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 $\stackrel{~}{E}$ be the reduction^{} of $E$ modulo $p$ (see bad reduction) which is a curve defined over ${\mathbb{F}}_{p}=\mathbb{Z}/p\mathbb{Z}$. 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 $E({\mathbb{F}}_{p})$ are related via the reduction map:
$${\pi}_{p}:E({\mathbb{Q}}_{p})\to \stackrel{~}{E}({\mathbb{F}}_{p})$$ 
$${\pi}_{p}(P)={\pi}_{p}([{x}_{0},{y}_{0},{z}_{0}])=[{x}_{0}\mathrm{mod}p,{y}_{0}\mathrm{mod}p,{z}_{0}\mathrm{mod}p]=\stackrel{~}{P}$$ 
Recall that $\stackrel{~}{E}$ might be a singular curve at some points. We denote ${\stackrel{~}{E}}_{\mathrm{ns}}({\mathbb{F}}_{p})$ the set of nonsingular points of $\stackrel{~}{E}$. We also define
$${E}_{0}({\mathbb{Q}}_{p})=\{P\in E({\mathbb{Q}}_{p})\mid {\pi}_{p}(P)=\stackrel{~}{P}\in {\stackrel{~}{E}}_{\mathrm{ns}}({\mathbb{F}}_{p})\}$$ 
$${E}_{1}({\mathbb{Q}}_{p})=\{P\in E({\mathbb{Q}}_{p})\mid {\pi}_{p}(P)=\stackrel{~}{P}=\stackrel{~}{O}\}=\mathrm{Ker}({\pi}_{p}).$$ 
Proposition 1.
There is an exact sequence of abelian groups^{}
$$0\u27f6{E}_{1}({\mathbb{Q}}_{p})\u27f6{E}_{0}({\mathbb{Q}}_{p})\u27f6{\stackrel{~}{E}}_{\mathrm{ns}}({\mathbb{F}}_{p})\u27f60$$ 
where the righthand 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 $m$torsion^{} of $G$, i.e. the points of order $m$.
Proposition 2.
Let $E\mathrm{/}\mathrm{Q}$ be an elliptic curve (as above) and let $m$ be a positive integer such that $\mathrm{gcd}\mathit{}\mathrm{(}p\mathrm{,}m\mathrm{)}\mathrm{=}\mathrm{1}$. Then:

1.
$${E}_{1}({\mathbb{Q}}_{p})[m]=\{O\}$$ 
2.
If $\stackrel{~}{E}({\mathbb{F}}_{p})$ is a nonsingular curve, then the reduction map, restricted to $E({\mathbb{Q}}_{p})[m]$, is injective^{}. This is
$$E({\mathbb{Q}}_{p})[m]\u27f6\stackrel{~}{E}({\mathbb{F}}_{p})$$ is injective.
Remark: Part $2$ of the proposition^{} is quite useful when trying 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 not divide $m$, $E(\mathbb{Q})[m]\u27f6\stackrel{~}{E}({\mathbb{F}}_{p})$ must be injective and therefore the number of $m$torsion points divides the number of points defined over ${\mathbb{F}}_{p}$.
Example:
Let $E/\mathbb{Q}$ be given by
$${y}^{2}={x}^{3}+3$$ 
The discriminant^{} of this curve is $\mathrm{\Delta}=3888={2}^{4}{3}^{5}$. Recall that if $p$ is a prime of bad reduction, then $p\mid \mathrm{\Delta}$. Thus the only primes of bad reduction are $2,3$, so $\stackrel{~}{E}$ is nonsingular for all $p\ge 5$.
Let $p=5$ and consider the reduction of $E$ modulo $5$, $\stackrel{~}{E}$. Then we have
$$\stackrel{~}{E}(\mathbb{Z}/5\mathbb{Z})=\{\stackrel{~}{O},(1,2),(1,3),(2,1),(2,4),(3,0)\}$$ 
where all the coordinates are to be considered modulo $5$ (remember the point at infinity!). Hence ${N}_{5}=\mid \stackrel{~}{E}(\mathbb{Z}/5\mathbb{Z})\mid =6$. Similarly, we can prove 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 $E(\mathbb{Q})[p]$ is nontrivial, 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 trivial torsion subgroup.
Notice that $(1,2)\in E(\mathbb{Q})$ is an obvious point in the curve. Since we have proved that there is no nontrivial torsion, this point must be of infinite order! In fact
$$E(\mathbb{Q})\cong \mathbb{Z}$$ 
and the group is generated by $(1,2)$.
Title  the torsion subgroup of an elliptic curve injects in the reduction of the curve 

Canonical name  TheTorsionSubgroupOfAnEllipticCurveInjectsInTheReductionOfTheCurve 
Date of creation  20130322 13:55:47 
Last modified on  20130322 13:55:47 
Owner  alozano (2414) 
Last modified by  alozano (2414) 
Numerical id  7 
Author  alozano (2414) 
Entry type  Theorem 
Classification  msc 14H52 
Related topic  EllipticCurve 
Related topic  BadReduction 
Related topic  MazursTheoremOnTorsionOfEllipticCurves 
Related topic  NagellLutzTheorem 
Related topic  ArithmeticOfEllipticCurves 