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!

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.