Talk:Profinite integer

Latest comment: 3 years ago by 93.132.116.137 in topic Abelian Galois Group of Q {\displaystyle \mathbb {Q} }

Someone should add: the abelianization map identifies the absolue Galois group G of Q with (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

Product formula

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I have a problem with the relation

 

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

 

because the   are (as complete sets) already uncountable, and   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

Abelian Galois Group of

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I think this statement in the last section is wrong:

 

The abelian Galois group of   is   and not  . 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