In Lemma 5.9 of Chapter II of his famous Eisenstein ideal paper, Mazur proved that
when $1/N$ is invertible in the ring $R$, if $\phi$ is a holomorphic modular form in $\omega^k$ over $\Gamma_0(N)$ and $\phi = f(q^N)$ for some $f \in R[[q]]$, then
$f$ is the $q$expansion of a holomorphic modular form over $\Gamma_0(1)$(again in $\omega^k$, and defined over $R$).
Can we generalize this lemma to the case that $\phi$ is a holomorphic modular form in $\omega^k$ over $\Gamma_0(Np)$ and $\phi=f(q^N)$ for some $f \in R[[q]]$?
In other words, is f the $q$expansion of a holomorphic modular form over $\Gamma_0(p)$?
Note that for any $\left(\begin{smallmatrix} a & b \\ cpN & d \end{smallmatrix} \right) \in \Gamma_0(pN)$, we have $$f\left(\frac{a\tau+bN}{cp\tau + d}\right) = \phi\left(\frac{a\tau+bN}{cpN\tau+Nd}\right) = (cpN(\tau/N)+d)^k \phi(\tau/N) = (cp\tau+d)^k f(\tau). $$
We conclude that $f(\tau)(d\tau)^{k/2}$ is invariant under the group generated by all $\left(\begin{smallmatrix} a & bN \\ cp & d \end{smallmatrix} \right)$ together with $T = \left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$. If you take any $\left(\begin{smallmatrix} a & b \\ cp & d \end{smallmatrix} \right) \in \Gamma_0(p)$, multiplication by a suitable power of $T$ will make the top right entry divisible by $N$. Thus, $f(\tau)(d\tau)^{k/2}$ is invariant under $\Gamma_0(p)$.

$\begingroup$ I don't think this answers the question, because this argument will only work if R embeds in $\mathbf{C}$. The question makes perfect sense if $R = \mathbf{F}_p$ with $p \nmid N$, for instance (and that sort of example is of fundamental importance in Mazur's work which the OP cites). $\endgroup$ – David Loeffler Oct 4 '16 at 6:56

$\begingroup$ @DavidLoeffler Thank you. That is a very reasonable point. I will try to revise my argument. $\endgroup$ – S. Carnahan♦ Oct 4 '16 at 13:09