Let $F$ be a ring and $f(x) = a_0 + a_1x + · · · + a_nx^n$ be in $F[x]$. Define $f'(x) = a_1 + 2a_2x + · · · + na_nx^n−1$ to be the derivative of $f(x)$.
we can define a homomorphism of abelian groups $D : F[x] \to F[x]$ by $(D(f(x)) = f'(x)$.
The kernel of $D$ when char $F=0$ is all the constants but I don't understand how to find the kernel when char $F =p$.
On
Since $p=0$, elements of the form $x^{pk}$ for integers $k$ are also in the kernel. In general, the derivative of a polynomial is given by $$na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1$$ This is 0 if and only if all coefficients are 0. If we consider integer representatives, all coefficients must be multiples of $p$. Since $p$ is prime, this can happen only if either the coefficient was already a multiple of $p$ or if the original exponent was a multiple of $p$. Thus the kernel is spanned by $x^{pk}$ for $k$ a nonnegative integer.
$f'(x)=0$ iff ($n a_n = 0$ for all $n$) iff ($n=0$ or $a_n=0$ for all $n$) iff ($p \nmid n \Rightarrow a_n = 0$ for all $n$) iff $f(x)$ is a polynomial in $x^p$.