If $P/I_kP$ are finitely-generated, is it true that $P/IP$ is finitely-generated where $I=\bigcap I_k$? I'm looking into some old results on "big projectives'', and trying to understand some steps.
Assume that $R$ is a (commutative) ring and $I_1,\ldots,I_n$ are ideals. Let $I=I_1\cap\cdots\cap I_n$ be their intersection. Finally suppose that $P$ is a module such that $P/I_kP = P\otimes R/I_k$ is finitely-generated for all $k$ (as an $R$-module or $R/I_k$-module, it doesn't matter).
Is it true that $P/IP$ is finitely-generated?

Here's a case where I can work it out: if the $I_k$'s were pairwise comaximal, then $R/I = \bigoplus R/I_k$ by the Chinese Remainder Theorem, so $P/IP = \bigoplus_{k=1}^n P/I_kP$ is certainly finitely-generated. In general I'm not sure how to write down a surjection from $\bigoplus_{k=1}^n P/I_kP$ to $P/IP$.

Here is some extra context I have in my case:


*

*$P$ is projective

*$R$ is noetherian

*$I_1,\ldots,I_n$ are the minimal primes over zero (by Noether's Theorem)


I suspect none of these additional assumptions are necessary, and that what I asked is true in general. Unless minimal primes are always pairwise comaximal, but I'm unaware of that ...

Here is a solution when $R$ is noetherian. Since all the $P/I_kP$'s are finitely-generated, so is their sum $\bigoplus P/I_kP$. This is finitely-generated over a noetherian ring, hence noetherian, so all of its submodules are finitely-generated. But we can view $P/IP$ as a submodule via $x+IP\mapsto (x+I_1P,\ldots,x+I_nP)$.
