# An ideal generated by a nontrivial idempotent is not a free module

Let $R = \Bbb QG$ with $G$ isomorphic to a cyclic group of order $2$, and $x$ its generator. I'm trying to show that $$\frac{1}{2}(1+x)R$$ is not a free $R$-module.

I found out that $\frac{1}{2}(1+x)\in\Bbb QG\;$ is idempotent but can't go further.

• You probably want to prove the following general fact: if $e$ is a nontrivial idempotent (meaning not equal to 0 or 1) in a ring $R$, then $eR$ is not a free $R$-module. However, in your case, it is enough to look at $\mathbb{Q}$-dimensions. – Jiangwei Xue Apr 24 '13 at 8:00
• @JiangweiXue, why not add a little and write down your comment as an answer? – DonAntonio Apr 24 '13 at 12:44

$$e(1-e)=0\implies\;\forall\;er\in eR\;,\;\;er\cdot(1-e)=e(1-e)r=0\implies$$
there cannot be any (free) basis for the $\,R$-module $\,eR\,$.