# 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.

# Examples

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.

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