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Denote $s_n$ be the $n^{th}$ root of unity. Let $K=\Bbb Q(s_1, s_2, s_3,\dots)$.

Then how to prove that $\operatorname{Aut}(K/\Bbb Q)$ is abelian?

Arnaud D.
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suliman
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  • $K = \mathbb{Q}(\zeta_m)$ for some $m$ – reuns Dec 28 '16 at 12:24
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    I don't think so, because K contains s_17, s_19, s_23, ... – suliman Dec 28 '16 at 12:30
  • $K$ is actually the largest abelian extension of $\mathbb Q$. That's the Kronecker–Weber theorem. – lhf Dec 28 '16 at 12:50
  • The question is discussed here. Take $k_0=\mathbb{Q}$. – Dietrich Burde Dec 28 '16 at 12:50
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    The answer is probably that the inverse limit of abelian groups is abelian. – lhf Dec 28 '16 at 12:51
  • I didn't get it contained all the roots of unity. Then yes the field is the "roots of unity" closure of $\mathbb{Q}$, and its Galois group is the inverse limit of those cyclic Galois groups, so it is locally cyclic and abelian. – reuns Dec 28 '16 at 12:58
  • Do you have any elementary proof without using inverse limit?... – suliman Dec 28 '16 at 13:24
  • Related: https://math.stackexchange.com/questions/142236, https://math.stackexchange.com/questions/191897/, https://math.stackexchange.com/questions/148660, https://math.stackexchange.com/questions/104004, https://math.stackexchange.com/questions/1265310, https://math.stackexchange.com/questions/884256/ – Watson Dec 28 '16 at 13:36
  • The compositum of any family of abelian extensions is abelian. That’s not hard to show as a direct application of the definitions. No inverse limits need apply.. – Lubin Dec 29 '16 at 22:02

1 Answers1

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That's really the definition. Your field is $K = \lim_{n \to \infty} \mathbb{Q}(\zeta_{n!})$ the limit of a tower of fields extensions $\mathbb{Q}(\zeta_{(n+1)!})/\mathbb{Q}(\zeta_{n!})$

  • if $\sigma,\sigma_2 \in Gal(K/\mathbb{Q})$ then $\sigma(\zeta_{n!}) = \zeta_{n!}^{k(n)},\sigma_2(\zeta_{n!}) = \zeta_{n!}^{l(n)}$ where $gcd(n!,k(n))=gcd(n!,l(n))=1$ and so $$\sigma \circ \sigma_2(\zeta_{n!}) = \sigma(\zeta_{n!}^{l(n)})=\zeta_{n!}^{l(n)k(n)}=\sigma_2 \circ \sigma(\zeta_{n!})$$

  • any element of $K$ can be written as $\sum_{m=0}^{n!} c_m \zeta_{n!}^m$ for some $n \in \mathbb{N}, c_m\in \mathbb{Q}$

  • by definition of the Galois group $\sigma \circ \sigma_2$ and $\sigma_2 \circ \sigma$ are elements of $Gal(K/\mathbb{Q})$

i.e. $$\sigma \circ \sigma_2(\sum_{m=0}^{n!} c_m \zeta_{n!}^m) = \sum_{m=0}^{n!} c_m\sigma \circ \sigma_2(\zeta_{n!}^m)=\sum_{m=0}^{n!}c_m\sigma_2 \circ \sigma(\zeta_{n!}^m)=\sigma_2 \circ \sigma(\sum_{m=0}^{ n!} c_m \zeta_{n!}^m)$$ $\implies$ $Gal(K/\mathbb{Q})$ is abelian.

reuns
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