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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?

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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

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