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

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.

• @DietrichBurde as I said; we don't know what cyclotomic polynomials are. Our professor asked us this knowing that fact. There must be some other way to prove it. – Hrit Roy Aug 16 '18 at 12:05
• I see. Still I think it's worth to look up what a cyclotomic polynomial is, and you asked anyway how we can get the result using that polynomial. – Dietrich Burde Aug 16 '18 at 12:09
• @DietrichBurde could you tell me how cyclotomic polynomials help us prove it? – Hrit Roy Aug 16 '18 at 15:40
• Hrit, the $\Bbb{Q}$-dimension of $\Bbb{Q}[\zeta_n]$ is by definition the degree of the minimal polynomial $\Phi_n(x)$ of $\zeta_n$, which obviously is $\phi(n)$; see the duplicate. – Dietrich Burde Aug 16 '18 at 18:15

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 $\mathbb Q [\zeta_n]$.