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.
Please help.
Best Answer
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 $\mathfrak p=(x,y)$.