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Jan 15, 2021 at 8:11 comment added Wilberd van der Kallen If everything is affine and $G$ is smooth, then it suffices to take for $B=R$.
Jan 14, 2021 at 17:52 comment added LSpice @RobPratt, I was just deliberating over whether "does" should be "do" in the title. I eventually decided that it's about the ring of invariants, not about the individual invariants, and so left it. But I think your edit does make the title read more smoothly. :-)
Jan 14, 2021 at 17:49 history edited RobPratt CC BY-SA 4.0
added 35 characters in body; edited title
Jan 14, 2021 at 15:48 comment added David E Speyer While I'm here, I just wanted to say that this is not at all a stupid question, and neither are any of your others!
Jan 14, 2021 at 15:48 answer added David E Speyer timeline score: 5
Oct 27, 2020 at 16:36 comment added David E Speyer Ah, found it. If (1) $k$ is infinite, (2) $G$ is connected and either (3a) $G$ is reductive or (3b) $k$ is perfect, then $G(k)$ is Zariski dense in $G$. mathoverflow.net/q/56192/297
Oct 27, 2020 at 15:52 comment added LSpice @DavidESpeyer is right; I definitely meant smooth connected, and there is a slight possibility I meant reductive.
Oct 27, 2020 at 15:20 comment added David E Speyer I agree that the key thing to say is that, if $G(k)$ is Zariski dense in $G$, then $A^{G(k)} = A^G$. But I don't think your finite field criterion is the right one. For example, let $G = \mu_3$, the group of $3$-rd roots of unity, and let $k = \mathbb{R}$. @LSpice
Oct 27, 2020 at 12:51 comment added LSpice A simple and unsatisfactory one: if $k$ is not an algebraic extension of a finite field, and $G$ is smooth, then $G(k)$ is Zariski dense, so $A^{G(k)} = A^G$.
Oct 27, 2020 at 8:50 history edited stupid_question_bot CC BY-SA 4.0
added 5 characters in body
Oct 27, 2020 at 8:45 history asked stupid_question_bot CC BY-SA 4.0