Having learned about separable polynomials today in class, I tried to do the following exercise concerning separable polynomials, namely:
Suppose $f$ is the minimal polynomial of $a$ over a field $F$ of prime characteristic. Let $K = F[a]$. Then $f$ is separable iff $F[a^p] = K$.
Now one direction I have proved, that is not separable implies $F[a^p] \subsetneqq F[a]$. For the other direction, I have some trouble at the end( which I will describe).
Suppose $f$ is separable and let $g$ be the minimal polynomial of $a$ over $F[a^p]$. We show that $g$ has degree one so that $a \in F[a^p]$, proving that $F[a] = F[a^p]$. Suppose we consider $g,f$ as polynomials in $\big(F[a]\big)[x]$. Then $g(a) = f(a) = 0$ in $\big(F[a]\big)[x]$. Since $[F[a]:F] > \bigg[F[a]:F[a^p] \bigg],$
this means that $g |f$. Now write $f(x) = (x-a)u$ where $u$ is some polynomial with coefficients in $F[a]$. Then we observe that since $a$ is algebraic over $F$, $F[a] = F(a)$ that is a field, hence trivially a UFD. Therefore the polynomial ring $\big( F[a] \big)[x]$ is a UFD so that $g$ being irreducible is actually prime. Now what I want to do now is to show that if $g|f$, then $g$ must divide $(x-a)$ forcing $g = (x-a)$ up to multiplication by a unit.
Suppose $g \nmid (x-a)$ so that $g|u$ by $g$ being a prime element in $\big(F[a]\big)[x]$. Since $f$ is separable its derivative is not zero, so that if $g$ divides f'(x) = (x-a)u' + u then I will have my desired contradiction. This is because we will have $g|u$ and g|u' implying that $u$ has a multiple root, contradicting $f$ having no multiple roots.
The problem now is how do I know that g|f'(x)? If it is not true here that g|f'(x), can the approach I have done above be salvaged?
For reference, the following theorem may be useful: Let $f$ be an irreducible polynomial in $F[x]$. The following are equivalent:
(1) $f(x)$ is not separable.
(2) f'(x) = 0
(3) $\operatorname{Char} F = p >0$ and $f$ is a polynomial in $x^p$