Reductive groups over non archimedean local fields. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T06:27:08Z http://mathoverflow.net/feeds/question/87191 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87191/reductive-groups-over-non-archimedean-local-fields Reductive groups over non archimedean local fields. Carlos De la Mora 2012-02-01T01:36:43Z 2012-02-01T03:33:07Z <p>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? </p> <p>Thank you</p> http://mathoverflow.net/questions/87191/reductive-groups-over-non-archimedean-local-fields/87196#87196 Answer by Alexander Braverman for Reductive groups over non archimedean local fields. Alexander Braverman 2012-02-01T03:09:09Z 2012-02-01T03:09:09Z <p>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 $X$). </p> http://mathoverflow.net/questions/87191/reductive-groups-over-non-archimedean-local-fields/87200#87200 Answer by Peter McNamara for Reductive groups over non archimedean local fields. Peter McNamara 2012-02-01T03:33:07Z 2012-02-01T03:33:07Z <p>This is true because G is unirational, eg Springer, Linear Algebraic Groups, Corollary 13.3.9(ii).</p>