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.

  • 2
    $\begingroup$ 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. $\endgroup$ – nfdc23 Mar 28 '17 at 3:27
  • $\begingroup$ 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). $\endgroup$ – nfdc23 Mar 28 '17 at 3:30
  • $\begingroup$ Thanks nfdc, the reference to the absolute Galois group was unnecessary and I've modified my conjectural description of $T$. $\endgroup$ – Evan O'Dorney Mar 29 '17 at 13:38

I've finally found the answer to my question by perusing Serre's Local Fields, Ch.XIII, specifically Propositions 10-12, which contain the functorial properties of the Artin symbol used below.

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.