# Computing the free-part

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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But am I going in right path sir ? . Thanks a ton, for answering my question. $+\infty$, but my ability is limited in only giving $+1$. –  Iyengar Feb 22 '12 at 16:53
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-Robledo –  Iyengar Feb 22 '12 at 17:24
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? –  Álvaro Lozano-Robledo Feb 22 '12 at 17:26