I am studying the proof of integrality of the conductors of Galois representations from these notes, and I have hit a roadblock in a step of the proof of Proposition 3.1.40 (page 57).
The setting is as follows: $K$ is a $p$-adic field, $\rho$ is a representation of the absolute Galois group $G_K$ of $K$ over a field $F$ with $p \in F^{\times}$. We assume that $\rho$ factors through the Galois group of some finite Galois extension $L/K$.
(I would be interested by arguments where $K$ is, more generally, a complete non-Archimedean discretely valued field with perfect residue field of characteristic $p$).
Let $V$ be the underlying vector space of $\rho$. For $u >-1$, let $V_u=V/V^{G_K^u}$ (where the exponent denotes ramification groups in “upper numbering”). The conductor exponent $n(\rho)$ (resp. the Swan exponent $S(\rho)$) is defined by the integral $\int_{-1}^{\infty}{\dim{V_u}\,du}$ (resp. the same integral from $0$ to $\infty$).
Let me recall, for the sake of clarity, the basic steps of the argument showing that $n(\rho)$ is integral.
- Show that we can reduce to the case where $L/K$ is totally ramified.
- Show that we can further reduce to the case where $\rho$ is irreducible.
- Prove (through Hasse-Arf) that $n(\rho)$ is an integer when $\rho$ is of dimension $1$.
- Show that if $\rho$ is induced from a representation $\rho_1$ of $G_M$ with $M/K$ finite, and $n(\rho_1)$ is an integer, so is $n(\rho)$.
- Show that under the hypotheses of 2., $\rho$ is induced by a character of a subgroup.
I am satisfied with steps 1-4, but encounter a problem in the proof of 5. The key claim (which is more or less the heart of the aforementioned Proposition 3.1.40) is as follows: assume that $\rho$ is faithful on $Gal(L/K)$ and irreducible with $L/K$ totally ramified, and $\rho$ does not have an abelian image. Then $\rho$ is induced by an irreducible representation from a proper subgroup.
The notes’ proof of this claim can be summarized as follows:
“By a lemma from Serre’s Linear Representations of Finite Groups, it is enough to show that $Gal(L/K)$ has a normal non-central abelian subgroup. Let $s \geq 0$ be minimal such that $Gal(L/K)_{s+1}$ is central. In fact, $s \geq 1$, and we have the exact sequence $1 \rightarrow Gal(L/K)_{s+1} \rightarrow Gal(L/K)_s \rightarrow Gal(L/K)_s/Gal(L/K)_{s+1} \rightarrow 1$ where the kernel is central and the cokernel is a $\mathbb{F}_p$-vector space; as the exact sequence splits, $Gal(L/K)_s$ is abelian normal non-central.”
Why can the exact sequence be split? That amounts to producing elements in $Gal(L/K)_s$ with order exactly $p$ such that they commute.
But perhaps we can be a little less specific: consider a subgroup $V \leq Q:=Gal(L/K)_s/Gal(L/K)_{s+1}$ and $W \leq Gal(L/K)_s$ its inverse image. We want $W$ to be abelian and normal.
Normality translates to a condition on $V$: namely, that it is stable under the natural action by conjugation of the Galois group – thus the [cyclic] tame inertia group – which is the $s$-th power of the inertia character according to Local Fields. But the only good criterion I know showing that $W$ is abelian is when $V$ is cyclic, and there is no reason why the inertia subgroup should have fixed $\mathbb{F}_p$-lines in $Q$!
If we distance ourselves further from the details of the method, the broad question becomes to find in $Gal(L/K)$ a normal abelian non-central subgroup. Is this always possible?
I know these exist for nilpotent groups (which $Gal(L/K)$ usually isn’t if there is tame ramification), and even if the group is supersolvable (but is this the case?).
Or perhaps this could still hold even though the group isn’t supersolvable because of other properties of the ramification filtration?
There are two possible workarounds but they don’t look very appealing:
a. Use Brauer’s theorem on induced characters. The issue is that [as far as I know] it only works really well in characteristic zero, and I’m interested in Artin and $\ell$-adic representations, and therefore I still need to work with representations mod $\ell$ with finite image.
b. Restrict to totally ramified extensions $L/K$ with no tame ramification. The hitch is that restricting a representation to the wild inertia group multiplies the Swan invariant by the tame ramification index, and therefore it’s not enough anymore to show that the restriction of the representation has integral Swan exponent.
So how can I proceed?
Thank you.