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".


1 Answer 1


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).

  • $\begingroup$ Thanks a lot, I was very confused. $\endgroup$
    – user028230
    Jun 9, 2017 at 20:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .