Is $\widehat{IM}=\widehat I \widehat M$ for any finitely generated module $M$, where $\widehat {(-)}$ denotes completion w.r.t. maximal ideal?

Question

Let $(R,\mathfrak m)$ be a Noetherian local ring, and let $(\widehat R, \widehat{\mathfrak m})$ be its $\mathfrak m$-adic completion. Let $M$ be a finitely generated $R$-module.

Then, is it true that $\widehat{IM}=\widehat I \widehat M? $

I know this is true if $I=\mathfrak m$ as explained here, but I'm not sure what happens for general $I$. Please help.

Answer 1

Yes. As far as I can see, the same proof as in the linked post works with $\mathfrak{a}$ replaced by $I$.

One thing to keep is mind is that $I$ is also finitely generated $R$-module so $\hat{I}\simeq I\otimes_{A}\hat{A}$ where the completions are $\mathfrak{m}$-adic.