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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?

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    $\begingroup$ If $G$ is compact it's automatically profinite, since $GL_n(\bar{K})$ is totally disconnected $\endgroup$
    – YCor
    Commented Aug 7, 2014 at 18:36
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    $\begingroup$ The question sounds reasonable to me in a general context: assume that $K$ is a (complete?) normed field (say ultrametric, maybe unnecessary) and $G$ is a compact subgroup of $GL_n(K)$; then is a conjugate of $G$ contained in $GL_n(L)$ for $L$ a closed locally compact subfield of $K$? $\endgroup$
    – YCor
    Commented Aug 7, 2014 at 18:47
  • $\begingroup$ Instead of a conjugate one may ask for merely an isomorphic copy... $\endgroup$
    – Pablo
    Commented Aug 7, 2014 at 18:51
  • $\begingroup$ Why is a compact subset contained in a finite extension? $\endgroup$
    – Pablo
    Commented Aug 7, 2014 at 18:54
  • $\begingroup$ no actually it's not true. $\endgroup$
    – YCor
    Commented Aug 7, 2014 at 18:57

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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.

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  • $\begingroup$ What do yo mean by $\hat{K}$? $\endgroup$
    – Pablo
    Commented Aug 8, 2014 at 4:49
  • $\begingroup$ I denote $\hat{K}$ the algebraic closure of $K$ because $\bar{K}$ is confusing (it looks like a closure). $\endgroup$
    – YCor
    Commented Aug 8, 2014 at 10:12
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    $\begingroup$ $F_p((t))$ has infinitely many extensions of degree $p$, see math.stackexchange.com/questions/353928/… $\endgroup$
    – YCor
    Commented Aug 8, 2014 at 10:30
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    $\begingroup$ No they construct countably many: as many as the cokernel of $P(x)=x-x^p$, with $P:K\to K$ ($K=F_p((t))$. The image of $P$ contains the open 1-ball, since for $|x|<1$ and $y=\sum_{n\ge 0} x^{p^n}$ we have $P(y)=x$. Hence $coker(P)$ is countable. $\endgroup$
    – YCor
    Commented Aug 8, 2014 at 10:59
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    $\begingroup$ Yes you miss something: it's an argument showing there are at least infinitely countably many extensions of degree $p$. It doesn't show there are at most countably many. $\endgroup$
    – YCor
    Commented Aug 8, 2014 at 17:26

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