$f$ admits multiple roots iff $\mathrm{gcd}(f,f')\neq 1_F$

Question

I want to check if this constitutes a proof for the following:

Let $f\in F[X]$ with $f=\sum_{k=0}^n a_kX^k$ and let $f'=\sum_{k=0}^na_kkX^{k-1}$ denote the formal derivative of $f$. Show that if $f$ has a non simple root in a field extension $E\supset F$ iff $\text{gcd}(f,f') \neq 1$.

Proof: let $f\in F[X]$ have a root $a$. Then with the euclidean algorithm $f(X)=(X-a)g(X)$ with $g\in F[X]$. If $a$ is not a simple root then $$(X-a) | g(X) \iff g(X)=(X-a)h(X),~h(X) \in F[X]$$

With the product rule $$f'(X)=g(X)+(X-a)g'(X)$$ and $$f'(X)=(X-a)h(X)+(X-a)g'(X)$$ So $(X-a)\ |\ f'$ and $\gcd(f,f')\neq 1$.

Answer 1

What you did proves that if $f$ has a multiple root, then $\gcd(f,f')\neq1$. In order to prove the reverse implication, you could observe that if all roots of $f$ are simple, then $f$ and $f'$ have no common root and that therefore (although this is not trivial) $f$ and $f'$ are relatively prime.

Answer 2

For the converse implication, say $L$ the splitting field of $F$.

Assume that $g(X):=\gcd(f(X),f'(X))\neq 1_L$. Suppose for contradiction that all the roots of $f(X)$ in $L$ are distinct. So, $f(X)=c(X-\alpha_1)\dotsb (X-\alpha_n)\in L[X]$ where $c\in L$ and $\alpha_1,\dots, \alpha_n\in L$ are all distinct. Then, $$f'(X)=\sum_{i=2}^n c(X-\alpha_1)\dotsb(X-\alpha_{i-1})(X-\alpha_{i}) \dotsb (X-\alpha_n)\in L[X].$$ Now, since $g(X)\neq 1_L$, it must be $\deg g(X)\geq 1$. So, any root of the common factor $g(X)$ must be a common root of $f(X)$ and $f'(X)$. Therefore, there is some $\alpha\in \{\alpha_1,\dots,\alpha_n\}$ such that $f'(\alpha)=0_L$. Without any loss of generality, say that $\alpha$ is $\alpha_1$. Then, $$f'(\alpha_1)=0_L\iff c(\alpha_1-\alpha_2)\dotsb (\alpha_1-\alpha_n)=0_L \iff \alpha_1\in \{\alpha_2,\dots,\alpha_n\}.$$ The latter is impossible, since we supposed all the roots are distinct. So $f(X)$ has repeated roots in $L$.

Hope that helps!