3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This happens already for specializations of Hirzebruch surfaces where the linear system on the generic fiber has degree $1$ on a fiber and has degree $0$ on the directrix (so that the linear system on the special fiber has negative degree on the directrix). $\endgroup$ Mar 30 at 14:38

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.