0
$\begingroup$

Let $(R,\mathfrak m)$ be a local, $\mathfrak m$-adically complete, Cohen-Macaulay ring of dimension $1$. Assume that there exists a finitely generated $R$-module $M$ of depth $0$ such that $M$, $\text{Hom}_R(M,M)$ and $\text{Ext}^1_R(M,M)$ all have finite injective dimension.

Then, is it true that $M$ has finite projective dimension, i.e., is it true that $R$ is Gorenstein?

My thoughts: As $R$ is complete local and Cohen-Macaulay, $R$ admits a canonical module, say $\omega$. By MCM approximation, we get an exact sequence $0\to X \to Y \to M \to 0$, where $Y$ is MCM and $X$ has finite injective dimension. Note that as $M$ is not MCM, so $X$ is non-zero. By depth lemma, we get $X$ is also MCM. Hence $X\cong \omega^{\oplus n}$ for some $n>0$. As $Y$ also have finite injective dimension, so $Y\cong \omega^{\oplus g}$ for some $g>0$. So, we have an exact sequence $0\to \omega^{\oplus n} \to \omega^{\oplus g} \to M\to 0$. Applying $\text{Hom}_R(\omega,-)$, and remembering $\text{Ext}^{>0}_R(\omega, \omega)=0$ , we get an exact sequence $$0\to R^{\oplus n}\to R^{\oplus g}\to \text{Hom}_R(\omega, M)\to 0$$ Hence, $\text{Hom}_R(\omega, M)$ has projective dimension $1$, so has depth $0$. Also, applying $\text{Hom}_R(-,M)$ to $0\to \omega^{\oplus n} \to \omega^{\oplus g} \to M\to 0$ , and remembering $\text{Ext}^{>0}_R(\omega, M)=0$ (since $\omega$ is MCM and $M$ has finite injective dimension), we get an exact sequence $$0\to \text{Hom}_R(M,M)\to \text{Hom}_R(\omega, M)^{\oplus g}\to \text{Hom}_R(\omega, M)^{\oplus n}\to \text{Ext}^1_R(M, M)\to 0$$

As $\text{Hom}_R(M,M)$ and $\text{Ext}^1_R(M, M)$ also have finite injective dimension, so we also have exact sequences $0\to \omega^{\oplus a} \to \omega^{\oplus b} \to \text{Hom}_R(M,M) \to 0$ and

$0\to \omega^{\oplus f} \to \omega^{\oplus h} \to \text{Ext}^1_R(M,M) \to 0$ , where $a,b,f,h$ are non-negative integers. And now I am stuck, and do not know how to proceed further. Please help.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy