# What is the quotient $E \!\times\! E^\prime / G$?

Consider a finite field $$\mathbb{F}_p$$ such that $$p \equiv 1 \ (\mathrm{mod} \ 3)$$, $$p \equiv 3 \ (\mathrm{mod} \ 4)$$, $$\mathbb{F}_{p^2}$$-isomorphic elliptic curves (of $$j$$-invariant $$0$$) $$E\!:y_1^2 = x_1^3 + b, \qquad E^\prime\!: y_2^2 + x_2^3 + b = 0,$$ where $$b \in \mathbb{F}_p^* \setminus (\mathbb{F}_p^*)^3.$$

Also, for $$i \in \mathbb{Z}/3$$ let $$P_i := (\zeta^i\sqrt{-b}, 0) \in E \cap E^\prime,$$ where $$\zeta^3 = 1$$, $$\zeta \neq 1$$. Note that $$\zeta \in \mathbb{F}_p$$.

Finally, denote by $$G \subset E \!\times\! E^\prime$$ the $$\mathbb{F}_p$$-invariant subgroup $$(\simeq \mathbb{Z}/2 \!\times\! \mathbb{Z}/2)$$ generated by the point $$(P_0, P_0)$$ (or, equivalently, by any point $$(P_i, P_i)$$).

What is the quotient $$E \!\times\! E^\prime / G$$? Is this Jacobian of genus 2 curve, the direct product of elliptic curves, or the Weil restriction (with respect to $$\mathbb{F}_{p^2}/\mathbb{F}_p$$) of an elliptic curve?

I claim that $$E\times E'/G$$ is the Weil restriction of $$E_{\mathbb{F}_{p^2}}$$ w.r.t. $$\mathbb{F}_{p^2}/\mathbb{F}_{p}$$. (I don't know about the product question, or the Jacobian question; the answers might depend on $$p$$).
Let $$A$$ be the Weil restriction in question (which is an abelian surface). For each $$\mathbb{F}_{p}$$-algebra $$R$$, write $$R\left[\sqrt{-1}\right]:=\mathbb{F}_{p^2}\otimes_{\mathbb{F}_{p}}R=R[t]/(t^2+1)$$.
By definition, $$A(R)=E\left(R\left[\sqrt{-1}\right]\right)$$. We have an obvious inclusion $$E\to A$$: in terms of $$R$$-points this comes from $$R\subset R\left[\sqrt{-1}\right]$$, so we may desribe it as $$(x_1,y_1)\mapsto(x_1,y_1)$$. There is also a morphism $$E'\to A$$ given by $$(x_2,y_2)\mapsto \left(x_2,\sqrt{-1}y_2\right)$$. Putting these together yields $$f:E\times E'\to A, \qquad((x_1,y_1),(x_2,y_2))\mapsto (x_1,y_1)\oplus\left(x_2,\sqrt{-1}y_2\right)$$ where $$\oplus$$ stands for the group law on $$E\left(R\left[\sqrt{-1}\right]\right)$$.
Let me show that $$\ker(f)=G$$. (It will follow that $$f$$ is surjective and $$E\times E'/G\cong A$$). If the RHS is zero, we must have $$x_1=x_2$$ and $$y_1=-\sqrt{-1}y_2$$ but the second relation forces $$y_1=y_2=0$$ because $$R\cap\sqrt{-1}R=\{0\}$$. In other words, $$((x_1,y_1),(x_2,y_2))\in G$$. QED
Of course you can do the same thing with any ellitic curve over a field of char. $$≠2$$ and its twist by a quadratic extension.