Let $R$ be a local, regular $\mathbb{C}$-algebra and $\mathfrak{m}$ be the maximal ideal. Let $M$ be a finitely generated torsion-free $R$-module. Suppose there exists $f \in \mathfrak{m}$ such that $M/fM$ is $R/(f)$-torsion-free in the sense that $M/fM$ is torsion-free as a $R/(f)$-module. Is it true that for all but countably many $g \in \mathfrak{m}$ (or $g \in \mathfrak{m}/\mathfrak{m}^2$) we have $M/gM$ is $R/(g)$-torsion-free?
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1$\begingroup$ What happens when $M=R/(f)$? Then it is torsion free mod $f$, but $M/gM = R/(f,g)$ is unlikely to be torsion free over $R/(g)$ if $(g)\neq (f)$. $\endgroup$– Piotr AchingerCommented May 13, 2022 at 19:49
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$\begingroup$ @PiotrAchinger Thanks I have added the condition that $M$ is torsion-free as well. $\endgroup$– ChenCommented May 14, 2022 at 4:39
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