# Are superelliptic curves singular?

It is an easy corollary of the Riemann-Hurwitz formula that smooth double covers of $$\mathbb{P}^1$$ can only be branched over an even number of points. Let $$F(x,z) \in \mathbb{C}[x,z]$$ be a homogeneous polynomial of degree $$2n+1 \geq 3$$, and consider the curve $$C$$ cut out by $$y^2 = F(x,z)$$ in the weighted projective space $$\mathbb{P}_{2,2n+1,2}$$ with coordinates $$[x : y : z]$$. Then there is a $$2$$-to-$$1$$ map $$\phi \colon C \to \mathbb{P}^1$$ given by sending $$[x : y : z] \mapsto [x : z]$$. It appears that $$\phi$$ realizes $$C$$ as a double cover of $$\mathbb{P}^1$$ branched over $$2n+1$$ points, namely the roots of $$F(x,z) = 0$$, so I'd guess that the curve $$C$$ has to be singular.

The weighted projective space $$\mathbb{P}_{2,2n+1,2}$$ can be covered by three affine patches: $$D_x = \{x \neq 0\}$$, $$D_y = \{y \neq 0\}$$, and $$D_z = \{z \neq 0\}$$. Recall that $$D_x = \operatorname{Spec} R_x$$, where $$R_x$$ is the $$0^{\mathrm{th}}$$-graded component of the ring $$\mathbb{C}[x,1/x,y,z]$$, and it is not hard to check that $$R_x = \mathbb{C}[y^2/x^{2n+1},z/x]$$. Thus, in the patch $$D_x$$, the curve $$C$$ has coordinate ring $$k[u,v]/(u - F(1,v)) \simeq k[v]$$, where $$u = y^2/x^{2n+1}$$ and $$v = z/x$$. (Note in particular that $$C$$ is rational and hence has geometric genus $$0$$.) So $$C \cap D_x$$ is smooth, and a similar argument shows that $$C \cap D_z$$ is smooth. Thus, any singular point on $$C$$ would have $$x = z = 0$$, but there is no such point on $$C$$, implying that $$C$$ is smooth.

I thus seem to have obtained a contradiction to the Riemann-Hurwitz formula, so clearly I made a mistake somewhere. What have I done wrong?

## 1 Answer

I think you made a mistake in claiming that $$\phi$$ is a 2:1 map. The two points $$[x:y:z]$$ and $$[x:-y:z]$$ are in fact equal in the weighted projective space - since $$(x,y,z) \sim ((-1)^2x, (-1)^{2n+1}y, (-1)^2z) = (x, -y, z)$$ by the definition of $$\mathbb{P}(2,2n+1,2)$$. So in fact $$\phi$$ is an isomorphism!