11
$\begingroup$

Let $\mathbf{F}_q$ be a finite field of odd characteristic. Let $X_t$ be the hyperelliptic curve over $\mathbf{F}_{q^2}(t)$ with affine equation $$y^2 = \left((x^{(q+1)/2}-(x-1)^{(q+1)/2})^2 - t\right) \left((x^{(q+1)/2}+(x-1)^{(q+1)/2})^2 - t\right).$$ This family of hyperelliptic curves is smooth of genus $q-1$ away from $t=0,1,\infty$. For theoretical reasons I expect the following claim to be true:

The Jacobians of $X_t$ and $X_{1-t}$ are isogenous. Equivalently, there is a nontrivial correspondence between $X_t$ and $X_{1-t}$.

It is easy to verify (with MAGMA for instance) that the zeta functions of $X_t$ and $X_{1-t}$ agree for small values of $q$ and specific values of $t$, to the point that I'm quite convinced of the claim. To prove it, one could try to construct the correspondence explicitly. There is some literature about pairs of hyperelliptic curves with isogenous Jacobians (for instance this paper of Mestre), but it seems to involve finding a congruence between bivariate polynomials, which comes out of nowhere.

Short of constructing the actual correspondence, is there an algorithm for deciding whether two hyperelliptic curves admit an isogeny of given degree between their Jacobians?

$\endgroup$
4
$\begingroup$

In the most easy case $q=3$, the curve $X_t$ is bielliptic (the bielliptic involution given by $x\mapsto 1-x$), and the Jacobian of $X_t$ is then $(2,2)$-isogenous to the product $E_{1,t}\times E_{2,t}$, where $E_{1,t}$ and $E_{2,t}$ are the elliptic curves $$E_{1,t}: y^2=x(x^2 - (t+1)x + (t- 1)^2)$$ and $$E_{2,t}: y^2=x(x^2 + x - (t^3 - 1))$$ Now, $E_{1,t}$ is 2-isogenous to $$E_{1,t}': y^2= x(x+1)(x-(t-1))$$ and one has that $$E_{1,t}'\cong E_{1,1-t}$$ by sending $x\mapsto 1-x$ and then twisting by $-1$, which is an isomorphic curve over $\mathbb{F}_9$ (this is the only place where we use we are over $\mathbb{F}_9$ and not over $\mathbb{F}_3$). while $E_{2,t}$ is 2-isogenous to $$E_{2,t}': y^2 = x(x^2 + x + t^3)$$ which in turn is clearly equal to $E_{2,1-t}$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you Xarles! I am attempting to see whether your construction generalizes. The Jacobian of $X_t$ is isogenous to a product of Jacobians of hyperelliptic curves $C_{1,t}$ and $C_{2,t}$. Evidence suggests that your pattern persists: the Jacobians of $C_{1,t}$ and $C_{1,1-t}$ are isogenous over $\mathbf{F}_{q^2}(t)$, and the Jacobians of $C_{2,t}$ and $C_{2,1-t}$ are isogenous already over $\mathbf{F}_q(t)$. $\endgroup$ – Jared Weinstein Feb 4 at 3:06
3
$\begingroup$

The Tate conjecture is known for function fields (Zarhin). To check whether the Tate modules are isomorphic you need to check that the image of Frobenius match for a finite set of places that can be bounded a priori (the argument is in Faltings's Mordell paper) although the bound is probably not very small. Checking that the image of Frobenius match at a place is essentially your computation of specializing $t$ and compute zeta functions. This gives, in principle, an algorithm to check that the Jacobians are isogenous for a given $q$. It's probably not practical except for very small $q$. It doesn't give any information of the degree of the isogeny, though.

| cite | improve this answer | |
$\endgroup$

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.