# Excellent rings

If $A$ is an excellent commutative ring and $G$ is a finite group of automorphisms of $A,$ is the invariant subring $A^G$ still excellent? I think this is false -- because if not it would probably be written in EGA IV, or in the recent Astérisque volume about Gabber's works on uniformisation and étale cohomology. But if anybody has a counter-example...

In fact, in the case I'm interested in, $A$ is not «any» excellent ring but an affinoid algebra over a non-Archimedean, complete field. Does one know something about $A^G$? I doubt that it is automatically affinoid -- if yes, it would be excellent.

• Your case of interest is treated affirmatively in the book by Bosch, Guntzer, Remmert, in Chapter 6. – grghxy May 29 '15 at 14:28
• BGR assumes $G$ acts over the ground field, as you also presumably are fine with assuming too. – grghxy May 29 '15 at 14:31
• Yes, in my case the action is over the ground field. – Antoine Ducros May 29 '15 at 14:34
• I suspect that in general, the universally catenary condition may be a source of trouble. – Laurent Moret-Bailly May 29 '15 at 15:34

Nagata's example of a group of order $p$ acting on the ring of dual numbers of a field is a counter example: the ring of invariants is not even Noetherian. You can find it as Example 12 in Koll\'ar's paper "Quotients by finite equivalence relations".
• Note that Example 13 in Koll\'ar's paper also gives an example with $A$ regular, but only in characteristic two. – Ariyan Javanpeykar Jun 23 '15 at 14:42