Let $R$ be a local, regular $\mathbb{C}$algebra and $\mathfrak{m}$ be the maximal ideal. Let $M$ be a finitely generated torsionfree $R$module. Suppose there exists $f \in \mathfrak{m}$ such that $M/fM$ is $R/(f)$torsionfree in the sense that $M/fM$ is torsionfree 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 $R/(g)$torsionfree?
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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 AchingerMay 13 at 19:49

$\begingroup$ @PiotrAchinger Thanks I have added the condition that $M$ is torsionfree as well. $\endgroup$– ChenMay 14 at 4:39
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