# Positive cone definition for ordered rings

There are two definitions of an ordered field: you can define it as a field with a total ordering and certain axioms relating the ordering to the field operations, or as a field with a "positive cone": a subset of the field satisfying certain axioms (see Orderered Field on Wikipedia). These definitions are equivalent: the total orders and the positive cones are in one-to-one correspondence.

I am wondering if the same equivalence holds for rings. I'm taking the definition of a totally ordered ring from Wikipedia, and I am taking the same definition for a positive cone as is given for fields: a positive cone on a ring $R$ is a subset $P$ of $R$ satisfying

1. $P$ is closed under $+$ and $\cdot$.
2. $x^2 \in P$ for all $x \in R$.
3. $-1 \notin P$.
4. either $x \in P$ or $-x \in P$ for all $x \in R$.

I have checked that the function $F$ taking a ring total order $\leq$ to the set $\{x \in R | x \geq 0\}$ and the function $G$ taking a positive cone $P$ to the relation $x \leq y \equiv y - x \in P$ are inverses.

I have also shown that if $\leq$ is a ring total order then $F(\leq)$ is a positive cone. However, I don't see how to show that if $P$ is a positive cone then $G(P)$ is a total order. In particular, I don't see how to show that $G(P)$ is anti-symmetric without using cancellation.

So I am wondering whether this equivalence holds or not. If not, is there a simple tweak to the positive cone style definition that makes it work? I'm interested because I think it is more natural to implement this style of definition in the algebra software I'm designing.

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The equivalence does not hold in $\:\mathbb{Z}\left[\sqrt0\hspace{.01 in}\right]\:$ with $\;\;\;\; P \;\; = \;\; \left\{ \:a+\left(b\cdot \sqrt0\right) \: : \: 0\leq a \: \right\} \;\;\;\;$.

Replacing 3 with $\;\;$ "For all $\:x\in R\:$,$\:$ if $\:x\in P\:$ and $\: -x \in P \:$ then $\:x=0\:$." $\;\;$ will make it work.
(Since that is exactly what you need for anti-symmetry.)

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Thanks! And that would be a suitable definition for fields as well since in a field (or an integral domain even) the two are equivalent. – mdgeorge Nov 29 '12 at 20:59
The definition has some problems still. First, it's not necessary to suppose that $x^2$ is in $P$ (in the strict ordering variant I describe in my answer, this doesn't quite hold). In any case, if $x\in P$, then we have $x^2\in P$ by (1). Otherwise, $-x\in P$, so $x^2 = (-x)^2 \in P$. In the weak ordering case, $0\in P$ so $0^2\in P$. In the strict ordering case, this doesn't apply. The second problem is that $-1$ only exists in a ring with unity. But in a (non-null) ring with unity, $1≠0$, so $1=1^2 \in P$. Thus $-1 \notin P$ by the antisymmetry condition. – dfeuer Feb 2 '13 at 19:33

I've been messing around with this notion for groups. Suppose you have a group $(G,\circ)$ with identity $e$ and a subset $C$ of $G$ such that for each $x,y\in G$:

$x,y \in C \implies x \circ y \in C$

$x\circ y \in C \implies y \circ x \in C$.

Then if you define a relation $\mathcal R$ by letting $x \mathrel{\mathcal R} y \iff y\circ x^{-1}\in C$, then that relation will be "compatible" with $\circ$ in the sense that if $x \mathrel{\mathcal R} y$ then $x \circ z \mathrel{\mathcal R} y \circ z$ and $z \circ x \mathrel{\mathcal R} z \circ y$. The relation will also be transitive.

If you impose the condition that $C \cap C^{-1} \subseteq \{e\}$, then the relation will be antisymmetric.

If you impose the condition that the group identity $e\in C \cap C^{-1}$, then the relation will be reflexive. Combined with the condition for antisymmetry, this gives a weak partial order. If instead you impose the condition that $e\notin C \cap C^{-1}$, then you get an irreflexive relation which, combined with the antisymmetry condition, gives a strict partial order. These combined conditions can be written as $C\cap C^{-1}=\{e\}$ and $C \cap C^{-1} = \varnothing$, respectively.

If you add to the weak partial order condition the condition that $C\cup C^{-1} = G$, you get a weak total order. If to the strict partial order condition you add the condition that $C\cup C^{-1}\cup\{e\}=G$, you get a strict total order.

All that is good for general groups. In a ring $(R,+,\cdot)$ you need to impose one of the above-described conditions for $(R,+)$, and add the condition that $x,y \in C \implies x\cdot y \in C$.

There are various definitions for a partially ordered field, but the definition that makes the most sense to me requires that $x \in C \implies x^{-1} \in C$. This added condition is automatically satisfied for a totally ordered field.

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