Consider an elliptic curve $E:y^2=x^3+ax+b,a,b\in\mathbb{Z}$ on $\mathbb{Q}$ with rank $r$ and trivial torsion group.(The curve is quasi-minimal,say,there are no primes $p$ such that $p^4\vert a$ and $p^6\vert b$)

Question:Suppose this curve has many integer points(say, more than $2^{r+1}$ integer points with $y>0$),is it possible to find $r$ integer points to generate the whole Mordell-Weil Group $E(\mathbb{Q})$?If not,is it possible to give an counterexample?

  • $\begingroup$ I'm not sure I understand your question : you want to extract a generating set from a big set of points? What if your big set of point is of the form P, 2P, 3P, 4P, etc? $\endgroup$ – Julien Puydt May 14 '12 at 13:17
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    $\begingroup$ It is not always possible to generate the Mordell-Weil group with integral points. For example $E=517c1$ satisfies $E(\mathbf{Q})=\mathbf{Z}$ but a generator is $P=(85/4,513/8)$. See Cremona's tables homepages.warwick.ac.uk/staff/J.E.Cremona/book/fulltext/… $\endgroup$ – François Brunault May 14 '12 at 14:35
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    $\begingroup$ @François Brunault:I use SAGE to calculate the integer points on $517c1$ and found no integer points on this curve. But my question is about a curve having many integer points. $\endgroup$ – Y. Zhao May 14 '12 at 15:01
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    $\begingroup$ That probably depends strongly on just what notion of "many" you use. Too large, and it's true vacuously (at least under ABC) because the number of integral points on a curve of rank $r$ is bounded by $C^r$. Too small, and you might find a curve with a subgroup of rank $r-1$ that has lots of integer points but also an $r$-th generator of such a large height that no equivalent integer point is expected. Of course, once $r$ itself is large enough we don't know how to find a rank-$r$ curve even without any hypothesis on integer points. $\endgroup$ – Noam D. Elkies May 14 '12 at 15:48
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    $\begingroup$ @Noam: What you say is true under ABC, but using the word "vacuously" gives the impression that the implication is trivial. The fact that ABC implies $#E(\mathbb{Z})\ll C^{rank E(\mathbb{Q}}$ (for quasi-minimal Weierstrass equations) is not so easy to prove. (At least, Marc Hindry and I didn't find it so easy!) $\endgroup$ – Joe Silverman May 14 '12 at 23:12

If I understand correctly, what you want is not possible in general.

Given a rank $x$ curve, using the group law find many rational points using only 1 generator. Then suitably scale $a$ and $b$ to get $E^'$ with the points integral and rank still $x$. The many known integral points will find only the used generator no matter how many they are.

The construction is similar to the first comment in this question.

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  • $\begingroup$ @joro:Thanks for your answer. I have slightly changed my question. $\endgroup$ – Y. Zhao May 14 '12 at 14:13

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