Finding minimum polynomial of $e^{i\pi/6 }$ Finding minimum polynomial of $e^{i\pi/6 }$:
I know it satisfies $t^6 +1 = 0$. I factorized $t^6+1 = (t^2+1)(t^4-t^2+1)$
obviously it does not satisfy $t^2 +1$ so it must satisfy $t^4-t^2 +1$. How do I show that this is indeed the minimum?
Is it enough to say: in modulo $2$ we have that $t^4 - t^2 + 1 \equiv t^4 +t^2 + 1$. In mod $2$ we only have 2 elements $0,1$. Obviously none of these satisfy $t^4+t^2+1 = 0$ hence it is irreducible in mod $2$ therefore in $\Bbb Q$?

Here's a slick way of showing irreducibility of $x^4- x^2+1$, using uniqueness of factorization in both $\Bbb Q[x]$ and $K[x]$, where $K$ is the field of Gaussian numbers, $\Bbb Q(i)$.
Note that $x^4-x^2+1=(x^2+ix-1)(x^2-ix-1)$, and that this is a factorization into irreducibles over $K$.  If the original quartic had any factorization over $\Bbb Q$, it would also be a factorization over $K$, but this product of quadratics is the only $K$-factorization that there is. So the original quartic is irreducible over the rationals. 
