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I came across the statement in a book:

Let $k$ be a number field and $K$ be a Galois extension of $\mathbb Q$ containing $k$, with Galois group $G=\operatorname{Gal}(K/\mathbb Q)$ and let $G_k:=\operatorname{Gal}(K/k)$. Let $\chi$ denote the character of the permutation representation of $G$ in $G/G_k$. Then the Artin $L$-function $L(s, \chi)$ is the Dedekind-Zeta function $\zeta_k$ of the extension $k / \mathbb Q$.

Now first of all, I wanted to confirm whether the "permutation representation" here is the 'standard' one given by $\rho: G \longrightarrow \operatorname{Sym}(G/G_k) : \rho(\sigma) (\tau G_k) := \sigma\tau G_k$ for every $\sigma, \tau \in G$ (i.e. the group action $\sigma \cdot (\tau G_k) = \sigma\tau G_k$).

Second, I could see that on account of the ring $\mathcal{O}_k$ of integers of $k$ being a Dedekind Domain, unique prime factorization of ideals holds and we may write the Dedekind Zeta function in the analogous Euler-Product representation: $$\zeta_k(s) \triangleq \sum_{\mathfrak a \lhd \mathcal O_k} \frac{1}{N(\mathfrak a)^s} = \prod_{\mathfrak{p} \in Spec(\mathcal{O}_k)} \frac{1}{1-N(\mathfrak{p})^{-s}}\text{ for }\Re(s)>1 \hspace{3mm} \cdots \hspace{2mm} (1)$$

But it is not clear to me from the definition of the Artin $L$-function why the above equality should hold. The definition of the Artin $L$-function that I am familiar with of a general character $\eta$ of a representation $\rho: G \longrightarrow GL(V)$ (for some complex vector space $V$) is the following one on Wikipedia: $$L(s, \chi) = \prod_{\mathfrak{p}} \frac{1}{\det[I - N(\mathfrak{p})^{-s}\rho(\sigma_{\mathfrak{p}})|V_{\mathfrak{p}, \rho}]} \hspace{3mm} \cdots \hspace{2mm} (2)$$ where the product on the left is over all prime ideals $\mathfrak p$ of $k$.

What am I missing?

Edit: I am sorry it wasn't clear about the question I was asking. First, I just wanted to confirm whether the "permutation representation" referred to in the problem statement is the one I wrote above or not. Second, the only part that is not clear to me is why $L(s, \chi) = \zeta_k(s)$, from the definition of the Artin $L$-function that I know (which is (2)). I am okay with (1) and (2), the only thing I don't understand is why $L(s, \chi) = \zeta_k(s)$.

Edit: We have in this case (as Will Sawin has pointed out) $$L(s, \chi) = \prod_p \frac{1}{\det[I-p^{-s}\rho(\sigma_p)]}$$ where the product on the left is over all integer primes $p$. I tried to show that this is equal to the product occurring on the right hand side of (1) for $\Re(s)>1$. We would therefore be done if could show, for all integer primes $p$ and for all such $s$, the identity $$\det[I-p^{-s}\rho(\sigma_p)|V_{p, \rho}] = \prod_{\mathfrak{p}|p} (1-N(\mathfrak{p})^{-s})$$ To show the last equality, I tried expanding the determinant on the left into a product of eigenvalues, but I'm not sure how to proceed from there.

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    $\begingroup$ What do you mean? You're not clear why you have convergence for Re(s) > 1? Or you don't understand how to prove the Euler product given convergence? In any case, my personal view is that math.stackexchange is a better fit for this as, if I understand correctly, I think you're asking about something that's treated in many standard (graduate) number theory textbooks. $\endgroup$ – Kimball May 18 at 23:09
  • $\begingroup$ Apologies for the misunderstanding, I have made the question clear now. $\endgroup$ – asrxiiviii May 18 at 23:35
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    $\begingroup$ 1. What have you tried so far? 2. Yes, the permutation representation is the standard one. 3. Because $\chi$ is a character of the Galois group of $K$ over $\mathbb Q$, the product is over prime ideals of $\mathbb Q$, not $k$. $\endgroup$ – Will Sawin May 18 at 23:40
  • $\begingroup$ Thank you for confirming. I have added the attempts I've made so far. $\endgroup$ – asrxiiviii May 19 at 0:14
  • $\begingroup$ You seem to have crossposted this at MSE now: math.stackexchange.com/q/3683157/11323 To be clear, this is generally discouraged (at least here) and I was suggesting that you should've just asked this question there and not here in the first place (the line is kind of blurry, I know). But then if you don't get an answer after a few days, you might try asking here, referencing your MSE post. $\endgroup$ – Kimball May 20 at 13:20
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It is not easy to give all the details so I'll give a sketch in the case of unramified prime

  • For $p$ an unramified prime number, $Q\subset O_K$ a prime ideal above $p$, those of $O_k$ are of the form $P_g =g(Q)\cap O_k$ for $g\in G$ with norm $N(P_g)=p^{f_g}$ for some integers $f_g$

    $\sigma$ a Frobenius such that $\forall a\in O_K,\sigma(a)\equiv a^p\bmod Q$

    The Frobenius of $O_K/g(Q)$ is $g^{-1}\sigma g$ and $f_g$ is the least integer such that $(g^{-1}\sigma g)^{f_g}\in H$ ie. such that $\sigma^{f_g} gH = gH$.

    (this is because if $f_g$ was smaller then $P_g$ would appear with multiplicity $>1$ in the factorization of $pO_k$)

  • With $\rho$ the representation of $G$ permuting $G/H$ then $\det(I-\rho(\sigma)p^{-s}) = \prod_{C \text{ orbits of } \langle \sigma \rangle \text{ on } G/H} (1-p^{-s|C|})$

    With $C= \langle \sigma \rangle g H$ then $|C|=f_g$

    Thus the unramified Euler factors of $\zeta_k(s)$ and $L(s,\rho,K/\Bbb{Q})$ are equal.

For the ramified primes it works similarly after taking the subfield fixed by the inertia subgroup.

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  • $\begingroup$ Thank you for your answer, I'll take some time to read and understand it. Can you give me a reference where I can find an accessible proof of it (preferably a proof where I'll need the least prerequisites)? $\endgroup$ – asrxiiviii May 19 at 0:30
  • $\begingroup$ Sorry I didn't understand why the formula $$\det(I-\rho(\sigma)p^{-s}) = \prod_{C \text{ orbits of } \langle \sigma \rangle \text{ on } G/H} (1-p^{-s|C|})$$ holds. Is it a standard result? Also by "frobenius of O_K/g(Q)" do you mean the Forbenius element of the prime $Q$ lying over $P_g$ (=$O_k \cap g(Q)$)? $\endgroup$ – asrxiiviii May 19 at 0:53
  • $\begingroup$ On each $C$, $\rho(\sigma)$ is permuting cyclically the $|C|$ elements of $C$ so on this subspace of dimension $|C|$ the matrix has order $|C|$. This is enough to say that its minimal/characteristic polynomial is $X^{|C|}-1$. $\endgroup$ – reuns May 19 at 1:03
  • $\begingroup$ Sorry for bringing this up after a long time, but could you please confirm the following: by "frobenius of $O_K/g(Q)$" do you mean the Frobenius automorphism of the residue field? Thanks. $\endgroup$ – asrxiiviii May 21 at 6:39
  • $\begingroup$ In the residue field there is a unique Frobenius $O_K/g(Q)$ which is $a\to a^p$, for $ K/Q$ Galois then it lifts to a unique $g\in Gal(K/Q)$ iff $p$ is unramified, when $p$ is ramified with $g$ a lift then the others are $I_{g(Q)} g$ (the inertia subgroup). This is what I meant with en.wikipedia.org/wiki/… is the main prerequisite. If the concepts are unclear to you then try with a few $p$ in $K=Q(i,\sqrt{2}),k=Q(i)$ @asrxiiviii $\endgroup$ – reuns May 21 at 17:55

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