3
$\begingroup$

I'm working on a paper which uses representation theory in order to compute some characters and deduce arithmetical statements about certain field extensions.

Let $\Delta$ be a group of order prime to $p$, $p$ a prime number. The claim is now:

The irreducible representations of $\Delta$ over the field $\mathbb{F}_p$ of $p$ elements are in bijection with the irreducible representations of $\Delta$ over $\mathbb{Q}_p$.

The author first translates into module language and afterwards uses the properties of the $cde$-triangle (as one can find in Serre's book) as follows: He first uses Maschke's theorem to say that in our case every $\mathbb{F}_p[\Delta]$-module $N$ is projective.

Then every projective $\mathbb{F}_p[\Delta]$-module $N$ has an (up to isomorphism) unique lift to a projective $\mathbb{Z}_p[\Delta]$-module $M$ for that $$N = M / pM$$ holds (Serre), what is still clear to me. Then he states, that two $\mathbb{Z}_p[\Delta]$-modules $M$ and $M'$ are isomorphic iff $M \otimes \mathbb{Q}_p$ and $M' \otimes \mathbb{Q}_p$ are isomorphic as $\mathbb{Q}_p[\Delta]$-modules (Serre again). He then says that finding the irreducible characters of an $\mathbb{F}_p[\Delta]$-module $M$ is the same as finding the absolutely irreducible characters of the $\mathbb{Q}_p[\Delta]$-module $M' \otimes \mathbb{Q}_p$, which i don't understand. Why absolutely irreducible? I thought a character is absolutely irreducible if it remains irreducible in an algebraic closure, but that would be $\mathbb{C}_p \otimes M$, right? The claim then follows translating back to representation language. But why should every representation of $\Delta$ over $\mathbb{Q}_p$ should have the form $M \otimes \mathbb{Q}_p$?

Also I was wondering if the following assertion is true: If one has a short exact sequence $$1 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 1$$ of $\mathbb{F}_p[\Delta]$-modules, is it true that the irreducible character of $B$ is the sum of the irreducible characters of $A$ and $C$?

Addition: What is the general point of lifting a representation over the field $\mathbb{F}_p$ to the field $\mathbb{Q}_p$? Are the latter simpler to calculate because of characteristic $0$?

Thank you again for your help :)

$\endgroup$
2
$\begingroup$

Your question has too many questions.

One thing you should probably keep in mind is that none of this is called the “modular” case. This is just ordinary representation theory of a finite group over a field whose characteristic does not divide the order of the group. Almost all proofs go over without change from the complex numbers, especially if you use Brauer characters (add up complex roots of unity instead of characteristic p roots of unity). There is a specific issue of rationality to address, but it could have been done at the character level.

Why absolutely irreducible? It seems better to me to work in a splitting field of $\Delta$ both times. The splitting field is found by adjoining a $\Delta$th root of unity to $\mathbb{F}_p$ and $\mathbb{Q}_p$, so in both cases the Galois group is generated by the Frobenius automorphism, and so in both cases one gets the same Galois conjugacy classes of characters. This means that by the end, you don't need absolutely irreducible characters, just irreducible, since the way in which the characters split into their absolutely irreducible constituents is the same.

Every $\mathbb{Q}_p[\Delta]$ module $V$ is of the form $M \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ for some $\mathbb{Z}_p[\Delta]$-module $M$ because $M$ can be taken to be the $\mathbb{Z}_p[\Delta]$-module generated by a $\mathbb{Q}_p$-basis of $V$. Since $\mathbb{Z}_p$ is a PID and $\Delta$ is finite, $M$ is free and finite rank as a $\mathbb{Z}_p$-module, and since $p$ does not divide $|\Delta|$, it is even projective (free) as a $\mathbb{Z}_p$-module (this is basically Maschke's theorem again, sometimes covered in a result called Higman's lemma on relative projectives).

It is true that characters of finite dimensional representations are additive as you want. This follows by writing down the matrices of the group elements in block triangular form. This is true both for ordinary characters and Brauer characters.

The point may be that absolutely irreducible ordinary character theory does not depend on the field, and this is more clear in characteristic 0, so one should lift to characteristic 0. If one is interested in “rationality” questions (which representations are available over $\mathbb{Z}/p\mathbb{Z}$ versus its algebraic closure), then one also has to worry about the rationality of the characteristic 0 representations. Luckily $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Q}_p$ contain the same $|\Delta|$th roots of unity, and so their characters have the same rationality properties. I believe the representations also have the same rationality properties (all Schur indices are 1).

$\endgroup$
  • $\begingroup$ Thank you for this post! I'm still trying to understand the details though. Could you give me some reference for the fact that every $\mathbb{Q}_p[\Delta]$-module $V$ is of the specified form above? I don't see how one could see that the $\mathbb{Z}_p$-module $M$ you choose is indeed projective. Thank you so much! $\endgroup$ – BIS HD Jul 24 '13 at 13:39
  • $\begingroup$ Corollary 29.19 in Curtis–Reiner vol 1, page 613 is one statement of this. Does the proof of Maschke's theorem not just work though? $\endgroup$ – Jack Schmidt Jul 24 '13 at 17:46
  • $\begingroup$ I solved the problem without Curtis-Reiner. The $\mathbb{Z}_p$-Module $M$ I've chosen is - as a submodule of the torsion free module $\mathbb{Q}_p^n$ - torsion free itself and finitely generated. Hence, since $\mathbb{Z}_p$ is a PID, $M$ is free and projective in particular. Do you have some reference for the fact that the Galois conjugacy classes over $\mathbb{F}_p$ and $\mathbb{Q}_p$ are the same? Thank you :) $\endgroup$ – BIS HD Aug 7 '13 at 16:02
  • 1
    $\begingroup$ Oh sorry on that first question. I thought you meant how to prove the $\mathbb{Z}_p[\Delta]$-module is projective. The $\mathbb{Z}_p$-module follows from exactly what you said. I'll look for a reference for the Galois conjugacy. It mostly is just asking about which roots of unity are in each field. "Maximal unramified extension" is the high powered version of this, math.stackexchange.com/questions/61900 $\endgroup$ – Jack Schmidt Aug 7 '13 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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