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Let $G$ be a finite group, and let $K$ be a finite field whose characteristic does not divide $|G|$. I am interested in the theory of finitely generated modules over $K[G]$. Of course many problems are not present here because $K[G]$ is semisimple and all modules are projective. My case is partly covered by Section 15.5 of Serre's book "Linear Representations of Finite Groups". However, Serre likes to assume that $K$ is "sufficiently large", meaning that it has a primitive $m$'th root of unity, where $m$ is the least common multiple of the orders of the elements of $G$. I do not want to assume this, so some Galois theory of finite extensions of $K$ will come into play. I do not think that anything desperately complicated happens, but it would be convenient if I could refer to the literature rather than having to write it out myself. Is there a good source for this?

[UPDATED]:

In particular, I would like to be able to control the dimensions over $K$ of the simple $K[G]$-modules. As pointed out in Alex Bartel's answer, these need not divide the order of $G$. I am willing to assume that $G$ is a $p$-group for some prime $p\neq\text{char}(K)$.

[UPDATED AGAIN]:

OK, here is a sharper question. Put $m=|K|$ (which is a power of a prime different from $p$) and let $t$ be the order of $m$ in $(\mathbb{Z}/p)^\times$. Let $L$ be a finite extension of $K$, let $G$ be a finite abelian $p$-group, and let $\rho:G\to L^\times$ be a homomorphism that does not factor through the unit group of any proper subfield containing $K$. Then $\rho$ makes $L$ into an irreducible $K$-linear representation of $G$, and every irreducible arises in this way. If I've got this straight, we see that the possible degrees of nontrivial irreducible $K$-linear representations of abelian $p$-groups are the numbers $tp^k$ for $k\geq 0$. I ask: if we let $G$ be a nonabelian $p$-group, does the set of possible degrees get any bigger?

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Maybe Geoff's answer to mathoverflow.net/questions/91132/… could help you? –  Someone May 16 '12 at 11:32
    
@Someone: thanks, that's a useful pointer. –  Neil Strickland May 16 '12 at 12:27
    
Could you clarify what you mean by "control the dimension"? The dimensions of the simple modules over $K$ are multiples of the dimensions of the simple modules over $\bar{K}$, but I am not even sure what "controlling the dimensions" over $\bar{K}$ would entail. –  Alex B. May 16 '12 at 15:04
    
@Alex: I have updated the question again. –  Neil Strickland May 16 '12 at 16:25
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2 Answers 2

up vote 8 down vote accepted

Your last statement is not true in general. Let $G=C_3$ and take your favourite finite field that does not contain the cube roots of unity, e.g. $K=\mathbb{F}_5$. Then the two non-trivial one-dimensional representations over $\bar{K}$ are not defined over $K$, but their sum is, since it's the regular representation minus the trivial.

In general, $K[G]$-modules for $K$ finite of characteristic co-prime to $|G|$ behave pretty much like modules over characteristic zero fields (the simple $K[G]$-modules are just sums over Galois orbits of the absolutely simple ones), the major simplification being that there are no Schur indices. I am not sure that there is much more to say about this. The second volume of Curtis and Reiner contains a whole chapter on rationality questions, i.e. fields that are not "sufficiently large" in the sense of Serre. Most of it is for characteristic zero, but as I say, a lot of it carries over.

Edit Re updated question: if $G$ is a finite $p$-group and $K$ is a finite field of characteristic different from $p$, then it is indeed true that any irreducible representation of $G$ over $K$ has dimension $tp^k$, for some $k$ and some $t\;|\;(p-1)$. Indeed, the absolutely irreducible representations have dimension a power of $p$, and the field of definition of any absolutely irreducible representation is $K$ adjoin $p^r$-th roots of unity for some $r$, so the Galois orbit of a representation has size dividing $(p-1)p^{r-1}$ for some $r$.

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The question you raise doesn't come up often enough to be dealt with explicitly in textbooks and such, I think. One way to extract an answer (possibly overkill) is to look more closely at the cde-triangle in the formulation of Serre or Curtis & Reiner. Changing your notation a bit, take $p$ not dividing $|G|$ and then form a triple $(K,A,k)$ with $A$ a complete d.v.r. (such as $\mathbb{Z}_p$) having $K$ as fraction field and $k$ as the finite residue field of characteristic $p$. Without assuming that these fields are "large enough", one knows that the decomposition homomorphism $d: \mathrm{R}_K(G) \rightarrow \mathrm{R}_k(G)$ is surjective: see my older question here.

EDIT: Looking back at what I wrote next, it seems too superficial. Maybe a more careful comparison of the behavior under field extensions is really needed. Back to the drawing board.

ADDED: Maybe I've missed something, but I think what Serre does in his Section 15.5 avoids any use of the assumption that the fields are large enough. So this should dispose of the original question asked, while equating dimensions of correlated simple modules over $K$ and $k$. (Serre tends to be careful about specifying where it matters that fields are large enough.) Even though simple modules may decompose further over field extensions, working with any fixed $p$-modular system seems to yield trivial Cartan and decomposition matrices. (In this situation, the fact that $d$ is surjective follows from the assumption that $p$ doesn't divide the group order.)

In any case Alex has addressed the modified questions well.

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