Question

For any nonzero ring k and any group G, show that the group ring kG is von Neumann regular iff k is von Neumann regular, G is locally finite, and the order of any finite subgroup of G is a unit in k.

Answer 1

First suppose $R = kG$ is von Neumann regular. As a quotient ring of $R$, $k$ is certainly von Neumann regular. Consider any finite subgroup $E \subseteq G$, say of order $m$. It suffices to show that any prime $p \mid m$ is a unit in $k$. To see this, we fix an element $\sigma \in E$ of order $p$ (which exists by Cauchy's Theorem). Taking an element $\alpha \in R$ such that $1 - \sigma = (1 - \sigma)\alpha(1 - \sigma)$, we can argue that $p \in \mathrm{U}(k)$. Finally, let $H$ be a subgroup of $G$ generated by a finite set of elements, say $h_1, \ldots, h_n$. It is easy to see that

$$I := \sum_{h \ni \in H} R \cdot (h - 1) = \sum_{i = 1}^{n} R \cdot (h_i - 1)$$

(using facts such as $h_i^{-1} - 1 = h_i^{-1}\big(1 - h_j\big)$, and $h_i h_j - 1 = h_i (h_j - 1) + h_i - 1$). Since $R$ is von Neumann regular, $I = Re$ for some idempotent $e$. Now $I$ is contained in the augmentation ideal of $R$, so $e \neq 1$ (since $k \neq 0$). Thus, $f := 1 - e \neq 0$. But for any $h \in H$, $(h - 1)f \in I \cdot f = R \cdot ef = 0 \Rightarrow f = hf$. Since $f$ involves only finitely many elements of $G$, this implies that $|H| < \infty$.

For the converse, let us assume the given conditions on $k$ and $G$. Consider any element $\alpha = a_1h_1 + \dots + a_nh_n \in R$ ($a_i \in k, h_i \in G$), for which we want to show $\alpha \in \alpha R\alpha$. Let $H$ be the (finite) subgroup of $G$ generated by $h_1, \ldots, h_n$. Since $\alpha \in kH$, we are done if we can show that $S := kH$ is von Neumann regular. Consider any principal left ideal, say $S \cdot \beta$, where $\beta \in S$. Viewing $S \cdot \beta \subseteq S$ as $k$-modules, we have $S \sim k^{|H|}$ and $S \cdot \beta = \sum_{h \in H} k \cdot (h\beta)$. Since $k$ is von Neumann regular, $S \cdot \beta$ is a direct summand of $S$ as $k$-modules. Then $S \cdot \beta$ is a direct summand of $S$ as $S$-modules. This checks that $S$ is von Neumann regular, as desired.