Suppose $f:X\rightarrow \mathbb{P}\_k^n$ is a proper smooth morphism, where $k$ is an algebraically closed field. If $f$ admits a rational section, can we prove that the fiber of $f$ above any codimension 1 point contains a geometrically integral open subscheme?



To prove the fiber above a codimension 1 point contains a geometrically integral open subscheme
This is true:


Because $f$ is smooth, any fibre $X\_y$ is smooth. If $y$ has codimension $1$, then the rational section $\sigma \colon \mathbb P^n\_k \dashrightarrow X$ is defined at $y$ by the valuative criterion of properness. Thus, $X\_y$ has a rational point $\sigma(y)$. If $U \subseteq X\_y$ is the connected component of $\sigma(y)$, then $U$ is a smooth connected $\kappa(y)$-scheme with a rational point, hence geometrically connected by [Tag 04KV](https://stacks.math.columbia.edu/tag/04KV).


