[proofplan]
We construct the map by sending an integer represented as a formal difference $a-b$ to the corresponding difference of $a$ copies and $b$ copies of $1_R$. The main point is well-definedness: equivalent formal differences must give the same element of $R$. Once this is checked, addition and multiplication follow from the recursive arithmetic of repeated sums of $1_R$, and uniqueness follows because every unital homomorphism is forced by the image of $1$.
[/proofplan]
[step:Define repeated sums of the unit in $R$]
Let $\mathbb{N} = \{1,2,3,\dots\}$ denote the set of positive integers, and let $\mathbb{N}_0 = \mathbb{N} \cup \{0\}$. Define a function
\begin{align*}
s:\mathbb{N}_0\to R
\end{align*}
recursively by $s(0)=0_R$ and $s(k+1)=s(k)+1_R$ for every $k\in\mathbb{N}_0$. Thus, for $k>0$, the element $s(k)$ is the sum of $k$ copies of $1_R$.
We record the two elementary identities needed below. For all $a,b\in\mathbb{N}_0$,
\begin{align*}
s(a+b)=s(a)+s(b).
\end{align*}
For all $a,b\in\mathbb{N}_0$,
\begin{align*}
s(ab)=s(a)s(b).
\end{align*}
The first identity follows by induction on $b$: the case $b=0$ is $s(a)=s(a)+0_R$, and the successor step uses
\begin{align*}
s(a+(b+1))=s((a+b)+1)=s(a+b)+1_R=(s(a)+s(b))+1_R=s(a)+s(b+1).
\end{align*}
The second identity follows by induction on $b$: the case $b=0$ is $s(0)=0_R=s(a)0_R$, and the successor step uses distributivity in $R$:
\begin{align*}
s(a(b+1))=s(ab+a)=s(ab)+s(a)=s(a)s(b)+s(a)1_R=s(a)(s(b)+1_R)=s(a)s(b+1).
\end{align*}
[/step]
[step:Construct the homomorphism from formal differences]
By [citetheorem:9700], every integer is represented by an equivalence class of a pair $(a,b)\in\mathbb{N}_0\times\mathbb{N}_0$, where $(a,b)$ represents the formal difference $a-b$, and
\begin{align*}
(a,b)\sim(c,d)\iff a+d=b+c.
\end{align*}
Define
\begin{align*}
\varphi:\mathbb{Z}\to R
\end{align*}
by
\begin{align*}
\varphi([(a,b)])=s(a)-s(b).
\end{align*}
This definition is independent of the chosen representative. Suppose $(a,b)\sim(c,d)$. Then $a+d=b+c$, so the additivity identity for $s$ gives
\begin{align*}
s(a)+s(d)=s(a+d)=s(b+c)=s(b)+s(c).
\end{align*}
Adding the additive inverses $-s(b)$ and $-s(d)$ to both sides gives
\begin{align*}
s(a)-s(b)=s(c)-s(d).
\end{align*}
Thus $\varphi$ is well-defined.
[/step]
[step:Verify that the constructed map preserves addition]
Let $m,n\in\mathbb{Z}$. Choose representatives $m=[(a,b)]$ and $n=[(c,d)]$ with $a,b,c,d\in\mathbb{N}_0$. In the formal-difference construction, addition is
\begin{align*}
[(a,b)]+[(c,d)]=[(a+c,b+d)].
\end{align*}
Therefore,
\begin{align*}
\varphi(m+n)=\varphi([(a+c,b+d)])=s(a+c)-s(b+d).
\end{align*}
Using the additivity identity for $s$,
\begin{align*}
s(a+c)-s(b+d)=(s(a)+s(c))-(s(b)+s(d)).
\end{align*}
Rearranging in the additive group of $R$ gives
\begin{align*}
(s(a)+s(c))-(s(b)+s(d))=(s(a)-s(b))+(s(c)-s(d)).
\end{align*}
Hence
\begin{align*}
\varphi(m+n)=\varphi(m)+\varphi(n).
\end{align*}
[guided]
We must show that the definition of $\varphi$ respects the addition law on formal differences. Let $m,n\in\mathbb{Z}$, and write $m=[(a,b)]$ and $n=[(c,d)]$, where $a,b,c,d\in\mathbb{N}_0$. The point of these representatives is that $[(a,b)]$ behaves like $a-b$ and $[(c,d)]$ behaves like $c-d$.
The addition rule in the construction of $\mathbb{Z}$ is
\begin{align*}
[(a,b)]+[(c,d)]=[(a+c,b+d)].
\end{align*}
Applying the definition of $\varphi$ to this sum gives
\begin{align*}
\varphi(m+n)=\varphi([(a+c,b+d)])=s(a+c)-s(b+d).
\end{align*}
Now we use the identity $s(x+y)=s(x)+s(y)$ for natural-number sums of the unit. It gives
\begin{align*}
s(a+c)=s(a)+s(c)
\end{align*}
and
\begin{align*}
s(b+d)=s(b)+s(d).
\end{align*}
Substituting these identities yields
\begin{align*}
\varphi(m+n)=(s(a)+s(c))-(s(b)+s(d)).
\end{align*}
Since the additive structure of a ring is an [abelian group](/page/Abelian%20Group), we may reassociate and commute the additive terms:
\begin{align*}
(s(a)+s(c))-(s(b)+s(d))=(s(a)-s(b))+(s(c)-s(d)).
\end{align*}
The two summands on the right are exactly $\varphi([(a,b)])$ and $\varphi([(c,d)])$. Therefore
\begin{align*}
\varphi(m+n)=\varphi(m)+\varphi(n).
\end{align*}
[/guided]
[/step]
[step:Verify that the constructed map preserves multiplication]
Let $m,n\in\mathbb{Z}$, with representatives $m=[(a,b)]$ and $n=[(c,d)]$ for $a,b,c,d\in\mathbb{N}_0$. In the formal-difference construction, multiplication is
\begin{align*}
[(a,b)][(c,d)]=[(ac+bd,ad+bc)].
\end{align*}
Therefore,
\begin{align*}
\varphi(mn)=s(ac+bd)-s(ad+bc).
\end{align*}
Using the additivity and multiplicativity identities for $s$ on $\mathbb{N}_0$,
\begin{align*}
\varphi(mn)=s(ac)+s(bd)-s(ad)-s(bc).
\end{align*}
Using $s(xy)=s(x)s(y)$ for each product of elements of $\mathbb{N}_0$ gives
\begin{align*}
\varphi(mn)=s(a)s(c)+s(b)s(d)-s(a)s(d)-s(b)s(c).
\end{align*}
By distributivity in $R$,
\begin{align*}
(s(a)-s(b))(s(c)-s(d))=s(a)s(c)-s(a)s(d)-s(b)s(c)+s(b)s(d).
\end{align*}
The right-hand side is the same expression as above, so
\begin{align*}
\varphi(mn)=\varphi(m)\varphi(n).
\end{align*}
[/step]
[step:Check the values at $0$, positive integers, and negative integers]
The integer $0$ is represented by $(0,0)$, hence
\begin{align*}
\varphi(0)=\varphi([(0,0)])=s(0)-s(0)=0_R.
\end{align*}
The integer $1$ is represented by $(1,0)$, so
\begin{align*}
\varphi(1)=\varphi([(1,0)])=s(1)-s(0)=1_R.
\end{align*}
Thus $\varphi$ is unital.
If $n>0$, then $n$ is represented by $(n,0)$, and therefore
\begin{align*}
\varphi(n)=s(n)=\underbrace{1_R+\cdots+1_R}_{n \text{ terms}}.
\end{align*}
If $n<0$, then $-n>0$, and $n$ is the additive inverse of $-n$ in $\mathbb{Z}$. Since $\varphi$ preserves addition and $\varphi(0)=0_R$,
\begin{align*}
0_R=\varphi(0)=\varphi(n+(-n))=\varphi(n)+\varphi(-n).
\end{align*}
Hence
\begin{align*}
\varphi(n)=-\varphi(-n).
\end{align*}
[/step]
[step:Prove uniqueness from the image of $1$]
Let
\begin{align*}
\psi:\mathbb{Z}\to R
\end{align*}
be any unital ring homomorphism. Then $\psi(1)=1_R$. Since $\psi$ preserves addition, induction on $n\in\mathbb{N}$ gives
\begin{align*}
\psi(n)=\underbrace{1_R+\cdots+1_R}_{n \text{ terms}}
\end{align*}
for every positive integer $n$. Since $\psi$ preserves additive identity, $\psi(0)=0_R$. If $n<0$, then $n+(-n)=0$, so
\begin{align*}
0_R=\psi(0)=\psi(n+(-n))=\psi(n)+\psi(-n).
\end{align*}
Thus $\psi(n)=-\psi(-n)$. Therefore $\psi$ agrees with $\varphi$ on positive integers, on $0$, and on negative integers. Hence $\psi=\varphi$, proving uniqueness.
[/step]
