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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Aug 23, 2013 at 7:00 vote accept Damien Robert
Aug 22, 2013 at 11:19 comment added Damien Robert Yes but as Will pointed out, connectivity is not a local property, so that's why I used locally noetherian instead. But this is a good point: if a local property fails for the global sections of a non affine scheme, is there any sort of additional global property that makes it work?
Aug 22, 2013 at 2:20 answer added Karl Schwede timeline score: 7
Aug 21, 2013 at 21:07 comment added Eric Wofsey It might be better to just assume all your schemes are connected for integrality. Then you don't need any noetherian hypothesis.
Aug 21, 2013 at 21:02 comment added Fred Rohrer I misunderstood the question and thus deleted my answer.
Aug 21, 2013 at 20:46 comment added Damien Robert Yes you are right of course! I just wanted to add another example than reduced, that's why I gave the integrality example. One could correct this as follows: a noetherian ring whose stalks are integral is a product of domain. This is a local condition, and so if I am not mistaken a "locally integral" locally noetherian scheme has global sections a product of domains also.
Aug 21, 2013 at 20:31 comment added Will Sawin Integrality is not local by your definition, I don't think. A disjoint union is an open cover, and a disjoint union of integral schemes need not be integral.
Aug 21, 2013 at 19:50 history edited Damien Robert CC BY-SA 3.0
Fix spelling in title
Aug 21, 2013 at 19:43 history asked Damien Robert CC BY-SA 3.0