I would like to know if there is something I can read to compute the following:


Let $H$ be a hyperelliptic curve of genus $2$ given by $y^2=x^5 + 10$ and let $J$ be its Jacobian. 


How can I prove for a fixed prime $p$ of good reduction that in $J(\mathbb{F}\_p)$ the element $D\_0:=[(-1,3)-\infty]$ is not divisible by $5$.


This question can go deeper. Since $\text{End}(J)=\mathbb{Z}(\zeta)$ where $\zeta=e^{2\pi i/5}$ and $\sqrt{5}=2\zeta+2\zeta^4 + 1$, and $\zeta$ acts on $J$ by multiplication by $\zeta$ in the $x$ coordinates of the supports of the divisors of $J$ (since this is an automorphism of $H$ which extends naturally to $J$), maybe it is easier to prove that there is no $D\in J$ such that $\sqrt{5}D=D\_0$.


In MAGMA I checked that the point $D\_0$ has infinite order over $\mathbb{Q}$ but I am working with certain families of primes which I would like to deduce if there is divisibility by 5 or not, or "divisibility by $\sqrt{5}$" or not.


A nice thing is that I know the cardinality and the structure of $J(\mathbb{F}\_p)\cong\mathbb{Z}/(\alpha)\times \mathbb{Z}/(\alpha)$ with the primes that I am using.


I tried in MAGMA to calculate the defining equations of the $\sqrt{5}$ endomorphism explicitly but I failed to construct the generic point of $J$.


Thanks



5-Descent or ($\sqrt{5}$-Descent?) on certain genus 2 Jacobians
Since you know the group structure of $J({\mathbb F}\_p)$
(I guess $p \equiv -1 \bmod 5$ and $\alpha = p+1$), the simplest way
would be to compute $(\alpha/5) \cdot (D\_0 \bmod p)$ and check if this
is the zero element in $J({\mathbb F}\_p)$. Your point is divisible by 5
if and only if it is. Computation of such multiples is reasonably
fast over finite fields.


