## Algebraic Geometry - Genus Formula

Statement of the Theorem : Suppose C is an irreducible algebraic curve of degree d with s singular points(all of which are ordinary double points). Suppose that f : C' -> C is its normalization, genus (C') = g. Then
g = (d - 1)(d - 2)/2 - s .

In the proof of this statement given in Griffith's Algebraic Curves book, we form two sets, namely C and E, which represents the points on the curve and the points whose partial derivatives with respect to y is zero. Then the number of intesection points of E and C is calculated by separating the intersection set into two parts: the points with vertical tangents and the singular points of C.Given the assumption that [0,0,1] is not on the curve and that C has no vertical tangents at ordinary double points, for the first part we proceed rather easily but my question is : why are all these intersection numbers are equal to the ramification points of the map
x: pi o f where pi represents the projection from [0,0,1] onto the x-axis? And why are all of them finite ? Thanks!