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If $G$ is a group scheme over $S$ acting on an $S$-scheme $X$, I'd like to understand the algebra of invariants $(\mathcal{O}_X)^G$. Specifically, I'd like to understand its relation to invariants $(\mathcal{O}_X)^{G(S)}$.

To simplify notation, say everything is affine: $G = \operatorname{Spec}R$, $X = \operatorname{Spec}A$, and $S = \operatorname{Spec}k$, where $k$ is an arbitrary ring (not necessarily a field). If it helps we can assume $G$ is smooth. We work in the category of $k$-schemes.

The action is given by a map $\sigma : G\times X\rightarrow X$. Let $p : G\times X\rightarrow X$ be the projection map. Then there is a natural bijection $A = \operatorname{Hom}(X,\mathbb{A}^1)$, and by definition the subalgebra of invariants $A^G$ is the set of $f\in A$ whose corresponding map $F : X\rightarrow\mathbb{A}^1$ satisfies $$F\circ\sigma = F\circ p$$ Via $\sigma$, the group $G(k)$ acts on $X(k)$, and for any $k$-scheme $T$, $G(k)$ maps to $G(T)$ and hence also acts on $X(T)$, so $G(k)$ acts on $X$. Thus, we may also consider the ring of invariants $A^{G(k)}$. Certainly we have $$A^G\subset A^{G(k)}$$ My main question is: What is the clearest way to express this relationship? I'm looking for a statement of the form $f\in A$ is $G$-invariant if and only if it is fixed by $G(k)$ and some other conditions.

I think one can say that $$A^G = \{f\in A| f\otimes_k 1\in A\otimes_k B \text{ is fixed by $G(B)$ for every $k$-algebra $B$}\}$$ Is this correct? Is it possible to further restrict the class of $B$'s that you have to consider? Are there other ways of thinking about this?

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    $\begingroup$ 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 $A^{G(k)} = A^G$. $\endgroup$ – LSpice Oct 27 '20 at 12:51
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    $\begingroup$ 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 $\endgroup$ – David E Speyer Oct 27 '20 at 15:20
  • $\begingroup$ @DavidESpeyer is right; I definitely meant smooth connected, and there is a slight possibility I meant reductive. $\endgroup$ – LSpice Oct 27 '20 at 15:52
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    $\begingroup$ 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 $\endgroup$ – David E Speyer Oct 27 '20 at 16:36
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    $\begingroup$ 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! $\endgroup$ – David E Speyer Jan 14 at 15:48
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Combining LSpice's (1 2) and my (1 2) comments into an answer: If $G(k)$ is Zariski dense in $G$, then $A^{G(k)} = A^G$. It is very common that $G(k)$ is Zariski dense in $G$: This happens whenever (1) $k$ is infinite and (2) $G$ is connected and either (3a) $G$ is reductive or else (3b) $k$ is perfect. See Density question in algebraic group.

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