Say I have an algebraic surface $S$ (smooth, projective over $\mathbb{C}$) and a morphism $f: S \to B$ to some curve $B$ (smooth, projective over $\mathbb{C}$) defined by the complete linear system corresponding to some divisor $D$ on $S$. We assume this linear system to be base point free.
Is it true that $D$ will be a sum of fibres of $f$ and if so, why?
In fact I'm just trying to understand the proof of Proposition IX.2.(b) in Beauville's book on algebraic surfaces. The proof goes as follows:
(Here $S$ is a smooth projective surface over $\mathbb{C}$ with $\kappa(S) = 1$. Moreover, $M$ is some effective divisor such that $|M|$ is base point free - corresponding to the mobile part of $|nK_S|$ for some $n$ such that the $n$-th plurigenus is at least $2$ - and $K \cdot M = 0$ where $K$ is the canonical divisor.)
"(...) It follows that $|M|$ defines a morphism $f$ from $S$ to $\mathbb{P}^N$ whose image is a curve $C$. Consider the Stein factorization $f: S \to B \to C \subseteq \mathbb{P}^N$, where $p: S \to B$ has connected fibres. Let $F$ be a fibre of $p$. Since $M$ is a sum of fibres of $p$ and $K \cdot M = 0$, we must have $K \cdot F = 0$. (...)"
There are three statements which are not clear to me:
*Why should $M$ be a sum of fibres of $p$?
*Why does $K \cdot M = 0$ imply that $K \cdot F = 0$ for ANY fibre $F$?
*Why on earth should the curve $B$ coming out of this construction be smooth? I mean, all we did is starting with a smooth, projective surface $S$ with $\kappa(S) = 1$, taking the mobile part of the linear system $|nK_S|$ for some $n$ such that the $n$-th plurigenus is $\geq 2$ (which is base point free), giving a curve $C$ which is the image of the corresponding pluricanonical map, and then taking the curve $B$ coming out of the Stein factorization. I see no reason why $B$ should be smooth, I'm really lost here.
On a related note, in the proof of Proposition IX.3:
"(...) Let $S$ be a minimal elliptic surface. Since $K_S \cdot F_b = 0$ for all $b \in B$, the maps contract the fibres $F_b$. It follows that the image of $\varphi_{nK}$ has dimension $\leq 1$. (...)"
Why is this true?