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 (:
 A: 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 $R \cong \mathbb{Z}$.
