Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 22 / 3 / ii

We use two precise facts about the formal group of an elliptic curve. At a prime $p$ of good reduction, its kernel of reduction of an elliptic curve is identified by the uniformizer $t=-x/y$ with the group $\hat{E}(p{\mathbb{Z}}_p)$. For odd $p$ this group is torsion-free. To see the second fact, the integral invariant differential of a formal group law has the form $(1+{\sum }_{j\ge 1}{c}_j{T}^j)dT$, with $c_j\in {\mathbb{Z}}_p$. Integrating constructs the formal logarithmIt is a group homomorphism to the additive group. For $0\ne t\in p{\mathbb{Z}}_p$ and $j\ge 2$,when $p$ is odd. Hence the series converges and $v_p({\log }_F(t))=v_p(t)$, so it is injective. The target has no nonzero torsion. Therefore reduction is injective on the entire rational torsion subgroup at an odd prime of good reduction, including its $p$-primary part.

For the present elliptic curve, $\Delta =64D^6$, so every odd $p\nmid D$ is a prime of good reduction. If $p\equiv 3(\mathrm{mod}4)$, the Legendre symbol of $-1$ is $-1$. The values $x$ and $-x$ cancel in the sum of the Legendre symbols of $x^3-D^{2}x$, yieldingThe rational torsion subgroup injects into each of these groups, so it is finite and its order $M$ divides every such $p+1$.

For any odd prime $\ell$, the Dirichlet theorem on primes in arithmetic progressions supplies infinitely many $p$ with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$. Discard the finitely many dividing $D$. Since $\ell \nmid p+1$, it cannot divide $M$. Similarly choose $p\equiv 3(\mathrm{mod}8)$, again avoiding $D$; then $v_2(p+1)=2$, so $M\mid 4$.

There are already four rational 2-torsion points,which are distinct because a squarefree integer $D$ is nonzero. ThusIts order is four. In fact the argument works for every nonzero integer $D$; squarefreeness is not needed for this torsion conclusion.