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

$\endgroup$
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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Aug 14 '17 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 Aug 15 '17 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 Aug 16 '17 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.