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

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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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