1. ## Cyclotomic field

Let $n>2$ and let $\mathbb Q_{2^n}$ be the

$2^n-$th cyclotomic field.
Show that $\mathbb Q_{2^n} \cap \mathbb R$ is a normal extension and its Galois group is cyclic of order $2^{n-2}.$

2. Originally Posted by ZetaX
Let $n>2$ and let $\mathbb Q_{2^n}$ be the

$2^n-$th cyclotomic field.
Show that $\mathbb Q_{2^n} \cap \mathbb R$ is a normal extension and its Galois group is cyclic of order $2^{n-2}.$
We know that $[\mathbb{Q}_{2^n}: \mathbb{Q}] = \phi(2^n) = 2^{n-1}$.
Define, $\sigma: \mathbb{Q}_{2^n}\to \mathbb{Q}_{2^n}$ by $\sigma (x) = \overline{x}$ i.e. $\sigma$ is complex conjugation.
Let $G = \text{Gal}(\mathbb{Q}_{2^n}/\mathbb{Q})$ and $H = \left< \sigma \right>$.
Therefore, by Galois theory $[\mathbb{Q}_{2^n}^{\left<\sigma\right>}:\mathbb{Q}] = [G: H] = \tfrac{2^{n-1}}{2} = 2^{n-2}$.
Now an element $x\in \mathbb{Q}_{2^n}$ is fixed by $\sigma$ if and only if $x\in \mathbb{Q}_{2^n}\cap \mathbb{R}$ (because $a+bi\mapsto a-bi$).
Therefore, $\mathbb{Q}_{2^n}^{\left< \sigma \right>} = \mathbb{Q}_{2^n}\cap \mathbb{R}$.
Thus, $[\mathbb{Q}_{2^n}\cap \mathbb{R} : \mathbb{Q}] = 2^{n-2}$.

This is a normal extension just notice that $G$ is an abelian group and so all its subgroups are normal.

To see that this subgroup is cyclic you need to understand group structure of $G$.
The group $G$ is isomorphic to $\mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2$.
It should be clear now the Galois extension is cyclic.

3. How do you know that G is isomorphic to $\mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2
$
?

4. Originally Posted by ZetaX
How do you know that G is isomorphic to $\mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2
$
?
It happens to be a known result from number theory that $\mathbb{Z}_{2^n}^{\times}$ is isomorphic to that group for $n>2$.