Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

On which classes of (non commutative) rings we have the following property: $aR=bR$ if and only if $Ra=Rb$ ?

While I googling around I found the notion of "Duo Ring" in which $aR=Ra$ for every $a\in R$. This is stronger that what I am looking for. Even for this, I don't know any example of duo ring.

share|improve this question
5  
A silly example: division rings work... –  Mariano Suárez-Alvarez Sep 12 '11 at 6:12
2  
math.rwth-aachen.de/~Florian.Eisele/ArithGrpRng/li.pdf discusses criteria for a group ring to be a duo ring, and gives the example $\mathbb{Q}Q_8$ as a duo ring, where $Q_8$ is the quaternion group on 8 elements. –  Ted Sep 12 '11 at 6:35
    
To expand on the division ring comment, any direct product of division rings is duo, and more generally any strongly von Neumann regular ring is duo. I've also read a few interesting papers by Weimin Xue on exotic duo rings. I sat here for a while trying to see if it was true for right-and-left Ore domains, but no luck. It's true at least that domains with central units have your property. –  rschwieb Apr 23 '12 at 19:47
add comment

1 Answer

One place to start would be the third question here, in which the property you are considering is referred to only as (*). Unfortunately, while skimming through articles that cited this paper, I didn't see anything else of particular relevance.

Nonetheless, the paper above proves:

Theorem. Let $R$ be a noetherian integrally closed duo domain. Then $aRbR = bRaR$ for all $a,b \in R$ and ideal multiplication is commutative.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.