Let $\mathbb{F}_p$ be a finite field such that $p \equiv 1 \ (\mathrm{mod} \ 3)$ and $p \equiv 3 \ (\mathrm{mod} \ 4)$. Consider the Jacobian of the hyperelliptic curve $C\!: y^2 = (x^3 + b)(x^3-b)$, where $b \in \mathbb{F}_p^* \setminus (\mathbb{F}_p^*)^3$. Is this Jacobian isogenous over $\mathbb{F}_p$ to the direct product $E\times E^\prime$ of the elliptic curves $$ E\!:y_1^2 = x_1^3 + b, \qquad E^\prime\!: y_2^2 = x_2^3 - b? $$


If $$y^2 = (x^3+b) (x^3-b) = x^6-b^2$$ then $$(y)^2 = (x^2)^3 - (b^2)$$ and $$(y x^{-3} b^{-1} )^2 = (- x^{-2})^3 + (b^{-2}) $$ giving two maps to elliptic curves, but not the elliptic curves you want. Specifically $+ b^{-2}$ is equivalent mod third powers to $+b$ and thus is equivalent mod sixth powers to $\pm b$, but $-b^2$ is not equivalent to the other one of $\pm b$.

We can tell these maps are not linearly dependent on the Jacobian because the first one is preserved by the involution $x\to-x$ while the second is negated by it.

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    $\begingroup$ "These aren't the elliptic curves you're looking for." $\endgroup$ – R. van Dobben de Bruyn Aug 22 '19 at 13:29

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