Finite generation of module of modular forms

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 $\mathbb{C}$)?

• Doesn't finite generation for all $R$ follow immediately from the $\mathbb Z[1/N]$ case? Oct 23 '17 at 15:44
• I think that in the case $R$ contains $\mathbb{Z}[\zeta_n,\frac{1}{n}]$, the space $M_*(\Gamma(n),R)$ is stable under the action of $SL_2(\mathbb{Z}/n\mathbb{Z})$. Therefore finite generation as an algebra should imply finite generation as a module in this case. Oct 23 '17 at 17:30
• @WillSawin Yes, indeed as $M_*(\Gamma_1(n); R)$ can differ from $M_*(\Gamma_1(n);\mathbb{Z}[\frac1n]) \otimes R$ only in degree $1$. Oct 23 '17 at 20:17
• @FrançoisBrunault: How does this last implication work? Oct 23 '17 at 20:18
• @LennartMeier For any $f \in M_k(\Gamma(n),R)$ we can form the polynomial $\prod_{\gamma \in SL_2(\mathbb{Z}/n\mathbb{Z})} X-f | \gamma$ which has coefficients in $M_*(SL_2(\mathbb{Z}),R)$. This shows that $M_*(\Gamma(n),R)$ is integral over $M_*(SL_2(\mathbb{Z}),R)$. I don't know whether this integrality property is true in general. Oct 24 '17 at 4:18

Edit: This is an answer to the wrong question. This explains the finite generation of $M_{k}(\Gamma_{1}(n), R)$ as an algebra, not as a module over the graded ring of level $1$ modular forms.
This result is already known. One good source for this is the paper of Nadim Rustom (Generators of graded rings of modular forms, published in the Journal of Number Theory in 2014, the arXiv version is here). This paper shows more, that the graded ring of modular forms for $\Gamma_{1}(n)$ is generated in weight at most $3$ for $n \geq 5$. This is fairly simple over $\mathbb{C}$ and follows from a lemma (dating back to Mumford, essentially) about surjectivity of the map $H^{0}(X,\mathcal{L}_{1}) \otimes H^{0}(X,\mathcal{L}_{2}) \to H^{0}(X, \mathcal{L}_{1} \otimes \mathcal{L}_{2})$ where $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are two line bundles on a curve $X$. However, it takes a bit of work to translate this result into the desired statement for an arbitrary commutative $\mathbb{Z}[1/n]$ algebra.
I would expect that there should be a much simpler way to see the finite generation of $M_{k}(\Gamma_{1}(n), R)$, but I don't know one.