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


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

y2+a1⁢x⁢y+a3⁢y=x3+a2⁢x2+a4⁢x+a6

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

πp:E⁢(ℚp)→E~⁢(𝔽p)
πp⁢(P)=πp⁢([x0,y0,z0])=[x0⁢mod⁡p,y0⁢mod⁡p,z0⁢mod⁡p]=P~

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

E0⁢(ℚp)={P∈E⁢(ℚp)∣πp⁢(P)=P~∈E~ns⁢(𝔽p)}
E1⁢(ℚp)={P∈E⁢(ℚp)∣πp⁢(P)=P~=O~}=Ker⁡(πp).
Proposition 1.

There is an exact sequence of abelian groupsMathworldPlanetmath

0⟶E1⁢(ℚp)⟶E0⁢(ℚp)⟶E~ns⁢(𝔽p)⟶0

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.
    E1⁢(ℚp)⁢[m]={O}
  2. 2.

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

    E⁢(ℚp)⁢[m]⟶E~⁢(𝔽p)

    is injective.

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

Example: 
Let E/ℚ be given by

y2=x3+3

The discriminantPlanetmathPlanetmathPlanetmath of this curve is Δ=-3888=-24⁢35. Recall that if p is a prime of bad reduction, then p∣Δ. Thus the only primes of bad reduction are 2,3, so E~ is non-singular for all p≥5.

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

E~⁢(ℤ/5⁢ℤ)={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 N5=∣E~⁢(ℤ/5⁢ℤ)∣=6. Similarly, we can prove that N7=13.

Now let q≠5,7 be a prime numberMathworldPlanetmath. Then we claim that E⁢(ℚ)⁢[q] is trivial. Indeed, by the remark above we have

∣E⁢(ℚ)⁢[q]∣⁢divides⁢N5=6,N7=13

so ∣E⁢(ℚ)⁢[q]∣ must be 1.

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

Notice that (1,2)∈E⁢(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⁢(ℚ)≅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 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