When looking at this I found the theorem
Given a surjective map of quasi-projective varieties $\phi:X\rightarrow Y$ and $y\in Y$, we have that dim$(\phi^{-1}(y))=$dim$(X)$-dim$(Y)$.
First I was trying to proof this but I couldn't seem to find a start. Then I tried to look it up online but no author bothered to proof the result.
Can one provide me with a hint how to start the proof, with a reference for the proof or even with the complete proof?
Thank you very much!
I can't give you a hint because the result is false !
For a counterexample take $Y=\mathbb P^2$ and blow-up some point $O\in Y$.
You will obtain a blown-up variety $X$ and a morphism $\phi:X\to Y$.
The fiber $\phi^{-1}(O)$ of isomorphic to $\mathbb P^1$ so that $$\operatorname {dim}(\phi^{-1}(O))=\operatorname {dim}(\mathbb P^1)=1\gt \operatorname {dim}(X)-\operatorname {dim}(Y)=2-2=0$$
The equality you are after is however generically true: there is an open non-empty subset $U\subset Y$ such that for every $y\in U$ one has indeed: dim$(\phi^{-1}(y))=$dim$(X)$-dim$(Y)$.
The (non-trivial) proof can be found in Shafarevich, page 76.