I'm given the curve $y^3=x^3-1$ and I want to find the genus. I have a ramified cover $\pi: Y \to \mathbb{P}^1$, where $Y$ is given by the curve. I know that $1, \omega, \omega^2$ are the $x$ values in $\mathbb{C}$ where $\pi$ is totally ramified and they each have ramification index of 3.
Then they consider the Newton polynomial $y^3-x^3+1$. This is where I get confused:
This exercise is [08.3] of these examples, and I'm wondering if someone can explain how they got $0, \infty, \infty, 1$ to be the vanishing order at each of the three zeros. Also, does ``at each of the three zeros'' mean that they plug in $x=1, \omega, \omega^2$?
I'm using section 8 of Paul Garrett's notes as a reference for notation.
My best guess is this: if we consider $y^3-x^3+1$ as a polynomial in $y$ with coefficients in in $\mathbb{C}[x]$. Then our polynomial is $y^3+0y^2+0y-x^3+1$. The coefficient $c_3=1$ means that the order of vanishing if the order of $1=(x-x_0)^n\frac{P(x)}{Q(x)},$ for $P$ and $Q$ are in $\mathbb{C}[x]$ and they are nonzero when $x=x_0$. $x_0$ is a root of $x^3-1=0.$ In this case, $n$ must be $0$, hence the first 0.
Then the second and their coefficients are 0, so by definition, we have vanishing order $\infty$.
Lastly, the coefficient $c_0=1-x^3$: we can factor this into $(1-x)(x^2+x+1)$, where $x=1$ is a root of $1-x^3$ (not sure if I need this. So the order of vanishing is $1$.