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The isomorphism Q<g>

by Somewheresafe
Tags: isomorphism, q<g>
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Somewheresafe
#1
Sep26-11, 07:02 AM
P: 4
HI everyone! Sorry to be bothering you again with another question. >.<

Anyway I think it's pretty well-known that

[itex]\mathbb{Q}\left\langle q\right\rangle\simeq\oplus_{d|n}\mathbb{Q}\left( \zeta_d \right)[/itex]

where n is the order of g (say in a group) and d the divisors of n.

I was kinda wondering if the same goes for

[itex]\mathbb{Z}\left\langle q\right\rangle\simeq\oplus_{d|n}\mathbb{Z}\left[\zeta_d \right][/itex]

What I do know is that, for the first isomorphism, the isomorphism was shown by using a lot of isomorphisms (first is regarding group rings over cyclic groups, second by the CRT, and lastly by Kronecker's Theorem, something like that). Can the second statement not be established by using these three isomorphisms? (I think it may fail for the third; ie, Kronecker's, since [itex]\mathbb{Z}[/itex] is not a field, but I'm not quite sure if there's a version of that theorem for rings which are not fields).

Thanks! :D
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