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I´m studying algebraic number theory from Neukirch´s book. I´m reading the Proposition 10.2 which says:

A $\mathbb{Z}$-basis of the ring $O$ of integers of $\mathbb{Q}(\zeta)$ is given by $1, \zeta, \zeta^2,..., \zeta^{d-1}$, with $d=\varphi(n)$, in other words, $O=\mathbb{Z}[\zeta]$.

In one step of the proof, the author mentions, if $n=l_1^{\nu_1}...l_r^{\nu_r}$ and $\zeta_i=\zeta^{n/l_i^{\nu_i}}$ is a primitive $l_i^{\nu_i}$-th root of unity, one has

$\mathbb{Q}(\zeta)=\mathbb{Q}(\zeta_1)...\mathbb{Q}(\zeta_r)$, with $\mathbb{Q}(\zeta_1)...\mathbb{Q}(\zeta_{i-1}) \cap \mathbb{Q}(\zeta_i)=\mathbb{Q}$.

This is the step I cannot see. Could anyone explain me why these equalities holds? I''ll really appreciate this help!

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  • $\begingroup$ I think one part of the argument is that you should write $1$ as a linear combination with integer coefficients of the $n/l_i^{\nu_i}$. $\endgroup$
    – Malkoun
    Oct 4, 2021 at 22:25

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For cyclotomic fields in general we have $\mathbb{Q}_a\mathbb{Q}_b=\mathbb{Q}_{\mathrm{lcm}(a,b)}$ and $\mathbb{Q}_a\cap\mathbb{Q}_b=\mathbb{Q}_{\gcd(a,b)}$. The first statement is fairly elementary, while the second is explained here for instance. For more than two fields, you use induction.

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