Ordered ring


by Dylan Hendrickson Jul 6 2016 updated Jul 7 2016

A ring with a total ordering compatible with its ring structure.

[summary: An ordered ring is a Ring that is totally ordered, where the ordering agrees with the ring operations. In particular, adding something to two elements doesn't change which of them is bigger, and the product of two [-positive] elements is positive.]

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$~$:

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

%%hidden(Show proof): 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$~$. %%

%%hidden(Show proof): 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 $~$0 \leq a \otimes b$~$. %%

%%hidden(Show proof): 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 $~$0 \leq a^2$~$. %%

%%hidden(Show proof): 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 [4zq $~$\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.