Ring of Integers Modulo Composite is not Integral Domain
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Theorem
Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Let $m$ be a composite number.
Then $\struct {\Z_m, +, \times}$ is not an integral domain.
Proof
Let $m \in \Z: m \ge 2$ be composite.
Then:
- $\exists k, l \in \N_{> 0}: 1 < k < m, 1 < l < m: m = k \times l$
Thus:
\(\ds \eqclass 0 m\) | \(=\) | \(\ds \eqclass m m\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass {k l} m\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass k m \times \eqclass l m\) |
So $\struct {\Z_m, +, \times}$ is a ring with zero divisors.
So by definition $\struct {\Z_m, +, \times}$ is not an integral domain.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $5$: Rings: $\S 18$. Definition of a Ring: Example $29$