I am reading separable extension. I don't quite understand why $\text{char} 0$ fields must be perfect. If $F$ is a $\text{char} 0$ field, $f(x)$ is an irreducible polynomial in $F[x]$. Then why $f$ must be separable? Assume $f$ has multiple roots, say $r$ is a multiple root of $f(x)$. Then $r$ is also a root of $f'(x)$. $r$ is in the splitting field of $f(x)$ over $F$, denote it as $E$. Then why $f$ must divide $f'$? I know $f$ is irreducible but I don't know how the argument goes on. Thanks for any help!
[Math] Irreducible polynomial is always separable in char 0 field
abstract-algebra
Best Answer
Suppose $f$ is an irreducible polynomial and $\xi$ is a root of $f$. If $g$ is any other polynomial such that $g(\xi)=0$, then $f\mid g$. Indeed, if we call $h$ the greatest common divisor of $f$ and $g$, then $h$ is not $1$, since it has $\xi$ a a root. As $h$ is then a non-contant polynomial which divides $f$, it must be a scalar multiple of $f$ itself. Since $h$ divides also $g$, we see that $f$ divides $g$.