Let $R$ be a commutative ring with unity, and let $M$ be an $R$-module. Then naturally $M[[x]]$ is an $R[[x]]$-module. Moreover, if $M$ is generated over $R$ by $b_1,...,b_n$, then $M[[x]]$ also is generated over $R[[x]]$ by $b_1,...,b_n$.
Now let $M$ be a finitely generated $R$-module and $P$ be a prime ideal of $R[[x]]$ such that the submodule $(PM)^*:= \{f(0) \in M : f(x)\in PM[[x]] \} $ of $M$ is finitely generated. Then is it true that $PM[[x]]$ also is a finitely generated submodule of $M[[x]]$?
I know the claim to be true for $M=R$ which is a Theorem in Kaplansky's commutative ring theory book. For the question I asked, I can only show that if $x\in P$, then $PM[[x]]$ is finitely generated by $a_1,...,a_n, xb_1,...,xb_n$ over $R[[x]]$, where $a_1,...,a_n$ generate $(PM)^*$ over $R$ and $b_1,...,b_n$ generate $M$ over $R$.
Please help. Thanks in advance.