Timeline for Q-curves and twisting
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2020 at 17:17 | comment | added | David Loeffler | This answer is really helpful -- thanks! I was hoping for examples over real quadratic fields, but if there are examples over imaginary quadratic fields then probably there will be examples over real fields as well. | |
| Jul 1, 2020 at 17:15 | vote | accept | David Loeffler | ||
| Jul 7, 2020 at 11:05 | |||||
| Jun 30, 2020 at 20:55 | comment | added | François Brunault | In any case, what I wrote about building blocks cannot be true, as David points out. | |
| Jun 30, 2020 at 20:52 | comment | added | Barinder Banwait | (c) above looks like the culprit to me. In the 1-dimensional case, it's saying "elliptic $\mathbb{Q}$-curves are completely defined over $K$", which is not true in general | |
| Jun 30, 2020 at 20:47 | comment | added | François Brunault | @DavidLoeffler Right, sorry these things always get me confused. I think that for this particular curve, the field of complete definition is $\mathbb{Q}(\sqrt{-2}, \sqrt{-3})$. This is explained in Section 3 of arxiv.org/abs/math/0611663 One would have to write down the isogeny, I haven't done that... | |
| Jun 30, 2020 at 20:25 | comment | added | David Loeffler | You are saying that (a) strongly modular $\Leftrightarrow$ completely def / K when K is quadratic; (b) the Guitart--Quer example is a building block that is not strongly modular, and (c) building blocks are completely def / K. Since the Guitart-Quer example is over a quadratic field, (a), (b), (c) can't all be true at once. | |
| Jun 30, 2020 at 20:23 | comment | added | David Loeffler | @FrançoisBrunault I'm sorry, that seems to be a contradiction? | |
| Jun 30, 2020 at 19:22 | comment | added | François Brunault | (However, the converse is true if $K$ is a quadratic field.) | |
| Jun 30, 2020 at 19:20 | comment | added | François Brunault | It is known that strongly modular implies completely defined over $K$ (I'm excluding CM to be safe), but the converse is not always true. I think that the example by Guitart and Quer is an instance of a building block which is not strongly modular, but building blocks are completely defined over $K$ if I understood correctly. | |
| Jun 30, 2020 at 17:31 | history | answered | Barinder Banwait | CC BY-SA 4.0 |