positive cone

Let $R$ be a commutative ring with 1. A subset $P$ of $R$ is called a pre-positive cone of $R$ provided that

1. $P+P\subseteq P$ ($P$ is additively closed)

2. $P\cdot P\subseteq P$ ($P$ is multiplicatively closed)

3. $-1\notin P$

4. $\mathrm{sqr}(R):=\{r^2\mid r\in R\}\subseteq P.$

As it turns out, a field endowed with a pre-positive cone has an order structure. The field is called a formally real (http://planetmath.org/FormallyRealField), orderable, or ordered field. Before defining what this “order” is, let’s do some preliminary work. Let $P_0$ be a pre-positive cone of a field $F$. By Zorn’s Lemma, the set of pre-positive cones extending $P_0$ has a maximal element $P$. It can be shown that $P$ has two additional properties:

1. $P\cup (-P)=F$

2. $P\cap (-P)=(0).$

Proof.

First, suppose there is $a\in F-(P\cup (-P))$. Let $\overline{P}=P+Pa$. Then $a\in \overline{P}$ and so $P$ is strictly contained in $\overline{P}$. Clearly, $\mathrm{sqr}(F)\subseteq \overline{P}$ and $\overline{P}$ is easily seen to be additively closed. Also, $\overline{P}$ is multiplicatively closed as the equation $(p_1+q_{1}a)(p_2+q_{2}a)=(p_1{p}_2+q_1{q}_2{a}^2)+(p_1{q}_2+q_1{p}_2)a$ demonstrates. Since $P$ is a maximal and $\overline{P}$ properly contains $P$, $\overline{P}$ is not a pre-positive cone, which means $-1\in \overline{P}$. Write $-1=p+qa$. Then $q(-a)=p+1\in P$. Since $q\in P$, $1/q=q{(1/q)}^2\in P$, $-a=(1/q)(p+1)\in P$, contradicting the assumption that $a\notin -P$. Therefore, $P\cup (-P)=F$.

For the second part, suppose $a\in P\cap (-P)$. Since $a\in -P$, $-a\in P$. If $a\ne 0$, then $-1=a(-a){(1/a)}^2\in P$, a contradiction.∎

A subset $P$ of a field $F$ satisfying conditions 1, 2, 5 and 6 is called a positive cone of $F$.A positive cone is a pre-positive cone. If $a\in F$, then either $a\in P$ or $-a\in P$. In either case, $a^2\in P$.Next, if $-1\in P$, then $1\in -P$. But $1=1^2\in P$, we have $1\in P\cap (-P)$, contradicting Condition 6 of $P$.

Now, define a binary relation $\le$, on $F$ by:

$$a\le b⟺b-a\in P$$

It is not hard to see that $\le$ is a total order on $F$. In addition, with the additive and multiplicative structures on $F$, we also have thefollowing two rules:

1. $a\le b\Rightarrow a+c\le b+c$

2. $0\le a$ and $0\le b\Rightarrow 0\le ab$.

Thus, $F$ is a field ordered by $\le$.

Remark. Positive cones may be defined for more general ordered algebraic structures, such as partially ordered groups, or partially ordered rings.