Let $M$ be a finitely generated module over a Noetherian local ring $(R,\mathfrak m)$. Let $x\in \mathfrak m$ be such that $\text{Tor}^R_1(M,R/xR)=0.$ Then, is it true that $x$ is $M$-regular?
I can prove this under the extra hypothesis that $x$ is $R$-regular. Indeed, in that case we have the exact sequence $0\to R \xrightarrow{\cdot x} R \to R/xR\to 0$. Tensoring with $M$ we get
$$\text{Tor}^R_1(M,R/xR)\to M \xrightarrow{\cdot x} M \to M/xM\to 0$$ Now remembering $\text{Tor}^R_1(M,R/xR)=0$ we get the exact sequence $0\to M \xrightarrow{\cdot x} M \to M/xM\to 0$ which gives $x$ is $M$-regular.
However, I'm not sure what happens if we do not assume $x$ is $R$-regular. Please help.
Best Answer
This is not true if $x$ is not assumed to be $R$-regular. Just take for $x$ any zero divisor and $M=R$ as in this case we always have $\operatorname{Tor}_1^R(M,R/xR)=0$. Alternatively take $x=0$ and $M$ any module.