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let $X$ be a smooth projective irreducible curve of genus $g$ over the complex numbers. Assume that $X$ comes with an action of $\mu_d$.

Is the quotient $Y:=X/\mu_d$ always smooth?

Let $\pi: X \to Y$ be the quotient map. Is it possible to calculate the genus of $Y$ by considering the map induced on jacobians $J(X) \to J(Y)$ and lifting the action of $\mu_d$ to an action on $J(X)$?

Thanks for your help

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    @Georges: I knew (from the function field side) that the quotient has a smooth model (corresponding to a model of the fixed field of a cyclic group of automorphisms). I was on some kind of autopilot thinking that something bad happens at a point of ramification. I totally forgot that you define the local structure in a way that steers clear from that kind of problems. IOW, had the question not been about complex structure, I would have gotten it. I like to think so at least :-) But I am a bit uncertain whenever differentiable structures rear their ugly head ...2012-07-01

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I have good news for you!

If $X$ is an arbitrary curve over an arbitrary algebraically closed field of any characteristic and if $G$ is an arbitrary finite group acting algebraically on $X$ with arbitrary stabilizing subgroups of points , the quotient $X/G$ exists.
The variety $X/G$ has the quotient topology inherited from $\pi:X\to X/G$ , the canonical morphism .
And most importantly we have the categorical property: any morphism $f:X\to Y$ of algebraic varieties that is constant on the orbits of $G$ factorizes through a morphism $\tilde f : X/G\to Y$, i.e. $f=\tilde f \circ f$

Moreover if $X$ is normal so is $X/G$.
Since for curves normality coincides with smoothness, this more than answers your question in the affirmative.

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    I had started writing this post before seeing QiL's answer. His answer is perfect and he also has priority in posting.I have just upvoted him.2012-07-01
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By construction, the quotient by a finite group $G$ of a normal quasi-projective variety $X$ is always normal (the ring of regular functions on $U/G$, when $U$ is an affine open subset of $X$ stable by $G$, is $O_X(U)^G$). For curve over a perfect field, normal is equivalent to smooth.

For the genus of $Y$, it is in general easier to use Riemann-Hurwitz formula.

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    @Brosn. Suppose that $X$ attains the Hurwitz bound, i.e., $\# $ Aut$(X) = 84(g-1)$. Then, if $G=\#$ Aut$(X)$, we have that $X/G$ is of genus zero. Also, $X\to X/G$ ramifies over only three points.2012-07-02