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Ordered ring

Edited by Dylan Hendrickson, Joe Zeng last updated 7th Jul 2016

An ordered ring is a ring R=(X,⊕,⊗) with a total order ≤ compatible with the ring structure. Specifically, it must satisfy these axioms for any a,b,c∈X:

  • If a≤b, then a⊕c≤b⊕c.
  • If 0≤a and 0≤b, then 0≤a⊗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≤0 if and only if 0≤−a.

Show proof

First suppose a≤0. Using the first axiom to add −a to both sides, a+(−a)=0≤−a. For the other direction, suppose 0≤−a. Then a≤−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≤0. From the first axiom, a+(−a)=0≤−a, and similarly 0≤−b. By the second axiom 0≤−a⊗−b. But −a⊗−b=a⊗b, so 0≤a⊗b.

  • The square of any element is nonnegative.

Show proof

Let a be such an element. Since the ordering is total, either 0≤a or a≤0. In the first case, the second axiom gives 0≤a2. In the second case, the previous property gives 0≤a2, since a is nonpositive. Either way we have 0≤a2.

  • The additive identity 1≥0. (Unless the ring is trivial, 1>0.)

Show proof

Clearly 1=1⊗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≤i, then, we have 0≤i×i=−1, which is false. Suppose that i≤0, then 0=i+(−i)≤0+(−i), but then we have 0≤(−i)×(−i)=−1, which is again false. Alternatively, i2=−1 is a square, so it must be nonnegative; that is, 0≤−1, which is a contradiction.

Parents:
Ring
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