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

Let E be an elliptic curveMathworldPlanetmath defined over and let p be a prime. Let


be a minimalPlanetmathPlanetmath Weierstrass equation for E/, with coefficients ai. Let E~ be the reductionPlanetmathPlanetmathPlanetmath of E modulo p (see bad reduction) which is a curve defined over 𝔽p=/p. The curve E/ can also be considered as a curve over the p-adics, E/p, and, in fact, the group of rational points E() injects into E(p). Also, the groups E(p) and E(𝔽p) are related via the reduction map:


Recall that E~ might be a singular curve at some points. We denote E~ns(𝔽p) the set of non-singular points of E~. We also define

Proposition 1.

There is an exact sequence of abelian groupsMathworldPlanetmath


where the right-hand side map is πp restricted to E0(Qp).

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

Proposition 2.

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

  1. 1.
  2. 2.

    If E~(𝔽p) is a non-singular curve, then the reduction map, restricted to E(p)[m], is injectivePlanetmathPlanetmath. This is


    is injective.

Remark: Part 2 of the propositionPlanetmathPlanetmathPlanetmath is quite useful when trying to compute the torsion subgroup of E/. As we mentioned above, E() injects into E(p). The proposition can be reworded as follows: for all primes p which do not divide m, E()[m]E~(𝔽p) must be injective and therefore the number of m-torsion points divides the number of points defined over 𝔽p.

Let E/ be given by


The discriminantPlanetmathPlanetmathPlanetmath of this curve is Δ=-3888=-2435. Recall that if p is a prime of bad reduction, then pΔ. Thus the only primes of bad reduction are 2,3, so E~ is non-singular for all p5.

Let p=5 and consider the reduction of E modulo 5, E~. Then we have


where all the coordinates are to be considered modulo 5 (remember the point at infinity!). Hence N5=E~(/5)=6. Similarly, we can prove that N7=13.

Now let q5,7 be a prime numberMathworldPlanetmath. Then we claim that E()[q] is trivial. Indeed, by the remark above we have


so E()[q] must be 1.

For the case q=5 be know that E()[5] divides N7=13. But it is easy to see that if E()[p] is non-trivial, then p divides its order. Since 5 does not divide 13, we conclude that E()[5] must be trivial. Similarly E()[7] is trivial as well. Therefore E() has trivial torsion subgroup.

Notice that (1,2)E() 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


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 2013-03-22 13:55:47
Last modified on 2013-03-22 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