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I'm trying to prove that $\operatorname{Gal}(\mathbb{Q}(\cup_{n \geq 1}\mu_n)/\mathbb{Q}) \cong \widehat{\mathbb{Z}}^\times = (\varprojlim (\mathbb{Z}/n\mathbb{Z}))^\times$, where $\varprojlim$ denotes the inverse limit as n increases (with $\mathbb{Z}/n\mathbb{Z}$ partially ordered by divisibility) and $\mu_n$ denotes the set of $n^{th}$ roots of unity.

I have managed to prove that $\widehat{\mathbb{Z}} \cong \prod_{p} \mathbb{Z}_p$, where $\mathbb{Z}_p$ denotes the p-adic integers (equivalently, $\mathbb{Z}_p = \varprojlim_{\,n} \mathbb{Z}/p^n \mathbb{Z} $ ), and the product runs over all primes. However, while I suspect this might be the easiest way to figure out the isomorphism between the Galois group and $\widehat{\mathbb{Z}}^\times$, I can't quite seem to formalise my argument.

Obviously any element of the Galois group must fix $\mathbb{Q}$ and send $n^{th}$ roots of unity to $n^{th}$ roots of unity; we can write a general element of $\mathbb{Q}(\cup_{n \geq 1}\mu_n)/\mathbb{Q}$ as a sum of elements of $\mu_1,\,\mu_2,\,\mu_3$ etc., but need to be careful to keep in mind the fact that any element of $\mu_m$ is an element of $\mu_n$ when $m \, | \, n$. I presume there is some clever way to work with just primes and prime powers when dealing with things of the form $e^{2 \pi i k / n} \in \mu_n$ which leads us to the required isomorphism, but I can't spot how to get to the result I want. Cold anyone help me with finishing off my proof please?

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I made the obvious correction in the title and in the body, writing \operatorname{Gal}. It told me the title was too long! So I took "Gal" out of the math environment in the title. – Michael Hardy Jan 30 '12 at 23:58
up vote 4 down vote accepted

The basic point is that for $n\gt 1$ you have a completely canonical morphism of groups $$ (\mathbb{Z}/n\mathbb{Z})^\times \stackrel {\simeq}{\to} Gal(\mathbb{Q}(\mu_n)/\mathbb{Q}) $$ It associates to $[a]\in (\mathbb{Z}/n\mathbb{Z})^\times$ the unique automorphism $\phi_a:\mathbb{Q}(\mu_n) \to \mathbb{Q}(\mu_n)$ satisfying $\phi_a(\zeta)=\zeta ^a$ for all $\zeta\in \mu_n$, primitive or not (this a subtle and important point).

The key remark now is that these isomorphisms are compatible with the projective limit.
In detail, if $n$ divides $N$, so that $N=nd$, and if $Z\in \mu_N$ then $Z^d=\zeta\in \mu_n$ and from $\Phi_a(Z)=Z^a$ you deduce $\Phi_a(\zeta )=\Phi_a(Z^d)=(\Phi_a(Z))^d=(Z^a)^d=(Z^d)^a=\zeta ^a=\phi_a(\zeta)$ so that $$\Phi_a:\mathbb{Q}(\mu_N) \to \mathbb{Q}(\mu_N)$$ with $[a]_N\in (\mathbb{Z}/N\mathbb{Z})^\times$ restricts to $$\phi_a:\mathbb{Q}(\mu_n) \to \mathbb{Q}(\mu_n)$$ with $[a]_n\in (\mathbb{Z}/n\mathbb{Z})^\times$.

So the $[a_n]$ glue to an $\alpha \in (\varprojlim (\mathbb{Z}/n\mathbb{Z}))^\times$ and the $\phi_a$ glue to an automorphism $\phi_\alpha \in Gal(\mathbb{Q}(\mu_\infty)/\mathbb{Q})$, and this yields the required isomorphism
$$ (\varprojlim (\mathbb{Z}/n\mathbb{Z}))^\times \stackrel {\simeq}{\to} Gal(\mathbb{Q}(\mu_\infty)/\mathbb{Q}):\alpha\mapsto \phi_\alpha $$

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Many thanks, that was just what I was trying to formalise. – Tom Jan 31 '12 at 18:52

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