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Let $\mathbb{Q}$($\zeta_n$) be some cyclotomic field, where $\zeta_n$ is a n-th root of unity.

I already managed to show that $\mathbb{Q}$($\zeta_n$) is an Galois extension, but now i struggle to show that $[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n)$, where $\varphi(n)$ is Euler's totient function.

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marked as duplicate by Dietrich Burde abstract-algebra Dec 19 '18 at 16:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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For any $\sigma \in Gal(\mathbb Q(\zeta_n)/\mathbb Q)),$ $\sigma(\zeta_n)$ must be a primitive $n$-th root of unity, because $\sigma$ is a $\mathbb Q$-automorphism. Since $\zeta_n$ generates $\mathbb Q(\zeta_n)/\mathbb Q$, it follows that this group has as many elements as the set of primitive $n$-th roots of unity, that is $\varphi(n)$, where $\varphi$ denotes Euler's totient function.

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    $\begingroup$ I have no idea why this is marked as correct. It's assuming without any comment that the cyclotomic polynomial is irreducible. $\endgroup$ – ancientmathematician Dec 19 '18 at 13:37