Question regarding separability of polynomials and perfect fields I have this definition
A field $F$ is perfect if it has characteristic $0$ or if it has characteristic $p$ and $F^p = F$
From Wikipedia, I have this fact about separable polynomials:
Irreducible polynomials over perfect fields are separable. 
I can show that if $p(x)$ is irreducible and $char(F) = 0$, then $p$ cannot have multiple zeros, i.e. it's separable. I can also show that, if $p(x)$ has multiple zeros, it has to be of the form $p(x) = q(x^p)$.
Now my question:
In the comment on Wikipedia, how does the second part follow? That is, that if $char(F)=p$ for $p$ prime and $f(x)$ is irreducible over $F$ then it cannot have multiple zeros?
Many thanks for your help.
 A: I'm expanding on this answer in response to a comment.
We'll use the following theorem:
$f \in F[x]$ has a multiple root in some extension $E$ of $F$ if and only if $f$ and $f^\prime$ have a common factor $d(x) \in F[x]$ of positive degree.
First, let $char(F) = 0$ and let $a_0 + a_1 x + \dots + a_n x^nf \in F[x]$ be irreducible over $F$. We prove by contradiction that $f$ cannot have a multiple root. So, assume $f$ has a multiple root. Then $f$ and $f^\prime$ have a common factor $d(x) \in F[x]$ of positive degree. $f$ is irreducible so the only divisor of positive degree is $f$ itself, so that $f$ divides $f^\prime$. A polynomial cannot divide a polynomial of smaller degree hence $f^\prime = 0$. But this means that $a_1 = 2 a_2 = \dots = n a_n = 0$. Since $char(F) = 0$ we get $a_k = 0$ and therefore $f = a_0$ which is not irreducible.
If $F$ has characteristic $p$ then $(x + y)^p = x^p + y^p$. Then if $f(x) = g(x^p)$ with coefficients in $F$ and $F^p = F$ then $f(x) = a_n^px^{pn} + \dots + a_0^p = (a_nx^n \dots + a_0)^p = g(x)^p$ because all $a_i \in F$ can be written as powers of $p$ by assumption. But then by assumption $f$ is irreducible over $F$ therefore contradiction.
