In J. Weinstein's note: Reciprocity laws and Galois representations: recent breakthroughs, section 3.3, a Galois representation that is unramified at a prime $\mathfrak{p}$ is defined in a different way.
Suppose $K$ is a number field with absolute Galois group $\text{Gal}(\bar{K}/K)$, and $F$ is a topological field, a Galois representation $\rho$ is a continous homomorphism
\begin{equation}
\rho:~\text{Gal}(\bar{K}/K) \rightarrow\text{GL}_n(F)
\end{equation}
$\rho$ is unramified at a prime $\mathfrak{p}$ of $K$ if it factors through $\text{Gal}(L/K)$, where $L/K$ is some (possibly infinite) algebraic extension which is unramified at $\mathfrak{p}$. The common definition in literature is that $\rho(I_{\mathfrak{p}})$ is trivial, where $I_{\mathfrak{p}}$ is the initial group at $\mathfrak{p}$.
Anyone knows how to show their equivalences?
Best Answer
$\require{AMScd}$ If $\rho$ factors through some $\overline{\rho}\colon \mathrm{Gal}(L/K) \to \mathrm{GL}_n(F)$ with $L/K$ unramified at $\mathfrak{p}$, then the image of $I_{\mathfrak{p}}$ under $\rho$ is \begin{equation*} \rho(I_{\mathfrak{p}}) = \overline{\rho}\circ \pi(I_{\mathfrak{p}}) = \overline{\rho}(I_{\mathfrak{p}}(L/K)) = \overline{\rho}(1)=1 \end{equation*} because $\mathfrak{p}$ is unramified in $L/K$ so it must have trivial inertia group there, and the image of the absolute inertia group in some intermediate extension is the inertia group of that extension.
Conversely, if $I_{\mathfrak{p}}$ is trivial then the statement is trivial as well. If on the contrary $I_{\mathfrak{p}}\neq 1$ then $\rho$ factors through $\mathrm{Gal}(L/K)$ where $K\subseteq L\subseteq \overline{K}$ is the maximal extension where $\mathfrak{p}$ is unramified. Indeed, the kernel of the projection is exactly $I_{\mathfrak{p}}$ which lies in the kernel of $\rho$.