 2022-01-14

Consider ${\stackrel{―}{F}}_{p}$, the algebraic closure of ${F}_{p}$. I want to see that: for every proper subfield $K\le \stackrel{―}{{F}_{p}},\frac{{\stackrel{―}{F}}_{p}}{K}$ is not a finite extension.
It is known that, and can be somewhat easily shown that ${\stackrel{―}{F}}_{p}={\cup }_{n\ge 1}{F}_{{p}^{n}}$.
Now, if any of the proper subfields have the form ${F}_{{p}^{n}}$, it is easy enough to see that ${\stackrel{―}{F}}_{p}\ne {F}_{{p}^{n}}\left({\alpha }_{1},\cdots ,{\alpha }_{m}\right)$ for some ${\alpha }_{i}$ by going high up enough, i.e, to some big enough m such that ${\alpha }_{i}\in {F}_{{p}^{m}}\subseteq \stackrel{―}{{F}_{p}}$
The problem is characterizing the proper subfields. Is every subfield of $\stackrel{―}{{F}_{p}}$ going to have this form? Can we have an infinite intermediate subfield? Annie Levasseur

Expert

Step 1
As Slade explained $\left(+1\right)$ this follows either from the known structure of the automorphism group of $\stackrel{―}{{F}_{p}}$ or from a theorem of Artin & Schreier stating that any finite extension $\frac{\stackrel{―}{K}}{K}$, with the bigger field algebraically closed, is of the form $\stackrel{―}{K}=K\left(\sqrt{-1}\right)$.
Step 2
An elementary argument can also be given. Assume that $\frac{\stackrel{―}{{F}_{p}}}{K}$ is a finite extension. Then that extension is Galois. It is obviously normal, and it is also separable because every element of $\stackrel{―}{{F}_{p}}$ belongs to a finite field, and hence is a zero of a separable polynomial over the ' field (and hence also over K). By basic Galois theory this implies that $\stackrel{―}{{F}_{p}}$ has an automorphism of a finite order. Daniel Cormack

Expert

The Galois group here is isomorphic to $G=\stackrel{^}{Z}$, the profinite integers. Since G is torsion-free, there are no finite index subfields.
This also follow from Artin-Schreier, though I don't know offhand of an extremely clean and short proof in the case of finite fields.

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