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Let $\mathcal{X}$ be an elliptic surface over $\mathbb{P}^1$ without a section and let $\mathcal{J}$ be an elliptic surface over $\mathbb{P}^1$ with a section. Assume we have the commutative diagram

\begin{array}{ccc} \mathcal{X} \times_{\mathbb{P}^1} C & \hspace{-0.2cm} \stackrel{\Large{\varphi}}{\leftarrow - \rightarrow} \hspace{-0.2cm} & \mathcal{J} \times_{\mathbb{P}^1} C \\ pr_{\mathcal{X}} \downarrow & \Box & \downarrow pr_{\mathcal{J}} \\ \mathcal{X} \longrightarrow \hspace{-1.6cm} & \mathbb{P}^1 & \hspace{-1.7cm} \longleftarrow \mathcal{J} \end{array}

In this diagram $C$ is a curve over $\mathbb{P}^1$, $\varphi$ is a birational isomorphism over $C$, and $\times_{\mathbb{P}^1}$ is the fiber product over $\mathbb{P}^1$.

Is $\mathcal{J}$ the Jacobian of $\mathcal{X}$ in that case?

The question arised from the following example. Let $\mathcal{X} = Q_1 \cap Q_2 \subset \mathbb{A}^4$ over an algebraically closed field $k$ of even characteristic, where

\begin{array}{l} Q_1\!: y^2 + y = t[(x^{\prime^2} + xx^\prime t + x^\prime t^2 + xt^3 + t^4) + (x^\prime + xt + t^2)], \\ Q_2\!: x^\prime = x^2. \end{array}

The projection on $t$ is the elliptic fibration, because a general fiber is an nonsingular intersection of two quadrics.

I know that $\mathcal{X}$ is rational. I can write the proof if necessary.

Consider the conic $C\!: s^2 = t \subset \mathbb{A}^2$. You can check that the surface $\mathcal{X} \times_{\mathbb{P}^1} C$ is birationally isomophic over $C$ to the surface $\mathcal{J}\times_{\mathbb{P}^1}C$ by the rational map $$ \varphi: \mathcal{J}\times_{\mathbb{P}^1} C \to \mathcal{X} \times_{\mathbb{P}^1} C, \qquad x := \frac{x_2 + s}{t^2}, \ y := \frac{y_2}{t^2} + (t^2 + 1 + x^2)s, \ x^\prime := x^2, $$ where $$ \mathcal{J}\!: y_2^2 + t^2y_2 = x_2^3 + t(t^2 + 1)x_2^2 + t(t^5 + t^3 + 1)x_2 + t^2(t^5 + t^3 + t^2 + 1). $$ It is clear that Kodaira-Néron model of this surface is the K3 surface. However in the book Enriques surfaces I by F. R. Cossec and I. V. Dolgachev the proposition 5.6.1.(ii) states that the Jacobian of an rational elliptic surface is again rational. Thus I have the contradiction, because a K3 surface is irrational. What is an error?

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  • $\begingroup$ 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. $\endgroup$ – Jason Starr Aug 1 '16 at 15:10
  • $\begingroup$ What arrows should I add to the diagram to define the Jacobian of $\mathcal{X}$? $\endgroup$ – Dima Koshelev Aug 1 '16 at 15:47
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    $\begingroup$ "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 $\mathbb{P}^1$. $\endgroup$ – Jason Starr Aug 1 '16 at 15:56
  • $\begingroup$ Thank you. Is it difficult to find the Jacobian of my $\mathcal{X}$? $\endgroup$ – Dima Koshelev Aug 1 '16 at 16:20
  • $\begingroup$ 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) $\endgroup$ – Will Sawin Aug 2 '16 at 18:36

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