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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jun 4, 2012 at 6:16 comment added Fred Rohrer Dear @Philipp, thank you very much for your answer. That the irreducible components are pairwise disjoint follows also from the general observation that irreducibility of the stalk at a point of a scheme $X$ is equivalent to this point lying in precisely one irreducible component of $X$ (cf. [EGA I.2.1.9]). (In particular, irreducibility of all stalks is equivalent to the irreducible components being pairwise disjoint.)
Jun 4, 2012 at 6:08 vote accept Fred Rohrer
Jun 2, 2012 at 13:25 history edited Will Sawin CC BY-SA 3.0
fixed latex!
Jun 2, 2012 at 13:23 history edited Dylan Thurston CC BY-SA 3.0
tex formatting issues
Jun 2, 2012 at 12:40 history answered Philipp Hartwig CC BY-SA 3.0