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

Question

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.

Answer 1

$$e(1-e)=0\implies\;\forall\;er\in eR\;,\;\;er\cdot(1-e)=e(1-e)r=0\implies$$

there cannot be any (free) basis for the $\,R$-module $\,eR\,$.