Timeline for Geometry of algebraic curve determined by point counts over all number fields?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Feb 28, 2020 at 0:59 | comment | added | Felipe Voloch | @WillSawin Yes, I think you are right. | |
| Feb 26, 2020 at 17:54 | history | edited | David Lampert | CC BY-SA 4.0 |
added (geometrically irreducible)
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| Feb 25, 2020 at 20:10 | comment | added | Will Sawin | @Asvin Given an elliptic curve $E$ defined by a cubic equation $y^2 = f_3(x)$ for a cubic polynomial $f_3$, the average rank of $E'$ over $\mathbb Q( \sqrt{ f_3(x)})$ should equal the rank of $E'$ over $\mathbb Q$ plus $1/2$ unless $E'$ is isogenous to $E$, and $3/2$ if $E'$ is isogenous to $E$.This follows from the $L$-function equidistribution conjecture of Sarnak, Shin, and Templier (at least the naive-height-based version). So this should indeed determine the isogeny class. | |
| Feb 25, 2020 at 18:47 | comment | added | Will Sawin | @FelipeVoloch I don't think this is true - given a field extension of $\mathbb Q(i)$, applying the nontrivial automorphism of $\mathbb Q(i)$ may produce a different extension. | |
| Feb 25, 2020 at 18:45 | answer | added | Will Sawin | timeline score: 7 | |
| Feb 25, 2020 at 15:27 | comment | added | Srks | small remark: the result of Caporaso Harris Mazur + Pacelli gives that if Lang conjecture is true then #C(L) is bounded only in terms of the genus of C and the degree $[L:\mathbb{Q}]$ so this should impose some sort of restriction on the informations you can recover? (note that there are unconditional results in this direction using Chabauty method by Stoll, Katz-Rabinoff-Zureick-Brown...) | |
| Feb 21, 2020 at 18:12 | comment | added | Asvin | @david that's right! So maybe the isogeny class is determined? | |
| Feb 21, 2020 at 18:07 | comment | added | David Lampert | @Asvin rank $E(K)$ is $K-$isogeny invariant | |
| Feb 21, 2020 at 17:23 | comment | added | Asvin | A variant: Suppose we start with an elliptic curve E over, say Q. Then the function that associates to a number field K the rank of the group E(K) makes sense. Does this function determine the elliptic curve? | |
| Feb 21, 2020 at 13:13 | comment | added | David Lampert | @damiano That's a nice plausible idea. Here are a couple more notions: (1) higher genus should mean maximum number of points in all extensions of discriminant $\leq d$ grows more slowly with $d$ (conjecturally the maximum over all extensions of bounded degree is finite); (2) extra automorphisms implies more points in generic extensions with smaller Galois groups. | |
| Feb 20, 2020 at 21:31 | comment | added | damiano | You can try to recover the gonality of the curve by looking at how the number of points behaves over extensions of $K$ of a fixed degree. For instance, lots of quadratic points should imply that the curve is hyperelliptic. | |
| Feb 20, 2020 at 19:39 | comment | added | David Lampert | Just a remark: the answer is no for affine curves: consider $C-\{P\}$ for a projective curve $C$ without automorphisms and 2 different $P \in C(K)$. | |
| Feb 20, 2020 at 19:11 | comment | added | David Lampert | Right, thanks @FelipeVoloch, I had $K=\mathbb{Q}$ in mind when I wrote the question. | |
| Feb 20, 2020 at 18:55 | comment | added | Felipe Voloch | The answer is no for a silly reason, the curves $y^2 = x^5 \pm ix +1$ have the same number of points over any extension of $\mathbb{Q}(i)$. But once one identifies Galois conjugate curves, the question comes back. | |
| Feb 20, 2020 at 17:47 | history | edited | GH from MO |
edited tags
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| Feb 20, 2020 at 16:53 | history | asked | David Lampert | CC BY-SA 4.0 |