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 $P: D \to \set {\mathrm T, \mathrm F}$.

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

It follows that $\struct {D, +, \times, \le}$ is a totally ordered ring.

Proof
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.

From Relation Induced by Strict Positivity Property is Compatible with Addition we have that $<$ is compatible with $+$.

From Relation Induced by Strict Positivity Property is Transitive we have that $<$ is transitive.

From Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive we have that $<$ is asymmetric and antireflexive.

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

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

From Reflexive Closure of Strict Ordering is Ordering we have that $\le$ is an ordering on $D$.

From Relation Induced by Strict Positivity Property is Trichotomy, and from the Trichotomy Law (Ordering), we have that $\le$ is a total ordering.

Hence the result.

Also see

 * Definition:Total Ordering Induced by Strict Positivity Property‎


 * Definition:Ordering Induced by Positivity Property