0

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.

  • 2
    There are many different proofs that the cyclotomic polynomials are irreducible, but none of them is easy https://www.lehigh.edu/~shw2/c-poly/several_proofs.pdf http://www-users.math.umn.edu/~garrett/m/algebra/notes/20.pdf – reuns Dec 19 '18 at 13:09
  • A duplicate of this. Not voting yet, because it is likely that a better target exists. – Jyrki Lahtonen Dec 19 '18 at 13:23

1 Answers1

-1

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.

A. Bailleul
  • 363