Let $\pi: \mathcal{X} \to \mathbb{P}^1$ be a pencil of projective $\mathbb{C}$-varieties such that a general fiber is smooth. Let $\mathbf{P}$ be one of the properties: rational, unirational, stably rational. My questions is: is there a property $\mathbf{P}$ such that if a general fiber of $\pi$ satisfies $\mathbf{P}$ then so does $\mathcal{X}$? Can we say anything if we add an extra assumption that a general fiber of $\pi$ is isomorphic to a fixed smooth projective variety $F$ (i.e., $\pi$ is an isotrivial family over an open subset of $\mathbb{P}^1$ with each fiber isomorphic to $F$), satisfying property $\mathbf{P}$?
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2$\begingroup$ According to a conjecture of Gino Fano, the total space of a sufficiently general conic bundle over a rational surface, say $\mathbb{P}^1\times\mathbb{P}^1$, is not unirational. This would give a counterexample to all three of your properties via projection to one of the two $\mathbb{P}^1$ factors. $\endgroup$– Jason StarrCommented Jun 2, 2021 at 16:13
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$\begingroup$ It is ok when $\mathcal{X}$ is a surface and $\pi$ has connected fibers. In this case you do not need the general fiber to be smooth, irreducible is enough. $\endgroup$– PuzzledCommented Jun 3, 2021 at 10:34
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For $\mathbf{P}=$ rational, the answer is no, even under your stronger assumption: take the Fermat cubic threefold $X_0^3+\ldots+X_4^3=0$, and consider the pencil of hyperplanes $X_1=\lambda X_0$. After blowing up the base locus you get a morphism $\mathscr{X}\rightarrow \mathbb{P}^1$ whose smooth fibers are isomorphic to the Fermat cubic surface, hence are rational, but $\mathscr{X}$ is not.
