# Equivalence of Definitions of Ordered Integral Domain

## Theorem

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

### Definition 1

An ordered integral domain is an integral domain $\struct {D, +, \times}$ which has a strict positivity property $P$:

 $(\text P 1)$ $:$ Closure under Ring Addition: $\ds \forall a, b \in D:$ $\ds \map P a \land \map P b \implies \map P {a + b}$ $(\text P 2)$ $:$ Closure under Ring Product: $\ds \forall a, b \in D:$ $\ds \map P a \land \map P b \implies \map P {a \times b}$ $(\text P 3)$ $:$ Trichotomy Law: $\ds \forall a \in D:$ $\ds \paren {\map P a} \lor \paren {\map P {-a} } \lor \paren {a = 0_D}$ For $\text P 3$, exactly one condition applies for all $a \in D$.

### Definition 2

An ordered integral domain is an ordered ring $\struct {D, +, \times, \le}$ which is also an integral domain.

That is, it is an integral domain with an ordering $\le$ compatible with the ring structure of $\struct {D, +, \times}$:

 $(\text {OID} 1)$ $:$ $\le$ is compatible with ring addition: $\ds \forall a, b, c \in D:$ $\ds a \le b$ $\ds \implies$ $\ds \paren {a + c} \le \paren {b + c}$ $(\text {OID} 2)$ $:$ Strict positivity is closed under ring product: $\ds \forall a, b \in D:$ $\ds 0_D \le a, 0_D \le b$ $\ds \implies$ $\ds 0_D \le a \times b$

## 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.

$\Box$

### $(2)$ 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 $D$:

 $(\text {OR} 1)$ $:$ $\le$ is compatible with $+$: $\ds \forall a, b, c \in D:$ $\ds a \le b$ $\ds \implies$ $\ds \paren {a + c} \le \paren {b + c}$ $(\text {OR} 2)$ $:$ Product of Positive Elements is Positive: $\ds \forall a, b \in D:$ $\ds 0_D \le x, 0_D \le y$ $\ds \implies$ $\ds 0_D \le x \times y$

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:

 $\ds 0_D + 0_D$ $\le$ $\ds x + y$ $\text {OR} 1$: Definition of Ordering Compatible with Ring Structure $\ds \leadsto \ \$ $\ds 0_D$ $\le$ $\ds x + y$ Preordering of Products under Operation Compatible with Preordering

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.

$\blacksquare$