# Galois group of $X^4 + 2X^2+4$

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?

• The question has no sense ! It should be "find the Galois group $Gal(E/\Bbb Q)$ where $E$ is the splitting field of $f(X)\in\Bbb Q[X]$" – idm Jan 7 '15 at 0:48
• The Galois group of a polynomial is a well-defined notion, which is exactly the one that you suggest by the way. I am not expert in notation but it seems that Galois theory was made first for polynomials and then for field extensions. en.wikipedia.org/wiki/Galois_theory – Jérémy Blanc Jan 7 '15 at 0:56
• @idm That's the definition of "Galois group of $f(X)$" – Marco Flores Jan 7 '15 at 0:56
• – Watson Dec 26 '16 at 13:03

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.