Let $X$ be a K3 surface and $D$ an effective divisor such that $h^0(D)\geq2$ and $h^1(D)=0$. 


Is this enough to show that $D$ is connected? 


Any reference would also be appreciated (I looked in Saint-Donat' thesis but did not get the answer)



Sufficient conditions for a divisor to be connected on a K3 surface
I think the answer is yes. Here is a sketch proof.


Keep in mind that by Serre duality that $h^2(D)=0$. 


Suppose that the divisor is not connected. Then there are two possibilities:


1) There is an isolated exceptional subdivsor in $D$ (by this I mean a divisor $E$ that is a connected collection of $-2$ curves with $E^2<0$). It is clear that if you remove this divisor from $D$ then $h^0(D)$ will not change. At the same time $D^2$ will increase. From Riemann Roch it would follow that $\chi (D)$ increase as well. But since $h^0$ and $h^2$
don't change, this means that $h^1$ decreased. Hence it could not be zero in the beginning.


2) $D$ is a union of $\ge 2$ fibres of an elliptic fibration. In this case $\chi(D)=2$,
at the same time if $D$ is not connected then $h^0(D)\ge 3$. This gives by Riemann-Roch that $h^1(D)\ge 1$.


