Is the Characteristic of a Field Detectable from the Topology of a Topological Vector Space?

Question

Motivation

A topological vector space is a vector space over a (topological) field, K, that carries a topology such that addition and scalar multiplication are continuous maps, e.g., all normed vector spaces. Since $\mathbb{C} \approx \mathbb{R}^2$ as topological spaces (with their standard topologies), it's clear that the actual field cannot be detected by the topology alone. Also, cardinality considerations show that any topology on a finite field cannot be homeomorphic to a vector space over a field of characteristic $0$. But that still says nothing about whether there are some topologies, such that, for instance $\mathbb{Z}_p(t) \approx \mathbb{Q}$. I can't find any topological invariants that will distinguish just the fields (as vectors spaces of dimension 1). I suspect that if this problem is solvable, it will be just be topological group considerations, but I can't figure out how to do it.

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So here are my two well-posed questions:

a) Suppose K and F are topological fields. If K and F are homeomorphic, then is it necessarily true that char K = char F?

b) Suppose K and F are topological fields. Further suppose that $V$ is a topological vector space over K and $W$ is a topological vector space over F. If $V$ and $W$ are homeomorphic, then is it necessarily true that char K = char F?

Obviously, an answer to b) implies an answer for a). I'm not sure how much of my thinking I should put here. But I do know that if char K = p, then we get a homeomorphism of $V$ onto itself by translation for each $v \in V$, which each have finite order p. These are mapped to self-homeomorphisms of $W$ which also have order p, but a priori, they don't "see" any of the algebraic structure of $W$, so I'm not sure this is actually an obstruction. Any thoughts (or references) to proofs or counterexamples would be greatly appreciated.

Answer 1

I think all non-archimedean locally compact fields are homeomorphic: Their rings of integers are compact, metric and totally disconnected and hence are all homeomorphic (to the Cantor set). The same is true for the units in those rings. The field is then topologically the disjoint union of the ring and a countable number of copies of the units.

Answer 2

Torsten Ekedahl answered your first question in a nontrivial set of cases. As a concrete example, the characteristic two field $\mathbb{F}_2((t))$ is homeomorphic to the characteristic zero field $\mathbb{Q}_2$ by the map $$\sum_{k \gg -\infty} a_k t^k \mapsto \sum_{k \gg -\infty} a_k 2^k.$$ This also provides a counterexample to your second question in dimension greater than zero.

The answer to the question in the title is a bit more subtle. That is, there are topological vector spaces that only allow one characteristic. This is obvious for finite dimensional vector spaces over finite fields, but it's also true that there are no fields of positive characteristic that are homeomorphic to $\mathbb{R}$, because the additive structure would allow one to produce a finite group of automorphisms of the order topology of size greater than 2.