# Find all fields $\mathbb{Q}\subset E \subset \mathbb{Q}(\zeta_{24})$

Let $\zeta_{24}$ be a $24$th primitive root of unity. Find all subfields $\mathbb{Q}\subset E \subset \mathbb{Q}(\zeta_{24})$.

I know that $\mathbb{Q}(\zeta_{24})/\mathbb{Q}$ is Galois with Galois group $\mathbb{Z}_{24}^*\cong\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$.
This group has $7$ subgroups of order $2$ and $7$ subgroups of order $4$.
Therefore we have $14$ such fields (not including the trivial $\mathbb{Q},\mathbb{Q}(\zeta_{24})$), half of degree $2$ above $\mathbb{Q}$ and half of degree $4$.

Here are the subfields that I quite easily "know" of degree 2:

• $\mathbb{Q}(\sqrt{2})$
• $\mathbb{Q}(\sqrt{2}i)$
• $\mathbb{Q}(\sqrt{3})$
• $\mathbb{Q}(\sqrt{3}i)$
• $\mathbb{Q}(i)$

(the first two from having all 8ths primitive roots of unity; the next two from having 6ths roots of unity; the last one from having 4th roots of unity).

And of of degree $4$:

• $\mathbb{Q}(\sqrt{2}, i)$
• $\mathbb{Q}(\sqrt{2}, \sqrt{3})$
• $\mathbb{Q}(\sqrt{3}, i)$

As you can see, I'm missing quite a few. Probably some unexpected ones either.

Can you help me catching them all?

• Consider the lattice of the integral divisors of the number 24. This gives you a starting point. – Wuestenfux Jul 23 '17 at 16:05

## 1 Answer

The extension is $\Bbb Q(i,\sqrt 2,\sqrt3)$. You have missed $\Bbb Q(\sqrt 6)$, $\Bbb Q(i,\sqrt6)$ and a few more.