Fundamental Theorem on Homomorphisms - Application

Question

I want to show

1.) For $n,m\in \mathbb Z$ with $m|n$ there exists a ring homomorphism between $\mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}$.

I know that there is the canonical homomorphism $f: \mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $n\mathbb{Z}$ is an ideal in $\mathbb Z$, to conclude the existence of a ring homomorphism $\mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}$. But here is the problem: $\mathbb{Z}/n\mathbb{Z}$ has to be a subset of $\ker(f)$ for that conclusion. Is that the case and if so: why?

2.) A restriction of the above Homomorphism on the unity groups $(\mathbb Z/n\mathbb Z)^* \rightarrow (\mathbb Z/m\mathbb Z)^*$ is a group homomorphism.

I believe that one can solve this easily if 1.) is solved.

Answer 1

Why not study the canonical map $\mathbb{Z}\xrightarrow{\varphi} \mathbb{Z}/m\mathbb{Z}$ given by the projection $n\mapsto [n]_m$? The first isomorphism theorem tells us that $\mathbb{Z}/\ker\varphi\cong \mathbb{Z}/m\mathbb{Z}$. Because $m|n$, we get that for $k\in n\mathbb{Z}$, i.e. $k=\ell n=\ell rm$ for some $k\in \mathbb{Z}$, we have $$\varphi(k) =[k]_m=[\ell r m]_m=0 \pmod{m}.$$ So, $n\mathbb{Z}\subseteq\ker\varphi$ and the factorization theorem tells us that we get an induced map $\widetilde{\varphi}:n\mathbb{Z}\to m\mathbb{Z}$. This is defined by $\widetilde{\varphi}(a+n\mathbb{Z})=\varphi(a)=[a]_m.$

Answer 2

If $f : G\to G'$ be a homomorphic and onto then $h : \dfrac{G}{K} \to G'$ is an isomorphism and onto where $K = \ker f$.