# $M/PM$ is torsion-free

Let $$R$$ be a PID and $$S$$ be a finite ring extension of $$R$$ which is a free module over $$R$$. Moreover, let $$S$$ be reduced and of dimension one. Let $$M$$ be a finitely generated $$S$$-module which is torsion-free. Thus $$M$$ is free as an $$R$$-module. Let $$P$$ be a minimal prime of $$S$$.

Is there a simple argument that shows that $$M/PM$$ is torsion-free as an $$R$$-module and hence free over $$R$$?

Note that $$P \cap R = 0$$.

• After few failed attempts I started to think that your claim might be wrong. We have to prove that the only associated prime of $M/PM$ (as $S$-module) is $P$. But I have an example where this doesn't hold. Unfortunately in my example $S$ is only a reduced ring of dimension one, and there is no PID $R$ such that $R\subset S$ is finite and free. – user26857 Nov 5 '18 at 20:42
• Anyway, an ambiguous way of asking a question. What is this supposed to mean: "Is there a simple argument..."? Do you have a complicated one? If yes, then why don't give a reference in order to evaluate how complicated is? – user26857 Nov 5 '18 at 21:14
• @user26857 At first glance I thought I was just missing something obvious. For I thought that this was easy. Unfortunately, I don't have any non-simple argument. – windsheaf Nov 6 '18 at 7:27
• Ok. Thanks for reply. But please let me know from where you taken this. Before starting to try again I want to be sure that this is true. Otherwise I'll focus on finding a better counter-example. – user26857 Nov 6 '18 at 8:01
• @user26857 I did not get this from anywhere, I came up with it as I needed it for some kind of argument in a proof of something else. So, it might be wrong. But I wasn't able to find a counter example. – windsheaf Nov 6 '18 at 8:18