Exercise 7.3.7

Proof.

(a)

Proposition 4.2.5, with $p=7$ prime, shows that $$f={\Phi }_7=x^6+x^5+x^4+x^3+x^2+x+1$$

is irreducible over $\mathbb{Q}$. $\zeta ={\zeta }_7=e^{2\mathit{i\pi }∕7}$ being a root of ${\Phi }_7=(x^7-1)∕(x-1)$, $f={\Phi }_7$ is the minimal polynomial of $\zeta$ over $\mathbb{Q}$.

The roots of $f$ are the roots of $x^7-1$ distinct of $1$, they are $\zeta ,{\zeta }^2,\cdots ,{\zeta }^6$. The splitting field of $f$ is so $\mathbb{Q}(\zeta ,{\zeta }^2,\cdots ,{\zeta }^6)=\mathbb{Q}(\zeta )$, since ${\zeta }^k\in \mathbb{Q}(\zeta )$ for all integers $k$.

Conclusion: $L=\mathbb{Q}(\zeta )$, where $\zeta =e^{2\mathit{i\pi }∕7}$, is the splitting field of $f=x^6+x^5+x^4+x^3+x^2+x+1$, and $f$ is the minimal polynomial of $\zeta$.

Therefore $\mathbb{Q}\subset \mathbb{Q}(\zeta )$ is a Galois extension, and

$$|\mathrm{Gal}(\mathbb{Q}(\zeta )∕\mathbb{Q})|=[\mathbb{Q}(\zeta ):\mathbb{Q}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=6.$$

(b)

The Exercice 6.2.4(f) shows that $G=\mathrm{Gal}(L∕\mathbb{Q})\simeq {(\mathbb{Z}∕7\mathbb{Z})}^{\text{∗}}$, the isomorphism $\phi$ being defined by

$\phi :\{\begin{array}{ccc}\hfill \mathrm{Gal}(\mathbb{Q}(\zeta )∕\mathbb{Q})\hfill & \hfill \to \hfill & \hfill {(\mathbb{Z}∕7\mathbb{Z})}^{\text{∗}}\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill [k]:\sigma (\zeta )={\zeta }_n^k\hfill \end{array}$

Let $\tilde{H}=\{-\bar{1},+\bar{1}\}\subset {(\mathbb{Z}∕7\mathbb{Z})}^{\text{∗}}$, and $H\subset G$ the corresponding subgroup. We compute its fixed field $L_H$.

Write $\tau$ the unique element of $G$ such that $\tau (\zeta )={\zeta }^{-1}$. We prove that $H=\{e,\tau \}$. As $\bar{\zeta }={\zeta }^6={\zeta }^{-1}\in \mathbb{Q}(\zeta )$, then $\chi :L\to L,z\mapsto \bar{z}$ is an automorphism of $L$ which is the identity on $\mathbb{Q}$, consequently $\chi \in \mathrm{Gal}(L∕\mathbb{Q})$. Since $\chi (\zeta )=\bar{\zeta }={\zeta }^{-1}=\tau (\zeta )$, $\tau =\chi$ is the complex conjugation restricted to $L$, $\phi (\tau )=[-1]$, and $H=\{e,\tau \}$.

For all $z\in L$,

$$z\in L_H⟺\bar{z}=z⟺z\in L\cap ℝ$$

$$L_H=\mathbb{Q}(\zeta )\cap ℝ.$$

$\zeta +{\zeta }^{-1}=2\cos <!--\; FUNCTION\; APPLICATION\; -->(2\pi ∕7)\in \mathbb{Q}(\zeta )\cap ℝ$, thus

$$\begin{array}{lll}\hfill \mathbb{Q}(\zeta +{\zeta }^{-1})\subset \mathbb{Q}(\zeta )\cap ℝ=L_H& & \hfill \text{(1)}\end{array}$$

Write $\alpha =\zeta +{\zeta }^{-1}$. Then

$$\begin{array}{llll}\hfill {\zeta }^2+{\zeta }^{-2}& ={(\zeta +{\zeta }^{-1})}^2-2={\alpha }^2-2\in \mathbb{Q}(\alpha ).& \hfill & \\ \hfill {\zeta }^3+{\zeta }^{-3}& =({\zeta }^2+{\zeta }^{-2})(\zeta +{\zeta }^{-1})-(\zeta +{\zeta }^{-1})=({\alpha }^2-2)\alpha -\alpha ={\alpha }^3-3\alpha \in \mathbb{Q}(\alpha ).& \hfill & \end{array}$$

As $f$ is irreducible over $\mathbb{Q}$, a basis of $L$ over $\mathbb{Q}$ is $(1,\zeta ,\dots ,{\zeta }^6)$.

Let $z=a_0+a_1\zeta +a_2{\zeta }^2+a_3{\zeta }^3+a_4{\zeta }^4+a_5{\zeta }^5+a_6{\zeta }^6,a_i\in \mathbb{Q},0\le i\le 6$, any element of $L$.

If $z\in L_H$, then $z=\tau (z)$, so

$$a_0+a_1\zeta +a_2{\zeta }^2+a_3{\zeta }^3+a_4{\zeta }^4+a_5{\zeta }^5+a_6{\zeta }^6=a_0+a_1{\zeta }^6+a_2{\zeta }^5+a_3{\zeta }^4+a_4{\zeta }^3+a_5{\zeta }^2+a_6\zeta ,$$

therefore $a_1=a_6,a_2=a_5,a_3=a_4$, so

$$z=a_0+a_1(\zeta +{\zeta }^{-1})+a_2({\zeta }^2+{\zeta }^{-2})+a_3({\zeta }^3+{\zeta }^{-3})\in \mathbb{Q}(\zeta +{\zeta }^{-1}),$$

thus $L_H\subset \mathbb{Q}(\zeta +{\zeta }^{-1})$, which gives, with the inclusion (1),

$$L_H=\mathbb{Q}(\zeta +{\zeta }^{-1})=\mathbb{Q}(\zeta )\cap ℝ.$$