Timeline for Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$?
Current License: CC BY-SA 2.5
15 events
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| Aug 18, 2010 at 16:47 | comment | added | David E Speyer | Write (p,q,r) for (x^2, xy, y^2). Let h(p,q,r) be a polynomial with smooth zero locus in A^2, not passing through the origin. The covers we are speaking about are given by adjoining u and v, subject to u^2=ph, uv=qh, v^2=rh, uq=vp and ur=vq. This is a two fold cover, branched over h=0. In particular, two such covers for h_1 and h_2 cannot factor through each other. Unless I am missing something, the conditions that h=0 be smooth and miss the origin makes the cover smooth. | |
| Aug 18, 2010 at 16:41 | comment | added | David E Speyer | I haven't figured out where BCnrd and Francesco arguments collide yet, but I've written out Francesco's covers explicitly. They do seem to work, and they are all formally of the form BCnrd predicted, so the difficulty should be in the (formal local)-->global transition. (continued...) | |
| Aug 18, 2010 at 1:27 | vote | accept | Anton Geraschenko | ||
| Aug 17, 2010 at 1:33 | comment | added | BCnrd | Hey guys, so maybe for my proposed argument which reaches a different conclusion, in the passage from formal to global cases there's something which goes awry when comparing things in different factor rings upstairs after completing on the base. Well, if Anton tries to write out the details (I only thought it through in a superficial way in my mind) then maybe an error will be found. Hopefully it won't be too bone-headed, and will be curious to see what geometric interpretation it has for the above counterexample (assuming that is right). | |
| Aug 16, 2010 at 19:07 | comment | added | David E Speyer | I see, I was misreading the original question. Still digesting your answer, but it is starting to look right to me. | |
| Aug 16, 2010 at 18:40 | comment | added | Francesco Polizzi | I pointed out that there are many curves $C$ in $Y$ such that there is a double cover of $Y$ branched on the vertex and $over$ $C$. So it seems to me that it is only when I take $C$ to be the boundary conic that I get a double cover of $S$ which is branched only at the vertex. In the other cases, I still obtain a double cover of $S$, which is branched at the vertex and over a smooth curve $C'$, obtained by removing from $C$ the intersection points with the boundary conic. I think this is ok, since the original question only asked for a finite morphism. But if I'm wrong, please correct me. | |
| Aug 16, 2010 at 17:34 | comment | added | David E Speyer | OK, but I think this makes your answer wrong. Let $Y$ be the projective closure of $S$. What you are pointing out is that there are many curves $C$ in $Y$ for which there are double covers of $Y \setminus C$. However, it is only when you take $C$ to be the boundary conic that you get a double cover of $S$, as originally requested. | |
| Aug 16, 2010 at 17:27 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Aug 16, 2010 at 17:10 | comment | added | Francesco Polizzi | @David You are right. I wrote the answer thinking about the projective situation, when you compactify and obtain a map $\mathbb{P}^2 \to$ Cone in $\mathbb{P}^3$. So the map $f_k$ must be considered as the restriction to the affine cone of the double covering defined as before. The map $f$ in particular corresponds to the case $k=1$ when the conic is contained in the hyperplane at infinity. But when $k \geq 2$ you always have a divisorial component in the branch locus also in the affine case. | |
| Aug 16, 2010 at 16:54 | comment | added | David E Speyer | "the morphism f \colon \mathbb{A}^2 \to S corresponds to a double cover branched on the vertex of the cone and on a smooth conic." I don't understand. Working with k algebraically closed and of characteristic not 2, every closed point has two preimages except the vertex. So where is the smooth conic you speak of? | |
| Aug 16, 2010 at 13:06 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Aug 16, 2010 at 12:22 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Aug 16, 2010 at 11:32 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Aug 16, 2010 at 10:57 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Aug 16, 2010 at 9:49 | history | answered | Francesco Polizzi | CC BY-SA 2.5 |