Timeline for Quotients of Tate modules
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 3, 2022 at 5:09 | answer | added | Adithya Chakravarthy | timeline score: 0 | |
| May 19, 2012 at 4:32 | comment | added | Lubin | Sorry, have I misread or seriously misunderstood the question? I don't see that there is such an exact sequence of ${\mathrm{Gal}}(\overline K\colon K)$-modules, because the kernel of reduction becomes isomorphic to the multiplicative (formal) group only after extension of the base to the complete maximal unramified extension of $K$. | |
| May 18, 2012 at 19:42 | answer | added | user23778 | timeline score: 0 | |
| May 18, 2012 at 6:48 | answer | added | Kevin Buzzard | timeline score: 5 | |
| May 17, 2012 at 23:24 | comment | added | user23778 |
Thank you, this indeed seems to be the case and should answer the question. My original goal was define a free $\mathbb{Z}_{p}$-module $T^{Q}_{p}A$ of rank $g$, together with a canonical surjection of $\mathbb{Z}_{p}$-modules: $f:T_{p}A \rightarrow T^{Q}_{p}A$. The canonical inclusion $Hom_{\mathbb{Z}_{p}}(T^{Q}_{p}A, \mathbb{\C}_{p}) \subset $Hom_{\mathbb{Z}_{p}}(T_{p}A, \mathbb{\C}_{p})$ should be equal to the image of $ H^{1}(A,O) \otimes _{K} \mathbb{C}_{p}$. Can such a module be constructed in the non-ordinary case? (The morphism f does not need to compatible with Galois)
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| May 17, 2012 at 21:45 | comment | added | David Loeffler | If $A$ is non-ordinary, then $T_p A$ is irreducible as a Galois representation. The submodule you are defining is Galois stable, isn't it? So it must be either zero or everything. | |
| May 17, 2012 at 20:22 | history | asked | user23778 | CC BY-SA 3.0 |