An ordered ring is a ring $R=(X,\oplus ,\otimes )$ with a total order $\le$ compatible with the ring structure. Specifically, it must satisfy these axioms for any $a,b,c\in X$:

• If $a\le b$, then $a\oplus c\le b\oplus c$.

• If $0\le a$ and $0\le b$, then $0\le a\otimes b$

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 $a$, $a\le 0$ if and only if $0\le -a$.

Show proof

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

• The product of nonpositive elements is nonnegative.

Show proof

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

• The square of any element is nonnegative.

Show proof

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

• The additive identity $1\ge 0$. (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 $0\le i$, then, we have $0\le i\times i=-1$, which is false. Suppose that $i\le 0$, then $0=i+(-i)\le 0+(-i)$, but then we have $0\le (-i)\times (-i)=-1$, which is again false. Alternatively, $i^2=-1$ is a square, so it must be nonnegative; that is, $0\le -1$, which is a contradiction.