Identifying the residue field of $\overline{\mathbb{Q}}_p$ with $\overline{\mathbb{F}}_p$ Could anyone epxlain to me how to identify the residue field of $\overline{\mathbb{Q}}_p$ with $\overline{\mathbb{F}}_p$ ? I am basically trying to go from a $p$-adic galois representation of $G_\mathbb{Q}$ to a representation of the type Serre uses...so my ultimate goal is to obtain a representation
$$
\rho: \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \rightarrow \operatorname{GL}_2(\overline{\mathbb{F}}_p)
$$
from a more natural one like $\rho: \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \rightarrow \operatorname{GL}_2(\mathbb{Q}_p)$. 
I was initially looking for an argument involving choosing a lattice in $V = \overline{\mathbb{Q}}_p^2$, but since then I read (and understood the proof that) the residue field of $\overline{\mathbb{Q}}_p$ is an algebraic closure of $\mathbb{F}_p$. Would that be an easier way to solve my problem? 
Thank you very much ! 
 A: The residue field of $\overline{\mathbf{Q}}_p$ is algebraically closed because every monic polynomial over the ring of integers $\overline{\mathbf{Z}}_p$ of $\overline{\mathbf{Q}}_p$ has a root in $\overline{\mathbf{Z}}_p$, and it is algebraic over $\mathbf{F}_p$ because every element in the ring of integers over $\overline{\mathbf{Z}}_p$ is integral over $\mathbf{Z}_p$. So the residue field of $\overline{\mathbf{Q}}_p$ is an algebraic closure of $\mathbf{F}_p$. It doesn't completely make sense to say that you want to "identify" the residue field with $\overline{\mathbf{F}}_p$, because there isn't really one algebraic closure of $\mathbf{F}_p$; any two are isomorphic, but non-canonically. 
Regarding how to take a Galois representation and reduce it modulo $p$, you have to prove that $G_\mathbf{Q}$ preserves an $\overline{\mathbf{Z}}_p$-lattice in a finite-dimensional $\overline{\mathbf{Q}}_p$-vector space on which it acts continuously. This is easier to see I guess for a representation on a finite-dimensional vector space $V$ over a finite extension $L$ of $\mathbf{Q}_p$. Start with any $\mathscr{O}_L$-lattice $T$ in $V$ (i.e. a finite $\mathscr{O}_L$-module that spans $V$ over $L$) and let $\Gamma$ be its stablizer in $\mathrm{GL}_n(L)$. This is an open subgroup of $\mathrm{GL}_n(L)$, so the inverse image $\Gamma^\prime$ is an open subgroup of $G_\mathbf{Q}$. Now take the sum of the translates of $T$ by the finitely many coset representatives for $G_\mathbf{Q}/\Gamma^\prime$. This is still a lattice, and it is $G_\mathbf{Q}$-stable by construction. 
What this means concretely is that you can conjugate $\mathrm{GL}_n(L)$ so that the image of $G_\mathbf{Q}$ is contained in $\mathrm{GL}_n(\mathscr{O}_L)$. Now you apply the natural reduction map $\mathrm{GL}_n(\mathscr{O}_L)\rightarrow\mathrm{GL}_n(\mathbf{F})$, where $\mathbf{F}$ is the residue field of $L$.
The reason you don't have to really think about representations over $\overline{\mathbf{Q}}_p$ when you do this is that the image of any continuous homomorphism $G_\mathbf{Q}\rightarrow\mathrm{GL}_n(\overline{\mathbf{Q}}_p)$ has image in $\mathrm{GL}_n(L)$ for some $L$ finite over $\mathbf{Q}_p$. This is a consequence of the Baire category theorem and works with $G_\mathbf{Q}$ replaced by any profinite group. I guess maybe the argument I gave above for the existence of a stable lattice still works with $\overline{\mathbf{Q}}_p$, but I worry about whether a sum of translates of a finite free $\overline{\mathbf{Z}}_p$-module spanning a finite-dimensional $\overline{\mathbf{Q}}_p$-vector space will still be free...anyway, you don't need it because you can always descend to a finite extension of $\mathbf{Q}_p$.
