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I want to write explicit equations for a curve $C$ that allows a 2-to-1 morphism onto a curve $E$ that is not rational, actually, $E$ is an elliptic curve. In more detail, I know that $E$ is an anticanonical curve of a degree 7 del Pezzo surface $S$, embedded in $\mathbb{P}^7$ by its anticanonical linear system; my curve $C$ is a canonical curve of genus 8, so, it should be possible to put all together in this projective space. What I did so far, is getting equations for $S$ and then taking a hyperplane section to get $E$, but then I don't know how to get equations for $C$. Any ideas/comments or even a similar example in which one gets equations for a curve allowing a 2-to-1 map to a non-rational curve would be useful.

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  • $\begingroup$ Why do you need these equations? $\endgroup$ Sep 29, 2016 at 21:41
  • $\begingroup$ I simply want to have as explicit as possible examples of certain varieties for testing some computer algebra tools. These curves also arise naturally as canonical curves of some interesting surfaces of general type. $\endgroup$
    – R. Jahvel
    Sep 29, 2016 at 22:31

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If you just want one example of a bielliptic curve $C$ of genus $8$, you can try the modular curve $X_0(101)$, which has an involution $w_{101}$ for which the quotient curve is the unique elliptic curve of conductor $101$. These days there are several packages that can give you a basis for the modular forms of weight $2$ for $\Gamma_0(101)$ (which are the holomorphic differentials on $C$) and the action of $w_{101}$ on those forms; these are represented by $q$-expansions long enough that you can then find generators for the space of quadratic relations by solving the simultaneous linear equations for the coefficients of those relations.

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Let $f_i$ be the equations of $S$ in $P^7$. Take a sufficiently general hyperplane $P^6 \subset P^7$. Then the restrictions of $f_i$ give you equations of $E$ in $P^6$. Choose one sufficiently general quadratic polynomial $q(x)$ on $P^6$. Add one more variable (say $y$), and write an additional equation $$ y^2 = q(x). $$ This gives $C$.

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