We know that the $m$th cyclotomic number field is given by $\mathbb{Q}(\xi_m)$, where $\xi_m$ is an $m$th primitive root of unity. We know also that if $\omega_m$ is another primitive root of unity, then $\mathbb{Q}(\xi_m)$ and $\mathbb{Q}(\omega_m)$ are isomorphic, since they have the same minimal polynomial $\Phi_m(X)$. My question is:

Can we say that $\mathbb{Q}(\xi_m)=\mathbb{Q}(\omega_m)$ ? (so that the definition of $m$'th cyclotomic number field is well defined)?

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    $\begingroup$ The general fact at play here is the following. If $L$ and $L'$ are 1) $\Bbb{Q}$-isomorphic, 2) both finite Galois extensions of $\Bbb{Q}$, and 3) subfields of $\Bbb{C}$, then we can conclude that $L=L'$. There are variants and generalizations of this result. As seen from KCd's answer you absolutely need to have a common "umbrella" field containing all the others. Undoubtedly you knew about the need for them to be Galois (easy to find plenty of counterexamples otherwise). Having them inside a common algebraically closed field is not enough: $\Bbb{F}_p(x)\simeq \Bbb{F}_p(x^2)$ etc. $\endgroup$ Jul 11, 2018 at 11:29

2 Answers 2


There is a subtlety here: are you trying to compare two subfields of a common field or two fields that are built separately?

If you have two primitive $m$th roots of unity in the same field of characteristic $0$ then they generate the same cyclotomic fields over $\mathbf Q$. But if your $\xi_m$ and $\omega_m$ are not in the same field then the fields they generate over $\mathbf Q$ are isomorphic but can't be equal.

For example, $\mathbf Q(i)$ inside $\mathbf C$ and $\mathbf Q[x]/(x^2+1)$ are isomorphic but definitely are not equal.


Yes, you can. Fix $n$ and positive integers $a,b$ relatively prime to $n$. Then if $\zeta_n = e^{2\pi i \frac{1}{n}}$, $\zeta_n^a$ and $\zeta_n^b$ are primitive $n^{th}$ roots of unity. We can solve the congruence $ax \equiv b \pmod{n}$. Then $(\zeta_n^a)^x = \zeta_n^{ax} = \zeta_n^b$, which means that $\zeta_n^b \in \mathbb{Q}(\zeta_n^a)$ and thus $\mathbb{Q}(\zeta_n^b) \subseteq \mathbb{Q}(\zeta_n^a)$. Of course, symmetry gives $\mathbb{Q}(\zeta_n^a) \subseteq \mathbb{Q}(\zeta_n^b)$.

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    $\begingroup$ Not all finite extensions of $\mathbf Q$ are in the complex numbers, so you are not always able to write $\zeta_n$ as $e^{2\pi i/n}$. For example, there are primitive 7th roots of unity in the 29-adic numbers $\mathbf Q_{29}$ but it is not natural to say any of them is $e^{2\pi i/7}$. $\endgroup$
    – KCd
    Jul 11, 2018 at 13:34

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