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Let $E/\mathbb{Q}$ be an elliptic curve, given by some minimal Weierstrass equation (say $Y^2 = X^3 + aX + b$ for some integer $a$ and $b$), and let $P$ be a rational point on $E$ which is not the infinite point. We assume that $P$ is not torsion.

The point $P$ can be written as $(\frac{x}{d^2}, \frac{y}{d^3})$ for some integers $x, y, d$ such that $d$ is prime to $x$ and $y$.

Now consider the sequence of points $P, 2P, 3P, \cdots$, we are led to three sequences $x_n, y_n, d_n$, such that $nP = (\frac{x_n}{d_n^2}, \frac{y_n}{d_n^3})$.

The question: does the limit $\lim_{n\rightarrow\infty} \frac{2\log(d_n)}{n^2}$ exist? Is it equal to the canonical height of $P$?

Same question for $x_n$ and $y_n$?


A related question:

Let us denote by $h(P)$ the natural height of $P$, i.e. $h(P) = \log(\max\{|x|, |d^2|\})$. Does the limit $\lim_{n\rightarrow\infty}\frac{h(nP)}{n^2}$ exist? (If it exists, then it of course is equal to the canonical height.)

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  • $\begingroup$ The answer to the second question is YES, and is certainly explained in books on elliptic curves such as Joe Silverman's : the canonical height of $P$, $\hat h(P)$, is the limit of $h(x(nP))/n^2$, when $n\to\infty$. $\endgroup$ – ACL May 12 '16 at 22:43
  • $\begingroup$ @ACL Could you tell me where in the book of Silverman is this result stated? The canonical height, by definition, is the limit of $h(2^n P)/4^n$, which is different from what you write. Besides, his Corollary 6.4 of Chapter VIII has a big O constant that depends on $m$. $\endgroup$ – WhatsUp May 12 '16 at 22:55
  • $\begingroup$ There is a uniform bound for $h(P)-\hat h(P)$; apply it to the point nP and use that $\hat h(nP)=n^2\hat h(P)$. $\endgroup$ – ACL May 12 '16 at 23:01
  • $\begingroup$ @ACL You are right. I just figured it out myself. Do you have idea for the denominator? $\endgroup$ – WhatsUp May 12 '16 at 23:05
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You need to use Siegel's theorem (which is quite deep and relies on Roth's theorem or some such). This is in Chapter IX of Arithmetic of Elliptic Curves, specifically Theoream IX.3.1. You'll need to unsort the definitions a bit, since it's stated for number fields, but in your notation, one has $$ \lim_{n\to\infty} \frac{2\log d_n}{n^2} = \lim_{n\to\infty} \frac{\log |x_n|}{n^2} = \hat h(P). $$ Hmmm.. Actually, the definitions are unsorted for you in Example IX.3.3, where you'll find the following formula (using your notation) in the middle of page 279 (of the 2nd edition): $$ \lim_{n\to\infty} \frac{\log|x_n|}{\log d_n^2} = 1. $$

BTW, the sequence $(d_n)_{n\ge1}$ is called the Elliptic Divisibility Sequence associated to the curve $E$ and point $P$. The fact that $\log d_n$ grows like a multiple of $n^2$ is an essential fact used to prove that elliptic divisibility sequences satisfy the Zsigmondy property: for all but finitely many $n$, there is a prime $p$ such that $p\mid d_n$ and $p\nmid d_m$ for all $m < n$.

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  • $\begingroup$ Thank you for the answer. It is now clear to me that the problem has something essential to do with diophantine approximations. The question I have in mind is the following: given a bound $M$, is it possible to determine all the $n$ such that $nP$ has denominator smaller than $M$? Of course the effective method of Baker gives a theoretical "Yes" answer, but since the expected number of $n$ is just $\sqrt(\log(M))$, I wonder if this can be done simpler, by proving e.g. that the sequence $d_n$ is strictly increasing for $n$ larger than an effectively computable bound. $\endgroup$ – WhatsUp May 13 '16 at 14:29
  • $\begingroup$ @WhatsUp Do you want to determine all $n$, or do you want to bound the number of such $n$? The latter can be done, since there are effective qualitative versions of Roth's theorem. In other words, I think that one can find an (effective) constant $C=C(E,P)$ so that for all $M\ge2$, $$\#\{n : d(nP)\le M\} \le C\sqrt{\log M}.$$ This should follow from the results in my paper "A quantitative version of Siegel's theorem", J. Reine Angew. Math. 378 (1987), 60-100. $\endgroup$ – Joe Silverman May 13 '16 at 14:42
  • $\begingroup$ My principle aim is actually to effectively determine all such $n$, for a big bound like $M = 1e6$. Is it possible, or something known to be hard? But this effective bound on the number also looks interesting to me! $\endgroup$ – WhatsUp May 13 '16 at 14:47

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