Show that there exists a ring homomorphism $f: Z/(m)\rightarrow Z/(n)$ sending 1 to 1 if and only if $n|m$.

Question

Let $f$ such that $f(a+(m))=a+(n)$
I do not know how to prove that is well defined and that $n|m$ ,Could anyone help me, please?

Answer 1

Assume $n\vert m$, if $a$ and $b$ are such that $a+(m)=b+(m)$, then $m\vert b-a$ and $n\vert b-a$. Therefore, $a+(n)=b+(n)$ and $f:a+(m)\mapsto a+(n)$ is well-defined. I let to you the fact that $f$ is a ring homomorphism.

Assume there exists $g$ a ring homomorphism from $\mathbb{Z}/(m)$ to $\mathbb{Z}/(n)$, one has: $$0+(n)=g(0+(m))=g(m+(m))=mg(1+(m))=m(1+(n))=m+(n).$$ Hence, $n\vert m$.