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?



Does a modular function primitive for $\Gamma$ generate the function field of $\mathcal{H}/\Gamma$?
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 $\Gamma$.


