Torsion point on elliptic curve over $\mathbb{Q},$ $E/\mathbb{Q}$

Question

Currently I am learning about elliptic curves. While playing with codes of elliptic curve in SageMath, I observed that if we take an elliptic curve $E: y^{2} = x^{3}+m,$ the torsion subgroup is trivial if $m$ is a prime. I just observed this for some prime numbers. I don't know whether it is true. If it is true, how to prove the same? Moreover, is there any generalization for torsion subgroup for elliptic curves over $\mathbb{Q}$?

Answer 1

The link given by Douglas Molin is very good, but I wanted to offer a direct “proof” (aka reduction to a tractable number of calculations).

By Nagell-Lutz, if $P=(x,y)$ is a rational torsion point, then $x,y$ are integers and $y=0$ or $y^2$ divides the discriminant of $x^3+m$ which is $-27m^2$. This means that $y$ is among $\pm 1,\pm 3, \pm m, \pm 3m$.

Assume $m|y$: then $m|x$, and $m^2|y^2-x^3=m$, absurd. So $y$ is $\pm 1,\pm 3$. Note that for similar reasons $x \neq 0$.

So $m$ is either $9-x^3$, or $1-x^3$. The latter case implies that $m=u^3+1$ for some positive $u$, but $u^3+1=(u+1)(u^2-u+1)$ and both factors are nontrivial if $u >1$ so the only remaining case is a point $(-1,\pm 1)$ on $y^2=x^3+2$.

In the former case, this means that $m=9+u^3$ for some positive $u$ and that the point is $P=(-u,\pm 3)$ which may or may not be torsion again.

I think (but do the computations yourself) that they’re not $3$-torsion points. Let’s now see why there is no $\mathbb{Q}$-rational $p$-torsion for $p>3$ (the argument is more abstract – I don’t know if we can make it simpler or entirely solve the question with the above and cleverer uses of Nagell-Lutz).

Let $P=(x,y) \in E(\mathbb{Q})$ have order $p$. So $x \neq 0$. Note that $u: (x,y) \longmapsto (jx,y)$ is an endomorphism of $E$ defined over $K$, and thus $u(P)$ is a point of $p$-torsion in $E(K) \backslash E(\mathbb{Q})$. In particular, it is independent from $P$ and $(P,u(P))$ is a basis of the $p$-torsion of $E$. It follows that the absolute Galois group of $K$ acts trivially on the $p$-torsion of $2$, and thus that the image of the Galois group of $\mathbb{Q}$ on $\mathrm{Aut}(E[p])$ has order at most $2$.

But it’s known (with the Weil pairing) that the determinant of that Galois representation is the surjective cyclotomic character, and it follows that $|\mathbb{F}_p^{\times}| \leq 2$, so $p \leq 3$.