Integer Divisor is Equivalent to Subset of Ideal

Theorem

Let $\Z$ be the set of all integers.

Let $\Z_{>0}$ be the set of strictly positive integers.

Let $m \in \Z_{>0}$ and let $n \in \Z$.

Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$.

Then:

$m \divides n \iff \ideal n \subseteq \ideal m$

Proof

The ring of integers is a principal ideal domain.

The result follows directly from Principal Ideals in Integral Domain.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.5$