# Are there “primitive” modular functions for $\Gamma(n)$ with Fourier coefficients in $\mathbb{Q}$?

The modular curve $Y(n)$ which over $\mathbb{C}$ is $\mathcal{H}/\Gamma(n)$ is often viewed as a curve defined over $\mathbb{Q}(\zeta_n)$. However, if one twists the moduli problem to be given by pairs $(E,\alpha)$ where $E$ is an elliptic curve over some $\mathbb{Q}$-scheme $S$, and $\alpha$ is an isomorphism: $$\alpha : E[n]\stackrel{\sim}{\longrightarrow}\mathbb{Z}/n\times\mu_n$$ "of determinant 1", then one obtains a model of $Y(n)$ over $\mathbb{Q}$.

The usual $q$-expansion arguments show that $Y(n)_{\mathbb{Q}(\zeta_n)}$ is isomorphic to the Spec of the ring of modular functions (modular forms of weight 0) for $\Gamma$ whose Fourier coefficients lie in $\mathbb{Q}(\zeta_n)$.

My question is: Is there an $n\ge 3$ such that $Y(n)_\mathbb{Q}$ isomorphic to the Spec of modular functions for $\Gamma$ whose Fourier coefficients lie in $\mathbb{Q}$? (ie, are there enough of these functions?)

My guess is that the answer is no, but how would one argue this?

I think this is in fact always true. Let $f$ be a rational function on this moduli space which is defined over $\mathbb Q$.
Recall that the Tate curve over is an elliptic curve $E / \mathbb Q((q))$ which is $q$-adically analytically and holomorphically isomorphic to the quotient $\mathbb G_m / \langle q \rangle$. So if we base change to $\mathbb Q ( q^{1/n})$, there is an isomorphism $E[n] \to \mathbb Z/n \times \mu_n$, which sends $q^{1/n}$ to $(1,0)$ and sends $\mu_n$ to $(0,\mu_n)$.
This defines a map $\operatorname{Spec} \mathbb Q((q)) \to Y(n)_{\mathbb Q}$ by the universal property of $Y(n)_{\mathbb Q}$. Pulling back $f$ along this map, we obtain a power series in $q^{1/n}$ with rational coefficients.
One can see that this is the same as the usual power series coefficients using the fact that the Tate curve $\mathbb G_m / \langle q \rangle$, is, for $q=e^{ 2 \pi i \tau}$, isomorphic to $\mathbb C / \langle 1,\tau\rangle$ under the map $z \to e^{ 2\pi i z}$.