Branching points for algebraic curves

I read a book on algebraic curves and Riemannian geometry (Prasolov, V. V., Schwarzman O.V), however, this in this book not all statements are rigorous.

To find the genus of an algebraic curve $C$ in $CP^2$ one considers some projection like: $$p: CP^2 \backslash [0:0:1] \rightarrow CP^1: [x:y:z] \mapsto [x:y].$$ Reduction of this projection on curve $C$ gives the cover map, which can have branching points.

The branching point is defined as follows: the inverse of $[x_0:y_0]$ under the cover map is a line in $CP^2$ - $[x_0:y_0:t], \, t\in \mathbb{C}$. Generally, this line has $n$ different intersections with $C$, where $n$ is the degree of the curve. If it has less different than $n$ intersections (i.e. there are multiplicities of the roots) then point $[x_0:y_0]$ is called a branching point.

Then there is a nice theorem of Riemann-Hurwitz, which gives the Euler number (and hence genus) of the curve.

1) Maybe my question is not correct, but using this definition it is not guaranteed that the number of branching points is finite. Is it finite?

p.s. The problem is that solving the equation: $$F(x_0,y_0,t) = 0,$$ where $F$ is an equation for the curve, I can obtain some mulitiplicities in $t$, but it does not imply that a curve has a singularity (which are finite by Bezout's theorem) at this point. It can be because of the direction of the line $[x_0:y_0:t]$.

2) Can one, for example, then turn the coordinate system to avoid such situations?

1) Yes it is : a point $b \in \Bbb CP^1$ is a branching point if there is a point $c \in C$ with $p(c) = b$ and $(p_{|C})'(c) = 0$. The set of such points is finite because it is a proper closed set of $C$, so in particular its image by $p$ is also finite but this is exactly the branch locus.
2) : No, $\Bbb CP^1$ is simply connected and unless $C$ is a line or a conic it will not be simply connected, so there is no unramified covering of $\Bbb CP^1$ by a curve of degree $\geq 3$.
• @FedorGoncharov 1) : it is not always true sorry, but it is true if $C$ is irreducible (notice that the closure is taken in Zariski topology). It's because if $x_0 \subset X_0 \subset C$ is a chain of closed irreducible set, then we should have $X_0 = C$ or $x_0 = X_0$. This shows that $X_0$ is a point. So in general, a proper closed set is a finite union of point. 2) It is not true that if $x$ is a ramified point then it is singular. For example, you can realize any torus as a ramified covering of $\Bbb CP^1$, branched at $4$ points. – Nicolas Hemelsoet Nov 1 '17 at 10:29