Find all subfields of $\mathbb{Q}(\mu_{24})$

Question

Problem: Let $\mu_{24} \in\mathbb{C}$ be a primitive 24'th root of unity and let $L = \mathbb{Q}(\mu_{24})$ be the 24'th cyclotomic extension of $\mathbb{Q}$. List all subfields of $L$ in the form $\mathbb{Q} = \mathbb{Q}(\alpha)$ or $\mathbb{Q} = \mathbb{Q}(\alpha, \beta)$.

I know that $L/\mathbb{Q}$ is a Galois extension with Galois group $G$ consisting of automorphisms $\sigma_{i}$, $i\in (\mathbb{Z}/24\mathbb{Z})^\times$, where $\sigma_{i}$ sends $\mu_{24}$ to $(\mu_{24})^i$. This group is isomorphic to $C_{2}\times C_{2} \times C_{2}$. Write $G = < \sigma_{5}, \sigma_{7}, \sigma_{13} >$.

By the Galois correspondence,

• The quadratic subfields are in 1-1 correspondence with the order-4 subgroups of $G$. There are 7 of these: each subgroup is generated by two elements of order 2; there are $7C2 = 21$ choices; but each subgroup arises from 3 distinct choices; hence there are $21/3 = 7$ subgroups.
• The quartic subfields biject with the order-2 subgroups of $G$, of which there are 7 also.

Instead of listing the subgroups and searching for their fixed fields, I first obtained information from the cyclotomic extensions contained in $L$.

• $\mathbb{Q}(\mu_{3})=\mathbb{Q}(\sqrt3)\subseteq L$, therefore $\sqrt3\in L$
• $\mathbb{Q}(\mu_{4})=\mathbb{Q}(i)\subseteq L$, therefore $i\in L$
• $\mathbb{Q}(\mu_{6})=\mathbb{Q}(3)$ so no new information
• $\mathbb{Q}(\mu_{8})=\mathbb{Q}(i,\sqrt2)\subseteq L$, therefore $\sqrt2 \in L$
• $\mathbb{Q}(\mu_{12})=\mathbb{Q}(i,\sqrt3)\subseteq L$ so no new information

Now I can obtain all 7 quadratic subfields:

1. $\mathbb{Q}(i)$
2. $\mathbb{Q}(\sqrt2)$
3. $\mathbb{Q}(\sqrt3)$
4. $\mathbb{Q}(i\sqrt2)$
5. $\mathbb{Q}(i\sqrt3)$
6. $\mathbb{Q}(\sqrt6)$
7. $\mathbb{Q}(i\sqrt6)$

and I can obtain 6 quartic subfields:

1. $\mathbb{Q}(i, \sqrt2)$
2. $\mathbb{Q}(i, \sqrt3)$
3. $\mathbb{Q}(\sqrt2, \sqrt3)$
4. $\mathbb{Q}(i, \sqrt6)$
5. $\mathbb{Q}(\sqrt2, i\sqrt3)$
6. $\mathbb{Q}(\sqrt3, i\sqrt2)$

I am missing one quartic subfield. What can I do to find it?

Answer 1

Identify the missing subgroup, and find its fixed field.

I reckon you are missing $\Bbb Q(\sqrt6,i\sqrt2)$.