quotients of curves by actions of roots of unity 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 
 A: 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.    
A: 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.
