Intermediate fields between between $\mathbb{Q}(\zeta)$ and $\mathbb{Q}$

Question

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\DeclareMathOperator{\Gal}{Gal}$>Let $\zeta$ be a primitive seventh root of unity. Find all the subgroups of $\Gal(\mathbb{Q}(\zeta))$ and the corresponding intermediate fields.

My Attempt:

I know from theorems: $\Gal(\mathbb{Q}(\zeta))\cong U(Z_7)$ Group of units in $Z_7$. That is $\{1,2,3,4,5,6\}$.

And by calculating the orders of elements one by one I saw that:
$\{1,2,4\}=\Span{2}$ and $\{1,6\}=\Span{6}$ are subgroups of $U(Z_7)$
Is there a better way to do that?

Then by the Galois correspondence theorem There are two intermediate fields, $H,K$ between $\mathbb{Q}(\zeta)$ and $\mathbb{Q}$, such that:
$[\mathbb{Q}(\zeta):H]=3$ and $[\mathbb{Q}(\zeta),K]=2$
Can we find these intermediate fields explicitly?

Appreciate your help

Answer 1

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\Q}{\mathbb{Q}}$$\DeclareMathOperator{\Gal}{Gal}$$\newcommand{\Z}{\mathbb{Z}}$First note that a generator of $G = \Gal(\Q(\zeta)/\Q)$ is $g : \zeta \mapsto \zeta^{3}$. By the Galois correspondence we have for the intermediate fields:

• $H$ is the set of fixed points of the subgroup $\Span{g^{2}}$ of order $3$
• $K$ is the set of fixed points of the subgroup $\Span{g^{3}}$ of order $2$

Since the minimal polynomial of $\zeta$ over $\Q$ is $1 + x + \dots + x^{6}$, the elements $1, \zeta, \dots, \zeta^{5}$ are independent over $\Q$, and thus also $$ \tag{indep} \text{$\zeta, \zeta^{2}, \dots, \zeta^{6}$ are independent over $\Q$} $$

Now note that

• $\alpha = \zeta + \zeta^{g^{2}} + \zeta^{g^{4}}$ is fixed by $g^{2}$, but it is not fixed by $g$ by (indep)
• $\beta = \zeta + \zeta^{g^{3}}$ is fixed by $g^{3}$, but it is not fixed by $g$ by (indep)

Hence $H = \Q(\alpha)$ and $K = \Q(\beta)$.