The first example: N=11, M=17

Consider the first example, in which N=11 and M=17. The spaces $S_2(\Gamma_0(11))$ and $S_2(\Gamma_0(17))$are spanned by newforms f and g, respectively, where

$$\begin{eqnarray*}f &=& q - 2q^2 - q^3 + 2q^4 + q^5 + 2q^6 - 2q^7 - 2q^9 - 2q^{10... ...2 - q^4 - 2q^5 + 4q^7 + 3q^8 - 3q^9 + 2q^{10} - 2q^{13} +\cdots. \end{eqnarray*}$$

We have

$$\begin{eqnarray*}(1 + M - a_M(f))(1 + M + a_M(f)) &=& 2^6\cdot 5,\\ (1 + N - a_N(g))(1 + N + a_N(g)) &=& 2^4\cdot 3^2. \end{eqnarray*}$$

To avoid reducibility we require that $p\,\vert\,\mbox{\rm num}((11-1)/12)=5$ and $q\,\vert\,\mbox{\rm num}((17-1)/12)=4$, so only the pair p=2, q=3 remains. The representation

$$\rho=\rho_{f,2}\times\rho_{g,3}:G_{\mathbf{Q}}\rightarrow\mbox{\rm GL}(2,\mathbf{Z}/6\mathbf{Z})$$

is irreducible mod 2 and mod 3. Choose newforms $h_1,h_2,\ldots, h_6$ from each Galois conjugacy class in $S_2(\Gamma_0(11\cdot 17))$, ordered as in Section . We find that $f\equiv h_2,h_3,h_4,h_5,h_6\pmod{2}$and $g\equiv h_1,h_4\pmod{3}$. Because h₄ occurs in both of these lists, $\rho$ is modular; $\rho$ arises from

$$h_4 = q + aq^2 + (-a - 1)q^3 - 2aq^4 + (a - 1)q^5 + (a - 2)q^6 - 2q^7+\cdots$$

where a² + 2a - 2=0. We have maximal success.