When our lecturer defined a seperable polynomial, he said:
A non-constant irreducible polynomial $f\in F[x]$ is called seperable if the two equivalent conditions hold:
- $f$ has no repeated roots in $\overline{F}$.
- $\gcd(f, f')=1$.
I'm not convinced that these two are always equivalent.
I know that "not 1" implies "not 2", therefore 2 implies 1 (here).
However, when I try proving the second direction, this happens:
$f$ is irreducible, hence either $\gcd(f, f')=1$ or $f|f'$. Assume [by contrary] that the second holds. Since $\deg f> \deg f'$, it implies that $f'=0$.
Now, if $char F=0$, we can conclude that $f$ is constant and that finishes (as seen here).
But what about the finite characteristic case?
Is it possible that $f'=0$, yet $f$ has no repeated roots in $\overline{F}$?
[I've tried $x^p-b$ as an example but it fails.]
Best Answer
In characteristic $p$, if $f'$ is the zero polynomial, then $f(x)=g(x^p)$ for some polynomial $g$, so $f(x)=(g(x))^p$, so repeated roots.