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Let $K$ be a finite extension of $Q_p$ and $H$ is the inertia group of $G_K$, then we have the well-known local Artin isomorphism $\widehat{K^\times}\cong G_K^{ab}$ where the completion is for the norm groups. But we also have $\widehat{K^\times}\cong O_K^*\times \hat{Z}$, so we have the isomorphism $O_K^*\times \hat{Z}\cong G_K^{ab}$.

Question: Define the composition map $H\hookrightarrow G_K\rightarrow G_K^{ab}\rightarrow O_K^*\times \hat{Z}$ by $f$, then what is the image $f(H)$ in $O_K^*\times \hat{Z}$ ? Do we have $f(H)\subseteq O_K^* \times \{1\} $ ?

Thanks for any answers!

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2 Answers 2

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In the Artin map $\newcommand{\Z}{\hat{\Bbb Z}}G_K\mapsto O_K^*\times\Z$ the induced map $G_K\to\Z$ corresponds to the action on the maximal unramified extension $H=K^{ur}$. In detail we have $$G_K\to\text{Gal}(K^{ur}/K)\cong\text{Gal}(k^{alg}/k)\cong\Z$$ where $k$ is the residue class field. By definition the inertia group consists of the elements of $G_K$ acting trivially on $K^{ur}$, so the image of the inertia group is contained in $O_K^*\times\{1\}$. Indeed it is equal to $O_K^*\times\{1\}$.

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In addition to your notations, let $I_K$ be the inertia subgroup of ${G_K}^{ab}$. By definition, $K^{ur}$ is the fixed field of $H$ (resp. $I_K$) in $G_K$ (resp. ${G_K}^{ab}$), so it is clear that $f(H)=I_K$, and we have only to show that the local reciprocity map $\theta: K^* \to {G_K}^{ab}$ induces an isomorphism ${O_K}^* \to I_K$. It is classically known that the composition $K^* \to {G_K}^{ab} \to G(K^{ur}/K) \cong \hat {\mathbf Z}$ is the valuation map $v$, so we have a commutative diagram built from the exact sequences $1\to {O_K}^*\to K^* \to {\mathbf Z}\to 0$ and $1\to I_K \to {G_K}^{ab} \to \hat {\mathbf Z} \to 0$, with vertical connecting maps naturally induced by $\theta$ and Id. The induced map $\theta :{O_K}^*\to I_K$ is continuous, with dense image (this is because the reciprocity law in an unramified extension $L/K$ is explicitly given by $x\to F^{v(x)}$, where $F$ is the Frobenius element of $G(L/K)$). As ${O_K}^*$ is compact, it follows that $\theta$ is surjective. The injectivity of $\theta$ is not trivial, it is one of the equivalent formulations of the so called existence theorem of local CFT.

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  • $\begingroup$ Thanks for your answers too! $\endgroup$
    – user685167
    Jul 17, 2019 at 10:04

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