Timeline for What is "the quotient of the universal ordinary Hecke algebra corresponding to an ordinary $\Lambda$-adic form"?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 27 at 23:34 | answer | added | Simone Maletto | timeline score: 4 | |
| Dec 6, 2021 at 15:05 | comment | added | Will Sawin | Just because the authors treat it as obvious and it's the only quotient of a Hecke algebra associated to a form that seems obvious to me. The quotient produced here is finite flat over $\Lambda$, and when you tensor with $\operatorname{Quot}(\Lambda)$ should be a field, so this is the quotient by a maximal ideal of the tensor product you write down. If I understand what you say correctly, I guess it must be the maximal ideal of one of the local rings. (Or if you really mean local fields, the quotient after tensoring should be one of these local fields.) | |
| Dec 6, 2021 at 6:40 | comment | added | Edward Evans | Hi @WillSawin, why surely? Does this have nothing to do with the local rings which occur in the decomposition of $\Bbb H_{Np^\infty}^\text{ord} \otimes_{\Lambda} \text{Quot}(\Lambda)$? | |
| Dec 5, 2021 at 15:51 | history | edited | LSpice | CC BY-SA 4.0 |
Name of "this"; DOI
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| Dec 5, 2021 at 14:20 | comment | added | Will Sawin | Surely this must be the quotient by the ideal of operators which send that eigenform to zero. | |
| S Dec 5, 2021 at 13:51 | review | First questions | |||
| Dec 5, 2021 at 16:03 | |||||
| S Dec 5, 2021 at 13:51 | history | asked | Edward Evans | CC BY-SA 4.0 |