In unpublished notes by Yi Hu (which appear to be no longer online), I found the following:



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> Corollary 2.4.5. Let the characteristic of $k$ is zero. Assume that a reductive group $G$ acts rationally on a finitely generated $k$-algebra $R$. Let $J$ be an ideal in $R$, invariant under $G$. Then $(R/J)^G = R^G /(J \cap R^G )$.
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Essentially, for affine varieties, we can exchange the operations "taking a subvariety" and "taking a quotient".


I have been unable to find a published source for this fact even though it seems fairly basic. I looked in Mumford & Fogarty's book but could not find anything so simple as an affine variety.



Reference on reductive group acting on quotient algebra
With some help, I now see that this result is not so hard. As Jason points out, we get a $k[G]$-module complement to $J$ in $R$, call it $C$. Obviously the projection $R \rightarrow C$ restricts to give a surjective map $R^G \rightarrow C^G$. But this is the same as the natural map $R^G \rightarrow (R/J)^G$, so that map is surjective, from which the claim is easy.


