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We work over a field $K=\overline{K}$, $ch(K)=0$. Let $C_1$ and $C_2$ be two hyperelliptic curves of the same genus, such that there are degree two morphisms $f_1:C_1\to\mathbb{P}^1$ and $f_2:C_2\to\mathbb{P}^1$ with the exact same branch points. How can I prove that $C_1$ and $C_2$ are isomorphic?

I would appreciate ideas as elementary as possible.

For instance I proved it for elliptic curves by showing that any elliptic curve is isomorphic to some $E_\lambda=\{y^2=x(x-1)(x-\lambda)\}\subset\mathbb{P}^2$, which clearly admits a degree 2 morphism to $\mathbb{P}^1$ with branch points $\{0,1,\infty,\lambda\}$. Say $C_1\cong E_\lambda$ and $C_2\cong E_{\lambda'}$. Then $\{0,1,\infty,\lambda\}$ is equivalent to $\{0,1,\infty,\lambda'\}$ under an automorphism of $\mathbb{P}^1$ and I gave an automorphism of $\mathbb{P}^2$ taking $E_\lambda$ to $E_\lambda'$.

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If you have a degree 2 map $f:C\to\mathbb{P}^1$, then $g(C)=g$ implies $f_*(\mathcal{O}_C)=\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(-g-1)$ and $C\cong\mathrm{Spec} f_*(\mathcal{O}_C)$. The algebra structure on $f_*\mathcal{O}_C$ is essentially given by the map $\mathcal{O}_{\mathbb{P}^1}(-g-1)^{\otimes 2}\to\mathcal{O}_{\mathbb{P}^1}$. The points where this map vanishes ($2g+2$ points) is the branch locus and thus if you have the same branch points, these rings are isomorphic. Thus $C_1\cong C_2$.

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  • $\begingroup$ Thank you. I don't have the knowledge to understand your proof, if you know of a more elementary approach I would appreciate it. $\endgroup$ Commented Apr 14, 2017 at 2:26
  • $\begingroup$ Do you know that two non-singular curves $C_1,C_2$ over $K$ are isomorphic if and only if $K(C_1)\cong K)C_2)$, where $K(C)$ denotes the rational functions on $C$? $\endgroup$
    – Mohan
    Commented Apr 14, 2017 at 2:59
  • $\begingroup$ Yes sure. I mean I don´t know how you got that the rings are isomorphic $\endgroup$ Commented Apr 14, 2017 at 5:36
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A hyperelliptic curve $C$ of the form you describe has an even number of branch points for the map $f$. If they are $a_1,\ldots,a_n$ with none of the $a_i$ infinity then $C$ is birational to $\{y^2=(x-a_1)\cdots(x-a_n)\}$. If some $a_i$ is infinity, say $a_n=\infty$, then $C$ birational to $\{y^2=(x-a_1)\cdots(x-a_{n-1})\}$.

Birationally equivalent projective curves are isomorphic.

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    $\begingroup$ How can I prove $C$ is biratonial to $\{y^2=(x-a_1)\cdots(x-a_n)\}$? $\endgroup$ Commented Apr 14, 2017 at 5:37

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