Polynomials over a field with characteristic $0$ is square free implies it's coprime with its formal derivative

Question

Let $F$ be a field with characteristic $0$, $f \in F[t]$ the polynomial ring over $F$. Show that $f$ is square free implies $ f, f'$ are relatively prime.

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I know this is actually an if and only if statement, but the other direction is easy. I get stuck on this direction, and I did some search but a lot of materials using extension fields and separable polynomials to get this. But our course haven't reached that far. So can anyone help me with an elementary proof? Thanks a lot.

Answer 1

Suppose that $f$ and $f'$ have a common irreducible factor $p$. Let $f(t)=p(t)q(t)$. Then $f'(t) = p(t)q'(t) + p'(t)q(t)$.

Since $p$ is irreducible and $F$ has characteristic zero, $p$ and $p'$ do not have a common factor, so $p(t)$ divides $q(t)$. In other words, $p^2 \mid f$.