Question about radicals. Let $A$ be a $K$-algebra, where $K$ is a algebraically closed field. Let $\text{rad} A$ be the radical of $A$, i.e. the intersection of all maximal right ideals of $A$. Let $g+\text{rad} A$ be an idempotent of $A/ \text{rad} A$. We have the following theorem. Theorem: there is an idempotent $e$ of $A$ such that $e-g\in \text{rad} A$. Suppose that there are only two idempotents $0, 1$ in $A$. How can we show that $A/ \text{rad} A$ has only two idempotents $\text{rad} A, A$? I am trying to prove this result. Let $g+\text{rad} A$ be an idempotent of $A/ \text{rad} A$ and $g \not\in \text{rad}A, g \neq 1$. By the theorem above, there is an idempotent $e$ of $A$ such that $e-g\in \text{rad} A$. If $e=0$, then $g\in \text{rad} A$. Therefore $e\neq 0$. If $e=1$, then $1-g\in \text{rad}A$. How can we obtain a contradiction? Thank you very much.

Instead of obtaining a contradiction try using $1 - g \in \mathrm{rad}A$ to prove that $g + \mathrm{rad}A = 1 + \mathrm{rad}A$.