Let $(R,\mathfrak m)$ be a Gorenstein local ring of dimension $1$. Let $M$ be an $R$-module (not finitely generated) such that $M\neq \mathfrak m M$ and there exists a non-zero-divisor $x\in \mathfrak m$ such that $x$ is also a non-zero-divisor on $M$. Then, is it true that $\operatorname{Ext}_R^{>0}(M,R)=0$ ?

I know this would be true if $M$ were finitely generated, and I have gone through the proof as I will explain now, so I can explain the difficulty of the same proof for non-finitely generated case: Put $\overline{(-)}:=(-)\otimes_R R/xR$. As $x$ is $R$- and $M$-regular, hence for all $i>0$, we have isomorphism $\operatorname{Ext}^i_R(M,\overline R)\cong \operatorname{Ext}_{\overline R}^i(\overline{M},\overline{R})=0$, where the last quantity is $0$ because $\overline R$ is an injective $\overline R$-module. We have an exact sequence $0\to R \xrightarrow{x} R \to R/xR \to 0$ , and applying $\operatorname{Hom}_R(M,-)$ to it and remembering $\operatorname{Ext}^i_R(M,\overline R)=0$ we get $\operatorname{Ext}_R^{i}(M,R)=x\mathrm{Ext}_R^{i}(M,R)$ for all $i>0$. Now if $M$ were finitely generated, I would be done at this point using Nakayama's lemma, but unfortunately, $M$ is not finitely generated. What happens now?

Let $(R,\mathfrak m)$ be a Gorenstein local ring of dimension $1$. Let $M$ be an $R$-module (not finitely generated) such that $M\neq \mathfrak m M$ and there exists a non-zero-divisor $x\in \mathfrak m$ such that $x$ is also a non-zero-divisor on $M$. Then, is it true that $\operatorname{Ext}_R^{1}(M,R)=0$ ?

I know this would be true if $M$ were finitely generated, and I have gone through the proof as I will explain now, so I can explain the difficulty of the same proof for non-finitely generated case: Put $\overline{(-)}:=(-)\otimes_R R/xR$. As $x$ is $R$- and $M$-regular, hence for all $i>0$, we have isomorphism $\operatorname{Ext}^i_R(M,\overline R)\cong \operatorname{Ext}_{\overline R}^i(\overline{M},\overline{R})=0$, where the last quantity is $0$ because $\overline R$ is an injective $\overline R$-module. We have an exact sequence $0\to R \xrightarrow{x} R \to R/xR \to 0$ , and applying $\operatorname{Hom}_R(M,-)$ to it and remembering $\operatorname{Ext}^i_R(M,\overline R)=0$ we get $\operatorname{Ext}_R^{i}(M,R)=x\mathrm{Ext}_R^{i}(M,R)$ for all $i>0$. Now if $M$ were finitely generated, I would be done at this point using Nakayama's lemma, but unfortunately, $M$ is not finitely generated. What happens now?

Let $(R,\mathfrak m)$ be a Gorenstein local ring of dimension $1$. Let $M$ be an $R$-module (not finitely generated) such that $M\neq \mathfrak m M$ and there exists a non-zero-divisor $x\in \mathfrak m$ such that $x$ is also a non-zero-divisor on $M$. Then, is it true that $\mathrm{Ext}_R^{>0}(M,R)=0$ ?

I know this would be true if $M$ were finitely generated, and I have gone through the proof as I will explain now, so I can explain the difficulty of the same proof for non-finitely generated case: Put $\overline{(-)}:=(-)\otimes_R R/xR$. As $x$ is $R$ and $M$-regular, hence for all $i>0$, we have isomorphism $\mathrm{Ext}^i_R(M,\overline R)\cong \mathrm{Ext}_{\overline R}^i(\overline{M},\overline{R})=0$, where the last quantity is $0$ because $\overline R$ is an injective $\overline R$-module. We have an exact sequence $0\to R \xrightarrow{x} R \to R/xR \to 0$ , and applying $\mathrm{Hom}_R(M,-)$ to it and remembering $\mathrm{Ext}^i_R(M,\overline R)=0$ we get $\mathrm{Ext}_R^{i}(M,R)=x\mathrm{Ext}_R^{i}(M,R)$ for all $i>0$. Now if $M$ were finitely generated, I would be done at this point using Nakayama's lemma, but unfortunately, $M$ is not finitely generated. What happens now?

Let $(R,\mathfrak m)$ be a Gorenstein local ring of dimension $1$. Let $M$ be an $R$-module (not finitely generated) such that $M\neq \mathfrak m M$ and there exists a non-zero-divisor $x\in \mathfrak m$ such that $x$ is also a non-zero-divisor on $M$. Then, is it true that $\operatorname{Ext}_R^{>0}(M,R)=0$ ?

I know this would be true if $M$ were finitely generated, and I have gone through the proof as I will explain now, so I can explain the difficulty of the same proof for non-finitely generated case: Put $\overline{(-)}:=(-)\otimes_R R/xR$. As $x$ is $R$- and $M$-regular, hence for all $i>0$, we have isomorphism $\operatorname{Ext}^i_R(M,\overline R)\cong \operatorname{Ext}_{\overline R}^i(\overline{M},\overline{R})=0$, where the last quantity is $0$ because $\overline R$ is an injective $\overline R$-module. We have an exact sequence $0\to R \xrightarrow{x} R \to R/xR \to 0$ , and applying $\operatorname{Hom}_R(M,-)$ to it and remembering $\operatorname{Ext}^i_R(M,\overline R)=0$ we get $\operatorname{Ext}_R^{i}(M,R)=x\mathrm{Ext}_R^{i}(M,R)$ for all $i>0$. Now if $M$ were finitely generated, I would be done at this point using Nakayama's lemma, but unfortunately, $M$ is not finitely generated. What happens now?