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. PlanetMath.
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:
- $[\overline{K}:K] = \infty$ and
- $\overline{K}$ 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?
Best Answer
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.