Timeline for Over which fields does every algebraic curve of genus one have a rational point?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2021 at 14:43 | comment | added | Evgeny Shinder | Welcome to Mathoverflow! This holds for finite fields. For an infinite field a necessary condition could be that every element has roots of arbitrary degree, otherwise you can construct a nontrivial torsor. See here for a discussion and references: mathoverflow.net/questions/377081/… and this paper for a sharp recent result: degruyter.com/document/doi/10.1515/crelle-2016-0037/html | |
| Sep 30, 2021 at 18:14 | comment | added | Gro-Tsen | (Disclaimer: I didn't check too carefully that the argument which works for Severi-Brauer varieties also works in your case, so don't take my word for it. This is why I'm writing this as a comment and not as an answer.) | |
| Sep 30, 2021 at 18:11 | comment | added | Gro-Tsen | Yes, each extension brings new curves, which is why the process must be iterated transfinitely (repeat with these new curves, and so on): but you run out of curves before you run out of ordinals. See arxiv.org/abs/math/0303168 theorem 2.3 and lemma 2.4 for a related situation (where we add points to Severi-Brauer varieties rather than curves of genus 1). | |
| Sep 30, 2021 at 13:57 | comment | added | Qasim | @Gro-Tsen I don't understand your construction. Each time you make an extension don't get new non-isotrivial curves? Why would they have a rational point? | |
| Sep 30, 2021 at 13:43 | comment | added | Gro-Tsen | Pseudo-algebraically closed fields are also examples. But I don't think there's a specific term. One construction which might interest you is this: start from any field $k_0$, let $E$ be a genus $1$ curve and $k_1 = k_0(E)$ its function field: then $E$ has a point over $k_1$; repeat transfinitely, taking inductive limits at limit steps: you will eventually get a field with the property you asked, in which $k_0$ is algebraically closed. | |
| Sep 30, 2021 at 13:29 | comment | added | Chris Wuthrich | Algebraically closed and finite fields are examples. You ask for which $K$ is $H^1(K,E)=0$ for all elliptic curves $E/K$. | |
| Sep 30, 2021 at 13:04 | history | asked | Qasim | CC BY-SA 4.0 |