# Is the $\mathfrak m$-adic completion of a radical ideal again a radical ideal?

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).

• I'm hampered a bit by not knowing about $C\{x\}$ . But maybe this will help: $C[[x]]$ is a chain ring with exactly two prime ideals $(0)$ and $(x)$ (the remaining ideals are $(x^i )$ and the whole ring, which are clearly not prime.) That means that the only radical ideals of $C[[x]]$ are those two prime ideals (since radical ideals are intersections of prime ideals.) Does it make sense that all radical ideals of $C\{x\}$ would get mapped to those two ideals? If not, then I think your question is decided. – rschwieb Aug 30 '13 at 13:19
• Sorry, i meant the multivariate case, i.e. $x = (x_{1},\dots, x_{n})$. – Sebastian Aug 30 '13 at 13:24
• OK: I guess it remains to be seen if requiring $n>1$ changes things so that it's possible. Seems unlikely though, if it doesn't work in the single variable case. – rschwieb Aug 30 '13 at 13:28
• But yes, this answers the case $n=1$ (which, however, is quite different from the general case) in the affirmative: $\mathbb{C}\{x\}$ is (for n=1) a principal ideal and its only ideals are $0$ and $<x^{n}>$, and the completion of $<x^{n}>$ is generated by $x^{n}$ in $\mathbb{C}[[x]]$. – Sebastian Aug 30 '13 at 13:29
• Yeah, this could be an edge case somehow. Maybe this ring really gets its act together after $n=1$ and magically gets better about where it sends ideals. Commutative algebra is pretty magical IMO :) – rschwieb Aug 30 '13 at 13:33

## 1 Answer

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).

• Thanks for the answer! Do I get this right that this holds for arbitrary ideals (of $\mathbb{C}\{x_{1},\dots,x_{n}\}$)? So that in the process of "completion" all the possible nil-potent element of $R/I$ vanish? – Sebastian Aug 30 '13 at 15:11
• Yes this holds for all $I$ in the right $R$. The completion for excellent local rings doesn't add nilpotent elements. But if you start with nilpotent elements in $R/I$, they will stay in the completion because the latter contains $R/I$. – Cantlog Aug 30 '13 at 16:56
• So the ideal $I$ has to be a radical ideal ( which is equivalent to $R/I$ being reduced) or otherwise $\widehat{R/I}$ will contain the nilpotent elements of $R/I$? – Sebastian Aug 30 '13 at 23:06
• @Sebastian: Yes, absolutely. – Cantlog Aug 31 '13 at 10:42