# 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

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Don't you get branch points at zero and infinity, e.g. when the action is multiplication on the projective line? – Jyrki Lahtonen Jul 1 '12 at 9:46
Assume the genus is at least 1... – brosn Jul 1 '12 at 9:59
The genus of $X$ is irrelevant: the quotient is smooth . Dear @Jyrki: you may indeed have branch points for the quotient morphism, but they don't prevent $Y$ from always being smooth. – Georges Elencwajg Jul 1 '12 at 12:20
@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 ... – Jyrki Lahtonen Jul 1 '12 at 13:49

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. – Georges Elencwajg Jul 1 '12 at 12:12
Thanks for your answer! – brosn Jul 1 '12 at 12:44

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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Hi, QiL, glad to see you back : to my taste there were too few answers from you here these last weeks! – Georges Elencwajg Jul 1 '12 at 12:15
Thanks for your answer ! Riemann-Hurwitz gives $2g-2=d(2g(Y)-2)+\sum_P (m_P-1)$ where $P$ are the fixed points of the action. Could one say something else in general? For instance, if $d$ is prime, then $m_p=d$. What does one get for $d=3$? – brosn Jul 1 '12 at 12:37
@GeorgesElencwajg:thanks for your kind words! I have been busy the last time and sometimes I just don't know the anwer. – user18119 Jul 1 '12 at 14:37
@brosn:if $G$ has prime order $d$, then $m_p=d$ as you said. Then $g$ depends on $g(Y)$ and the number of branched points in $Y$, and this number can vary. – user18119 Jul 1 '12 at 14:39
Thanks QiL! Do you know some examples where one can guarantee that $g(Y)=0$? – brosn Jul 1 '12 at 15:01