An ordered ring is a ring with a total order $\leq$ compatible with the ring structure. Specifically, it must satisfy these axioms for any $a, b, c \in X$ :

If $a \leq b$ , then $a \oplus c \leq b \oplus c$ .

If $0 \leq a$ and $0 \leq b$ , then $0 \leq a \otimes b$

An element $a$ of the ring is called "positive" if $0 < a$ and "negative" if $a < 0$ . The second axiom, then, says that the product of nonnegative elements is nonnegative.

An ordered ring that is also a field is an ordered_field.

Basic Properties

For any element $a$ , $a \leq 0$ if and only if $0 \leq - a$ .

Show proof

First suppose $a \leq 0$ . Using the first axiom to add $- a$ to both sides, $a + (- a) = 0 \leq - a$ . For the other direction, suppose $0 \leq - a$ . Then $a \leq - a + a = 0$ .

The product of nonpositive elements is nonnegative.

Show proof

Suppose $a$ and $b$ are nonpositive elements of $R$ , that is $a, b \leq 0$ . From the first axiom, $a + (- a) = 0 \leq - a$ , and similarly $0 \leq - b$ . By the second axiom $0 \leq - a \otimes - b$ . But $- a \otimes - b = a \otimes b$ , so $0 \leq a \otimes b$ .

The square of any element is nonnegative.

Show proof

Let $a$ be such an element. Since the ordering is total, either $0 \leq a$ or $a \leq 0$ . In the first case, the second axiom gives $0 \leq a^{2}$ . In the second case, the previous property gives $0 \leq a^{2}$ , since $a$ is nonpositive. Either way we have $0 \leq a^{2}$ .

The additive identity $1 \geq 0$ . (Unless the ring is trivial, $1 > 0$ .)

Show proof

Clearly $1 = 1 \otimes 1$ . So $1$ is a square, which means it's nonnegative.

Examples

The real numbers are an ordered ring (in fact, an ordered_field), as is any subring of $R$ , such as $Q$ .

The complex numbers are not an ordered ring, because there is no way to define the order between $0$ and $i$ . Suppose that $0 \leq i$ , then, we have $0 \leq i \times i = - 1$ , which is false. Suppose that $i \leq 0$ , then $0 = i + (- i) \leq 0 + (- i)$ , but then we have $0 \leq (- i) \times (- i) = - 1$ , which is again false. Alternatively, $i^{2} = - 1$ is a square, so it must be nonnegative; that is, $0 \leq - 1$ , which is a contradiction.