Let $K$ be a local field, say a finite extension of $\mathbb{Q}_p$ (which is the purpose of my interest).

Let $L$ be an unramified extension of $K$.

Local class field theory asserts that there exists a commutative diagram

$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$

$$ \begin{array}{c} \widehat{L^{\times}} & \ra{\varphi_L} & Gal(L^{ab}/L)\\ \da{Nm} & & \da{res} \\ \widehat{K^{\times}} & \ra{\varphi_K} & Gal(K^{ab}/K) \end{array} $$

where $res$ is the restriction map, $Nm$ is the norm and where $\varphi_L$ and $\varphi_K$ are the isomorphism given by local class field theory (here $\widehat{K^{\times}}$ denotes the completion of $K^{\times}$ with respect to the profinite topology, which is the same as the completion with respect to the norm topology).

If I am not mistaken, under $\varphi_L$ the image of the units of $L$, i.e. $\mathcal{O}_L^{\times}$ is $I_L^{ab}$, the inertia subgroup of $Gal(L^{ab}/L)$, and the same for $\mathcal{O}_K^{\times}$.

But those inertia subgroups are isomorphic because $L$ is unramified over $K$, whereas the norm map $Nm : \mathcal{O}_L^{\times} \to \mathcal{O}_K^{\times}$ is not because it is not an isomorphism when we reduce modulo the uniformizer of $K$.

What am I missing in this picture ?

  • 1
    $\begingroup$ They are abstractly isomorphic, but the restriction map is not an isomorphism. You should, as good practice, write down why this is true, but it's analagous to the fact that :$G_{\mathbb{F}_{p^r}}\to G_{\mathbb{F}_p}$ is not an isomorphism, even though the two groups are isomorphic (the map is multiplication by $r$). You'll find that your restriction map is surjective though, which is good since the norm surjects for unramified extensions. $\endgroup$ – Alex Youcis Apr 25 '15 at 10:37

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