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Jun 15, 2020 at 7:27 history edited CommunityBot
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Dec 23, 2011 at 8:57 answer added Fred Rohrer timeline score: 3
Sep 22, 2011 at 5:32 vote accept Fred Rohrer
Sep 18, 2011 at 19:39 answer added Pham Hung Quy timeline score: 3
Nov 24, 2010 at 7:39 comment added Fred Rohrer A ring with ITI with respect to every ideal is not necessarily Noetherian. Indeed, a ring with the property that every proper ideal is nilpotent has ITI with respect to every ideal, hence it suffices to exhibit a non-Noetherian local ring with nilpotent maximal ideal. This we do by taking the polynomial algebra in countably many indeterminates over a field modulo the ideal generated by all products of two indeterminates.
Nov 23, 2010 at 7:40 comment added Neil Strickland I do not know about these questions in particular, but Greenlees and May have worked quite hard to prove as much as possible about local (co)homology without Noetherian assumptions.
Nov 23, 2010 at 2:23 history asked Fred Rohrer CC BY-SA 2.5