Ring Epimorphism from Integers to Integers Modulo m

Theorem
Let $\left({\Z, +, \times}\right)$ be the ring of integers.

Let $\left({\Z_m, +_m, \times_m}\right)$ be the ring of integers modulo $m$.

Let $\phi: \left({\Z, +, \times}\right) \to \left({\Z_m, +_m, \times_m}\right)$ be the mapping defined as:
 * $\forall x \in \Z: \phi \left({x}\right) = \left[\!\left[{x}\right]\!\right]_m$

where $\left[\!\left[{x}\right]\!\right]_m$ is the residue class modulo $m$.

Then $\phi$ is a ring epimorphism, but specifically not a ring monomorphism.

The image of $\phi$ is $\left({\Z_m, +_m, \times_m}\right)$.

The kernel of $\phi$ is $m \Z$, the set of integer multiples of $m$.

Proof
Let $a, b \in \Z$.

Then:

Hence $\phi$ is a ring homomorphism.

Now let $\left[\!\left[{a}\right]\!\right]_m \in \Z_m$.

We have that $\left[\!\left[{a}\right]\!\right]_m = \left\{{x \in \Z: \exists k \in \Z: z = a + km}\right\}$.

Setting $k = 0$ we have that $\phi \left({a}\right) = \left[\!\left[{a}\right]\!\right]_m$ and so $\phi^{-1} \left({\left[\!\left[{a}\right]\!\right]_m}\right) \ne \varnothing$.

Thus $\phi$ is a surjection.

Now setting $k = 1$, for example, we have that $\phi \left({a + m}\right) = \left[\!\left[{a}\right]\!\right]_m$ and so $\phi \left({a}\right) = \phi \left({a + m}\right)$.

So $\phi$ is specifically not an injection.

It follows by definition that $\phi$ is a ring epimorphism, but specifically not a ring monomorphism.

Next we note that $\forall x \in \Z: \phi \left({x}\right) \in \Z_m$ and so $\operatorname{Im} \left({\phi}\right) = \Z_m$.

Finally, we have that the kernel of $\phi$ is:
 * $\ker \left({\phi}\right) = \left\{{x \in \Z: \phi \left({x}\right) = \left[\!\left[{0}\right]\!\right]_m}\right\}$

Let $\phi \left({x}\right) = \left[\!\left[{0}\right]\!\right]_m$.

Then $x = 0 + k m$ for some $k \in \Z$.

That is, $x \in m \Z$ and so $\ker \left({\phi}\right) \subseteq m \Z$.

Now let $x \in m \Z$.

Then $\exists k \in \Z: x = 0 + k m$ and so by definition $\phi \left({x}\right) = \left[\!\left[{0}\right]\!\right]_m$.

So $m \Z \subseteq \ker \left({\phi}\right)$.

Hence $\ker \left({\phi}\right) = m \Z$.