# Fixed part of a line bundle on a K3 surface

This question comes from Huybrechts' lecture notes on K3 surfaces, more specifically, chapter 2.

Let $$X$$ be a K3 surface (over an algebraically closed field $$k$$) and $$L$$ a line bundle on $$X$$. The base locus of the linear system $$|L|$$ is defined as a closed subscheme of $$X$$ by $$\text{Bs} (L) := \cap_{s \in H^0(X,L)} Z(s)$$ where $$Z(s)$$ is the zero locus of the section $$s$$.

On a surface $$X$$, the base locus may have components of dimension zero and one. Let $$F$$ be the one-dimensional part, called the fixed part of $$L$$.

(1) Why is $$F$$ a divisor on $$X$$?

I fail to see why $$F$$ should necessarily be a divisor, the one dimensional component may have something bad like embedded points. But provided that it is one,

(2) Why is $$h^0(X,F) = 1$$?

Huybrechts' doesn't really give any explanation and uses this 'fact' later to prove that $$F$$ is a sum of smooth rational curves. I would really appreciate any help.

(2) If $$F$$ is the fixed part, it means that every divisor in the linear system can be written as $$D = D' + F.$$ One can also assume that $$F$$ has no common components with $$D'$$.
If $$h^0(X,F) > 1$$ then $$F$$ is linearly equivalent to some $$F'$$, and then the original linear system also contains the divisor $$D' + F'.$$ But $$D' + F'$$ does not contain $$F$$, hence the fixed part of the original linear system is strictly smaller than $$F$$.
• Thank you for the answer. I had a question about the assumption of $D'$ and $F$ not containing a common component. This seems to rely on the following fact: if the field is infinite, then a finite dimensional vector space is never a finite union of proper subspaces. Or is there some other way to see this? – Ominusone Mar 6 at 21:42