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Jul 26, 2022 at 14:11 comment added Daniel Loughran @FrancescoPolizzi: The picture to have in mind is a family of varieties over $\mathbb{P}^1$ whose generic fibre is rationally connected, then trying to find a birational modification of the family all of whose singular fibres contain no multiple components.
Jul 26, 2022 at 13:15 comment added Jason Starr My intuition is that this is false, but I do not have a counterexample at hand. The result by Koll’ar in the case of a Fano fibration only gives a “canonical” irreducible component of multiplicity one (and over $\mathbb{C}$ that already follows from the work of Graber, Harris, Mazur and myself). The work of Hogadi-Xu is similar.
Jul 26, 2022 at 12:50 comment added Laurent Moret-Bailly @FrancescoPolizzi I think $X$ is defined over $\mathbb{C}((t))$, not $\mathbb{C}$.
Jul 26, 2022 at 11:44 comment added Francesco Polizzi But still I do not understand the question. There is always the trivial family $\mathcal{X}=X \times \operatorname{Spec}(R)$, right? Maybe you want some additional condition?
Jul 26, 2022 at 11:11 history edited Daniel Loughran CC BY-SA 4.0
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Jul 26, 2022 at 11:10 comment added Daniel Loughran Yes, added thanks!
Jul 26, 2022 at 11:06 comment added Francesco Polizzi You want the general fibre isomorphic to $X$, I presume.
Jul 26, 2022 at 10:47 history asked Daniel Loughran CC BY-SA 4.0