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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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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.

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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