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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?

amWhy
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Buh
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3 Answers3

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Yes, because $\Phi_6$ and $\Phi_3$ are actually $x^2-x+1$ and $x^2+x+1$ respectively. So $\Bbb Q(\zeta_6)$ and $\Bbb Q(\zeta_3)$ have both degree $2$ over $\Bbb Q$ and, since one obviosly contains the other, they are the same extension.

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    Alternatively, $\Phi_6(X)=\Phi_3(-X)$. Indeed, whenever $n>1$ is odd, then $\Phi_{2n}(X)=\Phi_n(-X)$. – Lubin Jul 02 '18 at 16:01
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Hint: Note that $\zeta_6=\zeta_3+1$.

Hagen von Eitzen
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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.

P Vanchinathan
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