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.

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    $\begingroup$ The formation of $A^G$ (ignore normality) commutes with flat base change on $A^G$, and normality is preserved by henselization, so WLOG $A^G$ is local henselian. Since $G$ is transitive on each fiber of Spec($A$) over Spec($A^G$) (Atiyah-MacDonald exercise), $A$ is semi-local. Since $A$ is the direct limit of module-finite subalgebras, each a direct product of local algebras (as $A^G$ is henselian) whose maximal ideals lift to those of $A$, clearly $A$ is the direct product of local algebras (so domains!), with $G$ transitive on the set of factors. That reduces to the domain case. QED $\endgroup$
    – user74230
    Mar 3, 2015 at 3:46
  • $\begingroup$ @user74230: Thank you, this is very helpful. $\endgroup$
    – Lisa S.
    Mar 3, 2015 at 4:19

1 Answer 1


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.


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