • $\begingroup$ If M is free then this is true ..so i was wondering under what condition this is true for other modules? $\endgroup$
    – user149914
    Dec 14, 2019 at 20:15
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    $\begingroup$ You clearly don't mean "=", since the two sides of the equality are sets of functions with different domains and codomains. Could you clarify what you do mean? $\endgroup$ Dec 14, 2019 at 20:16
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    $\begingroup$ Using a two-term free resolution $F_1\to F_0\to M$, we can write $\mathrm{End}_A(M)$ as the (middle) cohomology of a complex $\mathrm{Hom}(F_0,F_1)\to\mathrm{Hom}(F_0,F_0)\oplus\mathrm{Hom}(F_1,F_1)\to\mathrm{Hom}(F_1,F_0)$. This should lead to a proof since we know what the dual of this complex is. $\endgroup$
    – Kapil
    Dec 17, 2019 at 16:31
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    $\begingroup$ THe previous comment by me is missing a term. The first term of the complex should include $\{g\in\mathrm{Hom}(F_1,F_1)| d\circ g=9\}$. So it does not lead to a proof. $\endgroup$
    – Kapil
    Dec 17, 2019 at 22:27

1 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 $M$.


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