# Linear system corresponding to rational curves on a K3 surface

Assume we have a $K3$-surface $S$ over $\mathbb{C}$ which contains irreducible and smooth curves $C,D$ satisfying the properties: $C^2=D^2=-2$ and $C\cdot D=1$, i.e. two rational curves.

Now i'm intersted in the linear system $|C+D|$.

Since $(C+D)^2=-2$ we have $H^0(O_S(C+D))=1$, so the linear system has dimension 0. In spite of that, can anything interesting be said about this linear system? Especially:

What can be said about the base locus? Is $Bs(|C+D|)=C+D$? Does this linear system only consist of its fixed part? What about base points? (Maybe i am misunderstanding the notaion: are points in the fixed part also base points?)

Is $C+D$ another rational curve on $S$? It has the correct self-intersection, but is it smooth and irreducible?

Background: I'm trying to find points $p$ on $S$ which give a locally free extension $E$:

$0\rightarrow O_S \rightarrow E \rightarrow I_p\otimes O_S(C+D)\rightarrow 0$, here $I_p$ is the ideal sheaf of $p$.

And base points of the linear system $|C+D|$ are solutions, because they satisfy the Cayley-Bacharach property.

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If H^0(S,C+D)=1, there is a unique effective divisor in the linear system |C+D|, namely the reducible curve C union D. So yes, the base locus is C union D, and no, there cannot be a smooth curve in this linear system. –  user5117 Apr 30 '12 at 18:22

A base-point free linear system has non-negative self-intersection, so at least one of the two components of $C+D$ must lie in the base locus of $|C+D|$. But if one component were moving it would have non-negative self-intersection, so yes, the linear system only consists of its fixed part (in particular $h^0(C+D) =1$, as you said).
Therefore the only divisor in $|C+D|$ is the union of two rational smooth curves intersecting transversally in one point (in particular it is neither smooth nor irreducible).