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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?

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    $\begingroup$ Usually, for an infinite extension, one defines a prime to be unramified if its inertia group is trivial. For finite extensions, the equivalence follows by proving that the order of the inertia group is exactly the ramification index. Notice that the inertia group of $L/K$ is exactly the image of $I_\mathfrak p$ $\endgroup$ – Mathmo123 Nov 14 '16 at 11:05
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$\require{AMScd}$ 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 $\rho$.

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  • $\begingroup$ Thank you! I forget about the largest unramified extension! $\endgroup$ – Wenzhe Dec 4 '16 at 18:18
  • $\begingroup$ @Alessandro, why $ \overline{\rho}\circ \pi(I_{\mathfrak{p}}) = \overline{\rho}(I_{\mathfrak{p}}(L/K))$ ? How it is obtianed? Can you explain it shortly please? I am beginner $\endgroup$ – M. A. SARKAR Dec 1 '19 at 16:49

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