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In a book of algebraic geometry, the author says that equation $$ y^2=x(x^4-1) $$ defined a curve $S$ of genus $2$.

My problem is this: Because the curve has genus 2, it can be expressed as follows $$ y^2=(x-a_1)(x-a_2)...(x-a_6) $$ where $a_i$, $ 1\le i\le 6$, is the branch points of the canonical map $\varphi_{K}: S \longrightarrow \mathbb{P}^1$. Thus, $y^2=x(x^4-1)$ can not be a curve of genus $2$. A curve of genus $2$ has $6$ branch points, and not $5$ as suggested by equation $y^2=x(x^4-1)$.

Am I making a mistake in my thinking? Thank you!

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    $\begingroup$ $\infty$ is a branch point. $\endgroup$ – Lord Shark the Unknown Nov 13 '18 at 20:25
  • $\begingroup$ @LordSharktheUnknown, how did you realize that? $\endgroup$ – Manoel Nov 14 '18 at 1:43

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