Proof of Commutativity of the ring of endomorphisms over integers

Question

First I searched for some similar problems and studied some of the given references, but I was not able to understand them. So I pose the question, hopefully to reveal some aspects to a less advanced audience as myself. The problem is how to prove that the algebraic structure $ \operatorname{End}(\mathbb{Z},+,\circ)$ (The set of all Homomorphisms from $\mathbb{Z}$ into $\mathbb{Z}$ is a commutative ring, where $+$ and $\circ$ are binary operation of addition and composition of such functions respectively. I was able to prove that it is a ring, but I cannot prove how it is commutative. Specifically, how should I prove that for an arbitrary integer $a$ and for arbitrary $f,g \in \operatorname{End}(\mathbb{Z})$, we have $(f\circ g)(a)=(g\circ f)(a)$?

Answer 1

Every endomorphism of $\mathbb{Z}$ is determined by what it does to $1$, since $f(x + 1) = f(x) + f(1)$, so $f(n) = n f(1)$ by induction. Therefore $\mathrm{End}(\mathbb{Z}) \cong \mathbb{Z}$.

In general an endomorphism of $\mathbb{Z}^n$ is determined by what it does to the basis vectors $e_1, \dots, e_n$ by a similar argument. Hence $\mathrm{End}(\mathbb{Z}^n)$ is isomorphic to the ring $M_n(\mathbb{Z})$ of $n \times n$ matrices with integer entries. This is commutative iff $n = 1$.

Answer 2

If $f: \mathbb{Z}\rightarrow \mathbb{Z}$ is a unital ring homomorphism you get the following:

$$f(m)=f(1+\cdots +1)=f(1)+ \cdots +f(1)=mf(1)=m$$

hence a unital ring endomorphism of $\mathbb{Z}$ is the identity. Hence $End(\mathbb{Z})$ has only one element: $Id$