Timeline for Something like upper semicontinuity for finite fibres
Current License: CC BY-SA 3.0
14 events
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| Jul 5, 2013 at 22:09 | comment | added | Vivek Shende | @ Karl: the fibre over the generic point being a point is not enough to get birationality, and can't be: consider Frobenius. (So characteristic zero is a necessary hypothesis in the question.) | |
| Jul 5, 2013 at 16:02 | comment | added | Karl Schwede | One can also observe that the fraction field extension is a field extension. The degree and transcendence determines the size of the generic fibers. Here's a more geometric description. If the geometric fiber over the generic point is a finite set of points, so will the general fibers over an open set (with the same number of points). If the geometric fiber over the generic point is positive dimensional variety, the generic fibers will be likewise. Since your Zariski dense set will certainly intersect this open set, that should do it. | |
| Jul 5, 2013 at 14:03 | history | edited | Neil Strickland | CC BY-SA 3.0 |
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| Jul 5, 2013 at 9:55 | comment | added | Vivek Shende | Now the answer is yes. Here's a terrible proof: $Rf_* \mathbb{Q}$ is constructible, so it follows from your assumption that the fibre has the cohomology of a point over a (Zariski) open set in $Y$. By semicontinuity of fibre dimension, the fibre is a point in some open subset of that. By generic smoothness, I can conclude the map is birational. | |
| Jul 5, 2013 at 9:14 | history | edited | Hans | CC BY-SA 3.0 |
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| Jul 5, 2013 at 9:13 | comment | added | Hans | okay thank you. reading vivek's comment, i realized that the question was not stated correctly... | |
| Jul 5, 2013 at 9:08 | history | edited | Hans | CC BY-SA 3.0 |
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| Jul 5, 2013 at 9:06 | comment | added | Hans | thank you, i see. unfortunately, i did not pose the question, as i wanted to... i will edit now! | |
| Jul 5, 2013 at 8:57 | comment | added | John Salvatierrez | @Hans: sure, I just thought the answers there could help with cooking up a counterexample. (I said it was a comment :) | |
| Jul 5, 2013 at 8:55 | comment | added | Hans | Hi Carmelo, I am aware of this. When I say "birational morphism" i do not mean "isomorphism". A birational morphism is a morphism, which is invertible as a rational map. | |
| Jul 5, 2013 at 8:32 | review | First posts | |||
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| Jul 5, 2013 at 8:31 | comment | added | Vivek Shende | Consider $X = Y = E$ an elliptic curve, $f$ the map $\times 2$, and $U$ any infinite collection of points so that no two differ by 2-torsion. | |
| Jul 5, 2013 at 8:28 | comment | added | John Salvatierrez | (I don't have enough rep to comment) You might find this question to be of interest mathoverflow.net/questions/73321/…. If you allow Y to be non-normal then you can finde bijective morphisms which are not isomorphisms. | |
| Jul 5, 2013 at 8:13 | history | asked | Hans | CC BY-SA 3.0 |