Minimal polynomial of a twelth root of unity I am attempting to find the minimal polynomial for $\omega=\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}+i\frac{1}{2}$ over $\mathbb{Q}$.  I'm doing this in the context of cyclotomic field extensions.  The book gives $x^2-x+1$, the 6th cyclotomic polynomial, as an answer.  Is this correct?  When I do the calculation, it appears that $\omega$ isn't even a root.
I believe the answer is the 12th cyclotomic polynomial, $f(x)=x^4-x^2+1$.  We have $f(\omega)=0$, it is monic, and by the mod $p$ irreducibility test (or the irreducibility test of your choice), it is irreducible over $\mathbb{Q}$.  It also seems nicely intuitive that the minimal polynomial of a primitive $n$th root of unity would be the $n$th cyclotomic polynomial, since all the cyclotomic polynomials (I believe) are irreducible over $\mathbb{Q}$.  So is this the minimal polynomial, or am I missing something?
 A: Here are three ways to prove that the degree of $\zeta_{12}$ over $\Bbb{Q}$ is not 2. Two of them will actually prove to you that the degree of this over the rationals is actually 4.
$\textbf{Approach 1:}$ Let's do this in a very simple way and try to determine the degree of $\zeta_{12}$ over $\Bbb{Q}$. Notice that 
$$\Bbb{Q}(\zeta_{12}) \cong \Bbb{Q}(\sqrt{3},i) \cong \left(\Bbb{Q}(\sqrt{3})\right)(i). $$
Now the degree of $i$ over $\Bbb{Q}$ is 2 because the minimal polynomial of $i$ over $\Bbb{Q}$ is $x^2 +1$. Now we show that the degree of $\sqrt{3}$ over $\Bbb{Q}(i)$ is two as well. We already have a polynomial with coefficients in $\Bbb{Q}(i)$ with $\sqrt{3}$ as a root, namely the polynomial $x^2 - 3$. Now suppose that this polynomial is reducible, meaning that $\sqrt{3} \in \Bbb{Q}(i)$. Then we get that since as a vector space over $\Bbb{Q}$, $\Bbb{Q}(i)$ is two dimensional we can write 
$$\sqrt{3} = a + bi$$
for some $a,b \in \Bbb{Q}$. Then squaring both sides we get that $3 = a^2 - b^2 + 2abi$ and hence that $$\frac{3 - a^2 + b^2}{2ab} = i$$
in particular that $i$ is real. This is a contradiction so that the polynomial $x^2 - 3$ is irreducible, and hence that $[\Bbb{Q}(\sqrt{3},i) : \Bbb{Q}(i)] = 2$. It follows by the dimension counting formula that $[\Bbb{Q}(\sqrt{3},i):\Bbb{Q}] = 2 \times 2 = 4$ and hence the degree of the minimal polynomial of $\zeta_{12}$ over $\Bbb{Q}$ must be 4 and you are right. Therefore if you have found a monic degree polynomial of degree 4 with $\zeta_{12}$ as a root, it must be the case that that polynomial is irreducible otherwise this would contradict our result that $[\Bbb{Q}(\zeta_{12}):\Bbb{Q}] = 4$. 
$\textbf{Add-on to approach 1:}$ Suppose you did not know about cyclotomic polynomials (like me) and wanted to compute the minimal polynomial of $\zeta_12$ over $\Bbb{Q}$. You already know that it must be of degree 4 by the argument above. Once you find a monic one of degree 4, it must be unique by uniqueness of the minimal polynomial (exercise). Write
$$x = \frac{\sqrt{3} + i}{2}.$$
Then $2x = \sqrt{3} + i$ so that squaring both sides gives that $4x^2 = 3 - 1 + 2\sqrt{i}$ which implies that $4x^2 - 2 = 2\sqrt{3}i$. Hence 
$$\begin{eqnarray*} 2x^2 - 1 &=& \sqrt{3}i\\
\implies 4x^4 - 4x^2 + 1 &=& -3 \\
\implies 4x^4 - 4x^2 + 4 &=& 0 \\
\implies x^4 - x^2 +1 &=& 0 \\
\end{eqnarray*}$$
and voilà! This is exactly the cyclotomic polynomial of degree 4 that you found.
$\textbf{Approach 2: Galois Theory}$ Suppose that $[\Bbb{Q}(\zeta_{12}) : \Bbb{Q}] = 2$. Then this is a degree 2 extension of a field of characteristic zero and hence is a Galois Extension. It follows that the Galois Group $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q}) \cong C_2$, the cyclic group of order 2.
Now take some $\sigma \in G$. Then $\sigma$ must induce a permutation on the roots of the minimal polynomial of $\zeta_{12}$ over $\Bbb{Q}$. But then since $\sigma(\sqrt{3})^2 = \sigma(3) = 3$ which implies that $\sigma$ fixes $\sqrt{3}$ (similarly it fixes $i$) it follows that
$$\sigma(\zeta_{12}) = \zeta_{12}.$$
But then since $\sigma \in  \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ was arbitrary this means that every  $ \sigma \in  \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ is the identity permutation so that $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ is the trivial group. But then this contradicts the fact that $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q}) \cong C_2$ so it is not possible for $[\Bbb{Q}(\zeta_{12}):\Bbb{Q}]$ to be equal to 2. 
$\textbf{Edit:}$ The proof above on Galois Theory is incorrect. Namely because it is not true that $\sigma(\sqrt{3}) = \sqrt{3}$ alone, but rather $\sigma(\sqrt{3}) = \pm \sqrt{3}$ and similarly for $\sigma(i)$.
$\textbf{Approach 3:}$ There is this result that is useful: 
$$[\Bbb{Q}(\zeta_n) : \Bbb{Q}] = \varphi(n)$$
where $\varphi(n)$ is the Euler totient function. It counts the number of integers $k$ such that $1 \leq k \leq n$ with $k$ relatively prime to $n$. In your case, the integers relatively prime to 12 are $1,5,7,11$ which is $4$, exactly what you need.
A: If you know that $\prod_{d\mid n}\Phi_d(X) = X^n - 1$ (which is simple to prove), then it is easy to find $\Phi_{12}$, which has degree $\phi(12)=4$. One difficulty is proving that $\Phi_n$ is indeed irreducible,  a non-trivial result due to Gauss for general $n$, but it should be easy for $\Phi_{12}$.
Wikipedia lists $\Phi_{12}(x)=x^4-x^2+1$.
