Ordered ring

An ordered ring is a ring \(R=(X,\oplus,\otimes)\) 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 \(0<\) and “negative” if \(a<\). 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 \(0 \leq -a\).

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.

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.

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, \(1>0\).)

Clearly \(1 = 1 \otimes 1\). So \(1\) is a square, which means it’s nonnegative.


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

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, \(0 \leq -1\), which is a contradiction.