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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.

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  • $\begingroup$ @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. $\endgroup$ – Hrit Roy Aug 16 '18 at 12:05
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Aug 16 '18 at 12:09
  • $\begingroup$ @DietrichBurde could you tell me how cyclotomic polynomials help us prove it? $\endgroup$ – Hrit Roy Aug 16 '18 at 15:40
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Aug 16 '18 at 18:15
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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 $\mathbb Q [\zeta_n]$.

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  • $\begingroup$ @ArnaudD. Yeah, i found that. Thanks. $\endgroup$ – xbh Aug 16 '18 at 11:37

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