Given a commutative $\mathbb{Z}[\frac1n]$-algebra $R$, we can consider the ring of modular forms $M_*(\Gamma_1(n), R)$. If $R$ is a subring of $\mathbb{C}$, these can be defined as those (holomorphic) modular forms for the congruence subgroup $\Gamma_1(n)$ that have $q$-expansion with coefficients in $R$. More generally, we can define this ring as $H^0(\overline{\mathcal{M}}_1(n)_R; \underline{\omega}^{\otimes *})$, where $\overline{\mathcal{M}}_1(n)_R$ is the compactified moduli stack of elliptic curves with $\Gamma_1(n)$-level structure over $R$ and $\underline{\omega}$ is the pushforward of the sheaf of differentials on the universal elliptic curve.

Recently, I have investigated the structure of $M_*(\Gamma_1(n),R)$ as a graded module over $M_*(SL_2(\mathbb{Z}), R)$, the ring of modular forms without level. If $R = k$ is a field of characteristic bigger than $3$, it is quite easy to show that this graded module is free (by using that the compactified moduli stack $\overline{\mathcal{M}}_{ell}$ is in this case a weighted projective stacks line and vector bundles split into line bundles here); the same is actually true for field of characteristic $2$ or $3$, but significantly harder to show. From these results it follows that $M_*(\Gamma_1(n),R)$ is finitely generated (edit: as a graded module) over $M_*(SL_2(\mathbb{Z}), R)$ first in the case that $R$ is a field and one can deduce it also in the general case.

This seems to be a rather overkill way to show just this finite generation (and I know another overkill argument using topological modular forms). My question is the following:

Is it either an already known result that $M_*(\Gamma_1(n),R)$ is finitely generated (edit: as a graded module) over $M_*(SL_2(\mathbb{Z}), R)$ for all commutative $\mathbb{Z}[\frac1n]$-algebras $R$ or does it admit at least a simple/direct proof? (at least for subrings of $\mathbb{C}$)?

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