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 $P \cap R = 0$.