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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.

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    $\begingroup$ Your case of interest is treated affirmatively in the book by Bosch, Guntzer, Remmert, in Chapter 6. $\endgroup$
    – grghxy
    Commented May 29, 2015 at 14:28
  • $\begingroup$ BGR assumes $G$ acts over the ground field, as you also presumably are fine with assuming too. $\endgroup$
    – grghxy
    Commented May 29, 2015 at 14:31
  • $\begingroup$ Yes, in my case the action is over the ground field. $\endgroup$
    – Antoine Ducros
    Commented May 29, 2015 at 14:34
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    $\begingroup$ I suspect that in general, the universally catenary condition may be a source of trouble. $\endgroup$
    – Laurent Moret-Bailly
    Commented May 29, 2015 at 15:34

1 Answer 1

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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".

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  • $\begingroup$ Note that Example 13 in Koll\'ar's paper also gives an example with $A$ regular, but only in characteristic two. $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 23, 2015 at 14:42

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