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Let $R$ be a commutative Noetherian ring and let $M$ be a finitely generated $R$-module.

Definition. $M$ is called totally reflexive (or $G-\dim_RM = 0$) if it satisfies the following conditions:

(i) $\mathrm{Ext}_R^i(M, R) = 0$ for all $i >0$

(ii) $\mathrm{Ext}_R^i(M^*, R) = 0$ for all $i >0$, where $M^* = \mathrm{Hom}_R(M, R)$.

(iii) The canonical map $\delta_M : M \to M^{**}$ is an isomorphism, where $M^{**} = \mathrm{Hom}_R(M^*, R)$.

Totally reflexive modules have many properties similar to projective modules.

Note. (a) If $(R, \mathfrak{m})$ is a local ring and $M$ is projective, then $M$ is free.

(b) If $(R, \mathfrak{m})$ is a Gorenstein local ring and $M$ is totally reflexive, then $M$ is maximal Cohen-Macaulay.

Question. Suppose $(R, \mathfrak{m})$ is a local ring and $M$ is totally reflexive. Is it true that $\dim M = \dim R$?

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    $\begingroup$ A quick comment-if $R$ is a domain, any non-zero module $M$ for which $\delta_M$ is an isomorphism has annihilator zero and thus of maximal dimension. $\endgroup$
    – Mohan
    Sep 6, 2014 at 19:25
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    $\begingroup$ For that matter, if $R$ is a domain and $M^{**}\neq 0$, then $\dim M=\dim R$. $\endgroup$ Sep 6, 2014 at 20:09
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    $\begingroup$ In fact the formula $\text{Ass}_R(\text{Hom}_R(K,N))=\text{Ass}_R(N)⋂\text{Supp}_R(K)$ implies that the statement holds provided $R$ is unmixed. Thus at least the statement holds for a large class of rings including Cohen-Macaulay rings, quasi-Gorenstein rings and "$S_2$ rings with a canonical module". $\endgroup$
    – Aurora
    Sep 7, 2014 at 8:22

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