$\newcommand\Q {\Bbb Q} \newcommand\Z[1]{\Bbb Z/#1 \Bbb Z} \newcommand\Gal{\mathrm{Gal}} \newcommand\K[1]{\Bbb Q(\zeta_{#1})}$ Let $K/\Q$ be an abelian extension with Galois group $G$ and let $\rho : G \to \Bbb C^*$ be a character.

Let $\sigma_p \in \Gal(K/\Q)$ such that $\sigma_p(x)-x^p \in P$ for any $x \in \mathcal O_K$, where $P$ is any prime of $K$ above $p$.

Is it true that there is an integer $k \geq 1$ and a Dirichlet character $\chi : (\Z k)^{\times} \to \Bbb C^*$ such that $$\rho(\sigma_p)\Big\vert_{\Bbb C^{\,I_P}} = \chi(p)$$ for any rational prime $p$ ?

where $\Bbb C^{\,I_P}$ is the $0$ or $1$-dimensional subspace of $\Bbb C$ invariant under the inertia group $I(P/p) \leq \Gal(K/\Bbb Q)$.

Actually I want to show that there is an integer $k \geq 1$ and a Dirichlet character $\chi : (\Z k)^{\times} \to \Bbb C^*$ such that $L(\rho,s)=L(\chi,s)$ for any $s>1$, where the LHS is the Artin L function and the RHS is the Dirichlet L function.


— By Kronecker–Weber, there is an integer $k \geq 1$ such that $K \subset \K k$. Let $\pi : \Gal(\K k / \Q) \to \Gal(K/\Q)$ be the projection and define $\chi = \psi \circ \pi \circ \rho$, where $\psi$ is an isomorphism $(\Z k)^{\times} \to \Gal(\K k / \Q)$.

If $p$ is unramified, then $I(P/p) = 0$ for any $P$ above $p$, and we can take $\sigma_p$ to be the restriction of $\zeta_k \mapsto \zeta_k^p$ (where $(p,k)=1$). Then $\sigma_p$ is mapped to $p \pmod k$ via $\psi$, and $\chi(p) = \rho(\sigma_p)$ as desired.

But what happens for ramified primes? If $p$ is ramified, then $\chi(p) = 0$, so we would need $\Bbb C^{I_P}$ to be $0$. It doesn't seem to be true.

  • $\begingroup$ Notice that I defined : \newcommand\Q {\Bbb Q} \newcommand\Z[1]{\Bbb Z/#1 \Bbb Z} \newcommand\Gal{\mathrm{Gal}} \newcommand\K[1]{\Bbb Q(\zeta_{#1})} $\endgroup$ – Watson Dec 28 '16 at 20:17
  • $\begingroup$ @user1952009 : thank you for your comments! However, I don't understand, since here my character $\chi$ is fixed (or I have to find just one Dirichlet character), while you are talking about several characters. If $p$ is ramified, then $\chi(p)=0$ for me, even if $\chi$ is primitive. $\endgroup$ – Watson Dec 28 '16 at 20:38
  • $\begingroup$ @user1952009 : Yes, but I'm not talking about the trivial representation. I'm talking about the Dirichlet character $\chi : (\Z k)^{\times} \to \Bbb C^*$, and again if $p$ is ramified then $p$ divides $k$ and $\chi(p)=0$. $\endgroup$ – Watson Dec 28 '16 at 20:42
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    $\begingroup$ See there it makes a special case for ramification $\endgroup$ – reuns Dec 28 '16 at 21:00
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    $\begingroup$ Ah OK. I haven't checked all the details so I won't post this as an answer. I think the problem is that $\rho$ naturally corresponds to a subextension of $K$, and not to $K$ itself. By Galois theory, the $\ker\rho$ is $\mathrm{Gal}(K/L)$ for some subextension $L$. $\rho$ now factors through a representation $\phi$ of $\mathrm{Gal}(L/\mathbb Q)$. The point is that now $L$ is unramified at $p$ if and only if $\phi(I_{L,p})$ acts trivially on $\mathbb C$, and your argument with Kronecker-Weber plus the various compatibilities of Frobenius elements and inertia should give the result. $\endgroup$ – Mathmo123 Dec 29 '16 at 15:27

Thanks to Mathmo123's comments, I can provide an answer.

I will show that $L(\rho,s)=L(\chi,s)$, by showing that $$\det(1-\rho(\sigma_P)p^{-s}, \mathbb{C}^{I_P}) = 1-\chi(p)p^{-s}$$ for any prime $p$.

Without loss of generality, we can assume that $\rho$ is faithful. Let $H = \mathrm{ker}(\rho)$, and $\pi_H : G \to G/H$ the projection, where $G=\Gal(K/\Q)$. Then, from theory about Artin L functions, we have $$\mathcal L(K/\Q, \rho,s)=\mathcal L(K^H/\Q, \phi,s)$$ where $\rho = \phi \circ \pi_H$ (and $K^H/\Q$ is abelian). So we can replace $\rho$ and $K$ by $\phi$ and $K^H$ if necessary.

— If $p$ is not ramified in $K$, then $\sigma_p$ corresponds to $p$ as I said above. We get $$\det(1-\rho(\sigma_P)p^{-s}, \mathbb{C}^{I_P}) = 1-\chi(p)p^{-s}.$$

— If $p$ is ramified, then $I(P/p) \neq \{id\}$ and since $\rho$ is injective, we have $\rho(I(P/p)) \neq \{1\}$. Therefore $\mathbb{C}^{I(P/p)} = 0$ and $\det(1-\rho(\sigma_P)p^{-s}, \mathbb{C}^{I_P})=1$. But $p$ is also ramified in $\mathbb{Q}(\zeta_k) \supset K$, so $p$ divides $k$, so that $\chi(p)=0$. We also have $1-\chi(p)p^{-s}=1$ i.e. we also get $$\det(1-\rho(\sigma_P)p^{-s}, \mathbb{C}^{I_P}) = 1=1-\chi(p)p^{-s}$$

which concludes the proof.

  • $\begingroup$ Hi, your argument is not correct because $$\zeta_{\mathbb{Q}(\zeta_k)}(s) =\prod_{\chi \bmod k} L(s,\tilde{\chi}) \tag{1}$$ where $\tilde{\chi}$ is the primitive character underlying $\chi$. For $p$ ramified, that $\det(I-p^{-s}\rho(\sigma_p)|V^{\rho|_{I_p)}})= 1$ or $1-tr(\rho(\sigma_{\mathfrak{p}})) p^{-s}$ depends on that $\rho|_{I_p}$ is trivial or not. A solution could be to find $p,e(p),f(p),g(p),\sigma_p,I_p$ in $\mathbb{Q}(\zeta_k)$ in order to show $(1)$ is true, then to deduce what we get for $e,f,g,I_p$ in subfields and their representations. $\endgroup$ – reuns Nov 16 '17 at 7:28
  • $\begingroup$ A representation $\rho : \text{Gal}(K/\mathbb{Q}) \to GL_n$ is unramified at $p$ iff $\rho|_{I_p}$ is the trivial representation. In that case the local factor at $p$ is of degree $n$. $\endgroup$ – reuns Nov 16 '17 at 7:35
  • $\begingroup$ @reuns : I don't understand where exactly my above argument fails. I was able to show proposition 5.12 in Neukirch's Algebraic Number Theory book (which takes into account the ramified primes). $\endgroup$ – Watson Nov 16 '17 at 17:43
  • $\begingroup$ (Notice also the difference between $\rho\vert_{I_P}$ and $\rho(\sigma_p)\vert_{V^{I_P}}$). $\endgroup$ – Watson Nov 16 '17 at 17:43
  • $\begingroup$ I told you, if $p$ is unramified in $K$ then $\rho$ is unramified at $p$, but if $p$ is ramified in $K$ it doesn't mean $\rho$ is ramified at $p$. Here $\det(I-p^{-s}\rho(\sigma_p)|V^{\rho|_{I_p)}})=1$ iff $\rho(g) \ne 1$ for some $g \in I_P$. $V^{\rho|_{I_P}}$ is the subspace of $\mathbb{C}^n$ onto which $\rho|_{I_P}$ acts trivially. Another example : for $\rho$ the trivial representation $Gal(K/\mathbb{Q}) \to GL_1$ then $L(s,\rho) =\zeta_\mathbb{Q}(s)$ ie. $\rho$ is unramified at every prime. $\endgroup$ – reuns Nov 16 '17 at 23:00

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