Why is every l-adic Galois representation $$G_{\mathbb{Q}_p}\rightarrow GL_n(\mathbb{Q}_{l})$$ conjugate to one over the l-adic integers? $$G_{\mathbb{Q}_p}\rightarrow GL_n(\mathbb{Z}_{l})$$

## **closed** as off-topic by Venkataramana, Mikhail Bondarko, Olivier, Joël, Ryan Budney Oct 12 '15 at 19:13

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Recall that all ${\mathbb Z}_l$-lattices in the ${\mathbb Q}_l$-vector space ${\mathbb Q}_l^n$ are conjugate in ${\rm GL}_n ({\mathbb Q}_l )$. Since the ${\rm GL}_n ({\mathbb Q}_l )$ stabilizer of the standard lattice ${\mathbb Z}_l^n$ is ${\rm GL}_n ({\mathbb Z}_l )$, all lattices in ${\mathbb Q}_l^n$ have open stabilizer in ${\rm GL}_n ({\mathbb Q}_l )$. Moreover a finite sum of ${\mathbb Z}_l$-lattices is again an ${\mathbb Z}_l$-lattice.

Now take $\rho$ : $G_{{\mathbb Q}_p}\longrightarrow {\rm GL}_n ({\mathbb Q}_l )$ a continuous Galois representation. Let $L$ be any ${\mathbb Z}_l$-lattice in ${\mathbb Q}_l^n$. Then its stabilizer $H$ in $G_{{\mathbb Q}_p}$ is open and the coset set $G_{{\mathbb Q}_p}/H$ is finite, since $G_{{\mathbb Q}_p}$ is profinite. Now

$$ M := \sum_{g\in G_{{\mathbb Q}_p}/H} g.L $$

is a Galois stable lattice in ${\mathbb Q}_l^n$. We have proved that a conjugate of $\rho (G_{{\mathbb Q}_p})$ lies in ${\rm GL}_n ({\mathbb Z}_l )$.

Basic Number Theory) then implies that $V$ has a basis $(e_i)$ consisting of vectors of $\nu$-norm equal to $1$. It generates the sought-for lattice. $\endgroup$ – ACL Oct 11 '15 at 11:03