Let $\Gamma_1\le SL_2(\mathbb{Z})$ be a noncongruence subgroup of finite index.
Let $\Gamma_2\le SL_2(\mathbb{Z})$ be another subgroup of finite index.
Let $M_0(\Gamma_i)$ denote the ring of modular functions for $\Gamma_i$, by which I mean holomorphic functions on the upper half plane $\mathcal{H}$ which are invariant under $\Gamma_i$ - ie, regular functions on the affine modular curve $\mathcal{H}/\Gamma_i$ - ie, weakly holomorphic modular forms of weight 0 which may have poles at cusps. Since all modular curves are defined over a number field, I'm thinking of $M_0(\Gamma_i)$ as a $\overline{\mathbb{Q}}$-algebra.
- For any $\Gamma_1,\Gamma_2$ as above, is it true that $M_0(\Gamma_1\cap\Gamma_2)$ is generated by $M_0(\Gamma_1)$ and $M_0(\Gamma_2)$? (as $\overline{\mathbb{Q}}$-algebras)
That was a long shot, so if that's false, then
- Fixing an arbitrary finite index $\Gamma_1$, can we always find a torsion-free congruence subgroup $\Gamma_2$ such that $M_0(\Gamma_1\cap\Gamma_2)$ is generated by $M_0(\Gamma_1)$ and $M_0(\Gamma_2)$? (as $\overline{\mathbb{Q}}$-algebras)
I'd also appreciate pointers to references that specifically address questions about modular functions like this.
EDIT: Here's an example. Let $Y(1) := \mathcal{H}/SL_2(\mathbb{Z})$, and $Y(\Gamma_1) := \mathcal{H}/\Gamma_1$, then we may consider trying to set $\Gamma_2 := \Gamma_1(p)$ for some prime $p$. If $\pm\Gamma_1$ has index $d$ in $PSL_2(\mathbb{Z})$, then we may find a $p\ge 5$ such that $Y_1(p) := \mathcal{H}/\Gamma_1(p)$ and $Y(\Gamma_1)$ have coprime degrees over the $j$-line $Y(1)$. In this case, the tensor product $$M_0(\Gamma_1)\otimes_{\overline{\mathbb{Q}[j]}}M_0(\Gamma_1(p))$$ is an integral domain, and is a subring of $M_0(\Gamma_1\cap\Gamma_1(p))$ with the same fraction field. In this case, since such quotients of $\mathcal{H}$ are all nonsingular, we find that $M_0(\Gamma_1\cap\Gamma_1(p))$ must be the integral closure of $M_0(\Gamma_1)\otimes_{\overline{\mathbb{Q}[j]}}M_0(\Gamma_1(p))$ (inside their common fraction field). Thus, we can ask: can we find a $p$ such that $M_0(\Gamma_1)\otimes_{\overline{\mathbb{Q}[j]}}M_0(\Gamma_1(p))$ is already integrally closed? If not, can we do it with $\Gamma_1(p)$ replaced by another torsion-free congruence subgroup?
