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.


1 Answer 1


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 $\mathfrak p=(x,y)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.