# Is every well ordered commutative nontrivial ring with identity an well ordered integral domain?

$\mathbb Z$ is up to ring isomorphism the only well ordered domain, that is, $\mathbb Z$ is a integral domain and every nonempty subset of $\{n \in \mathbb Z: n\geq 0\}$ has a least element.

But what happens if we remove the integral property? i.e, is every well ordered commutative nontrivial ring with identity an well ordered integral domain? I was thinking about this but I came to no answer.

Edit: whacka's answer is right since by assuming the ring is ordered by $\leq$ we are assuming that the positive elements are closed under multiplication. But what if instead of assuming $a, b>0$ imples $ab>0$, we assume that if $a, b \geq 0$ then $ab\geq 0$? That was what I was assuming when I was trying to find the answer (:

• If $x\cdot y=0$, we may assume both of them are non-negative, and by the linearity of $\leq$ it means both are positive. But in an ordered ring the product of two positive elements is positive. No? I'm probably just yammering here. Apr 16, 2015 at 0:19
• Also math.stackexchange.com/questions/161329/… might be relevant. Apr 16, 2015 at 0:20
• Sorry, the standard exercise is that finite fields cannot be ordered. I thought there was another standard exercise with zero divisors but I can't seem to remember it correctly, apparently.
– anon
Apr 16, 2015 at 0:33
• But regardless to my answer below; aren't ordered rings usually axiomatized with strict orders rather than allowing equality? This should make my comment/the answer by @whacka valid again. Apr 16, 2015 at 1:14

Let $R$ be an ordered ring. I assume you are asking for the set of positive elements to be well-ordered. If so, let $\varepsilon > 0$ be the smallest positive element.
If $\varepsilon$ is invertible, then $\varepsilon^2 \neq 0$. So $0 < \varepsilon^2 \leq \varepsilon \cdot 1 = \varepsilon$ and minimality of $\varepsilon$ imply that $\varepsilon^2 = \varepsilon$. Cancellation gives $\varepsilon = 1$.
Assume for contradiction that $\varepsilon < 1$. Then $1 - \varepsilon > 0$. By the descending chain condition, there is a largest integer $n \geq 1$ such that $$1 > 1 - \varepsilon > \cdots > 1 - n \varepsilon > 0 \quad \mbox{and} \quad 1 - (n+1)\varepsilon \leq 0.$$ By minimality of $\varepsilon$ we have $1 - n\varepsilon \geq \varepsilon$, so that $1 - (n+1) \varepsilon \geq 0$. It follows that $1 - (n+1)\varepsilon = 0$, giving the contradiction that $\varepsilon$ is a unit with inverse $n+1$.
So $\varepsilon = 1$ is the smallest positive element. Now an argument as the one given by Bill Dubuque in the above link by Asaf Karagila shows that $R$ has no infinite elements, so that its positive elements are the positive integers and $R \cong \mathbb{Z}$.
• That was the thing missing from my answer, if $0<\varepsilon^2=\varepsilon$ then $\varepsilon=1$. So close... :-) Apr 16, 2015 at 9:40