# Vanishing of $\operatorname{Ext}_R^{1}(M,R)$ when $R$ is a Gorenstein local ring of dimension $1$ and $M$ is not finitely generated

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?

Take $$R=\mathbb{Z}_{(p)}$$ for some prime $$p$$, with $$x=p$$, and $$M=\mathbb{Q}\oplus R$$.
To show that this is a counterexample, the only nonobvious thing to show is that $$\operatorname{Ext}^{1}_{R}(\mathbb{Q},R)\neq0$$.
In fact, $$\operatorname{Ext}^{1}_{R}(\mathbb{Q},R)\cong\hat{\mathbb{Z}}_{p}/R$$. This is probably well known, but I couldn't find an explicit reference, so here is a proof.
Since $$\mathbb{Q}$$ is the union of the chain $$R\subseteq p^{-1}R\subseteq p^{-2}R\subseteq\cdots$$ of $$R$$-submodules, it follows from 3.5.10 in Weibel's An Introduction to Homological Algebra that there is a short exact sequence $$0\to\varprojlim{}\!^{1}\operatorname{Hom}_{R}(p^{-i}R,R)\to \operatorname{Ext}^{1}_{R}(R,\mathbb{Q})\to \varprojlim\operatorname{Ext}^{1}_{R}(p^{-i}R,\mathbb{Q})\to0.$$
Since $$p^{-i}R\cong R$$, so that $$\operatorname{Hom}_{R}(p^{-i}R,R)\cong p^{i}R$$ and $$\operatorname{Ext}^{1}_{R}(p^{-i}R,\mathbb{Q})=0$$, this gives $$\operatorname{Ext}^{1}_{R}(R,\mathbb{Q})\cong\varprojlim{}\!^{1}p^{i}R.$$
Since there is a short exact sequence of inverse systems $$0\to\{p^{i}R\}\to\{R\}\to\{R/p^{i}R\}\to0$$ and $$R/p^{i}R\cong\mathbb{Z}/p^{i}\mathbb{Z}$$, there is an exact sequence $$R=\varprojlim R\to \varprojlim\mathbb{Z}/p^{i}\mathbb{Z}=\hat{\mathbb{Z}}_{p} \to\varprojlim{}\!^{1}p^{i}R\to\varprojlim{}\!^{1}R=0,$$ and so $$\varprojlim{}\!^{1}p^{i}R\cong\hat{\mathbb{Z}}_{p}/R$$.