# What is ramification for z^(p/q)(1-z)^(r/s)?

1. Nov 19, 2013

### jackmell

May I ask what is the ramified branching geometry of the algebraic function:

$$w=z^{p/q}(1-z)^{r/s},\quad (p,q,r,s)\in \mathbb{Z}\backslash\{0\}$$

and is it computable in terms of the parameters p,q,r,s? The reason I ask is that it appears to be trivially predictable and I just want to know if that is indeed the case.

Ok thanks,
Jack

Last edited: Nov 19, 2013
2. Nov 23, 2013

### jackmell

Hi guys,

After studying some examples and their associated Newton polygons , I've reached the tentative conclusion that the ramified geometry is indeed simply predictable:

Given the algebraic function:

$$w=z^{r/s}(1-z)^{p/q},\quad (p,q,r,s)\in\mathbb{Z}\backslash\{0\}$$

and letting $d=LCD(s,q)$, then the ramified geometry about the origin is simply $d/s$ s-cycles, and $d/q$ q-cycles over $z=1$.

Take the simple example: $w=z^{5/2}(1-z)^{4/3}$. The lowest common denominator is of course 6. And when we expand the function about its singular points we obtain three 2-cycles about the origin and two 3-cycles about the point z=1.

I'm pretty sure that's the case for rational exponents in general. Can anyone confirm this for me?

Ok thanks,
Jack