Galois group of $x^8+x^4+1$ As the title suggests, my objective is to characterize the Galois group of the splitting field of the polynomial $x^8+x^4+1$ over $\mathbb{Q}$. That is, if $L$ is the splitting field of $x^8+x^4+1$, I want to know what $\mathrm{Gal}(L/\mathbb{Q})$ is.
I know that the roots of the polynomial are
\begin{equation}
\sqrt[4]{\frac{-1\pm\sqrt{-3}}{2}}\zeta_4^k,
\end{equation}
where $\zeta_4$ is the 4-th root of unity. And here is where I have the rub: To find the splitting field $L$, I need to adjoin some of the roots above. Clearly I need to adjoin $\zeta_4$, whose degree is 2. Question is: What are the others? Do I adjoin only $\sqrt[4]{\frac{-1+\sqrt{-3}}{2}}$ (whose degree is 8, so the resulting degree of $L$ over $\mathbb{Q}$ is 16), or $\sqrt[4]{\frac{-1-\sqrt{-3}}{2}}$ as well?
I am new to this kind of question, so please guide me.
Thanks!
 A: As Arthur mentioned, the roots are $12^{\text{th}}$ roots of $1$. Observe that $\zeta_{12}^8+\zeta_{12}^4+1=\zeta_{3}^2+\zeta_{3}+1=0$. So the four primitive $12^{\text{th}}$ roots of $1$ satisfy the polynomial. Similarly, 
$\zeta_{6}^8+\zeta_{6}^4+1=\zeta_{3}+\zeta_{3}^2+1=0$, and $\zeta_{3}^8+\zeta_{3}^4+1=\zeta_{3}^2+\zeta_{3}+1=0$, which account for the remaining roots. Thus, the field you are interested in is just $\mathbb Q(\zeta_{12})$.
The Galois group of this field can be understood by its action on the group of $12^{\text{th}}$ roots of $1$, which is isomorphic to $\mathbb Z/12$. Since the primitive $12^{\text{th}}$ roots of $1$ are all conjugate, we may send any generator this group of roots to any other, and thus the Galois group is $\operatorname{Aut}(\mathbb Z/12)\cong (\mathbb Z/12)^\times\cong \mathbb Z/2\times \mathbb Z/2$.
A: Another way of looking at this:
\begin{align}
X^8+X^4+1&=(X^4-X^2+1)(X^2-X+1)(X^2+X+1)\\
&=\Phi_{12}\Phi_6\Phi_3\,,
\end{align}
where $\Phi_n$ is the cyclotomic polynomial, each irreducible, with roots the primitive $n$-th roots of unity.
So the splitting field is $\Bbb Q(\zeta_{12})$, Galois group well described by @BrettFrankel.
