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Sep 6, 2010 at 15:29 vote accept Steven Sam
Oct 20, 2009 at 23:26 vote accept Steven Sam
Sep 6, 2010 at 15:29
Oct 20, 2009 at 23:26 comment added Steven Sam I think the graded Nakayama's lemma is much easier to prove. If the ring is positively graded, we can just use induction: say mM = M where m is the maximal ideal. Then M_0 = 0 since nothing maps to it, and hence M_1 = 0, etc. So maybe I will just reask if I run into a specific statement I am unsure about. Thanks!
Oct 20, 2009 at 1:58 history edited Greg Stevenson CC BY-SA 2.5
added stuff on more general gradings
Oct 20, 2009 at 1:42 history edited Greg Stevenson CC BY-SA 2.5
made it into more of a proper answer
Oct 19, 2009 at 5:25 history answered Greg Stevenson CC BY-SA 2.5