"Frobenius element" in cyclotomic extension for ramified primes Let $m = \prod\limits_{q \text{ prime}} q^{n(q)}\;$ be a positive integer, let $p$ be a prime factor of $m$.
$\def\Q {\Bbb Q}
\def\Gal{\mathrm{Gal}}
\newcommand\K[1]{\Bbb Q(\zeta_{#1})}$
Question:

How can I find an element $\sigma_P$ of $\Gal(\K m/\Q)$ such that $$\sigma(\zeta_m)-\zeta_m^p \in P,$$ where $P$ is a prime of $\K m$ above $p$ ?


Ideas:
— When $p$ is not a prime factor of $m$, we can take $\sigma(\zeta_m)=\zeta_m^p$, see here. But if $p\mid m$, this doesn't even define an element of $\Gal(\K m/\Q)$.
— This $\sigma_P$ lies in the decomposition group $D=D(P/p)$ and, modulo the inertia group $I=I(P/p)$, it generates $D/I = \langle \sigma_P \,I \rangle \cong
\Gal(\mathcal O_{\K m\,}\,/P     \;;\;    \Bbb F_p)$.
Because $p$ is ramified in $\K m$, such a $\sigma_P$ is only unique up to an element of $I$.
— I know that $|D/I| = f_{P/p} = f_p$ is the order of the class of $p$ mod $M=m/p^{n(p)}\,$ in $(\Bbb Z/M\Bbb Z)^{\times}$, and
$$p\mathcal O_{\K m\,} = p\Bbb Z[\zeta_m] = (P_1 \cdots P_r)^{\phi(p^{n(p)\;})},$$
each prime $P_j$ with inertia degree $f_p$.
— I don't know how to go further. Eventually I want to show that
$$\prod_{\chi} \det(1-\chi(\sigma_P)p^{-s}, \Bbb C^{I_P}) = 
\prod_{\mathfrak p \mid p} (1-N(\mathfrak p)^{-s})$$
where $\chi$ are the characters of $\Gal(\K m/\Q) \cong (\Bbb Z/m\Bbb Z)^{\times}$ and $\det(1-\chi(\sigma_P)T, \Bbb C^{I_P})$ the characteristic polynomial of the $0$ or $1$-dimensional matrix representing the restriction of $\chi(\sigma_P)$ to the subspace of $\Bbb C$ invariant under $I(P/p)$.
 A: Write $m = p^k n$ where $(p,n) = 1$. Now let $d$ be any integer such that
$d$ is $p$ modulo $n$, and $d$ is not divisible by $p$. By the Chinese remainder theorem, the number of possible such $d$ is the number of integers less than $p^k$ which are prime to $p^k$. Thus there are precisely $\phi(p^k)$ choices of $d$, which --- no co-incidentally --- is the order of the inertia group at $p$.
I claim that $\sigma(\zeta_m) = \zeta^d_m$ does the job, and that these are precisely the $d$ for which the desired congruence holds.
We would like to show that $\zeta^d_m - \zeta^p_m$ is trivial modulo one (or since the extension is abelian any) prime $P$ above $p$. But we have:
$$\begin{aligned} \zeta^d_m - \zeta^p_m  \equiv & \ 0 \mod P  \Leftrightarrow \\
( \zeta^d_m - \zeta^p_m)^{p^k} \equiv & \ 0 \mod P \Leftrightarrow \\
(\zeta^d_m)^{p^k} - (\zeta^p_m)^{p^k}  \equiv & \ 0 \mod P \Leftrightarrow \\
\zeta^d_n - \zeta^p_n  \equiv & \ 0 \mod P \Leftrightarrow \\
\zeta^p_n - \zeta^p_n \equiv & \ 0 \mod P.
 \end{aligned}$$


*

*The first line is equivalent to the second because taking $p$th powers (Frobenius) is an isomorphism over a finite field.

*The second line is equivalent to the third because Frobenius is a ring homomorphism.

*The third line is equivalent to the fourth because $m = p^k n$ implies that $\zeta^{p^k}_m = \zeta_n$.

*The fourth line is equivalent to the fifth because $d$ is $p$ mod $n$ and so $\zeta^d_n = \zeta^p_n$.

