Let $L$ be a finite-dimensional restricted Lie algebra over a field of characteristic $p>0$. An element $x$ of $L$ is called $p$-nilpotent if $x^{[p]^k}=0$ for some positive integer $k$. If $L$ is nilpotent as an ordinary Lie algebra and it has a basis consisting of $p$-nilpotent elements, can one conclude that every element of $L$ is $p$-nilpotent?



Restricted Lie algebras with a $p$-nilpotent basis
In general the answer is NO. For instance, let $F$ be a field of characteristic $2$ and consider the 3-dimensional Heisenberg algebra $H=Fx \oplus Fy \oplus Fz$ with $[x,y]=z$ and $[x,z]=[y,z]=0$ and power map defined by the conditions $x^{[2]}=y^{[2]}=0$ and $z^{[2]}=z$. Then $\mathcal{B}=\{x, y, x+y+z\}$ is a basis of $H$ consisting of $2$-nilpotent elements, however the element $z$ is not $2$-nilpotent.


On the other hand, it is easy to see that the question has positive answer when the nilpotency class of $L$ is less than $p$.


