# $Ra=Rb$ if and only if $aR=bR$

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.

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A silly example: division rings work... – Mariano Suárez-Alvarez Sep 12 '11 at 6:12
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

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.

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