Modular symbols and degeneracy maps

Question

Consider the space of (cohomological) modular symbols of level $N$,

$$ \operatorname{Symb}_{\Gamma_0(N)}(\mathbb{Z}) = \operatorname{Hom}_{\mathbb{Z}[\Gamma_0(N)]}(\mathrm{Div}^0(\mathbb{P}^1_\mathbb{Q}), \mathbb{Z}) \cong H^1_c(Y_0(N)(\mathbb{C}), \mathbb{Z}).$$

If I have a prime $q \nmid N$, then there's two injective maps $i_1, i_2: \operatorname{Symb}_{\Gamma_0(N)}(\mathbb{Z}) \hookrightarrow \operatorname{Symb}_{\Gamma_0(Nq)}(\mathbb{Z})$: one is just the natural inclusion, and the other is given by acting by $\begin{pmatrix} q & 0 \\\ 0 & 1 \end{pmatrix}$.

Say we choose a modular form $f$ of weight 2 and level $\Gamma_0(N)$ (with coefficients in $\mathbb{Z}$, for simplicity). Let $\operatorname{Symb}_{\Gamma_0(N)}(\mathbb{Z})[f]$ denote the rank 2 submodule where the Hecke operators act as they do on $f$. Similarly, let $\operatorname{Symb}_{\Gamma_0(Nq)}(\mathbb{Z})[f]$ be the subspace at level $Nq$ where the Hecke operators away from $q$ act as they do on $f$.

Do the images of $\operatorname{Symb}_{\Gamma_0(N)}(\mathbb{Z})[f]$ under $i_1$ and $i_2$ generate $\operatorname{Symb}_{\Gamma_0(Nq)}(\mathbb{Z})[f]$? This is certainly true after tensoring with $\mathbb{Q}$, but I'm worried that the sum of the images might not be saturated.

Answer 1

Please allow me to replace $\mathbb Z$ by $\mathbb Z_\ell$ in your question, as the problem of being saturated or not is local.

Assuming $f$ is a cusp form, the answer is yes if and only if your modular form f is not Eisentein mod $\ell$ (that is, if the Galois representation $\rho_f$ is still irreducible modulo $\ell$). This is (in the difficult sense) the famous Ihara's Lemma. Indeed, this lemma states that the map $H^1(X_0(N),\mathbb Z/\ell) \times H^1(X_0(N),\mathbb Z/\ell) \rightarrow H^1(X_0(Nq),\mathbb Z/\ell)$ is injective, cf Ribet's article at the ICM 1982/1983, page 10. Under the hypotheses $f$ not Eisenstein, one has $H^1(X_0(N),\mathbb Z/\ell)[f] = H^1_c(Y_0(N),\mathbb Z/\ell)[f]$ and the same result with $N$ replaced by $Nq$, and this implies what you want.

If $f$ is Eisenstein, this is always true.