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Let $G$ be a geometrically reductive algebraic group over an algebraically closed field $k$ . Let $X$ be an affine variety over $k$ on which $G$ acts regularly. Then $G$ acts on the coordinate ring $A$ of $X$ by automorphisms, and we denote by $A^G$ the subring consisting of invariant elements.

Problem: Prove that the inclusion $A^G\subset A$ induces a surjective map of the ring spectra.

Background: A reffined statement is that for a reducitve group action on an affine variety, the categorical quotient map is surjective.

In the book "Lectures on Invariant Theory" by Igor Dolgachev, it is proved that such a categorical quotient is "good" in the sense of G.I.T., but it seems that his proof is incomplete and the probleme above is the missing part.

flag	asked Aug 21 2011 at 19:28 Xin Nie 384●1●8

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I don't have an answer to your problem, but I checked Dolgachev's book and he seems to prove the surjectivity as well, not in theorem 6.1 but in proposition 6.2, to which he refers in the proof of the theorem (and which despite the numbering comes before the theorem.. weird) – Mattia Talpo Aug 22 2011 at 0:31

Once one knows that the image of a G-invariant closed set is closed, the surjectivity follows, because the map is dominant. This assumption is among the hypothesis of proposition 6.2, so there is nothing nontrivial in the proof. In the proof of theorem 6.1 he tried to verify the assumption above for any reducitve group action on affine variety. It is this argument which i found unsatisfactory. (the gap is, if one does not know surjectivity, then $p^{-1}(y)$ may be empty, and this dose not lead to a contradiction) – Xin Nie Aug 22 2011 at 6:27

Ok, I see what you mean now.. – Mattia Talpo Aug 22 2011 at 9:10

Does the proof of Theorem 1.2.4(ii) in www-fourier.ujf-grenoble.fr/~mbrion/… do the trick? – Bart Van Steirteghem Sep 29 2011 at 20:29

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