If we have $q= p^f$ and $\bar{\mathbb{F}_q}$ the algebraic closure of $\mathbb{F}_q$. A supernatural number n is defined in following way: $n = \prod _{p \ prime} p^{n_p}$ with $n_p \in \mathbb{N} \cup \{\infty\}$.
I want to show that we have a bijective map $n \to \mathbb{F}(n)$ where $\mathbb{F}(n) := \mathbb{F}_{p^n } $ with covariant relation $\mathbb{F}(m) \subset \mathbb{F}(n)$ if and only if $m | n$.
My last try: If $n$ is only natural so I can define $F(n) = \{a \in \bar{\mathbb{F}_q}| f_n(a) =0 \ for \ f_n(X) = X^{q^n}-X\}$ what obviously concerving the relation above. Is there a way to "transfer" this argumentation in appropriate way for to a arbitrary supernatural $n$?
For $p$ prime, $\mathbb{F}_{p^n}$ is defined as a finite field with $p^n$ elements, which implies $\mathbb{F}_{p^n}^\times$ is a multiplicative group with $p^n-1$ elements, so its elements satisfy $a^{p^n-1} = 1$ and the elements of $\mathbb{F}_{p^n}$ are roots of $X^{p^n}-X \in \overline{\mathbb{F}_p}[X]$.
But $X^{p^n}-X $ has at most $p^n$ roots in the field $\overline{\mathbb{F}_p}$ therefore $\mathbb{F}_{p^n}$ is exactly the splitting field of $X^{p^n}-X \in \mathbb{F}_p[X]$.
$X^{p^m-1}-1$ divides $X^{p^n-1}-1$ iff $p^m-1 | p^n-1$ iff $m | n$. Thus $\mathbb{F}_{p^m} \subset \mathbb{F}_{p^n}$ iff $m | n$.
For a formal infinite product $\mathfrak{M} = \prod q_i^{e_i}$ where the $q_i$ are distinct primes, we can define $$\mathbb{F}_{\textstyle p^\mathfrak{M}} = \bigcup_{j \to \infty} \mathbb{F}_{\textstyle p^{m_j}}, \qquad m_j = \prod_{i \le j} q_i^{e_i}$$
By the preceding $\mathbb{F}_{\textstyle p^\mathfrak{M}}$ is well-defined as a field, and it doesn't depend on which order we take the $q_i$.
We obtain $\mathbb{F}_{\textstyle p^\mathfrak{M}} \subset \mathbb{F}_{\textstyle p^\mathfrak{F}}$ iff $\forall i, e_i(\mathfrak{F}) - e_i(\mathfrak{M}) \ge 0$ which we can take as the definition of $\mathfrak{M}\ |\ \mathfrak{F}$