In the wikipedia article titled "topological vector space", there is a line saying the following.

"Let $K$ be a locally compact topological field, for example to real or complex numbers. A topological vector space over $K$ is locally compact if and only if it is finite-dimensional, that is, isomorphic to $K^n$ for some natural number $n$."

I am fine with the real or complex numbers. However, on the one hand if we take a finite field, say $\mathbb{F}_p$ for a prime number $p$, with the discrete topology, it seems to me that $\mathbb{F}_p$ satisfies all conditions of being a locally compact topological field. On the other hand it is well-known that the field $\mathbb{F}_p((t))$ is locally compact as well, and it is obviously not a finite dimensional vector space over $\mathbb{F}_p$. What am I missing? (Maybe one needs to add the condition that $K$ has to be non-discrete as well?) Thanks a lot.

EDIT: As Denis pointed out in the comment, $\mathbb{F}_p((t))$ is not equipped with the vector space topology corresponding to $\mathbb{F}_p$ (i.e., the product topology). This answers my question above, but it raises a new question. Note that the quoted wikipedia statement is used in a proof of the classification of local fields to conclude that any local field $F$ containing $\mathbb{Q}_p$ must be an algebraic extension of $\mathbb{Q}_p$. How do we know that there is no some other big local field which extends $\mathbb{Q}_p$ but does not carry the "right" vector space topology (product topology), in the same way as $\mathbb{F}_p((t))$ extends $\mathbb{F}_p$ but does not carry the right vector space topology?

  • 3
    $\begingroup$ The topology you are putting on $\mathbb{F}_p((t))$ is not the vector space topology (that is the product topology once you chose a basis). $\endgroup$ – Denis Nardin Mar 16 '17 at 15:32
  • 1
    $\begingroup$ @DenisNardin What you said is absolutely right. But my main concern is that people use this statement in the classification of local fields to conclude that the only local fields $F$ containing $\mathbb{Q}_p$ are algebraic extensions of $\mathbb{Q}_p$, which seems to only need $F$ to be some extension of $\mathbb{Q}_p$ that also extends the norm; but surely $\mathbb{F}_p((t))$ does the same thing to $\mathbb{F}_p$ as well. $\endgroup$ – Daps Mar 16 '17 at 15:41
  • 2
    $\begingroup$ This statement confuses me too. Suppose I take $\mathbb{F}_2$ which is a compact topological field. Then the countable product $\mathbb{F}_2^\omega$, with its product topology, is a topological vector space over $\mathbb{F}_2$, isn't it? But it's compact (it's homeomorphic to Cantor space). And it's certainly infinite dimensional. Unless I am missing something, I think that statement might simply be wrong. $\endgroup$ – Nate Eldredge Mar 16 '17 at 20:16
  • 2
    $\begingroup$ From my viewpoint, "non-discrete" is missing from the (quoted) requirements on the field. If that requirement is added, then everything works out as expected. $\endgroup$ – paul garrett Mar 16 '17 at 20:55
  • 1
    $\begingroup$ @AbdelmalekAbdesselam No the topology of $F_p((t))$ is not the topology inherited from $F_p^Z$ (the latter is not locally compact). $\endgroup$ – YCor Mar 17 '17 at 2:02

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.