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


*

*$P$ is closed under $+$ and $\cdot$.

*$x^2 \in P$ for all $x \in R$.

*$-1 \notin P$.

*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.
 A: 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.)
A: 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.
