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

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

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

To simplify notation, say every little thing is affine: $$G = operatorname{Spec}R$$, $$X = operatorname{Spec}A$$, and $$S = operatorname{Spec}ok$$, the place $$ok$$ is an capricious ring (not essentially a bailiwick). If it helps we are able to occupy $$G$$ is flush. We labor within the class of $$ok$$-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 $$A = operatorname{Hom}(X,mathbb{A}^1)$$, 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 $$G(ok)$$ acts on $$X(ok)$$, and for any $$ok$$-scheme $$T$$, $$G(ok)$$ maps to $$G(T)$$ and therefore too acts on $$X(T)$$, so $$G(ok)$$ acts on $$X$$. Thus, we could too deem the ring of invariants $$A^{G(ok)}$$. Certainly we’ve
$$A^Gsubset A^{G(ok)}$$
My important query is: What is the clearest route to specific this relationship? I’m in search of an announcement of the figure $$fin A$$ is $$G$$-invariant if and solely whether it is fastened by $$G(ok)$$ and another situations.

I cerebrate one can say that
$$A^G = {fin A| fotimes_k 1in Aotimes_k B textual content{ is fastened by G(B) for each ok-algebra B}}$$
Is this rectify? Is it workable to additional limit the category of $$B$$‘s that you need to deem? Are there different methods of occupied with this?

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