Ring of Integers Modulo Prime is Integral Domain/Proof 2

Theorem

Let $m \in \Z: m \ge 2$.

Let $\struct {\Z_m, +, \times}$‎ be the ring of integers modulo $m$.

Then:

$m$ is prime

if and only if:

$\struct {\Z_m, +, \times}$ is an integral domain.

Proof

From Ring of Integers Modulo m is Ring, $\struct {\Z_m, +, \times}$‎ is a ring.

It remains to be shown that $\struct {\Z_m, +, \times}$‎ has no proper zero divisors if and only if $m$ is prime.

$m$ Composite

Let $m$ be composite.

Then:

$m = m_1 m_2$

where $0 < m_1 < m, 0 < m_2 < m$.

Then:

$\eqclass {m_1} m \eqclass {m_2} m = \eqclass 0 m$

and so both $\eqclass {m_1} m$ and $\eqclass {m_2} m$ are proper zero divisors.

Hence if $m$ is not prime then $\struct {\Z_m, +, \times}$‎ is not an integral domain by definition.

$\Box$

$m$ Prime

Let $m$ be prime.

Let $\eqclass a m \eqclass b m = \eqclass 0 m$.

Then by Modulo Multiplication is Well-Defined:

$\eqclass {a b} m = \eqclass 0 m$

Thus either

$m \divides a$ or $m \divides b$

where $\divides$ denotes the divisibility relation.

If $m \divides a$ then $\eqclass a m = 0$.

If $m \divides b$ then $\eqclass b m = 0$.

Thus neither $a$ nor $b$ is a proper zero divisor.

Hence there are no proper zero divisors of $\struct {\Z_m, +, \times}$.

Hence, by definition, $\struct {\Z_m, +, \times}$‎ is an integral domain

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Integral Domains: $\S 6$. The Residue Classes: Theorem $6$