Questions about the formal derivative over $F[x]$ Let $F$ be a commutative ring and $f(x)=a_{0}+a_{1}x+.......+a_{n}x^n$ be in $F[x]$. Define $f'(x)=a_{1}+2a_{2}x+...+na_{n}x^{n-1}$ to be derivative of $f(x)$.

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*Prove that $(f+g)'(x)=f'(x)+g'(x)$, which concludes that $D:F[x]\to F[x]$ (defined by $D(f(x))=f'(x)$) is a homomorphism.
This is easy. I already proved it.


*Calculate the kernel of $D$ if $\operatorname{char}F=p$


*Suppose that we can factor a polynomial $f(x)\in F[x]$ into linear factors, say $f(x)=a(x-a_{1})(x-a_{2})...(x-a_{n})$. Prove that $f(x)$ has no repeated factors if and only if $f(x)$ and $f'(x)$ are relatively prime.
 A: (Not giving a full answers, only guidelines)
For (b), it should be easy if you write $f$ as a sum $f = \sum a_i x^i$, compute $f' = \sum (i+1) a_{i+1} x^i$, and then use the linear independence of the $x^i$.
For (c), we can do this by induction and elementary arithmetic in the principal ring $k[x]$. The polynomial $f$ has no repeated roots iff it is either irreducible or factors as $f = gh$ with $g, h$ coprime and both having no repeated roots. I leave the irreducible case as an easy exercise and show the induction step. We want to prove that $f, f'$ are coprime, which translates to the Bézout relation
\begin{equation}(1)\qquad 1 = p f + q f' = p\, gh + q\,(gh' +  g'h).\end{equation}
This is equivalent to $(g h' + g' h)$ having an inverse modulo $gh$. We consider this modulo $g$, $h$ respectively: since they are coprime, (1) is equivalent, by the Chinese theorem, to the two equations
$$\begin{split}
(2)\qquad 1 &\equiv q\,g'h \pmod{g},\\
(3)\qquad 1 &\equiv q\,g h'\pmod{h}.\\
\end{split}$$
Since $g$ has no repeated roots, $g, g'$ are coprime (induction hypothesis). Since $g, h$ are also coprime, $g' h$ is coprime to $g$, so that (2) has a solution. In the same way, (3) also has a solution.
