Degeneration of curves in smooth families

Heuristically, I want to know, given a smooth, projective morphism from a scheme to a discrete valuation ring, if the generic fiber can be 'covered' by a family of geometrically integral curves, is it true that the specialization of a general member of this family is also geometrically integral. More precisely,

Let $$\pi: \mathcal{X} \to \mathrm{Spec}(R)$$ be a smooth, projective morphism of relative dimension at least $$2$$ and $$R$$ is a discrete valuation ring with residue field isomorphic to $$\mathbb{C}$$. Let $$B$$ be an irreducible variety mapping surjectively to $$\mathrm{Spec}(R)$$. Consider a family of curves $$\mathcal{C} \subset \mathcal{X} \times_R B$$ parameterized by $$B$$ i.e., $$\mathcal{C}$$ is a closed subscheme in $$\mathcal{X} \times_R B$$ and the natural morphism from $$\mathcal{C}$$ to $$B$$ is flat and projective. Suppose that the natural morphism from $$\mathcal{C}$$ to $$\mathcal{X}$$ is surjective.

Denote by $$K$$ the fraction field of $$R$$ and $$\mathcal{C}_K:= \mathcal{C} \times_R \mathrm{Spec}(K)$$ the family of curves parameterized by the generic fiber $$B \times_R \mathrm{Spec}(K)$$. Similarly, denote by $$\mathcal{C}_k:= \mathcal{C} \times_R \mathrm{Spec}(k)$$ the family of curves parameterized by the special fiber $$B \times_R \mathrm{Spec}(k)$$. Note that, there are natural flat, projective morphisms from $$\mathcal{C}_K$$ (resp. $$\mathcal{C}_k$$) to $$B_K$$ (resp. $$B_k$$).

Is it true that if a general fiber of the natural morphism from $$\mathcal{C}_K$$ to $$B_K$$ is geometrically integral, then a general fiber for the morphism from $$\mathcal{C}_k$$ to $$B_k$$ is also geometrically integral?

Any hint/reference will be most welcome.

• Here is a counterexample. Let $\mathcal{X}$ be a family of Hirzebruch surfaces whose generic fiber is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ specializing to a Hirzebruch surface that is a minimal (crepant) resolution of a singuar quadric cone in $\mathbb{P}^3$. Consider the curve class whose specialization is the sum of a "fiber" curve class and the "directrix" curve class. Then $B$ is just $\mathbb{P}^1_R$. The curves on the generic fiber are integral (smooth, rational curves), but on the special fiber they are all reducible. Commented Jun 23, 2023 at 20:35

I am just writing my comments as one answer. Without further hypotheses, there are counterexamples. Even without a specific example of $$\mathcal{X}$$, there are plenty of examples of a $$K$$-scheme $$B_K$$ and a family of smooth, projective, geometrically connected relative curves $$\mathcal{C}_K\to B_K$$ such that for every fppf $$R$$-scheme $$B_R$$ that has $$B_K$$ as its generic fiber, for every flat, proper relative curve $$\mathcal{C}_R\to B_R$$ that extends the $$K$$-family, the geometric fibers over $$B_k$$ are not integral.
Indeed, for every proper, flat $$R$$-scheme $$B'_R$$ whose geometric fibers are integral, for every proper, flat morphism $$\mathcal{C}'_R\to B'_R$$ whose fibers are at-worst-nodal, connected curves with ample dualizing sheaf, for every "modification" of the family over a proper, flat $$R$$-scheme $$B_R$$ whose geometric fibers are integral, $$\mathcal{C}_R \to B_R$$, the "stabilization" of the geometric generic fiber of $$\mathcal{C}_k\to B_k$$ equals the geometric generic fiber of $$\mathcal{C}'_k\to B'_k$$: this is part of the uniqueness in "stable reduction". So if the geometric generic fiber of $$\mathcal{C}'_k\to B'_k$$ is reducible, the same is true for the geometric generic fiber of $$\mathcal{C}_k\to B_k$$.
Thus, to get a positive answer, you need to add the hypothesis that for the family $$\mathcal{C}_K\to B_K$$ whose geometric fibers are assumed to be integral, there exists an extension $$\mathcal{C}_R\to B_R$$ over an fppf $$R$$-scheme $$B_R$$ whose geometric generic fiber over $$B_k$$ is integral (just as a proper, flat family of abstract curves, with no morphism to $$\mathcal{X}_R$$ specified).
Even with this hypothesis, there are still counterexamples, e.g., the example in my comment where $$\mathcal{X}_K$$ is a Hirzebruch surface $$\mathbb{P}^1\times \mathbb{P}^1$$, yet $$\mathcal{X}_k$$ is a different Hirzebruch surface, e.g., the minimal (crepant) resolution of a singular quadric cone in $$\mathbb{P}^3$$. The usual way to deal with this is to change the question slightly: does there exist an extension $$\mathcal{C}_R \to B_R\times_{\text{Spec}(R)} \mathcal{X}_R$$ and an irreducible component $$\mathcal{C}_{k,i}$$ of $$\mathcal{C}_k$$ satisfying all of your conditions. If you begin with a $$K$$-family of curves $$\mathcal{C}_K\to B_K\times_{\text{Spec}(K)} \mathcal{X}_K$$ that is constructed in a "sufficiently general" way, e.g., the family of all complete intersection curves of sufficiently very ample divisors in given linear equivalence classes on $$\mathcal{X}_K$$, then this new question has a positive answer if and only if there exists an irreducible component $$\mathcal{X}_{k,i}$$ of $$\mathcal{X}_k$$ that is geometrically irreducible.