Let $F$ be a field and let $f(x)$ be irreducible polynomial of degree $p$ "prime" in $F[x]$ having splitting field $K$. Let $[K,F]=pt$, for some integer $t$, then:
1) $f(x)$ is irreducible over $F$
2) If $t$ > $1$, then $f(x)$ is separable over $F$
My question: In part (2), we prove it by contradiction, so we suppose $f(x)$ is not separable, and since it is irreducible, its derivative is zero. and so $f(x)= a_px^{p}+a_0$. What I don't understand is the derivative of $f(x)$ being zero and the coefficients between $a_n$ and $a_0$ being zeros. Could someone explain?
If $K/F$ is not seperable, then $f(x)$ is not. Now if $f(x)$ is irreducible and inseperable then $F$ is an infinite field of prime characteristic, say $q$, avoiding confusion with $p$. And so, you can find a polynomial $g(x)$ in $\mathbb{F}[x]$ such that $f(x)=g(x^{q})$. Setting the degrees equal, $q$ divides $p$ so $q=p$. Now ( I will be assuming $f(x)$ is monic), we can write $f(x)=x^{p}-a$ for some $a \in \mathbb{F}$. Now let $\alpha \in K$ such that $\alpha^{p}=a$. So, $f(x)=(x-\alpha)^{p}$ (char $p$), so $K=\mathbb{F}(\alpha)$ which has degree $p$ over $F$ so $t=1$.