What is the Galois group of $f = X^4 - 3X^2+3 \in \mathbb{Q}[X]$?

Question

The polynomial $f$ is an irreducible Eisenstein polynomial with $p = 3$.

Its roots are easy to find using the substitution $Y = X^2$ and then the $abc$-formula: $\{\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}}, -\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}}, \sqrt{\frac{3}{2}-\frac{\sqrt{-3}}{2}}, -\sqrt{\frac{3}{2}-\frac{\sqrt{-3}}{2}} \}$.

One automorphism is the complex conjugation, but I have no clue on other maps between these roots.

I think the order of the Galois group is 8. As we have $[\mathbb{Q(\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}})} : \mathbb{Q}] = 4$. Now if I could prove that $\sqrt{\frac{3}{2}-\frac{\sqrt{-3}}{2}}\not \in \mathbb{Q(\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}})}$, then I know that $\sqrt{\frac{3}{2}-\frac{\sqrt{-3}}{2}}$ has a minimal polynomial of degree 2 over $\mathbb{Q(\sqrt{\frac{3}{2}+\frac{\sqrt{-3}}{2}})} $.

Answer 1

Preliminaries:

• $Y=X^2$ gives $Y^2 - 3Y + 3 = 0$, whose discriminant is $-3$, so $\Bbb Q(\sqrt{-3})$ is a subfield.
• Quartic, so subgroup of $S_4$, so order is $1, 2, 3, 4, 6, 8, 12, 24$.
• Irreducible, so acts transitively, so at least $4$ elements, so $C_4$ or $V_4$ or $D_8$ or $A_4$ or $S_4$.
• $Y = \dfrac {3 \pm \sqrt{-3}} 2$

Let $A,B,C,D$ be the four roots. Under an appropriate order, $A+B=C+D=0$, and $AB=-\dfrac{3+\sqrt{-3}}2 = -A^2 = -B^2$, and $CD = -\dfrac{3-\sqrt{-3}}2 = -C^2 = -D^2$, and $AC=\sqrt3$.

So $A$ generates everything in $\Bbb Q(\sqrt 3)$.

$$\Bbb Q \subset \Bbb Q(\sqrt{3}) \subset \Bbb Q(\sqrt{3}, A)$$

So this is an extension of degree $8$, so the Galois group has order $8$, so it is $D_8$.

Answer 2

Consider the roots that you list as $a,b,c,d$. It is easy to see that $\Bbb Q(a)\cap \Bbb R=\Bbb Q$, i.e. real numbers in the field are rational. Your question whether $c$ belongs to $\Bbb Q(a)$ is easily answered now since $ac=\sqrt{3}$; so the splitting field $\Bbb Q(a,c)$ has degree 8.