Exercise 1.12

Proof. Let $A$ be our local ring; by Cor.  $1.5$, $\Re =A\setminus A^{\text{×}}$. If $e$ is idempotent, one of $e,1-e$ is a unit for otherwise $1=e+(1-e)\in \Re$, a contradiction. Since $e$ is idempotent, $e(1-e)=0$. So, if $e$ is a unit, then $0=0e^{-1}=(1-e)e{e}^{-1}=1-e$ and $e=1$; if $1-e$ is a unit, then $0=0{(1-e)}^{-1}=e(1-e){(1-e)}^{-1}=e$ and $1-e=1$. □