# 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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