Prove that $Gal(K/k) \cong \hat{\mathbb{Z}}$

Question

Let $K$ be the algebraic closure of a finite field $k$. Prove that $Gal(K/k) \cong \hat{\mathbb{Z}}$.

From the definition in the book, here is how $\hat{\mathbb{Z}}$ is defined: Let $D = Cr(\mathbb{Z}_{p} | \; p \; prime)$, let $\delta: \mathbb{Z} \rightarrow D$ be the map taking $x \in \mathbb{Z}$ to the vector with all coordinates equal to $x$. Then the group $D$ together with the map $\delta$ is the profinite completion of $\mathbb{Z}$, denoted $\hat{\mathbb{Z}}$.

There seem to be many sources online that cite this result as true, but I'm having trouble finding anywhere that shows a proof. This question is from Profinite Groups (Wilson), so I doubt that the solution is all that straight-forward. Could anyone offer me a solution or perhaps some insight on how to tackle this problem?

Answer 1

A detailed proof is, for example, given in James S. Milne's lecture notes on Fields and Galois Theory, EXAMPLE 7.16., page $97$. Ingredients are the canonical Frobenius element $\sigma:a\mapsto a^p$, the profinite completion of $\mathbb{Z}$, and the isomorphism $\widehat{\mathbb{Z}}\rightarrow Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)$.

Answer 2

For any field $k$ and a fixed algebraic closure $K$ of $k$, practically by definition, $K$ is the inductive (=direct) limit of its subextensions $L/k$ of finite degree. By (infinite) Galois theory, $Gal(K/k)$ is then the projective (=inverse) limit of its subgroups of finite index. If $k$ is a finite field, for any integer $n$, we know that $K$ admits a unique subextension $L/k$ of degree $n$, and this extension is cyclic. Hence $Gal(K/k)$ is the projective limit of the $(Z/nZ, +)$, i.e. $(\hat Z, +)$ .