Timeline for Is there an elliptic curve over a number field with a point of order 64 and Mordell-Weil rank zero?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 2, 2023 at 15:02 | comment | added | David McKinnon | @ChrisWuthrich Well, to be fair I don't know. But my thought had been to see if anyone had proven an average rank bound that would apply, so that some reasonable number of those curves would have to have rank zero. It is definitely a problem that the family is high genus, so that the field of definition of the curves veers all over the place. | |
| Nov 2, 2023 at 14:58 | comment | added | Chris Wuthrich | @DavidMcKinnon The universal elliptic curve over $X_1(p^k)$ is precisely such a family. For each $K$ it has finitely many $K$-rational fibres. But how would you show that there is one of rank $0$? | |
| Nov 2, 2023 at 13:24 | comment | added | David McKinnon | I feel like there should be some world in which there's a family of curves, all with rational n-torsion, where the 0/1 50-50 applies, and I can pick my curve out from there. But I can't figure out such a family. | |
| Nov 2, 2023 at 13:24 | comment | added | David McKinnon | Also, the 64 thing is a minimal thing, not a crucial thing -- all I need is a torsion point of order at least 64. If 2 is bad (because, let's be honest, it probably is!), then 3^k would be just fine. | |
| Nov 2, 2023 at 13:22 | comment | added | David McKinnon | Thank you all for those comments! I should have been clearer in my original question that I don't need an explicit example of a curve like this -- a proof that one exists will suffice. The whole 50/50 thing for rank 0/1 has been in my head, but I haven't been able to make a rigorous proof out of it yet. | |
| Nov 2, 2023 at 9:43 | answer | added | Chris Wuthrich | timeline score: 5 | |
| Nov 1, 2023 at 23:52 | comment | added | Chris Wuthrich | I agree, if $E$ is a curve that is not coming from a lower field, having a $n$-torsion point, I bet chances are good that the rank is $0$ or $1$. On the other hand for certain Galois groups the parity results coming from Brauer relations impose rank growth. I have not thought about $n=64$. But in any case, all I can offer so far is speculation that this should exist, but no idea how to construct an example or how to verify it. | |
| Nov 1, 2023 at 22:46 | comment | added | Joe Silverman | @StanleyYaoXiao In any case, I'd expect that over a fixed number field and with elliptic curves ordered by some reasonable height, 100% of them should have rank 0 or 1. (Probably 50% of each, but since I haven't thought about it, maybe there's something that affects the sign of the functional equation to create unequal probabilities.) As for David's question, better probably to start with a curve with an 8-torsion point over Q. But even then, adjoining a 64-torsion point gives an extension of huge degree. Can we do a 2-descent in that field, even with the 2-torsion point to help? | |
| Nov 1, 2023 at 20:15 | comment | added | Stanley Yao Xiao | @DrorSpeiser I am actually not sure what to believe on that front. I am not sure if the Katz-Sarnak conjecture extends to arbitrary number fields (that is, if one fixes a number field $K$, and one orders elliptic curves over $K$ with respect to some reasonable height, whether the average rank should be $1/2$). Certainly for a fixed elliptic curve, one can find a tower of number fields $K_0 \subset K_1 \subset \cdots K_j \subset \cdots$ such that the rank of $E$ increases as $j$ increases, but usually the construction of such a tower is contrived. | |
| Nov 1, 2023 at 19:53 | comment | added | Dror Speiser | @StanleyYaoXiao Don't we expect the average rank to increase with number field degree? | |
| Nov 1, 2023 at 17:56 | comment | added | Stanley Yao Xiao | One strategy is to find an elliptic curve $E$ over $\mathbb{Q}$ with full 2-torsion and rank $0$, then consider its $64$-torsion field $K_{64}$. Since there’s no reason for $E$ to have a non-torsion rational point over $K_{64}$ other than luck, at least for some $E$ the rank will stay $0$ over $K_{64}$. The choice of $E$ to have full 2-torsion over $\mathbb{Q}$ is simply to reduce the degree of $K_{64}$ over $\mathbb{Q}$, which minimizes the chances $E$ will pick up new points. | |
| Nov 1, 2023 at 15:29 | history | asked | David McKinnon | CC BY-SA 4.0 |