# Something special for general finite extension with top field being algebraically closed

Let $K/F$ be a finite extension with $K$ algebraically closed.

How can I show that $\mathrm{char}(F)=0$ and $K=F((-1)^{1/2})$ ?

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You should add the assumption that $K \neq F$, otherwise $K = F = \overline{\mathbb F_p}$ is a counterexample with characteristic $p$. – marlu Apr 18 '12 at 19:39
This is the Artin-Schreier Theorem. The link is to an exposition of it by Keith Conrad. – Arturo Magidin Apr 18 '12 at 19:39

You could show that $F$ is real closed, see http://en.wikipedia.org/wiki/Real_closed_field for various equivalent definitions. $\operatorname{char}(F) = 0$ follows from the fact that the field is ordered, and $K = F(\sqrt{-1})$ is one of the equivalent definitions.
Edit: As marlu points out, you need $F \neq K$.