the torsion subgroup of an elliptic curve injects in the reduction of the curve

Let $E$ be an elliptic curve defined over $\mathbb{Q}$ and let$p\in \mathbb{Z}$ be a prime. Let

$$y^2+a_{1}xy+a_{3}y=x^3+a_2{x}^2+a_{4}x+a_6$$

be a minimal Weierstrass equation for $E/\mathbb{Q}$,with coefficients $a_i\in \mathbb{Z}$. Let $\tilde{E}$ be the reductionof $E$ modulo $p$ (see bad reduction) which is a curve definedover ${𝔽}_p=\mathbb{Z}/p\mathbb{Z}$. The curve $E/\mathbb{Q}$ can also be considered as a curve over the $p$-adics, $E/{\mathbb{Q}}_p$, and, in fact, the group of rational points $E(\mathbb{Q})$ injects into $E({\mathbb{Q}}_p)$. Also, the groups $E({\mathbb{Q}}_p)$ and $E({𝔽}_p)$ are related via the reductionmap:

$${\pi }_p:E({\mathbb{Q}}_p)\to \tilde{E}({𝔽}_p)$$

$${\pi }_p(P)={\pi }_p([x_0,y_0,z_0])=[x_0\mathrm{mod}p,y_0\mathrm{mod}p,z_0\mathrm{mod}p]=\tilde{P}$$

Recall that $\tilde{E}$ might be a singular curve at somepoints. We denote${\tilde{E}}_{\mathrm{ns}}({𝔽}_p)$ the set ofnon-singular points of $\tilde{E}$. We also define

$$E_0({\mathbb{Q}}_p)=\{P\in E({\mathbb{Q}}_p)\mid {\pi }_p(P)=\tilde{P}\in {\tilde{E}}_{\mathrm{ns}}({𝔽}_p)\}$$

$$E_1({\mathbb{Q}}_p)=\{P\in E({\mathbb{Q}}_p)\mid {\pi }_p(P)=\tilde{P}=\tilde{O}\}=\mathrm{Ker}({\pi }_p).$$

Proposition 1.

There is an exact sequence of abelian groups

$$0⟶E_1({\mathbb{Q}}_p)⟶E_0({\mathbb{Q}}_p)⟶{\tilde{E}}_{\mathrm{ns}}({𝔽}_p)⟶0$$

where the right-hand side map is ${\pi }_p$ restricted to$E_{\mathrm{0}}\mathit{}\mathrm{(}{\mathrm{Q}}_p\mathrm{)}$.

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

Proposition 2.

Let $E\mathrm{/}\mathrm{Q}$ be an elliptic curve (as above) and let $m$ be apositive integer such that $\gcd \mathit{}\mathrm{(}p\mathrm{,}m\mathrm{)}\mathrm{=}\mathrm{1}$. Then:

1. $$E_1({\mathbb{Q}}_p)[m]=\{O\}$$

2. If$\tilde{E}({𝔽}_p)$ is a non-singular curve, then thereduction map, restricted to $E({\mathbb{Q}}_p)[m]$, is injective. This is

   $$E({\mathbb{Q}}_p)[m]⟶\tilde{E}({𝔽}_p)$$

   is injective.

Remark: Part $2$ of the proposition is quite useful whentrying to compute the torsion subgroup of $E/\mathbb{Q}$. As we mentioned above, $E(\mathbb{Q})$ injects into $E({\mathbb{Q}}_p)$. The proposition can be reworded as follows: for all primes $p$ which do notdivide $m$, $E(\mathbb{Q})[m]⟶\tilde{E}({𝔽}_p)$ must be injective and therefore thenumber of $m$-torsion points divides the number of points definedover ${𝔽}_p$.

Example: 
Let $E/\mathbb{Q}$ be given by

$$y^2=x^3+3$$

Thediscriminant of this curve is $\mathrm{\Delta }=-3888=-2^4{3}^5$. Recall thatif $p$ is a prime of bad reduction, then $p\mid \mathrm{\Delta }$. Thus theonly primes of bad reduction are $2,3$, so $\tilde{E}$ isnon-singular for all $p\ge 5$.

Let $p=5$ and consider the reduction of $E$ modulo $5$,$\tilde{E}$. Then we have

$$\tilde{E}(\mathbb{Z}/5\mathbb{Z})=\{\tilde{O},(1,2),(1,3),(2,1),(2,4),(3,0)\}$$

where all the coordinates are to beconsidered modulo $5$ (remember the point at infinity!). Hence$N_5=\mid \tilde{E}(\mathbb{Z}/5\mathbb{Z})\mid =6$. Similarly, we canprove that $N_7=13$.

Now let $q\ne 5,7$ be a prime number. Then we claim that$E(\mathbb{Q})[q]$ is trivial. Indeed, by the remark above we have

$$\mid E(\mathbb{Q})[q]\mid \text{divides}{N}_5=6,N_7=13$$

so $\mid E(\mathbb{Q})[q]\mid$ must be 1.

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

Notice that $(1,2)\in E(\mathbb{Q})$ is an obvious point in the curve.Since we have proved that there is no non-trivial torsion, thispoint must be of infinite order! In fact

$$E(\mathbb{Q})\cong \mathbb{Z}$$

and the group is generated by $(1,2)$.