I would like to say that the Riemann surface S defined by $y=x^n$ has genus 0 but I don't know if I make a mistake anywhere, could you check if I'm right?
First option: $Y: S \to \mathbb{P}^1$, $(x, y) \to y$ has degree n and has 2 branching points: 0 and infinity, both of order n, so $\chi(S)=n * \chi(\mathbb{P}^1) - (n-1 + n-1)$ which leads to a genus of zero.
Second option: $\mathbb{P}^1 \to S$ , $z \to (z, z^n)$ is an isomorphism so the genus is preserved.
I'm asking especially because I would imagine the genus would be n or something different and if I'm missing something. Thanks!