TITLE: Describe all ring homomorphisms
QUESTION [28 upvotes]: Describe all ring homomorphisms of:
a) $\mathbb{Z}$ into $\mathbb{Z}$
b) $\mathbb{Z}$ into $\mathbb{Z} \times \mathbb{Z}$
c) $\mathbb{Z} \times \mathbb{Z}$ into $\mathbb{Z}$
d) How many homomorphisms are there of $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$ into $\mathbb{Z}$
Note: These were past homework questions and my professor already gave out answers. I just need someone to help me understand and approach this type of problem.
Thank you.

REPLY [13 votes]: There is only one ring homomorphism from $\mathbb{Z}$ to any ring $S$ (assuming that ring homomorphisms preserve $1$).
This homomorphism is uniquely determined because a ring homomorphism $f\colon R\to S$ has the property that
$$f(nx) = n f(x)$$
for all $x\in R$ and $n\in\mathbb{Z}$ (easy induction). Since in $\mathbb{Z}$ we have $n=n\cdot1$, for $f\colon\mathbb{Z}\to S$ we must have $f(n)=nf(1)=n\cdot1$. The assignment
$$f(n)=n\cdot1$$
indeed defines a ring homomorphism from $\mathbb{Z}$ to any ring $S$. Its kernel is an ideal of the form $k\mathbb{Z}$ $(k\ge0)$ and $k$ is the characteristic of $S$.
This answers your first two questions.
Let now $f\colon \mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$ be a ring homomorphism. Since
$$(1,0)(0,1)=(0,0)$$
we have that $f(1,0)f(0,1)=0$, so either $f(1,0)=0$ or $f(0,1)=0$. Not both, because $(1,1)=(1,0)+(0,1)$ is the identity on $\mathbb{Z}\times\mathbb{Z}$ and we want $f(1,1)=1$. Moreover we see that
$$f(0,1)=1-f(1,0).$$
If $f(1,0)=1$ and $f(0,1)=0$, then $f$ is completely determined: we must have
$$f(m,n)=f\bigl(m(1,0)+n(0,1))=mf(1,0)+nf(0,1)=m$$
It's easy to check that this indeed defines a ring homomorphism. Similarly
$$g(m,n)=n$$
defines another ring homomorphism and the list is complete.
For $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ the reasoning is similar. Consider
$$(1,0,0)+(0,1,0)+(0,0,1)=(1,1,1)$$
and the fact that two among $f(1,0,0)$, $f(0,1,0)$ and $f(0,0,1)$ must be $0$.