Timeline for Is there an R=T type result for modular forms with additive reduction?
Current License: CC BY-SA 2.5
11 events
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| Apr 12, 2010 at 5:44 | comment | added | Soroosh | @Emerton: Thanks. That clarifies the situation a lot. | |
| Apr 12, 2010 at 5:44 | vote | accept | Soroosh | ||
| Apr 12, 2010 at 3:19 | history | edited | Emerton | CC BY-SA 2.5 |
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| Apr 12, 2010 at 3:17 | comment | added | Emerton | In short, one has to look at the inertial action on the $D_{pst}$ of the Tate module. | |
| Apr 12, 2010 at 3:13 | history | edited | Emerton | CC BY-SA 2.5 |
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| Apr 12, 2010 at 3:07 | comment | added | Emerton | Naively, no, but non-naively, yes! I've edited to explain what I mean. | |
| Apr 11, 2010 at 14:55 | comment | added | Soroosh | @Emerton: Can $\ell=p$ in above? That is, is the image of $I_p$ on the automorphism group of $T_pE$ the same as $T_\ell E$? | |
| Apr 10, 2010 at 18:56 | comment | added | Kevin Buzzard | Aah thanks Emerton, that's the point. | |
| Apr 10, 2010 at 18:48 | comment | added | Emerton | Both $R^{[0,1],\tau}$ and $\mathbb T$ are defined to be torsion-free. (But as I commented, the down-side is that one doesn't have such an explicit interpretation of the Artinian-valued points of the deformation ring.) | |
| Apr 10, 2010 at 16:13 | comment | added | Kevin Buzzard | @Emerton: do you really get an R=T theorem and not an "R=T up to torsion" theorem? | |
| Apr 10, 2010 at 15:59 | history | answered | Emerton | CC BY-SA 2.5 |