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I have two "elementary" questions. Let $\mathbb{F}_p$ be a finite field with $p$ elements, with $p$ odd prime. Let $\overline{\mathbb{F}}_p$ denote its algebraic closure, and $K_p$ denote its quadratic closure (closure under the operation of taking square roots). Is any of these two fields, $\overline{\mathbb{F}}_p$ and $K_p$, a quadratic extension of one of its subfields?

I am not sure what the answer is, though I have some guesses. I know that $K_p$ is the direct limit of $\mathbb{F}_{q_n}$, where $q_n = p^{2^n}$, as $n$ goes to $\infty$, and that, at each "finite" stage, the $n$'th term of this sequence of fields is a quadratic extension of the previous term. But what happens at the limit, namely at $K_p$?

This is I am sure well known to the experts. I am not even sure what the various Galois groups look like, though I can look that part up, of course. Does that amount (in the case of $K_p$) to the question of whether or not $\operatorname{Gal}(K_p/\mathbb{F}_p)$ has a subgroup of index $2$?

Edit: see Eric Wofsey's answer below, the question actually amounts to whether the corresponding Galois group has a subgroup of order $2$ (and not, as I had initially thought, of index $2$).

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  • $\begingroup$ Closure in the sense here implicitly implies an infinity union of fields, in particular $$\overline{\mathbb{F}}_p= \large\cup_{n\ge 1}\mathbb F_{p^n}$$ so "intuitively" the answer is no. $\endgroup$ Commented Mar 10, 2018 at 18:12
  • $\begingroup$ @Piquito, I was expecting the answer to be no in that case, but I wasn't sure how to prove it, since I did not know what the Galois groups were, before Eric Wofsey posted his answer. $\endgroup$ Commented Mar 10, 2018 at 18:16
  • $\begingroup$ It is possible to say more. The are no fields $K$ such that $[\overline{\Bbb{F}}_p:K]<\infty$. This follows from the identification of the Galois group as $\hat{\Bbb{Z}}$ (See Eric's +1 answer). I tried to give a more elementary argument for that fact here. $\endgroup$ Commented Mar 10, 2018 at 21:11
  • $\begingroup$ @JyrkiLahtonen, thank you. Yes, I have been away from characteristic $p$ for a very long time. Thank you. $\endgroup$ Commented Mar 10, 2018 at 21:18

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In terms of Galois theory, the question is whether the Galois group has a subgroup of order $2$, not index $2$. The Galois groups in question have simple descriptions: $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)=\widehat{\mathbb{Z}}$, the profinite completion of $\mathbb{Z}$, and $Gal(K_p/\mathbb{F}_p)=\mathbb{Z}_2$, the $2$-adic integers. In general, the Galois group of an infinite extension is just the inverse limit of the Galois groups of finite subextensions. In this case, the Galois groups of finite subextensions are just cyclic groups (or cyclic $2$-groups for the quadratic extensions), with the maps between them being the usual maps, and so the inverse limit is $\widehat{\mathbb{Z}}$ (or $\mathbb{Z}_2$ in the quadratic case).

So, to answer your question: no, because these Galois groups are torsion-free, and in particular have no subgroups of order $2$.

You can also just see this directly by looking at what a subfield can look like. For simplicity, let's just consider the case of $K_p$ (the case of $\overline{\mathbb{F}_p}$ is similar but a bit messier). Any subfield $L$ of $K_p$ is a union of finite subfields, since any element generates a finite subfield. So $L$ is some union of fields of the form $\mathbb{F}_{p^{2^n}}$. If this union involves only finitely many such fields, then $L$ is finite, so $[K_p:L]$ is infinite. If the union involves infinitely many such fields, then $L$ is all of $K_p$, since it contains $\mathbb{F}_{p^{2^n}}$ for arbitrarily large $n$. So $K_p$ has no proper subfield $L$ such that $[K_p:L]$ is finite.

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  • $\begingroup$ Thank you so much for your answer. I have learned a lot! $\endgroup$ Commented Mar 10, 2018 at 17:38

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