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Let $p$ be a prime number. Consider for all $k,r\in\mathbb{N}_0$ with $k|r$ the field $\mathbb{F}_{p^k}$ as a subfield of $\mathbb{F}_{p^r}$. Define

$$ \overline{\mathbb{F}_p}:=\bigcup_{i\in\mathbb{N}_0}\mathbb{F}_{p^i}.$$

I want to show that $\overline{\mathbb{F}_p}$ is the algebraic closure of $\mathbb{F}_p$. I already managed to show that $\overline{\mathbb{F}_p}$ is a field and that it is algebraic over $\mathbb{F}_p$, but I can't seem to show that it is also algebraically closed.

I know that for every $i\in\mathbb{N}_0$, $\mathbb{F}_{p^i}$ is a decomposition field of the polynomial $X^{p^i}-X$. Which means this polynomial can be expressed as $$X^{p^i}-X=\prod_{a\in\mathbb{F}_{p^i}}(X-a),$$

and I thought that this could help me prove that every irreducibel polynomial in $\overline{\mathbb{F}_p}[X]$ has degree $1$ but so far it hasn't.

Any help would be appreciated.

TheHunter
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    An "innocent" question. What do you mean by the union here? For example, let $p=2$. Then $\Bbb{F}4=\Bbb{F}_2[x]/(x^2+x+1)$, $\Bbb{F}_8=\Bbb{F}_2/(x^3+x+1)$, $\Bbb{F}{16}=\Bbb{F}2[x]/(x^4+x+1)$, $\Bbb{F}{32}=\Bbb{F}2[x]/(x^5+x^2+1)$, $\Bbb{F}{64}=\Bbb{F}2x$, $\Bbb{F}{128}=\Bbb{F}_2[x]/(x^7+x^3+1)$ etc. What does their union look like? Keep in mind that some of these have non-trivial intersections, and the injective embeddings of subfields are not unique. – Jyrki Lahtonen Jan 26 '21 at 19:57
  • My attempt at answering that is to form a nested chain, when the (set theoretic) direct limit is simple to manage. – Jyrki Lahtonen Jan 26 '21 at 19:58
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    I think this is a duplicate of this recent question. Because I have no less than three applicable dupehammers, my vote would take immedate effect. So I will wait for a few minutes at least so that people can object. – Jyrki Lahtonen Jan 26 '21 at 20:02
  • @JyrkiLahtonen That is indeed a duplicate, I guess it didn't pop up as a suggestion when creating my question. – TheHunter Jan 26 '21 at 20:10

1 Answers1

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If $q(x) \in \mathbb{F}_{p^k}[x]$ is of degree $n$ and irreducible, $q$ has a non zero root in the rupture field $\mathbb{F}_{p^k}[x]/(q)$ which is isomorphic to $\mathbb{F}_{(p^k)^n}= \mathbb{F}_{p^{kn}}$.

By induction, you can prove that $q$ splits in one of the $\mathbb{F}_{p^m}$. Which implies the desired result.

mathcounterexamples.net
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