Determining irreducible polynomial of $\zeta_n$ In a homework I did I had to determine the irreducible polynomials of some $\zeta_n$ functions over $\mathbb{Q}$. In $\zeta_6$, I set $\zeta_6=x$ and I know that $x^6=1 \rightarrow x^6-1=0$. So,when I factor it out I get $x^6-1=(x^3-1)(x^3+1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$ From there I know the expressions $(x^2-x+1)$ and $(x^2+x+1)$ are irreducible over $\mathbb{Q}$ but I don't know which one of those is the right answer. The solutions to the homework say its $(x^2+x+1)$  but I don't see why. Any hints? Thanks!
 A: You can think of the correct answer as:
$$\frac{(x^6-1)(x-1)}{(x^3-1)(x^2-1)}$$
Basically we "remove" the roots of $x^3-1$ and $x^2-1$. But that divided by $(x-1)$ twice, so you bave to multiply that one back in once.
That gives a sense of what the general answer looks like, which can be proven by the multiplicative version of Möbius inversion, so that the minimal polynomial for $\zeta_n$ is:
$$\Phi_n(x) = \prod_{d\min n} (x^d-1)^{\mu(n/d)}$$
which follows by first showing:
$$x^n-1=\prod_{d\mid n} \Phi_d(x)$$
See Cyclotomic polynomials.
A: It is sometimes helpful to write primitive roots of unity in their trigonometric form, and use De Moivre's theorem.
In this case, we may take $\zeta_6 = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})$, although the complex-conjugate would work as well (this is also the multiplicative inverse).
Then it is clear to see that $(\zeta_6)^3 = \cos(\pi) + i\sin(\pi) = -1 + i0 = -1$, that is:
$\zeta_6^3 + 1 = 0$.
So $\zeta_6$ is a root of $x^3 + 1$, not a root of $x^3 -1$.
We can also compute $f(\zeta_6)$ directly, where $f(x) = x^2 - x + 1$.
$\zeta_6^2 - \zeta_6 + 1 = (\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))^2 - \cos(\frac{\pi}{3}) - i\sin(\frac{\pi}{3}) + 1$
$= \cos(\frac{2\pi}{3}) - \cos(\frac{\pi}{3}) + 1 + i(\sin(\frac{2\pi}{3}) - \sin(\frac{\pi}{3}))$
$= -\frac{1}{2} - \frac{1}{2} + 1 + i(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2})$
$= 0 + i0 = 0$.
In fact:
$(x - \zeta_6)(x - \zeta_6^{\ast}) = x^2 - 2\mathfrak{Re}(\zeta_6)x + \zeta_6(\zeta_6)^{\ast}$
$= x^2 - 2(\frac{1}{2})x + |\zeta_6|^2 = x^2 - x + 1$.
The above factorization shows that the splitting field of $x^3 + 1$ (indeed, even of $x^6 - 1$ since any cube root of unity is the square of a sixth root of unity) is indeed $\Bbb Q(\zeta_6)$, which therefore has degree $2$ over the rationals.
$2 = \phi(6)$, so this is not surprising. The sole non-trivial automorphism of $\Bbb Q(\zeta_6)$ which fixes $\Bbb Q$ is $\zeta_6 \mapsto \zeta_6^5 = (\zeta_6)^{-1} = \zeta_6^{\ast}$, that is, it is complex-conjugation restricted to $\Bbb Q(\zeta_6)$, and thus the Galois group is cyclic, of order $2$.
