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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 $I(P/p)$.

Watson
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    If $m = p$ then $X^p-1 = (X-1)^p \bmod p, \Phi_p(X) = (X-1)^{p-1} \bmod p$ so that $p\mathcal{O}_K = (p\mathcal{O}_K+(\zeta-1)\mathcal{O}_K)^{p-1}$ and $ \zeta^{p-1}-\zeta^{p} \equiv 0 \bmod (p,\zeta-1)$ i.e. the Frobenius element is $\sigma(\zeta) = \zeta^{p-1}$. And you can probably use the tower of cyclotomic extensions for generalizing to $p | m$ – reuns Dec 27 '16 at 20:04
  • @user1952009 : Thank you very much! What do you mean by "use the tower of cyclotomic extensions for generalizing to $p \mid m$" ? – Watson Dec 27 '16 at 20:34
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    If $gcd(p,r) = 1$ and $m = pr$ then write $\zeta_m = \zeta_p^a \zeta_r^b$ and I think $\sigma( \zeta_p) = \zeta_p^{p-1}, \sigma(\zeta_r)= \zeta_r^p$ so that $\sigma(\zeta_m) = \sigma(\zeta_p^a \zeta_r^b) = \zeta_p^{a(p-1)} \zeta_r^{bp}$ – reuns Dec 27 '16 at 23:08
  • "I know that $|D/I|=f_{P/p}=f_p$ is the order of the class of p mod M" (Neukirch, ANT, p. 61). – Watson Dec 28 '16 at 11:03

1 Answers1

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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:

$$\begin{aligned} \zeta^d_m - \zeta^p_m \equiv & \ 0 \mod P \Leftrightarrow \\ ( \zeta^d_m - \zeta^p_m)^{p^k} \equiv & \ 0 \mod P \Leftrightarrow \\ (\zeta^d_m)^{p^k} - (\zeta^p_m)^{p^k} \equiv & \ 0 \mod P \Leftrightarrow \\ \zeta^d_n - \zeta^p_n \equiv & \ 0 \mod P \Leftrightarrow \\ \zeta^p_n - \zeta^p_n \equiv & \ 0 \mod P. \end{aligned}$$

  1. The first line is equivalent to the second because taking $p$th powers (Frobenius) is an isomorphism over a finite field.
  2. The second line is equivalent to the third because Frobenius is a ring homomorphism.
  3. The third line is equivalent to the fourth because $m = p^k n$ implies that $\zeta^{p^k}_m = \zeta_n$.
  4. The fourth line is equivalent to the fifth because $d$ is $p$ mod $n$ and so $\zeta^d_n = \zeta^p_n$.
The Potter's Vessel
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  • Thank you for your answer! I think there is a small typo in the 3rd paragraph : $\zeta^n_m - \zeta^p_m$ [is trivial modulo…] should be $\zeta^d_m - \zeta^p_m$, shouldn't it? – Watson Dec 28 '16 at 10:46
  • Moreover, can we justify the existence of an integer d which is p modulo n, and d is not divisible by p, without using Dirichlet arithmetic progression theorem? – Watson Dec 28 '16 at 10:57
  • [Just for myself : the set

    $${[d]_m \mid d \equiv p \pmod n, (d,p)=1 }$$ is equal to

    $${\phi([d_0]n, [u]{p^k}) \mid [u]_{p^k} \in (\Bbb Z/p^k)^{\times}}$$

    where $d_0 \equiv p \pmod n$ is fixed, such that $(d_0,p)=1$, and $\phi : \Bbb Z/p^k \times \Bbb Z/n \to \Bbb Z/m$ is an isomorphism.]

     – Watson Dec 28 '16 at 11:01
  • One more question: the inertia subgroup $I_p$ (which only depends on $p$ since the extension is abelian) is $(\Bbb Z/p^k)^{\times}$, when $\mathrm{Gal}(\K m / \Bbb Q)$ is identified with $(\Bbb Z/m)^{\times}$, right? – Watson Dec 28 '16 at 12:07
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    First comment: typo corrected. Second comment: I'm not sure why you would think of Dirichlet's theorem, this is much easier; it is just the Chinese remainder theorem: solve for $d$ in the equations $d \equiv p \mod n$ and $d \equiv u \mod p^k$ for any choice of $u \in (\mathbf{Z}/p^k \mathbf{Z})^{\times}$ (this is implicit in your third comment anyway). Fourth comment: yes, where $(\mathbf{Z}/p^k \mathbf{Z})^{\times}$ is identified as the kernel of $(\mathbf{Z}/m \mathbf{Z})^{\times} \rightarrow (\mathbf{Z}/n \mathbf{Z})^{\times}$. – The Potter's Vessel Dec 28 '16 at 18:14