This should be a simple property of the generic fibre, but I cannot formulate the argument properly.
Let $X$ be a projective surface over an algebraically closed field $k$ with a morphism onto a smooth curve $\pi : X \rightarrow B$ with generic fiber $X_{\eta}$ isomorphic to $\mathbb{P}^1_K$, where $K$ is the function field of $B$.
Then $X$ is birational to $B \times \mathbb{P}^1_k$.
How do I show this cleanly?
My vague idea would be that the isomorphism $f: X_{\eta} \rightarrow \mathbb{P}^1_K $ extends to $X_{\eta} \times_B \operatorname{Spec} k(x) \rightarrow \mathbb{P}^1_K \times_B \operatorname{Spec} k(x) $ for all closed points in $x \in B$ with residue field $k(x)$. Then one sees that $\mathbb{P}^1_K \times_B \operatorname{Spec} k(x) = \mathbb{P}^1_k$ by going to an affine open neighborhood of $x$ and using that $k(x) = k$, because $k$ is algebraically closed. Moreover, $X_{\eta} \times_B \operatorname{Spec} k(x) $ should be a dense open of the fiber $X_x$. This would give compatible birational maps to $\mathbb{P}^1_k$ on all fibers showing the claim.
Does this argument work? If so, any help in making it cleaner and more rigorous would be highly appreciated. In particular, I do not feel safe with "compatible birational maps on all fibers".