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I hope my question is clear.

In summary, If $\phi:E\to E'$ is an isogeny and $P\in E$ is not divisible by $2$, under which conditions $\phi(P)\in E'$ is also not divisble by $2$. Here is the detail with some computations over finite fields with supersingular isogenous elliptic curves with $j$ invariant 1728.

Let $E_1$ be the elliptic curve given by $y^2 = x^3 - (1+t^2)x =f_1(x)$ and consider the curve $E_2$ given by $y^2 = x^3 + 4(1+t^2)x=f_2(x)$. Both $E_1$ and $E_2$ are isogenous via

$\phi:E_1\to E_2$

$(x,y)\mapsto (\tfrac{y^2}{x^2},\tfrac{-(1+t^2)-x^2}{x^2})$

according to Silverman's X.6 when $-(1+t^2)$ is fourth-power free.

Let $p\equiv 3\bmod 4$ and fix $t\in\mathbb{F}_p^\times$. Consider the curves $E_1(\mathbb{F}_p)$ and $E_2(\mathbb{F}_p)$, both have $p+1$ points and when $f_1(x)$ is irreducible is easy to see that it must be cyclic.

The point $P_1:=(-1,t)\in E_1(\mathbb{F}_p)$ is never divisible by $2$.

Under which conditions the point $\phi(P_1)=(t^2,t^3+2t)\in E_2$ is also not divisible by $2$ ?

I have special primes $p$ for which I need to generate the 2-Sylow subgroup of $E_1(\mathbb{F}_p)$, thats why I am concerned with a point not divisible by $2$ and I use the cyclicity of $E_1$ to know the order of this 2-Sylow Subgroup. If $E_1(\mathbb{F}_p)$ it is not cyclic, I want to look at $E_2$ which also is supersingular and must be cyclic but unfortunately $\phi(-1,t)\in E_2$ sometimes IS divisble by two according to some experiments, I would like to get a point on $E_2$ that is not divisible by two that generates the 2-Sylow-subgroup on $E_2$ using the information I know of $E_1$.

Thanks

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