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Let $X$ be a (smooth complex algebraic) surface. Suppose $\theta$ is an automorphism of order $2$ of $X$, such that its fixed locus is a disjoint union of smooth curves. I am trying to prove that the quotient $$Y=X/\langle\theta\rangle$$ is in fact a smooth surface. (First of all: is this true/does it follow from some general result?)

My attempt:

Since the quotient is clearly smooth away from the fixed points, we can localize the question around a fixed point $p=(0,0)\in \Bbb{C}^2$. Up to a chart change we can suppose that the fixed curve is the $z$-axis and $$\theta(z,w)=(z,-w)$$ $\Bbb{C}[z,w]^\theta$ has two generators $Z=z^2$ and $W=w$. Now the smoothness of the quotient amounts to show that in fact there are no relations between $Z$ and $W$ and thus the quotient around $p$ is locally the spec of $\Bbb{C}[z_1,z_2]$. But I have no idea how to do this.

A secondary question is: is it always $\rho(Y)=\rho(X)$ (Picard number) ?

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    $\begingroup$ $\mathbb{C}[z^2, w] \cong \mathbb{C}[z^2][w]$. It is obvious that $\mathbb{C}[z^2]$ is a polynomial ring, so $\mathbb{C}[z^2, w]$ is a polynomial ring, too. Hence $\textrm{Spec}\,(\mathbb{C}[z^2, w])$ is smooth. $\endgroup$
    – Francesco Polizzi
    Commented Nov 29, 2013 at 16:11
  • $\begingroup$ Any nontrivial relation between $X$ and $Y$ would, by substitution, immediately give a nontrivial relation between $z$ and $w$, which does not exist. $\endgroup$
    – Will Sawin
    Commented Nov 29, 2013 at 17:30
  • $\begingroup$ Thanks @WillSawin. You also made me notice the terrible mistake I made by mixing notation for $X$ and $Y$ (now $Z$ and $W$). Let me edit and fix that. $\endgroup$
    – Heitor
    Commented Nov 29, 2013 at 17:37

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Francesco gave the answer to your first question. As for the second one, look at my answer to this post.

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  • $\begingroup$ Dear abx, if I understand correctly what you are saying is that in general $S(\theta)=\pi^\ast S_Y$, where $S_Y$ is the Picard group of $Y$ and $S(\theta)$ are the $\theta$-fixed elements of $S_X$. Could you please explain me why is this true? $\endgroup$
    – Heitor
    Commented Nov 29, 2013 at 17:32
  • $\begingroup$ This is true is up to multiplication by 2 : there is a homomorphism $Nm: \mathrm{Pic}(X)\rightarrow \mathrm{Pic}(Y)$ such that $\pi ^*Nm(L)=L\otimes \theta ^*L$. Thus if $L$ is $\theta $-invariant, $L^2$ belongs to $\pi ^*S_Y$. $\endgroup$
    – abx
    Commented Nov 29, 2013 at 17:48

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