the torsion subgroup of an elliptic curve injects in the reduction of the curve TorsionSubgroupOfAnEllipticCurveInjectsInTheReductionOfTheCurve 2003-09-04 14:14:20 2008-01-23 18:30:06 Theorem torsion compute torsion subgroup elliptic curve % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newtheorem{thm}{Theorem} \newtheorem{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} % Some sets \newcommand{\Nats}{\mathbb{N}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\Reals}{\mathbb{R}} \newcommand{\Complex}{\mathbb{C}} \newcommand{\Rats}{\mathbb{Q}} 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 $$E_0(\Rats_p)=\{ P\in E(\Rats_p) \mid \pi_p(P)=\widetilde{P}\in \widetilde{E}_{\operatorname{ns}}(\mathbb{F}_p)\}$$ $$E_1(\Rats_p)=\{ P\in E(\Rats_p) \mid \pi_p(P)=\widetilde{P}=\widetilde{O}\}= \operatorname{Ker}(\pi_p).$$ \begin{prop} There is an exact sequence of abelian groups $$0\longrightarrow E_1(\Rats_p)\longrightarrow E_0(\Rats_p)\longrightarrow \widetilde{E}_{\operatorname{ns}}(\mathbb{F}_p)\longrightarrow 0 $$ where the right-hand side map is $\pi_p$ restricted to $E_0(\Rats_p)$. \end{prop} Notation: Given an abelian group $G$, we denote by $G[m]$ the $m$-torsion of $G$, i.e. the points of order $m$. \begin{prop} Let $E/\Rats$ be an elliptic curve (as above) and let $m$ be a positive integer such that $\gcd(p,m)=1$. Then: \begin{enumerate} \item $$E_1(\Rats_p)[m]=\{ O \}$$ \item If $\widetilde{E}(\mathbb{F}_p)$ is a non-singular curve, then the reduction map, restricted to $E(\Rats_p)[m]$, is injective. This is $$E(\Rats_p)[m] \longrightarrow \widetilde{E}(\mathbb{F}_p)$$ is injective. \end{enumerate} \end{prop} {\bf 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 $\mathbb{F}_p$. {\bf Example}:\quad\\ 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 $(1,2)$.