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Following on from the question Does quotient commute with localization?, I'm interested in doing the same sort of thing but over non-commutative rings. Is there a non-commutative analogue of the exact sequences result used in that question? References (commutative or non-commutative) would be appreciated.

Assume we've replaced the multiplicative set with a denominator set. Do concerns about zero-divisors need to play any part in the proof, or does it all work regardless? (In practice I'm only interested in denominator sets generated by finitely many normal regular elements that remain regular in the quotient, but I'd like to start as general as possible and then specialize later.)

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Eithil
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  • The answer to your first question is yes. In particular, if $S$ is a right Ore set of a ring $A$, then $AS^{-1}$ is flat over $A$, that is, passing to the localization preserves exact sequences. One good resources are Michael Artin's notes http://www-math.mit.edu/~etingof/artinnotes.pdf. For a more in-depth look at noncommutative localization, you might want to look at Allen Bell's notes https://pantherfile.uwm.edu/adbell/www/Research/locnotes.pdf. – J. Gaddis Nov 23 '13 at 20:55
  • Excellent, thank you! Bell's notes even cover the case where $S$ might contain non-regular elements, which is perfect. Now to try and visualise a ring where that happens... – Eithil Nov 25 '13 at 11:10

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