Assume we are working over $\mathbb{C}$, and we have a projective morphism with connected fibers $f: X \rightarrow Z$ whose geometric generic fiber $X_\overline{\eta}$ is isomorphic to a Hirzebruch surface $\mathbb{F}_n$.

Thus, $X_\overline{\eta}$ admits a morphism to $\mathbb{P}^1$, and this is defined over some finite extension of $K(Z)$. So, we know that, up to a generically finite base change and birational modification of the main component of the fiber product, we get a morphism $f': X' \rightarrow Z'$ that factors as $g': X' \rightarrow Y'$ and $h': Y' \rightarrow Z'$, where $g'$ and $h'$ are both generically $\mathbb{P}^1$-bundles.

My question is the following. Given the setup above, are there cases when we know that the finite base extension is not needed? My naive hopes rely on two facts. First, the morphism $\mathbb{F}_n \rightarrow \mathbb{P}^1$ is defined over $\mathrm{Spec}(\mathbb{Z})$, and so we can base change it to $K(Z)$. Second, if the base $Z$ is a curve, by the theorem of Graber-Harris-Starr we know that $X_\eta$ has a $K(Z)$-point. For instance, is it reasonable to get something in the direction I want if the base is a curve?

  • $\begingroup$ I am not sure, but how about to run K-MMP to get a Mori fiber space over Z? $\endgroup$
    – Chen Jiang
    Sep 29, 2018 at 1:38
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    $\begingroup$ @ChenJiang I thought about it. My fear is that the MFS could have fibers of dimension 2: Imagine that we are over a curve and, the fibers are $\mathbb{P}^1 \times \mathbb{P}^1$ and the two rulings are interchanged if we follow a loop in the base. Then, the fibration would have relative Picard 1 to start with. $\endgroup$
    – Stefano
    Sep 29, 2018 at 2:37
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    $\begingroup$ I think that if $n = 1$ and $Z$ is a curve, then you blow down the $(-1)$-curve and then project away from your favourite rational point to get the fibration defined over $K(Z)$. I don't know if you can make a similar strategy work for general $n$. $\endgroup$ Sep 30, 2018 at 1:31

1 Answer 1


If $n>0$, and if you have a rational section (for instance when $Z$ is a curve), then you do not need the finite extension. The reason is that the field $K(Z)$ is perfect (as you work in characteristic zero), and that the Galois group acts on $\mathbb{F}_n$ preserving the exceptional curve (unique curve of negative self-intersction) and the fibration (the fibres are the only curve of self-intersection $0$). The morphism to the curve is then defined over $K(Z)$, and the base is rational as it has a point, so the generic fibre is already isomorphic to $\mathbb{F}_n$ over $K(Z)$.

If $n=0$, then you really need the finite extension in general. Take for instance the variety

$X=\{([w:x:y:z],t)\in \mathbb{P}^3 \times \mathbb{A}^1\mid xy-z^2-w^2p(t)=0\}$

for a polynomial $p$ of large degree with no multiple root and consider the morphism $f\colon X\to Z=\mathbb{A}^1$ given by the second projection (you can obtain a projective example by putting this in the suitable $\mathbb{P}^2$-bundle over $\mathbb{P}^1$). The generic fibre is isomorphic to a smooth quadric, not isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$, but after a finite extension that consists of adding a square root of $f$, you get a fibre which is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$.

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    $\begingroup$ If you construct the fibration just by noting that the fibres are the only curves of self-intersection $0$, how do you know the base of the fibration will be a $\mathbb P^1$ (as opposed to some twist)? $\endgroup$ Sep 30, 2018 at 1:30
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    $\begingroup$ I am a bit confused. Wouldn't Graber-Harris-Starr imply that the generic fiber in the example admits a rational point? $\endgroup$
    – Stefano
    Sep 30, 2018 at 15:17
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    $\begingroup$ In fact, $\mathbb C(t)$ is a $C_1$ field by Tsen's theorem, i.e. every homogeneous polynomial of degree $d$ in $n > d$ variables has a nontrivial root. This in particular implies that a quadric in $\mathbb P^3$ has a rational point. (You don't need the full strength of GHS for this.) $\endgroup$ Sep 30, 2018 at 19:19
  • $\begingroup$ @R. van Dobben de Bruyn: yes, you are right, if you do not have any point, you could a priori have a form of a Hirzebruch surface. This is however avoided if you have a rational point, so as it is said is true if $Z$ is a curve. I edited the answer accordingly. $\endgroup$ Oct 1, 2018 at 20:59
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    $\begingroup$ @JérémyBlanc: the confusion comes from your sentence "the generic fibre is not isomorphic to $\mathbb P^1 \times \mathbb P^1$, but to a quadric with no point [...]". This is different than what you describe in your comment. $\endgroup$ Oct 2, 2018 at 0:58

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