# 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 $P$:

 $(P \, 1)$ $:$ Closure under Ring Addition: $\displaystyle \forall a, b \in D:$ $\displaystyle \map P a \land \map P b \implies \map P {a + b}$ $(P \, 2)$ $:$ Closure under Ring Product: $\displaystyle \forall a, b \in D:$ $\displaystyle \map P a \land \map P b \implies \map P {a \times b}$ $(P \, 3)$ $:$ Trichotomy Law: $\displaystyle \forall a \in D:$ $\displaystyle \map P a \lor \map P {-a} \lor a = 0_D$ For $P \, 3$, 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 $\struct {D, +, \times}$:

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

## 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 $D$:

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

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:

$0_D \le x \land 0_D \ne x$
$0_D \le y \land 0_D \ne y$

$(P \, 1)$:

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

We have:

 $\displaystyle 0_D + 0_D$ $\le$ $\displaystyle x + y$ $OR \, 1$: Definition of Ordering Compatible with Ring Structure $\displaystyle \leadsto \ \$ $\displaystyle 0_D$ $\le$ $\displaystyle x + y$ 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 $P \, 1$.

$(P \, 2)$:

From $OR \, 2$: 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 $P \, 2$.

$(P \, 3)$:

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

$\blacksquare$