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I'm hoping I can get some assistance with a revision problem and also a notational issue I'm not sure about (although it may not be standard). I seem to remember going over this or something similar somewhere online a few months ago, but I can't find any such link for the life of me I'm afraid. If anyone has seen it feel free to point me in the right direction.

Suppose $A_n \,(n \in \mathbb{N})$ are sets, groups, rings etc., and we have homomorphisms $\pi_n: A_{n+1} \to A_n$. We define the inverse limit of $A_n$ by $\lim \limits_{\longleftarrow}A_n = \{(x_n)_{n \geq 1}: x_n \in A_n,\, \pi_n(x_{n+1} = x_n \forall n\}$. More generally, we can index by some partially ordered set $I$ with transition maps $f_{ij}:A_j \to A_i$ whenever $i \leq j$, such that $f_{ii} = id_{A_{i}}$. We will write $\mathbb{Z}_p$ to denote the p-adic integers (not $\mathbb{Z}/p\mathbb{Z}$).

Now first of all, a few of the problems I'm coming across refer to the object $\widehat{\mathbb{Z}}$, and I'm wondering whether this is standard notation for some sort of completion or other operation on $\mathbb{Z}$: I know that $\widehat{\mathbb{Z}}$ is isomorphic to $\lim \limits_{\longleftarrow} \mathbb{Z}/n\mathbb{Z}$ with a partial ordering by divisibility. I have proved it is also isomorphic to $\prod_p \mathbb{Z_p}$, the product being over all primes $p$, and to a few other things too.

However, I have been proving these isomorphisms by proving an isomorphism to $\prod_p \mathbb{Z}_p$. Is there perhaps some more "canonical" meaning of $\widehat{\mathbb{Z}}$, for which these isomorphisms would make sense? Of course as I say, this may not be standard so if nothing obvious comes to mind then don't worry.

Now onto the main problem, another inverse limit: I wish to prove that

$Gal(\mathbb{Q}(\bigcup \limits_{n \geq 1} \mu_n)/\mathbb{Q}) \cong \widehat {\mathbb{Z}}^\times$,

where $\mu_n$ is the group of $n$-th roots of unity, and we can take $\widehat{\mathbb{Z}}$ to be whichever isomorphism I mentioned above is easiest to work with. Note that this is $\widehat {\mathbb{Z}}^\times$ however, rather than $\widehat {\mathbb{Z}}$ as we had in the earlier isomorphisms (this is why I am keen to figure out what the hat represents).

I have the result that for $K$ a perfect field and $\bar{K}$ its algebraic closure, $Gal(\bar{K}/K) =\lim \limits_{\longleftarrow} Gal(L/K) $ where the inverse limit is over $L/K$ finite, and partially ordered by inclusion of fields. So, it's very possible that this Galois group just needs to be set up as a clever limit of Galois groups, perhaps each of which is $\mathbb{Z}/n\mathbb{Z}$ (as long as we're sure the LHS is a perfect field), but I can't quite figure out how: it doesn't help that I'm not entirely sure what I want to prove an isomorphism to! Any thoughts or general constructive suggestions you have would be very gratefully received.

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+1 for the effort and just a few comments. $\hat{\mathbb{Z}}$ for me is defined as $\varprojlim \mathbb{Z}/n\mathbb{Z}$, if for you it is just isomorphic to the inverse limit above what is your definition? The second observation is that every field of $char=0$ is perfect, so this is your case. The last point is that the Galois group of the $n$-th cyclotomic extension $\mathbb{Q}(\mu_n)/\mathbb{Q}$ is the group of $n$-th roots of unity and it is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ for any $n$. –  Giovanni De Gaetano May 23 '12 at 9:55
    
Well, $\hat{\mathbb{Z}}$ is a ring, and so it has a group of units $\hat{\mathbb{Z}}^\times$, which happens to be isomorphic to the product of $\mathbb{Z}_p^\times$ over all primes. –  Zhen Lin May 23 '12 at 9:57
    
@Giovanni: Indeed, this is precisely what I was wondering; I can't find a definition for $\widehat{\mathbb{Z}}$ anywhere, so I suppose the statement that it's equal to $\lim \limits_{\longleftarrow} \mathbb{Z}/n\mathbb{Z}$ must have actually just been a poorly worded definition. Regarding your last comment; I guess we're simply taking the limit of $\mathbb{Q}(\mu_n)$ for successively larger $n$ and then equating that to the union over all $n$, which we then say is the algebraic closure? There's no issue with the fact we want the units $\widehat{\mathbb{Z}}^\times$, as Zhen Lin said? –  Spyam May 23 '12 at 10:33
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1 Answer

The main ingredient is the canonical group isomorphism $$f_n:Gal(\mathbb Q(\mu_n)/\mathbb Q)\stackrel {\cong }{\to} (\mathbb Z/n\mathbb Z)^\times : \sigma\mapsto \alpha_n=\bar a_n\quad (\star)$$ where $\alpha_n$ is determined by the condition $\sigma(\zeta)=\zeta^{a_n}$ for all $\zeta\in \mu_n$.

The great thing about this ultra-canonical isomorphism is that it is compatible with divisibility in $\mathbb N.$ Namely if $n=rm$, we can write any $\omega \in \mu_m$ as $\omega=\zeta^r$ with $\zeta\in \mu_n$ and then for $\sigma$ as above we get $$\sigma (\omega)=\sigma(\zeta^r)=\sigma (\zeta)^r=\zeta^{a_nr}=\omega^{a_n}$$
so that $f_m(\sigma \mid \mathbb Q(\mu_m))=\bar a_n\in (\mathbb Z/m\mathbb Z)^\times$ .

We can thus take the projective limit of the isomorphisms $(\star)$ and obtain the required isomorphism $$f:Gal(\mathbb Q(\mu_\infty)/\mathbb Q)\stackrel {\cong }{\to} (\widehat {\mathbb Z})^\times =\lim \limits_{\longleftarrow} \;(\mathbb Z/n\mathbb Z)^\times $$

Note that $\mathbb Q(\mu_\infty)=\mathbb Q^{ab}$ is also the largest abelian extension of $\mathbb Q$, according to a celebrated theorem of Kronecker-Weber.

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