Is the endomorphism ring of a module over a non-commutative ring always non-commutative? I was reading the "Undergraduate Commutative Algebra". It formalises the definition of Module.
Consider M, an A-module where A is a ring.
It defines $\mu_f : M \to M$ for the map $m \mapsto fm $, where $f \in A$.
Then the text claims that $f \mapsto \mu_f$ is a ring homomorphism $A \to \operatorname{End}(M)$ from A to the noncommutative ring of endomorphisms of M.
So, am I correct to think that in this case $\operatorname{End}(M)$ is a noncommutative ring because $A$ is not commutative?
$\operatorname{End}(M)$ seems to commutative if $A$ is commutative.
--update
Sorry, I made a mistake. I was thinking about $\operatorname{End}(M)$.
 A: 
So, am I correct to think that in this case
  $End M$
  is a noncommutative ring because
  $A$
  is not commutative?

No, not necessarily. There isn’t a connection. 
You can have $End(M)$ noncommutative and $A$ commutative ($A=\mathbb Z$ and $M=C_2\times C_2$)
You can also have $A$ noncommutative and $End(M)$ commutative (for this you can take a ring $A$ which isn’t commutative, but which has a unique maximal right ideal $I$ such that $A/I$ is commutative, and let $M=A/I$.) 
A: It is true that $f \mapsto \mu_f$ defines a ring homomorphism
$$\begin{align*}\mu : A &\to \operatorname{End}_{\mathbb Z}(M) \\
f & \mapsto \mu_f
\end{align*}$$
Thus when $A$ is noncommutative and $\mu$ is injective, $\operatorname{End}_{\mathbb Z}(M)$ is noncommutative. The point is that $\mu$ is not necessarily injective.
The module $M$ is called "faithful" when $\mu$ is injective. So, a priori, there could exist examples of non-faithful modules over noncommutative rings with commutative endomorphism ring.
A silly example is the following: take any ring $A'$ and an $A'$-module $M'$ with commutative endomorphism ring. Take any noncommutative ring $B$. Define $A = A' \times B$ and define the $A$-module structure on $M$ by restricting scalars through the projection $A' \times B \to A'$. Then $A$ is noncommutative, even though $\operatorname{End}_{\mathbb Z}(M)$ is commutative.
Concretely, take e.g. $A = \mathbb Z \times M_2(\mathbb Z)$ and $M = \mathbb Z$, where $(a, b) \in A$ acts on $n \in M$ by $(a, b)n = an$. Then $\operatorname{End}_{\mathbb Z}(M) \cong \mathbb Z$ is commutative.

Conversely, for modules with noncommutative endomorphism rings over commutative rings: There are examples which you've certainly encountered before: let $k$ be a field, and $M$ an $n$-dimensional $k$-vector space. The endomorphism ring $\operatorname{End}_{\mathbb Z}(M)$ contains the ring $\operatorname{End}_{k}(M) \cong M_n(k)$. So if $\operatorname{End}_{\mathbb Z}(M)$ is commutative, so is the subring $M_n(k)$. However, the latter is not commutative for $n > 1$.
Thus $\operatorname{End}_{\mathbb Z}(M)$ is not commutative for $n > 1$, even though $k$ is commutative.
A: You can start by proving that:
$$ \mu_{g}\circ\mu_{f} = \mu_{gf},\ \forall\, f,g\in A $$
and use that $M$ is a $A$-module to argue that:
$$ (a_{1}a_{2})m=a_{1}(a_{2}m),\  \forall\,m\in M,\forall\, a_{1},a_{2}\in A$$
A: For a specific counterexample, take $k$ to be any field, $A = k\{x,y\}$ (polynomials over $k$ in two non-commuting variables), and $M = k$, considered as an $A$-module, where for $a \in k$ and $p \in k\{x,y\}$, we take $pa$ to be $ap(0,0)$. Then $A$ is certainly non-commutative, but you can show $\mathrm{End}(M) \cong k$, which is commutative.
