TITLE: Excellent rings
QUESTION [13 upvotes]: 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.

REPLY [7 votes]: 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".