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?

  • 2
    $\begingroup$ 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 $\endgroup$ – Paultje May 20 '19 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.


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