Timeline for About complete residues on curves
Current License: CC BY-SA 3.0
13 events
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| S Mar 31, 2016 at 19:52 | history | bounty ended | CommunityBot | ||
| S Mar 31, 2016 at 19:52 | history | notice removed | CommunityBot | ||
| S Mar 23, 2016 at 17:56 | history | bounty started | Dubious | ||
| S Mar 23, 2016 at 17:56 | history | notice added | Dubious | Improve details | |
| Mar 3, 2016 at 22:30 | history | edited | Dubious | CC BY-SA 3.0 |
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| Mar 3, 2016 at 18:22 | comment | added | Dubious | @t3suji, this seems a very good explanation. If you write a quite complete answer, I will accept it. | |
| Mar 3, 2016 at 15:28 | comment | added | t3suji | Yes, it is possible to extend it, for instance because $\Omega$ is the completion of $\Omega_{K|F}$ with respect to the topology defined by $v_P$, and $res_P$ is continuous in this topology (as follows immediately from the explicit formula). | |
| Mar 3, 2016 at 12:06 | history | edited | Dubious | CC BY-SA 3.0 |
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| Mar 3, 2016 at 11:45 | history | edited | Dubious | CC BY-SA 3.0 |
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| Mar 3, 2016 at 11:39 | history | edited | Dubious | CC BY-SA 3.0 |
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| Mar 3, 2016 at 11:33 | comment | added | Dubious | You are right $K_P\otimes_K\Omega^1_{K|F}$ is a more reasonable module. Is it possible to extend the residue map to this module? | |
| Mar 3, 2016 at 11:24 | comment | added | Jason Starr | The field extension $K_P/F$ has infinite transcendence degree. So the $K_P$-module of relative differentials $\Omega^1_{K_P|F}$ is infinitely generated. Are you sure you would not prefer to work with he $K_P$-module $K_P\otimes_K \Omega^1_{K|F}$? That still includes elements of the form $g\ dt$ for $t\in K$ a uniformizing element at $p$ and $g\in K_P$ arbitrary. | |
| Mar 3, 2016 at 10:00 | history | asked | Dubious | CC BY-SA 3.0 |