4
$\begingroup$

Let $K$ be a $p$-adic field with algebraic closure $\overline{K}$. Then if $K^\text{nr}$ is the maximal unramified extension of $K$ in $\overline{K}$, there is an explicit invariant map: $$ H^2(\Gamma_{K^\text{nr}/K}, (K^\text{nr})^{\times}) \to \mathbb{Q}/ \mathbb{Z} $$ given by the composition $$ H^2(\Gamma_{K^\text{nr}/K}, (K^\text{nr})^{\times}) \to H^2(\widehat{\mathbb{Z}}, \mathbb{Z}) \xrightarrow{\delta^{-1}} \operatorname{Hom}(\widehat{\mathbb{Z}}, \mathbb{Q}/\mathbb{Z}) \xrightarrow{\operatorname{ev}(1)} \mathbb{Q}/ \mathbb{Z}. $$

For a finite Galois extension $L/K$ in $\overline K$, I can define an invariant map via the composition $$ H^2(\Gamma_{L/K}, L^{\times}) \hookrightarrow H^2(\Gamma_{\overline{K}/K}, \smash{\overline{K}}^{\times}) \cong H^2(\Gamma_{K^\text{nr}/K}, (K^\text{nr})^{\times}), $$ and then composing with the previously defined map. However it's not so obvious to me on the level of cocycles what this latter map does because the isomorphism at the end is not very explicit.

Does anyone know an explicit way to make sense of this invariant map so that given a cocycle in $H^2(\Gamma_{L/K}, L^{\times})$, one could reasonably compute what class in $\mathbb{Q}/\mathbb{Z}$ it corresponds to?

EDIT: For instance suppose that $L=K(\zeta_{p^n})$. What should the invariant map look like in this case?

$\endgroup$
4
  • 3
    $\begingroup$ Have a look at Section 1.1 of Serre, "Local class field theory", in: Algebraic Number Theory (J. W. S. Cassels and A. Fröhlich, eds.) pp. 128–161. Academic Press, London; Thompson Book Co., Inc., Washington, D.C., 1967. Hopefully this will help. $\endgroup$ Commented Jul 15, 2021 at 14:26
  • $\begingroup$ @MikhailBorovoi Dear Professor Borovoi, yes I agree that section has many useful facts. I think that essentially what I want is an explicit version of the inverse of the map in Serre's thm 1. Given an $\overline{K}$-valued cocycle, how to construct a $K^{nr}$-valued cocycle that corresponds under inflation. Or maybe there is another way to think about it? $\endgroup$
    – Alexander
    Commented Jul 15, 2021 at 14:40
  • 2
    $\begingroup$ I guess your extension $L/K$ should be Galois? $\endgroup$
    – LSpice
    Commented Jul 15, 2021 at 14:42
  • 1
    $\begingroup$ @LSpice yes, thank you. $\endgroup$
    – Alexander
    Commented Jul 15, 2021 at 15:33

0

You must log in to answer this question.