Trivialization of flat proper morphisms with sections

Question

Let $f:X \to Y$ be a flat proper morphism of noetherian schemes. Assume both $X$ and $Y$ are regular, irreducible and $\dim Y=1$. Suppose there exists a section to the morphism $f$ i.e., a morphism $s:Y \to X$ such that $f \circ s:Y \to Y$ is identity. Is this enough to conclude that for every closed point $y \in Y$, there exists an open neighbourhood $U$ of $Y$ for which $f^{-1}(U) \cong U \times S$ for some scheme $S$ or do we need further assumptions? Any reference in this direction will be most welcome.

Answer 1

No, this is certainly not the case.

For example, one could take $X$ to be an elliptic surface over $Y=\mathbf P^1$, and $f: X \rightarrow \mathbf P^1$ the elliptic fibration. There are many kinds of such fibrations, including ones with sections.

The most basic problem is that $f$ can have some singular fibres. (For example if $X$ is an elliptic $K3$ surface, there will generally be 24 singular fibres; if $X$ is rational and $f$ is minimal, there will be generally be 12.) If $U \subset \mathbf P^1$ is an open set in which there is a point with a singular fibre, then certainly $f^{-1}(U)$ is not isomorphic to a product over $U$, because a singular fibre isn't isomorphic to a smooth one.

So you might try to restrict the question a bit more and assume $f$ is a smooth morphism. That doesn't really help, though. As long as $f$ is not isotrivial, which is true in the examples mentioned above, the fibres over points of $U$ will not all be isomorphic for any open subset $U \subset \mathbf P^1$.

In both cases, it makes no difference whether "open" refers to the Zariski or the etale topology.