# Is a completion of an algebraically closed field with respect to a norm also algebraically closed?

Assume we have an algebraically closed field $F$ with a norm (where $F$ is considered as a vector space over itself), so that $F$ is not complete as a normed space. Let $\overline F$ be its completion with respect to the norm.

Is $\overline F$ necessarily algebraically closed?

Thanks.

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Google Ax Sen Tate Theorem. –  Plop Mar 29 '11 at 13:30
The Ax Sen Tate theorem is not appropriate for a solution to this problem. It is quite deep compared to the question that is asked. –  KCd Mar 29 '11 at 22:07

The answer is yes, and the issue is discussed in detail in $\S 3.5-3.6$ of these notes from a recent graduate number theory course.

It is very much as Akhil suggests: the key idea is Krasner's Lemma (introduced and explained in my notes). However, as Krasner's Lemma pertains to separable extensions, there is a little further work that needs to be done in positive characteristic: how do we know that $\overline{F}$ is algebraically closed rather than just separably closed?

The answer is given by the following fact (Proposition 27 on p. 15 of my notes):

A field which is separably closed and complete with respect to a nontrivial valuation is algebraically closed.

The idea of the proof is to approximate a purely inseparable extension by a sequence of (necessarily separable) Artin-Schreier extensions. I should probably also mention that I found this argument in some lecture notes of Brian Conrad (and I haven't yet found it in the standard texts on the subject).

Note that Corollary 28 (i.e., the very next result) is the answer to your question.

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Your notes are very fast becoming a standard text on the subject, so... –  Mariano Suárez-Alvarez Mar 29 '11 at 16:55
@Mariano: Seriously? I hadn't heard. If so, I should really get around to putting in the proof of Hensel's Lemma! –  Pete L. Clark Mar 29 '11 at 17:39
Google seems to think so! It keeps taking me to them :) –  Mariano Suárez-Alvarez Mar 29 '11 at 19:14
Pete: include in your answer a link to B's handout on that, which is math.stanford.edu/~conrad/248APage/handouts/algclosurecomp.pdf. As for standard texts which contain a proof, I suspect an argument is in Bosch Guntzer Remmert's Non-archimedean Analysis, although I haven't checked it before writing this comment. –  KCd Mar 29 '11 at 22:03
@K: Okay, I included the link. As for BGR: it's a standard text for something I guess, but not for basic graduate number theory. (It's also so @$(%*!#^ expensive that I haven't been able to bring myself to purchase a copy, even with university and/or government funds.) I agree that it seems to be quite good for basic valuation theory: I've been referred to it before... – Pete L. Clark Mar 29 '11 at 22:41 Yes, (assuming the field is nonarchimedean) this follows from Krasner's lemma. The point is that if some element$\alpha$is algebraic over$\overline{F}$, then it must satisfy a monic polynomial equation with coefficients in$\overline{F}$; hence it must be very close to satisfying a monic polynomial equation with coefficients in$F$, and in particular must be very close to a root of this second monic polynomial equation (which is in$F$). Now Krasner's lemma implies (if we take this second polynomial really close to the first one) that$\alpha \in F\$.

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Do you assume F is non-archimedean? –  GEdgar Mar 29 '11 at 14:40
@GEdgar: Whoops, yes, I assumed that. Thanks for the correction. –  Akhil Mathew Mar 29 '11 at 15:34