Separability by surjectivity of Frobenius Endomorphism

Question

I'm trying to prove the following statement:

Let $F$ be a finite field of prime characteristic $p$ and let $E$ be the field generated by $F$ and the elements $\{t^{1/p},n \geq 1\}$, where $t$ is an indeterminate. Then, any algebraic extension of $E$ is separable.

I've already proved that the Frobenius Endomorphism on $E$ is surjective. I'm trying to use the following fact:

Given $\alpha$ an element of an algebraic extension of $E$, then $\alpha$ is separable if, and only if, the derivate of $Irr(\alpha,E)$ is not zero.

Then, I know that the irreducible polynomial can be expressed function of $x^p$,

$$Irr(\alpha, E) = p(x^p)\in F[x].$$

However, I got stuck here and I don't know what else to do. Could someone give me and advice?

Answer 1

Write $P(X) = \sum_{k=0}^n a_k X^k$ with $a_k \in E$. Since you know that the Frobenius on $E$ is surjective, you can find elements $b_k \in E$ such that $b_k^p = a_k$.

It follows that $$P(x^p) = \sum_{k=0}^n a_k X^{pk} = \sum_{k=0}^n b_k^p X^{pk} = \Big(\sum_{k=0}^n b_k X^k\Big)^p$$ which is a contradiction, since $P(x^p)$ is supposed to be irreducible.