Let $A$ be a normal ring (in the sense that its localizations at prime ideals are normal domains), and suppose that a finite group $G$ acts on $A$ by ring automorphisms. Form the subring $A^G \subset A$ of $G$-invariant elements. Is the ring $A^G$ also normal? What if $A$ is Noetherian (so that it is a product of Noetherian normal domains, which, however, need not be preserved by the $G$-action)?


I know that the answer is 'yes' if $A$ is an integral domain, but I am curious about the general case.



Does the ring of invariants inherit normality?
This is the answer of user74230 using localization instead of henselization. Namely, if $\mathfrak p$ is a prime of $A^G$, then we can replace $A^G$ by $(A^G)\_\mathfrak p$ and $A$ by $A\_\mathfrak p$ (same explanation as in the answer of user74230). Thus we may assume $A^G$ is local. Then $A$ has finitely many maximal ideals, each lying over the maximal ideal of $A^G$ (same reason as in the answer of user74230). Since $A$ is normal, each of the local rings has a unique mimimal prime. Hence $A$ has finitely many minimal primes. A normal ring with finitely many minimal primes is a finite product of normal domains. The case of a product of normal domains is easy.


