Recall that if $A, B$ are rings with unity then a ring homomorphism $f: A \to B$ is a map satisfying

$f(x+y)=f(x)+f(y)$

$f(xy)=f(x)f(y)$

$f(1_A)=1_B$

for all $x, y \in A$ and $1_A, 1_B$ are unity elements of $A$ and $B$, respectively.

Proof.

We claim that there is one and only one ring homomorphism from $\Z$ to $R$.

Let us first remark that there is at least one ring homomorphism $\Z \to R$.
Define the map $f_0:\Z\to R$ by
\[f(n)=n.\]
Then it is clear that $f_0$ is a ring homomorphism from $\Z$ to $R$.
We want to prove that this is the only ring homomorphism.

Suppose that $f:\Z\to R$ is a ring homomorphism.
By definition, we must have
\[f(1)=1_R.\]
Using property (1) with $x=y=0$, we see that
\[f(0)=f(0)+f(0).\]
Thus, we have $f(0)=0$.
Next, we apply (1) with $x=1, y=-1$ and obtain
\[0=f(0)=f(1+(-1))=f(1)+f(-1).\]
Thus we have
\[f(-1)=-f(1)=-1_R.\]

We want to determine the value $f(n)$ for any $n\in \Z$.
If $n$ is a positive integer, then we can write it as
\[n= \underbrace{1+\cdots+1}_{n\text{ times}} \]
By property (a) applied repeatedly, we have
\begin{align*}
f(n) &= \underbrace{f(1)+\cdots+f(1)}_{n\text{ times}} \\
&=\underbrace{1_R+\cdots+1_R}_{n\text{ times}}=n.
\end{align*}
If $n$ is a negative integer, we express it as
\[n=\underbrace{(-1)+(-1)+\cdots+(-1)}_{n\text{ times}}\]
and obtain
\begin{align*}
f(n)&=\underbrace{f(-1)+f(-1)+\cdots+f(-1)}_{n\text{ times}}\\
&=\underbrace{(-1_R)+(-1_R)+\cdots+(-1_R)}_{n\text{ times}}=n.
\end{align*}
Therefore, we have proved that
\[f(n)=n\]
for any $n\in \Z$. Hence any ring homomorphism from $\Z$ to $R$ is the ring homomorphism $f_0$ that we saw at the beginning of the proof.

In conclusion, there is exactly one ring homomorphism from $\Z$ to $R$, which is given by
\[f_0(n)=n\]
for any $n\in\Z$.

Comment.

In category theory, we say that the ring of integers $\Z$ is an initial object in the category of rings with unity.

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