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I wanted to ask about some existing algorithms for computing points over elliptic curves.

Background : We know that the famous theorem of Mordell and Weil says that " Group of rational points on an elliptic curve is finitely generated " . i.e. $E(\mathbb{Q})=\mathbb{Z}^{r}\oplus E_{\rm{torsion}}(\mathbb{Q})$ . So we can find the torsion points a bit easily by using theorem of Nagell and Lutz.

My main question is that

  • How can we find the free part ? ( I mean points of infinite order ) . I have read that we can just use the homogeneous spaces ( Torsors ) and just dig for some points with infinite order. But its well known that they are not such efficient methods that apply universally and work every time. So are there any other known ( Efficient ) methods for computing the points of infinite order ? , apart from using Homogeneous spaces.

P.S : Apart from this, giving rational points on the original elliptic curve corresponds to giving sections of the projection $S \to C$ ( where $S$ is elliptic surface over curve $C$ ). To determine such sections essentially amounts to determining all the curves lying on $S$ that can be defined over $\mathbb F$ ( where $F$ is a finite field where $C$ lies ).

So is there any anagolous way of finding the points of infinite order by considering the associated surfaces. Is there any work done in that direction ? ( Any good references to read further )

Thank you.

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The method of descent is the only "algorithm" we know that will determine the points of infinite order on an elliptic curve. It is not an algorithm until someone proves that Sha(E/Q) is finite... and even if someone proves this, the algorithm is not very efficient on curves with large conductor (i.e., it may take an enormous amount of time for the method to work, and it might require several consecutive descents, which is hard to implement - and not implemented in the generality that it would be needed for the method to work every time).

The method of descent (and a discussion about its efficiency and dependency on Sha) is very nicely described in Silverman's "The arithmetic of elliptic curves".

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    But, I have some idea in my mind. Can we use the coefficient predicted by Birch and Swinnerton-Dyer to compute the Regulator and later compute the heights of points and then search for homogeneous spaces that have the obtained height and then obtain the corresponding point of finite order. Sir, I need to learn some methods for computing coefficients and then try this. I realized that B.S.D apart from giving the cardinality of Mordell-Weil group, also should suggest some good way of computing points of infinite order. And some constraints have to be fulfilled to do so, like the one you said.2012-02-22
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    But am I going in right path sir ? . Thanks a ton, for answering my question. $+\infty$, but my ability is limited in only giving $+1$.2012-02-22
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    BS-D is a conjecture so, at best, you would get an "algorithm" until BS-D is proved. However, if the rank > 1, how do you get the height of each generator point from the regulator?2012-02-22
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    Yes sir, it would indeed fail if rank$\gt1$ , and many considerations like parity conjecture etc, come into play. But will my idea atleast work when rank$=1$ ( and if B.SD is assumed to be true ) ? , if so, I wanted to do some calculations, and submit my work to a journal. But does idea makes sense ? , and is new ? . Thank you. @Alvaro Lozano-Robledo2012-02-22
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    This is not a new idea. But just for fun, let $E: y^2=x^3-157^2x$. This curve has rank $1$. Can you find the generator using your method?2012-02-22
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    Lozano-Robledo : Sir, I am trying my level, to attempt a solution, but I dont have any sophisticated environment, and even proper books and teachers to guide me, I have been learning these things myself. If my answer doesn't make any sense, please forgive me. I know that $157$ is a congruent number, I don't know routines to compute the things, but I have been trying to compute things by hands, B.S.D predicts the coefficient to be $\large \frac{|Sha(E/\mathbb{Q})| . \Omega . \prod_p c_p. Reg(E/\mathbb{Q})}{ E_{\rm{torsion}}(\mathbb{Q})}$ contd...2012-02-22
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    But I don't know how to find all those things, but if there is some way to find that coefficient, we can just find out regulator, by sending all terms to the other side. As $Reg(E/\mathbb{Q})$ is nothing but twice the canonical height, we can obtain the height and later on if we can make a database of all heights ( I think it takes some years to make so ) we can search for homogeneous spaces and match the height and get the coordinates of points. But I don't know how far it works, I recently heard here (at MO ) that there is some software called Sage, where we can compute these things. Contd..2012-02-22
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    But I don't know how can one learn all those things, and I even doubt the computation of coefficient, are there any machine codes to find the taylor expansion sir. If I am provided with all those things I can surely compute a coordinate, but I dont know how far I would be successful, Ball now is in your court, its you who needs to judge. @Alvaro Lozano-Robledo2012-02-22
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    My hand computations take months for just computing small expansions of L-functions , I am looking forward for some computational help sir. My place is very unsophisticated, only few people know what is a computer, and I am currently at Internet cafe, so I can install some softwares @Alvaro Lozano-Robledo2012-02-22
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    But is 157 a congruent number ? , I have read this somewhere in a short material of Andrew Wiles2012-02-22
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    Lozano-Robledo : Am I in correct path ? , and is it possible to make such computations in a finite time ? .2012-02-23
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    No, the correct path is to read Silverman's book first, cover to cover. Study that book well!2012-02-23
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    Lozano-Robledo : But you didn't comment about the method I described above, will it work ?2012-02-23
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    Yes, it will work but it will be *very* inefficient and slow for curves of large conductor, or curves with a generator of high height.2012-02-23
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    Lozano-Robledo : Thank you for your valuable interaction, I duly follow your advice and start mastering the Silverman's book . Thanks a lot !!! sir .2012-02-23