Take the tour ×  Why is it that cubics are separable over fields
that are not of characteristic $2$ or $3$?  This is the starting
point for some a discussion of the Galois group of a cubic, but
I seem to be stuck right off the bat.  share|improve this
question add comment  2 Answers  up vote 4 down vote accepted  A
polynomial has multiple roots if and only if it is not
relatively prime to its formal derivative.  Suppose that $f(x) =
x^n + a_{n-1}x^{n-1}+\cdots+a_0$ is an irreducible polynomial of
degree $n\gt 0$ over a field $\mathbf{F}$. Then $f'(x) =
nx^{n-1}+(n-1)a_{n-1}x^{n-1}+\cdots + a_1$. Because $f(x)$ does
not have any proper factors in $\mathbf{F}[x]$, then the only
possible common factors of $f(x)$ and $f'(x)$ are, up to units,
$1$ and $f(x)$. But if $f'(x)\neq 0$, then it is of degree
strictly smaller than $f(x)$, so it cannot be divisible by
$f(x)$. So either $f'(x)=0$, or else $\gcd(f,f')=1$. But the
only way in which we can have $f'(x)=0$ is if $n=0$ in
$\mathbf{F}$; that is, the characteristic