Let $E$ be an elliptic curve defined over $\Rats$ and let $p\in\Ints$ be a prime. Let $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ be a minimal Weierstrass equation for $E/\Rats$ , with coefficients $a_i\in\Ints$ . Let $\widetilde{E}$ be the reduction of $E$ modulo $p$ (see bad reduction) which is a curve defined over $\mathbb{F}_p=\Ints/p\Ints$ . The curve $E/\Rats$ can also be considered as a curve over the $p$ -adics, $E/\Rats_p$ , and, in fact, the group of rational points $E(\Rats)$ injects into $E(\Rats_p)$ . Also, the groups $E(\Rats_p)$ and $E(\mathbb{F}_p)$ are related via the reduction map: $$\pi_p \colon E(\Rats_p) \to \widetilde{E}(\mathbb{F}_p)$$ $$\pi_p(P)=\pi_p([x_0,y_0,z_0])=[x_0 \operatorname{mod} p,y_0 \operatorname{mod} p,z_0\operatorname{mod} p]=\widetilde{P}$$

Recall that $\widetilde{E}$ might be a singular curve at some points. We denote $\widetilde{E}_{\operatorname{ns}}(\mathbb{F}_p)$ the set of non-singular points of $\widetilde{E}$ . We also define $$E_0(\Rats_p)=\{ P\in E(\Rats_p) \mid \pi_p(P)=\widetilde{P}\in \widetilde{E}_{\operatorname{ns}}(\mathbb{F}_p)\}$$ $$E_1(\Rats_p)=\{ P\in E(\Rats_p) \mid \pi_p(P)=\widetilde{P}=\widetilde{O}\}= \operatorname{Ker}(\pi_p).$$

Notation: Given an abelian group $G$ , we denote by $G[m]$ the $m$ -torsion of $G$ , i.e. the points of order $m$ .

Remark: Part $2$ of the proposition is quite useful when trying to compute the torsion subgroup of $E/\Rats$ . As we mentioned above, $E(\Rats)$ injects into $E(\Rats_p)$ . The proposition can be reworded as follows: for all primes $p$ which do not divide $m$ , $E(\Rats)[m] \longrightarrow \widetilde{E}(\mathbb{F}_p)$ must be injective and therefore the number of $m$ -torsion points divides the number of points defined over $\mathbb{F}_p$ .

Example:    
Let $E/\Rats$ be given by $$ y^2=x^3+3$$ The discriminant of this curve is $\Delta=-3888=-2^43^5$ . Recall that if $p$ is a prime of bad reduction, then $p\mid \Delta$ . Thus the only primes of bad reduction are $2,3$ , so $\widetilde{E}$ is non-singular for all $p\geq 5$ .

Let $p=5$ and consider the reduction of $E$ modulo $5$ , $\widetilde{E}$ . Then we have $$\widetilde{E}(\Ints/5\Ints)=\{ \widetilde{O}, (1,2), (1,3), (2,1), (2,4),(3,0) \}$$ where all the coordinates are to be considered modulo $5$ (remember the point at infinity!). Hence $N_5=\mid \widetilde{E}(\Ints/5\Ints)\mid=6$ . Similarly, we can prove that $N_7=13$ .

Now let $q\neq 5,7$ be a prime number. Then we claim that $E(\Rats)[q]$ is trivial. Indeed, by the remark above we have $$\mid E(\Rats)[q] \mid \text{divides}\ N_5=6,N_7=13$$ so $\mid E(\Rats)[q] \mid$ must be 1.

For the case $q=5$ be know that $\mid E(\Rats)[5] \mid$ divides $N_7=13$ . But it is easy to see that if $E(\Rats)[p]$ is non-trivial, then $p$ divides its order. Since $5$ does not divide $13$ , we conclude that $E(\Rats)[5]$ must be trivial. Similarly $E(\Rats)[7]$ is trivial as well. Therefore $E(\Rats)$ has trivial torsion subgroup.

Notice that $(1,2)\in E(\Rats)$ is an obvious point in the curve. Since we have proved that there is no non-trivial torsion, this point must be of infinite order! In fact $$E(\Rats)\cong \Ints$$ and the group is generated by $(1,2)$ .