4
$\begingroup$

Let $G$ be a finite group, let $M$ be a $\mathbb{Z}[G]$-module whose underlying abelian group is finite-rank free abelian, and let $\mathbb{F}$ be a field of characteristic $0$. For some prime $p$ dividing $|G|$, assume that $M \otimes \overline{\mathbb{F}}_p$ is an irreducible $\overline{\mathbb{F}}_p[G]$-module. It then follows that $M \otimes \mathbb{F}$ is an irreducible $\mathbb{F}[G]$-module.

I need this result for a side comment in a paper I'm writing. I know how to derive it from stuff in Part III of Serre's "Linear representations of finite groups", but I don't want to have to explain this and would just prefer to give a clean citation for it. Does anyone know one?

$\endgroup$
1
  • 1
    $\begingroup$ @GeoffRobinson I was about to write the same thing: Consider the augmentation ideal in $\mathbb{Z}C_3$; this is irreducible over $\mathbb{F}_p$ if p is congruent to 2 mod 3, but not over $\mathbb{C}$. You need to either say absolutely irreducible or assume that the fields are algebraically closed. $\endgroup$
    – Ben Webster
    Commented Jun 28, 2021 at 19:17

1 Answer 1

1
$\begingroup$

I don't know of just a citation, but here is a pretty quick way to deduce this from the literature (I imagine that this might be the argument you had in mind, in which case apologies for telling you things that you already know):

If $\mathcal{O}$ denotes the integral closure of $\mathbb{Z}$ in $\mathbb{F}$, then one has $$ M\otimes\bar{\mathbb{F}}_p=((M\otimes \mathcal{O})\otimes_{\mathcal{O}}\mathcal{O}/\mathfrak{p})\otimes_{\mathcal{O}/\mathfrak{p}} \bar{\mathbb{F}}_p, $$ where $\mathfrak{p}$ is an arbitrarily chosen maximal ideal of $\mathcal{O}$ above $p$. Now use the fact that if $N$ and $N'$ are two $\mathcal{O}$-lattices in the same $\mathbb{F}[G]$-module $V$ (i.e. finitely generated $\mathcal{O}[G]$-submodules of $V$ that generate $V$ over $\mathbb{F}$), then the semi-simplifications of $N\otimes\bar{\mathbb{F}}_p$ and $N'\otimes\bar{\mathbb{F}}_p$ are isomorphic. This is Theorem 32 in Serre. If the $\mathbb{F}[G]$-module $M\otimes \mathbb{F}$ was reducible, then you would be able to find a decomposable lattice inside it, and the reduction of that lattice modulo $\mathfrak{p}$ would be reducible, in which case the reduction of $M\otimes \mathcal{O}$, having the same semi-simplification, would also be reducible. Finally, because of the displayed equality, $M\otimes \bar{\mathbb{F}}_p$ would then also be reducible.

$\endgroup$
1
  • 1
    $\begingroup$ This is basically homotopic to the proof I had in mind. $\endgroup$
    – Gina
    Commented Jun 29, 2021 at 17:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .