# Artin map restricted to base field

Let $M/L/K$ be a tower of local fields such that $M/L$ is abelian with Galois group $G$. The Artin map $\psi_{M/L}$ restricted to $K^\times$ is a continuous map to $G$ and thus corresponds to some abelian extension $T/K$ with an embedding $\operatorname{Gal}(T/K) \hookrightarrow G$. Challenge: Find $T$. (I am primarily interested in the mixed characteristic case, i.e. $K$ extends $\mathbb{Z}_p.$)

I conjecture the following Galois-theoretic description. Let $\tilde M$ be the normal closure of $M$. Then there is an embedding $G_1 = \operatorname{Gal}(\tilde M/K) \hookrightarrow S_n \wr G$ (where $n = [L:K]$). Let $H$ be the intersection of $G_1$ with the subgroup $$\{(\sigma,g_1,\ldots,g_n) \in S_n \wr G : \sum g_i = 0 \}.$$ Then $T = \tilde M^H$ is the $T$ we seek, and $\operatorname{Gal}(T/K) \cong G_1/H$ has a natural embedding into $G$.

Using Kummer theory, I was able to prove this when $\mu_m \subseteq K$, where $m$ is the exponent of $G$, or when $m$ is squarefree. Also the global analogue, where $M/L/K$ is a tower of number fields and we restrict $\psi_{M/L}$ to ideals of $K$, yields readily to manipulation of Frobenius elements. But there we are aided by having to consider only ideals composed of unramified primes. Indeed, the local unramified case is not hard.

• In what way does your conjectural description "find $T$"? The intersection $K^{\times} \cap {\rm{N}}_{M/L}(M^{\times})$ inside $L^{\times}$ is an open subgroup of $K^{\times}$ with finite index and hence as such "corresponds" to a finite abelian extension $T$ of $K$. But short of explicit local reciprocity, in what sense do you want to "find $T$"? Referring to a map $\Gamma \rightarrow G$ with such abstract $\Gamma$ doesn't seem any more informative about $T$ than forming $K^{\times} \cap {\rm{N}}_{M/L}(M^{\times})$, so it would help to clarify features about $T/K$ that you seek to know. – nfdc23 Mar 28 '17 at 3:27
• As an analogy, one could contemplate the maximal abelian extension $K'/K$ such that $LK' \subset M$, but that is a rather abstract recipe that doesn't really "find" such $K'$ (but maybe it good enough, depending on what it is one wishes to know). – nfdc23 Mar 28 '17 at 3:30
• Thanks nfdc, the reference to the absolute Galois group was unnecessary and I've modified my conjectural description of $T$. – Evan O'Dorney Mar 29 '17 at 13:38

We describe $\left.\phi_{M/L}\right|_K$ in terms of $\phi_{\tilde M / K}$: \begin{alignat*}{5} \text{Gal}(\tilde M / K)^{\text{ab}} &\xrightarrow{\text{Ver}} \text{Gal}(\tilde M / L)^{\text{ab}} &&\to \text{Gal}(M/L) \\ \phi_{\tilde M / K}(x) & \longmapsto \phi_{\tilde M / L}(x) && \mapsto \phi_{M/L}(x). \end{alignat*} But these maps between Galois groups can be described as the suitable restrictions of the maps of abstract groups $$(G \wr S_n)^{\text{ab}} \xrightarrow{\text{Ver}} \left(G \times (G \wr S_{n-1})\right)^{\text{ab}} \xrightarrow{\pi_1} G.$$ Thus $T$ arises from a certain map $G \wr S_n \to G$ which, upon computation, is none other than $$(g_1,\ldots,g_n,\sigma) \mapsto \sum g_i$$ as desired.