Defining a subgroup of elliptic curves with specific characteristics
Hey,
is there a way, to define a subgroup of an elliptic curve with two or more characteristics? I would like to take an elliptic curve over a finite field of order p and p4, define the r-torsion subgroup (where r is a prime, too) and reduce those to the set of points, which also lays in the Frobenius-eigenspace.
For example:
p= 13
r=5
R=GF(p)
_.<x> = PolynomialRing(R)
R4.<x> = R.extension(x^4 - 2, 'x')
_.<y> = PolynomialRing(R)
b= x^-1
E = EllipticCurve(R, [1,0]) # y^2 = x^3+x
E4 = EllipticCurve(R4, [b,0])Well, it is easy to find a point on Q∈E4, such that r∗Q=(0:1:0), use
Q=ZZ(E4.order()/r *Q, but checking, if (x(Q)p,y(Q)p)=π(Q)=pQ is hard. I only need one point of that group at all, but my p is even larger (~340 bits), so brute-forcing would be an option, if I could start it 6-12 month ago :)
Furthermore, if I concider to evaluate the secant or tangent on E and let me return a point on that curve, it will have projective coordinates, with z(P)≠1. Shall I apply π to all three coordinates?
What is the projection π?
The order of
E4is not divisible byr = 5:Please edit the question, so that these two sensible points become clear.
The "real case" has to do with a prime
plike - say -p = ZZ(10^50).next_prime()? (Or bigger...)Ok, I do. :) π is the p-Frobenius, that means: π(P)=(x(P)p,y(P)p) and the prime is around 340 bits of size. You can find that in the edit also.
Things are still somehow unclear for me. I'll try to write sentences, please correct me where i am going wrong...
Let K=Fp4 for some prime p. Let F be the Frobenius morphism, Fx=xp, on K, x∈K.
Let E be the elliptic curve over K defined by the equation y2=x3+bx for some suitable b not in the prime field Fp inside K.
So we fix a point Q=(xQ,yQ) in E(K), it satisfies: y2Q=x3Q+bxQ . Applying F on the above, we get a point πQ=(xpQ,ypQ) on the curve E′, y2=x3+bpx . And bp≠b.
On the other side, the point pQ is on the curve E. An equality πQ=pQ leads to bpxQ=bxQ...
Please give a ...(more)
Ok. You got it right. I might produce an example in the next hours. For instance the parameters "p=13" and "r=5" should do it. Consider E′: y2=x3+(2−1/4)x over Fp4. This curve becomes isomorphic over Fp16 to E: y2=x3+x, where E(Fp)[r]=E(Fp16)[r]∩P∈E: π(P)=O. (Twist)
I'm on generating an example from now on.