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, $P = R_{\ge 0}$.

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 $P = R_{\ge 0}$.

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

Proof

Necessary Condition

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

Hence:

$(OR1): \quad \le$ is compatible with $+$

$(OR2): \quad \forall x, y \in R: 0_R \le x, 0_R \le y \implies 0_R \le x \circ y$.

$(1)$: 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.

Sufficient Condition

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

By item $(OR2)$ 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 $P = R_{\ge 0}$, 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

By $(2)$:

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

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 $y \le x$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 23$: Theorem $23.12$