22
$\begingroup$

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.

$\endgroup$
3
  • 5
    $\begingroup$ A silly example: division rings work... $\endgroup$
    – Mariano Suárez-Álvarez
    Sep 12, 2011 at 6:12
  • 2
    $\begingroup$ 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. $\endgroup$
    – Ted
    Sep 12, 2011 at 6:35
  • $\begingroup$ 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. $\endgroup$
    – rschwieb
    Apr 23, 2012 at 19:47

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .