$R$-modules are also $\mathbb Z$-modules Simple question: 
Why all $R$-modules are also $\mathbb Z$-modules and $$End_R(V) \subseteq End_{\mathbb Z}(V)$$ is a subring? 

If $R$ is a ring with 1, then there is a unique homomorphism $\phi\colon \mathbb Z \to R$.
The action of $\mathbb Z$ on a (unital) $R$-module $V$ is realized via this map. 
That is, for $v \in V$ and $n \in \mathbb Z$,
$$
  n \cdot v = \phi(n) \cdot v.
$$
The fact that $\operatorname{End}_R(V) \subset \operatorname{End}_{\mathbb Z} (V)$
follows since $\phi(\mathbb Z) \subset R$.
Easier way of saying this: every module is an abelian group.
