Timeline for "Étalification" of a scheme
Current License: CC BY-SA 3.0
16 events
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| Mar 10, 2013 at 18:41 | comment | added | Will Sawin | On the upvotes: Perhaps they are based on the idea of hitting functions pulled back from $Z$ with differential operators pulled back from $X$ to distinguish every pair of points. On the highly non-separatedness: I think this will be caused by $X$ being non-normal. Indeed, considering the case of two lines meeting at a point was the inspiration for my current (attempted) answer. But if $X$ is normal and $Z \to X$ is separated, I think $\bar{Z}$ might be separated. | |
| Mar 10, 2013 at 18:33 | comment | added | Martin Brandenburg | @André: For me your answer is still just a reformulation of my question. You write down the sheaf represented by the scheme we are looking for, and then claim that it is a scheme. You don't offer any proof. Meanwhile I wonder about all the upvotes ... does this mean that I am too blind to see the answer? | |
| Mar 10, 2013 at 18:22 | history | edited | André Henriques | CC BY-SA 3.0 |
added 7 characters in body
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| Jun 25, 2012 at 22:30 | comment | added | Will Sawin | I think so. The higher directional derivatives are really coming from a section of the sheafification of $Z$, and they define a map from the local ring at a point of $Z$ to the etale local ring at the profinite completion of the local ring of the point of $X$ it lies over, which must pass through the etale local ring. Agreement of maps to the etale local ring gives local agreement of sections, and local agreement means they're glued together in the espace etale, so they give the same point in the algebraic space. Does that sound good? | |
| Jun 25, 2012 at 8:39 | comment | added | André Henriques | Ok. I think that I have an argument about why there are enough functions on $\tilde Z$. The map $\tilde Z\to Z$ is maybe not a monomorphism, but it has enough features of a monomorphism. More precisely, given two points $x,y\in \tilde Z$, there exists a function $f\in\mathcal O(U)$ (defined on some Zariski open $U\subset Z$) such that the higher directional derivatives of $f$ distinguish $x$ and $y$. I think that this at least deals with the case $\mathbb A^m\to \mathbb A^n$. Can the argument be adapted to work in general? | |
| Jun 25, 2012 at 8:24 | comment | added | André Henriques | @Will Sawin: point taken. It's not a monomorphism. One could try to argue the other way around, and try to find a map from an algebraic space (not a scheme) to $\mathbb A^m$, such that the projection to $\mathbb A^n$ ($n<m$) is everywhere étale. If such an example exists, then that would answer in the negative the original question of Martin. But somehow, I don't believe that such a thing exists... | |
| Jun 24, 2012 at 22:21 | comment | added | Will Sawin | I was trying to get an argument of the following type to work: Assume $X$ is normal, then etale covers of $X$ are ireducible. Fix a point $P\in Z$, and look at all the maps over $X$ from an etale cover of $X$ to $Z$ whose generic point maps to $P$. If you can show that all of these factor through a single etale cover of $X$, take the disjoint union of those and you're done. If that's true it should be relatively elementary, but my argument for it was becoming too messy. I'm not sure how this argument would extend to the non-normal case. | |
| Jun 24, 2012 at 22:17 | comment | added | Will Sawin | If Z is $\mathbb A^2$ with the obvious projection to $X=\mathbb A^1$ over, for simplicity, an algebraically closed field $k$, then a closed point of the espace etale is an $x$ coordinate, a $y$ coordinate, and a polynomial $f(x,y)$ with those roots whose $y$-derivative is nonzero at that point. A closed point of $Z$ is just an $x$ coordinate and a $y$ coordinate. How is that a monomorphism? If it is, why is it obvious that the other Zariski-local sections exist? | |
| Jun 24, 2012 at 20:09 | comment | added | André Henriques | I see your point about needing to show that my $\tilde Z$ is a scheme and not just an algebraic space. Schemes can be thought of as algebraic spaces with enough Zariski-local sections of their structure sheaf (enough to separate points). Since my algebraic space $\tilde Z$ comes equipped with a monomorphism to the scheme $Z$, and the latter has enough Zariski-local sections of its structure sheaf, I would guess that $\tilde Z$ also has enough Zariski-local sections of its structure sheaf, and hence is a scheme. Is this an argument, or am I fooling myself? | |
| Jun 24, 2012 at 17:47 | comment | added | David Carchedi | @Will: This is certainly true, and it made me wary of the answer. However, perhaps this is just an abuse of terminology, and "espace etale" doesn't mean what it usually does (as it only applies to the open cover topology of a locale). Perhaps it's also customary to call an algebraic space etale over $X$ which represents an etale sheaf, the sheaf's "espace etale". If so, I'll wait for Andre to explain why in this case, its not only an algebraic space, but a scheme. | |
| Jun 24, 2012 at 17:00 | comment | added | Will Sawin | @David: Taking the espace etale of the Zariski sheaf certainly gives the wrong answer. If $X$ is a nontrivial etale cover of $Z$ then it has no Zariski sections, yet obviously a functor that takes it to the empty set is not adjoint. | |
| Jun 24, 2012 at 16:35 | comment | added | Martin Brandenburg | So what is the definition of $\tilde{Z}$? | |
| Jun 24, 2012 at 15:11 | comment | added | David Carchedi | @Martin: I think Andre's point is that the sheaf "sections of $X \to Z$" is a Zariski-sheaf, i.e. a sheaf over the underlying topological space of $X,$ and then one can form the \'etal\'e space of this sheaf of sets, which is space over $X$ via a local homeomorphism, and one can pull the structure sheaf of $X$ back along this local homeomorphism, to get a locally ringed space locally isomorphic to $X,$ hence it is also a scheme. | |
| Jun 24, 2012 at 14:33 | comment | added | Martin Brandenburg | @André: This is basically just a reformulation of the question. Why is your sheaf represented by an étale scheme? | |
| Jun 24, 2012 at 14:20 | comment | added | David Carchedi | Every sheaf on the small etale site is representable by an algebraic space etale over the base. Is it true that if the original sheaf is the sheaf of sections for a morphism of schemes that this algebraic space is a scheme? | |
| Jun 24, 2012 at 12:04 | history | answered | André Henriques | CC BY-SA 3.0 |