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Someone should add: the abelianization map identifies the absolue Galois group G of Q with
Z
^
∗
{\displaystyle {\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) [ reply ]
I have a problem with the relation
Z
^
⊂
∏
n
Z
/
n
Z
,
{\displaystyle {\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
Z
^
=
∏
p
Z
p
,
{\displaystyle {\widehat {\mathbb {Z} }}=\prod _{p}\mathbb {Z} _{p},}
because the
Z
p
{\displaystyle \mathbb {Z} _{p}}
are (as complete sets) already uncountable, and
Z
^
=
lim
←
Z
/
n
Z
{\displaystyle {\widehat {\mathbb {Z} }}=\varprojlim \mathbb {Z} /n\mathbb {Z} }
is, of course, OK . –Nomen4Omen (talk ) 07:04, 30 May 2021 (UTC) [ reply ]
Solved! The right side is an infinite direct product . –Nomen4Omen (talk ) 19:19, 30 May 2021 (UTC) [ reply ]
I think this statement in the last section is wrong:
A
Q
×
/
Q
×
≅
(
R
×
Z
^
)
/
Z
=
lim
←
(
R
/
m
Z
)
=
lim
x
↦
x
m
S
1
=
Z
^
{\displaystyle {\begin{aligned}\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{aligned}}}
The abelian Galois group of
Q
{\displaystyle \mathbb {Q} }
is
Z
^
×
{\displaystyle {\hat {\mathbb {Z} }}^{\times }}
and not
Z
^
{\displaystyle {\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) [ reply ]