I'm proving that $\overline{\mathbb{F}}_p = \bigcup\limits_{i=1}^{\infty} \mathbb{F}_{p^i}$ is an algebraic closure of $\mathbb{F}_p$ where $p$ is a prime. I think I've gotten down how to prove that $\overline{\mathbb{F}}_p$ is a field and that it is algebraic over $\mathbb{F}_p$.

I have some difficulties with proving that $\overline{\mathbb{F}}_p$ is algebraically closed.

My attempt is as following:

Suppose $f$ is a non-constant polynomial in $\overline{\mathbb{F}}_p [X]$. If $\overline{\mathbb{F}}_p$ contains a root of $f$, then it is algebraiclly closed. Per definition there must exist a $\mathbb{F}_{p^k}$ for a certain positive integer $k$ that contains all the coefficients of $f$. Take a root $\alpha$ of $f$ and consider the extension $\mathbb{F}_{p^k}(\alpha)$. How is this now a field of the form $\mathbb{F}_{p^l}$ for a certain positive integer $l$?

  • 4
    $\begingroup$ Well, $\mathbb{F}_{p^k}(\alpha)$ is still a finite field... $\endgroup$
    – Aphelli
    Dec 31, 2020 at 11:16
  • $\begingroup$ So you can write it as a similar union but finite? $\endgroup$ Dec 31, 2020 at 11:33
  • 3
    $\begingroup$ No, @SamoGrecco, but the extension $\mathbb F_{p^k}(\alpha)$ is a finite extension of a finite field, therefore a finite field itself, and, as it is of characteristic $p$, it must be isomorphic to exactly one of $\mathbb F_{p^l}$. $\endgroup$ Dec 31, 2020 at 11:49
  • $\begingroup$ See 1, 2. Nothing very detailed, I'm afraid. $\endgroup$ Dec 31, 2020 at 14:46
  • $\begingroup$ There is also this cute related, but different question. $\endgroup$ Dec 31, 2020 at 14:47

1 Answer 1


Take a polynomial $p(x) \in \overline{\mathbb{F}}_p [x]$. Then, $p(x) \in \mathbb{F}_{p^k}[x]$ for some $k$. The splitting field is a finite extension of characteristic $p$, so it is isomorphic to $\mathbb{F}_{p^l}$ for some $l$ (by characterisation of finite fields). Hence, $p(x)$ must split over $\overline{\mathbb{F}}_p$.


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