Suppose $(R, +, \times, 0, 1)$ is a ring with unity and $a \in \mathbb{R}\setminus \{0\}$ is a non-zero nilpotent element. Can it be an idempotent?
We call $a$ nilpotent if $\exists n \in \mathbb{N}$, such that $a^n = 0$. We call $a$ idempotent, if $a^2 = a$.
If $p$ is idempotent, then $p^2=p.$ Show by induction , that $p^n=p$ for all $n \in \mathbb N, n \ge 1.$
What can you conclude, if $p$ is nilpotent ?