If $F$ is a field of characteristic $0$ and $p(x) \in F[x]$ is irreducible, then $p(x)$ has no repeated roots. (Hint: Consider $(p(x),\;p'(x))$).
Let $F$ be a field and $f(x) \in F[X]$, then $f(x)$ has repeated roots if there is a field $E$ containing $F$ and a factorisation in $E[x]$ of the form
$$f(x)=(x-a)^2h(x).$$
My attempt: If $f(x)$ has repeated roots then $d=\gcd(f,f')\neq 1$ then $0<\deg(d)<n$ and thus $f$ is reducible.
Is my argument correct and if it is correct then where did I used that $F$ has char $0$ ?
Best Answer
You mistyped the inequality, that should have read $0<\deg(d)<n$, but otherwise your argument works; being divisible by such a $d$ (even if of degree$~1$, which is actually quite likely) is sufficient for being reducible. Where you (implicitly) used that the characteristic is $0$ is in assuming that $f'\neq 0$: the (formal) derivative of a polynomial of degree $n>0$ has degree $n-1$ (hence is nonzero) in characteristic$~0$, but this can fail in characteristic$~p$ (namely when $p$ divides $n$).