# Algebraic closure of $\mathbb F_p$ [duplicate]

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 $$l$$?

• Well, $\mathbb{F}_{p^k}(\alpha)$ is still a finite field... Dec 31, 2020 at 11:16
• So you can write it as a similar union but finite? Dec 31, 2020 at 11:33
• 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 $\mathbb F_{p^l}$. Dec 31, 2020 at 11:49
• See 1, 2. Nothing very detailed, I'm afraid. Dec 31, 2020 at 14:46
• There is also this cute related, but different question. Dec 31, 2020 at 14:47

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 $$\overline{\mathbb{F}}_p$$.