# About the Stein factorisation of a morphism defined by a complete linear system on a surface, whose image is a curve

Let $X$ be an algebraic complex projective surface, and let $D$ be an effective divisor on $X$ with empty base locus. Assume that the morphism $\varphi = \varphi_D : X \to \mathbb P^{h^0(D)-1}$ has image a curve $C$ (a sufficient condition is $D^2=0$). Let \begin{equation} X \to \widetilde{C} \to C \end{equation} be the Stein factorisation of $\varphi : X \to C$. Now, assume that $h^1(\mathcal O_X) = 0$. Then

necessarily $\widetilde{C}$ is isomorphic to $\mathbb P^1$, and $\widetilde{C}=C$ since $\widetilde{C} \to \mathbb P^{h^0(D)-1}$ is defined by a complete linear system.

Could anyone help me explain (in detail!) this argument?

1. Why is $\widetilde C$ isomorphic to $\mathbb P^1$?
2. How do you prove that $\widetilde C = C$, using that $\widetilde{C} \to \mathbb P^{h^0(D)-1}$ is defined by a complete linear system?
3. Why is $\widetilde{C} \to \mathbb P^{h^0(D)-1}$ defined by a complete linear system? What linear system defines that map?

The quotation above is taken from the proof of Theorem 3.8 (b) in Chapters on Algebraic surfaces by M. Reid, page 70. I find it very difficult to understand it.

Thank you in advance.

EDIT: I found some helpful details in Lectures on K3 surfaces by D. Huybrechts, Proposition 3.10, page 31. Nevertheless, the arguments seem to me quite involved. I wonder if there is some more "elementary" approach.

• For 1, $\tilde{C}$ can't have any 1-forms, since they would pull-back to give 1-forms on $X$. For 2, every complete linear system on $\mathbb{P}^1$ is very ample, so defines an embedding. For 3, it's the (unique!) complete linear system on $P^1$ which pulls back to $|D|$. Hope that helps! – user64687 Sep 4 '13 at 14:40
• You can use Leray spectral sequence for (1). – Cantlog Sep 7 '13 at 15:48
• How can you use this spectral sequence? It is mentioned in Huybrechts's Paper, also... – Francesco Genovese Sep 9 '13 at 10:26