Splitting field of $x^6+x^3+1$ over $\mathbb{Q}$ I am trying to find the splitting field of $x^6+x^3+1$ over $\mathbb{Q}$.
Finding the roots of the polynomial is easy (substituting $x^3=t$ , finding the two roots of the polynomial in $t$ and then taking a 3-rd root from each one). The roots can be seen here [if there is a more elegant way of finding the roots it will be nice to hear]
Is is true the that the splitting field is $\mathbb{Q}((-1)^\frac{1}{9})$ ?
I think so from the way the roots look, but I am unsure.
Also, I am having trouble finding the minimal polynomial of $(-1)^\frac{1}{9}$, it seems that it would be a polynomial of degree 9, but of course the degree can't be more than 6...can someone please help with this ?
 A: You've got something wrong: the roots of $t^2+t+1$ are the complex cubic roots of one, not of $-1$: $t^3-1 = (t-1)(t^2+t+1)$, so every root of $t^2+t+1$ satisfies $\alpha^3=1$). That means that you actually want the cubic roots of some of the cubic roots of $1$; that is, you want some ninth roots of $1$ (not of $-1$). 
Note that 
$$(x^6+x^3+1)(x-1)(x^2+x+1) = x^9-1.$$
So the roots of $x^6+x^3+1$ are all ninth roots of $1$. Moreover, those ninth roots should not be equal to $1$, nor be cubic roots of $1$ (the roots of $x^2+x+1$ are the nonreal cubic roots of $1$): since $x^9-1$ is relatively prime to $(x^9-1)' = 9x^8$, the polynomial $x^9-1$ has no repeated roots. So any root of $x^9-1$ is either a root of $x^6+x^3+1$, or a root of $x^2+x+1$, or a root of $x-1$, but it cannot be a root of two of them. 
If $\zeta$ is a primitive ninth root of $1$ (e.g., $\zeta = e^{i2\pi/9}$), then $\zeta^k$ is also a ninth root of $1$ for all $k$; it is a cubic root of $1$ if and only if $3|k$, and it is equal to $1$ if and only if $9|k$. So the roots of $x^6+x^3+1$ are precisely $\zeta$, $\zeta^2$, $\zeta^4$, $\zeta^5$, $\zeta^7$, and $\zeta^8$. They are all contained in $\mathbb{Q}(\zeta)$, which is necessarily contained in the splitting field. Thus, the splitting field is $\mathbb{Q}(\zeta)$, where $\zeta$ is any primitive ninth root of $1$.
A: This polynomial is $\Phi_9(x)$, the ninth cyclotomic polynomial whose roots are precisely the primitive ninth roots of unity. A $\mathbb{Q}$-basis for this extension is $\{\zeta_1,\zeta_2,\zeta_4,\zeta_5,\zeta_7,\zeta_8\}$. So you have your splitting field.
