prove that $2 < [\bar{F} : F] < \infty$ is impossible

Question

Let $F$ be a Field, and $\bar{F}$ be an algebraic closure of $F$

I want to prove that $2 < [\bar{F} : F] < \infty$ is impossible.

For $[\bar{F}:F]=2$, I know there is Artin-Schreire theorem which characterized this special case and classify such $F$

It seems to me that $[\bar{F} : F]=\infty$ gives $\bar{F} \simeq F^{\infty}$... How this is possible?

Answer 1

This is exactly the Artin–Schreier theorem:

If a field $F$ is not algebraically closed but its algebraic closure $C$ is a finite extension of $F$, then $C = F (i)$ with $i^2=-1$.

For a proof, see an exposition by Keith Conrad.