This is exercise 7.21 from Fultons "Algebraic Curves".
Let $X$ be a nonsingular projectiv curve, $P \in X$. Show that there is a projective plane curve $C$ with only orinary multiple points and a birational morphism $f:X \rightarrow C$ such that $f(P)$ is simple on $C$.
I do have a birational morphism from $X$ to a proj. plane curve with only ordinary multiple points, but i dont know how to make $f(P)$ simple. Fulton gives the hint to do quadratic transformation centered at $f(P)$ but i dont understand how this leads to $f(P)$ becoming simple.
I don't even really understand where the point $f(P)$ is after the transformation since "centered at $f(P)$" means that after a coordinate transformation $f(P)$ is the point $(0,0,1)$ on which the quadratic transformation is not defined.
I would really appreciate some help with this.
Ok, after reading some additional stuff it seems that the point $f(P)$ is "replaced" by points $P_1,\ldots,P_r$ on the transformed curve with multiplicities $s_1,\ldots,s_r$ and $\sum s_i \leq r$ if $r$ is the multiplicity of $f(P)$. And in "Lectures on Curves,Surfaces and Projective Varieties, A Classical View on Algebraic Geometry" they say that this way the multiplicities can be reduced to $1$ by successive quadratic transformation. But I don't understand why there cant just be one point with multiplicity $r$.
