[proofplan] Let $\mathcal{O}_K$ be the valuation ring of $K$, let $\lambda$ be the chosen uniformizer, let $k := \mathcal{O}_K/\lambda\mathcal{O}_K$ be the residue field, and let $k[T]_{\le n}$ denote the finite set of polynomials in $k[T]$ of degree at most $n$. The proof first observes that the residual characteristic polynomial functions are continuous maps from the profinite group $G$ to the finite discrete set $k[T]_{\le n}$, so equality on a [dense subset](/page/Dense%20Subset) implies equality on all of $G$. Since the residual images are finite, both residual representations factor through a finite quotient of $G$. We then apply the finite-group Brauer–Nesbitt theorem over the finite residue field $k$, which says that finite-dimensional semisimple representations are determined by the characteristic polynomials of all group elements. This identifies the semisimplifications; the fact that the lattices lie inside the same $K$-representation is used only to produce the two residual representations with the stated characteristic-polynomial hypothesis, and the final argument needs only that hypothesis. [/proofplan] [step:Define the residual characteristic polynomial functions] For each $i \in \{1,2\}$, define the residual characteristic polynomial function \begin{align*} c_i: G &\to k[T]_{\le n} \\ g &\mapsto \det(TI_n - \bar{\rho}_i(g)), \end{align*} where $k[T]_{\le n}$ denotes the finite set of polynomials in $k[T]$ of degree at most $n$, and $I_n$ denotes the identity matrix after choosing any $k$-basis of the $n$-dimensional [vector space](/page/Vector%20Space) $L_i/\lambda L_i$. The characteristic polynomial is independent of this choice of basis. Let $GL_{\mathcal{O}_K}(L_i)$ denote the group of $\mathcal{O}_K$-linear automorphisms of the lattice $L_i$. The map $\bar{\rho}_i$ is continuous because it is the composition of the continuous action of $G$ on the stable lattice $L_i$, the restriction map \begin{align*} G &\to GL_{\mathcal{O}_K}(L_i), \end{align*} and the reduction modulo $\lambda$ map \begin{align*} GL_{\mathcal{O}_K}(L_i) &\to GL_k(L_i/\lambda L_i). \end{align*} The reduction map is continuous, and $GL_k(L_i/\lambda L_i)$ is finite with the discrete topology. The characteristic polynomial map from $GL_k(L_i/\lambda L_i)$ to $k[T]_{\le n}$ is a map between finite discrete sets. Therefore $c_i$ is continuous. [guided] For each stable lattice $L_i$, the action of $G$ on $L_i$ descends to an action on the finite-dimensional $k$-vector space $L_i/\lambda L_i$. This gives the residual representation \begin{align*} \bar{\rho}_i: G &\to GL_k(L_i/\lambda L_i). \end{align*} We package the characteristic polynomial of each residual operator into a function \begin{align*} c_i: G &\to k[T]_{\le n} \\ g &\mapsto \det(TI_n - \bar{\rho}_i(g)). \end{align*} Here $k[T]_{\le n}$ is the finite set of polynomials over $k$ of degree at most $n$, and $I_n$ denotes the identity matrix after choosing any $k$-basis of the $n$-dimensional vector space $L_i/\lambda L_i$. This choice does not affect the determinant, because characteristic polynomials are invariant under change of basis. The continuity point matters because the hypothesis only gives equality on a dense subset. Let $GL_{\mathcal{O}_K}(L_i)$ denote the group of $\mathcal{O}_K$-linear automorphisms of the lattice $L_i$. Since $L_i$ is $G$-stable, the representation $\rho$ restricts to a continuous homomorphism \begin{align*} G &\to GL_{\mathcal{O}_K}(L_i). \end{align*} The residual representation is obtained by composing this restricted lattice action with the reduction modulo $\lambda$ map \begin{align*} GL_{\mathcal{O}_K}(L_i) &\to GL_k(L_i/\lambda L_i). \end{align*} The reduction map is continuous, and its target is finite with the discrete topology. Composing with the characteristic polynomial map gives a continuous function $c_i: G \to k[T]_{\le n}$. [/guided] [/step] [step:Extend equality of characteristic polynomials from the dense subset to all of $G$] By hypothesis, $c_1(g) = c_2(g)$ for every $g \in D$. Since $k[T]_{\le n}$ is finite and discrete, the set \begin{align*} E := \{g \in G : c_1(g) = c_2(g)\} \end{align*} is closed in $G$. The subset $D$ is contained in $E$, and $D$ is dense in $G$, hence $E = G$. Therefore \begin{align*} \det(TI_n - \bar{\rho}_1(g)) = \det(TI_n - \bar{\rho}_2(g)) \end{align*} in $k[T]$ for every $g \in G$. [guided] We want to replace equality on the dense subset $D$ by equality everywhere. Define \begin{align*} E := \{g \in G : c_1(g) = c_2(g)\}. \end{align*} The set $E$ is closed because it is the preimage of the diagonal \begin{align*} \Delta := \{(P,P) : P \in k[T]_{\le n}\} \end{align*} under the continuous map \begin{align*} G &\to k[T]_{\le n} \times k[T]_{\le n} \\ g &\mapsto (c_1(g),c_2(g)). \end{align*} The codomain is finite and discrete, so every subset of it is closed. The hypothesis says $D \subset E$. Since $D$ is dense in $G$ and $E$ is closed, the closure of $D$ is contained in $E$, hence $G \subset E$. Therefore $E = G$, which means that for every $g \in G$, \begin{align*} \det(TI_n - \bar{\rho}_1(g)) = \det(TI_n - \bar{\rho}_2(g)) \end{align*} as polynomials in $k[T]$. [/guided] [/step] [step:Pass to a finite quotient carrying both residual representations] For each $i \in \{1,2\}$, the image of $\bar{\rho}_i$ is finite because $GL_k(L_i/\lambda L_i)$ is finite. Define the closed [normal subgroup](/page/Normal%20Subgroup) \begin{align*} N := \ker(\bar{\rho}_1) \cap \ker(\bar{\rho}_2) \subset G. \end{align*} Then $G/N$ is a finite group, and both $\bar{\rho}_1$ and $\bar{\rho}_2$ factor through representations \begin{align*} \tilde{\rho}_i: G/N &\to GL_k(L_i/\lambda L_i) \end{align*} satisfying $\bar{\rho}_i = \tilde{\rho}_i \circ q$, where \begin{align*} q: G &\to G/N \end{align*} is the quotient map. The equality of characteristic polynomials for all $g \in G$ implies that, for every $h \in G/N$, \begin{align*} \det(TI_n - \tilde{\rho}_1(h)) = \det(TI_n - \tilde{\rho}_2(h)). \end{align*} Indeed, choose any $g \in G$ with $q(g)=h$ and use $\bar{\rho}_i(g)=\tilde{\rho}_i(h)$. [guided] The residual representations have finite image. This is because $k$ is a finite field and $L_i/\lambda L_i$ is an $n$-dimensional $k$-vector space, so $GL_k(L_i/\lambda L_i)$ is finite. Define \begin{align*} N := \ker(\bar{\rho}_1) \cap \ker(\bar{\rho}_2). \end{align*} This subgroup is normal because it is an intersection of kernels of group homomorphisms. It is closed because both residual representations are continuous and their targets are discrete. Since both residual images are finite, the quotient $G/N$ embeds into the finite product of the two residual image groups, so $G/N$ is finite. Let \begin{align*} q: G &\to G/N \end{align*} be the quotient map. Since $N$ is contained in both kernels, each $\bar{\rho}_i$ factors through a representation \begin{align*} \tilde{\rho}_i: G/N &\to GL_k(L_i/\lambda L_i) \end{align*} with $\bar{\rho}_i = \tilde{\rho}_i \circ q$. Now take any $h \in G/N$. Choose $g \in G$ such that $q(g)=h$. The equality already proved on $G$ gives \begin{align*} \det(TI_n - \bar{\rho}_1(g)) = \det(TI_n - \bar{\rho}_2(g)). \end{align*} Using $\bar{\rho}_i(g)=\tilde{\rho}_i(h)$, this becomes \begin{align*} \det(TI_n - \tilde{\rho}_1(h)) = \det(TI_n - \tilde{\rho}_2(h)). \end{align*} Thus the two finite-group representations have the same characteristic polynomial at every element of $G/N$. [/guided] [/step] [step:Apply Brauer–Nesbitt to identify the semisimplifications] Since $k = \mathcal{O}_K/\lambda\mathcal{O}_K$ is a finite field, we may use the finite-group Brauer–Nesbitt theorem over $k$ in the form needed here: if $H$ is a finite group and $V,W$ are finite-dimensional semisimple $k[H]$-modules whose representing operators have the same characteristic polynomial for every element of $H$, then $V$ and $W$ are isomorphic as $k[H]$-modules. Apply this result with $H := G/N$ and with $V_i^{\mathrm{ss}}$ denoting the semisimplification, as a $k[G/N]$-module, of $V_i := L_i/\lambda L_i$. The corresponding semisimple representations are \begin{align*} \tilde{\rho}_1^{\mathrm{ss}}: G/N &\to GL_k(V_1^{\mathrm{ss}}), \\ \tilde{\rho}_2^{\mathrm{ss}}: G/N &\to GL_k(V_2^{\mathrm{ss}}). \end{align*} Semisimplification preserves the multiset of Jordan–Hölder factors, hence it preserves the characteristic polynomial of every group element. Therefore the characteristic polynomial equality for $\tilde{\rho}_1$ and $\tilde{\rho}_2$ also holds for $\tilde{\rho}_1^{\mathrm{ss}}$ and $\tilde{\rho}_2^{\mathrm{ss}}$. Brauer–Nesbitt gives \begin{align*} \tilde{\rho}_1^{\mathrm{ss}} \cong \tilde{\rho}_2^{\mathrm{ss}}. \end{align*} Pulling this isomorphism back along $q: G \to G/N$ gives \begin{align*} \bar{\rho}_1^{\mathrm{ss}} \cong \bar{\rho}_2^{\mathrm{ss}} \end{align*} as representations of $G$. This is the desired conclusion. [guided] The remaining input is Brauer–Nesbitt over the residue field $k$. Since $k = \mathcal{O}_K/\lambda\mathcal{O}_K$ is finite, the finite-group form over $k$ applies: if $H$ is a finite group and $V,W$ are finite-dimensional semisimple $k[H]$-modules whose representing operators have the same characteristic polynomial at every element of $H$, then $V$ and $W$ are isomorphic as $k[H]$-modules. We apply it with $H := G/N$. The representations \begin{align*} \tilde{\rho}_1: G/N &\to GL_k(L_1/\lambda L_1), \\ \tilde{\rho}_2: G/N &\to GL_k(L_2/\lambda L_2) \end{align*} need not be semisimple. Let $V_i := L_i/\lambda L_i$ be regarded as a $k[G/N]$-module through $\tilde{\rho}_i$, and let $V_i^{\mathrm{ss}}$ denote its semisimplification as a $k[G/N]$-module. The induced semisimple representations are \begin{align*} \tilde{\rho}_1^{\mathrm{ss}}: G/N &\to GL_k(V_1^{\mathrm{ss}}), \\ \tilde{\rho}_2^{\mathrm{ss}}: G/N &\to GL_k(V_2^{\mathrm{ss}}). \end{align*} The characteristic polynomial of a group element is unchanged by semisimplification. This follows from choosing a composition series: the operator induced by a group element is block upper triangular with diagonal blocks acting on the composition factors, so its characteristic polynomial is the product of the characteristic polynomials of those diagonal blocks. The semisimplification has exactly the same diagonal blocks, now arranged as a direct sum, and therefore the same product. Thus, for every $h \in G/N$, \begin{align*} \det(TI_n - \tilde{\rho}_1^{\mathrm{ss}}(h)) = \det(TI_n - \tilde{\rho}_2^{\mathrm{ss}}(h)). \end{align*} Brauer–Nesbitt applies and yields \begin{align*} \tilde{\rho}_1^{\mathrm{ss}} \cong \tilde{\rho}_2^{\mathrm{ss}}. \end{align*} Finally, the original residual representations are obtained from the $\tilde{\rho}_i$ by pullback along \begin{align*} q: G &\to G/N. \end{align*} Pulling the isomorphism of $G/N$-representations back along $q$ gives an isomorphism of $G$-representations: \begin{align*} \bar{\rho}_1^{\mathrm{ss}} \cong \bar{\rho}_2^{\mathrm{ss}}. \end{align*} This proves the theorem. The conclusion is only about semisimplifications because Brauer–Nesbitt determines the semisimple representation, not the extension classes between its simple constituents. [/guided] [/step]