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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 $P \cap R = 0$.

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  • $\begingroup$ 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. $\endgroup$ – user26857 Nov 5 '18 at 20:42
  • $\begingroup$ 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? $\endgroup$ – user26857 Nov 5 '18 at 21:14
  • $\begingroup$ @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. $\endgroup$ – windsheaf Nov 6 '18 at 7:27
  • $\begingroup$ 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. $\endgroup$ – user26857 Nov 6 '18 at 8:01
  • $\begingroup$ @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. $\endgroup$ – windsheaf Nov 6 '18 at 8:18

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