3
$\begingroup$

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

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Indeed, my prime is of that form. Thank you Michael. One last question would be the second, if there is something like $\sqrt{5}$ descent ? I am using the $\text{End}(J)$ module structure of the Jacobian. On elliptic curves, is quite explicit to calculate, since the endomorphisms are of "known shape" (The Function Field is "small") . Here, with $\sqrt{5}$, how is usually solved pragmatically ? $\endgroup$
    – Eduardo R. Duarte
    Commented Aug 14, 2017 at 7:58
  • $\begingroup$ @EduardoRuizDuarte If you know how to multiply by $\sqrt{5}$ in $J({\mathbb F}_p)$, then you can multiply by $(p+1)/\sqrt{5}$ instead. Here is another approach that gives at least a necessary condition: if $P \in J({\mathbb F}_p)$ is divisible by $\sqrt{5}$, then it must be divisible by $1 - \zeta_5$ in $J({\mathbb F}_{p^2})$ (using that ${\mathbb F}_{p^2} = {\mathbb F}_p(\zeta_5)$ when $p \equiv -1 \bmod 5$). This in turn is equivalent to $3 - \sqrt{10}$ being a fifth power in ${\mathbb F}_{p^2}$. $\endgroup$
    – Michael Stoll
    Commented Aug 15, 2017 at 20:16
  • $\begingroup$ Thank you Michael, Yes, I know how to multiply by $\sqrt{5}$ using the $\zeta_5\in \text{Aut}(H)$ and then extending this automorphism to $J$. The thing is that $\zeta_5\in \mathbb{F}_p^2$ but its trace $\tfrac{1\pm\sqrt{5}}{2}$ lives in $\mathbb{F}_p$. I will use what you say and I will try to calculate $[\sqrt{5}] \in \text{End}_{\mathbb{F}_p}(J)$ explicitly through Mumford Representation. $\endgroup$
    – Eduardo R. Duarte
    Commented Aug 16, 2017 at 12:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .