# Local Cohen-Macaulay ring over which every finitely generated module of finite injective dimension also has finite projective dimension

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.

• 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. – Mohan Nov 14 '19 at 14:21
• @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 ? – uno Nov 14 '19 at 20:59
• @Mohan: am I making a valid point ? – uno Nov 15 '19 at 1:58

## 2 Answers

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.

• 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 ... – uno Nov 15 '19 at 1:55
• @uno yes it is sup. Sorry. I don't think lemma 2 need $x$ is $R$ regular. Where do we use this? – Jian Nov 15 '19 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.