If $X$ and $Y$ are Riemann surfaces (not necessarily compact), and $f:X\to Y$ is a holomorphic function, then it is obvious that the ramification points of $f$ in $X$ form a discrete subset of $X$. Is the same true of the branch points of $f$ (the set made up of the images of the ramification points)?
[Math] Basic question about branch points on Riemann surfaces
riemann-surfaces
Best Answer
Dear Robert, there exists a holomorphic function $X\to Y $ having non discrete and even dense set of branch points, with $X=\mathbb C^\ast \setminus \{0\}$ and $Y=\mathbb C$.
Consider an enumeration $(q_n)$ of $\mathbb Q$ and the polynomials $P_n(z)=q_n + (z-1/n)^2$.
A theorem due to Mittag-Leffler says that there exists a holomorphic function $f:\mathbb C^\ast \setminus \{0\} \to \mathbb C$ whose Taylor development at $1/n$ is $P_n(z)$. The $q_n=f(1/n)$ , that is all of $\mathbb Q$, are then branch points of $f$.
Bibliography and comments The version of Mittag-Leffler used above is not so easy to find in the literature (I just checked). It is proved in Ash-Novinger's Complex Variables ( theorem 6.3.3 ) where they deduce from it some algebraic properties ( due to Helmer) of the ring $\mathcal O(D)$ of holomorphic functions on an open connected $D\subset \mathbb C$. It is a non-noetherian domain, not a UFD but any collection of elements of $\mathcal O(D)$ has a GCD and all its finitely generated ideals are principal.