# 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 $$\sqrt[4]{\frac{-1\pm\sqrt{-3}}{2}}\zeta_4^k,$$ 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!

• This is the cyclotomic field generated by the $24$-th roots of unity. – Angina Seng Jul 12 '18 at 18:57
• No, the cyclotomic poly for the 24th root of unity is $x^8-x^4+1$. – user134070 Jul 12 '18 at 19:01
• The zeroes are all 12th roots of unity, excluding $\pm1$ and $\pm i$. – Arthur Jul 12 '18 at 19:08

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.
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 $\operatorname{Aut}(\mathbb Z/12)\cong (\mathbb Z/12)^\times\cong \mathbb Z/2\times \mathbb Z/2$.