I am looking for examples of invertible sheaves in smooth, projective families such that the associated base locus (i.e., the intersection of all the effective divisors in the complete linear system) jumps. More precisely, take a discrete valuation ring $R$ and $\pi: X \to \mathrm{Spec}(R)$ be a smooth, projective morphism of relative dimension at least $2$. Denote by $X_K$ (resp. $X_k$) the generic (resp. special) fiber of $\pi$. I am looking for examples of invertible sheaves $L$ on $X$ such that the base locus $B_K$ of $L|_{X_K}$ over the generic fiber satisfies the property: for the closure $\overline{B}_K$ of $B_K$ in $X$ we have $\overline{B}_K \cap X_k$ does not contain the base locus of the invertible sheaf $L|_{X_k}$ on the special fiber. If necessary assume that the fibers of $\pi$ are geometrically irreducible and the underlying field is $\mathbb{C}$. Any reference / idea will be appreciated.
Algebraic Geometry – Examples of Jumping Base Locus of Complete Linear Systems
ag.algebraic-geometrydeformation-theoryinvertible-sheaves
Related Solutions
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.
Best Answer
Take $X$ to be $\mathbb P^2$ blown up at three $R$-points which are colinear on the special fiber and not on the generic, and take $L$ to be $\mathcal O(2)$ minus the three exceptional divisors. The line containing the three points will be the base locus of the special divisor, because its intersection number with $L$ is $-1$, but there are no base points on the generic fiber, as one can check using the three sections whose vanishing loci are the strict transforms of two lines each through two of the points.