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Yes, this is yet another "foundational" question in valuation theory.

Here's the background: it is a well known classical fact that the dimension (in the purely algebraic sense) of a real Banach space cannot be countably infinite. The proof is a simple application of the Baire Category Theorem: see e.g.

http://planetmath.org/encyclopedia/ABanachSpaceOfInfiniteDimensionDoesntHaveACountableAlgebraicBasis.html

Suppose now that $(K,| \ |)$ is a complete non-Archimedean (edit: nontrivial) normed field. One has the notion of a $K$-Banach space, and the Baire Category Theorem argument works verbatim to show that such a thing cannot have countably infinite $K$-dimension.

Now let $\overline{K}$ be an algebraic closure of $K$. Then $\overline{K}$, by virtue of being a direct limit of finite-dimensional normed spaces over the complete field $K$, has a canonical topology, and indeed a unique multiplicative norm which extends $|\ |$ on $K$.

My question is: does there exist a complete normed field $(K, | \ |)$ such that:
(i) $[\overline{K}:K] = \infty$ and
(ii) $\overline{K}$ is complete with respect to its norm?

As with a previous question, it is not too hard to see that this does not happen in the most familiar cases. Indeed, by the above considerations this can only happen if $[\overline{K}:K]$ is uncountable. But $[\overline{K}:K]$ will be countable if $K$ has a countable dense subfield $F$ [to be absolutely safe, let me also require that $F$ is perfect]. Indeed, the algebraic closure of any infinite field has the same cardinality of the field, so $\overline{F}$ can be obtained by adjoining roots of a countable collection of separable polynomials $P_i(t) \in F[t]$. It follows from Krasner's Lemma that by adjoining to $K$ the roots of these polynomials one gets $\overline{K}$.

What about the general case?

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    $\begingroup$ yes: let K be the rationals and |.| the trivial norm (|0|=0, |x|=1 otherwise). $\endgroup$ Feb 11 '10 at 7:46
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    $\begingroup$ More helpfully: if the norm is (non-trivial and) discrete then probably an easy argument (analogous to one of the proofs that Q_p-bar isn't complete) gives that it can't happen. For the norm on K-bar won't be discrete (take roots of a uniformiser pi of K) so you can build some awkward x in the completion of K-bar with v(x)=1, v(x-pi)=3.5, v((x-pi)^2-pi^7)=10!+1/3 and so on and so on, meaning that the valuation on K(x) induced by K contains numbers with arbitrarily large denominator, so K(x) can't be discrete, so x can't be algebraic over K. $\endgroup$ Feb 11 '10 at 7:48
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No, there exists no such field (with a non-trivial norm). A proof can be found in Bosch, Güntzer, Remmert: Non-Archimedean Analysis, Lemma 1, Section 3.4.3.

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    $\begingroup$ So in fact the proof is the same idea as the one in my comment above. Replace "has valuation n+1/d" by "is close to an element that generates an extension of K of dgree d" and then use Krasner's Lemma. $\endgroup$ Feb 11 '10 at 9:16

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