Algebraic closure of $\mathbb F_p$ [duplicate]

Question

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 $l$?

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