Which rings arise as a group ring?

Question

Let $R$ be an arbitrary associative ring with identity. When does there exist a group $G$ and a field $F$ such that $F[G] = R$?

Do we obtain more rings as $F[G]$ if we loosen the condition that $F$ be a field? When can we get $G$ to be a finite group?

There seem to be some obvious restrictions: The characteristic of $F$ is equal to that of $R$, for instance. Also, when $F$ and $G$ are finite, then so is $R$. Also, all elements of $G$, considered as a subset of $R$, are units in $R$. If $R$ is commutative, $G$ has to be abelian. But I fail to see much more.

Answer 1

The simplest case, which is already quite complicated, is if $F$ is algebraically closed of characteristic $0$ and $G$ is finite, or equivalently if $R$ is finite dimensional over $F$ (note that $R$ is necessarily an $F$-algebra). Then $F[G]$ is semisimple by Maschke's theorem, and furthermore splits up as a finite product of matrix algebras $M_n(F)$, one for every irreducible representation of $G$ over $F$ of dimension $n$. So in this special case the question reduces to asking:

Which tuples $(n_1, \dots n_k)$ arise as the dimensions of the irreducible representations of a finite group $G$ over $F$?

I don't think there's any hope of a simple answer to this question. Here are some necessary conditions. Note that $|G| = \sum n_i^2$.

1. At least one of the $n_i$ must be equal to $1$ (since the trivial representation is always irreducible), and in fact the number of $n_i$ equal to $1$ must divide $|G|$ (since it's the order of the abelianization).
2. If $|G|$ is prime then each of the $n_i$ must be equal to $1$ (since in this case $G$ is a cyclic group of prime order).
3. Each of the $n_i$ must divide $|G|$.

For example, $F \times M_2(F)$ is not a group algebra (because $1^2 + 2^2 = 5$ and the only group of order $5$ is $C_5$), but $F \times F \times M_2(F)$ is (it's the group algebra of $S_3 \cong D_3$).

Answer 2

If your group $G$ contains a non-trivial element of finite order then your group ring $R:=SG$ contains a zero-divisor (here, $S$ is just a ring, not necessarily a field): If $g^n=1$ then $$\begin{align*} &(1-g)(1+g+g^2+\cdots+g^{n-2}+g^{n-1})\\ &=(1+g+g^2+\cdots+g^{n-2}+g^{n-1})-(g+g^2+g^3+\cdots+g^{n-1}+1)\\ &=0.\end{align*}$$

Note that in his PhD thesis Higman proved an "opposite" result using locally indictable groups (groups where every non-trivial, finitely generated subgroup maps onto $\mathbb{Z}$): G. Higman, The units of group rings, Proc. London Math. Soc., vol. 46 (1940), pp. 231- 248.