Compute genus via Riemann-Hurwitz and monodromy

[jackmell]

Hi guys,

I new to Algebraic Geometry and was wondering if someone could help me with this problem:

Homework Statement

Given the algebraic curve $w(z)$ represented by $w^5+w^2+z^2=0$, show that the genus is one by employing the Riemann-Hurwitz formula.

Homework Equations

For the mapping $f:X\to S$, the Riemann-Hurwitz formula is given by:
$$2g(X)-2=\text{deg}(f)\left[2g(S)-2\right)+\sum (e_w-1)$$

which I assume the expression $f:X\to S$ as applied to my problem can be written as $w(z):\mathbb{C}\to S$, that is, the function w(z) maps the (five-sheeted) complex plane to the Riemann surface given by $S$ and it is this surface that is topologically equivalent to a Riemann sphere with one handle thus having genus one. If that's not the correct way to state that, could someone correct that for me please?

The Attempt at a Solution

Homework Statement

I believe I understand somewhat well, direct construction of the Riemann surface given by the expression $w^2-p(z)=0$ and the subsequent determination of its genus directly from the union of two Riemann spheres. However I'm not able to deduce that via Riemann-Hurwitz.

But back to the problem above:

I'm taking $g(X)$ to be the genus of the complex plane which is zero.

The degree of $f$ I'm taking to be the highest degree of $f(z,w)$ in $w$ which is five.

The sum is where I'm having the problem. This I believe can be computed from the monodromy group which I assume represents the ramifications around each branch-point which I take to mean the branching around each branch-point including infinity. Here's my analysis of that:

The function has six finite branch points and the surface $S$ ramifies around each one as a two-valued branch and three single-valued branches. I think each monodromy group about these branch-points can be written in terms of $\left\{(n,m)\right\}$ with $n$ being the sheet number and $m$, the branch order as:

$$((1,2),(3,1),(4,1),(5,1))$$

Maybe though that's not a valid way to write that. However, around the branch-point at infinity, the function ramifies into a single five-valued branch so the monodromy group is $((1,5))$

So my (incorrect) analysis of the sum would be $\sum (e_w-1)=(2-1)+(1-1)+(1-1)+(1-1)$ for each finite branch-point and $\sum (e_w-1)=5-1$ for the point at infinity. Plugging all this into the Riemann-Hurwitz formula I obtain:

$$-2=5\left[2g-2\right]+10$$

and solving for g I obtain $g=-1/5$ which is obviously not correct.

I don't know, maybe I'm not even qualified to even ask the question. Still I would like to see how it's solved at the very least if someone could help me with that and as I study more the subject, I'm sure it will come together eventually for me.
Thanks,
Jack

 

[jackmell]

Let $f(z,w)=z^2+w^2+w^5$

then it's much easier to compute it via the formula:

$$p=\sum_{i=1}^M \frac{r_i-1}{2}-n+1$$

with $n$ the degree of the function and $r_i$ the ramifications over each critical point including infinity.

The finite critical points are zeros to $R(f,f_w)=0$ where R is the resultant since the function has no poles. There are seven critical points, one of which is the origin. At the origin, the function splits into five single-cycles. At the other six, it ramifies into a 2-cycle and three single cycles. At infinity, it fully ramifies into a 5-cycle. So then we have:

$$p=\sum_{i=1}^8 \frac{r_i-1}{2}-n+1=1/2(1+1+1+1+1+1+4)-5+1=1$$