Let $R$ be a commutative ring with unity and $I \subset R$ an ideal such that $R$ is $I$-adically separated and denote by $R^I$ the $I$-adic completion of $R$ with respect to $I$. Let $\overline{I}$ be the closure of $I$ in $R^I$:
Is $R^I$ $\overline{I}$-adically complete?
The answer seems to be yes. I have taken this result from a paper I am trying to understand. However I can't prove the result, which has been bugging me for a while now. Any help would be appreciated, thank you!
What do you mean by $\bar{I}$ exactly? I will assume you see $\hat{R}$ as an $R$-module with the induced $I$-adic topology and want $\bar{I}$ to be the closure of $I \hat{R}$ with regard to that topology; which is then just $\bigcap_{n \in \mathbb{N}} (I \hat{R} + I^n \hat{R}) = I \hat{R}$. At least that is the notation that makes most sense to me here (e.g. one could also consider the closure of the submodule $\mathrm{im}(IR) \subseteq \hat{R}$, which should in general not be an ideal; or you could equip $\hat{R}$ with the limit topology, see here.)
If $I$ is finitely generated, what you ask is true. For example, see here.
If $I$ is not finitely generated, there are counterexamples. For example, see the discussion here