Timeline for Potentially good, semi-stable reduction => good reduction ?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2013 at 7:22 | vote | accept | Matthieu Romagny | ||
| Jun 16, 2013 at 21:05 | comment | added | Will Sawin | This is a special fact about the etale cohomology of a curve. $H^0$ is trivial, $H^2$ is a Tate twist of trivial by Poincare duality, and $H^1$ is the Tate module of the Jacobian, so since the Jacobian of a torsor for $E$ is just $E$, it's the same. | |
| Jun 16, 2013 at 20:15 | comment | added | Matthieu Romagny | They are like in the link; namely, E is an elliptic curve over $\mathbb{Q}_p$ with good reduction, and P is an E-torsor with no $\mathbb{Q}_p$-rational point. Then P does not have good reduction (for, the smooth reduction would have a rational point lifting to a rational point of P, a contradiction). But Matthew Emerton claims that the étale cohomology of P is crystalline since it is isomorphic to that of E, but my first guess would be that it is isomorphic to the étale cohomology of E only as a $\text{Gal}(\bar{K}/K')$-representation, for some finite K'/K. | |
| Jun 16, 2013 at 17:04 | comment | added | Will Sawin | What are P and E? | |
| Jun 16, 2013 at 15:31 | comment | added | Matthieu Romagny |
Thank you Will. Now I'm a bit confused with the Galois representation counterpart, e.g. like in Matthew Emerton's example answering this MO question. Namely, in that example Matt Emerton claims that $P$ and $E$ have the same étale cohomology but I would have thought that the Galois action on the cohomologies match only after restriction to a finite extension $K'/K$. Am I wrong?
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| Jun 16, 2013 at 4:26 | comment | added | Matt | Oops. Of course. In the one-dimensional case the set of torsors is identified with the Brauer group of the field under the standard connecting homomorphism, which in this case is $C_1$. Sorry. The other part of the argument is the same. I was trying to remember why genus $0$ curves over finite fields always have points. | |
| Jun 16, 2013 at 2:58 | comment | added | Will Sawin | If it had good reduction you could lift a point over the residue field using Hensel's lemma, and all genus $0$ curves over finite fields have points. Alternately, by a well-known fact there are two genus $0$ curves over a local field, one with good reduction and rational points and one with bad reduction and without rational points. | |
| Jun 16, 2013 at 2:31 | comment | added | Matt | Maybe I'm being really dumb, but why does no rational points imply bad reduction? I know the standard argument for genus $1$, but it involves Lang's theorem. | |
| Jun 16, 2013 at 1:03 | history | answered | Will Sawin | CC BY-SA 3.0 |