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Aug 11, 2016 at 11:39 comment added Friedrich Knop That $R^G$ is a direct summand of $R$ follows from complete reducibility. The projection $R\to R^G$ is sometimes called the Reynolds operator. The purity of the inclusion of a direct summand is considered "obvious" by Hochster-Roberts in section 6 of their famous paper "Rings of invariants [$\ldots$] are Cohen-Macaulay".
Aug 11, 2016 at 1:01 comment added Avi Steiner Do you have a reference for this fact about direct summands?
Aug 11, 2016 at 0:53 comment added Friedrich Knop The injection $R^G\hookrightarrow R$ is always pure since $R^G$ is a direct summand of $R$ as an $R^G$-module. So "pure injective" is no restriction at all, at least in characteristic zero.
Aug 11, 2016 at 0:49 comment added Avi Steiner Do you know if there's anything known about the less restrictive condition that the inclusion of rings $R^G\to R$ is pure injective?
Aug 11, 2016 at 0:46 vote accept Avi Steiner
Aug 11, 2016 at 0:44 history answered Friedrich Knop CC BY-SA 3.0