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Exercise I.F-8 from "Arbarello, Cornabla, Griffiths, Harris: Geometry of algebraic curves" states that for a complex algebraic genus $g$ curve and its automorphism $\varphi$ of order $n$ the number of fixed points $\alpha$ of $\varphi$ satisfies

$\alpha\leq \frac{2g-2+2n}{n-1}$

with inequality only if the quotient $C/\varphi$ has genus $0$ (i.e. $\mathbb{P}^1$).

So the inequality itself is an easy part (it follows from Riemann-Hurwitz). But I can't see why the number of fixed points is exactly that number if $g(C/\varphi)\ne 0$. Can you help me?

Edited: in fact this number can be non-integer and it's confusing.

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The map $\pi:C\to C/\varphi$ is totally ramified over the fixed points of $\varphi$. So if equality holds in the given inequality, we obtain that the degree of the ramification divisor of $\pi$ is $\geq \alpha.(n-1)=2g-2+2n$. On the other hand, it is an easy observation that for a degree $n$ morphism of curves $f:X\to Y$, the degree of the ramification divisor is maximal if $Y=\mathbb{P}^1$ and in this case is equal to $2g-2+2n$.

So, $\mbox{deg} R=2g-2+2n$ and by Riemann Hurwitz $g(C/\varphi)=0$.

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