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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.

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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. –  ACL May 14 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 $S$.

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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).

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