i hope you can help me with a problem I discovered while dealing with the resolution of singularities for an algebraic curve in $\mathbb{P}^2$.
A resolution means for me, that for an algebraic curve $C$ you've got a compact Riemann surface $S$ and a proper map $\pi:S \rightarrow C$, such that $\pi|_{S-Sing(C))}$ is a biholomorphism.
Now I want to proof, that $S$ is unique up to biholomorphic equivalence. The only thing I'm not sure in that proof is, that for $x \in Sing(C)$ the preimage $\pi^{-1}(\{x\})$ is discrete and therefore since $\pi$ is a proper map finite.
Am I able to proof this assertion? And if I should be how can I do so?
Well, $\pi^{-1}(x)$ is (Zariski) closed and $S$ is irreducible. Therefore $\pi^{-1}(x)$ is either 0 or 1 dimensional. If it were $1$ dimensional, then it would be equal to $S$.
Another way of showing that $S$ is unique is the following way: If $\pi':S'\to C$ is another non-singular curve that satisfies the same thing, then by taking away all singular points in $C$ and in the preimage of $\pi$ and $\pi'$, we get a birational map $(\pi')^{-1}\circ\pi:S\to S'$. But the place where a birational map between non-singular projective varieties is not defined is of codimension greater than or equal to 2. Therefore, this map is defined everywhere. Since a map between two non-singular curves has a degree that is well defined, and the birational map above is generically 1-1, we have that it is actually an isomorphism.