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It is stated in Geometry of Algebraic Curves by Harris that there are two equivalent definitions of a hyperelliptic curve $X$:

(i) There exists a degree $2$ meromorphic global function of $X$.

(ii) $X$ is expressible as a $2$-sheeted cover of $\mathbb{P}^1$.

Question: How are these equivalent?

I have introductory knowledge of embeddings of curves in projective space, and I understand that the equivalence should follow somewhat as follows:

Condition $(i) \Rightarrow$ we have a degree $2$ divisor $D$ on $X$ such that $\dim(H^0(O_C(D)))\geq 2$

$\Rightarrow$ we have a map $\phi_D:X\longrightarrow \mathbb{P}^1=\mathbb{P}H^0(O_C(D))^*$ given by $p\mapsto H_p\subset H^0(O_C(D))$, where $H_p$ is the hyperplane in the vector space given by $\sigma\in H^0(O_C(D))|(\sigma)\geq p$, $(\bullet)$ denoting the divisor.

How do we get from here that the map is degree $2$ and thus condition $(ii)$? Also, how do we show the converse?

Tejas Rao
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  • I'd think of it in more naive terms: A meromorphic function on a complex curve is a holomorphic map to the projective line (i.e., there are no points of indeterminacy), and a non-constant holomorphic map on a connected curve is a branched covering. But perhaps you're working over other fields than $\mathbf{C}$...? – Andrew D. Hwang Dec 04 '21 at 02:55
  • Since $C$ is a compact Riemann surface, you may realize $C \subseteq \mathbb{P}^n_k$ as an algebraic curve in projective n-space over the field $k$ of complex numbers. Any non-constant rational function $s$ defines a finite surjective morphism $\phi_s: C \rightarrow \mathbb{P}^1_k$. You must prove that $\phi_s$ has degree $2$. https://math.stackexchange.com/questions/4255500/hartshorne-chapter-1-exercise-6-4-maps-of-curves-and-function-fields – hm2020 Dec 04 '21 at 16:04

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