Algebraicity of the completion of a field? Finiteness?

Question

At the end of my 8410 class today (see http://alpha.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?

Answer 1

(I'll delete this if your student came up with the same answer.)

Choose a ring-theoretic automorphism of the complex numbers that doesn't fix the reals (I'm pretty sure any nontrivial automorphism other than complex conjugation will work), and consider the image of the reals in it. A similar trick should work for any real closed field with transcendence degree at least 1 over Q. I'm not sure what I was thinking with the last sentence, but it's clearly false.

However, a similar trick should work for any finite Galois extension of complete normed fields such that the overfield has a discontinuous automorphism. For example, if we hit $\mathbb{C}((t))$ with some discontinuous non-$\mathbb{C}$-linear automorphism, I think the subfield $\mathbb{C}((t^3))$ is sent to a dense subfield.

Answer 2

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.