TITLE: Is $\mathbb Q(\zeta_6)=\mathbb {Q}(\zeta_3)$?
QUESTION [7 upvotes]: I got myself confused over the following: 
We have $$\mathbb Q(\zeta_3)=\mathbb Q(\exp(2\pi i/3))=\mathbb Q\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)=\mathbb Q\left(-\frac{1}{2}+\frac{i\sqrt 3}{2}\right)=\mathbb Q(i\sqrt 3),$$ but also $$\mathbb Q(\zeta_6)=\mathbb Q(\exp(2\pi i/6))=\mathbb Q\left(\cos\frac{2\pi}{6}+i\sin\frac{2\pi}{6}\right)=\mathbb Q\left(\frac{1}{2}+\frac{i\sqrt 3}{2}\right)=\mathbb Q(i\sqrt 3).$$
So the fields are absolutely identical? $\Phi_6$ splits in $\mathbb Q (\zeta_3 )$ and vice versa?

REPLY [3 votes]: Fix an odd integer $n > 2$.  If $\alpha$ is a primitive $n$th root of unity
then it is easily checked that $-\alpha$ is a primitive $2n$th root of unity.
Also for any algebraic number $\beta$ the number field generated by $\beta$ and  $-\beta$ are one and the same.
Now answer for your question can be easily deduced from the above statements which are themselves easy to verify.