Quotient Epimorphism from Integers by Principal Ideal

Theorem
Let $m$ be a strictly positive integer.

Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$.

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

In particular, $\struct {\Z, +, \times} / \ideal m$ 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:
 * $\map {q_m} {x \times_m y} = \map {q_m} {x y} = \map {q_m} x \map {q_m} y$

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

Let $a \in \Z$.

Then:
 * $\exists q, r \in \Z: a = q m + r: 0 \le r < m$

so:
 * $\map {q_m} a = \map {q_m} r \in q_m \sqbrk {\N_m}$

Therefore:
 * $\Z / \ideal m = q_m \sqbrk \Z = q_m \sqbrk {\N_m}$

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

If $0 < r < m$, then $r \notin \ideal m$ and thus $\map {q_m} r \ne 0$.

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 $\Z / \ideal m$.