Locally compact vector spaces over non-Archimedean field I'm in a $p$-adic analysis class right now and we were tasked with proving the following:
Suppose $(F,\|\cdot\|_F)$ is a non-Archimedean field, say $\mathbb{Q}_p$. By this I mean a field with a norm such that the induced distance satisfies a stronger form of the triangle inequality, namely:
$$\|x-y\|\leq\max\{\|x\|, \|y\|\}$$
Let $(V,\|\cdot\|_V)$ be a normed vector space over $F$. I want to show that $V$ is locally compact if and only if the unit ball in $V$ is compact. Now my question is pretty simple, does the fact that $F$ is non-archimedean change anything here as opposed to an Archimedean field?
The proof of this exercise I had in mind is just using the additive structure of the vector space to transfer the balls around. For instance if the closed unit ball is compact, then you can just translate it to a point $x$ to get a compact neighborhood of $x$. Is that correct or am I missing a subtlety due to the non-Archimedean underlying field? I think it should be correct because I'm not using the triangle inequality but vector spaces over non-Archimedean fields are still a bit exotic to me.
 A: Indeed, let $(F, \lvert \cdot \rvert)$ be any field with an absolute value, and $(V, \|\cdot\|)$ be a normed $K$-vector space with respect to the that absolute value (whether that norm and/or absolute value satisfy the strong triangle inequality or not). Then w.r.t. the metric induced by the norm, we have that

$\bar B_1(0) :=\{v \in V: \| v \| \le 1\}$ is compact $\Rightarrow  V$ is locally compact.

Because as you suggest, using that for any $v_0 \in V$ the translation map $x \mapsto v_0+x$ is a homeomorphism (which can be proved right away, or follows just from the fact that the underlying additive group of $V$ is a topological group w.r.t. the considered structure), we get that $v_0 + \bar B_1(0)$ is a compact neighbourhood of $v_0$.
For the converse direction

$V$ is locally compact $\Rightarrow \bar B_1(0) :=\{v \in V: \| v \| \le 1\}$ is compact

we need only one little extra restriction, namely we have to assume that the absolute value $\lvert \cdot \vert$ is not trivial, i.e. there is some $c \in F$ such that $\lvert c \rvert \neq 0,1$.
Namely, even without that assumption, from local compactness we can readily find some closed ball of radius $r > 0$, $\bar B_{r}(0) :=\{v \in V: \| v \| \le r\}$ around $0 \in V$ which is compact; now we need that assumption to further say w.l.o.g. that $r = \lvert a^{-1} \rvert$ for some scalar $a \in F$; and again, quite generally scaling $x \mapsto ax$ is a homeomorphism for any $a \in F^\times$, and will map that closed ball into and onto the closed unit ball.
(And in that trivial case that the absolute value and hence the norm is trivial, say $\lvert v \rvert = \begin{cases} c \text{ if } v \neq 0 \\ 0 \text{ if } v=0 \end{cases}$ for some $c > 0$, just notice that such a $V$ is always locally compact, whereas closed balls $\bar B_{r}(0)$ are singletons (hence compact) for $c > r$, but the entire $V$ for $c \le r$; the last one is compact iff it is finite iff either $V=0$ or ($F$ is finite and $\dim_F(V)$ is finite).)
Note now that the equivalence hold in great generality, but obviously neither of the statements is true in many settings. E.g. if $F$ itself is not locally compact, like $\mathbb Q$ with either the standard archimedean or any $p$-adic value, both statements are wrong e.g. if $\dim_F(V) \in \mathbb N_{\ge 1}$ (and for many infinite-dimensional $V$ as well).
