Let $f(z)$ be a modular form of some integral weight $k \geq 0$ and level $\Gamma_1(N)$ (I insist I want $\Gamma_1(N)$, not $\Gamma_0(N)$ or $\Gamma(N)$). Thus for any $d \in (\mathbb Z/N\mathbb Z)^\ast$ one has he diamond operator $\langle d \rangle$ which applied to $f$ gives rise to an other modular form $\langle d \rangle f$.
Is it true that if the $q$-expansion of $f$ at infinity has its coefficients in $\mathbb Z[1/N]$, then the $q$-expansion of $\langle d \rangle f$ for $d \in (\mathbb Z/N\mathbb Z)^\ast$ also have their $q$-expansion in $\mathbb Z[1/N]$?
I'd like an argument (or reference) or a counter-example. Thanks a lot (I am a little bit ashamed to ask this, but I am pretty confused right now).
The answer is yes at least if $N>4$. Let $\Gamma=\Gamma_1(N)$ and for any ring $A$ let $\mathbf{M}_k(\Gamma;A)=\mathbf{M}_k(\Gamma;\mathbf{Z})\otimes_\mathbf{Z}A$, where $\mathbf{M}_k(\Gamma;\mathbf{Z})$ is the inverse image in $\mathcal{M}_k(\Gamma)$ under the $q$-expansion map of $\mathbf{Z}[[q]]$. If $A$ is a subring of $\mathbf{C}$, then $\mathbf{M}_k(\Gamma;A)$ can be identified with the inverse image under the $q$-expansion map of $A[[q]]\subseteq\mathbf{C}[[q]]$. This follows from the $q$-expansion principle (see the discussion following Theorem 12.3.4 in Diamond and Im's article on modular forms and modular curves). We also have $\mathbf{M}_k(\Gamma;A)=\mathcal{M}_k(\Gamma;A)$ for $A$ any flat $\mathbf{Z}$-algebra, where the target is the geometrically defined $A$-module of modular forms of level $\Gamma$ and weight $k$ (see Theorem 12.3.7 in loc. cit.).
The action of the diamond operators can be defined on $\mathcal{M}_k(\Gamma;A)$ geometrically and when $A=\mathbf{C}$, we get the usual diamond operators via $\mathcal{M}_k(\Gamma)\simeq\mathcal{M}_k(\Gamma;\mathbf{C})$. This is enough to conclude that if $A$ is any subring of $\mathbf{C}$, $\mathbf{M}_k(\Gamma;A)$ is preserved by the diamond operators.
EDIT: To be slightly more detailed (I think this would be clearer if I could produce the commutative diagram I have in mind, but I can't), we have diamond operators on $\mathcal{M}_k(\Gamma;A)$ and on $\mathcal{M}_k(\Gamma;\mathbf{C})$. Since $\mathbf{M}_k(\Gamma;A)=\mathcal{M}_k(\Gamma;A)$, we get diamond operators on the "naive" module $\mathbf{M}_k(\Gamma;A)$ of modular forms with coefficients in $A$. Under $\mathcal{M}_k(\Gamma;\mathbf{C})=\mathcal{M}_k(\Gamma)$ (the LHS is the geometrically defined space and the RHS is the analytically defined space), the geometrically defined diamond operators on the source correspond to the usual diamond operators on $\mathcal{M}_k(\Gamma)$. Finally, the composite $\mathbf{M}_k(\Gamma;A)=\mathcal{M}_k(\Gamma;A)\hookrightarrow\mathcal{M}_k(\Gamma;\mathbf{C})=\mathcal{M}_k(\Gamma)$ is the usual inclusion of the source into the target, and by construction, the diamond operators on $\mathcal{M}_k(\Gamma)$ induce the ones on $\mathbf{M}_k(\Gamma;A)$. This shows that this $A$-submodule is stable under the diamond operators.