$F$ field, $\alpha$ separable on $F$. Is $F(\alpha)$ a separable extension of $F$?

Question

Let $F$ be a field, and let $\alpha$ be algebraic and separable over $F$. Is $F(\alpha)$ a separable extension of $F$?

By "$\alpha$ is separable" I mean that its minimum polynomial over $F$ is separable.

I'm pretty sure this is true, but I can't find a proof.

Thank you

UPDATE: @leo proved that the statement is true. By the way, I posted this question because I needed it to prove the "$\Longleftarrow$" of this exercise (Lang, Algebra):

Let $F$ a field of characteristic $p$. Let $\alpha$ be algebraic over $F$. Show that $\alpha$ is separable if and only if $F(\alpha)=F(\alpha^{p^n})$ for all positive integers $n$.

It seems to me that the theorem used by @leo comes after this exercise. Otherwise, the solution is trivial. So, the question is: is there a way to prove my original statement by using something less powerful?

Thank you again

Answer 1

First notice that the result it's true if $F$ has characteristic $0$. Otherwise:

Theorem. Let $F \subseteq E$ fields of characteristic $p$. If $E / F$ is a separable extension then $E = F(E^p)$. If $E / F$ is finite and $E = F(E^p)$, then $E / F$ is a separable extension.

Corollary Let $F \subseteq E$ fields of characteristic $p$. If $E / F$ is an algebraic extension and $a\in E$ is separable, then $F(a) = F(a^p)$ and $F(a) / F$ is separable.

Answer 2

Let me give another proof for this fact. The proof requires the following lemma:

Let $K$ be a field with characteristic $p \neq 0$ then $\exists n \ge 0. \alpha$ $K(\alpha^{p^n})/K$ is separable.

I did in fact reuse the proof here by Arturo Magidin. The proof needs some further comments and I do them here.

At some point he makes the following distinction:

1. If $deg(h) = 1 \implies \alpha^{p^l} \in K$.
2. If $deg(h) > 1 \implies \alpha^{p^m}$ is separable over $K$.

But I want to point out that when the polynomial $h$ has degree 1 we would have that except a constant (which is a unit) $h(x) = x^{p^l} - \alpha^{p^l} = (x - \alpha)^{p^l}$ and it follows that the polynomial is not separable while in his development it is. So you can rule out this case.

Using this lemma we can proof this fact as follows:

$\alpha$ is separable over $K \iff K(\alpha) = K(\alpha^2) = \ldots$

I denote $Irr(\beta,Q)$ the minimal polynomial of $\beta$ over $Q$ throughout my development. Take $K(\alpha^{p^n})$ to be the separable extension provided by the lemma.

$\Leftarrow)$ Observe that $Irr(\alpha,K(\alpha^{p^n})) = x - \alpha$ this happens because $x- \alpha$ is irreducible since it is linear, $\alpha$ is a root and $\alpha \in K(\alpha^{p^n})[X]$ since by hypothesis $\alpha \in K(\alpha^{p^n})[X]$

This tells that the minimal polynomial of $\alpha$ has no multiple roots and this says that $\alpha$ is separable over $K(\alpha^{p^n})$. Therefore, the whole extension $K(\alpha)/K(\alpha^{p^n})$ is separable (because the generators are) and by the lemma we had that $K(\alpha^{p^n})/K$ is separable. Therefore, $K(\alpha)/K$ is separable.

$\Rightarrow)$ If $K(\alpha)/K$ is separable then you have that $K(\alpha)/K(\alpha^{p^n})$,$K(\alpha^{p^n})/K$ are separable and therefore $Irr(\alpha,K(\alpha^{p^n})$ does not have multiple roots.

Considering polynomial $x^{p^n} - \alpha^{p^n}$ we can apply the properties of the Frobenius homomorphism to have $x^{p^n} - \alpha^{p^n} = (x-\alpha)^{p^n}$ and then you can reason that any irreducible factor of this polynomial (reason that the polynomial cannot be irreducible by separability) needs to be of degree one so that $x-\alpha \in K(\alpha^{p^n})$ and therefore $\alpha \in K(\alpha^{p^n})$. Now multiplying by inverses you can see that the whole chain is true $K(\alpha) = K(\alpha^2) = \ldots$.