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Jul 12, 2011 at 14:16 history edited Michal Zydor CC BY-SA 3.0
added 49 characters in body
Jul 12, 2011 at 14:00 history edited Michal Zydor CC BY-SA 3.0
changed "regular scheme" to "scheme"
Jul 12, 2011 at 13:52 vote accept Michal Zydor
Jul 12, 2011 at 13:44 history edited Michal Zydor CC BY-SA 3.0
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Jul 12, 2011 at 8:45 answer added S. Carnahan timeline score: 3
Jul 12, 2011 at 6:19 comment added Martin Brandenburg Do you really mean $k[[\epsilon]]$? If you are writing $k$-valued points above, this does not give the correct algebraic set. Perhaps you mean $k((\epsilon)) := \text{colim}_n \text{Spec}(k[x_{-n},...,x_0,x_1,...])$?
Jul 12, 2011 at 5:31 comment added S. Carnahan I think the easiest example is the completion of $\mathbb{A}^1_k$ at a $k$-point. As a locally ringed space, it is a topological point whose ring of functions is $k[[t]]$ - obviously not a scheme.
Jul 11, 2011 at 22:03 comment added Michal Zydor Yes, that's what I'm trying to do. Just hoping for an "elementary" proof, since this seems to be the easiest example (maybe alongside with infinite dimensional vector space).
Jul 11, 2011 at 21:27 vote accept Michal Zydor
Jul 11, 2011 at 22:01
Jul 11, 2011 at 21:03 comment added Martin Brandenburg Do you want a prove that this limit does not exist in the category of schemes? In this case you may find mathoverflow.net/questions/65506/… helpful. Often people just "argue" in this case that there does not seem to be any natural model of this limit, thus it does not exist ... perhaps motivated by some sort of "good limits" in the same spirit as GIT uses good quotients.
Jul 11, 2011 at 20:13 answer added Jason Starr timeline score: 5
Jul 11, 2011 at 19:27 history edited Michal Zydor CC BY-SA 3.0
added 210 characters in body; edited title
Jul 11, 2011 at 18:24 history asked Michal Zydor CC BY-SA 3.0