Timeline for Which curves have infinitely many rational points
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2010 at 17:57 | answer | added | Minhyong Kim | timeline score: 7 | |
| Nov 21, 2010 at 16:21 | vote | accept | Tim Dokchitser | ||
| Nov 21, 2010 at 13:49 | answer | added | Pete L. Clark | timeline score: 16 | |
| Nov 21, 2010 at 13:30 | comment | added | Tim Dokchitser | @Qiaochu: Very nice, thank you! I did not know about this paper. In fact Bjorn does indeed discuss this (bottom of p.9: "It is known that if Sha(E) is finite, then in principle, there is an algorithm for determining whether X(Q) is nonempty: ..."), although I still retain hope for a reference with a detailed proof. | |
| Nov 21, 2010 at 13:16 | comment | added | Qiaochu Yuan | Bjorn Poonen has a nice survey article on the subject: www-math.mit.edu/~poonen/papers/millennial.pdf . Hopefully he takes a look at this question. | |
| Nov 21, 2010 at 13:15 | comment | added | Tim Dokchitser | Thank you, David, nice summary! (I suppose for genus 0 you've got to check the primes above 2 as well, if there is more than one.) | |
| Nov 21, 2010 at 13:07 | comment | added | David E Speyer | By Falting's Theorem en.wikipedia.org/wiki/Faltings%27_theorem , the answer is always "no" for curves of genus $\geq 2$. For a curve of genus $0$, embed the curve as a conic in $\mathbb{P}^2$ by it's anticanonical embedding and use the Hasse-Minkowski theorem en.wikipedia.org/wiki/Hasse%E2%80%93Minkowski_theorem ; you only need to check the primes dividing the determinant of the bilinear form. So all the interest is in genus 1. Hopefully, an expert will come along soon to explain what is known in this case. | |
| Nov 21, 2010 at 13:00 | answer | added | user19475 | timeline score: 2 | |
| Nov 21, 2010 at 12:55 | history | asked | Tim Dokchitser | CC BY-SA 2.5 |