Suppose we have a noncommutative ring $R$ and multiplicatively closed set that is both right Ore, and right reversible, i.e. it is a right denominator set. Now, we can localize $R$ at $S$ to form $RS^{-1}$. If $R$ is commutative and $S$ is a prime ideal, then I know $RS^{-1}$ is local, but when is this true for $R$ noncommutative and $S$ as above? Is it ever true for a general class of rings, such as when $R$ is Noetherian?

  • $\begingroup$ I haven't played around with noncommutative localization in a while, but it seems like nothing should be different for a prime ideal $P$ in $R$ such that $S=R\setminus P$ is a right denominator set. Did you try apply the commutative proof to the resulting ring of fractions? It seems like nothing is different. $\endgroup$ – rschwieb Dec 2 '16 at 13:36
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    $\begingroup$ Also asking "when is $RS^{-1}$ local?" for arbitrary sets of denominators might be too ambitious(?) I don't know. Maybe a commutative algebraist will tell me it's well known which denominator sets do this. $\endgroup$ – rschwieb Dec 2 '16 at 13:39

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