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?

$\begingroup$ If you relax "divisor", you can take $D$ to be a union of two planes in $\mathbb P^4$. $\endgroup$ – Allen Knutson Jan 4 '15 at 23:10

$\begingroup$ I don't see the complete argument, but I'd like to say that since $X$ is irreducible and $D$ moves, that its deformations cover $X$, and hence the linear system has no basepoints. Then use Bertini on the image of the map to projective space, to show that the general deformation of $D$ is reducible, absent the case $\dim X = 1$. $\endgroup$ – Allen Knutson Jan 4 '15 at 23:12

$\begingroup$ I'm not sure I understand. What if $X$ is $\mathbb P^3$ blown up at two points, and $D$ the strict transform of a plane through the points? Its deformations cover $X$, but there's a curve in the base locus. $\endgroup$ – user47305 Jan 4 '15 at 23:21

$\begingroup$ Oh indeed. The thing I was missing was that $D$ moves, but its deformations all intersect in that curve. (Not sure why you blew up two points instead of one.) $\endgroup$ – Allen Knutson Jan 5 '15 at 4:12

$\begingroup$ If you only blow up one, it's linearly equivalent to any other plane through the point, which may or may not contain the line... $\endgroup$ – user47305 Jan 5 '15 at 5:00
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 blowup of $\mathbb{P}^1\times \mathbb{P}^1$ at one point, and let $D$ be the union of the exceptional divisor and the pullback 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.

$\begingroup$ I don't think 2) is movable, though. $\endgroup$ – user47305 Jan 4 '15 at 22:26


1$\begingroup$ I think the notation isn't entirely consistent, but usually $\text{Mov}(X)$ is the cone ($\mathbb R$)spanned by divisors whose base locus has codimension 2. In the example the exceptional divisor is fixed. (Do we call a divisor "movable" if some multiple has this property? I don't think everyone agrees.) $\endgroup$ – user47305 Jan 4 '15 at 23:58

$\begingroup$ Yes, it seems to be the right definition. In my mind, the divisor $D$ was moving but one part was not. But the correct definition is certainly the one that you give. So we still do not have an example (if it exists). $\endgroup$ – Jérémy Blanc Jan 5 '15 at 0:02
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 basepointfree 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 basepointfree linear system with the required property.
For a basepointfree system, first notice that if the Kodairadimension of the linear system is at least $2$, then a general member of the very ample linear system whose pullback 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 $1$Kodairadimensional linear system. (By Kodairadimension 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 basepointfree, 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 pullback of $L$ on $X$. Now, any member of $L$ which is not supported at a single point pullsback to a reducible divisor. So do those supported at the $P_i$'s because of the blowup. Finally, since the fibers of the morphism $X\to Y$ are connected, all the members of $\mathfrak d$ are actual pullbacks (just look at $h^0$ of the corresponding line bundles), so we have accounted for all the members.

1$\begingroup$ Your answer seems right to me  nice! Just a couple remarks on "movable divisor"  I think the notion of having small base locus at least deserves a name. For example, if X is CalabiYau (or MDS), then Mov(X) is (conjecturally) the union of the strict transforms of the nef cones of all the small modifications. It's true, as you say, that this isn't closed under pullbacks, but it is closed under pushforward via any map that extract no divisors (for example, an arbitrary run of the MMP, where we might want our $\Delta$ to lie in this class). $\endgroup$ – user47305 Jan 5 '15 at 4:59

2$\begingroup$ The notation of $\text{Mov}(X)$ for a cone of divisors, while perhaps unfortunate, is not so uncommon. Off the top of my head, you can find it in the HuKeel MDS paper, Matsuki's intro MMP book, or several papers on the KawamataMorrison cone conjecture (which is in part a conjecture on the structure of the movable cone of divisors on a CY). $\endgroup$ – user47305 Jan 5 '15 at 5:00

$\begingroup$ Mark, thanks for the explanation. This makes sense. I had actually looked at Matsuki's book. He defines the movable cone indeed (which I forgot by the time I wrote the answer), but I couldn't find his definition of a movable divisor... Anyway, what you are saying makes perfect sense and indeed, for CYs, or more generally for flips and flops this notion is invariant. I'm convinced. Thanks. $\endgroup$ – Sándor Kovács Jan 5 '15 at 5:14

$\begingroup$ Mark, you are absolutely correct. Kawamata defines a movable divisor as you suggested. Unfortunately, it still seems to be different from the definition of Lazarsfeld (and others) of a movable curve. So on a surface a curve could be movable as a divisor, but not as a curve. :( $\endgroup$ – Sándor Kovács Jan 5 '15 at 5:26
Let $X = \mathbb P^1 \times \mathbb P^1$ and let $D$ be the sum of two elements of the same ruling.


$\begingroup$ I think you beat me to it! I posted a more complicated example first and then replaced it before reloading. (so please accept his answer!) $\endgroup$ – user47305 Jan 4 '15 at 22:07

$\begingroup$ I think this is not quite what was asked (or I misunderstood it); otherwise just a curve would do for $X$. Here, the linear system contains curves that are irreducible, although not reduced. $\endgroup$ – Alex Degtyarev Jan 4 '15 at 22:07

$\begingroup$ Good point. In this case, I do not know if such example really exists. $\endgroup$ – Jérémy Blanc Jan 4 '15 at 22:12

$\begingroup$ Hmm, me neither. $\endgroup$ – user47305 Jan 4 '15 at 22:13