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Consider the algebraic curve $X$ defined by $y^2=x^5+1$; it's a smooth plane curve (so a Riemann Surface) and it's an hyperelliptic curve since it has a map to $\mathbb{P}^1$ of degree 2 (its projection to the $x$ axis). $$$$By Plucker's formula it should have genus 6, since it is the zero locus of a polynomial of degree 5. But I also know that since $X$ is an hyperelliptic curve defined by the equation $y^2=f(x)$ with $f$ having degree $d=5$, its genus $g$ should satisfy the condition $d=2g+1$, so it should be $g=2$. $$$$ So I'm clearly making some confusion here, but I can't understand where's the problem.

user720386
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    By "Plucker's formula" you mean the degree-genus formula for smooth plane curves? If so, I don't think it applies here because if you homogenize the equation and then analyze the chart at infinity, you'll find a singular point. – Tabes Bridges Sep 05 '24 at 16:25
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    We really ought to have a canonical duplicate for this sort of question - this error is a common one :) – KReiser Sep 05 '24 at 16:33
  • Oh, you're right @TabesBridges. It's definitely not a smooth projective curve. – user720386 Sep 05 '24 at 16:33
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    Here's a related question, maybe a duplicate: https://math.stackexchange.com/questions/4896561/projective-curve-covered-by-two-affine-pieces . And I'll continue proselytizing that the natural ambient space of a hyperelliptic curve of genus $g$ is the weighted projective space $\mathbb{P}(1,g+1,1)$. The closure of the affine curve $y^2 = f(x)$ in this space is projective and smooth. – Viktor Vaughn Sep 05 '24 at 23:09

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