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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.

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  • $\begingroup$ 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. $\endgroup$
    – mercio
    May 26, 2014 at 13:29
  • $\begingroup$ 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? $\endgroup$ May 26, 2014 at 13:32
  • $\begingroup$ 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. $\endgroup$
    – mercio
    May 26, 2014 at 13:35
  • $\begingroup$ Do you mean $(\mathbb{Z}/2\mathbb{Z})^2$? $\endgroup$ May 26, 2014 at 13:39
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    $\begingroup$ No, the other two points of order $2$ are not real, so they are not rational. $\endgroup$
    – mercio
    May 26, 2014 at 13:40

1 Answer 1

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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.

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

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