TITLE: An identity in Lie algebras over fields of positive characteristic
QUESTION [10 upvotes]: Let $L$ be a Lie algebra over a field of characteristic $p>0$ and $D$ a derivation of $L$. For every $x\in L$ denote by $\mathrm{ad} x$  the adjoint map $\mathrm{ad}x: L \rightarrow L, a\mapsto [x,a]$.   Is the following relation true?

$$D^{p-1}((\mathrm{ad} x)^{p-1}(D(x)))=(\mathrm{ad} x)^{p-1}(D^p(x))$$ 

I proved it for small values of $p$, but I am not able to find a general argument. In the case the result is well-known, a reference will be very welcome.

REPLY [5 votes]: Yes, this identity holds, and both sides are equal to $-D^p(x^p)$.
We need the following facts, all of which are straightforward to verify, where $D$ is an arbitrary derivation, still working over characteristic $p>0$: $$(ad_x)^k(y)=\sum_{m=0}^{k}(-1)^{m+1} {k\choose m}x^m y x^{k-m}$$
$$D(x^k)=\sum_{m=0}^{k-1}x^m D(x)x^{k-1-m}$$
$$D^p(xy)=D^p(x)y+xD^p(y)$$
$$(-1)^m {p-1\choose m}=1 \text{mod p}$$
Then we have:
\begin{align*}
D^{p-1}((ad_x)^{p-1}(D(x))&=D^{p-1}\bigg(\sum_{m=0}^{p-1}(-1)^{m+1}{p-1\choose m}x^mD(x)x^{p-1-m}\bigg)\\
&=-D^{p-1}\bigg(\sum_{m=0}^{p-1}x^mD(x)x^{p-1-m}\bigg)\\
&=-D^{p}(x^p)\\
&=-\sum_{m=0}^{p-1}x^mD^p(x)x^{p-1-m}\\
&=\sum_{m=0}^{p-1}(-1)^{m+1}{p-1\choose m}x^mD^p(x)x^{p-1-m}\\
&=(ad_x)^{p-1}(D^p(x))
\end{align*}