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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 $$P_i := (\zeta^i\sqrt[3]{-b}, 0) \in E[2] \cap E^\prime[2],$$ where $\zeta^3 = 1$, $\zeta \neq 1$. Note that $\zeta \in \mathbb{F}_p$.

Finally, denote by $G \subset E[2] \!\times\! E^\prime[2]$ 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?

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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.

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  • $\begingroup$ Honestly speaking, I have already understood the answer myself. I agree with you that this is the Weil restriction. However, thank you very much for your answer! $\endgroup$ – Dima Koshelev Aug 24 '19 at 15:18

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