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Aug 26, 2014 at 17:01 history edited David E Speyer CC BY-SA 3.0
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Jul 22, 2014 at 15:48 vote accept rrr
Jul 22, 2014 at 15:48 comment added rrr Many Thanks. It is exactly what i want.
Jul 22, 2014 at 9:04 comment added Simon Wadsley That the maximal right ring of quotients of $Mat_n(R)$ is $Mat_n(q.f(R))$ (in your notation) follows immediately from Corollary 3.1.6 of McConnell and Robson's book `Noncommutative Noetherian rings': ams.org/bookstore-getitem/item=GSM-30
Jul 22, 2014 at 5:00 comment added rrr Thanks a lot. in fact i more like a reference for such result. q.f.(R) is the skew field of fraactions. Since the isomorphisms of matrix rings over division rings will imply the isomorphisms of division rings. I wonder whether we can find somewhere that the maximal right ring of quotient of $Mat_n(R)$ is $Mat_n(q.f.(R))$?
Jul 21, 2014 at 19:43 history answered David E Speyer CC BY-SA 3.0