I want to prove the following:
Let $R$ be a commutative ring and $M$ is an abelian group then $M$ is an $R$-module iff there exists a ring homomorphism $f: R \rightarrow End_Z(M).$
My professor said that its proof is elementary but I can not see that, could someone push me in the right direction of the proof please?
EDIT:
An abelian group $M$ can be given the structure of $R$-module if and only if there exists a ring homomorphism $$ \varphi\colon R\to\operatorname{End}_{\mathbb{Z}}(M) $$ If $M$ is given a structure of $R$-module, then the homomorphism $\varphi$ is defined by $\phi(r)\colon x\mapsto rx$. Conversely, given the ring homomorphism $\varphi$, we can define $rx=\varphi(r)(x)$.
Is this a valid proof?
I will give you the corresponding maps/actions and you can then verify that they satisfy the desired properties:
Let $M$ be an $R$-Module. Define $f:R\to \operatorname{End}_{\Bbb Z}(M)$ by $f(r)=g_r$ where $g_r:M\to M$ is given by multiplication by $r$, i.e. $g_r(m)=r\cdot m$. Using the module properties check that this $f$ gives a well-defined homomorphism.
Conversely given such a morphism $f:R\to \operatorname{End}_{\Bbb Z}(M)$, define the $R$-module structure on $M$ by $r\cdot m= f(r)(m)$ for $r\in R,m\in M$. Here as well we can check that this indeed defines an $R$-module structure.
Furthermore these two constructions are inverse to each other.
Edit: For $r,s\in R$ we have: $$(r+s)\cdot m:=f(r+s)(m)=(f(r)+f(s))(m)=f(r)(m)+f(s)(m)=r\cdot m+s\cdot m$$ and $$(rs)\cdot m=f(rs)(m)=(f(r)\circ f(s))(m)=f(r)(f(s)(m))=f(r)(s\cdot m)=r\cdot (s\cdot m)$$ The other properties can be proved similarly.