# Localization of finitely generated torsion-free module over Noetherian ring (not necessarily an integral domain)

For a finitely generated module $$M$$ over a Commutative Noetherian ring $$R$$, we call $$M$$ is torsion-free if for every non zero-divisor $$r\in R$$ and for every $$0\ne m\in M$$, it holds that $$rm\ne 0$$ . Note that this is equivalent to saying that the natural map $$M\to Q(R)\otimes_R M\cong S^{-1}M$$ is injective, where $$S=$$set of non zero-divisors of $$R=R\setminus \bigcup_{P\in\operatorname{Ass}(R)} P$$ , and $$Q(R)=S^{-1}R$$. Moreover, it is also equivalent to saying $$\cup_{\mathfrak p \in Ass(M)}\mathfrak p\subseteq \cup_{P\in Ass(R)} P$$ .

Now my question is: If $$M$$ is a finitely generated torsion-free $$R$$-module, where $$R$$ is Noetherian, then is it true that $$M_P$$ is also torsion-free $$R_P$$-module for every prime ideal $$P$$ of $$R$$ ?

I can show this when $$R$$ is an integral domain by using the canonical map $$M\to Q(R)\otimes_R M$$ is imjective and then localizing and remembering that if $$R$$ is a domain then $$Q(R)=Q(R)_P=Q(R_P)$$.

In general, for arbitrary Noetherian rings, from $$M$$ torsion-free, I can only show that $$M_P$$ is torsion-free over $$R_P$$ when $$P\in \operatorname{Ass}(R)$$ , because in that case trivially $$R_P$$ is a local ring whose only non zero-divisors are its units.

Apart from these, I have no idea in general.

Let $$R$$ be a noetherian ring and $$\mathfrak p\subset R$$ a prime ideal. Then $$M=R/\mathfrak p$$ is torsion-free iff $$\mathfrak p$$ is contained in an associated prime ideal of $$R$$. (If $$\mathfrak p$$ is maximal this is equivalent to $$\mathfrak p\in\mathrm{Ass}(R)$$.)
Now suppose that $$\mathfrak p$$ is contained in an associated prime, but $$\mathfrak p\notin\mathrm{Ass}(R)$$. Then $$M_{\mathfrak p}$$ is not torsion-free.
For a concrete counterexample consider $$R=K[X,Y,Z]/(X^2,XY,XZ)$$ and $$\mathfrak p=(x,y)$$.