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?