positive cone
Let $R$ be a commutative ring with 1. A subset $P$ of $R$ is called a prepositive 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 prepositive 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 prepositive cone of a field $F$. By Zorn’s Lemma, the set of prepositive cones extending ${P}_{0}$ has a maximal element^{} $P$. It can be shown that $P$ has two additional properties:

5.
$P\cup (P)=F$

6.
$P\cap (P)=(0).$
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 prepositive 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 prepositive 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:
$$a\le b\u27faba\in P$$ 
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 the following 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.
References
 1 A. Prestel, Lectures on Formally Real Fields, Springer, 1984
Title  positive cone 

Canonical name  PositiveCone 
Date of creation  20130322 14:46:54 
Last modified on  20130322 14:46:54 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 13J25 
Classification  msc 12D15 
Related topic  PositivityInOrderedRing 
Related topic  FormallyRealField 
Defines  prepositive cone 