Movable Divisors

Question

Let $X$ be a projective variety. Does anyone know an example of a movable reducible divisor $D\in Mov(X)$ such that any element in the linear system $|D|$ of $D$ is reducible?

Answer 1

1) Take $X=\mathbb{P}^1\times \mathbb{P}^1$, which satisfies $\mathrm{Pic}(X)=\mathbb{Z}f_1+\mathbb{Z} f_2$, where $f_1,f_2$ are fibres of the two projections. Choose $D=2f_1$. Then $\lvert D\rvert$ is movable but any element corresponds to the union of two fibres (since $D\cdot f_1=0$, it is contained in fibres).

EDIT: I do not know if this example is what you want, as some of the members are not reduced but irreducible (as pointed by Alex Degtyarev).

2) Another example: Take $X$ to be the blow-up of $\mathbb{P}^1\times \mathbb{P}^1$ at one point, and let $D$ be the union of the exceptional divisor and the pull-back of one fibre. Any member of $\lvert D\rvert$ is the union of the exceptional divisor and a smooth fibre, or the union of the singular fibre and the exceptional divisor. In any case it is reducible.

Answer 2

Just so I can use this later, let me start by saying that obviously $X$ has to be irreducible for this to be interesting.

EDIT: In an earlier version of this answer I ruminated on what a movable divisor might mean as it seems to conflict with the definition of a movable curve. Following Mark's advice in the comments (thanks!) I unearthed a definition in a paper of Kawamata which agrees with what Mark is suggesting in another comment. It still seems an unfortunate overlap between the two notions, but probably we can't help it now. In any case, a basepoint-free divisor should be considered movable by any definition (and it satisfies the one by Mark et al), so I will give an example of a basepoint-free linear system with the required property.

For a basepoint-free system, first notice that if the Kodaira-dimension of the linear system is at least $2$, then a general member of the very ample linear system whose pull-back is our original linear system will be irreducible and hence so will be a general member of our linear system. So the only chance is with a $1$-Kodaira-dimensional linear system. (By Kodaira-dimension I mean the dimension of the image of $X$).

So, finally, here is an example. Allen's suggestion of two points on a curve is going the right direction, but it doesn't work. If the genus is at least $2$, then it will not move, if it is at most $1$, then its linear system contains a double point. I suppose the next idea is to find an example of a complete linear system on a curve which is basepoint-free, but has no member which is supported at a single point. I would expect that such linear systems exist and perhaps one could construct one with the clever use of étale covers or other tricks. However, I think, one can get an example in a cheaper way:

Let $Y$ be a smooth projective curve (say over an algebraically closed field) of some high genus and $L$ a very very ample linear system. No matter what, there will be only finitely many (zero is finite!) points, say $\{P_1,\dots,P_m\}\subset Y$ such that if $L$ contains a member supported on a single point, then that point is one of the $P_i$'s. Next take an arbitrary projective surjective morphism $f:Z\to Y$ with connected fibers and pick points $\{Q_1,\dots,Q_m\}\subset Z$ such that $f(Q_i)=P_i$. Finally, let $X$ be the blow up of $Z$ along $\{Q_1,\dots,Q_m\}$ and consider the linear system $\mathfrak d$ which is the pull-back of $L$ on $X$. Now, any member of $L$ which is not supported at a single point pulls-back to a reducible divisor. So do those supported at the $P_i$'s because of the blow-up. Finally, since the fibers of the morphism $X\to Y$ are connected, all the members of $\mathfrak d$ are actual pull-backs (just look at $h^0$ of the corresponding line bundles), so we have accounted for all the members.

Answer 3

Let $X = \mathbb P^1 \times \mathbb P^1$ and let $D$ be the sum of two elements of the same ruling.