# Hyperelliptic curves express by branch points

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!

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