Towards the end of ch. VII §10 of Neukirch's Algebraische Zahlentheorie, he proves that the Artin $L$-series of Abelian extensions coincide with Hecke $L$-series, thereby proving Artin's conjecture for Abelian extensions.

He then mentions in passing that this also settles the Artin conjecture for all soluble extensions. I have been trying to convince myself why this is the case.

Let $E/K$ be a soluble extension of algebraic number fields, and let $G':=\textrm{Gal}(E/K)$. Then there exists a subgroup $N \vartriangleleft G'$ such that $G:=G'/N$ is Abelian. Let $M$ be the subfield fixed by $N$. Then we have $G \cong \textrm{Gal}(M/K)$, and for any non-trivial simple character $\chi$ of $G$, we have, by inflation: $$L(E/K,\chi',s) = L(M/K,\chi,s)$$ $\chi ' = \chi \circ \pi$, where $\pi:G' \to G \cong G'/N$ is the canonical projection.

As $M/K$ is an Abelian extension, this proves that $L(E/K,\chi',s)$ is holomorphic on $\mathbb{C}$.

The problem is that not every simple character of $\textrm{Gal}(E/K)$ can be expressed as $\chi \circ \pi$, where $\chi$ is a simple character of $\textrm{Gal}(M/K)$.

So why does this prove the Artin conjecture for soluble groups?

Thank you for your attention.

$\textbf{Addendum:}$ I thought I had a solution. I include it here, as it might contain some useful idea. I begin by listing three properties, the third of which is not true:

1) The "converse of inflation" (deflation?) mentioned here: If $G:=\textrm{Gal}(E/K)$ and $\chi$ is a simple character of $G$, then $L(E/K,\chi,s) = L(E_{\chi}/K,\chi',s)$, where $\chi': G/\textrm{Ker}(\chi) \cong \textrm{Gal}(E_{\chi}/K) \to \mathbb{C}^{\times}$ is a faithful simple character given by $\chi'(g\ \textrm{Ker}(\chi)):=\chi(g)$.

2) That every normal subgroup $N \vartriangleleft G$ can be expressed as the kernel of some simple character of $G$.

3) The quotient of a soluble group by a normal subgroup is always Abelian (NB: This is not true. Take $\textit{e.g.}$ $S_4/(C_2 \times C_2) \cong S_3$).

We then have that as $G$ is soluble, there exists a simple character $\chi$ of $G$, so that $G/\textrm{Ker}(\chi)$ is Abelian, and by the above $$ L(E/K,\chi,s) = L(E_{\chi}/K,\chi',s) $$ is entire.

My next idea was to replace the erroneous property 3) with the fact that the quotient of a soluble group by a normal subgroup is a soluble group, and then apply the process iteratively, but this does not seem to work.

$\textbf{Addendum II:}$ I have still not found a solution and hence add a bounty.

$\textbf{Addendum III:}$ I think I have a solution. Correct me if I am wrong.

I think the answer is just to keep "quotienting".

By 1) above, for any character $\chi$ of $\textrm{Gal}(E/K)$, we get a simple character $\chi'$ of $\textrm{Gal}(E'/K)$ and $L(E/K,\chi,s) = L(E'/K,\chi',s)$, where $E'$ is the subfield fixed by $\textrm{Ker}(\chi)$.

Applying this again w.r.t. the group $\textrm{Gal}(E'/K)/\textrm{Ker}(\chi') = \textrm{Gal}(E''/K)$, we get a simple character $\chi''$, and $L(E/K,\chi,s) = L(E'/K,\chi',s) = L(E''/K,\chi'',s)$, where $E''$ is the subfield of $E'$ fixed by $\textrm{Ker}(\chi')$.

Applying this iteratively, we find that $L(E/K,\chi,s) = L(E^{(n)}/K,\chi^{(n)},s)$, where $E^{(n)}/K$ is an Abelian extension for some $n \in \mathbb{N}$. Note that this process must terminate after a finite number of iterations, as the groups in question are finite - by "terminate", I mean that eventually the resulting quotient group is Abelian.

$\textbf{Addendum IV:}$ The above is not a solution. See the comment by Lukas Heger below.

$\textbf{Addendum V:}$ Would the Artin conjecture for all soluble extensions not trivially imply the proofs of Langlands and Tunnell of the Artin conjecture for tetra- and octahedral representations? What I mean is the following:

What they proved is that if $(V,\rho)$ is a degree 2 representation of a finite group $G$ and $\rho(G)/Z(\rho(G))$ is isomorphic to either $A_4$ or $S_4$, then the corresponding Artin $L$-series is entire.

But we know that if $G/Z(G)$ is soluble, then so is $G$. And $A_4$ and $S_4$ are soluble groups!

On the other hand, $\rho(G)$ soluble does not necessarily imply that $G$ is soluble, so I suppose the work of Langlands and Tunnell is only non-trivial for degree 2 representations of insoluble extensions whose images in the projective general linear group are isomorphic to either $A_4$ or $S_4$. In particular, these representations must be unfaithful.

This is of course assuming that we have a proof of the Artin conjecture for all soluble extensions.

I should also add that when I wrote $\textrm{Ker}(\chi)$ above, I meant $\textrm{Ker}(\rho)$, where $(V,\rho)$ is the representations corresponding to the character $\chi$. I realise this is poor notation, but hopefully unambiguous.

$\textbf{Addendum VI:}$ The same is true for unfaithful representations!

We assume the Artin conjecture for soluble extensions.

Let $G$ be insoluble and $(V,\rho)$ an unfaithful representation of $G$ so that $\rho(G)$ is soluble. Let $E_{\rho}$ be the subfield fixed by the kernel of $\rho$. We then have $\rho(G) \cong G/\textrm{Ker}(\rho) \cong \textrm{Gal}(E_{\rho}/K)$, and:

$$ L(E/K,\rho,s) = L(E_{\rho}/K,\rho',s) $$

where $\rho' = \rho \circ \pi$. As the RHS is an $L$-series of a soluble extension, it is entire by our assumption.

Thus the Artin conjecture for all soluble extensions does indeed imply the proofs of Langlands and Tunnell.

My conclusion: Neukirch was mistaken, and proving the Artin conjecture for all soluble extensions is not as simple as he seems to have imagined.

It seems that the Artin conjecture for soluble extensions is in fact an open problem.

Addendum VII: I should add that the credit for discovering the falseness of the above claim goes entirely to my friend and colleague O. Justinussen.

If anyone is able to prove him wrong by providing an elementary proof of the Artin conjecture for all soluble extensions, please let me know.

  • 2
    $\begingroup$ The process will always terminate after one step, since $\chi'$ is injective. Thus it only terminates at an abelian group if the image of $\chi$ is abelian. $\endgroup$ Jan 14, 2020 at 7:42
  • $\begingroup$ Do you use that G is solvable? $\endgroup$
    – eduard
    Jan 14, 2020 at 10:55
  • $\begingroup$ @LukasHeger You are completely right. Back to the drawing board. $\endgroup$ Jan 14, 2020 at 12:30
  • $\begingroup$ @eduard In my "solution" I used the property that the quotient of a soluble group by a normal subgroup is again a soluble group. $\endgroup$ Jan 14, 2020 at 12:31

1 Answer 1


The following is just a partial answer.

An example: $G = \mathfrak S_3$

Let $\mathfrak S_3$ denote the symmetric group of three elements and let $C_3=\mathfrak A_3$ be the alternate subgroup. The group $\mathfrak S_3$ has 3 irreducible characters.

  1. The trivial character
  2. The sign, i.e. the unique non-trivial group homomorphism $\mathfrak S_3\rightarrow \pm 1$.
  3. The character attached to the induced representation $\rho =Ind^{\mathfrak S_3}_{C_3}\chi$ of the non-trivial character $\chi:C_3 \rightarrow \mathbb C^\times$.

The character defined in 3. is the smallest non-linear character and looks like a good starting point.

Let $E/K$ be a $\mathfrak S_3$-Galois extension (of number and let $M$ the fixed field by $C_3$. It is a well know property that $$ L(E/M, \chi) = L(E/K,\rho).\qquad (1) $$

Thus the Artin conjecture holds for $\mathfrak S_3$- Galois extensions, since $\chi$ is a non-trivial (degree 1) character.

Immediate generalization

Equation (1) holds for general induced representations. Hence the above discussion applies to any dihedral group. Indeed, one can check that every irreducible character of $D_n$ is either degree 1 or induced from a (degree 1) character of $C_n$. A nice reference here is 5.3, Serre's book on Linear Representations of finite groups.

Some remarks on representation theory

Let G be a group. A well known theorem of Brauer predicts that every irreducible character of G is a linear combination (in integer coefficients) of (characters of) induced representations $Ind_H^G \chi$ where $H$ ranges over subgroups of $G$ and $\chi$ ranges over linear characters of $H$.

Hence, if the linear combination is of positive coefficients then Artin Conjecture holds.

See this wikipedia article or this one for further remarks.


Here is an inductive proof for super supersolvable groups exploiting equation (1).

  • $\begingroup$ Many thanks for your insightful answer. I am aware of the fact that the Artin conjecture holds for all supersoluble groups, as every character of a supersoluble group is monomial. But unfortunately, not all soluble groups are supersoluble. Moreover, I do not think this was what Neukirch was referring to when he casually mentioned that "this also settles the Artin conjecture for every soluble extension". I am afraid that I cannot accept this as a complete answer. $\endgroup$ Jan 19, 2020 at 18:42

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