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?



Breaking a morphism with generic fiber $\mathbb{F}_n$
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 $\mathbb{P}^1\times \mathbb{P}^1$.


