# 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. 1.

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

2. 2.

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

3. 3.

$-1\notin P$

4. 4.

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

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. 5.

$P\cup(-P)=F$

2. 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, $\operatorname{sqr}(F)\subseteq\overline{P}$ 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 $a\neq 0$, 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  $\leq$, on $F$ by:

 $a\leq b\Longleftrightarrow b-a\in P$

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

1. 1.

$a\leq b\Rightarrow a+c\leq b+c$

2. 2.

$0\leq a$ and $0\leq b\Rightarrow 0\leq ab$.

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

## References

• 1 A. Prestel, Lectures on Formally Real Fields, Springer, 1984
Title positive cone PositiveCone 2013-03-22 14:46:54 2013-03-22 14:46:54 CWoo (3771) CWoo (3771) 10 CWoo (3771) Definition msc 13J25 msc 12D15 PositivityInOrderedRing FormallyRealField pre-positive cone