The p-th power of a derivation on $\mathbb{F}_p[X]$ is still a derivation I come across the following problem in Kostrikin's algebra textbook.
Every derivation of the polynomial ring $P[X]$ has the form $T_u:f \mapsto uf'$, $u\in P[X]$. Suppose that char$P=p>0$, then $(T_u)^p$ is also a derivation.
It seems to me that we need find out the coefficients of the monomial like $u^{p-1}u^{(p-k)}f^{(k)}$ etc. But I have no idea on how to do that.
 A: A useful characterization of these "derivatives" is that they are exactly the $P$-linear maps $T:P[X]\to P[X]$ which satisfy $T(fg)=T(f)g+fT(g)$ (such maps are called derivations).  Indeed, given a derivation $T$, define $u=T(X)$, and then you can prove by induction on $n$ that $T(X^n)=T_u(X^n)$ for all $n$ (for instance, $T(X^2)=T(X)X+XT(X)=2uX=T_u(X^2)$) and thus $T=T_u$ by linearity.  Conversely, it is easy to see that $T_u$ is a derivation for any $u$.
So it suffices to show that if $T$ is a derivation, then $T^p$ is also a derivation.  This is not hard: for any $n$, we have $$T^n(fg)=\sum_{k=0}^n \binom{n}{k}T^k(f)T^{n-k}(g)$$ (you can prove this by induction on $n$ or by a combinatorial argument; it's basically the same as the proof of the binomial theorem; try computing $T^n(fg)$ directly for small values of $n$ to see how it works).  When $n=p$, the binomial coefficients $\binom{p}{k}$ are divisible by $p$ for all $k\neq 0,p$ and hence $0$ in $P$, so this sum reduces to just $$T^p(fg)=T^p(f)g+fT^p(g).$$  Thus $T^p$ is a derivation.
A: Thanks to the kind and nice answers above.
The most crucial thing in the proof is that we need $T(fg)=T(f)g+T(g)f$, which is easy to check. Also, $u=TX$. We thus have, by induction, that $T^p(f)=\sum_{k=1}^{p}\binom{p}{k}(T^{p-k}f')(T^kX)$, which indeed reduces to $T^p(f)=(T^pX)f'$ when char$P=p$. We thus conclude that $T^p$ is also a derivation.
