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Jan 10, 2010 at 14:19 history edited Akhil Mathew
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Dec 27, 2009 at 8:55 answer added Hailong Dao timeline score: 4
Dec 13, 2009 at 2:39 history edited TJCM CC BY-SA 2.5
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Dec 12, 2009 at 21:57 comment added Mariano Suárez-Álvarez If "$I$-adic" implies complete in your statement, could you add it explicitly?
Dec 12, 2009 at 18:20 comment added Ben Webster I don't think that's a very good assumption.
Dec 12, 2009 at 18:14 comment added Akhil Mathew Yeah that wasn't clear to me; if $A$ is local and $I$ the maximal ideal, though, the answer is yes. (I had assumed in the comment he meant $I$ was just some ideal).
Dec 12, 2009 at 18:11 comment added Ben Webster Then Nakayama's lemma shows that $M=0$, I believe (I guess it depends what $I$-adic Noetherian ring means).
Dec 12, 2009 at 18:00 comment added Akhil Mathew What if $IM = M$ but $M$ is not free?
Dec 12, 2009 at 17:20 history asked TJCM CC BY-SA 2.5