A question on the map associated to a divisor on an algebraic surface. Let $S$ be a K3 surface and $E\subset S$ be a genus 1 smooth curve. By Riemann-Roch, $h^0(S,\mathcal{O}(E))=2$ and hence there is a map $\phi_{E}:S\rightarrow \mathbb{P}^1$. How do we know that this is an elliptic fibration with $E$ as a fiber?
More generally, is it true that for any algebraic surface $T$ and any smooth curve $C\subset T$ with $h^0(T,\mathcal{O}(C))=2$, the map $\phi_{C}:T\rightarrow \mathbb{P}^1$ gives a fibration with $C$ as a fiber? 
Thank you very much. 
 A: To answer the first question, one should first note that not only does $O(E)$ have 2 linearly independent sections, but also that their zero sets are disjoint. This follows from the adjunction formula, relating $K_S$, $K_E$ and $E^2$. So $O(E)$ is basepoint-free. (Without this, we only know that $\phi_E$ is a rational map, not a morphism.)
Now the conclusion follows almost from the definition of $\phi_E$: it is a morphism such that the sections of $O_{\mathbf{P}_1}(1)$ pull back to give the sections of $O(E)$. In particular the section $s_E$ corresponding to $E$ itself pulls back from a section $t$ of $O_{\mathbf{P}^1}(1)$. But now considering zero sets of $s_E$ and $t$, this says that $E$ is the preimage of a point in $\mathbf{P}^1$, i.e. a fibre.
The same argument works in general, as you want, IF you know that $O(C)$ is basepoint-free. (On the other hand, it could happen that $O(C)$ has 2 sections, but their zero sets intersect, so that $\phi_C$ is just a rational map.)
Edit: Here's an example that shows that my dire warning in the last paragraph is necessary. Take two cubic curves $C_1$, $C_2$ in the plane intersecting in 9 distinct points. Blow up 8 of the points to get a surface $X$, and let $E$ be the proper transform on $X$ of $C_1$ say. Then $h^0(X,O(E))=2$, but since any cubic through 8 of the basepoints must also pass through the ninth, the line bundle $O(E)$ on $X$ still has 1 basepoint.
