Geometry of contracted divisors

Question

Let $f:\mathbb{P}^3\dashrightarrow\mathbb{P}^2$ be a dominant rational map defined over a field $k$ (not necessarily algebraically closed) of characteristic zero.

Consider a resolution $\widetilde{f}:X\rightarrow\mathbb{P}^2$ of $f$ and assume that a general fiber of $\widetilde{f}$ is the strict transform of a conic in $\mathbb{P}^3$ while $S = \widetilde{f}^{-1}([1:0:0])$ is a surface.

Under these hypotheses can we say something on the birational type of $S$? For instance assuming that $S$ has a point defined over the base field $k$ how far can $S$ be from being rational over $k$?

Answer 1

The surface $S$ is uniruled. Indeed, considering $X$ as a family of conics in $\mathbb{P}^3$ parameterized by an open subset of $\mathbb{P}^2$, we obtain a rational map $$ \phi \colon \mathbb{P}^2 \dashrightarrow \mathrm{Hilb}(\mathbb{P}^3) $$ to the Hilbert scheme of conics in $\mathbb{P}^3$. Resolving this map by an appropriate blowup $\pi \colon Y \to \mathbb{P}^2$, we obtain a morphism $$ \tilde\phi \colon Y \to \mathrm{Hilb}(\mathbb{P}^3). $$ Moreover, if $Z \subset Y \times \mathbb{P}^3$ is the corresponding family of conics, the projection $\pi$ induces a surjective morphism $Z \to X$, such that $$ Z_0 := Z \times_Y \pi^{-1}(0) $$ dominates $S = X \times_{\mathbb{P}^2} \{0\}$. So, it remains to note that $Z_0 \to \pi^{-1}(0)$ is a family of conics.