One way to define an ordered ring is as a ring with a total order $\leq$, satisfying
- $x\leq y\implies x+z\leq y+z$;
- $0\leq x$ and $0\leq y\implies 0\leq xy$.
Does it hurt to replace 1. with the condition
- $x<y\implies x+z<y+z$?
What about for fields?
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2 Answers
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If you define $$x < y ⇔ (x ≤ y ~\text{and}~ x≠ y),$$ note that $$(A)~~ x = y ⇒ x + z = y + z \quad \text{and}\quad (B)~~x ≠ y ⇒ x + z ≠ y + z.$$ Now “(A) and (3) ⇒ (1)” and “(1) and (B) ⇒ (3)” are straightforward by case differentiation.
(A) is always true and (B) (being the converse to (A)) is true if addition is right-cancellable (which always holds for rings).
If you think about semi-rings, consider $ℕ_0 = (ℕ_0,·,\max,≤)$ with the usual notion of “$≤$”. This is a semi-ring. Note that $0 < 1$, but $0·0 = 0·1$, where (1) and (2) certainly hold.
(If you want an multiplcative identity, extend by $∞$, defining $0·∞ = 0$ and $n·∞ = ∞$ else. This should work – not sure, though.)
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$\begingroup$ Of course! Thanks for pointing out right-cancellability. Now how about for semirings, in which addition might not be right-cancellable? $\endgroup$ Commented Dec 23, 2014 at 15:14
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$\begingroup$ Thanks, again. I'm pretty green when it comes to ordered algebraic structures, hence the soft questions. $\endgroup$ Commented Dec 23, 2014 at 15:41
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Certainly, if you know (3), then (1) is an easy consequence. On the other hand, if you know (1), you might worry that you have a case when $x < y$ but $x + z = y + z$ (this is the only way you could possibly not satisfy (3)). But this is impossible---if $x + z = y + z$, then $x = y$, and therefore you couldn't have had $x < y$ (by virtue of $<$ being a total order).
answered Dec 23, 2014 at 15:05