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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.

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(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 $F$.

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  • $\begingroup$ 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? $\endgroup$ – Ominusone Mar 6 at 21:42

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