Why does $F(X)$ divide $F'(X)$?

Question

Let $\Bbb K$ be a field of characteristic zero or a finite field, let $F(X) \in \Bbb K[X]$ be a monic irreducible polynomial, and let $F(X)= \prod_{i=1}^n(X-x_i)$ be its decomposition in an extension $\Bbb K'$ of $\Bbb K$. Prove that the $n$ roots are all distinct.

The proof in the book reads: "If not, $F(X)$ has a root in common with its derivative $F'(X)$. Therefore, $F(X)$ divides $F'(X)$". The proof is very clear afterwards, but why is that?

Answer 1

Call $a \in K'$ a common root of $F, F'$. Since $F$ is irreducible, it is the minimal polynomial of $a$ over $K$. Since $F'(a)=0$, you must have that $F$ divides $F'$ (it is a property of the minimal polynomial).

Answer 2

A polynomial $F$ has multiple roots if and only if $\gcd(F,F')$ is not $1$. Since $F$ is irreducible and the gcd can be computed in $K$, the gcd has to be $F$ or $1$.

Answer 3

If, say, $x_2= x_1$, then $F(X)=(X-x_1)^2G(X)$ and $$F'(X)=2(X-x_1)G(X)+(X-x_1)^2G'(X),$$ hence $F'(x_1)=0=F(x_1)$.

It is well known that $F$ has no multiple root is equivalent to $F(X)$ and $F'(x)$ are coprime.