Exercise 4.3.2

Proof.

(a)

Note that $\sqrt[4]{2}$ is a root of $p=x^4-2\in \mathbb{Q}[x]$, and $p$ is irreducible over $\mathbb{Q}$ by Exercise 4.2.8 (or Schönemann-Eisenstein Criterion for the prime 2). Thus $$[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=4.$$

$i$ is a root of $x^2+1$, which has no root in $ℝ$, a fortiori in $\mathbb{Q}[\sqrt[4]{2}]$. As its degree is 2, it is irreducible over $\mathbb{Q}[\sqrt[4]{2}]$, thus

$$[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}(\sqrt[4]{2})]=2.$$

Moreover $\mathbb{Q}(i,\sqrt[4]{2})=\mathbb{Q}(\sqrt[4]{2},i)$. The Tower Theorem gives

$$[\mathbb{Q}(i,\sqrt[4]{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{2},i):\mathbb{Q}(\sqrt[4]{2})]\times [\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=8.$$

(b)

$\sqrt[3]{2}$ is irrational, so $f=x^3-2$ has no root in $\mathbb{Q}$, and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=3$, thus $f$ is irreducible over $\mathbb{Q}$ and $f$ is the minimal polynomial over $\mathbb{Q}$ of $\sqrt[3]{2}$, and so $$[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3.$$

The roots of $x^2-3$ are $\pm \sqrt{3}$ and are irrational. As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(x^2-3)=2$, and as $x^2-3$ has no root in $\mathbb{Q}$, $x^2-3$ is irreducible over $\mathbb{Q}$. It is the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}$, thus

$$[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=2.$$

Moreover

$$\begin{array}{llll}\hfill [\mathbb{Q}(\sqrt{3},\sqrt[3]{2}):\mathbb{Q}]& =[\mathbb{Q}(\sqrt{3},\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})]\times [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q})]& \hfill & \\ \hfill & =[\mathbb{Q}(\sqrt{3},\sqrt[3]{2}):\mathbb{Q}(\sqrt{3})]\times [\mathbb{Q}(\sqrt{3}):\mathbb{Q}],& \hfill & \end{array}$$

thus, if we write $d=[\mathbb{Q}(\sqrt{3},\sqrt[3]{2}):\mathbb{Q}]$, then $2\mid d,3\mid d$, with $2\wedge 3=1$, thus $6\mid d$, $6\le d$.

$\sqrt{3}$ is a root of $x^2-3$, and the degree of $x^2-3$ is 2. Its coefficients are in $\mathbb{Q}$, a fortiori in $\mathbb{Q}(\sqrt[3]{2})$. Thus the minimal polynomial $p$ of $\sqrt{3}$ over $\mathbb{Q}(\sqrt[3]{2})$ divides $x^2-3$. Its degree $\delta =\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)=[\mathbb{Q}(\sqrt{3},\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})]$ satisfies then $\delta \le 2$.

As $d=3\delta \ge 6$, and so $\delta \le 2$, $d\le 6$. Therefore $d=6$.

$$[\mathbb{Q}(\sqrt{3},\sqrt[3]{2}):\mathbb{Q}]=6.$$

(c)

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

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

$\alpha$ is a root of

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

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

Conclusion: $f=x^4-4x^2+2$ is irreducible over $\mathbb{Q}$. $f$ is the minimal polynomial of $\alpha =\sqrt{2+\sqrt{2}}$, thus

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

(d)

$x^2+1$ has no real root, a fortiori no root in $\mathbb{Q}(\sqrt{2+\sqrt{2}})$, and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(x^2+1)=2$. Thus $x^2+1$ is irreducible over $\mathbb{Q}(\sqrt{2+\sqrt{2}})$, it is the minimal polynomial of $i$ over $\mathbb{Q}(\sqrt{2+\sqrt{2}})$, thus $$\left[\mathbb{Q}\left(i,\sqrt{2+\sqrt{2}}\right):\mathbb{Q}\left(\sqrt{2+\sqrt{2}}\right)\right]=2.$$

Consequently

$$\left[\mathbb{Q}\left(i,\sqrt{2+\sqrt{2}}\right):\mathbb{Q}\right]=\left[\mathbb{Q}\left(i,\sqrt{2+\sqrt{2}}\right):\mathbb{Q}\left(\sqrt{2+\sqrt{2}}\right)\right]\times \left[\mathbb{Q}\left(\sqrt{2+\sqrt{2}}\right):\mathbb{Q}\right]=8.$$