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If a curve is nonsingular then the arithmetic genus and the geometric genus should be the same.

Using the genus-degree formula, I obtain that the genus of this curve should be 10.

However, the book I'm reading states that this curve has genus 2.

Simon M
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  • If the curve is the zero set of an irreducible polynomial - as we have here - then the arithmetic genus of the curve is $\frac{(n-1)(n-2)}{2}$ where $n$ is the degree of the polynomial – Simon M Jan 16 '24 at 02:57
  • If it helps to figure out what's going on here, the book also claims that any curve $y^2=(x^2-1)...(x^2-n^2)$ has genus $n-1$ – Simon M Jan 16 '24 at 03:03
  • why genus 2? Any idea? – sti9111 Jan 16 '24 at 03:19
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    This model is indeed singular: if you projectivise in the obvious way, then the point $(0:1:0)$ is singular. The easiest way to compute the genus is by using Riemann-Hurwitz with the degree 2 cover to $\mathbb{P}^1$ with $n$ ramification points. – Alex B. Jan 16 '24 at 03:19
  • I think, that It is convenient to consider the curve as a double cover of a plane curve of degree 6 with 6 branched points. – sti9111 Jan 16 '24 at 03:22
  • @AlexB. That's an answer - please consider recording it as such below. – KReiser Jan 16 '24 at 03:31
  • So, in summary, there are singularities at infinity that I hadn't yet considered. I also encourage @AlexB to consider posting the point he made as answer. I'd be more than glad to accept the answer – Simon M Jan 16 '24 at 03:37
  • In the future I will be more careful to make sure to watch out for singularities at infinity. A lesson learned – Simon M Jan 16 '24 at 03:41

1 Answers1

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This model is indeed singular: if you projectivise in the obvious way, then the point $(0:1:0)$ is singular.

The easiest way to compute the genus is by using Riemann-Hurwitz with the degree $2$ cover to $\mathbb{P}^1$ that sends a point to its $x$-coordinate, which has $6$ ramification points, namely the zeros of the right-hand polynomial.

Alex B.
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