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ag.algebraic geometry – How does the invariants of a bunch strategy motion examine to the invariants of the group motion by international sections

If $G$ is a bunch strategy over $S$ performing on an $S$-scheme $X$, I’d love to grasp the algebra of invariants $(mathcal{O}_X)^G$. Specifically, I’d love to grasp it is relation to invariants $(mathcal{O}_X)^{G(S)}$.

To simplify notation, say all the things is affine: $G = textual content{Spec }R$, $X = textual content{Spec }A$, and $S = textual content{Spec }okay$, the place $okay$ is an capricious ring (not essentially a bailiwick). If it helps we will occupy $G$ is flush. We labor within the class of $okay$-schemes.

The motion is given by a map $sigma : Gtimes Xrightarrow X$. Let $p : Gtimes Xrightarrow X$ breathe the projection map. Then there’s a unaffected bijection $A = Hom(X,mathbb{A}^1)$, and by definition the subalgebra of invariants $A^G$ is the clique of $fin A$ whose corresponding map $F : Xrightarrowmathbb{A}^1$ satisfies

$$Fcircsigma = Fcirc p$$

Via $sigma$, the group $G(okay)$ acts on $X(okay)$, and for any $okay$-scheme $T$, $G(okay)$ maps to $G(T)$ and therefore too acts on $X(T)$, so $G(okay)$ acts on $X$. Thus, we could too deem the ring of invariants $A^{G(okay)}$. Certainly we now have

$$A^Gsubset A^{G(okay)}$$

My principal query is: What is the clearest route to specific this relationship? I’m in search of a press release of the figure $fin A$ is $G$-invariant if and solely whether it is fastened by $G(okay)$ and another situations.

I cerebrate one can say that

$$A^G = {fin A| fotimes_k 1in Aotimes_k B textual content{ is fastened by $G(B)$ for each $okay$-algebra $B$}}$$

Is this rectify? Is it workable to additional prohibit the category of $B$‘s that it’s a must to deem? Are there different methods of fascinated by this?

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