If $\zeta_n$ is a primitive $n$th root of unity, why is $\text{dim}_{\Bbb Q}\Bbb Q[\zeta_n]=\phi(n)$?

Question

I have no idea what cyclotomic polynomials are and how we can get the result using that. Is there another way to prove it? Any hint is appreciated.

Answer 1

If $\zeta_n$ is primitive, then $\zeta_n^j\neq \zeta_n^k$ whenever $n \nmid (k-j)$. Knowing this you could write $\mathbb Q [\zeta_n]$ concretely. The major point is $(\zeta_n^j)_{\gcd(j,n)=1}$ is a $\mathbb Q$-basis of $\mathbb Q [\zeta_n]$.