Prove that Group Ring RG is not a Division Ring for R commutative and G a finite group How do I prove that a group ring RG is not a division ring, assuming that R is commutative and G=(g_1,g_2,...g_n) finite?
I understand that I need to show there exists some element of RG that is not a unit. But how do I find such element? Would the element g_1+g_2+g_3+....+g_n work?
 A: Assume that $|G| > 1$ (with identity element $e$) and $R \neq \{0\}$. Consider the map
$$RG \to R, \ \ \ \ \sum_{g \in G} a_g g \mapsto \sum_{g \in G} a_g.$$
It is a ring homomorphism. Its kernel contains all elements of the form $e-g$ for $g \in G$. The kernel is a two-sided ideal in $RG$, and the map is non-trivial (because $G$ and $R$ are non-trivial). Hence $RG$ cannot be a division ring, since a divison algebra only contains the trivial two-sided ideals.
A: Let $G$ and $R$ be nontrivial.
Let $\alpha = \sum_{g\in G}g$. We claim that $\alpha$ is not a unit in $RG$. To see this, suppose that $x = \sum_{h\in G}r_h h$ is any element of $RG$. We will show that $\alpha x\neq 1$. We have
$$
\alpha x = \sum_{g,h \in G}r_h\cdot gh = \sum_{s\in G}\Big[(\sum_{h\in G}r_h)\cdot s\Big],
$$
where the second equality makes the substitution $s = gh$. We see that each element of $G$ has the same coefficient, namely $\sum_h r_h$, in the expansion of $\alpha x$. However, $1$ has different coefficients for the identity and non-identity elements, so we cannot have $\alpha x = 1$.
