Let $k$ be a field, let $G$ be a linear algebraic group over $k$ and let $A$ be a commutative $k$-algebra which is acted on by $G$.

We say that an $A$-module $M$ is a $(G,A)$-module if it satisfies the following two properties:

1) $M$ is a rational $G$-module (over $k$)

2) The multiplication map $A \otimes M \rightarrow M$ is a morphism of rational $G$-modules.

If we let $X=\text{Spec}(A)$ and view $M$ as a quasi-coherent sheaf on $X$, then giving $M$ the structure of a $(G,A)$-module should be equivalent to giving it the structure of a $G$-equivariant sheaf on $X$ (see https://en.wikipedia.org/wiki/Equivariant_sheaf). Is there an easy, explict way to see why these two definitions coincide?

  • $\begingroup$ No, because you only give one definition. What you could do is give a definition of $G$-equivariant sheaf that matches. $\endgroup$ Jun 9, 2016 at 7:18
  • $\begingroup$ @WilberdvanderKallen : en.wikipedia.org/wiki/Equivariant_sheaf $\endgroup$ Jun 9, 2016 at 10:28
  • $\begingroup$ @Wilberdvanderkallen: I apologize, I should linked the Wikipedia article... $\endgroup$
    – gustav101
    Jun 9, 2016 at 12:53
  • $\begingroup$ Here are some thoughts, which are not entirely rigorous. I would like to write these as a comment but I do not have enough reputation. The kernel of the surjective multiplication map $A\otimes M\rightarrow M$ is both a $G$-module and a sub $A$-module. Therefore, the kernel is a $(G,A)$-module (The $(G,A)$-module structure on $M\otimes A$ comes from pulling back the $(G,k)$-module $M$ over Spec $k$). In my understanding we should be able to give a quotient of $(G,A)$-modules a natural structure of a $(G,A)$-module. I gathered these based on ch3 in d-nb.info/98519670X/34. $\endgroup$ Jun 13, 2022 at 19:38


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