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Suppose $G$ is a connected reductive group defined over a field $F$ of characteristic $0$. Does every maximal torus contain a regular semisimple element defined over $F$?

I know that over an algebraically closed field this is true because being regular corresponds to being in the intersection of the complements of the kernels of the roots. Since the complement of each kernel is open and dense, we can pick an element in the intersection.

But what about if $F$ is not algebraically closed?

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A torus is unirational. The set of regular elements is open, hence also unirational. Over an infinite field, any unirational variety has (a Zariski dense set of) F-points. This answers the question in the affirmative.

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  • $\begingroup$ Great, thanks. I suspected this sort of thing would work but didn't know the right way to do it. $\endgroup$
    – Alexander
    Aug 9, 2018 at 2:28
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    $\begingroup$ There's no denying the validity of this argument, but I've always had the impression that unirationality was a rather deep theorem. Is it not possible to do this by, say, taking the norm of a "suitably regular" element defined over a splitting field? (I'm not saying that I know how to do it, just that I'd be a little surprised if such a hands-on approach doesn't work.) $\endgroup$
    – LSpice
    Aug 9, 2018 at 3:28

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