Show that $ \mathbb{Q}[x]/(x^4 - 6x^2 + 4) \simeq \mathbb{Q}(\sqrt{2}, \sqrt{5})$

Question

I have a field $E = \mathbb{Q}[x]/(x^4 - 6x^2 + 4)$ and the computer has told me that the degree of the extension is $[E:F]= 2$ that is a quadratic extension of $F = \mathbb{Q}(\sqrt{5})$. If I read correctly the polynomials is polynomial $y^2 - 2 = 0$ over $F$, so that $E = F[x]/(x^2 - 2)$. That is an interesting thought: we'd have that: $$ \mathbb{Q}[x]/(x^4 - 6x^2 + 4) \simeq \mathbb{Q}(\sqrt{2}, \sqrt{5})$$ I've also been told, for what it's worth that $E/\mathbb{Q}$ is a Galois extension.

• normal: every polynomial that is irreducible over E either has no root in $\mathbb{Q}$ or splits into linear factors

• separable: for every $\alpha \in E$, the minimal polynomial of $ \alpha$ over $\mathbb{Q}$ is a separable polynomial (the number of distinct roots is equal to the degree of the polynomial).

and so the Galois group is either $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ or $\mathbb{Z}_4$.

Why should I believe anything the computer has told me thus far? How do we prove any of this?

Answer 1

We can explicitly solve the quartic equation. Replace $t=x^2$ to get $t^2 - 6t + 4 = 0$, whose solutions are:

$$t_{1/2} = \frac{6 \pm \sqrt{36-16}}{2} = 3 \pm \sqrt{5}$$

Now to compute the roots of the quartic polynomial you can use the following identity:

$$\sqrt{a\pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2-b}}{2}}$$

Hence:

$$x_{1} = \sqrt{t_1} = \sqrt{3 + \sqrt{5}} = \sqrt{\frac{3 + 2}{2}} + \sqrt{\frac{3-2}{2}} = \sqrt{\frac 52} + \sqrt{\frac 12}$$ $$x_{2} = -\sqrt{t_1} = - \sqrt{\frac 52} - \sqrt{\frac 12}$$ $$x_{3} = \sqrt{t_2} = \sqrt{\frac 52} - \sqrt{\frac 12}$$ $$x_{4} = -\sqrt{t_2} = -\sqrt{\frac 52} + \sqrt{\frac 12}$$

Now:

$$ \mathbb{Q}[x]/(x^4 - 6x^2 + 4) \simeq \mathbb{Q}(x_1)$$

On the other side we have that $x_1 \in \mathbb{Q}(\sqrt{2},\sqrt{5})$ and as both extensions are of fourth degree we conclude that $\mathbb{Q}(x_1) = \mathbb{Q}(\sqrt{2},\sqrt{5})$. Hence the proof.

Additionally as the extension has two quadratic extension we must have that the Galois group is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$