Let $l, p$ be distinct primes and let $K$ be a local field of mixed characteristic $(0,p)$. Let $G=G(K^s/K)$, $I_K$ be the inertia subgroup and $V$ a finite dimensional vector space over $\mathbb Q_l$ with a continuous representation $\rho:G \to GL(V)$. The condition of $V$ having potentially good reduction is defined by the condition that $I_K \subseteq \textrm{ker}\,\rho$. I need to show that this condition is equivalent to there being a finite extension $K'$ of $K$ inside $K^s$ such that as $\rho(I_{K'})$ is trivial.
I am unable to see either direction.
The direction $\rho(I_K)< \infty \implies \rho(I_{K'})=1$ for a finite extension:
$\rho(I_K)< \infty$ means that $I_K \cap \textrm{ker}\rho$ has finite index in $I_K$. But we also have $I_K \cap \textrm{ker} \rho$ as a closed subset of $I_K$, hence is open in $I_K$. Therefore there is an open set $G_{K'}=G(K^s/K')$ of $G_K$ given by a finite extension $K'/K$ such that $I_K \cap G_{K'} \subseteq I_K \cap \textrm{ker} \rho$. But $I_K \cap G_{K'}=I_{K'}$ and by the above containment we have $\rho(I_{K'})=1$.
The reverse direction:
But since $G_{K'}$ has finite index in $G_K$, $I_{K'}$ has finite index in $I_K$. By the condition $\rho(I_{K'})=1$, we have $I_{K'} \subseteq \textrm{ker}\rho \cap I_K \subseteq I_K$. So $[I_K: \textrm{ker}\rho \cap I_K]= \rho(I_{K})< \infty$.