Equivalence of Definitions of Ordered Integral Domain

Proof
Let $\struct {D, +, \times}$ be a integral domain whose zero is $0_D$ and whose unity is $1_D$.

$(1)$ implies $(2)$
Let $\struct {D, +, \times \le}$ be a ordered integral domain by definition 1.

By Strict Positivity Property induces Total Ordering, $P$ induces this total ordering $\le$ on $D$.

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

$(2)$ implies $(1)$
Let $\struct {D, +, \times \le}$ be a ordered integral domain by definition 2.

That is, $\struct {D, +, \times}$ has a relation $\le$ which is compatible with the ring structure of $D$:

Let $P$ be the set of elements of $D$ which fulfil the conditions:
 * $P = \set {x \in D: 0_D \le x \land 0_D \ne x}$

We check the (strict) positivity property axioms as follows.

Let $x, y \in P$.

That is:
 * $0_D \le x \land 0_D \ne x$
 * $0_D \le y \land 0_D \ne y$

$(\text P 1)$:

Because $\le$ is an ordering, it is a fortiori a preordering.

We have:

But as $x, y \ne 0_D$ it follows that:
 * $0_D \ne x + y$

That is:
 * $x + y \in P$

So it is seen that $P$ fulfils (strict) positivity property $\text P 1$.

$(\text P 2)$:

From $\text {OR} 2$: Product of Positive Elements is Positive:


 * $0_D \le x \times y$

We have that:
 * $0_D \ne x$

and:
 * $0_D \ne y$

As $\struct {D, +, \times}$ is an integral domain, it has no proper zero divisors by definition.

It follows that:
 * $0_D \ne x \times y$

and so:
 * $x \times y \in P$

So it is seen that $P$ fulfils (strict) positivity property $\text P 2$.

$(\text P 3)$:

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