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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?

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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!

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