Automorphism of $\mathbb{Q}({\zeta_n})/\mathbb{Q}$

Question

I came across the theorem where, for $n=p^{a_1}\cdots p^{a_m}$:

$Gal(\mathbb{Q}({\zeta_n})/\mathbb{Q})\simeq$ $Gal(\mathbb{Q}({\zeta_{p^{a_1}}})/\mathbb{Q})\times Gal(\mathbb{Q}({\zeta_{p^{a_2}}})/\mathbb{Q)}\times \cdots\times Gal(\mathbb{Q}(\zeta_{p^{a_m}})/\mathbb{Q})$.

Do we have a corresponding property for $Aut(\mathbb{Q}({\zeta_n})/\mathbb{Q})?$

In particular is it true that:

$Aut(\mathbb{Q}({\zeta_n})/\mathbb{Q})\simeq$ $Aut(\mathbb{Q}({\zeta_{p^{a_1}}})/\mathbb{Q})\times Aut(\mathbb{Q}({\zeta_{p^{a_2}}})/\mathbb{Q})\times\cdots \times Aut(\mathbb{Q}({\zeta_{p^{a_m}}})/\mathbb{Q})$

Can we apply a similar trick as something like Chinese remainder theorem and say something in terms of $\mathbb{Z}_n^{\times}?$

Answer 1

It is Chinese Remainder theorem. When we have a co-prime factorization $n=ab$, then we have an isomorphism of rings: $\mathbf{Z}/n\mathbf{Z}\cong \mathbf{Z}/a\mathbf{Z}\times \mathbf{Z}/b\mathbf{Z}$. So correspondingly the group of inverible elements are also isomorphic, which are obtained component wise. What has been given above can be extended to prime power factorization of $n$ which I avoided to minimize my typing work!