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Let $p_1=2, p_2 = 3,\ldots,$ be the prime numbers, and define $n_i = \prod_{j=1}^i p_j$. Moreover, let $E_i $ be the elliptic curve defined by $y^2 = x^3 + n_i$.

Can one compute the torsion group $E_i(\mathbb Q)_{tors}$ in terms of $i$? Or at least its cardinal?

I expect it to be trivial. I checked this for all $i\leq 1000$.

I was thinking about considering the reduction of $E_i$ modulo $p_{i+1}$ and showing that $E_i(\mathbb F_{p_{i+1}}) = \{ 0\}$. That would imply the torsion group is trivial.

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up vote 8 down vote accepted

Exercise 10.19 of Chapter X of Silverman's Arithmetic of elliptic curves gives a classification of the torsion subgroup of $E : y^{2} = x^{3} + D$ if $D$ is a sixth-power free integer. The torsion subgroup is trivial unless $D$ is a square, a cube, or $-432$.

One way to prove this is to show that if $p \equiv 2 \pmod{3}$ is prime and $p \nmid 6D$, then $\# E(\mathbb{F}_{p}) = p+1$. The idea then is to use Dirichlet's theorem on primes in arithmetic progressions. (A similar argument for the curves $y^{2} = x^{3} - n^{2} x$ is given in Koblitz's "Introduction to Elliptic Curves and Modular Forms" Proposition I.9.17.)

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A full proof can also be found here: – René May 29 '14 at 14:59
Thank you for your answer. – user1234 May 29 '14 at 16:10

An alternative to the "reduction mod $p$ and Dirichlet's theorem" approach described by Jeremy Rouse is to do a height computation. It is not that hard to find explicit absolute constants $c_1>0$ and $c_2$ so that every point $P=(x,y)\in E_D(\mathbb{Q})$ with $xy\ne0$ satisfies $$ \hat h(P) \ge c_1 \log\bigl|\hbox{Disc}(\mathbb{Q}(\sqrt[6]{D})/\mathbb{Q})\bigr| - c_2. $$ In particular, this shows that for all 6'th power free $D$ larger than some absolute bound, the only possible torsion points have $xy=0$. Finally one notes that $xy=0$ leads to 2-torsion or 3-torsion when $D$ is, respectively, a cube or a square.

To me, this height proof seems more elementary than the mod $p$ proof because it doesn't invoke Dirichlet's theorem; although I guess Dirichlet's theorem for $p\equiv2\pmod3$ probably has an elementary proof. Another advantage of the height proof is that it generalizes quite well to families of twists of abelian varieties of arbitrary dimension.

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Something closer to the full strength of Dirichlet's theorem is needed for the other proof, namely given a prime $\ell > 3$, the existence of some $p \equiv 2 \pmod{3}$ with $\ell \nmid p+1$. – Jeremy Rouse May 29 '14 at 15:37
Thank you for your answer. – user1234 May 29 '14 at 16:09
Yes, Dirichlet for $p \equiv 2 \bmod 3$ is elementary: consider $3P-1$ where $P$ is the product of the already-known primes of the form $3k-1$. – Noam D. Elkies Nov 13 '15 at 17:47

Suppose that $(x,y)$ is a torsion point on the curve given by $Y^2=X^3+n$, where $n\ne\pm1$ is a squarefree integer. We can't have $y=0$. So assume $y\ne0$. Then, by Nagell-Lutz, $y^2$ divides $-27n^2$, so $y$ divides $9n$. If $p>3$ is a divisor of $y$, then (note that $n$ is square free) $y^2=x^3+n$ implies that $p$ divides $x^3$, but $p^2$ does not, which is absurd. So $y$ is, up to a sign, a power of $3$, which yields the same contradiction unless $y=\pm1$.

So suppose that $y=1$ (thanks to Jeremy Rouse for pointing out that this case was missing). Then $2(x,y)=(x',y')$ with $y'=(-27n^2 + 18n + 1)/8\ne0$. Again by Nagell-Lutz, $y'$ divides $9n$, which fails for size reasons except for a few small values of $n$ which can be checked by hand.

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Jeremy and I were answering a more general question, but you're right, the specific question is easier (although Nagell-Lutz is only marginally easier than the height argument). More generally, you're argument should work for $y^2=x^3+D$ with $D$ square-free. But it's not clear (to me) if it can be made to work for 6'th power free $D$. – Joe Silverman May 29 '14 at 15:55
It takes a little bit of work to handle the case that $y = 1$. – Jeremy Rouse May 29 '14 at 16:06
Thank you for your answer. – user1234 May 29 '14 at 16:10
@Jeremy Rouse: Thanks for remarking that $y=1$ isn't covered by the original answer. I fixed it now. – Peter Mueller May 29 '14 at 19:18

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