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Let P=(x1,y1) be a non torsion point on an elliptic curve y^2=x^3+Ax+B. Let (xn,yn)=P^{2^n}. xn,yn are rationals with heights growing rapidly. Can {xn} {yn} stays bounded ?

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EDIT: this answer is wrong. I misread the question as looking at the group generated by P, not the points obtained by repeated doubling. I would be OK if the subset of S^1 generated by taking a non-torsion point and repeatedly doubling came arbitrarily close to the origin---but it may not, as the comments below show. As I write, this question is still open. If a correct answer appears I might well delete this one.

Original answer:

"Bounded" in what sense? You mention heights, that's why I ask. But in fact the answer is "no" in both cases. The height will get bigger because of standard arguments on heights. And the absolute values of x_n and y_n will also be unbounded: think topologically! The real points on the curve are S^1 or S^1 x Z/2Z and if the point isn't torsion then the subgroup it generates will be dense in the identity component and hence will contain points arbitrarily close to the identity, which, by continuity, translates to "arbitrarily large absolute value" in the affine model.

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I mean the absolute. Thanks. You've answered my question. – defgh Mar 28 '10 at 10:30
Kevin, repeated doubling of a point doesn't produce the subgroup generated by that point. What is needed is a theorem to the effect that if $\xi$ is irrational then the set of $2^n\xi$ in $\mathbb{R}/\mathbb{Z}$ meets every neighbourhood of the origin in $\mathbb{R}/\mathbb{Z}$. I'm sure this is true but can't see an immediate proof. – Robin Chapman Mar 28 '10 at 10:32
@Robin: you're right---I misread the question. Thanks. – Kevin Buzzard Mar 28 '10 at 10:57
However, $2^n \xi$ doesn't need to meet every neighborhood of the origin if you only assume $\xi$ is irrational. That $\xi$ is irrational just means the "digits" of the binary expansion aren't preperiodic, not that $\xi$ is normal base 2. – Douglas Zare Mar 28 '10 at 11:22
In that case should the ``elliptic logarithm'' of a rational point have this property then the answer to the original question would be `yes'. I don't think this can be the case but proving it suddenly looks like hard work :-( – Robin Chapman Mar 28 '10 at 11:31

From the discussion above it looks like the answer is yes (EDIT: if you allow real numbers; the OP was unclear, perhaps they wanted a rational point, in which case I'm uncertain. Does anybody know anything about the binary expansion of complex numbers with rational Weierstrass p-values?). Let the origin of your group be the point at infinity in the curve in $\mathbb{RP}^2$, and pick a topological group isomorphism of $S^1$ to the component of the identity to $S^1\cong \mathbb{R}/\mathbb{Z}$. The doublings of a point are given by truncating off the first $m$ digits of the base 2 expansion of your point. Thus the doublings of a point stay bounded if and only if the length of a consecutive string of 0's and 1's in this expansion is bounded above (there are plenty of irrational numbers with this property).

There's a similar answer for putting the origin somewhere else: you can never allow too much of the beginning of the expansion of the point at infinity to show up in the expansion of your point.

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P seems to be rational. – S. Carnahan Apr 5 '10 at 19:33

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