At the end of my 8410 class today (see [http://alpha.math.uga.edu/~pete/MATH8410.html](http://alpha.math.uga.edu/%7Epete/MATH8410.html) if you care), one of my students asked me the following very interesting question:


Let $(K,|\ |)$ be a [normed field](http://eom.springer.de/n/n067360.htm), 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?



Algebraicity of the completion of a field?  Finiteness?
(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.


