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I want to know if connected reductive groups over non archimedean local fields have a dense countable subset. I was thinking that this should be true because if $G(\mathbb{F})$ is such group where $\mathbb{F}$ is a non archimedian local field then there exist an embedding of $G(\mathbb{F})$ into some matrix group GL$_n(\bar{\mathbb{F}})$. I then suppose that this map is continuous and open, right? and since $\bar{\mathbb{F}}$ has a dense countable set, or not? then $G(\mathbb{F})$ has a dense countable set. Can somone give me a reference of this result if it is true?

Thank you

flag	asked Feb 1 2012 at 1:36 Carlos De la Mora 113●5

Alexander Braverman · Answer 1 · 2012-02-01 03:09:09Z UTC

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I think this is true for any affine variety $X$ over $F$: by Noether normalization lemma it can be represented as a finite cover of an affine space, for which the statement is clearly true (then take pre-image in $X$).

answered Feb 1 2012 at 3:09

Alexander Braverman
3,333●6●25

It can't be true for any affine X since X may not even have any F-points. – Peter McNamara Feb 1 2012 at 3:33

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Well, but in that case it still has a dense countable subset by definition (I thought countable means actually no more than countable) – Alexander Braverman Feb 1 2012 at 4:04

Peter McNamara · Answer 2 · 2012-02-01 03:33:07Z UTC

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This is true because G is unirational, eg Springer, Linear Algebraic Groups, Corollary 13.3.9(ii).

answered Feb 1 2012 at 3:33

Peter McNamara
3,236●6●18

Reductive groups over non archimedean local fields.

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