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

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

(Theorem)

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

$\displaystyle y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$

be a minimal Weierstrass equation for $E/\mathbb{Q}$ , with coefficients $a_i\in\mathbb{Z}$ . Let $\widetilde{E}$ be the reduction of

modulo

(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

-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:

$\displaystyle \pi_p \colon E(\mathbb{Q}_p) \to \widetilde{E}(\mathbb{F}_p)$

$\displaystyle \pi_p(P)=\pi_p([x_0,y_0,z_0])=[x_0 \operatorname{mod} p,y_0 \operatorname{mod} p,z_0\operatorname{mod} p]=\widetilde{P}$

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

$\displaystyle E_0(\mathbb{Q}_p)=\{ P\in E(\mathbb{Q}_p) \mid \pi_p(P)=\widetilde{P}\in \widetilde{E}_{\operatorname{ns}}(\mathbb{F}_p)\}$

$\displaystyle E_1(\mathbb{Q}_p)=\{ P\in E(\mathbb{Q}_p) \mid \pi_p(P)=\widetilde{P}=\widetilde{O}\}= \operatorname{Ker}(\pi_p).$

Proposition 1 There is an exact sequence of abelian groups

$\displaystyle 0\longrightarrow E_1(\mathbb{Q}_p)\longrightarrow E_0(\mathbb{Q}_... ...ongrightarrow \widetilde{E}_{\operatorname{ns}}(\mathbb{F}_p)\longrightarrow 0$

where the right-hand side map is $\pi_p$ restricted to $E_0(\mathbb{Q}_p)$ .

Notation: Given an abelian group , we denote by the -torsion of , i.e. the points of order .

Proposition 2 Let $E/\mathbb{Q}$ be an elliptic curve (as above) and let be a positive integer such that $\gcd(p,m)=1$ . Then:

$\displaystyle E_1(\mathbb{Q}_p)[m]=\{ O \}$
If $\widetilde{E}(\mathbb{F}_p)$ is a non-singular curve, then the reduction map, restricted to $E(\mathbb{Q}_p)[m]$ , is injective. This is
$\displaystyle E(\mathbb{Q}_p)[m] \longrightarrow \widetilde{E}(\mathbb{F}_p)$
is injective.

Remark: Part 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 which do not divide , $E(\mathbb{Q})[m] \longrightarrow \widetilde{E}(\mathbb{F}_p)$ must be injective and therefore the number of -torsion points divides the number of points defined over $\mathbb{F}_p$ .

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

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

The discriminant of this curve is $\Delta=-3888=-2^43^5$ . Recall that if

is a prime of bad reduction, then $p\mid \Delta$ . Thus the only primes of bad reduction are

, so $\widetilde{E}$ is non-singular for all $p\geq 5$ .

Let and consider the reduction of modulo , $\widetilde{E}$ . Then we have

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

where all the coordinates are to be considered modulo

(remember the point at infinity!). Hence $N_5=\mid \widetilde{E}(\mathbb{Z}/5\mathbb{Z})\mid=6$ . Similarly, we can prove that

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

$\displaystyle \mid E(\mathbb{Q})[q] \mid$ divides $\displaystyle \ N_5=6,N_7=13$

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

For the case be know that $\mid E(\mathbb{Q})[5] \mid$ divides . But it is easy to see that if $E(\mathbb{Q})[p]$ is non-trivial, then divides its order. Since does not divide , 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 non-trivial torsion, this point must be of infinite order! In fact

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

and the group is generated by

"the torsion subgroup of an elliptic curve injects in the reduction of the curve" is owned by alozano.

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Keywords:

torsion, compute torsion subgroup, elliptic curve

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Cross-references: generated by, infinite order, torsion, easy to see, infinity, coordinates, discriminant, divide, torsion subgroup, proposition, injective, non-singular, integer, positive, order, restricted, side, abelian groups, exact sequence, non-singular points, singular, map, points, rational, group, curve, bad reduction, reduction, coefficients, Weierstrass equation, minimal, prime, elliptic curve
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This is version 4 of the torsion subgroup of an elliptic curve injects in the reduction of the curve, born on 2003-09-04, modified 2008-01-23.
Object id is 4688, canonical name is TorsionSubgroupOfAnEllipticCurveInjectsInTheReductionOfTheCurve.
Accessed 2569 times total.

Classification:

AMS MSC:

14H52 (Algebraic geometry :: Curves :: Elliptic curves)

Pending Errata and Addenda

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