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.

  • 3
    $\begingroup$ 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 '15 at 3:46
  • $\begingroup$ @user74230: Thank you, this is very helpful. $\endgroup$
    – Lisa S.
    Mar 3 '15 at 4:19

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.