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The problem is

Let $E$ be a finite extension of $F$ and suppose $E$ is algebraically closed. Show that $F$ is perfect.

I know that a field $F$ is called perfect if every irreducible polynomial in $F[x]$ is separable, and a field $F$ is perfect if and only if it has characteristic $0$, or it has characteristic $p$ and $F=F^p$; then, fields of characteristic $0$ and finite fields are perfect.

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    $\begingroup$ If you’re willing to call in the big guns, if $E$ is algebraically closed, and $[E:F]<\infty$, then the characteristic is zero, and the degree is $2$. That’s Artin-Schreier. Since $F$ is of characteristic zero, it’s automatically perfect. $\endgroup$
    – Lubin
    Dec 12, 2014 at 18:11
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    $\begingroup$ One can also exclude positive characteristic $p$ with the following easy lemma: If $F$ is a non-perfect field of characteristic $p$, then so is $F^{p^{-1}} \neq F$. Consequently, $F \subsetneq F^{p^{-1}} \subsetneq F^{p^{-2}} \subsetneq ...$ is an infinite tower of fields within $E$, contradicting $E|F$ being finite. $\endgroup$
    – Torsten Schoeneberg
    Jan 26, 2017 at 22:51
  • $\begingroup$ Related: If L∣K is a finite extension of fields then K is perfect iff L is perfect $\endgroup$
    – nowhere dense
    Feb 3, 2019 at 23:20

1 Answer 1

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You can prove:

  • If every extension of F splits, then F is perfect.
  • Let F ⊆ K ⊆ E a tower of fields. If E/F is algebraic and splits, then K/F splits.

And finally use the definition of algebraically closed.

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    $\begingroup$ I do not understand this answer. What does it mean here for an extension of $F$ to split? $\endgroup$
    – Torsten Schoeneberg
    Jan 26, 2017 at 22:39

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