# Element in Integral Domain is Divisor iff Principal Ideal is Superset

## Theorem

Let $\struct {D, +, \circ}$ be an integral domain.

Let $\ideal x$ denote the principal ideal of $D$ generated by $x$.

Let $x, y \in \struct {D, +, \circ}$.

Then:

$x \divides y \iff \ideal y \subseteq \ideal x$

where $x \divides y$ denotes that $x$ is a divisor of $y$.

## Proof

Let that $x \divides y$.

Then by definition of divisor:

 $\ds x \divides y$ $\leadsto$ $\ds \exists t \in D: y = t x$ Definition of Divisor of Ring Element $\ds$ $\leadsto$ $\ds y \in \ideal x$ Definition of Principal Ideal of Ring $\ds$ $\leadsto$ $\ds \ideal y \subseteq \ideal x$ Definition of Principal Ideal of Ring: $\ideal y$ is the smallest ideal containing $y$

Conversely:

 $\ds \ideal y \subseteq \ideal x$ $\leadsto$ $\ds y \in \ideal x$ as $y \in \ideal y$ $\ds$ $\leadsto$ $\ds \exists t \in D: y = t x$ Definition of Principal Ideal of Ring $\ds$ $\leadsto$ $\ds x \divides y$ Definition of Divisor of Ring Element

So:

$x \divides y \iff \ideal y \subseteq \ideal x$

$\blacksquare$