[proofplan] We begin with Deligne's $\lambda$-adic Galois representation attached to the newform $f$. Since $G_{\mathbb{Q}}$ is compact and the representation is continuous, its image stabilizes an $\mathcal{O}_{f,\lambda}$-lattice after conjugation. Reducing the action on this stable lattice modulo $\lambda$ gives a two-dimensional representation over $k_\lambda$, and taking its semisimplification gives the required semisimple residual representation. The trace, determinant, and ramification properties are inherited by reducing Deligne's Frobenius characteristic polynomial modulo $\lambda$. [/proofplan]

[step:Start from Deligne's $\lambda$-adic representation attached to $f$]Let \begin{align*} \rho_{f,\lambda}:G_{\mathbb{Q}}\longrightarrow GL_2(K_{f,\lambda}) \end{align*} be Deligne's $\lambda$-adic representation attached to the normalized cuspidal newform $f$ (citing a result not yet in the wiki: Deligne's Galois representation attached to a newform). Thus $\rho_{f,\lambda}$ is continuous, unramified outside the primes dividing $N\ell$, and for every rational prime $p\nmid N\ell$ one has \begin{align*} \operatorname{tr}\rho_{f,\lambda}(\operatorname{Frob}_p)&=a_p(f),\\ \det\rho_{f,\lambda}(\operatorname{Frob}_p)&=\varepsilon(p)p^{k-1}. \end{align*} Equivalently, the characteristic polynomial of $\rho_{f,\lambda}(\operatorname{Frob}_p)$ is \begin{align*} X^2-a_p(f)X+\varepsilon(p)p^{k-1}\in \mathcal{O}_{f,\lambda}[X]. \end{align*} The coefficients lie in $\mathcal{O}_{f,\lambda}$ because Fourier coefficients of normalized newforms are algebraic integers and $\varepsilon(p)$ is a root of unity.[/step]

[guided]The input is the standard $\lambda$-adic representation constructed by Deligne. In the present notation, this is a continuous group homomorphism \begin{align*} \rho_{f,\lambda}:G_{\mathbb{Q}}\longrightarrow GL_2(K_{f,\lambda}), \end{align*} where $K_{f,\lambda}$ is the local field obtained by completing the coefficient field $K_f$ at the prime $\lambda$. Deligne's theorem gives exactly the two properties we need before reduction: first, $\rho_{f,\lambda}$ is unramified outside the primes dividing $N\ell$; second, for every rational prime $p\nmid N\ell$, the arithmetic Frobenius element $\operatorname{Frob}_p$ satisfies \begin{align*} \operatorname{tr}\rho_{f,\lambda}(\operatorname{Frob}_p)&=a_p(f),\\ \det\rho_{f,\lambda}(\operatorname{Frob}_p)&=\varepsilon(p)p^{k-1}. \end{align*} Therefore the characteristic polynomial of $\rho_{f,\lambda}(\operatorname{Frob}_p)$ is \begin{align*} X^2-a_p(f)X+\varepsilon(p)p^{k-1}. \end{align*} Why can this polynomial be reduced modulo $\lambda$? The coefficient $a_p(f)$ is an algebraic integer because $f$ is a normalized Hecke eigenform, and $\varepsilon(p)$ is a root of unity, hence also an algebraic integer. Since $p\nmid \ell$, the element $p^{k-1}$ is a unit in $\mathcal{O}_{f,\lambda}$. Thus both coefficients belong to $\mathcal{O}_{f,\lambda}$, so reduction modulo the maximal ideal $\lambda\mathcal{O}_{f,\lambda}$ is defined.[/guided]

[step:Choose a $G_{\mathbb{Q}}$-stable lattice in the $\lambda$-adic representation space]Let $V_\lambda=K_{f,\lambda}^2$ denote the two-dimensional $K_{f,\lambda}$-[vector space](/page/Vector%20Space) on which $G_{\mathbb{Q}}$ acts through $\rho_{f,\lambda}$. Since $G_{\mathbb{Q}}$ is profinite, it is compact; since $\rho_{f,\lambda}$ is continuous, the image \begin{align*} H_\lambda:=\rho_{f,\lambda}(G_{\mathbb{Q}})\subset GL_2(K_{f,\lambda}) \end{align*} is compact. A compact subgroup of $GL_2(K_{f,\lambda})$ stabilizes an $\mathcal{O}_{f,\lambda}$-lattice in $V_\lambda$ (citing a result not yet in the wiki: compact subgroups of $GL_n$ over a non-archimedean local field stabilize lattices). Hence there exists a free $\mathcal{O}_{f,\lambda}$-submodule \begin{align*} \Lambda\subset V_\lambda \end{align*} of rank $2$, spanning $V_\lambda$ over $K_{f,\lambda}$, such that \begin{align*} \rho_{f,\lambda}(\sigma)(\Lambda)=\Lambda \end{align*} for every $\sigma\in G_{\mathbb{Q}}$.[/step]

[guided]To reduce a representation modulo $\lambda$, we need matrices whose entries lie in the valuation ring $\mathcal{O}_{f,\lambda}$. A representation into $GL_2(K_{f,\lambda})$ does not initially provide such matrices. The standard way to create integral matrices is to choose a stable lattice. Let $V_\lambda=K_{f,\lambda}^2$ be the representation space. An $\mathcal{O}_{f,\lambda}$-lattice in $V_\lambda$ means a free $\mathcal{O}_{f,\lambda}$-submodule \begin{align*} \Lambda\subset V_\lambda \end{align*} of rank $2$ such that the natural scalar extension map \begin{align*} K_{f,\lambda}\otimes_{\mathcal{O}_{f,\lambda}}\Lambda\longrightarrow V_\lambda \end{align*} is an isomorphism. We now verify why a stable lattice exists. The absolute [Galois group](/page/Galois%20Group) $G_{\mathbb{Q}}$ is profinite, hence compact. The representation \begin{align*} \rho_{f,\lambda}:G_{\mathbb{Q}}\longrightarrow GL_2(K_{f,\lambda}) \end{align*} is continuous, so its image \begin{align*} H_\lambda:=\rho_{f,\lambda}(G_{\mathbb{Q}}) \end{align*} is a compact subgroup of $GL_2(K_{f,\lambda})$. The lattice-stabilization theorem for compact subgroups of $GL_n$ over a non-archimedean local field says that such a compact subgroup stabilizes an $\mathcal{O}_{f,\lambda}$-lattice. Applying it with $n=2$ gives a lattice $\Lambda\subset V_\lambda$ satisfying \begin{align*} \rho_{f,\lambda}(\sigma)(\Lambda)=\Lambda \end{align*} for every $\sigma\in G_{\mathbb{Q}}$. This is the integrality step: once the action preserves $\Lambda$, the matrix of each $\rho_{f,\lambda}(\sigma)$ in an $\mathcal{O}_{f,\lambda}$-basis of $\Lambda$ lies in $GL_2(\mathcal{O}_{f,\lambda})$.[/guided]

[step:Reduce the stable lattice modulo $\lambda$ and semisimplify]Define the two-dimensional $k_\lambda$-vector space \begin{align*} \overline{\Lambda}:=\Lambda/\lambda\Lambda. \end{align*} Because $\Lambda$ is stable under $G_{\mathbb{Q}}$, the action of $G_{\mathbb{Q}}$ on $\Lambda$ descends to a representation \begin{align*} \tilde{\rho}_{f,\lambda}:G_{\mathbb{Q}}&\longrightarrow GL(\overline{\Lambda})\cong GL_2(k_\lambda),\\ \sigma&\longmapsto \left(v+\lambda\Lambda\mapsto \rho_{f,\lambda}(\sigma)v+\lambda\Lambda\right). \end{align*} The representation $\tilde{\rho}_{f,\lambda}$ is continuous because it is the composite of the continuous integral representation \begin{align*} G_{\mathbb{Q}}\longrightarrow GL_2(\mathcal{O}_{f,\lambda}) \end{align*} with the reduction map \begin{align*} GL_2(\mathcal{O}_{f,\lambda})\longrightarrow GL_2(k_\lambda), \end{align*} whose target is finite and discrete. Let \begin{align*} \bar{\rho}_{f,\lambda}:G_{\mathbb{Q}}\longrightarrow GL_2(k_\lambda) \end{align*} be the semisimplification of $\tilde{\rho}_{f,\lambda}$.[/step]

[guided]The lattice $\Lambda$ is a free rank-$2$ module over the valuation ring $\mathcal{O}_{f,\lambda}$. Reducing it modulo the maximal ideal $\lambda\mathcal{O}_{f,\lambda}$ gives \begin{align*} \overline{\Lambda}:=\Lambda/\lambda\Lambda, \end{align*} which is a two-dimensional vector space over the residue field \begin{align*} k_\lambda=\mathcal{O}_{f,\lambda}/\lambda\mathcal{O}_{f,\lambda}. \end{align*} Because the lattice is stable, the action of $G_{\mathbb{Q}}$ preserves the submodule $\lambda\Lambda$. Therefore the rule \begin{align*} v+\lambda\Lambda\longmapsto \rho_{f,\lambda}(\sigma)v+\lambda\Lambda \end{align*} is well-defined for every $\sigma\in G_{\mathbb{Q}}$. This defines a representation \begin{align*} \tilde{\rho}_{f,\lambda}:G_{\mathbb{Q}}&\longrightarrow GL(\overline{\Lambda})\cong GL_2(k_\lambda),\\ \sigma&\longmapsto \left(v+\lambda\Lambda\mapsto \rho_{f,\lambda}(\sigma)v+\lambda\Lambda\right). \end{align*} Continuity follows from the construction. After choosing an $\mathcal{O}_{f,\lambda}$-basis of $\Lambda$, the action gives a continuous homomorphism \begin{align*} G_{\mathbb{Q}}\longrightarrow GL_2(\mathcal{O}_{f,\lambda}). \end{align*} Composing with the reduction homomorphism \begin{align*} GL_2(\mathcal{O}_{f,\lambda})\longrightarrow GL_2(k_\lambda) \end{align*} gives $\tilde{\rho}_{f,\lambda}$. Since $k_\lambda$ is finite, $GL_2(k_\lambda)$ is finite and carries the discrete topology, so the composite is continuous. The representation $\tilde{\rho}_{f,\lambda}$ need not be semisimple. To obtain the residual representation appearing in the theorem, we define \begin{align*} \bar{\rho}_{f,\lambda}:G_{\mathbb{Q}}\longrightarrow GL_2(k_\lambda) \end{align*} to be its semisimplification. Semisimplification preserves the trace and determinant of each group element, because it replaces a representation by the direct sum of its Jordan-Hölder constituents while keeping the same characteristic polynomial for every acting operator.[/guided]

[step:Reduce Deligne's Frobenius formulas modulo $\lambda$]Let $p\nmid N\ell$ be a rational prime. Since $\rho_{f,\lambda}$ is unramified at $p$, the inertia subgroup at $p$ acts as the identity on $V_\lambda$; hence its induced action on $\overline{\Lambda}$ is also the identity, so $\tilde{\rho}_{f,\lambda}$ is unramified at $p$. Moreover, the matrix of $\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)$ is obtained by reducing the matrix of $\rho_{f,\lambda}(\operatorname{Frob}_p)$ modulo $\lambda$ in an $\mathcal{O}_{f,\lambda}$-basis of $\Lambda$. Therefore \begin{align*} \operatorname{tr}\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv \operatorname{tr}\rho_{f,\lambda}(\operatorname{Frob}_p)\pmod{\lambda},\\ \det\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv \det\rho_{f,\lambda}(\operatorname{Frob}_p)\pmod{\lambda}. \end{align*} Using Deligne's formulas gives \begin{align*} \operatorname{tr}\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv a_p(f)\pmod{\lambda},\\ \det\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv \varepsilon(p)p^{k-1}\pmod{\lambda}. \end{align*} Since semisimplification preserves trace and determinant, the same congruences hold for $\bar{\rho}_{f,\lambda}$: \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 \varepsilon(p)p^{k-1}\pmod{\lambda}. \end{align*}[/step]

[guided]Fix a rational prime $p\nmid N\ell$. Deligne's representation is unramified at $p$, so the inertia subgroup at $p$ acts as the identity on $V_\lambda$. Since the reduced representation is obtained from the same action after passing to $\Lambda/\lambda\Lambda$, the inertia subgroup also acts as the identity on $\overline{\Lambda}$. Hence $\tilde{\rho}_{f,\lambda}$ is unramified at $p$. Now choose an $\mathcal{O}_{f,\lambda}$-basis of the stable lattice $\Lambda$. In this basis, the matrix of \begin{align*} \rho_{f,\lambda}(\operatorname{Frob}_p) \end{align*} lies in $GL_2(\mathcal{O}_{f,\lambda})$, and the matrix of \begin{align*} \tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p) \end{align*} is obtained by reducing each entry modulo $\lambda$. Trace and determinant are polynomial functions in the matrix entries, so reduction modulo $\lambda$ commutes with taking trace and determinant. Therefore \begin{align*} \operatorname{tr}\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv \operatorname{tr}\rho_{f,\lambda}(\operatorname{Frob}_p)\pmod{\lambda},\\ \det\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv \det\rho_{f,\lambda}(\operatorname{Frob}_p)\pmod{\lambda}. \end{align*} Substituting Deligne's formulas gives \begin{align*} \operatorname{tr}\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv a_p(f)\pmod{\lambda},\\ \det\tilde{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv \varepsilon(p)p^{k-1}\pmod{\lambda}. \end{align*} Finally, $\bar{\rho}_{f,\lambda}$ is the semisimplification of $\tilde{\rho}_{f,\lambda}$. Passing to semisimplification does not change the trace or determinant of any group element: after choosing a composition series, the acting matrix becomes block upper triangular, and its semisimplification has the same diagonal blocks. Trace and determinant depend only on these diagonal blocks. Hence \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 \varepsilon(p)p^{k-1}\pmod{\lambda}. \end{align*}[/guided]

[step:Inherit the ramification bound and conclude the construction] For every rational prime $p\nmid N\ell$, the representation $\rho_{f,\lambda}$ is unramified at $p$. Hence the inertia subgroup at $p$ acts as the identity on $\Lambda$ and therefore on $\Lambda/\lambda\Lambda$. Thus $\tilde{\rho}_{f,\lambda}$ is unramified at $p$, and its semisimplification $\bar{\rho}_{f,\lambda}$ is also unramified at $p$. Consequently $\bar{\rho}_{f,\lambda}$ is unramified outside the primes dividing $N\ell$. Together with continuity, semisimplicity by construction, and the Frobenius trace and determinant congruences proved above, this is the desired residual representation attached to $f$ modulo $\lambda$. [/step]