$\def\fd{{\frak d}}\def\fa{{\frak a}}\def\fb{{\frak b}}\def\fm{{\frak m}}\def\fp{{\frak p}}\def\fP{{\frak P}}\def\ff{{\frak f}}\def\fii{{\frak i}}$
$\def\a{\alpha}\let\b\beta \let\ss\subset$
$\let\f\frac \let\r\sqrt$
$\def\H{{\Bbb H}} \def\Z{{\Bbb Z}}\def\Q{{\Bbb Q}}\def\K{{\Bbb K}}$
$\def\cD{{\cal D}} \def\cE{{\cal E}}\def\cA{{\cal A}}\def\cO{{\cal O}}$
Hello
I have at my disposal the Artin reciprocity theorem (ideal version) as :
$$I(\fm)\Big/P_{\frak m}N_K(\tilde \fm)\simeq G(K/F)$$
in the case where $P_{\fm}\ss\ker\cA$ (which is a possible definition of admissible module) where:
$F\ss K$ is an extension of number fields;
$\fm$ a $F-$module;
$I(\fm)$ fractionnary ideals of $F$ prime with $\fm$.
$P_{\fm}$ fractionnary principal with a generator verifying $\a=^*1\mod\fm$ (multiplicativ congruence as usual).
$\tilde\fm$ the "lift" of $\fm$ in $K$
and at last $\cA$ Artin morphism.
I have to prove this result stay true if $\fm$ is such that $\cE_{F,\fm}\ss N(\cE_K)$ (another definition for admissible module) where $\cE_K$ is the idèles of $K$ and $\cE_{F,\fm}$ idèles of $F$ such that $\forall \fp,\:a_\fp=^*1\mod\fm$. In particular the conductor $\ff(K/F)$ satisfies Artin reciprocity.
I have no idea how to pass from this hypothese to $P_\fm\ss\ker\cA$.
Thank you for any help
off chance it is exercise 5.7 p 123 of N. Childress "Class Field Theory"