TITLE: Minimal polynomial of a twelth root of unity
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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 $\Phi_{12}(x)=x^4-x^2+1$.