To prove given elliptic curve is Jacobi variety of given genus $1$ curve 　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?
 A: $E$ 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 $\Bbb{Q}$.
