Timeline for Which curves have infinitely many rational points
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17 events
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| Nov 24, 2010 at 11:43 | comment | added | Tim Dokchitser | @Ariel: Ah, I see your point. I agree, in practice this is what I'd do as well! For this question I was more concerned with the existence of any kind of algorithm, however impractical. (And at the moment it might be really impractical, I'm afraid. Constructing Jacobians of arbitrary genus 1 curves is the point I am mostly worried about.) | |
| Nov 24, 2010 at 1:19 | comment | added | A. Pacetti | Tim as I said, it only works to prove that there are only finitely many in the case the Jacobian has rank 0. Even assuming the BSD conjecture, knowing that the L-series vanishes implies the rank is positive, you are talking of the rank of the Jacobian, so all the above discussion applies. My only suggestion was first to try the L-series before computing any Jacobian whatsoever... | |
| Nov 24, 2010 at 1:04 | comment | added | Tim Dokchitser | @Ariel: I think the problem is that if L(E,1)=0, this does not prove anything (if the zero is of order $\gt 1$). And I don't want to assume BSD, only finiteness of Sha. | |
| Nov 23, 2010 at 19:43 | comment | added | A. Pacetti | Just a stupid comment on your question, cannot you just compute the L-function? (for the first 4 parts of Pete algorithm, which avoids computing the Jacobian in the first place). If you assume modularity of the Ell. Curve, it satisfies a F.E. which you can just check numerically to see if the L-function vanishes or not (at 1). If it doesn't, the rank of the Jacobian is zero and you are done. If it does, then compute the Jacobian and keep going... | |
| Nov 21, 2010 at 20:48 | comment | added | Felipe Voloch | @Pete: I guess your comment is not so much directed at mine but a general question of how to compute the jacobian of a curve of genus 1. I'm sure this has been looked at. I recall a paper of Greg Anderson with some very complicated formulas for this. | |
| Nov 21, 2010 at 20:44 | comment | added | Felipe Voloch | @Pete: I thought, since we are already on step 5, that we already know that the Jacobian of C and C_i is E. | |
| Nov 21, 2010 at 18:42 | comment | added | Pete L. Clark | @Felipe: thanks, I think I understand a little better now. Still it puts more emphasis on the other weak point in the algorithm: certainly in order to compute the pairing of $C$ and $C_i$ we need to have some way of identifying the Jacobians of these two curves. Don't you think? | |
| Nov 21, 2010 at 16:21 | vote | accept | Tim Dokchitser | ||
| Nov 21, 2010 at 16:10 | comment | added | Felipe Voloch | @Pete: The beauty of Bjorn's suggestion is that you don't need to identify which element of Sha the curve C is. Once you have a list C_i of the elements of Sha, you compute <C,C_i> for all i and you get 0 always if and only if C is trivial. Computing the pairing of a pair of explicitly given curves is a calculation on divisors on their product, I believe and is in principle effective. (Also for Tim) I don't have a reference but my guess is that the original definitions are computable. In any case it's best to ask Bjorn. | |
| Nov 21, 2010 at 14:37 | comment | added | Pete L. Clark | Okay, slightly less ridiculously, you can compute the set of primes of bad reduction for $C$, a finite set which has to contain the set of primes of bad reduction for the Jacobian $E$. This leaves you with a finite set of isomorphism classes. If you are over $\mathbb{Q}$ and can make use of the Cremona-Stein tables, it doesn't sound too bad. | |
| Nov 21, 2010 at 14:28 | comment | added | Pete L. Clark | It is a definitely a question worth thinking about though (and perhaps asking here if an answer is not immediately forthcoming). I guess the most convenient setup is that you postulate the existence of a rational divisor $D$ on $C$ of a certain degree $n$ and compute the canonical embedding (I assume $n \geq 3$; the other cases are very classical) into $\mathbb{P}^{n-1}$ as the (excess) intersection of (some quadratic in $n$ that I forget at the moment) quadric hypersurfaces. | |
| Nov 21, 2010 at 14:24 | comment | added | Pete L. Clark | @Tim: Yes, good point. I'm afraid the only answer I can think of at the moment that works in general is the ridiculous one you mentioned in your comment above. For small indices -- i.e., if you know that $C$ is endowed with a rational divisor of degree $n$ for $1 \leq n \leq 5$, this is well-understood in terms of (neo-)classical invariant theory: see especially the papers of Tom Fisher. But in general...I'm afraid ridiculous is the best I can do at the moment. | |
| Nov 21, 2010 at 14:14 | comment | added | Tim Dokchitser | @Pete: Excellent!! Let me think about it for a bit more, but I am sure I'll accept it. Incidentally, how do you compute the Jacobian E of C? (Just don't tell me you'll try all elliptic curves until you find one with an element in Sha[n] isomorphic to C for some n.) @Felipe: Is there a way to compute the Cassels-Tate pairing if C is not already given in some sort of canonical form? (Forgive my ignorance here) | |
| Nov 21, 2010 at 14:11 | comment | added | Pete L. Clark | @Felipe: sounds good. Do you have a reference for explicit computation of the Cassels-Tate pairing? More precisely, I'm assuming $C$ is given by a system of equations in projective space. How do I know which element of Sha(K,E)[n] to identify it with? | |
| Nov 21, 2010 at 14:05 | comment | added | Felipe Voloch | (As I learned from Poonen) The best way to decide if an element of Sha is trivial or not is to compute the Cassels-Tate pairing on it and, if a given element pairs trivially with a complete list of elements, then it is trivial. This takes care of your step 5. without doing anything ridiculous. | |
| Nov 21, 2010 at 13:56 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Nov 21, 2010 at 13:49 | history | answered | Pete L. Clark | CC BY-SA 2.5 |