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I'm trying to generalize one aspect of the accepted answer in: $\text{Hom}(\mathbb{F}_p G, M)$ and $H^1(G,M)$

Is the following statement correct?

If:

  1. $R_1$ and $R_2$ are rings;
  2. $F:R_1-\text{mod}\rightarrow R_2-\text{Mod}$ is an additive functor;
  3. $M \in R_1-\text{Mod}$.

Then:

  1. $F(M)$ is a $\text{Hom}_{R_1}(M,M)$-module where
  2. the action of $\text{Hom}_{R_1}(M,M)$ on $F(M)$ is given by $f\cdot x=F(f)(x)$.
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1 Answer 1

up vote 1 down vote accepted

Yes. You can see this easily by noting that the action of $F$ on endomorphism rings is a ring homomorphism, by virtue of the fact that $F$ is an additive functor: so in particular we get a ring homomorphism $\textrm{End}_{R_1} (M) \to \textrm{End}_{R_2}(F M)$. In fact more is true: the action of $\textrm{End}_{R_1} (M)$ on $F M$ will be $R_2$-linear. (Recall the fact that a left $R$-module $M$ is the same thing as an abelian group $M$ and a ring homomorphism $R \to \textrm{End}_{\mathbb{Z}}(M)$.)

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