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?
smooth quotient
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algebraic-geometry
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0What is $\mu_3$? The cyclic group with $3$ elements? Whatever it is, don't you also make an assumption about the genus of $X$? – 2012-06-25
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0Yes, $\mu_3$ 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 $g=1$? – 2012-06-25
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0Have you tried Riemann-Hurwitz formula? http://en.wikipedia.org/wiki/Riemann%E2%80%93Hurwitz_formula – 2012-06-25
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0Isn'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. – 2012-06-27