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Let $p$ be a prime number and $\mathbb{F}_{p^n}$ denote the finite field with $p^n$ elements. Note that $\overline{\mathbb{F}_p} = \cup_{n\geq 1}\mathbb{F}_{p^n}$ is the algebraic closed field of $\mathbb{F}_p$.

I am wondering if every proper sub-field of $\overline{\mathbb{F}_p}$ is finite.

Yan Zhu
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  • Recheck the algebraic closure. You want $\overline{ \Bbb F}=\bigcup_m\Bbb F_{p^{nm}}$. – calc ll Jul 04 '22 at 21:36
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    Well, @Cpc , as long as both formulas give you an algebraically closed field, the two formulas are equivalent. I don’t think that OP meant for “$\Bbb F_{p^n}$” to be a particular, chosen finite field. – Lubin Jul 05 '22 at 00:58
  • @Cpc See my updates. – Yan Zhu Jul 05 '22 at 02:16
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    Your main question was already answered. But I want to point out that forming that union is a little bit problematic. For example, if $\Bbb{F}4=\Bbb{F}_2[x]/\langle x^2+x+1\rangle$ and $\Bbb{F}_8=\Bbb{F}_2[x]/\langle x^3+x+1\rangle$ what does $\Bbb{F}_4\cup\Bbb{F}_8$ mean even? When you add $\Bbb{F}{16}=\Bbb{F}_2[x]/\langle x^4+x+1\rangle$ to the mix you run into the problem that there are two different embedding of $\Bbb{F}_4$ into that, and to form the union you need to pick one of them. I explained one way of sidestepping such issues here. – Jyrki Lahtonen Jul 05 '22 at 04:52
  • (cont'd) That approach is, in a way, based on having the union transformed into a direct limit (which is less problematic set theoretically). Using the sequence of factorials as extension degrees makes it a nested union, which at least to me feels simpler still. – Jyrki Lahtonen Jul 05 '22 at 04:57

1 Answers1

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You’re asking whether the algebraic closure of a finite field has any subfields that are not finite as sets.

You may know that $\Bbb F_{p^n}\subset\Bbb F_{p^m}$ if and only if $n|m$, and if this happens, the field extension degree is $[\Bbb F_{p^m}:\Bbb F_{p^n}]=m/n$. Granting this, even if you didn’t know it before, consider the union of the fields $k_\ell=\Bbb F_{p^{2^\ell}}$. You see that $k_\ell\subset k_{\ell+1}$, with $[k_{\ell+1}:k_\ell]=2$, so that we have an ascending chain of $2$-extensions of the prime field $k_0=\Bbb F_p$.

Then, according to the principle at the beginning of the preceding paragraph, $\Bbb F_{p^3}$ is not contained in any $k_\ell$. In particular, $\bigcup_\ell k_\ell$ is an infinite field not equal to the algebraic closure.

Lubin
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  • You're right. I found the result. It would have to be the algebraic closure of the prime subfield. And that can be shown to be $\bigcup_n\Bbb F_{p^n}$. Thanks. – calc ll Jul 05 '22 at 01:53