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Sep 15, 2017 at 9:24 comment added Fred Rohrer @Mariano: Have a look at Bourbaki, Algèbre VIII.9.3 Théorème 2 Corollaire 1. (I hope it is ok that the book is not about commutative algebra...)
Oct 9, 2014 at 21:56 comment added Mariano Suárez-Álvarez Someone should write a commutative algebra textbook ASAP and include this.
Jul 27, 2010 at 15:38 comment added Emerton Dear Georges, Of course! Regards, Matt
Jul 27, 2010 at 15:23 comment added Georges Elencwajg Dear Emerton: I find it very cold to address someone as friendly as you by surname. May I use the opportunity to ask for permission to call you Matt? ( I do this automatically if the first name is part of the username)
Jul 27, 2010 at 14:52 comment added Emerton Dear Georges, Thanks for making this important point clear!
Jul 27, 2010 at 14:43 comment added Georges Elencwajg This is, needless to say, perfectly correct. There is a caveat though: to say that $\mathfrak m$ is nilpotent means indeed that some power of this ideal is zero. This is stronger (for non-noetherian rings) than just asserting that $\mathfrak m$ consists of nilpotent elements: the terminology for this last condition is "nil ideal".
Jul 27, 2010 at 14:33 vote accept ashpool
Jul 27, 2010 at 14:27 comment added Keenan Kidwell I didn't know this was true. Thanks!
Jul 27, 2010 at 14:25 history answered Emerton CC BY-SA 2.5