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small technical correction
Neil Strickland
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Modular representations with unequal characteristic - reference request

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?

Neil Strickland
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