TITLE: Is the $\mathfrak m$-adic completion of a radical ideal again a radical ideal?
QUESTION [8 upvotes]: Let $(R,\mathfrak m)$ be a local (Noetherian) ring and let $\hat{R}$ be its $\mathfrak m$-adic completion. Let $I$ be an ideal of $R$ which is a radical ideal. Is it then also true that $\hat{I}$  (the $\mathfrak m$-adic closure of $I$ in $\hat{R}$) is a radical ideal?

I am asking this question being interested mainly in the case where $R$ is $\mathbb{C}\{x_1,\dots,x_n\}$ - the ring of convergent power series over $\mathbb{C}$ (and hence $\hat{R} = \mathbb{C}[[x_1,\dots,x_n]]$, the ring of formal power series).

REPLY [7 votes]: As  $\hat{R}/\hat{I}\simeq \widehat{R/I}$, you are asking whether $R/I$ is analytically reduced. This is true when $R$ is excellent. 
It is known that $R=\mathbb C\{ x_1,\dots, x_n\}$ is excellent (EGA IV.7.8.4.v), so the answer to your question is yes in this case. But beware that there exist non excellent noetherian local rings (even discrete valuation rings).