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 $$\Z / \left({m}\right)$$.