In unpublished notes by Yi Hu (which appear to be no longer online), I found the following:

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 $(R/J)^G = R^G /(J \cap R^G )$.

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.

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    $\begingroup$ Welcome new contributor. Appendix A to Chapter 1 in "Geometric Invariant Theory" recalls that geometrically reductive groups in characteristic $0$ are linearly reductive, i.e., every rational representation is completely reducible. Thus, the invariant subspace is a summand that is complemented, and every subrepresentation is the direct sum of its intersection with these two summands. That is, of course, meant for finite dimensional representations. But it is a standard argument (included in the proof on the next page) that $R$ is an increasing union of finite dimensional representations. $\endgroup$ – Jason Starr Sep 2 '18 at 1:31

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.


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