$f(x)=(x-a)(x-a_2)...(x-a_n)\in F[x]$ where $F$ is a field and $a_j\in $ for $j=1,2,...,n$ has no repeated roots iff gcd$(f(x),f'(x))=1\in F[x]$

Question

This makes sense to me if $a_j\ne a_k$ for $j\ne k$ as $(x-a_j)=0 \implies a_j$ is a root of $f(x)$. So if all $a_j$ are different, then all the roots will be different. Do I have to somehow show this using the fact that gcd$(f(x),f'(x))=1$, because if so, I don't know how to

Answer 1

Compute the derivative: $$f'(x)=(x-a_2)\cdots(x-a_n)+(x-a_1)(x-a_3)\cdots(x-a_n)+\ldots+(x-a_1)\cdots(x-a_{n-1})$$ Suppose $f$ has a multiple root; wlog assume $a_1=a_2$. Then in the sum above, each term is divisible by at least one of $(x-a_1)$ and $(x-a_2)$. Hence $f'(a_1)=0$, and thus $(x-a_1)\mid\mathrm{gcd}(f,f')$.

Conversely, suppose $f$ has no multiple roots. Then $$f'(a_i)=(a_i-a_1)\cdots(a_i-a_{i-1})(a_i-a_{i+1})\cdots(a_i-a_n)$$ Since $f$ has no multiple roots, all these factors $(a_j-a_i)$ are nonzero. Hence $f'(a_i)\ne0$, thus no factor $(x-a_i)$ of $f$ divides $f'$, which means that the gcd is $1$.