Different definitions of ordered ring?

Question

My book defines an ordered ring as a ring $R$ with total order $\leq$ such that

$x<y\implies x+z<y+z$
$x,y>0\implies xy>0$.

The ordered ring definition from Wikipedia uses

$x\leq y\implies x+z\leq y+z$
$x,y\geq 0\implies xy\geq 0$.

It seems to me that these definitions are not equivalent. The first definition implies the second, but the second doesn't imply the first. In the second definition, we could have $x,y > 0$ and $xy=0$, and in the first definition this is impossible. For an ordered field I believe the definitions are equivalent, but for rings they are not. Is this true?

score 3 · Answer 1 · answered Sep 12 '20 at 02:58

You're right, these definitions aren't equivalent. Below I'll give a counterexample which has a multiplicative identity (for me rings always have multiplicative identity).

Let me rephrase both definitions in a way that I personally find easier to think about. A strict ordered ring (your first definition) is a ring $R$ equipped with a subset $R_{+}$, the positive elements of $R$, satisfying the following axioms:

$R_{+}$ is disjoint from $R_{-} = - R_{+}$, neither of them contains $0$, and $R$ is the disjoint union $R_{+} \sqcup \{ 0 \} \sqcup R_{-}$ (trichotomy).
$R_{+}$ is closed under addition (transitivity).
$R_{+}$ is closed under multiplication.

(We recover the total order from here by defining $x < y$ iff $y - x \in R_{+}$.)

An ordered ring (Wikipedia's definition) is a ring $R$ equipped with a subset $R_{\ge 0}$, the nonnegative elements of $R$, satisfying the following axioms:

The intersection of $R_{\ge 0}$ and $R_{\le 0} = - R_{\ge 0}$ is exactly $\{ 0 \}$, and $R = R_{\ge 0} \cup R_{\le 0}$.
$R_{\ge 0}$ is closed under addition (transitivity).
$R_{\ge 0}$ is closed under multiplication.

Which of these definitions seems more natural to you will depend on whether you find it more natural to think about total orders in terms of $<$ or in terms of $\le$. The first two axioms turn out to be equivalent (once we translate between "positive" and "nonnegative" by adding resp. removing $0$), but in the third axiom the difference is that the second definition allows the product of two positive elements to be zero, and so allows zero divisors, while the first definition implies that $R$ is an integral domain.

Here's an example of a non-strict ordered ring that seems pretty natural to me: consider the ring $R = \mathbb{R}[\varepsilon]/\varepsilon^2$, equipped with the ordering in which the nonnegative elements are $x + y \varepsilon$ such that either $x > 0$ or $x = 0$ and $y \ge 0$. Then $\varepsilon$ is positive but $\varepsilon^2 = 0$. This ring is a natural setting for comparing the growth rates of, say, two polynomials near a point.

+1 for giving a much more conceptual answer. – Bill Dubuque Sep 12 '20 at 18:59 — Bill Dubuque, Sep 12 '20 at 18:59

score 2 · Accepted Answer · answered Sep 12 '20 at 02:39

2

Take the (additive) group $Z$ of integers and define the product as $xy=0$ for every $x,y$. Then $Z$ becomes an associative ring. It has a total order which satisfies the second definition but not the first. This is the natural order on integers.

answered Sep 12 '20 at 02:39

markvs

19,653

Thanks for the example – Lilypad Sep 12 '20 at 02:49
Would you have an example with a multiplicative identity? – J.-E. Pin Sep 12 '20 at 05:46
One can add multiplicative identity and extend the order lexicographically. Essentially that is done in anoither answer. – markvs Sep 12 '20 at 07:34
The emvedding is standard: The unital ring is $\Bbb {Z}+Z$ where $Z$ is the ring from my answer. $$(n+z)(m+z')= ( nm)+(nz'+mz+zz').$$ I think it is called ring $Z$ with identity adjoint. – markvs Sep 12 '20 at 07:51
@J.-E.Pin It is known as the Dorroh adjunction of $1$. – Bill Dubuque Sep 13 '20 at 01:09

Different definitions of ordered ring?

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