$\bar{\mathbb{F}}_p$ is not a finite degree extension of any proper subfield.

Question

Consider $\bar{\mathbb{F}}_p$, the algebraic closure of $\mathbb{F}_p$. I want to see that: for every proper subfield $K \leq \bar{\mathbb{F}_p}$, $\bar{\mathbb{F}}_p/K$ is not a finite extension.

It is known that, and can be somewhat easily shown that $\bar{\mathbb{F}}_p = \cup_{n \geq 1}\mathbb{F}_{p^n}$

Now, if any of the proper subfields have the form $\mathbb{F}_{p^n}$, it is easy enough to see that $\bar{\mathbb{F}}_p \neq \mathbb{F}_{p^n}(\alpha_1, \cdots, \alpha_m)$ for some $\alpha_i$, by going high up enough, i.e, to some big enough $m$ such that $\alpha_i \not \in \mathbb{F}_{p^m} \subseteq \bar{\mathbb{F}_p}$

The problem is characterizing the proper subfields. Is every subfield of $\bar{\mathbb{F}_p}$ going to have this form? Can we have an infinite intermediate subfield?

Surely $\cup_{k\geqslant 1}\mathbb{F}_{q^k}$ where $q=p^s$ is proper and not a finite extension. — ancient mathematician, Nov 24 '21 at 10:33
Finite extension as in finite degree not finite cardinality. I'm referring to the index of the extension. — Cruise Control, Nov 24 '21 at 10:34
But yes, there are infinite intermediate fields which I have now realized, if that's what you're referring to. — Cruise Control, Nov 24 '21 at 10:34
Yes, but I'm looking at if $\bar{\mathbb{F}_p}$ is a finite degree extension of any subfield $K$ or if they are all finite. — Cruise Control, Nov 24 '21 at 10:37
ancient's field is indeed infinite, which is what is asked for in the last sentence of the question. But is it really proper? — Andrew Dudzik, Nov 24 '21 at 10:39
I think it is proper, if we take some $t$ prime to $s$ for example. But even so, even with infinite intermediate field, is $\bar{\mathbb{F}_p}$ a finite extension of ancient's field? Or is degree still finite. — Cruise Control, Nov 24 '21 at 10:41
My comment only referred to the last question you asked. I am not claiming to have an answer to the first part. — ancient mathematician, Nov 24 '21 at 10:42
@ancientmathematician That looks good to me. The Galois group here is the profinite integers, if $q=p$ this looks like picking out just the $\mathbb{Z}_2$ factor. — Andrew Dudzik, Nov 24 '21 at 10:46
The fields you're discussing are not proper. Any finite field of characteristic $p$ has the same algebraic closure. — Smn, Nov 24 '21 at 10:46
@Mentos So what? No one has said my example is the algebraic closure of anything. — ancient mathematician, Nov 24 '21 at 10:48
Your first example is the algebraic closure of $\mathbb{F}_q$, hence equal to $\overline{\mathbb{F}_p}$. I misread the second as $q^{2k}$, I agree with the last example — Smn, Nov 24 '21 at 10:58
Correct. $\bigcup_{k\ge1}\Bbb{F}_{q^{2^k}}$ (with the union taken inside a fixed algebraic closure $\overline{\Bbb{F}_q}$) is infinite. But it is, of course, a long way from $\overline{\Bbb{F}_q}$. It only consists of the zeros of irreducible polynomials $\in\Bbb{F}_q[x]$ of degree that is a power of two. A "complementary" example would consist of the zeros of all odd degree irreducible polynomials $\in\Bbb{F}_q[x]$. — Jyrki Lahtonen, Nov 24 '21 at 12:01

score 4 · Answer 1 · answered Nov 24 '21 at 11:02

The Galois group here is isomorphic to $G=\hat{\mathbb{Z}}$, 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.

Jyrki Lahtonen · Accepted Answer · 2021-11-24T12:21:34.727

3

As Slade explained (+1) this follows either from the known structure of the automorphism group of $\overline{\Bbb{F}_p}$ or from a theorem of Artin & Schreier stating that any finite extension $\overline{K}/K$, with the bigger field algebraically closed, is of the form $\overline{K}=K(\sqrt{-1})$.

An elementary argument can also be given. Assume that $\overline{\Bbb{F}_p}/K$ is a finite extension. Then that extension is Galois. It is obviously normal, and it is also separable because every element of $\overline{\Bbb{F}_p}$ belongs to a finite field, and hence is a zero of a separable polynomial over the prime field (and hence also over $K$). By basic Galois theory this implies that $\overline{\Bbb{F}_p}$ has an automorphism of a finite order. But I shamelessly link to an old elementary answer of mine explaining that there are no such automorphisms.

edited Nov 24 '21 at 12:21

answered Nov 24 '21 at 11:54

Jyrki Lahtonen

140,891

Very nice! I was hoping for a more elementary proof! – Cruise Control Nov 24 '21 at 12:18
@CruiseControl If you get a chance to study Galois theory of infinite algebraic extensions, you will learn that there is a topology on the Galois group $G$ such that the intermediate fields correspond bijectively with the closed subgroups of $G$. That gives you a better description of the intermediate fields in terms of the closed subgroups of $\hat{\Bbb{Z}}$. – Jyrki Lahtonen Nov 24 '21 at 13:20

$\bar{\mathbb{F}}_p$ is not a finite degree extension of any proper subfield.

2 Answers2