TITLE: Galois group of $X^4 + 2X^2+4$
QUESTION [5 upvotes]: Find the Galois group of $f(X) = X^4 + 2X^2+4$ over $\mathbb{Q}$.

Let $L$ be the splitting field of $f$ over $\mathbb{Q}$.  Finding the roots of this polynomial, I got $$X^2 = \frac{-2\pm \sqrt{4-16}}{2} = -1 \pm \sqrt{3}i$$ so the roots are $\alpha_1 = \sqrt{-1+\sqrt{3}i}, \alpha_2 =\sqrt{-1-\sqrt{3}}i, \alpha_3 =  -\sqrt{-1+\sqrt{3}i}$ and  $ \alpha_4 -\sqrt{-1-\sqrt{3}}i$.  Now $L = \mathbb{Q}(\alpha_1, \alpha_2)$, and since $$\alpha_1 \alpha_2  = \sqrt{(-1 + \sqrt{3}i)(-1-\sqrt{3}i)} = \sqrt{1+3} = 2$$ we see that $L = \mathbb{Q}(\alpha_1) = \mathbb{Q}(\alpha_2)$.  It's not difficult to see that $[\mathbb{Q}(\alpha_1) : \mathbb{Q}] = 4$, so the Galois group of $L/\mathbb{Q}$ over $\mathbb{Q}$ is is either $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.  
My guess is that this Galois group is cyclic, since I think that $\mathbb{Q}(\sqrt{3}i)$ (a cyclotomic extension by a cubic root of unity) is the only intermediate subfield of $L/\mathbb{Q}$.  How can I know for sure?

REPLY [3 votes]: You can also write the solutions in the following way: 
$$\frac{1}{2}\left(\pm \sqrt{2}\pm \sqrt{6}i\right)$$
and then find that $L=\mathbb{Q}[\sqrt{2},\sqrt{3}i]$. The Galois group is then $(\mathbb{Z}/2\mathbb{Z})^2$.
One generator corresponds to the conjugation $z\mapsto \bar{z}$ and the second is $\alpha\mapsto -\alpha$, where $\alpha$ is one of your roots; this is an automorphism since $L=\mathbb{Q}[\alpha]$ and because $-\alpha$ is also a root.