Let $F$ be a finite field. Let $\rho:G\to GL_2(\overline F)$ be a continuous representation from a profinite group $G$, where $\overline F$ is an algebraic closure of $F$. Suppose that for all $g\in G$, we have $tr(g),det(g)\in F$. Does it follow that $\rho$ can be conjugated to a representation to $GL_2(F)$?
I was wondering about this when I was reading p.31 of this note https://www.ma.imperial.ac.uk/~tsg/Index_files/ArizonaWinterSchool2013.pdf, in which the author constructed a Galois representation associated to a maximal ideal of a Hecke algebra. It was stated there that the Brauer group of a finite field is trivial, but I don't see why it is relevant.
Take $F=\mathbb{R}$ and the canonical representation of the quaternions $G\longrightarrow{\rm GL}_2(\mathbb{C})$ seen as a subgroup of ${\rm GL}_2(\mathbb{C})$. All the quaternion matrices have real trace and determinant but they can't be seen as a (conjugated) subgroup of ${\rm GL}_2(\mathbb{R})$. The reason behind this is that ${\rm Br}(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$ and the only non-trivial element of the Brauer group of $\mathbb{R}$ is the obstruction of your problem. More generally, suppose $\rho:G\longrightarrow{\rm GL}_2(\overline{F})$ is an irreducible representation, for each $\sigma\in\mathscr{G}_F:={\rm Gal}(\overline{F}/F)$, let $\rho^{\sigma}:=\sigma\circ\rho:G\longrightarrow{\rm GL}_2(\overline{F})$ then ${\rm tr}\rho^{\sigma}(g)={\rm tr}\rho(g)$ for all $g\in G$ so, $\rho$ being irreducible, $\rho$ and $\rho^{\sigma}$ are conjugated, say $\rho^{\sigma}=A_{\sigma}\rho A_{\sigma}^{-1}$. Now, one can show that, up to conjugacy, this implies that $$ A_{\sigma\circ\tau}=a_{\sigma,\tau}A_{\sigma}\sigma(A_{\tau}) $$ where $a$ is a $2$-cocycle, that is an element of $H^2(\mathscr{G}_F,\overline{F}^*)={\rm Br}(F)$. If one has ${\rm Br}(F)=0$ (which is the case for finite fields), then $a$ is a $2$-coboundary, say $a=\partial\alpha$ where $\alpha$ is a $1$-cocycle and so $\sigma\mapsto\alpha_{\sigma}^{-1}A_{\sigma}$ is a $1$-cocycle and it is a $1$-coboundary by Hilbert $90$. Therefore, there exists a matrix $B$ such that $$ \forall\sigma\in\mathscr{G}_F,A_{\sigma}=\alpha_{\sigma}\sigma(B)B^{-1} $$ so that $\rho^{\sigma}\sigma(B^{-1})=\rho B^{-1}$, which means that $\rho^{\sigma}=\rho$. Thus $\rho$ can be seen as a representation of $G$ into ${\rm GL}_2(F)$ (up to conjugacy).