$\def\fp{{\frak p}}\def\fP{{\frak P}} \def\bfm#1{{\mathbf#1}}\def\cE{{\cal E}}\def\fm{{\frak m}}\let\ss\subset $ I would like to make a link between idelic norm and "normal" norm.
For $L/F$ an extension of number field, one defines admissible module $\fm$ by $ \cE_\fm\ss N(\cE_L)$
(with $\cE_L=$ inversible ideles of $L$ and $\cE_\fm$ inversible ideles of $K$ which are $=1\mod\fm$)
and i would like to know if it is an idelic version of the ideal one, generaly writen for the $\fp|\fm$. as $\bfm U_\fp^{(b)}\ss N(\bfm U_\fP)$ where $\bfm U_\fp^{(b)}=1+\fp^b$, and for $\fP|\fp$, $\bfm U_\fP$ the units of $L_\fP$
My problem is that if $(\bfm a_\fp)$ is an idele of $K$ and $(\bfm b_\fP)$ an idele of $L$, $\bfm a=N(\bfm b)$ means $$a_\fp=\prod_{\fP|\fp}N(b_\fP)$$ and i would like to get rid of the product.
Is it possible ? Is it true ? All this is of course around Artin map.
$\def\fp{{\frak p}}\def\fP{{\frak P}} $Hum ! I think i got it but i have some doubts.
Actually, $L_\fP=\tau L\,K_\fp$ for a $K$-imbedings $\tau$ into a closure of $K_\fp$. So two $L_\fP$ are isomorphe with a $K_\fp$ morphism so the pol. char. of the product by an element $x\in L_{\fP_1}$ is the same than the one for $\tau_2\circ\tau_1^{-1}(x)$ because the matrix invoques coefficients in $K_\fp$ and the base of $\tau L$ is just the image of a base of $L$ over $K$.
Consequently $N_{L_1}(x)=N_{L_2}(\tau_2\circ\tau_1^{-1}(x))$ with $L_i=L_{\fP_i}$
Am I right ?
So regarding my question all norms in the second member could be reduce to a norm in just one $L_i$. I think i say something wrong, it's too simple, but i don't see where