Ring of Integers Modulo m cannot be Ordered Integral Domain

Theorem

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

Let $\left({\Z_m, +, \times}\right)$‎ be the ring of integers modulo $m$.

Then $\left({\Z_m, +, \times}\right)$ cannot be an ordered integral domain.

Proof

First note that from Ring of Integers Modulo Prime is an Integral Domain, $\left({\Z_m, +, \times}\right)$‎ is an integral domain only when $m$ is prime.

So for $m$ composite the result holds.

If $m$ is prime, and $\left({\Z_m, +, \times}\right)$ is therefore an integral domain, its order is finite.

The result follows from Finite Integral Domain cannot be Ordered.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 2.7$: Example $11$