I'm new to modules and I'm trying to see the following equivalence:
"If $A$ is a commutative ring, $R$ an $A$-algebra (i.e. a ring $R$ with a homomorphism $i : A \rightarrow Z(R)$) and $M$ an Abelian group, giving $M$ an $R$-module structure is equivalent to giving it an $A$-module structure with an $A$-algebra homomorphism $f : R \rightarrow \text{End}_A(M)$."
Now I've seen such a question for the more obvious case where an $R$-module structure is equivalent to a homomorphism $\psi : R \rightarrow \text{End}(M)$. Here it's easy to show that given a module structe $(M, \phi)$ where $\phi : R \times M \rightarrow M$ satisfies the module axioms, you can simply take $\psi : r \mapsto (m \mapsto \phi(r,m))$ and check that this is indeed a homomorphism. Conversely you can define $\phi : (r,m) \mapsto (\psi(r))(m)$ and check that it satisfies the module axioms. I even have a certain 'visualisation' with this. The sentence: "given an $r \in R$, I can make any $m \in M$ into another element of $M$, depending on $r$" seems to describe both cases.
In the current case however, this visualisation is lost on me, and I have no more intuition for what's happening exactly. Let alone proving the equivalence.
Any help is greatly appreciated.