Order of group $\text{Aut}(E/\Bbb Q)$ Let $E$ be splitting field for $x^4+x^2+1$ over $\Bbb Q$, what is order of group $\text{Aut}(E/\Bbb Q)$?
I know that $[E:\Bbb Q] = 2$ since $(x^2-1)(x^4+x^2+1) = x^6-1$  and somewhere I seen in book that $|\text{Aut}(E/\Bbb Q)| \le [E:\Bbb Q]$ (probably for irreducible polynomial). But $x^4+x^2+1$ has four roots. I think the order of the group must be $2$ but I am not sure. Can anyone guide me to correct answer?
 A: For Galois extensions the size of the automorphism group is equal to the degree of the field extension. In your case you know that your extension is bi-quadratic and is generated by $\zeta_6$ a primitive $6^{th}$ root of unity (see this by noting that the roots of $y^2+y+1$ are the primitive $3^{rd}$ root of unity, and note $y=x^2$ gives $\pm\sqrt{\zeta_3}$. In fact your polynomial factors as $(x^2+x+1)(x^2-x+1)$. And so since the degree is just $2$ of the splitting field, the automorphism clearly being given by complex conjugation, you're all set!
You can say even more than this in fact: all quadratic extensions are Galois--and therefore $|Aut(E/\Bbb Q)|$ for such $E$ is always $2$. To see this merely note that since they are generated by $\sqrt{a}$ for some $a\in\Bbb Q$, and since fields are closed under subtraction, $0-\sqrt{a}=-\sqrt{a}$ also squares to $a$ and is a distinct square root (since $-1\ne 1$). So the automorphism $\sqrt{a}\mapsto -\sqrt{a}$ is non-trivial, showing that $|\text{Aut}(E/\Bbb Q)|\ge 2$.
