# 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...)

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.

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)