Talk:Profinite integer

Someone should add: the abelianization map identifies the absolue Galois group G of Q with $$\widehat{\mathbb{Z}}^*$$ (class field theory?) Put in another way the non-abelian-ness of profinite integers are hidden in the commutator subgroup of G (this stuff is beyond me). -- Taku (talk) 02:55, 27 April 2015 (UTC)

Product formula
I have a problem with the relation
 * $$\hat{\mathbb{Z}} \subset \prod_n \mathbb{Z}/n\mathbb{Z} ,$$

because on the left side there is an uncountable set and on the right side there is a countable product of finite sets.

The mentioned problem does not exist with
 * $$\widehat{\mathbb{Z}} = \prod_p \mathbb{Z}_p ,$$

because the $$\mathbb{Z}_p$$ are (as complete sets) already uncountable, and $$\widehat{\mathbb{Z}} = \varprojlim \mathbb{Z}/n\mathbb{Z} $$ is, of course, OK. –Nomen4Omen (talk) 07:04, 30 May 2021 (UTC)


 * Solved! The right side is an infinite direct product. –Nomen4Omen (talk) 19:19, 30 May 2021 (UTC)

Abelian Galois Group of $$\mathbb{Q}$$
I think this statement in the last section is wrong:

$$\begin{align} \mathbb{A}_\mathbb{Q}^\times/\mathbb{Q}^\times &\cong (\mathbb{R}\times \hat{\mathbb{Z}})/\mathbb{Z} \\ &= \underset{\leftarrow}{\lim} \mathbb({\mathbb{R}}/m\mathbb{Z}) \\ &= \underset{x \mapsto x^m}{\lim} S^1 \\ &= \hat{\mathbb{Z}} \end{align}$$

The abelian Galois group of $$\mathbb{Q}$$ is $$\hat\mathbb{Z}^\times$$ and not $$\hat\mathbb{Z}$$. Furthermore, this is an isomorphism of abstract groups, but not an isomorphism of topological groups.

93.132.116.137 (talk) 20:13, 2 July 2021 (UTC)