Prove there is a bijection between the set of $R[x]$-module structures on $M$ and $\mathrm{End}_{R-\mathsf{Mod}}(M)$. Let $R$ a commutative ring and $M$ a $R$-module. Prove there is a bijection between the set of $R[x]$-module structures on $M$ and $\mathrm{End}_{R-\mathsf{Mod}}(M)$.
I really don't even know how to start with this question.
 A: Hint:
If $f\colon M\longrightarrow M$ is any  $R$-endomorphism of $M$, show the map
\begin{align}
R[X]\times M&\longrightarrow M\\
(P(X),m)&\longmapsto P(f)(m)
\end{align}
turns $M$ (with its group structure) into an $R[X]$-module.
A: A little late, but I want to give another perspective on this fact...
Let $R$ be a unital commutative ring. It is well-known that if $S$ is a unital (not necessarily conmmutative) ring, and $\mu : R \to S$ is a fixed ring homomorphism, then for every $s \in S$ such that $$(\forall r \in R) \quad s\mu(r) = \mu(r)s$$ there exists a unique ring homomorphism $\mu_s : R[x] \to S$ that extends $\mu$ and sends $x$ to $s$. This is the universal property of the polynomial rings.
In other words, there is a bijection $$\{\nu \in \operatorname{Hom}_{\textsf{Ring}}(R[x],S) : \nu|_R = \mu\} \longrightarrow \{s \in S : (\forall r \in R) \ s\mu(r) = \mu(r)s\}$$ given by $\nu \mapsto \nu(x)$ (with inverse $s \mapsto \mu_s$).
Now, when we consider an $R$-module $M$, let $S = \operatorname{End}_{\textsf{Ab}}(M)$ (recall, with point-wise addition and composition as multiplication) and $\mu : R \to S$ such that $\mu(r)(m) = rm$ for all $r \in R$ and $m \in M$. In this case, $\{\nu \in \operatorname{Hom}_{\textsf{Ring}}(R[x],S) : \nu|_R = \mu\}$ is (in bijection with) the set of all $R[x]$-module structures on $M$ extending the given $R$-module structure, and $\{s \in S : (\forall r \in R) \ s\mu(r) = \mu(r)s\}$ is exactly the same as $\operatorname{End}_{R\textsf{-Mod}}(M)$.
