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

As it turns out, a field endowed with a pre-positive cone has an order structure. The field is called a formally real, 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:

- 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, $\operatorname{sqr}(F)\subseteq\overline{P}$ 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. $a\leq b\Rightarrow a+c\leq b+c$

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

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

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

## Mathematics Subject Classification

13J25*no label found*12D15

*no label found*

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## Comments

## Difference in definitions

Again we have difference in definition. I think you need to add a reference here. Your definition of positive cone is not for instance in agreement with the definition found in

"Real Algebraic Geometry" by Bochnak, Coste and Roy. Springer 1998.

In modern real algebra, as far as I know, we do not consider P \cup -P to be the entire field. Maybe like your definition of partially ordered rings, you have to add a reference here.

## Re: Difference in definitions

Check out Lectures on Formally Real Fields by Alexander Prestel, published by Springer in 1984. I will add the reference shortly...

The condition P \cup - P = F* basically renders the partial order linear. So the entire book deals with linearly ordered fields (and rings).