mg.metric geometry - Algebraicity of the completion of a field? Finiteness?

I had to think for a while to understand Scott's answer (or at least, what I suspect he meant by his answer), and in the end there were enough details to sort out that I thought they were worth posting. It ended up being too long to post as a comment, so here it is as a separate answer. Unless it's all nonsense, of course....

Let {$x_{alpha}$} be a transcendence basis of $mathbb{R}$ over $mathbb{Q}$, and let $L$ be the intermediate field that they generate, so that $mathbb{C}$ is the algebraic closure of $L$ in $mathbb{C}$. Take also a collection of open disks $D_{alpha}$ in $mathbb{C}$ such that any collection of points $y_{alpha} in D_{alpha}$ is dense in $mathbb{C}$ in the usual topology. Now for each $alpha$, take $x_alpha$ and multiply it by an appropriate root of unity and a rational number so that the result $y_alpha$ lies in $D_alpha$. The collection {$y_{alpha}$} is still algebraically independent over $mathbb{Q}$, because a dependence gives an algebraic dependence of {$x_alpha$} over some finite extension of $mathbb{Q}$, which implies the existence of an algebraic dependence over $mathbb{Q}$ as well.

So there exists $sigma : L to mathbb{C}$ sending $x_{alpha} mapsto y_{alpha}$. Now by the usual fact that field embeddings into algebraically closed fields can be extended across algebraic extensions, $sigma$ extends to a map $mathbb{C} to mathbb{C}$. But note that by construction $sigma$ is surjective! The image contains each $y_alpha$, and it contains all the roots of unity, so it contains all the $x_alpha$'s; thus the image is an algebraic closure of $L$ in $mathbb{C}$, hence all of $mathbb{C}$.

In particular $mathbb{C}$ is a quadratic extension of $sigma(mathbb{R})$, obtained by adjoining $sigma(i)$. But finally $sigma(mathbb{R})$ is dense in $mathbb{C}$ since its image contains all the $y_alpha$'s, and so giving $sigma(mathbb{R})$ the norm induced from the usual norm on $mathbb{C}$, we get a normed field $sigma(mathbb{R})$ whose completion is exactly $mathbb{C}$, i.e., a quadratic extension of it. Thus the answer to your second question actually yes.

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