Strict Positivity Property induces Total Ordering

Theorem

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

Let $D$ be endowed with a (strict) positivity property $P: D \to \set {\T, \F}$.

Then there exists a total ordering $\le$ on $\struct {D, +, \times}$ induced by $P$ which is compatible with the ring structure of $\struct {D, +, \times}$.

$(\text P 1)$	$:$	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 $a \in D$.

Let us define a relation $<$ on $D$ as:

$\forall a, b \in D: a < b \iff \map P {-a + b}$

Setting $a = 0$:

$\forall b \in D: 0 < b \iff \map P b$

demonstrating that (strictly) positive elements of $D$ are those which are greater than zero.

Thus by definition, $<$ is a strict ordering.

Let the relation $\le$ be defined as the reflexive closure of $<$.

Hence the result.

$\blacksquare$

1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order

\((\text P 1)\)	$:$	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 $a \in D$.