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Aug 2, 2016 at 18:36 comment added Will Sawin If $C$ is a Galois covering of $\mathbb P^1$, the descent data for the descent of $\mathcal J \times_{\mathbb P^1} C$ to $\mathcal X$ will consist of automorphisms of $\mathcal J \times_{\mathbb P^1} C$. Each automorphism will be some combination of automorphisms of the elliptic curve group and translations along that group. Ignoring the translation part and taking only the group homomorphisms, you obtain descent data for the Jacobian. (This follows from the fact that the Jacobian of an elliptic surface is itself, and from knowing how automorphisms of that elliptic surface act on its Jacobian)
Aug 1, 2016 at 16:20 comment added Dimitri Koshelev Thank you. Is it difficult to find the Jacobian of my $\mathcal{X}$?
Aug 1, 2016 at 15:56 comment added Jason Starr "What arrows should I add to the diagram to define the Jacobian of $\mathcal{X}$?" The least number of arrows is $\mathcal{J}\times_{\mathbb{P}^1} \mathcal{X} \to \mathcal{X}$, defining an action of $\mathcal{J}$ on $\mathcal{X}$ as a principal homogeneous space over $\mathbb{P}^1$.
Aug 1, 2016 at 15:47 comment added Dimitri Koshelev What arrows should I add to the diagram to define the Jacobian of $\mathcal{X}$?
Aug 1, 2016 at 15:10 comment added Jason Starr In general $\mathcal{J}$ need not be the Jacobian of your elliptic fibration. For instance, perhaps the Jacobian of your elliptic fibration is constant, say $E\times (\mathbb{P}^1\setminus \Delta)$ for some elliptic curve $E$ and discriminant divisor $\Delta$, yet $\mathcal{J}$ is an isotrivial family of elliptic curves that is not isomorphic to $E\times (\mathbb{P}^1\setminus \Delta)$. Since the pullback of $\mathcal{J}$ over some finite cover of $\mathbb{P}^1$ becomes trivial, there will be a diagram as above even though $\mathcal{J}$ is not the Jacobian.
Aug 1, 2016 at 15:05 history asked Dimitri Koshelev CC BY-SA 3.0