Showing that $\varphi:\mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/m\mathbb{Z}$ is a well-defined surjective ring homomorphism

I have to show that $$\varphi:\mathbb{Z}/n\mathbb{Z}\rightarrow \mathbb{Z}/m\mathbb{Z},a+n\mathbb{Z}\mapsto a+m\mathbb{Z}$$

is a well-defined and surjective ring homomorphism for $$m|n$$.

My idea was to look at the map $$\psi :\mathbb{Z}\rightarrow\mathbb{Z}/m\mathbb{Z},a\mapsto a+m\mathbb{Z}$$ which is clearly surjective and then to use the fundemental theorem on homomorphisms.

Because $$m|n$$, the ideal $$n\mathbb{Z}$$ is a subset of kernel of $$\psi$$. That tells me that $$\varphi$$ is surjective because of the above mentioned theorem.

But is this enough to say that $$\varphi$$ is well-defined or do I have to show that differently?

• No using the fundamental theorem is enough to show that the induced map is well-defined but in your case you don't know that the induced map is equal to $\varphi$. So you need to show that first but it is a one liner – Paultje May 20 at 16:01

It suffices to show that $$n\mathbb{Z}\subseteq \ker(\psi)$$ because then $$\psi$$ factors through a unique (well-defined) homomorphism $$\phi:\mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/m\mathbb{Z}$$ given by $$a+n\mathbb{Z}\mapsto a+m\mathbb{Z}$$.
On the other hand, you can also check directly. Suppose that $$a\equiv b \pmod{n}$$. Then $$n\mid (a-b)$$ and hence $$m\mid(a-b)$$. So, $$a\equiv b \pmod{m}$$. So, $$\phi(a)=a\equiv b=\phi(b)\pmod{m}.$$ Hence, $$\phi$$ is well-defined.