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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

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  • $\begingroup$ yes. And the proof is long $\endgroup$ – leo Nov 23 '13 at 18:28
  • $\begingroup$ Don't know, what's the order of exposition in the text by Lang, but the exercise follows using the Corollary and the Theorem in my answer and induction. Are you trying to prove the $(\impliedby)$ of the first question or of the exercise? $\endgroup$ – leo Nov 24 '13 at 15:20
  • $\begingroup$ I must prove both the implications of the exercise. I managed to prove the "$\Longrightarrow$", no problem. I also managed to prove the "$\Longleftarrow$" IF I can prove my original statement (i.e. the first question). I could use your theorem, of course, but then the exercise seems too simple to me. Anyway, I checked Lang, and that theorem is present (before the exercise). So, if I can't find another solution, I think I am allowed to use it...Thank you then $\endgroup$ – aerdna91 Nov 24 '13 at 16:08
  • $\begingroup$ Glad to help you $\endgroup$ – leo Nov 24 '13 at 16:20
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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.

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  • $\begingroup$ Thanks for your answer. By "$E/F$ is finite" you mean that $E$ has finite degree over $F$, right? By the way, take a look to my question update if you want. $\endgroup$ – aerdna91 Nov 24 '13 at 15:03
  • $\begingroup$ @aerdna91 Yes, that's what it means. $\endgroup$ – leo Nov 24 '13 at 15:05
  • $\begingroup$ ok, I updated the question. Check it if you want. $\endgroup$ – aerdna91 Nov 24 '13 at 15:13
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Let f(x) be the mini. poly. over F, since alpha is separable, let E be the splitting field of f(x), and by Galois theory, E/F is Galois extension, thus separable. E-F(alpha)-F, as we see, E/F is separable, then E/F(alpha) and F(alpha)/F are all separable. Just check by definition.

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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 $K(\alpha) = K(\alpha^2) = \ldots$.

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