I need some help, to find the strategy for solving the following problem: Given an analytic surface $S^0$, its compacitfication $S$ and a horizontal Divisor $D^0$ on $S^0$, I have to continue $D^0$ to a Divisor $D$ on $S$. The problem is, that to continue $D^0$ as a smooth Curve in $S$ by its Limes can only be half of the answer. $D$ should have some vertical components lying over the points of $S\setminus S^0$. But I don't no, how to find them.
Being more precise about the situation: I have an atlas $(U_\alpha)_{\alpha \in I}$ for $S^0$ and holomorphic functions $g_{\alpha\beta} : S^0 \to \mathbb C\setminus\{ 0\}$, which fulfill the properties: $g_{\alpha\beta} \cdot g_{\beta\alpha} = 1$ and $g_{\alpha\beta} \cdot g_{\beta\gamma} \cdot g_{\gamma\alpha} = 1$. Therefore these functions define a line-bundle on $S^0$, having the $g_{\alpha\beta}$ as transition-functions vor the charts $U_\alpha$ and $U_\beta$.
I also have maps $s_\alpha : U_\alpha \to \mathbb C$, which fulfill: $s_\beta = g_{\alpha\beta} \cdot s_\beta$. Hence I get a Divisor $D^0$ on $S^0$ by taking
$D^0:= \sum_C ord_C ( {s_{\alpha}}_{|U_\alpha\cap C} ) \cdot C $ where $C$ is running over all irreducible Curves in $S^0$.
For continuing $D^0$, I thought about looking at the limites of the functions $s_\alpha$ in the points of $S\setminus S^0$. But my transitionfunctions $g_{\alpha\beta}$ do have poles in this points. Therefore the order of $s_\alpha$ is different to the one of $s_\beta$.
So how do I compute the vertical components then?
Thanks for any help. Greyfox
