Quotient Epimorphism from Integers by Principal Ideal

Theorem
Let $m$ be a strictly positive integer.

Let $\left({m}\right)$ be the principal ideal of $\Z$ generated by $m$.

The restriction to $\N_m$ of the canonical epimorphism $q_m$ from the ring $\left({\Z, +, \times}\right)$ onto $\left({\Z, +, \times}\right) / \left({m}\right)$ is an isomorphism from the ring $\left({\N_m, +_m, \times_m}\right)$ of integers modulo $m$ onto the quotient ring $\left({\Z, +, \times}\right) / \left({m}\right)$.

In particular, $\left({\Z, +, \times}\right) / \left({m}\right)$ has $m$ elements.

Proof
Let $x, y \in \N_m$.

By the Division Theorem:

Then $x +_m y = r$ and $x \times_m y = s$, so:

and similarly $q_m \left({x \times_m y}\right) = q_m \left({x y}\right) = q_m \left({x}\right) q_m \left({y}\right)$.

So the restriction of $q_m$ to $\N_m$ is a homomorphism from $\left({\N_m, +_m, \times_m}\right)$ into $\left({\Z / \left({m}\right), +_{\left({m}\right)}, \times_{\left({m}\right)}}\right)$.

Let $a \in \Z$.

Then $\exists q, r \in \Z: a = q m + r: 0 \le r < m$, so $q_m \left({a}\right) = q_m \left({r}\right) \in q_m \left({\N_m}\right)$.

Therefore $\Z / \left({m}\right) = q_m \left({\Z}\right) = q_m \left({\N_m}\right)$.

Therefore the restriction of $q_m$ to $\N_m$ is surjective.

If $0 < r < m$, then $r \notin \left({m}\right)$ and thus $q_m \left({r}\right) \ne 0$.

Thus the kernel of the restriction of $q_m$ to $\N_m$ contains only zero.

Therefore by the Quotient Theorem for Group Epimorphisms, the restriction of $q_m$ to $\N_m$ is an isomorphism from $\N_m$ to $\Z / \left({m}\right)$.