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?
Best Answer
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 $\varprojlim{}\!^{1}p^{i}R\cong\hat{\mathbb{Z}}_{p}/R$.