# Torsion on $y^2=x^3+d$

A question that I am stuck on is: prove that the $\mathbb{Q}$-torsion subgroup of the elliptic curve $y^2=x^3+d$ has order dividing 6. Any hints on how to start would be nice. I tried saying something about the reduced curve, but the lack of information about $d$ was a problem. I guess it amount to trying to say something about the Jacobi symbol $\big( \frac{x^3+d}{p}\big)$ for $p\nmid 6d$, but I can't see it.

• if you pick $d = a^6$ then you have the six points $(-a^2,0), (0,\pm a^3), (2a^2,\pm 3a^3)$ and the point at infinity. May 26, 2014 at 13:29
• That's 5 points and the point at infinity? $\pm Y$ involution on first point is identity since $y=0$. Also it's not clear to me that the $(0,\pm a^3),(2a^2,\pm 3a^3)$ are torsion? May 26, 2014 at 13:32
• Yes, $6$ in total. I was just looking for curves where there are $6$ "obvious" points. That aside, $\mathcal E(\Bbb R)$ is a circle so the $2$-torsion is either trivial or $\Bbb Z/2\Bbb Z$. If you can prove there can't be a rational point of order $4$ or $5$, you are done I believe. May 26, 2014 at 13:35
• Do you mean $(\mathbb{Z}/2\mathbb{Z})^2$? May 26, 2014 at 13:39
• No, the other two points of order $2$ are not real, so they are not rational. May 26, 2014 at 13:40

You can see it in this way: for all but finitely many primes $$p$$, there is an injective morphism of groups $$E(\mathbb Q)_{tors}\to E(\mathbb F_p)$$. Now take $$p\equiv -1\bmod 6$$. Then there are no elements of order $$3$$ in $$\mathbb F_p^*$$ and this tells you that $$x\to x^3+d$$ is a bijection of $$\mathbb F_p$$. Therefore for such $$p$$'s you must have $$|E(\mathbb F_p)|=p+1$$. Now since $$m=|E(\mathbb Q)_{tors}|\mid p+1$$ for all these $$p$$'s, you must have that all but finitely many primes $$\equiv -1\bmod 6$$ are $$\equiv -1\bmod m$$, and this is possible iff $$m\mid 6$$ because of Dirichlet's theorem on primes in arithmetic progression.

• I thought we want to show that $m$ divides $6$?
– Marc
Jul 10, 2015 at 14:21