Let $(R,\mathfrak m,k)$ be a local Cohen-Macaulay ring. If every finitely generated $R$-module that has finite injective dimension also has finite projective dimension, then is it true that $R$ is Gorenstein ?

My approach: Let $d=\dim R(=\operatorname {depth } R)$ . Let $\{x_1,...,x_d\}$ be a system of parameters, then it is also a regular sequence of maximal length. Let $I=(x_1,...,x_d)$. So $N:= \operatorname {Hom}_R (R/I, E_R(k))$ is a finitely generated module (even has finite length) of finite injective dimension (where $E_R(-)$ denotes injective hull as $R$-module) . So by assumption $N$ has finite projective dimension. We also notice that $N$ has depth $0$. So by Auslander-Buchsbaum formula, $pd_R (N)=$depth $(R)-$depth$(N)=$depth $(R)$. I'm not sure if this would be useful or not.

Please help.

  • 1
    $\begingroup$ The canonical module $\omega_R$ has finite injective dimension and maximal depth, so by your hypothesis it has finite projective dimension and thus by Auslander-Buchsbaum, it is free. This is equivalent to Gorenstein. $\endgroup$
    – Mohan
    Commented Nov 14, 2019 at 14:21
  • $\begingroup$ @Mohan: since Cohen-Macaulay rings need not necessarily admit canonical modules, you've to pass to the completion ... one thing I'm not sure about though is that if $M$ is an $\hat R$-module, hence also an $R$-module by the canonical map $R\to \hat R$ , then is it true that $M$ has finite projective (resp. injectice ) dimension as an $\hat R$-moduke iff it has so as an $R$-module ? $\endgroup$
    – uno
    Commented Nov 14, 2019 at 20:59
  • $\begingroup$ A Noetherian local ring is Gorenstein iff its completion is $\endgroup$
    – math54321
    Commented Apr 4, 2023 at 17:58

2 Answers 2


Your argument is very useful. I follow your argument. Since $R/I$ has finite length, we know $$R/I\cong \mathrm{Hom}_R(\mathrm{Hom}_R(R/I,E_R(k)),E_R(k))=\mathrm{Hom}_R(N,E_R(k)).$$ You have showed that $\mathrm{pd}_RN<\infty$, so we know $\mathrm{id}_R(R/I)<\infty$.

Lemma 1. Let $(R,m,k)$ be a Noetherian local ring. If $M$ is a finitely generated $R$-module, then $\mathrm{id}_RM=\sup\{i\mid \mathrm{Ext}^i_R(k,M)\neq 0\}$.

By lemma 1, and by Nakayama we know:

Lemma 2. Let $(R,m,k)$ be a Noetherian local ring, $M$ is a finitely generated $R$-module, and $x\in m$ is a regular element on $M$. Then $\mathrm{id}_R(M)=\mathrm{id}_R(M/xM)$.

So we know $\mathrm{id}_R(R)=\mathrm{id}_R(R/I)<\infty$, since $I$ is generated by a regular sequence.

More generally, Foxby showed:

If $R$ is a Noetherian local ring, if there exists a finitely generated module with finite injective dimension and finite projective dimension, then $R$ is Gorenstein.

  • $\begingroup$ In the hypothesis of your theorem saying $id_R(M)=id_R(M/xM)$ you should have $x\in \mathfrak m$ is both $R$ and $M$-regular ... $\endgroup$
    – uno
    Commented Nov 15, 2019 at 1:55
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    $\begingroup$ @uno yes it is sup. Sorry. I don't think lemma 2 need $x$ is $R$ regular. Where do we use this? $\endgroup$
    – Jian
    Commented Nov 15, 2019 at 2:00

This has been proved that if there exists a finitely generated $R$-module of finite projective dimension and finite injective dimension, then $R$ is Gorenstein (see Corollary 4.4 of "Isomorphisms between complexes with applications to the homological theory of modules" , Foxby). So the answer is positive.


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