Let $a \in A$ be a nilpotent element. There is an integer $k \geq 1$ such that $a^k = 0$.
Let $m$ be the minimum of these integers.
By the euclidian division of $m$ with the number $3$, there exist two integers $q$ and $r$ such that $m = 3q + r $, with $0 \leq r \leq 2$. Thus
$$a^m = a ^{3q+r}=a^{q+r}=0$$
If $q \neq 0$, then $q+r < 3q+r = m$, and so $a = 0$.
If $q=0$, then there are three cases:
- if $r=0$, then $m=0$, and so $a^m=1$; a contradiction.
- if $r=1$, then $m=1$, and so $a^m=a^1=a=0$;
- if $r=2$, then $a^2=0$
That's as far as I've gotten. I can't find a way to exclude the case $q=0, r=2$, so I think there's a different way to approach this.
Any thoughts?
More generally, $a$ is called potent when $a^m=a$ for some $m > 1$. The only nilpotent potent element is $0$.
Proof: Assume $a^m = a$ and $ a$ is nilpotent. The relation $a^m = a$ implies $a^n = a^{n+m-1}$ for all $n \geq 1$. Choose a minimal $n \geq 1$ with $a^n = 0$. If $n \leq m$, then $a^m=0$ and hence $a=0$. If $n > m$, then $0 = a^n = a^{n-m+1}$ with $n-m+1 < n$, contradiction to minimality.