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Let $A$ be a ring with a an action of a group $G$. Is the morphism $A^G\rightarrow A$ faithfully flat? If not, is it true under some reasonable conditions?

My motivation for this problem is that the morphism on topological spaces $Spec(A)\rightarrow Spec(A)/G$ seems to be surjective, open with relatively well behaved fibers.

Thanks!

xlord
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    In general no. If the action is free and thus give a covering, then yes. For a simple counterexample, consider the action of $\mathbb{Z}/2\mathbb{Z}$ on affine plane by the action, $x\mapsto -x, y\mapsto -y$. – Mohan Dec 02 '20 at 15:43

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