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Jun 1, 2015 at 11:26 vote accept Konan
Jun 1, 2015 at 11:18 comment added Jason Starr @Konan: Yes, that is what prevents the functor from being effectively prorepresentable. Thus the functor is not an algebraic space.
Jun 1, 2015 at 11:14 comment added Konan Thank you very mcuh for this example. I have one question to reassure myself I understood correctly. The values of the functor restricted to the sequence are singletons, whereas the value on the formal scheme \wideht{Spec} $\mathbb Z_p$ is the empty set, right? This is what obstructs the functor being effetively pro-representable?
S Jun 1, 2015 at 10:56 history answered Jason Starr CC BY-SA 3.0
S Jun 1, 2015 at 10:56 history made wiki Post Made Community Wiki by Jason Starr