# 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 $p^4$, 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 $Q\in E4$, such that $r*Q = (0:1:0)$, use

Q=ZZ(E4.order()/r *Q


, but checking, if $( x(Q)^p, y(Q)^p )=\pi(Q) = pQ$ 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 $z(P)\neq1$. Shall I apply $\pi$ to all three coordinates?

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What is the projection $\pi$?

The order of E4 is not divisible by r = 5:

sage: E4.order().factor()
2 * 14401


Please edit the question, so that these two sensible points become clear.

The "real case" has to do with a prime p like - say - p = ZZ(10^50).next_prime()? (Or bigger...)

( 2017-11-22 16:55:19 +0200 )edit

Ok, I do. :) $\pi$ is the p-Frobenius, that means: $\pi(P)=(x(P)^p, y(P)^p)$ and the prime is around 340 bits of size. You can find that in the edit also.

( 2017-11-23 12:57:14 +0200 )edit

Things are still somehow unclear for me. I'll try to write sentences, please correct me where i am going wrong...

Let $K=\mathbb F_{p^4}$ for some prime $p$. Let $F$ be the Frobenius morphism, $Fx=x^p$, on $K$, $x\in K$.

Let $E$ be the elliptic curve over $K$ defined by the equation $$y^2 = x^3+bx$$ for some suitable $b$ not in the prime field $\mathbb F_p$ inside $K$.

So we fix a point $Q = (x_Q,y_Q)$ in $E(K)$, it satisfies: $$y_Q^2 = x_Q^3 +bx_Q\ .$$ Applying $F$ on the above, we get a point $\pi Q=(x_Q^p,y_Q^p)$ on the curve $E'$, $$y^2 = x^3+b^p x\ .$$ And $b^p\ne b$.

On the other side, the point $pQ$ is on the curve $E$. An equality $\pi Q=pQ$ leads to $b^p x_Q = bx_Q$...

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 $E':\ y^2=x^3+ (2^{-1/4})x$ over $\mathbb F_{p^4}$. This curve becomes isomorphic over $\mathbb F_{p^{16}}$ to $E:\ y^2=x^3+x$, where $E(\mathbb F_p)[r]=E(\mathbb F_{p^{16}})[r]\cap {P\in E:\ \pi (P)= \mathcal O }$. (Twist)