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

$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}\DeclareMathOperator{\Gal}{Gal}$$>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

1 Answer

$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}\newcommand{\Q}{\mathbb{Q}}\DeclareMathOperator{\Gal}{Gal}\newcommand{\Z}{\mathbb{Z}}$$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 $$K = \Q(\beta)$$.