Positivity Property induces Total Ordering

Theorem

Let $\left({D, +, \times}\right)$ be an ordered integral domain where $P$ is the positivity property.

Then there exists an ordering $\le$ on $\left({D, +, \times}\right)$ which is compatible with the ring structure of $\left({D, +, \times}\right)$.

It follows that $\left({D, +, \times, \le}\right)$ is a totally ordered ring.

Proof

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

$\forall a, b \in D: a < b \iff P \left({-a + b}\right)$

Informally, we can see where we're going with this. We have defined what we are going to demonstrate is a strict ordering based on the notion that:

$a$ is less than $b$ iff $b$ minus $a$ is positive.

First, note that setting $a=0$ we have that:

$\forall b \in D: 0 < b \iff P \left({b}\right)$

thus demonstrating that positive elements of $D$ are those which are greater than zero.

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

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

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

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

Now we form the relation $\le$ defined as $< \cup \Delta_D$, where $\Delta_D$ is the diagonal relation.

From Ordering is Strict Ordering Union Diagonal Relation we have that $\le$ is an ordering on $D$.

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

Hence the result.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 2.7$