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.

score 4 · Accepted Answer · answered Jul 21 '21 at 06:47

4

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.

answered Jul 21 '21 at 06:47

Evans Gambit

2,043

And the proof works for $A$ Noetherian, not necessarily local and completed at some ideal $\mathfrak{a}$ (not necessarily maximal). – Elías Guisado Villalgordo Sep 19 '23 at 12:40
An interesting particular case is $M=I$, i.e., in $\mathfrak{a}$-adic completions, one has the equality $\widehat{I^n}=\hat{I}^n$ of ideals of $\hat{A}$. – Elías Guisado Villalgordo Feb 21 '24 at 16:04

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

1 Answers1