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.

Solution 1:

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.

Solution 2:

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 $\alpha \in F$.