How do you get that additive answer when the action on R isn’t trivial?

Well my thinking was that the G action has been trivialised on R/G, so the G-action on Map(R, R/G) [from acting on the first R] is also trivial. But I'm not really convinced.

@HarryGindi Harry -- though I gather that Denis has given you Clark's thesis by now, it is worth pointing out that most universities subscribe to ProQuest, and on that website you can download most PhD theses that are not too old (certainly almost everything post 2000 in the US is available). Here's a link to Clark's thesis, which I just downloaded: drive.google.com/file/d/1ABAFll1bvuAonCo6M1bUs42-R0xBI3TH/…

(I'll now return to lurking; I drop by here occasionally while wasting time doing other things)

@AndyPutman Thanks! Denis sent it to me yesterday though. Also I'm completely puzzled how to access journals here because all of the instructions are in German…

@TomBachmann but presumably the action on R is not R-linear, otherwise it would be trivial

@DylanWilson Why would that make it trivial?

Ah, sorry- I thought in the original statement G was acting by ring endomorphisms, in which case adding on R-linearity would make it trivial.

I think it's more like a character acting in the natural way.