[proofplan] We compare the residual representation attached to $f$ with the direct sum representation $1\oplus \bar{\chi}_\ell$ by evaluating both on Frobenius elements away from $N\ell$. The congruence hypothesis gives equality of traces, while weight $2$ and nebentypus equal to $1$ give equality of determinants. Since these characteristic polynomials agree on Frobenius elements at a density-one set of primes, Chebotarev density and Brauer-Nesbitt identify the semisimplifications. [/proofplan] [step:Define the comparison representation and the Frobenius set] Let $K_f$ denote the coefficient field of $f$, let $k_\lambda$ denote the residue field at $\lambda$, let $G_{\mathbb{Q}}:=\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ denote the absolute [Galois group](/page/Galois%20Group) of $\mathbb{Q}$, and let $GL_2(k_\lambda)$ denote the group of invertible $2\times 2$ matrices with entries in $k_\lambda$. Let \begin{align*} \bar{\rho}_{f,\lambda}: G_{\mathbb{Q}} &\to GL_2(k_\lambda) \end{align*} denote the residual Galois representation attached to $f$, normalized so that arithmetic Frobenius at every prime $p\nmid N\ell$ has trace congruent to $a_p(f)$ and determinant congruent to $p$ modulo $\lambda$. Define the comparison representation \begin{align*} \sigma: G_{\mathbb{Q}} &\to GL_2(k_\lambda) \\ g &\mapsto \begin{pmatrix} 1 & 0 \\ 0 & \bar{\chi}_\ell(g) \end{pmatrix}. \end{align*} For each prime $p\nmid N\ell$, let $\operatorname{Frob}_p\in G_{\mathbb{Q}}$ denote an arithmetic Frobenius element at $p$. Both $\bar{\rho}_{f,\lambda}$ and $\sigma$ are unramified at every such prime. [guided] We first put the two representations that must be compared into the same ambient language. Let $K_f$ be the coefficient field of $f$, let $k_\lambda$ be the residue field at the prime $\lambda$ of $K_f$, let $G_{\mathbb{Q}}:=\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ be the absolute Galois group of $\mathbb{Q}$, and let $GL_2(k_\lambda)$ be the group of invertible $2\times 2$ matrices over $k_\lambda$. The residual representation attached to $f$ is a continuous map \begin{align*} \bar{\rho}_{f,\lambda}: G_{\mathbb{Q}} &\to GL_2(k_\lambda), \end{align*} normalized with arithmetic Frobenius, meaning that for every prime $p\nmid N\ell$ the element $\operatorname{Frob}_p$ has trace congruent to $a_p(f)$ and determinant congruent to $p$ modulo $\lambda$. This is the same Frobenius convention under which the mod $\ell$ cyclotomic character sends $\operatorname{Frob}_p$ to $p$ modulo $\ell$. The representation predicted by the theorem is the direct sum of the one-dimensional constant character and the mod $\ell$ cyclotomic character, so we define \begin{align*} \sigma: G_{\mathbb{Q}} &\to GL_2(k_\lambda) \\ g &\mapsto \begin{pmatrix} 1 & 0 \\ 0 & \bar{\chi}_\ell(g) \end{pmatrix}. \end{align*} This declaration matters because Brauer-Nesbitt applies to two finite-dimensional representations over the same field. For each prime $p\nmid N\ell$, let $\operatorname{Frob}_p\in G_{\mathbb{Q}}$ denote an arithmetic Frobenius element at $p$. The standard construction of the Galois representation attached to a weight $2$ eigenform gives that $\bar{\rho}_{f,\lambda}$ is unramified at such $p$, and the mod $\ell$ cyclotomic character is also unramified away from $\ell$. Hence both representations have well-defined characteristic polynomials at $\operatorname{Frob}_p$ for every prime $p\nmid N\ell$. [/guided] [/step] [step:Match traces and determinants at every prime away from $N\ell$] For every prime $p\nmid N\ell$, the defining property of the Galois representation attached to $f$ gives \begin{align*} \operatorname{tr}(\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)) &\equiv a_p(f) \pmod{\lambda}, \\ \det(\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)) &\equiv p \pmod{\lambda}, \end{align*} because $f$ has weight $2$ and nebentypus equal to $1$. The cyclotomic character satisfies \begin{align*} \bar{\chi}_\ell(\operatorname{Frob}_p) \equiv p \pmod{\ell}. \end{align*} Therefore \begin{align*} \operatorname{tr}(\sigma(\operatorname{Frob}_p)) &= 1+\bar{\chi}_\ell(\operatorname{Frob}_p) \equiv 1+p \pmod{\lambda}, \\ \det(\sigma(\operatorname{Frob}_p)) &= \bar{\chi}_\ell(\operatorname{Frob}_p) \equiv p \pmod{\lambda}. \end{align*} By the assumed Eisenstein congruence $a_p(f)\equiv 1+p\pmod{\lambda}$, the two representations have equal trace and determinant at $\operatorname{Frob}_p$. [guided] The next goal is to compare characteristic polynomials. For a two-dimensional representation, the characteristic polynomial of an element is determined by its trace and determinant, so we compute both. The standard Frobenius compatibility for the Galois representation attached to $f$ gives, for every prime $p\nmid N\ell$, \begin{align*} \operatorname{tr}(\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)) &\equiv a_p(f) \pmod{\lambda}, \\ \det(\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)) &\equiv p \pmod{\lambda}. \end{align*} The determinant formula is where the hypotheses on the weight and nebentypus enter: for a weight $2$ eigenform with nebentypus equal to $1$, the determinant of the associated $\lambda$-adic representation is the $\ell$-adic cyclotomic character, whose reduction sends arithmetic Frobenius at $p$ to $p$ modulo $\ell$. For the comparison representation $\sigma=1\oplus \bar{\chi}_\ell$, the matrix is diagonal by definition. Hence \begin{align*} \operatorname{tr}(\sigma(\operatorname{Frob}_p)) &= 1+\bar{\chi}_\ell(\operatorname{Frob}_p) \equiv 1+p \pmod{\lambda}, \\ \det(\sigma(\operatorname{Frob}_p)) &= \bar{\chi}_\ell(\operatorname{Frob}_p) \equiv p \pmod{\lambda}. \end{align*} The assumed congruence $a_p(f)\equiv 1+p\pmod{\lambda}$ now identifies the trace of $\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)$ with the trace of $\sigma(\operatorname{Frob}_p)$, and the determinant computation identifies the determinants. [/guided] [/step] [step:Convert matching traces and determinants into matching characteristic polynomials] Let $T$ be an indeterminate. For every prime $p\nmid N\ell$, the characteristic polynomial of $\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)$ is \begin{align*} T^2-\operatorname{tr}(\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p))T+\det(\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)), \end{align*} and the characteristic polynomial of $\sigma(\operatorname{Frob}_p)$ is \begin{align*} T^2-\operatorname{tr}(\sigma(\operatorname{Frob}_p))T+\det(\sigma(\operatorname{Frob}_p)). \end{align*} The trace and determinant equalities from the preceding step therefore imply equality of these two characteristic polynomials in $k_\lambda[T]$ for every prime $p\nmid N\ell$. [/step] [step:Apply Chebotarev density and Brauer-Nesbitt to identify semisimplifications] The set of primes $p\nmid N\ell$ has Dirichlet density $1$, since only finitely many primes divide $N\ell$. By the [Chebotarev Density Theorem](/page/Chebotarev%20Density%20Theorem), the Frobenius conjugacy classes $\operatorname{Frob}_p$ for these primes are dense in $G_{\mathbb{Q}}$ in the sense needed to test continuous semisimple representations by their characteristic polynomials on Frobenius elements. Since $k_\lambda$ is finite, the group $GL_2(k_\lambda)$ is finite; hence the continuous representations $\bar{\rho}_{f,\lambda}$ and $\sigma$ have finite image and factor through the finite quotient of $G_{\mathbb{Q}}$ by the intersection of their kernels. Their characteristic polynomial functions are therefore locally constant, and equality on the dense Frobenius set gives equality on this finite quotient. Applying the [Brauer-Nesbitt Theorem](/page/Brauer-Nesbitt%20Theorem) to the resulting two-dimensional representations over $k_\lambda$ gives \begin{align*} \bar{\rho}_{f,\lambda}^{\mathrm{ss}} \cong \sigma^{\mathrm{ss}}. \end{align*} The representation $\sigma=1\oplus \bar{\chi}_\ell$ is already a direct sum of one-dimensional representations, so $\sigma^{\mathrm{ss}}\cong 1\oplus \bar{\chi}_\ell$. Therefore \begin{align*} \bar{\rho}_{f,\lambda}^{\mathrm{ss}}\cong 1\oplus \bar{\chi}_\ell, \end{align*} as required. [guided] We now pass from equality on Frobenius elements to equality of semisimplified representations. The primes excluded by the condition $p\nmid N\ell$ are precisely the finitely many prime divisors of $N\ell$, so the set of remaining primes has Dirichlet density $1$. The [Chebotarev Density Theorem](/page/Chebotarev%20Density%20Theorem) says that Frobenius conjugacy classes are sufficiently dense in the absolute Galois group to determine continuous semisimple representations through their characteristic polynomials, once the equality is known on a density-one set of unramified primes. The hypotheses needed for this comparison are satisfied here. Both \begin{align*} \bar{\rho}_{f,\lambda}: G_{\mathbb{Q}} &\to GL_2(k_\lambda) \end{align*} and \begin{align*} \sigma: G_{\mathbb{Q}} &\to GL_2(k_\lambda) \end{align*} are continuous finite-dimensional representations over the same finite field $k_\lambda$. Because $k_\lambda$ is finite, $GL_2(k_\lambda)$ is finite. Thus both images are finite, and both representations factor through the finite quotient \begin{align*} G_{\mathbb{Q}}/(\ker \bar{\rho}_{f,\lambda}\cap \ker \sigma). \end{align*} This finite-quotient reduction is the point that makes the usual Brauer-Nesbitt comparison applicable over the finite coefficient field: the characteristic polynomial of each group element is a function on a finite group after quotienting. The preceding step proved that the characteristic polynomials agree at $\operatorname{Frob}_p$ for every prime $p\nmid N\ell$, hence on a density-one set of Frobenius classes. Chebotarev density then extends this equality to the finite quotient, and the [Brauer-Nesbitt Theorem](/page/Brauer-Nesbitt%20Theorem) identifies the semisimplifications: \begin{align*} \bar{\rho}_{f,\lambda}^{\mathrm{ss}} \cong \sigma^{\mathrm{ss}}. \end{align*} Finally, $\sigma$ was defined as the diagonal direct sum $1\oplus \bar{\chi}_\ell$, so its semisimplification is itself: \begin{align*} \sigma^{\mathrm{ss}}\cong 1\oplus \bar{\chi}_\ell. \end{align*} Combining the two isomorphisms gives \begin{align*} \bar{\rho}_{f,\lambda}^{\mathrm{ss}}\cong 1\oplus \bar{\chi}_\ell. \end{align*} This is exactly the asserted reducibility of the semisimplified residual representation. [/guided] [/step]