Definition. Let $A$ be a commutative ring and $E$ an associative, unital algebra. A left $E$-module is a left module over the underlying
ring of $E$.
If $M$ is an $E$-module, the homomorphism $\eta:A\rightarrow E$ then
defines on $M$ an $A$-module structure, said to be underlying the
$E$-module structure on $M$; for $\alpha\in A$, $s\in E$, $x\in M$,
$$\alpha(sx)=s(\alpha x)=(\alpha s)x,$$ so that for all $s\in E$, the
homothety $h_s:x\mapsto sx$ of $M$ is an endomorphism of the
underlying $A$-module structure. Conversely, being given an $E$-module
structure on $M$ is equivalent to being given an $A$-module structure
on $M$ and an $A$-algebra homomorphism $s\mapsto h_s$ of $E$ into
$\text{End}_A(M)$.
I am not sure what equivalence is being asserted here.
Let $M$ be an $E$-module. Then clearly $\alpha. x:=\eta(\alpha)x$, $\alpha\in A$ and $x\in M$, defines an $A$-module structure on $E$. Moreover, writing $h_s(x):=sx$, $s\in E$ and $x\in M$, we get an $A$-algebra homomorphism from $E$ into $\text{End}_A(M)$. So this shows that from the datum of an $E$-module structure on $M$ we can extract an $A$-module structure and an $A$-algebra homomorphism.
Now, what about the other direction? Suppose we are given an $A$-module on $M$ and an $A$-algebra homomorphism $\phi:E\rightarrow\text{End}_A(M)$. Obviously, $\phi$ is also a ring homomorphism from $E$ into $\text{End}_{\mathbb{Z}}(M)$ and so we have an $E$-module structure on $M$. But is there a relation between the $A$-module that we have been given and the $A$-module defined by $\alpha.x:=\eta(\alpha)x$? Shouldn't they be the same?
Best Answer
Yes, they should. That is because you asked that $\phi$ is an $A$-algebra morphism, and the $A$-algebra structure on $\operatorname{End}_A(M)$ is defined with reference to the pre-established $A$-module structure on $M$.
This means you can't just pick any ring morphism $\phi$, you have to pick one such that the $A$-module structure it induces on $M$ agrees with the one you had chosen.