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).

  • $\begingroup$ 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. $\endgroup$ – rschwieb Aug 30 '13 at 13:19
  • $\begingroup$ Sorry, i meant the multivariate case, i.e. $x = (x_{1},\dots, x_{n})$. $\endgroup$ – Sebastian Aug 30 '13 at 13:24
  • $\begingroup$ 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. $\endgroup$ – rschwieb Aug 30 '13 at 13:28
  • 1
    $\begingroup$ 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 $\mathbb{C}[[x]]$. $\endgroup$ – Sebastian Aug 30 '13 at 13:29
  • $\begingroup$ 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 :) $\endgroup$ – rschwieb Aug 30 '13 at 13:33

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).

  • $\begingroup$ 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? $\endgroup$ – Sebastian Aug 30 '13 at 15:11
  • $\begingroup$ 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 $R/I$. $\endgroup$ – Cantlog Aug 30 '13 at 16:56
  • $\begingroup$ 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 $R/I$? $\endgroup$ – Sebastian Aug 30 '13 at 23:06
  • $\begingroup$ @Sebastian: Yes, absolutely. $\endgroup$ – Cantlog Aug 31 '13 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.