I got this question while I am studying infinite Galois correspondence.
Let $\overline{\mathbb{F}}_p$ the algebraic closure of $\mathbb{F}_p$. It's a fact [ http://assets.press.princeton.edu/chapters/s9103.pdf ] that this is an infinite Galois extension, and $\text{Gal}(\overline{\mathbb{F}}_p \; / \; \mathbb{F}_p) \cong \hat{\mathbb{Z}}$.
Now suppose we want to characterize each sub-extension of $\overline{\mathbb{F}}_p \; / \; \mathbb{F}_p$ (equivalently each closed subgroup of $\hat{\mathbb{Z}}$). Is there a way to solve this problem using the Fundamental Theorem for infinite Galois extensions?
I'll try to explain my doubt better in the following: If $q$ is a prime, define $L_q := \bigcup_{n \in \mathbb{N}} \mathbb{F}_{p^{q^n}}$. We can show that $[L_q : \mathbb{F}_p] = [\overline{\mathbb{F}}_p : L_q] = \infty$ and Gal$(L_q/\mathbb{F}_p) \cong \mathbb{Z}_q$, where $\mathbb{Z}_q$ denotes the ring of p-adic integers.
Let $\mathbb{F}_p \subseteq F \subseteq L_q$ a field, and consider $q^{n_F} = $ sup$\lbrace \mathbb{F}_p(x) : \mathbb{F}_p \vert x \in F \rbrace$, with $0 \leq n_F \leq \infty$.
It's not hard to show that $F = \mathbb{F}_{p^{n_F}}$ (with the convention that $L_q = \mathbb{F}_{p^{q^{\infty}}}$) and so every closed subgroup of $\hat{\mathbb{Z}}$ can be seen as Gal$(L_q/\mathbb{F}_{p^{q^n}})$, for some $n \in \mathbb{N} \cup \lbrace \infty \rbrace$.
Consider now the Frobenius of $L_q / \mathbb{F}_p$ and denote it as $\phi$. We have $\mathbb{Z}_q \cong Gal(L_q / \mathbb{F}_p) = \overline{\langle \phi \rangle}$, and also $\overline{\langle \phi ^ {q^n} \rangle}$ fixes $\mathbb{F}_{p^{q^n}}$, but doesn't fix $\mathbb{F}_{p^{q^{n+1}}}$, and so $Gal(L_q / \mathbb{F}_{p^{q^n}}) = \overline{\langle \phi^{q^n} \rangle}$. But if $\varphi: \overline{\langle \phi \rangle} \rightarrow \mathbb{Z}_q$ is the isomorphism such that $\phi \mapsto 1$, then the restriction $\varphi_{|\overline{\langle \phi \rangle}}$ gives us an isomorphism $$\varphi_{|\overline{\langle \phi^{q^n} \rangle}} : \overline{\langle \phi^{q^n} \rangle} \rightarrow q^n \mathbb{Z}_q$$ such that $\phi^{q^n} \mapsto q^n$. So it's clear that we can carachterize each closed subgroup of $\mathbb{Z}_q$ as $q^n \mathbb{Z}_q$, with $n \in \mathbb{N} \cup \lbrace \infty \rbrace$ and the convention that $q^{\infty}\mathbb{Z}_q = \lbrace 0 \rbrace$. I would like to apply this particular fact to the initial problem and prove that each closed subgroup of $\hat{\mathbb{Z}}$ is in the form $\prod_{q} q^{n_q} \mathbb{Z}_q$, for a certain $n_q \in \mathbb{N} \cup \lbrace \infty \rbrace$.
Thanks
Suppose we want to find each sub-extension of $\overline{\mathbb{F}}_p/\mathbb{F}_p$, equivalently each closed subgroup of $\hat{\mathbb{Z}}$.
Let $q$ a prime, and denote $L_q := \bigcup_{n \in \mathbb{N}} \mathbb{F}_{p^{q^n}}$. It's easy to verify that $L_q$ is a union of concentric extensions, $\vert L_q : \mathbb{F}_p \vert = \infty = \vert \overline{\mathbb{F}}_p : L_q \vert$ and $\text{Gal}(L_q : \mathbb{F}_p) \cong \mathbb{Z}_q$;
Let $\mathbb{F}_p \subseteq F \subseteq L_q$ an intermediate field and let $q^{n_F} := \text{sup}_{x \in F} \lbrace \vert \mathbb{F}_p(x) : \mathbb{F}_p \vert \rbrace$, with $0 \leq n_F \leq \infty$. Claim: $F = \mathbb{F}_{p^{q^{n_F}}}$.
[$\subseteq$] If $n_F = \infty$ then the inclusion is obvious. Suppose $n_F < \infty$; let $x \in F$, so $\vert \mathbb{F}_p(x) : \mathbb{F}_p \vert \leq q^{n_F}$, but for the costruction of $L_q$ we have $\vert \mathbb{F}_p(x) : \mathbb{F}_p \vert \; | \; q^{n_F}$, and from the well-know theory about finite field we immediately obtain $\mathbb{F}_p(x) \subseteq \mathbb{F}_{p^{q^{n_F}}}$.
[$\supseteq$] If $n_F = \infty$ there exist elements of arbitrary big degree over $\mathbb{F}_p$, but for each $x \in \mathbb{F}_{p^{q^n}}$ the dimension $\vert \mathbb{F}_p(x) : \mathbb{F}_p \vert$ is finite; again, from the construction of $L_q$ we have $\mathbb{F}_{p^{q^n}} \subseteq F$. Suppose $n_F < \infty$, then there exists $x \in F$ such that $\vert \mathbb{F}_p(x) : \mathbb{F}_p \vert = q^{n_F}$ and so $\mathbb{F}_{p^{q^{n_F}}} = \mathbb{F}_p(x)$
We can now use the Foundamental theorem for infinite galois extensions and carachterize each closed subgroup of $\mathbb{Z}_q$ as $\text{Gal}(L_q / \mathbb{F}_{p^{q^n}})$, with $n \in \mathbb{N} \cup \lbrace \infty \rbrace$.
Claim: each closed subgroup of $\mathbb{Z}_q$ is in the form $q^n\mathbb{Z}_q$.
Let $\phi$ the Frobenius of $L_q / \mathbb{F}_p$. We have $\mathbb{Z}_q \cong \text{Gal}(L_q / \mathbb{F}_p) \cong \overline{\langle \phi \rangle}$. For the same reason $\text{Gal}(L_q / \mathbb{F}_{p^{q^m}}) \cong \overline{\langle \phi^{{q^m}} \rangle}$, but if we consider the isomorphism $$\varphi: \overline{\langle \phi \rangle} \rightarrow \mathbb{Z}_q$$ such that $\phi \mapsto 1$, which bind $\overline{\langle \phi \rangle}$ and $\text{Gal}(L_q / \mathbb{F}_p)$, and denote $1 := \lbrace \big[ 1 \big]_{q^n} \rbrace_{n \in \mathbb{N}}$, we have that the restriction $$\varphi_{|_{\overline{\langle \phi^{{q^m}} \rangle}}} : {\overline{\langle \phi^{q^m} \rangle}} \rightarrow {q^m\mathbb{Z}_q}$$ that is the map such that $\phi^{q^m} \mapsto q^m$, define an isomorphism between $\overline{\langle \phi^{{q^m}} \rangle}$ and $q^m\mathbb{Z}_q$. Again we have denoted $q^m := \lbrace \big[ q^m \big]_{q^n} \rbrace_{n \in \mathbb{N}}$.
In conclusion each closed subgroup of $\mathbb{Z}_q$ can be seen as $q^m\mathbb{Z}_q$, for some $m \in \mathbb{N} \cup \lbrace \infty \rbrace$.
Consider now the general case, that is a generic intermediate field of $K$ of$\overline{\mathbb{F}}_p / \mathbb{F}_p$. Leq $q$ a generic prime and denote $K^{(q)} := K \cap L_q$. We'll show that $$K = \prod_{q}K^{(q)}$$ where $\prod_{q}K^{(q)}$ is the smallest subfield of $\overline{\mathbb{F}_p}$ that contains each $K^{(q)}$.
[$\subseteq$] Let $x \in K$. we say that $\vert \mathbb{F}_p(x) : \mathbb{F}_p \vert = m$, with $m = q_1^{e_1} \cdot \cdots \cdot q_r^{e_r}$; so, for the cinese remainder theorem we obtain that $x \in \mathbb{F}_{p^m} = \prod_{i}\mathbb{F}_{p^{q_i^{e_i}}} \subseteq \prod_{i}K^{(q_i)}$.
[$\supseteq$] obviously because each $K^{(q)}$ lives in $K$.
For what we have said is clear that $K^{(q)} = \mathbb{F}_{p^{q^{n_q}}}$, for some $n_q \in \mathbb{N} \cup \lbrace \infty \rbrace$, so $$K = \prod_{q}\mathbb{F}_{p^{q^{n_q}}}$$ For each $q$ consider the morphism $\theta: \text{Gal}(\overline{\mathbb{F}_p}/K) \rightarrow \text{Gal}(L_q / K^{(q)})$ such that $\sigma \mapsto \sigma_{|_{L_q}}$ . note that the choice of the codomain is correct, since if $\sigma \in \text{Gal}(\overline{\mathbb{F}_p}/K) \subseteq \text{Gal}(\overline{\mathbb{F}_p}/ \mathbb{F})$, so we have $\sigma_{|_{L_q}} \in \text{Gal}(L_q/ \mathbb{F})$ and it is an automoprhism of $L_q$ which clearly fixes $K^{(q)}$. All these morphisms generate a map $$\theta: \text{Gal}(\overline{\mathbb{F}_p}/K) \rightarrow \prod_{q}\text{Gal}(L_q / K^{(q)})$$ such tath $\sigma_{|_{L_q}} \mapsto \sigma$
But the extensions of $ K^{(q)} = \mathbb{F}_{p^{q^{n_q}}}$ have trivial intersection, i.e $\mathbb{F}_p$, and so it's easy to verify that $\theta$ is a bijective morphism, that is $$\text{Gal}(\overline{\mathbb{F}_p} / K) \cong \prod_{q}\text{Gal}(L_q / K^{(q)}) \cong \prod_{q}q^{n_q}\mathbb{Z}_q$$ Finally, all closed subgroups of $\hat{\mathbb{Z}}$ are in the form $ \prod_{q}q^{n_q}\mathbb{Z}_q$, with $n_q \in \mathbb{N} \cup \lbrace \infty \rbrace$.