# Do partially ordered rings that aren't totally-ordered, which nonetheless satisfy $x+y \geq 0 \rightarrow x \geq 0 \vee y \geq 0,$ exist?

(All my rings are commutative and unital. I include $1 \geq 0$ as a partially-ordered ring axiom.)

Let $R$ denote a partially-ordered ring. Observe that if $R$ is totally-ordered, then $R$ satisfies the following implication: $$(*) \qquad x+y \geq 0 \rightarrow x \geq 0 \vee y \geq 0.$$

Interestingly, the partially-ordered ring $\mathbb{R} \times \mathbb{R},$ which obviously fails to be totally-ordered, also fails to satisfy $(*).$ For example $(1,-1)+(-1,1) \geq (0,0)$, but $(1,-1) \not\geq (0,0)$ and $(-1,1) \not\geq (0,0).$ In fact, this will happen if we replace $\mathbb{R}$ by any totally-ordered ring. Nonetheless, I expect that partially ordered rings satisfying $(*)$ exist, which nonetheless aren't totally-ordered. Any ideas how to construct such a thing?

• "Linear" is a confusing word to use in this context because it might or might not have anything to do with some compatibility with addition. Better to use "total." – Qiaochu Yuan Mar 24 '15 at 17:38
• @QiaochuYuan, I've been trying to wean myself off using "total" though because when you're simultaneously talking about partial functions, it all gets very confusing. I agree that in this context, total is better. – goblin Mar 24 '15 at 21:10

No; in fact this axiom is equivalent to the order being total. To see this, set $y = -x$. Then $x + (-x) = 0 \ge 0$, hence either $x \ge 0$ or $-x \ge 0$, and then setting $x = a - b$ shows that either $a \ge b$ or $b \ge a$.
• Nice one! And the construction for semirings is easy; $\mathbb{N} \times \mathbb{N}$ satisfies $(*)$ vacuously, but certainly isn't totally-ordered. In fact, for any non-trivial partially-ordered semiring $N$ with $0$ as least element, $N \times N$ will be a not-totally-ordered semiring satisfying $(*)$. So everything is wrapped up in a neat little package. – goblin Mar 24 '15 at 21:07