Im currently trying to construct the algebraic closure of $\mathbb{F}_p$. I know that this has been resolved in some other posts, but I want to understand the solution completely and therefore I have some questions:
Let $p$ be prime and $\mathbb{F}_{p}^{a}/\mathbb{F}_p$ be the algebraic closure of $\mathbb{F}_p$. Show that \begin{equation} \mathbb{F}_{p}^{a}=\bigcup_{n=1}^{\infty}\mathbb{F}_{p^n}. \end{equation} Here $\mathbb{F}_{p^n}:=\mathbb{F}_p(V_n)$, where $V_n=\{x\in\mathbb{F}_{p}^a | X^{p^n}-X=0\}$.
I know that for every polynomial from $\mathbb{F}_{p}[X]$, we need to find a $n\in\mathbb{N}$ s.t. the polynomial splits over $\mathbb{F}_{p^n}$. Why is there always such an $n$? I know that $\mathbb{F}_{p^n}$ is unique up to an isomorphism, but I don't know why a polynomial of $\mathbb{F}_{p}[X]$ eventually splits in it.
Now, if there's always such an $n$, then we just need to unite every extension of $\mathbb{F}_{p}$. I know that a finite field $\mathbb{F}_{p^k}$ is a subfield of $\mathbb{F}_{p^n}$ iff $k|n$. So for example $\mathbb{F}_{p}\subset\mathbb{F}_{p^2}\subset\mathbb{F}_{p^4}\subset\mathbb{F}_{p^8}\subset\cdots$. Would this mean that the union of $\mathbb{F}_{p^n}$ for all $n\in\mathbb{N}$ would cover every possible extension and must therefore be the algebraic closure of $\mathbb{F}_{p}$?
I think I have some fundamental problems understanding finite fields, so any elaboration and help would be greatly appreciated.
First of all, the words "the algebraic closure" are very problematic. An algebraic closure is unique only up to isomorphism (not even a natural one), not unique in general. It is common to use the word "the" here because most of the time we fix a specific algebraic closure and always work with it. However, when you just begin studying field theory you have to be extremely careful with it.
In the same way, the finite fields $\mathbb{F}_{p^n}$ are not unique in general, but only up to isomorphism. What is true is the following: if we fix an algebraic closure $\mathbb{F}^a_p$ (which exists as we know) then this specific algebraic closure contains a unique subfield of order $p^n$. This subfield is defined as follows:
$\mathbb{F}_{p^n}=\{\alpha\in\mathbb{F}^a_p: \alpha^{p^n}=\alpha\}$
And the statement that $k|n$ is equivalent to $\mathbb{F}_{p^k}\subseteq\mathbb{F}_{p^n}$ is true for these subfields, inside our fixed algebraic closure. (not for any finite fields of order $p^k$ and $p^n$ in general, that would make no sense. In this case we would only have an injection, not inclusion)
Now, you asked to prove that $\mathbb{F}^a_p=\bigcup\limits_{n=1}^{\infty}\mathbb{F}_{p^n}$. Note that one inclusion is trivial by definition. Conversely, let $\alpha\in\mathbb{F}^a_p$. Let $[\mathbb{F}_{p}(\alpha):\mathbb{F}_p]=n$, this is a finite number because $\alpha$ is algebraic. Then $\mathbb{F}_p(\alpha)$ is a subfield of $\mathbb{F}^a_p$ of order $p^n$, and since $\mathbb{F}_{p^n}$ is the unique subfield of $\mathbb{F}^a_p$ of order $p^n$, it follows that $\mathbb{F}_p(\alpha)=\mathbb{F}_{p^n}$. (again, remember that $\mathbb{F}_{p^n}$ here is a subfield inside our fixed algebraic closure, not any other finite field) Thus, $\alpha\in\mathbb{F}_{p^n}$, and so $\alpha$ is in the union.