Positive Elements of Ordered Ring

Theorem
Let $$\left({R, +, \circ, \le}\right)$$ be an ordered ring whose zero is $$0_R$$ and whose unity is $$1_R$$.

Let $$P$$ be the set of positive elements of $$R$$, that is, $$P = R_+$$.

Then:


 * $$(1) \quad P + P \subseteq P$$


 * $$(2) \quad P \cap \left({- P}\right) = \left\{{0_R}\right\}$$


 * $$(3) \quad P \circ P \subseteq P$$

If $$\le$$ is a total ordering, that is, if $$\left({R, +, \circ, \le}\right)$$ is a totally ordered ring, then:


 * $$(4) \quad P \cup \left({- P}\right) = R$$

The converse is also true:

Let $$\left({R, +, \circ}\right)$$ be a ring.

Let $$P \subseteq R$$ such that $$(1)$$, $$(2)$$ and $$(3)$$ are satisfied.

Then there is one and only one ordering $$\le$$ compatible with the ring structure of $$R$$ such that $$P = R_+$$.

Also, if $$(4)$$ is also satisfied, then $$\le$$ is a total ordering.

Proof
First, suppose that $$\le$$ is compatible with the ring structure of $$R$$.

Hence:
 * 1) $$\le$$ is compatible with $$+$$;
 * 2) $$\forall x, y \in R: 0_R \le x, 0_R \le y \implies 0_R \le x \circ y$$.


 * $$(1)$$: Let $$x, y \in R: 0_R \le x, 0_R \le y$$.

Then $$0_R + 0_R \le x + y$$ by the fact that $$\le$$ is compatible with $$+$$.

Thus $$0_R \le x + y$$ and thus $$x + y \in P$$.


 * $$(2)$$: By Properties of an Ordered Ring item 4, $$-P = \left\{{x \in R: x \le 0_R}\right\}$$.

Let $$x \in P \cap \left({- P}\right)$$. Then $$x \le 0_R$$ and $$0_R \le x$$.

So from the antisymmetric nature of $$\le$$, $$x = 0_R$$.


 * $$(3)$$: This is equivalent to $$0_R \le x, 0_R \le y \implies 0_R \le x \circ y$$ which is one of the properties of being compatible with the ring structure of $$R$$.


 * $$(4)$$: Now if $$\le$$ is a total ordering, then $$\forall x \in R$$, either $$x \le 0_R$$ or $$0_R \le x$$, and the result follows.


 * Now to prove the converse.

Let $$P \subseteq R$$ such that $$(1)$$, $$(2)$$ and $$(3)$$ are satisfied.

By item 2 of Properties of an Ordered Ring, we have:
 * $$x \le y \iff 0 \le y + \left({-x}\right)$$

so there is at most one ordering on $$R$$ compatible with the ring structure of $$R$$ such that $$P = R_+$$, namely, the one that satisfies:


 * $$x \le y \iff y + \left({-x}\right) \in P$$

Now we need to show that $$\le$$ thus defined has the required properties.


 * Reflexivity: $$\forall x \in R: x \le x$$ because $$x + \left({-x}\right) = 0_R \in P$$ by $$(2)$$.


 * Antisymmetry: Let $$x \le y$$ and $$y \le x$$.

Then $$y + \left({-x}\right) \in P$$ and $$- \left({y + \left({-x}\right)}\right) = x + \left({-y}\right) \in P$$.

Thus by $$(2)$$, $$y + \left({-x}\right) = 0_R$$ and thus $$y = x$$.


 * Transitivity: If $$x \le y$$ and $$y \le z$$, then $$y + \left({-x}\right) \in P$$ and $$z + \left({-y}\right) \in P$$.

But as $$z + \left({-x}\right) = z + \left({-y}\right) + y + \left({-x}\right)$$, we have that $$z + \left({-x}\right) \in P$$ from $$(1)$$.

Hence $$x \le z$$.


 * Let $$x \le y$$.

Then $$z + x \le z + y$$ since $$\left({z + y}\right) + \left({- \left({x + x}\right)}\right) = y + \left({-x}\right) \in P$$.


 * Finally, Ordering Compatible with Ring holds because of $$(3)$$.


 * If $$(4)$$ holds, then $$\forall x, y \in R$$, either $$y + \left({-x}\right) \in P$$ or $$x + \left({-y}\right) = -\left({y + \left({-x}\right)}\right) \in P$$, that is, either $$x \le y$$ or $$y \le x$$.