It's an exercise (N. Childress Ex. 6-8 p. 152)
I have to prove that two notions of conductor are in fact the same one. $\let\ss\subset \let\f\frac\let\r\sqrt\let\w\wedge\let\imp\Longrightarrow\let\fa\forall \def\Z{{\Bbb Z}}\def\Q{{\Bbb Q}}\def\K{{\Bbb K}} \def\ff{{\frak f}} \let\z\zeta \def\cE{{\cal E}}\def\fp{{\frak p}}\def\cu{{\cal U}} \def\fm{{\frak m}}\def\cA{{\cal A}} \def\artin#1#2{\genfrac{(}{)}{}{}{#1}{#2}} $
Let $\Q\ss\K=\Q]\r d[.$ a first notion of conductor is the least $m$ such that $\K\ss\Q[\z_m]$.
In idèles, Artin map, class field, another definition is given : the smallest ideal $f\Z=\ff\ss\Z$ such that $$\cE_\ff^+\ss\Q^*N(J_K)$$ where $J_K$ are idèle of $K=\Q[\r d]$ and $\cE_\ff^+$ units idèle $=\{\fa\fp,\: a_\fp\in\cu_\fp,a_\fp>0,\:{\rm for}\:\fp {\rm\ real},\:a_\fp=1\mod\fp^{v(\fm)}\}$.
I know $f|m$.
If it is suppose to be known that $\Q[\z_f]$ is the class field of the sub-group $\Q^*\cE_\ff^+$ I think the result follows (if I don't make a mistake) because $\Q^*N(J_{\Q[\z_f]})=\Q^*\cE_\ff^+\ss\Q^*N(J_K)$ because of definition and using decreasing correspondance (group)-(class field)
$\K\ss\Q[\z_f]$.
I'm looking for an independant proof (actualy this exercise is before the result about class field)
I began to say
as $f|m$, $\quad\Q[\z_f]\ss\Q[\z_m]$ and actualy i have to prove $\K\ss\Q[\z_f]$ that is equivalent to $\Q^*N(J_{\Q[\z_f]}\ss\Q^*N(J_\K)$ or $N(J_{\Q[\z_f]})\ss\Q^*N(J_\K)=\ker\cA_d$, the Artin map of $\Q[\r d]$, $\cA_d=\artin{}{\Q[\r d]/\Q}$.
this could be writen (with for short $J_f=J_{\Q[\z_f]}$) $\artin{N(J_f)}{\Q[\r d]/\Q}=1$. But I know that in such "diamond" situation ($\Q$ at the bottom, $\Q[\z_f],\K$ in the middle, $\Q[\z_m]$ on top) $$\artin{N(J_f)}{\Q[\r d]/\Q}=\artin{J_f}{\Q[\z_m]/\Q[\z_f]}\Big|_{\Q[\r d]}=1 \qquad (?)$$ and i have to prove this latter or equivalently it's in Galois group of $\Q[\z_m]/\Q[\r d]$ (may be, surely ? something stupid)
At this point i cannot see any way out !
Thank you for any help