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Let $F$ be a complete metric topological field. Suppose there is a subfield $F_1 \subset F$, algebraically closed and topoolgically dense in $F$. Must $F$ itself be algebraically closed?

We can find this in the literature in case $F$ is a valued field.

I asked it this way, rather than just starting with a metric algebraically closed field and taking its completion. As Laurent Moret-Bailly remarked in a comment HERE, the metric completion of a metric topological field need not even be a field.

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    $\begingroup$ This is an excellent question. I wanted to post it myself a while ago, but first I wanted to make sure that such a field - completely metrizable, but not topologically isomorphic to a valued field - exists at all. This is why I posted the following question: mathoverflow.net/q/238809/89334 $\endgroup$
    – Uri Bader
    Commented Apr 3, 2021 at 21:03
  • $\begingroup$ When you say "complete metric", do you assume the metric and the field topology define the same uniform structure? Or even that the metric is translation-invariant? $\endgroup$
    – Laurent Moret-Bailly
    Commented Apr 4, 2021 at 6:38
  • $\begingroup$ @LaurentMoret-Bailly Yes, I assume a translation-invariant metric, so that the metric topology is the uniform topology. $\endgroup$
    – Gerald Edgar
    Commented Apr 4, 2021 at 10:44
  • $\begingroup$ Is there a counterexample without the metric restriction? $\endgroup$
    – Laurent Moret-Bailly
    Commented Apr 5, 2021 at 8:23
  • $\begingroup$ @LaurentMoret-Bailly Here is a stupid example, regarding topological fields. Take $F=\bar{\mathbb{Q}}(\pi)\subset \mathbb{C}$ with the induced topology and $F_1=\bar{\mathbb{Q}}$. This shows that you better ask for some sort of completeness. $\endgroup$
    – Uri Bader
    Commented Apr 5, 2021 at 8:37

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