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I'm trying to parse a page in Milne's CFT notes. The local reciprocity law gives us isomorphisms $$\phi_{L/K}:K^\times/Nm(L^\times)\to \textrm{Gal}(L/K)$$ for all abelian extensions $L$ of a nonarchimedean local field $K$. Taking limits we get by infinite Galois theory the isomorphism $$\phi_K:\widehat{K}^\times\to \textrm{Gal}(K^{ab}/K).$$

If we fix a uniformizer/prime element of $K$, call it $\pi$, then we have the canonical decomposition $$K^\times \simeq \pi^\mathbb{Z}\times U_K,$$ where $U_K$ are the units in the valuation ring of $K$. When we take limits, then we apparently get the decomposition $$\widehat{K}^\times \simeq \pi^\widehat{\mathbb{Z}}\times U_K.$$

My question is really how one should think about $\pi^\widehat{\mathbb{Z}}$. For example $\phi_K(\pi)$ acts as the Frobenius element on $K^{un}$ and it's easy to see how $\phi(\pi^a)$ acts for $a\in\mathbb{Z}$. My questions are the following:

  1. How should one think of general exponents in $\widehat{\mathbb{Z}}$ instead of $\mathbb{Z}$?

  2. How would one show that the fixed field of $\phi_K(\pi)$ i.e. $K_\pi$ is also fixed by $\pi^{\widehat{\mathbb{Z}}}$, so that infinite Galois theory gives us $K^{ab}=K_\pi\cdot K^{un}$?

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2 Answers 2

up vote 5 down vote accepted
  1. Here $\pi^\hat{\mathbb{Z}}$ is a notation to denote the group $\hat{\mathbb{Z}}$, but instead of writing an element $n \in \hat{\mathbb{Z}}$ we write $\pi^n$.

In a more general context : let $G$ be a profinite group and $\alpha \in G$. Then we have the following : the map $\mathbb{Z} \rightarrow G, n \mapsto \alpha^n$ extends constinuously to a map $\hat{\mathbb{Z}} \rightarrow G$, and its image is denoted $\pi^\hat{\mathbb{Z}}$ and the image of $n \in \hat{\mathbb{Z}}$ is written $\alpha^n$. Moreover $\pi^\hat{\mathbb{Z}}$ is the closed subgroup generated by $\alpha$.

  1. When $\pi^{\hat{Z}}$ is seen as a subgroup of $Gal(\overline{K}/K)$, then it is the closed subgroup generated by the Frobenius (= image of $\pi$). Since the stabalizer of a subfield is a closed subgroup, this answers your question.
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user10676 has basically answered your question. Maybe I can just add that $K_{\pi}$ is a particular totally ramified abelian extension of $K$ obtained by adjoining the coordinates of all the $p$-power division points in the Lubin--Tate formal group corresponding to $\pi$.

E.g. if $K = \mathbb Q_p$ and $\pi = p$, then the corresponding Lubin--Tate formal group is just the multiplicative formal group, and $K_{\pi}$ is the extension of $\mathbb Q_p$ obtained by adjoining all $p$-power roots of unity.

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Thanks. I guess my confusion still lies in how this particular totally ramified extension works i.e. how freely we can pick them and still get the same thing. –  pki Nov 6 '11 at 0:51
    
@pki: Dear pki, The totally ramified extension depends very much on the choice of $\pi$, as does the splitting of $K^{\times}$. But the dependence on the choice of $\pi$ is "the same" on both sides, so that the local Artin map is well-defined independent of any choices. Regards, –  Matt E Nov 6 '11 at 1:29

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