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Let $\alpha = e^{\frac{i\pi}{6}}$. Compute $[\mathbb{Q}(\alpha):\mathbb{Q}]$ and find the minimal polynomial of $\alpha$, $m_{\mathbb{Q}}(\alpha)$.

I can see clearly that $\alpha^6+1=0$ but I have no idea how to see if $p(x)=x^6+1$ is irreducible over $\mathbb{Q}$. For the degree of the extensional field, my guess is that $\mathbb{Q}(\alpha)=\mathbb{Q}(\cos(\frac{i\pi}{6}),i\sin(\frac{i\pi}{6}))$ so $[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}(\cos(\frac{i\pi}{6})][\mathbb{Q}(\cos(\frac{i\pi}{6}):\mathbb{Q}]=4$. But I don't know how to prove this. Any idea?

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    $\begingroup$ Note that $\alpha$ is a primitive root of unity of order $12$, and $\varphi(12)= \varphi(3)\varphi(4) = 2\times 2=4$. $\endgroup$ – Pedro Tamaroff Apr 23 '16 at 0:14
  • $\begingroup$ it is for answering to the problem of the minimal polynomial of $e^{2 i \pi / n}$ that the en.wikipedia.org/wiki/Cyclotomic_polynomial were created $\endgroup$ – reuns Apr 23 '16 at 0:34
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    $\begingroup$ $t^3 + 1 = (t+1)(t^2 - t + 1)$ $\endgroup$ – Will Jagy Apr 23 '16 at 1:08
  • $\begingroup$ @PedroTamaroff: Could you explain what $\varphi$ means here? $\endgroup$ – user112358 Apr 23 '16 at 3:24
  • $\begingroup$ @Lewis It's Euler's totient function. $\endgroup$ – Pedro Tamaroff Apr 23 '16 at 3:47
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For any $n$ the cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\Bbb Q$ (this is a non-trivial fact, and I won't prove it here). This is the minimal polynomial of $e^{2\pi i/n}$.

Fortunately for you, there is a way to calculate this if you know all the "smaller" cyclotomic polynomials:

$\Phi_n(x) = \dfrac{x^n - 1}{\prod\limits_{d|n, d< n}\Phi_d(x)}$

Your problem is $n = 12$, so the $\Phi_d(x)$ you must calculate first are for $d = 1,2,3,4,6$.

For example, $\Phi_1(x) = x - 1, \Phi_2(x) = \dfrac{x^2 - 1}{x - 1} = x+1$

so $\Phi_4(x) = \dfrac{x^4 - 1}{(x - 1)(x + 1)} = x^2 + 1$.

$\Phi_6(x)$ is sort of "tricky", but I have faith in you. $\Phi_3(x)$ should be easy.

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