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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?

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    What is your definition of $\hat{\mathbb{Z}}$? $Gal(K/k)$ is given by $\lim_{n \in \mathbb{N}}(\mathbb{Z}/n\mathbb{Z})$, which is the definition of $\hat{\mathbb{Z}}$ according to some. – 2017-02-16
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    $\hat{\mathbb{Z}}$ is the profinite completion of the integers. It might be worth mentioning that this chapter of the book is where the link between galois groups and profinite groups is first introduced. – 2017-02-16
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    More precisely: The group $D = Cr(\mathbb{Z}_{p} |\; p \; prime)$ together with the map $\delta: \mathbb{Z} \rightarrow D$ is the profinite completion of $\mathbb{Z}$. The map $\delta$ is the map taking $x \in \mathbb{Z}$ to the vector with all coordinates equal to $x$. – 2017-02-16
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    Sure, I know what $\hat{\mathbb{Z}}$ denotes. My point is that the profinite completion of a group $G$ is defined by some to be the inverse limit of $G/N$ over all normal subgroups $N$ of finite index, which is exactly what's written above. Does this agree with your definition? – 2017-02-16
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    Hmm, I'm not sure what's meant by $Cr(\mathbb{Z}_{p} \mid p \text{ prime})$. Is this supposed to be the infinite product $\prod_{p \text{ prime}} \mathbb{Z}_{p}$ of all of the $p$-adic integers? Regardless, I think it'd be helpful if you updated your question with explicit definitions of the objects you are using. – 2017-02-16
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    It's the cartesian product of all $\mathbb{Z}_{p}$ such that $p$ is prime. You're probably right, I'll edit it now. – 2017-02-16

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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 $\widehat{\mathbb{Z}}\rightarrow Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)$.

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    Is it correct to start with $\rho = \lim_{n \to \infty} \sigma^{k_n }$ where $k_{n+1} \equiv k_n \bmod n!$ ? – 2017-02-16
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    Much appreciated, thank you! – 2017-02-16
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    Milne’s given name is James. – 2017-02-17
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    @Lubin I apologize! – 2017-02-17
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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. $(\hat Z, +)$ .