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 $C\!: y^2 = (x^3 + b)(x^3b)$, 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 $$ E\!:y_1^2 = x_1^3 + b, \qquad E^\prime\!: y_2^2 = x_2^3  b? $$
1 Answer
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If $$y^2 = (x^3+b) (x^3b) = x^6b^2$$ 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 $x\tox$ while the second is negated by it.

6$\begingroup$ "These aren't the elliptic curves you're looking for." $\endgroup$ Aug 22, 2019 at 13:29