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Let $X$ be a smooth projective curve over the field of complex numbers. Assume it comes with an action of $\mu_3$. Could someone explain to me why is the quotient $X/\mu_3$ a smooth curve of genus zero?

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What is $\mu_3$? The cyclic group with $3$ elements? Whatever it is, don't you also make an assumption about the genus of $X$? – Hagen Knaf Jun 25 '12 at 11:48
Yes, $\mu_3$ is the group of roots of unity of order $3$. I know the result is true for genus bigger or equal than $2$. Is it also true for $g=1$? – guelph Jun 25 '12 at 12:07
Have you tried Riemann-Hurwitz formula? – Ehsan M. Kermani Jun 25 '12 at 15:53
Isn't there something missing in the statement of the problem? Suppose I start with some smooth projective curve $Y$ of genus $>0$ and I take the normalization $X$ of $Y$ in a cyclic extension $F/K(Y)$ of degree $3$ of the function field of $Y$. Then $X$ is smooth and the normalization map $X\rightarrow Y$ is a cyclic cover of degree $3$. Thus $X/\mu_3 =Y$ and we have a counter example to the original statement. – Hagen Knaf Jun 27 '12 at 10:47

No, the result is not true for genus $1$.

Consider an elliptic curve $E$ (which has genus one) and a non-zero point $a\in E$ of order three .
Translation by $a$ is an automorphism $\tau_a:E\to E: x\mapsto x+a$ of order $3$ of the Riemann surface $E$.
It generates a group $G=\langle \tau_a\rangle\cong \mu_3$ of order $3$ acting freely on $E$ and the quotient $E/G$ is an elliptic curve, a smooth curve of genus one and not of genus zero.

a) There are eight choices for $a$, since the $3$-torsion of $E$ is isomorphic to $\mathbb Z/3\mathbb Z\times \mathbb Z/3\mathbb Z$.
b) The quotient morphism $E\to E/G$ is a non-ramified covering

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