Exercise 5.1.6

Proof.

(a)

Let $\alpha =\sqrt{2+\sqrt{2}}$.

Then ${\alpha }^2=2+\sqrt{2},{\alpha }^2-2=\sqrt{2},{({\alpha }^2-2)}^2-2=0,{\alpha }^4-4{\alpha }^2+2=0$.

So $\alpha$ is a root of

$$f=x^4-4x^2+2.$$

The calculation of the roots in $ℂ$ of $f$ gives (cf Ex. 4.3.2) :

$$\begin{array}{llll}\hfill f(\beta )=0& ⟺{({\beta }^2-2)}^2=2& \hfill & \\ \hfill & ⟺{\beta }^2=2+𝜀\sqrt{2},𝜀\in \{-1,1\}& \hfill & \\ \hfill & ⟺\beta ={𝜀}^{\prime }\sqrt{2+𝜀\sqrt{2}},𝜀,{𝜀}^{\prime }\in \{-1,1\}& \hfill & \\ \hfill & ⟺\beta \in \left\{\sqrt{2+\sqrt{2}},-\sqrt{2+\sqrt{2}},\sqrt{2-\sqrt{2}},-\sqrt{2-\sqrt{2}}\right\}.& \hfill & \end{array}$$

Thus

$$\begin{array}{lll}\hfill f=\left(x-\sqrt{2+\sqrt{2}}\right)\left(x+\sqrt{2+\sqrt{2}}\right)\left(x-\sqrt{2-\sqrt{2}}\right)\left(x+\sqrt{2-\sqrt{2}}\right).& & \hfill \text{(1)}\end{array}$$

We show that $f$ is irreducible over $\mathbb{Q}$. The Schönemann-Eisenstein Criterion, with $p=2$ applies to the polynomial $f=x^4-4x^2+2=a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0$ ( $2\nmid a_4=1,2\mid a_3=0,2\mid a_2=-4,2\mid a_1=0,2\mid a_0=2,2^2\nmid a_0=2$). $f$ is so irreducible over $\mathbb{Q}$.

Consequently, $f$ of degree 4, is the minimal polynomial of $\alpha =\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$, so,

$$\left[\mathbb{Q}\left(\sqrt{2+\sqrt{2}}\right):\mathbb{Q}\right]=4.$$

(b)

The splitting field of $f$ over $\mathbb{Q}$ is $$K=\mathbb{Q}\left(\sqrt{2+\sqrt{2}},-\sqrt{2+\sqrt{2}},\sqrt{2-\sqrt{2}},-\sqrt{2-\sqrt{2}}\right)=\mathbb{Q}\left(\sqrt{2+\sqrt{2}},\sqrt{2-\sqrt{2}}\right).$$

Let $\alpha =\sqrt{2+\sqrt{2}},\gamma =\sqrt{2-\sqrt{2}}$. Then $K=\mathbb{Q}(\alpha ,\gamma )$.

Note that $\mathit{\alpha \gamma }=\sqrt{4-2}=\sqrt{2}$

Moreover ${\alpha }^2=2+\sqrt{2},{\gamma }^2=2-\sqrt{2}$, thus ${\alpha }^2-{\gamma }^2=2\sqrt{2}=2\mathit{\alpha \gamma }$.

$${\alpha }^2-\frac{2}{{\alpha }^2}=2\mathit{\alpha \gamma }.$$

So

$$\gamma =\frac{1}{2}\left(\alpha -\frac{2}{{\alpha }^3}\right)\in \mathbb{Q}(\alpha ).$$

Consequently $\mathbb{Q}(\alpha )=\mathbb{Q}(\alpha ,\gamma )$ is the splitting field of $f$ over $\mathbb{Q}$. The splitting field of $x^4-4x^2+2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{2+\sqrt{2}})$, of degree 4 over $\mathbb{Q}$.

Note: With Sage, we obtain $\gamma =\frac{1}{2}\left(\alpha -\frac{2}{{\alpha }^3}\right)={\alpha }^3-3\alpha$, and the factorization

$$x^4-4x^2+2=(x-\alpha )(x+\alpha )(x-{\alpha }^3+3\alpha )(x+{\alpha }^3-3\alpha ).$$