Let $F$ be a non archimedean local field and let $G$ be linear algebraic group over $F$. I do not have a lot of experience with linear algebraic group, but it seems very obvious that $G$ inherits the property of being locally profinite from $F$. Moreover in some texts, they work in this situation without explanation, so I am sure enough that it is true and probably obvious. Some hints for the proof?
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$\begingroup$ You mean, the group of $F$-points. The answer is yes, by an old theorem of Van Dantzig: every totally disconnected locally compact group has a compact (hence profinite) open subgroup. Van Dantzig's theorem can be found at many places and can also be proved as an exercise (once it is granted that there is a basis of clopen compact neighborhoods of the identity). $\endgroup$– YCorCommented Sep 8 at 9:51
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$\begingroup$ @YCor , Thank you very much for your answer. It is not clear to me how the Van Danzig theorem proves that the $F$-points of $G$ form a locally profinite group. $\endgroup$– MarioCommented Sep 8 at 10:44
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$\begingroup$ A slightly sharper formulation of van Dantzig's theorem is that every totally disconnected locally compact group has a neighbourhood base of compact open subgroups. As a compact totally disconnected Hausdorff group is the same thing as a profinite group, you are done. $\endgroup$– AntoniusCommented Sep 9 at 3:43
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