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For $n=1,2,3,4$ I found it to be $1,1,2,2$ respectively. So I hypothesize that $\dim_{\Bbb Q}\Bbb Q[\zeta_{2n}]=n$ and $\dim_{\Bbb Q}\Bbb Q[\zeta_{2n+1}]=n+1$. But how to prove it? When we have $2n$, the roots lie on the vertices of a regular $2n-$gon centred at $0$. The roots occur in pairs; placed on the extremities of a diagonal through $0$. So the dimension is no more than $n$, since half of them are simply scalar multiples of the other half.

But this is all I got. Any help is appreciated.

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  • $\begingroup$ "The $n$th root of unity" does not make sense. $1$ is a $n$th root of unity for any $n$. You probably meant "a primitive $n$th root of unity". $\endgroup$ – fkraiem Aug 16 '18 at 5:09
  • $\begingroup$ Try $n=5$...... $\endgroup$ – lhf Aug 16 '18 at 10:26
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When $n=6$, the sixth roots of unity are the cube roots of unity and their negatives, so that $\Bbb Z(\zeta_6)=\Bbb Z(\zeta_3)$ which has degree $2$ over $\Bbb Q$. That rather puts the kibosh on your conjecture.

The minimum polynomial of $\zeta_n$ is called the $n$-th cyclotomic polynomial.

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