TITLE: Which rings arise as a group ring?
QUESTION [10 upvotes]: 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.

REPLY [2 votes]: 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.