$R$-modules are also $\mathbb Z$-modules

Question

Simple question:

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

Answer 1

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.

Answer 2

Why are all $R$-modules also $\Bbb Z$-modules?

Any abelian group is a $\Bbb Z$-module in a canonical way. This is why:

1. For any abelian group $V$, its set of group endomorphisms $\mathrm{End}_{Ab}(V)$ is a unital ring under addition and composition.
2. To specify a left $R$-module structure on $V$ is to specify a ring homomorphism $\rho:R\to\mathrm{End}_{Ab}(V)$ (sending $1_R$ to $\mathrm{id}_{V}$ if $R$ is unital).
3. The ring of integers $\Bbb Z$ has a special property: for any unital ring $S$ there is a unique ring homomorphism $\Bbb Z\to S$, which, by definition of a homorphism of unital rings, must send $1_{\Bbb Z}$ to $1_S$. (One says that $\Bbb Z$ is an initial object in the category of unital rings with ring homomorphisms.)

As a consequence, any abelian group $V$ admits a unique $\Bbb Z$-module structure specified by the unique homomorphism of unital rings $\Bbb Z\to\mathrm{End}_{Ab}(V)$. Furthermore, any endomorphism of the abelian group $V$ is an endomorphisms of $V$ equipped with its canonical $\Bbb Z$-module structure, that is $$\mathrm{End}_{\Bbb Z}(V)=\mathrm{End}_{Ab}(V)$$ Indeed, by definition, for an $R$-module $V$ (with structure map $\rho:R\to\mathrm{End}_{Ab}(V)$), $$\mathrm{End}_{R}(V)=\left\lbrace\phi\in\mathrm{End}_{Ab}(V)\mid\forall r\in R,\rho(r)\circ\phi=\phi\circ\rho(r)\right\rbrace\,.$$ In the case of the canonical $\Bbb Z$-module structure $\mathrm{End}_{\Bbb Z}(V)$ is the set of all group endomorphisms of $V$ that commute to the identity map, which is to says that $\mathrm{End}_{\Bbb Z}(V)$ actually equals $\mathrm{End}_{Ab}(V)$.

Why is $\mathrm{End}_{R}(V)$ is a subring of $\mathrm{End}_{\Bbb Z}(V)$?

We just brought attention to the fact that $\mathrm{End}_{\Bbb Z}(V)=\mathrm{End}_{Ab}(V)$, so the question may be rephrased as

Why is $\mathrm{End}_{R}(V)$ is a subring of $\mathrm{End}_{Ab}(V)$?

It follows from the definition: $\mathrm{End}_{R}(V)=\left\lbrace\phi\in\mathrm{End}_{Ab}(V)\mid\forall r\in R,\rho(r)\circ\phi=\phi\circ\rho(r)\right\rbrace$ is the commutant of the subset (actually subring) $\rho(R)\subset\mathrm{End}_{Ab}(V)$. Indeed, it is an easy exercise to show that the commutant $\mathrm{Comm}(X)$ of any subset $X\subset S$ is a subring of $S$, where $S$ is an arbitrary ring and the commutant of $X$ is the subset of elements $s$ of $S$ that commute to $X$, i.e. the subset of those $s\in S$ such that for all $x\in X,\,sx=xs$.