Let $E\supseteq F$ be a splitting field of $f(x)=x^6+1\in F[x]$. Find $[E:F]$. Let $E\supseteq F$ be a splitting field of $f(x)=x^6+1\in F[x]$. Find $[E:F]$ for $F=\mathbb{Z}_2$ and $F=\mathbb{Q}$.
I think in $\mathbb{Z}_2$, we can rewrite it as $f(x)=x^6-1=(x^3-1)(x^3+1)=(x^3-1)^2=(x-1)^2(x^2+x+1)^2$. Since $1$ are in $\mathbb{Z}_2$, we only need to care about $g=(x^2+x+1)^2$. But the roots are $\frac{-1\pm \sqrt{3}i}{2}$. I don't know what to do next...
I have similar idea about $\mathbb{Q}$. I wonder whether there is a better way instead of calculating all the roots then adjoining them. 
Thanks a lot!
 A: In $\mathbf F_2[x]$, we have $\;x^6+1=(x^3+1)^2$, hence the splitting field of $\;x^6+1$ is that of $x^3+1=(x+1)(x^2+x+1)$ and finally the splitting field of the irreducible polynomial $\;x^2+x+1$, so
$$[E:F]=2.$$
A: The answer over $\mathbb{Q}$ is 4. One of the primitive twelfth roots of unity is a root of $f(x)$, namely $\zeta = e^{\pi i/6}$. So the splitting field contains $\mathbb{Q}(\zeta)$. But then $f(x)$ splits over $\mathbb{Q}(\zeta)$. That can be seen from calculating all the roots. They are $e^{\pi i \theta}$, for $\theta = 1/6, 1/2, 5/6, 7/6, 3/2, 11/6$. Thus, aside from the four primitive 12th roots of unity, $f(x)$ has $i$ and $-i$ as roots, with $i=e^{\pi i/2}=\zeta^3$ and $-i=e^{3\pi i/2}=\zeta^9$.
The degree of $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ is then equal to the degree of the 12th cyclotomic polynomial (the minimal polynomial of $\zeta$). This is the same as the number of primitive twelfth roots of unity, which is 4.
I’m not sure if this answer takes the form that you were looking for, but hopefully it’s straightforward enough that it helps.
