Saturday 28 July 2007

mg.metric geometry - Can the algebraic closure of a complete field be complete and of infinite degree?

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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