Equivalence of Definitions of Well-Ordered Integral Domain

Theorem

The following definitions of the concept of Well-Ordered Integral Domain are equivalent:

Let $\struct {D, +, \times \le}$ be an ordered integral domain whose zero is $0_D$.

Definition 1

$\struct {D, +, \times \le}$ is a well-ordered integral domain if and only if the ordering $\le$ is a well-ordering on the set $P$ of (strictly) positive elements of $D$.

Definition 2

$\struct {D, +, \times \le}$ is a well-ordered integral domain if and only if every subset $S$ of the set $P$ of (strictly) positive elements of $D$ has a minimal element:

$\forall S \subseteq D_{\ge 0_D}: \exists x \in S: \forall a \in S: x \le a$

where $D_{\ge 0_D}$ denotes all the elements $d \in D$ such that $\map P d$.

Proof

$(1)$ implies $(2)$

Let $\struct {D, +, \times \le}$ be a well-ordered integral domain by definition 1.

Then by definition the ordering $\le$ is a well-ordering on the set $P$ of (strictly) positive elements of $D$.

Let $S \subseteq P$.

Thus by definition of well-ordering, $S$ has a minimal element.

Thus $\struct {D, +, \times \le}$ is a well-ordered integral domain by definition 2.

$\Box$

$(2)$ implies $(1)$

Let $\struct {D, +, \times \le}$ be a well-ordered integral domain by definition 2.

Then by definition every subset $S$ of the set $P$ of (strictly) positive elements of $D$ has a minimal element:

$\forall S \subseteq D_{\ge 0_D}: \exists x \in S: \forall a \in S: x \le a$

where $D_{\ge 0_D}$ denotes all the elements $d \in D$ such that $\map P d$.

Thus by definition $\le$ is a well-ordering of $P$.

Thus $\struct {D, +, \times \le}$ is a well-ordered integral domain by definition 1.

$\blacksquare$