It's exercise 6.4 in N. Childress - Class F. Th. $q$ is a prime number $F\subset L$ an cyclic extension $[L:F]=q$; $\zeta$ a $q$-root of 1.
One have shown ${\cal A}_{L/F}({\bf i}(F_{{\frak p}}))=1$ for a place ${\frak p}$, ${\bf i}$ the imbeding of $F_{{\frak p}}$ in idele (shorly : the idelic artin map is trivial on local field).
One have to show the same stay true for the extension:
$${\cal A}_{L(\zeta)/F(\zeta)}({\bf i}(F(\zeta)_{{\frak P}}))=1$$ where ${\frak P}$ is above ${\frak p}$.
And of course, if i am here, i don't see why.
I just have the level to say that the restriction to $L$ is indeed trivial as it is the Artin map of the norm witch is trivial by the preceding result. But i cannot see how to get information about the image of $\zeta$. I know $\sigma(\zeta)=\zeta^{N({\frak P})}\mod{\frak P}$, so what ?
Can someone help me ? Thanks I think that $[L\zeta):F(\zeta)]=q$ (but not sure)
I think the answer is completely stupid.
$\def\artin#1#2{\genfrac{(}{)}{}{}{#1}{#2}}\def\fp{{\frak p}}\def\fP{{\frak P}} $
We have a diamond situation (sorry but there is no topics on help to explain how to draw commutative diagram and xymatrix doesn't work here)
$\begin{alignat}{5} &&L(\zeta)\\ &\swarrow&&\searrow\\ L & & &&F(\zeta), \fP\\ &\searrow&&\swarrow\\ &&F, \fp \end{alignat}$ $\def\bi{\mathbf{i}}$
$$\artin{\bi\big(F(\zeta)_\fP\big)}{L(\zeta)/F(\zeta)}\Big|_L=\artin{N(\bi(F(\zeta)_\fP))}{L/F}=1$$ because it is assumed that $\artin{\bi(F_\fp))}{L/F}=1$ but $\artin{\bi(F(\zeta)_\fP)}{L(\zeta)/F(\zeta)}\in{\rm Gal}\big(L(\zeta)/F(\zeta)\big)$ so they fix $\zeta$
Fix $\zeta$ + Fix $L$ $\Longrightarrow$ Fix $L(\zeta)$