I would like to understand what exactly is the motivation for defining the notion of a ramification divisor of a function.

As I see the definition,

If $f$ is a meromrophic function between two Riemann surfaces - say $X$ and $X'$ then let $\nu_p(f)$ be the ramification (or order) of the function $f$ at $p$. Basically if one is working in local coordinates such that $z(p)=0$ then $f$ in a neighbourhood of $p$ looks like $f=z^{\nu_p}h(z)$ where $h(z)$ is a holomorphic function which is never $0$ in a neighbourhood of $p$.

In the above definition of ramification, can the function $h$ be always set to unity? By choosing coordinate in$X'$ such that $f(p)=0$? (...I am not sure..)

Does anything in the above definition depend on $X$ or $X'$ being compact?

Now for a similar map $f$ one defines its ramification divisor ($R_f$) as $R_f = \sum _{p \in X} (\nu_f(p) - 1)p$

Its not clear to me whether people define ramification divisors for meromorphic functions too since i almost seem to see the texts exclusively using it in the case of non-constant holomorphic functions. I would be glad if someone can clarify this...may be I am missing something very basic.

Also this definition almost exclusively seems to be used when $X$ and $X'$ are compact Riemann surfaces. Is that somehow necessary?

{I guess in all this discussion one has to keep in mind that a holomorphic function on a Riemann surface and a holomorphic function between two Riemann surfaces are defined "differently" - as i see it. I guess there is no analogue of Liouville's theorem in the later case.}

Why that "-1" in the definition? Is $\nu_p(f)$ always greater than $1$ ?

Let $q \in X'$ and let $p_1$ be a pre-image of $q$ under $f$ with multiplicity of $m_1$. Then I guess one will say that $\nu_f(p_1) = m_1$. Now is it obvious that any "small" perturbation of $q$ can only "split" $p_1$ into $m_1$ points each with $\nu_f = 1$? That nothing else can happen? For "large" enough perturbation to $q$ isn't it possible for many of its pre-images to "join up" and have larger ramifications than initially?

consider this set, $p \in X' \vert f^{-1} (p)$ has all points with $\nu_f(p)=1$ (called "simple points"?). Is this set open and dense in $X'$?

Finally a curiosity - is there a "simple" way to see the Riemann-Hurwitz formula

*without*using the Poincare-Hopf formula?

usuallyis just 1. Only at $p$ where there is ramification do you have $\nu_p(f) > 1$, so in the definition of $R_f$ you get a finite sum. If the coefficient was $\nu_f(p)$ then the formal sum would have uncountably many nonzero terms! $\endgroup$ – KConrad Oct 5 '11 at 2:32