Timeline for Colimits of schemes
Current License: CC BY-SA 2.5
14 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 19, 2015 at 7:18 | comment | added | user40276 | Sorry, but why $W$ is not empty. Maybe I misunderstood what $V$ and $V'$ means. I'm assuming that $V$ is the connected component of the inverse image of the map $A^1 \coprod A^1 \rightarrow P$. | |
| May 9, 2010 at 0:31 | comment | added | Anton Geraschenko | The prototypical bug-eyed cover I know of is the quotient of $\mathbb A^1$ by the ℤ/2 action given by x↦-x, except you "remove the action at 0" from the relation. The pushout in the category of algebraic spaces is "the right one", but there is also a pushout in the category of schemes. If there were an example which is an algebraic space, it would also answer David Brown's question mathoverflow.net/questions/4587/…. | |
| May 9, 2010 at 0:19 | comment | added | Martin Brandenburg | The proof above shows more generally: Let $X$ be an integral scheme with non-closed generic point $\eta$, such that the closed points are dense in $X$ (for example a variety over a field). Then the coequalizer of $\eta \rightrightarrows X \sqcup X$ does not exist. I think a slogan for the proof is that identifying the two generic points implies that some neighborhoods should be identified as well, but closed points are not identified because you can use test schemes $X$ and $X$ with a doubled point. | |
| May 9, 2010 at 0:11 | comment | added | Anton Geraschenko | @BCnrd: Awesome! Thanks. @Martin: yes, you're right ... even when I try to get these things backwards, I get them backwards. | |
| May 9, 2010 at 0:10 | history | edited | Anton Geraschenko | CC BY-SA 2.5 |
inserted Brian's proof
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| May 9, 2010 at 0:08 | comment | added | BCnrd | @VA: can you explain your quotient suggestion more fully, so I can see how it is different from the line with a doubled point? Maybe I am not seeing what $\mathbf{Z}/2\mathbf{Z}$-action you have in mind. I did consider trying to do something more ordinary like this, but got stuck on not removing enough closed points. I agree it is much better if the example will be an algebraic space. | |
| May 8, 2010 at 23:52 | comment | added | VA. | An easier example is to divide the "double-headed snake" $\mathbb A^1\cup \mathbb A^1$ by $\mathbb Z_2$. That is not a scheme either, but it is an algebraic space. In other words, replace your $Spec\ k(t)$ by $Spec\ k[t,1/t]$ . That feels more ordinary. Koll'ar called this algebraic space this a "bug-eyed cover" of 1. He has a paper with "bug-eyed" in the title on these. | |
| May 8, 2010 at 23:23 | comment | added | Martin Brandenburg | @Anton: The problem here is to represent a covariant functor Sch -> Set, right? | |
| May 8, 2010 at 23:12 | vote | accept | Martin Brandenburg | ||
| May 8, 2010 at 23:11 | comment | added | Martin Brandenburg | Great teamwork! Thank you very much, Anton and Brian :) | |
| May 8, 2010 at 22:03 | comment | added | Anton Geraschenko | I have one answer to the last question. If F is a contravariant functor from Schemes to Sets such that F(Spec(ℤ)) is more than one point, then F cannot be corepresentable. | |
| May 8, 2010 at 22:02 | history | edited | Anton Geraschenko | CC BY-SA 2.5 |
added 114 characters in body
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| May 8, 2010 at 21:55 | history | answered | Anton Geraschenko | CC BY-SA 2.5 |