# Representation theory of finite groups on locally free $R$-modules

Let $R$ be a commutative ring with 1, and let $G$ be a finite group of order $n$. What assumptions do we need to put on $R$ in order to have a reasonable theory of representations of finite groups on finitely generated locally free $R$-modules $M$?

By a 'reasonable theory', to be precise I would like the following to be true:

1. Two representations of $G$ are isomorphic if and only if their characters agree (ie, traces of elements).
2. Every representation $M$ is a direct sum of irreducible representations $I$ with multiplicity equal to the rank of $M\otimes_R I^\vee$.
3. Characters of distinct irreducible representations are orthogonal under the pairing $\langle \chi_1,\chi_2\rangle = \frac{1}{n}\sum_{g\in G}\chi_1(g)\chi_2(g^{-1})$. Further, every irreducible character has norm 1 under this pairing.
4. The irreducible characters form a basis for the set of class functions $G\rightarrow R$.

Certainly one would want $n$ to be invertible in $R$. Is this enough? Can we recover any of the above statements in this situation? What if we further assume that $R$ contains a primitive $n$-th root of unity - what changes?

Is there a reference for representation theory of finite groups on (locally free modules over) rings where the order of the group is invertible? Most representation theory notes I see only work over fields.

• Expecting characters to work nicely over a ring is waaaay to optimistic. Even the representation theory of finite groups on free abelian groups is very complicated! – Mariano Suárez-Álvarez Mar 4 '17 at 20:25
• Way too optimistic even under my assumption that $n\in R^\times$? – user355183 Mar 4 '17 at 20:27
• You want all representations to be direct sum of simple ones. In oarticular, this should be true of the trivial representation, and from that it follows that your ring R is semisimple. Wedderburn's theorem then tells you that it is a direct product of fields and using maschke's theorem you can check at once that each of those fields is of characteristic not dividing the order of the group. You can easily work with each direct factor of R separately, and then everything else you want follows ... – Mariano Suárez-Álvarez Mar 4 '17 at 21:32
• ...as usual (characters are not usually treated except in the complex case, but things work almost exactly the same; see the book by Curtis and Reiner, for example) – Mariano Suárez-Álvarez Mar 4 '17 at 21:32
• But, as you see, nothing new happens, as every thing reduces to the nonmodular situation over a field. The interesting theory is nonsemisimple, like over the integers. – Mariano Suárez-Álvarez Mar 4 '17 at 21:34