Let $n \in \mathbb{N}$, $K$ a (nonarchimedean) local field, $\overline{K}$ its algebraic closure. Take a compact subgroup $G \leq \text{GL}\_n(\overline{K})$. Must there be a finite extension $F$ of $K$ such that $G \leq \text{GL}\_n(F)$?


What if in addition $G$ is profinite? can that make a difference?


Does this generalize to higher local fields?



Compact subgroups of linear groups over nonarchimedean fields
Here's an argument. It works under the assumption that $K$ is a complete normed field and has countably many finite extension up to $K$-isomorphism. This holds in particular when $K$ has finitely many extensions in each given degree, e.g., $p$-adic fields. 


Since given a finite extension it has finitely many embeddings into $\hat{K}$ (I avoid $\bar{K}$ which is confusing), this implies that we can write $\hat{K}=K\_n$, where $K=K\_0\subset K\_1\subset\dots$. Since $K\_n$ is complete, it is closed in $\hat{K}$. Now let $G$ be a compact subgroup of $\mathrm{GL}\_n(\hat{K})$. Then $G=\bigcup G\_n$ where $G\_n=G\cap\mathrm{GL}\_n(K\_n)$. By Baire's theorem, some $G\_n$ has non-empty interior, hence is an open subgroup of $G$, hence by compactness has finite index in $G$. Hence for some possibly larger $n$, $G\_n=G$.


Note that the conclusion is false when to consider instead the completion $\bar{\hat{K}}$ of $\hat{K}$. Indeed, just pick $x\in \bar{\hat{K}}\smallsetminus\hat{K}$ and consider the group of order 2, generated by the matrix $\begin{pmatrix}1 & x \\ 0 & -1\end{pmatrix}$. On the other hand, it might necessarily be conjugated into a finite extension, but I can't prove it.


