Element in Integral Domain is Divisor iff Principal Ideal is Superset
Jump to navigation
Jump to search
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:
- $x \divides y \iff \ideal y \subseteq \ideal x$
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 62.3$ Factorization in an integral domain: $\text{(i)}$