Let E be an elliptic curve over Q of non-zero rank. Let S be the union of the primes of bad reduction of E with a Chebotarev set [1]. Suppose additionally that S has density strictly less than one.

What can one say about the set of S integral points of E(Q)? Is this set finite or infinite, and is there any explicit way to describe this set?

[1] We say that a subset S of rational primes is a Chebotarev set if there is a finite Galois field extension K/Q and a union C of conjugacy classes of its Galois group s.t. a prime p is in S if and only if $Frob_{p}$ is in C. (E.g. the set of primes congruent to 1 mod 4.)

[2] A point of E(Q) is S integral if it does not reduce to the identity mod any prime outside of the set S.

  • 3
    $\begingroup$ I would look at the problem from the opposite direction. Take a point $P\in E(\mathbf Q)$ and study the orders of its reductions $P \mod p$, when $p$ varies. This question is known as the order problem and has been studied recently. A first reference could be the papers of Antonella Perrucca, who quotes earlier results of Khare, Prasad, Pink, Silverman, etc. $\endgroup$
    – ACL
    May 14, 2014 at 13:33

2 Answers 2


This answer is related to the comment by ACL.

The set of $S$-integral points can be infinite, regardless of the size of the density of $S$. For example, assume that $E(\mathbb{Q})$ is infinite and cyclic and generated by $P$. Fix a prime number $\ell$. Let $S$ be the set of primes $p$ for which $|E(\mathbb{F}_{p})|$ is a multiple of $\ell$, but for which there is no point $Q \in E(\mathbb{F}_{p})$ such that $\ell Q = P$. This set is a Chebotarev set (with density about $1/\ell$) and if $p \in S$ then the order of $P$ in $E(\mathbb{F}_{p})$ is a multiple of $\ell$. If $m$ is coprime to $\ell$, then $mP \equiv 0 \pmod{q}$ implies that the order of $P$ modulo $q$ is coprime to $\ell$ and hence $q \not\in S$. Thus, $mP$ is $S$-integral.

The set of $S$-integral points can also be finite and non-empty if $S$ has density $1/2$. For example, let $E : y^{2} = x^{3} - 47x$. Then, $E(\mathbb{Q}) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$ and is generated by $T = (0,0)$ and $P = (64/9,137/27)$. Let $S$ be the set containing $2$, $47$ and all primes $p$ such that $\left(\frac{p}{47}\right) = -1$. The descent homomorphism (from Silverman and Tate, for example) shows that the coordinates of $mP+T$ are related to the equation $-47m^{4} + e^{4} = n^{2}$ and this shows that if $p$ is a prime that divides a denominator of $mP + T$, then $\left(\frac{p}{47}\right) = 1$ and hence $mP+T$ is not $S$-integral unless it is $\{2,47\}$-integral. Now, the denominator of the $x$-coordinate of $P$ is $9$, and so the denominator of the $x$-coordinate of $mP$ for all $m$ will also be a multiple of $3$ and so $mP$ is not $S$-integral either. Hence, the only $S$-integral points are the $\{2,47\}$-integral points, and Magma says that $(0,0)$ and $(-423/64,-2397/512)$ are the only ones.

In each of these cases, the set $S$ is chosen in a very specific way in reference to the elliptic curve. I talked about this same question today with rlo, and he plans to post a heuristic that for a set $S$ with "no relation to the curve", the answer depends on the density of $S$.


Here's a heuristic that suggests that the density might matter. It's not always right, as Jeremy has pointed out.

Fundamental assumption: Let $\mathrm{den}(P)$ denote the denominator of $P$, and assume that $\mathrm{den}(mP)$ is essentially a random integer of its size, subject to the condition that $\mathrm{den}(Q)\mid\mathrm{den}(kQ)$ for any point $Q$ and integer $k$. This can be broken by choosing the set $S$ to depend on the given elliptic curve (c.f. Jeremy's answer), but if the set $S$ has no relation to the curve, then this might be plausible.

I also want to assume, given some point $P$, that $\log \mathrm{den}(mP) \asymp m^2$. This is reasonable by a consideration of heights. I'm also going to ignore the divisibility condition, since this also shows that its contribution to $\log(\mathrm{den}(mP))$ should be smaller by a factor of a constant.

For simplicity, let's assume that $E(\mathbb{Q})$ is infinite cyclic, and let $Q$ denote a generator. For a number of size $N$, the probability that $N$ is composed only of primes in $S$ is $1/(\log N)^{1-\alpha}$, where $\alpha$ is the density of $S$. If we look at primes up to some point $X$, then, believing that $\mathrm{den}(pQ)$ is random of its size, the expected number of $m\leq X$ for which $\mathrm{den}(mQ)$ is $S$-integral should be $$ \sum_{m\leq X} \frac{1}{(\log \mathrm{den}(mQ))^{1-\alpha}} \asymp \sum_{m\leq X} \frac{1}{m^{2(1-\alpha)}}. $$ Notice that if $\alpha>1/2$, this sum diverges as $X\to\infty$, indicating we should expect there to be infinitely many $S$-integral points. On the other hand, if $\alpha<1/2$, the sum converges, and we should expect only finitely many $S$-integral $pQ$'s. In the unlucky situation that $\alpha=1/2$, though the sum diverges, what I've said is fuzzy enough that I wouldn't be comfortable making a guess here (and perhaps, both behaviors can occur, even if the assumption holds).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.