Let $R = \mathbb{Z}/4\mathbb{Z} = \{0, 1, 2, 3\}$ and the group $G = \mathbb{Z}/2\mathbb{Z} = \{e, a\}$. Consider the group ring $R[G]$.
I have read somewhere (1) that $\frac{e + a}{2}$ and $\frac{e - a}{2}$ are a complete set of orthogonal idempotents for some criteria of $R$.
This does not apply for the $R$ we defined since $2$ is not invertible as an element of $R$. However, if I "pretend" that $2$ is invertible, I can show that $e_0 = \frac{e + a}{2}$ and $e_1 = \frac{e - a}{2} = \frac{e + 3a}{2}$ are a complete set of orthogonal idempotents.
- $e_0e_1 = \frac{1}{4}(e^2 - a^2) = 0$.
- $e_0^2 = \left(\frac{e + a}{2}\right)^2 = \frac{e + 2a + e}{4} = e_0$
- $e_1^2 = \left(\frac{e - a}{2}\right)^2 = \frac{e - 2a + e}{4} = e_1$
- $e_0 + e_1 = \frac{e + a}{2} + \frac{e - a}{2} = \frac{2e}{2} = e$
Of course, pretending that $2$ is invertible is nonsense and so everything above (a priori) shouldn't work. Is there something missing which makes this work? Was it just a lucky coincidence?
(1) From another stackexchange question but I lost the link.
The ring $R[G]$ consists of elements $\alpha e+\beta a$, with $\alpha,\beta\in\mathbb Z_4$. If $\alpha e+\beta a$ is idempotent, then $$ \alpha e+\beta a = (\alpha^2+\beta ^2) e + 2\alpha\beta a. $$ This forces $\alpha=\alpha^2+\beta^2$ and $\beta=2\alpha\beta$. The only possible solutions are $\beta=0$, $\alpha=0$ or $\alpha=1$.
So the only idempotents are $0$ and $e$.
The computations you did will work only if $R$ is big enough so that $2$ is invertible.