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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?

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If $\hat K/K$ is finite then $K$ is a closed $K$-subvector space of $\hat K$, for it is a finite dimensional subspace, so in particular $K$ is complete, no? –  Mariano Suárez-Alvarez Jan 29 '10 at 6:39
    
I was thinking along these lines. But unlike in the case of usual Banach spaces we don't have a complete field "at the bottom", so is it obvious that a finite-dimensional subspace is closed? –  Pete L. Clark Jan 29 '10 at 6:42
    
A finite dimensional $K$-subspace is closed, because it is the kernel of a linear map to some other space, I guess. –  Mariano Suárez-Alvarez Jan 29 '10 at 6:46
    
I can't tell whether you are speculating or answering. If the latter, could you say a little more? –  Pete L. Clark Jan 29 '10 at 6:51
3  
OK, here's a counterexample to the automatic continuity claim: consider Q(\sqrt{2}) as a vector space over Q with the standard Archimedean norm. There are two ways to extend this norm to Q(\sqrt{2}), differing by Galois conjugation. Therefore Galois conjugation is a Q-linear map which is not continuous for either norm. More plainly, the sequence (\sqrt{2}-1)^n converges to zero in one topology but not the other. –  Pete L. Clark Jan 29 '10 at 8:14

2 Answers 2

up vote 10 down vote accepted

(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.

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My (apparently MO-shy) student came up with the example of a subextension L of R/Q such that L/Q is purely transcendental and R/L is (certainly nontrivial) algebraic. As I pointed out in today's problem session, the Artin-Schreier theorem guarantees that R/L is infinite, so this does not answer the second question. In fact, your very nice answer to the second question prompts a third question: can we have $2 < [\hat{K}:K] < \infty$? –  Pete L. Clark Jan 29 '10 at 21:24
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Thanks again, Scott. Your new example sounds reasonable; I'll have to think about it. Now what about $[\hat{K}:K] \geq 4$? (Just kidding.) –  Pete L. Clark Jan 29 '10 at 22:02
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Ah, I see, it's way easier than what I wrote below, isn't it? If an automorphism of C doesn't fix the reals, then the image of R contains a non-real z (R has no nontrivial automorphisms, so if an automorphism doesn't fix the reals then it doesn't preserve them), and then the Q-vector-subspace of the image generated by 1 and z is dense, so the whole image is dense. –  D. Savitt Jan 29 '10 at 23:48
    
@Dave: Yes, that's right. –  Pete L. Clark Jan 30 '10 at 0:18

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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+1 to you too, Dave. –  Pete L. Clark Jan 29 '10 at 21:25
    
Does this answer depend on the axiom of choice? –  Dror Speiser Jan 30 '10 at 0:01
    
Boy oh boy, does it ever! But probably Scott's does too, since you need to construct non-obvious automorphisms of C? –  D. Savitt Jan 30 '10 at 0:11
    
Yes, certainly. The entire theory of transcendental field extensions as we know it depends heavily on AC. It is consistent with ZF that the only automorphisms of the complex numbers are the identity and complex conjugation. I've used Zorn's Lemma several times in my lectures already. –  Pete L. Clark Jan 30 '10 at 0:21

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