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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?

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  • $\begingroup$ Any field of characteristic 0 is perfect. $\endgroup$ Commented Sep 3, 2021 at 2:46
  • $\begingroup$ @MartinSkilleter Huh? That does not seem to be a hint? $\endgroup$
    – SmoothKen
    Commented Sep 3, 2021 at 2:48
  • $\begingroup$ Consider your field $F$. It either has characteristic 0 or characteristic $p$. Use my above comment to conclude. $\endgroup$ Commented Sep 3, 2021 at 2:56
  • $\begingroup$ @MartinSkilleter No. It is not that simple.You may want to read the question again..Why does it have to be $p$? $\endgroup$
    – SmoothKen
    Commented Sep 3, 2021 at 2:58
  • $\begingroup$ My apologies, you're right. I missed that the two primes were meant to be the same. $\endgroup$ Commented Sep 3, 2021 at 3:03

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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.

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    $\begingroup$ Why does there exists an $\alpha$ that is not a $q$th-power? $\endgroup$
    – SmoothKen
    Commented Sep 3, 2021 at 3:10
  • $\begingroup$ @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? $\endgroup$ Commented Sep 3, 2021 at 3:13
  • $\begingroup$ 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..... $\endgroup$
    – SmoothKen
    Commented Sep 3, 2021 at 3:16
  • $\begingroup$ precisely! :) ${}{}$ $\endgroup$ Commented Sep 3, 2021 at 3:17

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