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.



Fixed part of a line bundle on a K3 surface
(1) I guess, as a scheme, the base locus might have embedded points. But the fixed part is defined as the pure 1-dimensional part of the base locus scheme.


(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 $F$.


