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Consider an ordinary elliptic curve $E_b\!: y^2 = x^3 + b$, of $j$-invariant $0$ over a finite field $\mathbb{F}_q$, such that $\sqrt{b} \not\in \mathbb{F}_q$.

Question. What are some examples of hyperelliptic $\mathbb{F}_q$-curves $C$ of geometric genus $2$ whose Jacobians $J_C$ are $\mathbb{F}_q$-isogenous to the direct square $E_b^2$?

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  • $\begingroup$ $p=13,b=2, y^2 = x^6+4x^4+1$ $\endgroup$
    – Felipe Voloch
    Commented Apr 13, 2021 at 4:31
  • $\begingroup$ Could you clarify ? Is your example generalized to infinitely many $p$ ? $\endgroup$
    – Dimitri Koshelev
    Commented Apr 13, 2021 at 8:43
  • $\begingroup$ Don't know. First example computer spat out. $\endgroup$
    – Felipe Voloch
    Commented Apr 13, 2021 at 19:48

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Edit: this answer originally ignored an important twist.

Take $C: y^2 = x^6 + b$. There is a $(2,2)$-isogeny $J_C \to E_b\times E_{b^2}$ with kernel corresponding to the factorization $\{x^2 - \beta_1, x^2 - \beta_2, x^2 - \beta_3\}$ of $x^6 + b$, where the $\beta_i$ are the roots of $x^3 + b$. While these $\beta_i$ may only be defined over some extension of $\mathbb{F}_q$, the set of factors is rational over $\mathbb{F}_q$, and therefore the isogeny is defined over $\mathbb{F}_q$. The curve $E_{b^2}$ is a cubic twist of $E_b$, so if $b$ is a cube in $\mathbb{F}_q$ then you can compose to get an $\mathbb{F}_q$-isogeny to $E_b^2$.

I think a good reference for the isogeny construction (over number fields) is Section 3 of Howe, Leprevost, and Poonen's Large torsion subgroups of split Jacobians of curves of genus two or three.

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  • $\begingroup$ hyperellcharpoly(Mod(1,13)*(x^6+2)) = x^4 + 12*x^3 + 61*x^2 + 156*x + 169 hyperellcharpoly(Mod(1,13)*(x^3+2))^2 = x^4 + 10*x^3 + 51*x^2 + 130*x + 169 $\endgroup$
    – Felipe Voloch
    Commented Apr 13, 2021 at 22:36
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    $\begingroup$ Thanks @FelipeVoloch, I forgot about the possible twist! The couple of examples I tried as a sanity check had b nonsquare but (unluckily) cubic. $\endgroup$
    – Ben Smith
    Commented Apr 14, 2021 at 10:06
  • $\begingroup$ Thanks @BenSmith. What about $(n,n)$-isogenies ($n > 2$) over $\mathbb{F}_q$ if $\sqrt{b}, \sqrt[3]{b} \not\in \mathbb{F}_q$ ? $\endgroup$
    – Dimitri Koshelev
    Commented Apr 14, 2021 at 11:48

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