[proofplan] Choose one finite set of primes outside which both representations are unramified and the Frobenius trace equality is known. Pass from $G_{\mathbb{Q}}$ to the quotient $G_{\mathbb{Q},S}$ classifying extensions unramified outside that set; the unramifiedness hypothesis makes both representations factor through this quotient. Chebotarev density says that Frobenius conjugacy classes at primes outside $S$ are dense in $G_{\mathbb{Q},S}$, so the descended trace difference vanishes everywhere by continuity. Pulling back to $G_{\mathbb{Q}}$ gives equality of characters, and Brauer–Nesbitt identifies the two semisimple representations. [/proofplan] [step:Enlarge the exceptional set so all relevant Frobenius classes are available] Let $S$ be a finite set of rational primes containing $T$ and containing every prime at which at least one of $\rho_1$ or $\rho_2$ is ramified. Then for every rational prime $p\notin S$, both representations are unramified at $p$, and the hypothesis gives \begin{align*} \operatorname{tr}\rho_1(\operatorname{Frob}_p)=\operatorname{tr}\rho_2(\operatorname{Frob}_p). \end{align*} Because trace is invariant under conjugation in $GL_n(K)$, this equality is independent of the representative chosen from the Frobenius conjugacy class. [/step] [step:Convert Frobenius equality into equality of continuous trace functions] Let $\mathbb{Q}_S$ denote the maximal algebraic extension of $\mathbb{Q}$ unramified outside $S$, and let \begin{align*} \pi_S: G_{\mathbb{Q}}&\longrightarrow G_{\mathbb{Q},S}:=\operatorname{Gal}(\mathbb{Q}_S/\mathbb{Q}) \end{align*} be the natural continuous quotient homomorphism. Equivalently, $\ker \pi_S$ is the closed [normal subgroup](/page/Normal%20Subgroup) generated by the inertia subgroups at rational primes $p\notin S$. Since each $\rho_i$ is unramified outside $S$, each $\rho_i$ is trivial on those inertia subgroups and therefore on their closed normal closure. Hence there are unique continuous representations \begin{align*} \bar{\rho}_i: G_{\mathbb{Q},S}&\longrightarrow GL_n(K) \end{align*} such that $\rho_i=\bar{\rho}_i\circ \pi_S$ for $i\in\{1,2\}$. Define the descended trace functions \begin{align*} \bar{\chi}_i: G_{\mathbb{Q},S}&\longrightarrow K\\ \gamma&\longmapsto \operatorname{tr}\bar{\rho}_i(\gamma) \end{align*} for $i\in\{1,2\}$, and define their difference \begin{align*} \bar{\chi}: G_{\mathbb{Q},S}&\longrightarrow K\\ \gamma&\longmapsto \bar{\chi}_1(\gamma)-\bar{\chi}_2(\gamma). \end{align*} Since each $\bar{\rho}_i$ is continuous and the trace map $GL_n(K)\to K$ is continuous for the $l$-adic topology, the functions $\bar{\chi}_1$, $\bar{\chi}_2$, and $\bar{\chi}$ are continuous. The equality from the previous step says that \begin{align*} \bar{\chi}(\operatorname{Frob}_p)=0 \end{align*} for every rational prime $p\notin S$, where $\operatorname{Frob}_p$ denotes the Frobenius conjugacy class in $G_{\mathbb{Q},S}$. The function $\bar{\chi}$ is a continuous class function, because for all $\gamma,\delta\in G_{\mathbb{Q},S}$ and $i\in\{1,2\}$, \begin{align*} \bar{\chi}_i(\delta\gamma\delta^{-1}) &=\operatorname{tr}\bigl(\bar{\rho}_i(\delta)\bar{\rho}_i(\gamma)\bar{\rho}_i(\delta)^{-1}\bigr)\\ &=\operatorname{tr}\bar{\rho}_i(\gamma)\\ &=\bar{\chi}_i(\gamma), \end{align*} using invariance of trace under conjugation in $GL_n(K)$. By Chebotarev's density theorem for the maximal extension of $\mathbb{Q}$ unramified outside $S$ (citing a result not yet in the wiki: Chebotarev density theorem), the union of Frobenius conjugacy classes $\operatorname{Frob}_p$ with $p\notin S$ is dense in $G_{\mathbb{Q},S}$. Since $\bar{\chi}$ vanishes on this [dense subset](/page/Dense%20Subset) and $\{0\}\subset K$ is closed, continuity implies \begin{align*} \bar{\chi}(\gamma)=0 \end{align*} for every $\gamma\in G_{\mathbb{Q},S}$. Finally define \begin{align*} \chi_i: G_{\mathbb{Q}}&\longrightarrow K\\ g&\longmapsto \operatorname{tr}\rho_i(g) \end{align*} for $i\in\{1,2\}$. Since $\rho_i=\bar{\rho}_i\circ \pi_S$, for every $g\in G_{\mathbb{Q}}$ we have \begin{align*} \chi_1(g)-\chi_2(g) =\bar{\chi}(\pi_S(g)) =0. \end{align*} Thus \begin{align*} \operatorname{tr}\rho_1(g)=\operatorname{tr}\rho_2(g) \end{align*} for every $g\in G_{\mathbb{Q}}$. [guided] We now turn equality at almost all Frobenius elements into equality everywhere. The important point is that Chebotarev is a statement about a [Galois group](/page/Galois%20Group) unramified outside the chosen finite set, so we first pass to that quotient. Let $\mathbb{Q}_S$ denote the maximal algebraic extension of $\mathbb{Q}$ unramified outside $S$, and let \begin{align*} \pi_S: G_{\mathbb{Q}}&\longrightarrow G_{\mathbb{Q},S}:=\operatorname{Gal}(\mathbb{Q}_S/\mathbb{Q}) \end{align*} be the natural continuous quotient homomorphism. Its kernel is the closed normal subgroup generated by the inertia subgroups at rational primes $p\notin S$. Why is this the right quotient? Because a representation unramified outside $S$ is precisely insensitive to inertia at all primes outside $S$. Since each $\rho_i$ is unramified outside $S$, it is trivial on those inertia subgroups, and hence trivial on their closed normal closure. Therefore there are unique continuous representations \begin{align*} \bar{\rho}_i: G_{\mathbb{Q},S}&\longrightarrow GL_n(K) \end{align*} such that \begin{align*} \rho_i=\bar{\rho}_i\circ \pi_S \end{align*} for $i\in\{1,2\}$. Define the descended trace functions \begin{align*} \bar{\chi}_i: G_{\mathbb{Q},S}&\longrightarrow K\\ \gamma&\longmapsto \operatorname{tr}\bar{\rho}_i(\gamma) \end{align*} for $i\in\{1,2\}$, and define their difference \begin{align*} \bar{\chi}: G_{\mathbb{Q},S}&\longrightarrow K\\ \gamma&\longmapsto \bar{\chi}_1(\gamma)-\bar{\chi}_2(\gamma). \end{align*} The maps $\bar{\rho}_i$ are continuous by construction from the quotient, and the trace map $GL_n(K)\to K$ is continuous in the $l$-adic topology. Hence $\bar{\chi}_1$, $\bar{\chi}_2$, and $\bar{\chi}$ are continuous. For every $p\notin S$, the prime $p$ is unramified in $\mathbb{Q}_S/\mathbb{Q}$, so it has a Frobenius conjugacy class $\operatorname{Frob}_p$ in $G_{\mathbb{Q},S}$. The equality from the hypothesis, rewritten through the factorizations $\rho_i=\bar{\rho}_i\circ \pi_S$, gives \begin{align*} \bar{\chi}(\operatorname{Frob}_p) = \operatorname{tr}\bar{\rho}_1(\operatorname{Frob}_p) - \operatorname{tr}\bar{\rho}_2(\operatorname{Frob}_p) =0. \end{align*} Trace is invariant under conjugation, so this value is independent of the chosen representative of the Frobenius conjugacy class. We verify the class-function condition needed to use the density of conjugacy classes. For every $\gamma,\delta\in G_{\mathbb{Q},S}$ and each $i\in\{1,2\}$, \begin{align*} \bar{\chi}_i(\delta\gamma\delta^{-1}) &=\operatorname{tr}\bar{\rho}_i(\delta\gamma\delta^{-1})\\ &=\operatorname{tr}\bigl(\bar{\rho}_i(\delta)\bar{\rho}_i(\gamma)\bar{\rho}_i(\delta)^{-1}\bigr)\\ &=\operatorname{tr}\bar{\rho}_i(\gamma)\\ &=\bar{\chi}_i(\gamma), \end{align*} where the third equality is invariance of trace under conjugation in $GL_n(K)$. Thus $\bar{\chi}=\bar{\chi}_1-\bar{\chi}_2$ is a continuous class function on $G_{\mathbb{Q},S}$. The density input is Chebotarev's density theorem for the maximal extension of $\mathbb{Q}$ unramified outside $S$ (citing a result not yet in the wiki: Chebotarev density theorem). In the precise form used here, it says that the union of Frobenius conjugacy classes $\operatorname{Frob}_p$ with $p\notin S$ is dense in the profinite group $G_{\mathbb{Q},S}$. Since $\bar{\chi}$ is continuous, vanishes on that dense union, and $\{0\}$ is closed in the Hausdorff topological field $K$, continuity forces \begin{align*} \bar{\chi}(\gamma)=0 \end{align*} for every $\gamma\in G_{\mathbb{Q},S}$. Finally we pull the equality back from the quotient to the original absolute Galois group. Define \begin{align*} \chi_i: G_{\mathbb{Q}}&\longrightarrow K\\ g&\longmapsto \operatorname{tr}\rho_i(g) \end{align*} for $i\in\{1,2\}$. For every $g\in G_{\mathbb{Q}}$, \begin{align*} \chi_1(g)-\chi_2(g) &=\operatorname{tr}\bar{\rho}_1(\pi_S(g))-\operatorname{tr}\bar{\rho}_2(\pi_S(g))\\ &=\bar{\chi}(\pi_S(g))\\ &=0. \end{align*} Therefore \begin{align*} \operatorname{tr}\rho_1(g)=\operatorname{tr}\rho_2(g) \end{align*} for every $g\in G_{\mathbb{Q}}$. [/guided] [/step] [step:Apply Brauer–Nesbitt to recover the semisimple representation from its character] View $\rho_i$ as a representation of $G_{\mathbb{Q}}$ on the $K$-[vector space](/page/Vector%20Space) $V_i=K^n$ for $i\in\{1,2\}$. The previous step proves that the two characters \begin{align*} g\longmapsto \operatorname{tr}\rho_i(g) \end{align*} are equal as functions $G_{\mathbb{Q}}\to K$. Since $K$ has characteristic $0$ and both representations are finite-dimensional and semisimple, the Brauer–Nesbitt theorem (citing a result not yet in the wiki: Brauer–Nesbitt theorem) implies that $\rho_1$ and $\rho_2$ are isomorphic as semisimple $K$-representations of $G_{\mathbb{Q}}$. This is exactly the desired conclusion. [/step]