1
$\begingroup$

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

  • 0
    Don't you get branch points at zero and infinity, e.g. when the action is multiplication on the projective line?2012-07-01
  • 0
    Assume the genus is at least 1...2012-07-01
  • 1
    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.2012-07-01
  • 0
    @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

2 Answers 2