TITLE: Algebraicity of the completion of a field?  Finiteness?
QUESTION [11 upvotes]: At the end of my 8410 class today (see http://www.math.uga.edu/~pete/MATH8410.html if you care), one of my students asked me the following very interesting question:
Let $(K,|\ |)$ be a normed field, with completion $(\hat{K},| \ |)$.  Suppose $\hat{K}$ is algebraic over $K$.  Must we then have $\hat{K} = K$?
As I have mentioned here before, I feel very lucky to be getting such penetrating questions.  This one I was not able to answer on the spot, although I remarked that it is true in all of the most familiar examples and that the (possible) lack of algebraicity of the completion is a key motivation for considering the Henselization instead.
Edit: the answer is no, as I have just heard from one of my students.  I have encouraged him to come to this site and register the answer.
To make the question more interesting, suppose we ask whether $\hat{K}/K$ can be finite and nontrivial?

REPLY [8 votes]: 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.