I really don't understand how to calculate ramification points for a general map between Riemann Surfaces. If anyone has a good explanation of this, would they be prepared to share it? Disclaimer: I'd prefer an explanation that avoids talking about projective space!
I'll illustrate my problem with an example. The notion of degree for a holomorphic map is not well defined for non-compact spaces, such as algebraic curves in $\mathbb{C}^2$. I've had advice from colleagues not to worry about this and to use the notion of degree anyway, because it works in a different setting (I don't know which). In particular consider the algebraic curve defined by
$$p(z,w)=w^3-z(z^2-1)$$
and the first projection map
$$f(z,w)=z$$
In order to find the ramification points of this we know that generically $v_f(x)=1$ and clearly when $z\neq0,\pm 1$ we have $|f^{-1}(z)|=3$ so the 'degree' should be $3$. Thus $z=0,\pm1$ are ramification points with branching order 3. I've had feedback that this is correct. Why did this work?
Now let's look at an extremely similar example. Consider the algebraic curve defined by
$$p(z,w)=w^2-z^3+z^2+z$$
and the second projection map
$$g(z,w)=w$$
Now again we see the 'degree' of $g$ should be $3$. Now $g^{-1}(i)=\{(1,i),(-1,i)\}$. So by the degree argument exactly one of these is a ramification point, of branching order 2. Is this correct? If so, how do I tell which one it is?
Finally in more generality, does this method work for the projection maps of all algebraic curves in $\mathbb{C}^2$? Sorry for the long exposition!
Edit: Here's an idea I just had. If our map $f$ is proper then we don't need $X$ to be compact for $\deg(f)$ to be well defined. Now the projection map is clearly proper (I think) so that's why this works. Am I right? This of course raises the natural question - 'what standard maps are proper'? I guess I should ask this in a separate question though!