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Apr 13, 2017 at 12:58 history edited CommunityBot
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Feb 4, 2010 at 11:43 vote accept Anweshi
Feb 4, 2010 at 11:30 comment added Kevin Buzzard @Pace: yeah but he explicitly said "B non-zero", for precisely this reason.
Feb 3, 2010 at 21:26 comment added Pace Nielsen *make that "minimal ideal"
Feb 3, 2010 at 21:24 comment added Pace Nielsen Pedantic ring theorist speaking: All elements (even nilpotents) can map to units. Just map to the zero ring. Pedantry aside: first of all we want every ideal $I$ of $R$, which contains the left and right annihilators of $a$, to NOT contain $a$ itself. This encompasses the situation of nilpotents and full idempotents.
Feb 3, 2010 at 20:26 comment added Kevin Buzzard Yes and thanks to Matt too: I mean (1 0;0 0). Oh dear what a disastrous comment! I should have made it an answer, then I could have edited it!
Feb 3, 2010 at 20:24 comment added Kevin Buzzard Yes, thanks darij. The argument is still OK but the "typo" is rather grotesque :-)
Feb 3, 2010 at 20:02 answer added Emerton timeline score: 8
Feb 3, 2010 at 19:23 answer added darij grinberg timeline score: 1
Feb 3, 2010 at 19:11 comment added darij grinberg I think "a can't be contained in any prime ideal" should be "a is contained in every prime ideal" in your answer.
Feb 3, 2010 at 19:06 answer added Mariano Suárez-Álvarez timeline score: -1
Feb 3, 2010 at 19:06 comment added Kevin Buzzard In the commutative case this is easy, right? a can't be contained in any prime ideal (consider A-->Frac(A/P)) and hence a is nilpotent by a standard result in ring theory. In the non-commutative case though I'm not so sure: what about (0 1;0 0) in M_2(C)? M_2(C) has no non-trivial 2-sided ideals so f must be injective and although this isn't a proof, I'm surely not far off.
Feb 3, 2010 at 19:05 history edited Anweshi CC BY-SA 2.5
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Feb 3, 2010 at 18:56 history asked Anweshi CC BY-SA 2.5