Let $L$ be a finitedimensional 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?
In general the answer is NO. For instance, let $F$ be a field of characteristic $2$ and consider the 3dimensional 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 $p$.


$\begingroup$ Because in this specific case you have $(x+y+z)^{[2]}=(x+y)^{[2]}+z^{[2]}=x^{[2]}+y^{[2]}+[x,y]+z^{[2]}=2z=0$. $\endgroup$ – Salvatore Siciliano Jul 25 '16 at 15:17

$\begingroup$ Just a remark: $H$ is just $sl(2,F)$. $\endgroup$ – Friedrich Knop Jul 25 '16 at 15:45

$\begingroup$ @ Fredrich Knop: You are right. In characteristic two, $sl(2,F)$ with the natural power mapping is indeed isomorphic as restricted Lie algebra to the 3dimensional Heisenberg algebra with the power mapping that I considered. $\endgroup$ – Salvatore Siciliano Jul 25 '16 at 17:26
