Ring of Integers Modulo m cannot be Ordered Integral Domain
Jump to navigation
Jump to search
Theorem
Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then $\struct {\Z_m, +, \times}$ cannot be an ordered integral domain.
Proof
First note that from Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain only when $m$ is prime.
So for $m$ composite the result holds.
If $m$ is prime, and $\struct {\Z_m, +, \times}$ is therefore an integral domain, its order is finite.
The result follows from Finite Integral Domain cannot be Ordered.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $11$