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.

  • 2
    $\begingroup$ 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, 2020 at 23:09
  • $\begingroup$ Apologies for the misunderstanding, I have made the question clear now. $\endgroup$
    – asrxiiviii
    May 18, 2020 at 23:35
  • 2
    $\begingroup$ 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, 2020 at 23:40
  • $\begingroup$ Thank you for confirming. I have added the attempts I've made so far. $\endgroup$
    – asrxiiviii
    May 19, 2020 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, 2020 at 13:20

1 Answer 1


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.

  • $\begingroup$ 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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.