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The Dedekind zeta function of an abelian extension $E$ of $\mathbb{Q}$ factors as a product of Artin L function $L(s, \chi)$, where the product runs over all irreducible representations $\chi$ of $Gal(E : \mathbb{Q})$ .

Question: What is known for irreducible representation $ \sigma$ of $G(F) = Gal(\overline{Q}, F)$. How does the Artin $L$ function decompose? Something like $$L_F(s, \sigma) = \prod\limits_{\sigma' \subset Ind_{G(F)}^{G(E)} \sigma} L_E(s, \sigma'),$$ where $F$ is a finite extension of $E$?

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    $\begingroup$ Surely the fact that the Dedekind zeta function of $E$ factorises is simply a restatement of the fact that the regular representation of a group $G$ factorises as a product over all the irreducible representations of $G$? Artin reciprocity is more the statement that any L-function associated to a one-dimensional representation is in fact a Hecke L-function. $\endgroup$ Commented Jun 8, 2011 at 9:42
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    $\begingroup$ Yes, this is true, and is a standard consequence of the inductivity properties of Artin L-functions. $\endgroup$ Commented Jun 8, 2011 at 10:21
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    $\begingroup$ Anywhere Artin L-functions are sold. $\endgroup$ Commented Jun 8, 2011 at 10:26
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    $\begingroup$ For example, try taking a look at Chapter 2 of M. Ram Murty & V. Kumar Murty, Non-vanishing of L-functions and applications, Birkhauser 1997. $\endgroup$
    – Stefano V.
    Commented Jun 8, 2011 at 11:55
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    $\begingroup$ For $F$ a number field, the above product is incorrect. For example it would imply that $1=\prod_{\sigma′\in Ind^{G(E)}_{G(F)} \setminus \{\sigma\}} L(s,\sigma′,F)$, which is false unless $F=\mathbb{Q}$. What is true however is that $L(s,\sigma,F) = L(s, Ind^{G(E)}_{G(F)}(\sigma), E)$ for any $E\subset F.$ $\endgroup$
    – JSpecter
    Commented Jun 8, 2011 at 13:24

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You should define what you mean by a decomposition of an Artin $L$-function. If you assume standard conjectures of Langlands and Selberg, then the Artin $L$-function of an irreducible representation of $G(\mathbb{Q})$ is a primitive function in the Selberg class, hence it has no nontrivial decomposition there (or among Artin $L$-functions for that matter). If you start with an irreducible representation $\sigma$ of $G(F)$, then $L_F(s,\sigma)=\prod_\rho L_\mathbb{Q}(s,\rho)^{m(\rho)}$, where $\rho$ runs through the irreducible representations of $G(\mathbb{Q})$ and $m(\rho)$ denotes the multiplicity of $\rho$ in the induced representation of $\sigma$ from $G(F)$ to $G(\mathbb{Q})$. This should be the unique maximal factorization into $L$-functions over $\mathbb{Q}$. In particular, if $F/\mathbb{Q}$ is Galois, then $L_F(s,\sigma)$ should be "irreducible". For a reference I recommend Murty's paper here.

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    $\begingroup$ The example I looked in detail is $F = K(\alpha),K= \mathbb{Q}(\sqrt{\Delta})$ the splitting field of a rational cubic whose discriminant $\Delta < 0$. Then the irreducible representation $\rho : \text{Gal}(F/\mathbb{Q}) \simeq S_3 \to GL_2$ is induced by $\psi : \text{Gal}(F/K) \to GL_1$ which is (by Artin reciprocity and because $F/K$ is unramified) just a character of the ideal class group of $K$. From this we obtain that $L(s,\rho)= L(s,\psi)$ comes from a modular form $f \in S_1(\Gamma_0(\Delta),(\frac{.}{\Delta}))$ $\endgroup$
    – reuns
    Commented Oct 8, 2017 at 17:50
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Thinking from the automorphic point of view, it should be an $L$-function over $\mathbb Q$ corresponding to the automorphic induction of the Hecke character to a representation $\Pi$ of $GL_N(\mathbb A_{\mathbb Q})$ where $N=deg(E:\mathbb Q)$. However, automorphic induction is only known for cyclic (whence solvable) and non-normal cubic extensions.

Depending on your extension and what you induce, it may be that $\Pi$ is cuspidal, but in general it should break up as an isobaric sum $$ \Pi = \pi_1 \boxplus \cdots \boxplus \pi_r,$$ where the $\pi_i$'s are cuspidal. ($r=1$ and $\pi_1 = \Pi$ if $\Pi$ is cuspidal). Then your desired $L$-function decomposition over $\mathbb Q$ is $$ L(s,\pi_1)L(s,\pi_2) \cdots L(s,\pi_r).$$

In the case of the Dedekind zeta function for an abelian extension, these $\pi_i$'s correspond to the irreducible characters of Gal$(E/\mathbb Q)$. For more general extensions, one knows this when each irreducible representation of Gal$(E/\mathbb Q)$ is known to be modular (corresponding to a cuspidal automorphic representation).

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  • $\begingroup$ Thanks for this answer and the automorphic picture. I edited the question, since i am here interested in a purely galois theoretic formulation of the decomposition. $\endgroup$
    – Marc Palm
    Commented Jun 8, 2011 at 10:13

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