Theorem. Let $\ell$ be prime, and $Q, R \ge 1$ such that $(\ell, Q, R)$ are pairwise coprime. Let $N = QR$ and for simplicity assume $N \ge 4$. Then $W_Q$ preserves $M_k(\Gamma_1(N), \mathbf{Z}[1/N, \zeta_Q])$.
Proof. Let $M_k^{\mathrm{wk}}(\Gamma_1(N), A)$ denote the space of weakly modular forms (possibly meromorphic at the cusps) with $q$-expansions in the ring $A$. If $A$ is a subring of $\mathbf{C}$ then
$$M_k(\Gamma_1(N), A) = M_k^{\mathrm{wk}}(\Gamma_1(N), A) \cap M_k(\Gamma_1(N), \mathbf{C})$$
so it suffces to show that $M_k^{\mathrm{wk}}(\Gamma_1(N), \mathbf{Z}[1/N, \zeta_Q])$ is stable under $W_Q$.
If $A$ is a $\mathbf{Z}[1/N]$-algebra, then elements of $M_k^{\mathrm{wk}}(\Gamma_1(N), A)$ can be interpreted as rules sending "test objects" $(E, P_Q, P_R, \omega) / B$ to elements of $B$. Here $B$ is a $A$-algebra, $E$ is an elliptic curve over $B$, $P_Q$ and $P_R$ are points of exact order $Q$, $R$ respectively, and $\omega$ is a global differential on $E$. These have to satisfy various conditions (the main ones are compatibility with base change and homogeneity in $\omega$ of weight $k$).
So it suffices to show that $W_Q$ makes sense on test objects if $A = \mathbf{Z}[1/N, \zeta_Q]$. The map will send $(E, P_Q, P_R, \omega) / B$ to $(E', P_Q', P_R', \omega')/B$, where all but one of these objects are simple to define:
- $E' = E/\langle P_Q \rangle$
- $P_R' = \pi(P_R)$, where $\pi : E \to E / \langle P_Q\rangle$ is the quotient map
- $\omega' = \pi_*(\omega)$
The hard one is $P_Q'$: one checks that $E[Q] / \langle P_Q\rangle$ has a unique generator $P_Q'$ characterised by the Weil pairing $\langle P_Q, P_Q'\rangle = \zeta_Q$ (and since it is unique, its formation is compatible with base-change). $\square$
(Caveat: this construction gives an operator whose square is something like $\langle Q\rangle_R Q^k$ where $\langle -\rangle_R$ denotes the diamond operator for something that is 1 (mod Q) and Q (mod R). Some people prefer to normalise away the $Q^k$, but this may not be possible if $Q$ is odd without introducing an extraneous $\sqrt{Q}$ into your ring.)