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I have an explicit genus one curve $E$ with two points $p_1$ and $p_2$ on it and am looking for an explicit degree seven cover $X\to E$ with ramification precisely over $p_i$, with a single preimage point over each (by Hurwitz formula, $g(X)=7$). This should be a non-Galois extension, and the splitting field should have the Galois group ${\rm PSL}(2,7)$.

Literature search did not yield anything useful: there is a fair amount of work on covers of genus one curves with one ramification point (called "origami") and even that seems nontrivial. Similarly, there does not appear to be a simple general theory of ${\rm PSL}(2,7)$ Galois covers. Any suggestions will be appreciated.

Added for clarity. My question is what algorithms/tricks can be used to explicitly compute the cover $X$, given explicit formulas for $E$, $p_1$ and $p_2$.

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    $\begingroup$ What is your question? $\endgroup$
    – Will Chen
    Commented Apr 17, 2022 at 19:40
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    $\begingroup$ Are you asking whether the group $\mathbf{PSL}_2(\mathbb{Z}/7\mathbb{Z})$ has a subgroup of index $7$? $\endgroup$
    – Jason Starr
    Commented Apr 17, 2022 at 23:50
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    $\begingroup$ There is a degree $7$ cover of the projective line with Galois group $\textbf{PSL}_2(\mathbb{Z}/7\mathbb{Z})$ with $2$ branch points with branching equal to a $7$-cycle and with $2$ branch points with branching equal to a product of two $2$-cycles. Pullback this cover by any degree $2$ cover of the projective line by a genus $2$ curve whose four branch points include the two branch points with $(2,2)$-branching. $\endgroup$
    – Jason Starr
    Commented Apr 18, 2022 at 0:04
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    $\begingroup$ One approach would be to calculate explicit equations for the modular curve $X_0(11) \times_{X(1)} X(7)$ as a cover of $X_0(11)$ using modular forms methods. It's not so hard to check that this cover has all the properties you want. You can replace $X_0(11)$ by any modular curve of genus $1$, with level prime to $7$, with no elliptic points and two cusps. $\endgroup$
    – Will Sawin
    Commented Apr 18, 2022 at 0:04
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    $\begingroup$ It does, but the ramification should all be canceled by the ramification of $X_0(11) $, which is totally ramified over $i$ and $\rho$ (since $11$ in inert in $\mathbb Q(i)$ and $\mathbb Q(\rho)$). However, I didn't fully process the specific elliptic curve aspect of this, and I'm not very optimistic that deforming the equation will be enlightening. But I don't disagree it's worth a shot... $\endgroup$
    – Will Sawin
    Commented Apr 18, 2022 at 0:42

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