In Algebra Chapter 0 Aluffi states that the ring $\Bbb{Z}$ of integers with usual addition and multiplication is initial in the category Ring. That is for a ring $R$ with identity $1_{R}$ there is a unique ring homomorphism $\phi:\Bbb{Z}\rightarrow{R}$ defined by $\phi(n)\mapsto{n\bullet1_{R}}$ $(\forall{n\in\Bbb{Z}})$which makes sense for rings such as $\Bbb{Q},\Bbb{R},\Bbb{C}$ which have $\Bbb{Z}$ as a subring but I fail to see how $\phi$ holds when the codomain is a ring which doesn't contain $\Bbb{Z}$. If someone could provide examples of ring homomorphisms from $\Bbb{Z}$ to rings other than the rings mentioned above I would appreciate it.
[Math] Why is the ring of integers initial in Ring
category-theoryring-theory
Best Answer
In every ring (with unit) $R$ you have a $1$. And you have thus $1+1$, and $1+1+1$, &c. You can consider the collection of all these elements and their additive inverses, $-(1+1), -(1+1+1)$, &c. For $n>0$ call these elements $n1$ and set $(-n)1=-(n1)$, if $n=0$; $n1=0$. You should convince yourself that $m1+n1=(m+n)1$ for any pair of integers $n,m$, and $n1\cdot m1=(nm)1$. Of course $1\cdot 1=1$. This means the collection of such elements is a subring of $R$. It looks like $\Bbb Z$, since every element is just a sum of $1$ (or difference), but note that these elements needn't be distinct. For example, in $\Bbb Z/3\Bbb Z$, $2\cdot 1=-1\cdot 1$.
At any rate, now we can define a function $f:\Bbb Z\to R$ for any ring $R$ with unit that sends the integer $n$ to $n1$, as defined above. What you checked above is precisely the claim that $f(n)f(m)=f(mn)$ and $f(n+m)=f(n)+f(m)$, $f(1)=1$. This means $f$ is a morphism of rings. Since $R$ was arbitrary, we have shown that every ring $R$ admits a morphism of rings $f:\Bbb Z \to R$.
It remains to see that $f$ is unique. Now if $g:\Bbb Z\to R$ is another morphism of rings, we know that $g(1)=1$ (i.e. $g$ sends the unit of $\Bbb Z$ to the unit of $R$), hence $g(-1)=-1$ (why?). But any nonzero integer is $n$ is $1+\dots+1$ ($n$ times) or $-1-\cdots-1$ ($n$ times), so using $g$ is $\Bbb Z$-linear ($g(a+b)=g(a)+g(b)$) gives $g$ sends $n$ to $f(n)$ as defined above, i.e. $f=g$.