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Apr 17, 2015 at 1:52 comment added Jonathan Gleason I think the map $A\rightarrow k=\mathrm{End}\, _{\mathsf{Vect}_k}(k)$ into a field should be thought of as a representation of $A$, so that in the general non-commutative case, $\rho :A\rightarrow E$ with $E$ not necessarily a field, we don't want $E$ itself to be a division ring, but rather, 'maps that commute $\rho$' to form a division ring. (I'm being imprecise about this on purpose because I'm not sure what the 'correct' notion of morphism should be.)
Jul 25, 2012 at 17:23 comment added Rasmus For a non-commutative ring $A$, one might replace Field be some meaningful full subcategory of test objects in Ring. A naive first thought would be skewfields.
Oct 25, 2010 at 11:29 comment added Martin Brandenburg Interesting answer, but it does not deal with noncommutative rings (which was the question).
Feb 21, 2010 at 11:05 vote accept Shizhuo Zhang
Feb 15, 2010 at 13:22 history answered Tom Leinster CC BY-SA 2.5