# If $E/F$ is a finite extension and $E$ is algebraically closed, then $F$ is perfect.

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.

• 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. – Lubin Dec 12 '14 at 18:11
• 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. – Torsten Schoeneberg Jan 26 '17 at 22:51
• – nowhere dense Feb 3 '19 at 23:20

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