# A surface $\pi : X \rightarrow B$ over a smooth curve with generic fiber $\mathbb{P}^1_K$ is birational to $B \times \mathbb{P}^1_k$

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".

I don't understand what you are trying to do, however I don't think it can work because $X_\eta\times_B\operatorname{Spec} k(x)$ is actually empty, because the fibers of $x$ and $\eta$ do not intersect. You can also verify that it is empty by showing that it does not have $L$-points for any field extension $L/k$ (easy exercise).
The way I would do this is to reduce to the affine case. Take $\operatorname{Spec}R\subseteq B$ a dense open subscheme of $B$, so that $R$ is a domain with field of fraction $K$. Let $\operatorname{Spec} A\subseteq X$ be a dense open subscheme of $X$. We know that $\operatorname{Spec}(K\otimes_RA)$ is birational to $\mathbb{P}^1_K$ and we want to prove that $\operatorname{Spec}A$ is birational to $\mathbb{P}^1_R$.
The hypothesis means that $(K\otimes_RA)_g\cong K[t]_f$ for some non-zero $f\in K[t]$ and $g\in K\otimes_RA$. The claim means that $A_a\cong R[t]_r$ for some $a\in A, r\in R[t]$ (again non-zero). Now use that $A$ is an $R$-algebra of finite type, i.e. $$A=R[x_1,\dots,x_h]/(f_1,\dots,f_s),$$ so $$K\otimes_RA= K[x_1,\dots,x_h]/(f_1,\dots,f_s).$$ It is now clear that the isomorphism of the hypothesis becomes an isomorphism of $A$ with $R[t]$, after inverting a suitable element (the product of all the denominators appearing in the given isomorphism).