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This is probably well known, but I'm trying to understand "equivariant maps" over different rings. Namely, let $R,S$ be rings and let $\phi:R\to S$ be a ring homomorphism. Let $M$ be an $R$-module with action $\rho:R\otimes M\to M$, and $M'$ an $S$-module with $\rho':S\otimes M'\to M'$. Then consider maps $f:M\to M'$ such that the following diagram commutes: $$ \require{AMScd} \begin{CD} R\otimes M @> \rho > > M\\ @V \phi\otimes f V V @V V f V\\ S\otimes M' @> > \rho' > M' \end{CD} $$

For $R=S$, $\phi=\text{id}_R$, we get regular intertwiners for modules over $R$; but one could as well consider some other non-trivial isomorphism of $R$. For $M=M'$ we could think about base change from $S$-mod to $R$-mod, but those don't seem to have anything to do with such an equivariance.

Question:

Are these non-trivial and/or interesting objects? How do you call them, and where can I know more about them?


Just came across slight variations of this question:

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These maps are called $R$-module homomorphisms, where $M'$ is regarded as an $R$-module via restriction along the homomorphism $\phi$. They are of course non-trivial and very interesting objects, in general.

For instance, Frobenius reciprocity is the statement that the obvious map $$\mathrm{Hom}_R(M, \mathrm{Res}(M')) \to \mathrm{Hom}_S(S \otimes_R M, M')$$ is an isomorphism. The domain of this map is the set of maps you are looking at.

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  • $\begingroup$ Cool, after a little thinking, this is exactly what I was looking for. Thank you. $\endgroup$
    – Carlos
    Apr 27, 2018 at 17:04
  • $\begingroup$ The existence of a ring homomorphism $\phi:R\to S$ automatically makes $S$ into an $(R,S)$-bimodule, and that's what trivally get us the above Frobenius adjunction between $R$-mod and $S$-mod. Furthermore, any ring $R$ automorphism induces an automorphism of its category of $R$-modules. Of course, for a given module $M$ there might exist no non-trivial $R$-module homomorphisms between $M$ and its restriction $\text{Res}_\phi M$ along $\phi$ (cf. example here). $\endgroup$
    – Carlos
    Apr 27, 2018 at 17:07
  • $\begingroup$ It would be interesting to characterize $\text{Hom}_R(M,\text{Res}_\phi M)$. Indeed, there's a short exact sequence of groups $$ 1\to \text{Aut}_R(M)\hookrightarrow \text{Hom}_R(M,\text{Res}_\phi M)\to \text{Aut}(R)\to 1\,, $$ where the first map is given by considering $\phi=\text{id}_R$, and the last map forgets $f$. $\endgroup$
    – Carlos
    Apr 27, 2018 at 17:08

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