Quotient Epimorphism from Integers by Principal Ideal

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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:

\(\ds \exists q, r \in \Z: \, \) \(\ds x + y\) \(=\) \(\ds m q + r\) for $0 \le r < m$
\(\ds \exists p, s \in \Z: \, \) \(\ds x y\) \(=\) \(\ds m p + s\) for $0 \le s < m$


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

\(\ds \map {q_m} {x +_m y}\) \(=\) \(\ds \map {q_m} r\)
\(\ds \) \(=\) \(\ds \map {q_m} {m q} + \map {q_m} r\)
\(\ds \) \(=\) \(\ds \map {q_m} {m q + r}\)
\(\ds \) \(=\) \(\ds \map {q_m} {x + y}\)
\(\ds \) \(=\) \(\ds \map {q_m} x + \map {q_m} y\)


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 $\Z / \ideal m$.





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