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Let $X$ be a smooth projective rationally connected variety over $\mathbb{C}((t))$ and $R= \mathbb{C}[[t]].$

Does there exist a proper regular scheme $\mathcal{X} \to \mathrm{Spec}(R)$ whose special fibre is reduced and whose generic fibre is isomorphic to $X$? I.e. the special fibre is a sum of divisors on $\mathcal{X}$, and I want there to be no multiple components appearing.

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  • $\begingroup$ You want the general fibre isomorphic to $X$, I presume. $\endgroup$
    – Francesco Polizzi
    Commented Jul 26, 2022 at 11:06
  • $\begingroup$ Yes, added thanks! $\endgroup$
    – Daniel Loughran
    Commented Jul 26, 2022 at 11:10
  • $\begingroup$ 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? $\endgroup$
    – Francesco Polizzi
    Commented Jul 26, 2022 at 11:44
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    $\begingroup$ @FrancescoPolizzi I think $X$ is defined over $\mathbb{C}((t))$, not $\mathbb{C}$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Jul 26, 2022 at 12:50
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    $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented Jul 26, 2022 at 13:15

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