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Problem statement: Given field $F$, if for any field extension $M/F$, $[M:F]$ is divisible by a fixed prime $p$, show that $F$ is either perfect or have characteristic $p$.

Previously in this question Extension degree must be power of prime, I see that $[K:F]$ is a power of $p$ through Galois closure. I also know that irreducible but inseparable polynomials must have certain form. But is it possible to continue from here without the notion of separable closure?

SmoothKen
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Suppose that $F$ is not perfect; then we have $q:=\operatorname{char}F>0$, and there exists an element $\alpha\in F$ that is not a $q$-th power. Now the polynomial $x^q-\alpha$ is irreducible in $F$, so the ring $E:=F[x]\big/\langle x^q-\alpha\rangle$ is a field extension of $F$ of degree $q$. On the other hand, by the problem hypotheses, $[E:F]$ must also be divisible by $p$, so this forces $q=p$, as desired.

Atticus Stonestrom
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    Why does there exists an $\alpha$ that is not a $q$th-power? – SmoothKen Sep 03 '21 at 03:10
  • @SmoothKen oh, apologies, that's the definition of a perfect field that I use :) (namely, a field is perfect iff either its characteristic is $0$ or its characteristic is $p>0$ and every element of it is a $p$-th power.) ... what characterization are you familiar with? – Atticus Stonestrom Sep 03 '21 at 03:13
  • perfect if and only if char 0 or Frobenius is surjective, which means imperfect and char > 0 implies Frobenius not surjective.....I see your magic..... – SmoothKen Sep 03 '21 at 03:16
  • precisely! :) ${}{}$ – Atticus Stonestrom Sep 03 '21 at 03:17