3
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 7
    $\begingroup$ 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 '14 at 18:11
  • 2
    $\begingroup$ 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 '17 at 22:51
  • $\begingroup$ Related: If L∣K is a finite extension of fields then K is perfect iff L is perfect $\endgroup$ – yamete kudasai Feb 3 at 23:20
-1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • 3
    $\begingroup$ I do not understand this answer. What does it mean here for an extension of $F$ to split? $\endgroup$ – Torsten Schoeneberg Jan 26 '17 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.