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 $\text P 2$.
$(\text P 3)$:
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Thus $\struct {D, +, \times \le}$ is a ordered integral domain by definition 1.
$\blacksquare$