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Let $f$ be a modular function (that is, a meromorphic modular form of weight 0) holomorphic on $\mathcal{H}$ which is invariant under $\Gamma\le SL_2(\mathbb{Z})$ (not necessarily congruence!), and not invariant under any larger group.

Is $\mathbb{C}(j)[f]$ precisely the function field $\mathbb{C}(\Gamma)$ of the modular curve $\mathcal{H}/\Gamma$?

Certainly if $\mathbb{C}(j)[f]$ is strictly smaller than $\mathbb{C}(\Gamma)$ then it couldn't be the function field of any modular curve. Is this possible?

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  • $\begingroup$ If $\Gamma$ is a congruence subgroup, then the answer is yes, because the extension $\mathbb{C}(\Gamma(N))/\mathbb{C}(\Gamma(1))$ is Galois with Galois group $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})/\pm 1$, so every intermediate subfield is of the form $\mathbb{C}(\Gamma)$ for some $\Gamma$. $\endgroup$ Oct 15, 2015 at 7:16

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I think the answer to your question is yes.

If $\Gamma$ is a finite index subgroup of $\Gamma_1 = \mathrm{PSL}_2(\mathbb{Z})$, then there exists a finite index normal subgroup $\tilde{\Gamma}$ of $\Gamma_1$ such that $\tilde{\Gamma} \subset \Gamma \subset \Gamma_1$. Now the extension of function fields $\mathbb{C}(\tilde{\Gamma})/\mathbb{C}(\Gamma_1)$ is Galois with Galois group $\Gamma_1/\tilde{\Gamma}$, so every intermediate subfield is of the form $\mathbb{C}(\Gamma')$ for some finite index subgroup $\Gamma'$ of $\Gamma_1$.

In particular $\mathbb{C}(j)[f]$ has to be of the form $\mathbb{C}(\Gamma')$ for some $\Gamma' \supset \Gamma$, and your assumption on $f$ implies that $\Gamma'$ cannot be bigger than $\Gamma$.

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