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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?

edited May 14 2012 at 14:57

asked May 14 2012 at 13:11

zy
289●5

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? – Julien Puydt May 14 2012 at 13:17

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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/… – François Brunault May 14 2012 at 14:35

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@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. – zy May 14 2012 at 15:01

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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. – Noam D. Elkies May 14 2012 at 15:48

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@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!) – Joe Silverman May 14 2012 at 23:12

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joro · Answer 1 · 2012-05-14 13:38:48Z UTC

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.

Integer points on Elliptic Curves and Mordell-Weil Group

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