# Is the Jacobian isogenous over $\mathbb{F}_p$ to the direct product of the elliptic curves?

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 $$E\!:y_1^2 = x_1^3 + b, \qquad E^\prime\!: y_2^2 = x_2^3 - b?$$

## 1 Answer

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.

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