# 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 $\widetilde{E}$ 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:

 $\pi_{p}\colon E(\mathbb{Q}_{p})\to\widetilde{E}(\mathbb{F}_{p})$
 $\pi_{p}(P)=\pi_{p}([x_{0},y_{0},z_{0}])=[x_{0}\operatorname{mod}p,y_{0}% \operatorname{mod}p,z_{0}\operatorname{mod}p]=\widetilde{P}$

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

 $E_{0}(\mathbb{Q}_{p})=\{P\in E(\mathbb{Q}_{p})\mid\pi_{p}(P)=\widetilde{P}\in% \widetilde{E}_{\operatorname{ns}}(\mathbb{F}_{p})\}$
 $E_{1}(\mathbb{Q}_{p})=\{P\in E(\mathbb{Q}_{p})\mid\pi_{p}(P)=\widetilde{P}=% \widetilde{O}\}=\operatorname{Ker}(\pi_{p}).$
###### Proposition 1.

 $0\longrightarrow E_{1}(\mathbb{Q}_{p})\longrightarrow E_{0}(\mathbb{Q}_{p})% \longrightarrow\widetilde{E}_{\operatorname{ns}}(\mathbb{F}_{p})\longrightarrow 0$

where the right-hand side map is $\pi_{p}$ restricted to $E_{0}(\mathbb{Q}_{p})$.

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

1. 1.
 $E_{1}(\mathbb{Q}_{p})[m]=\{O\}$
2. 2.

If $\widetilde{E}(\mathbb{F}_{p})$ is a non-singular curve, then the reduction map, restricted to $E(\mathbb{Q}_{p})[m]$

 $E(\mathbb{Q}_{p})[m]\longrightarrow\widetilde{E}(\mathbb{F}_{p})$

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 $m$, $E(\mathbb{Q})[m]\longrightarrow\widetilde{E}(\mathbb{F}_{p})$ must be injective and therefore the number of $m$-torsion points divides the number of points defined over $\mathbb{F}_{p}$.

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

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

The discriminant   of this curve is $\Delta=-3888=-2^{4}3^{5}$. 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

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

where all the coordinates are to be considered modulo $5$ (remember the point at infinity!). Hence $N_{5}=\mid\widetilde{E}(\mathbb{Z}/5\mathbb{Z})\mid=6$. Similarly, we can prove that $N_{7}=13$.

Now let $q\neq 5,7$ be a prime number  . Then we claim that $E(\mathbb{Q})[q]$ 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 $\mid E(\mathbb{Q})\mid$ divides $N_{7}=13$. But it is easy to see that if $E(\mathbb{Q})[p]$ is non-trivial, then $p$ divides its order. Since $5$ does not divide $13$, we conclude that $E(\mathbb{Q})$ must be trivial. Similarly $E(\mathbb{Q})$ 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 non-trivial 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 TheTorsionSubgroupOfAnEllipticCurveInjectsInTheReductionOfTheCurve 2013-03-22 13:55:47 2013-03-22 13:55:47 alozano (2414) alozano (2414) 7 alozano (2414) Theorem msc 14H52 EllipticCurve BadReduction MazursTheoremOnTorsionOfEllipticCurves NagellLutzTheorem ArithmeticOfEllipticCurves