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.