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　Let $E$ be an elliptic curve $X^3+Y^3+60Z^3=0$ and $C$ be a genus $1$ curve given by $C:3X^3+4Y^3+5Z^3=0$.

　I want to prove elliptic curve $E$ is Jacobi variety of $C$. What I should to prove is that $Pic^0(C) \cong E$ as a group. But I cannot come up with explicit isomorphism.

　Cassel's lectures, chapter $20$ deal with this as an exrcise, but does not follow this way. Maybe $C$ is $E$ torsor ($E$ acts simply transitively on $C$) has something to do with the proof of isomorphism?

Poitou-Tate
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1 Answers1

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$E$ is an elliptic curve over $\Bbb{Q}$ thanks to the base point $O=[1:-1:0]$ whereas $C$ doesn't have any $\Bbb{Q}$-point.

$$f([X:Y:Z]) = [3 (4/3)^{1/3} X:4 (3/4)^{1/3}Y:Z]$$ is an isomorphism $E\to C$ defined over $\Bbb{Q}((4/3)^{1/3})$.

Let $$h:E\times C\to C, \quad h(P,Q)=f(P+f^{-1}(Q))$$ I didn't check but I think $h$ is defined over $\Bbb{Q}$, that is $$\forall \sigma \in G_\Bbb{Q}, \quad \sigma h(P,Q)=h(\sigma P,\sigma Q)$$ Then $$Pic^0(C) =Div^0(C)/Prin(C)$$ where $Prin(C) = f(Prin(E))=\{ \sum_j [A_j]-[B_j]\in Div^0(C), \sum_j f^{-1}(A_j)-f^{-1}(B_j)=O\in E\}$.

The isomorphism $i:E\to Pic^0(C)$ is $$i(P)= [f(P)]-[f(O)]= [h(P,f(O))]-[f(O)]$$ At first it is defined over $\Bbb{Q}((4/3)^{1/3})$.

To show that it is defined over $\Bbb{Q}$ take $\sigma\in G_\Bbb{Q}$ then $$i(\sigma P)-\sigma i(P) = [h(\sigma P,f(O))]-[f(O)]-[\sigma h( P,f(O))]+[\sigma f(O)]$$ $$=[h(\sigma P,f(O))]-[f(O)]-[ h(\sigma P,\sigma f(O))]+[\sigma f(O)]$$ which is a principal divisor as $$f^{-1}(h(\sigma P,f(O)))-f^{-1}(f(O))- f^{-1}(h(\sigma P,\sigma f(O)))+f^{-1}(\sigma f(O))$$ $$ = (\sigma P+O)-O-(\sigma P+f^{-1}(\sigma f(O)))+f^{-1}(\sigma f(O))=O\in E$$

So $i(\sigma P)-\sigma i(P)=0$ in $Pic^0(C)$, $i$ commutes with the Galois action, it is defined over $\Bbb{Q}$.

reuns
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