Exercise 5.1.8

Proof.

(a)

The set of the roots of $x^n-2\in \mathbb{Q}[x]$ is $S=\{{\zeta }^k\sqrt[n]{2},k=0,\cdots ,n-1\}$, where $\zeta =e^{2\mathit{i\pi }∕n}$: the splitting field $L$ of $x^n-2$ over $\mathbb{Q}$ is so $\mathbb{Q}(S)$.

As $\zeta =\zeta \sqrt[n]{2}∕\sqrt[n]{2}\in \mathbb{Q}(S)$, $L=\mathbb{Q}(\zeta ,\sqrt[n]{2})$.

(b)

Suppose that $n$ is prime.

As $f=x^n-2$ is irreducible over $\mathbb{Q}$, $f$ is the minimal polynomial of $\sqrt[n]{2}$ over $\mathbb{Q}$, so $[\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=n$.

As $n$ is prime, $1+x+\cdots +x^{n-1}$ is irreducible over $\mathbb{Q}$, thus $[\mathbb{Q}(\zeta ):\mathbb{Q}]=n-1$.

From the Tower Theorem,

$$\begin{array}{llll}\hfill [L:\mathbb{Q}]& =[L:\mathbb{Q}(\sqrt[n]{2})][\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q}]=n[L:\mathbb{Q}(\sqrt[n]{2})]& \hfill & \\ \hfill [L:\mathbb{Q}]& =[L:\mathbb{Q}(\zeta )][Q(\zeta ):\mathbb{Q}]=(n-1)[L:\mathbb{Q}(\zeta )]& \hfill \text{(1)}\end{array}$$

Thus $n\mid [L:\mathbb{Q}]$ and $n-1\mid [L:\mathbb{Q}]$.

As $n,n-1$ are relatively prime,

$$\begin{array}{lll}\hfill n(n-1)\mid [L:\mathbb{Q}].& & \hfill \text{(2)}\end{array}$$

Moreover, the minimal polynomial $p$ of $\sqrt[n]{2}$ over $\mathbb{Q}(\zeta )$ divides $x^n-2\in \mathbb{Q}[x]\subset \mathbb{Q}(\zeta )[x]$, thus $[L:\mathbb{Q}(\zeta )]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)\le n$. By (1), $[L:\mathbb{Q}]\le n(n-1)$, and by (2) $n(n-1)\mid [L:\mathbb{Q}]$, thus

$$[L:\mathbb{Q}]=n(n-1).$$