# Cyclotomic field

• Apr 13th 2009, 04:12 AM
ZetaX
Cyclotomic field
Let $\displaystyle n>2$ and let $\displaystyle \mathbb Q_{2^n}$ be the

$\displaystyle 2^n-$th cyclotomic field.
Show that $\displaystyle \mathbb Q_{2^n} \cap \mathbb R$ is a normal extension and its Galois group is cyclic of order $\displaystyle 2^{n-2}.$
• Apr 13th 2009, 09:50 AM
ThePerfectHacker
Quote:

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

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

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

This is a normal extension just notice that $\displaystyle G$ 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 $\displaystyle G$.
The group $\displaystyle G$ is isomorphic to $\displaystyle \mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2$.
It should be clear now the Galois extension is cyclic.
• Apr 13th 2009, 02:35 PM
ZetaX
How do you know that G is isomorphic to $\displaystyle \mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2$?
• Apr 14th 2009, 10:22 AM
ThePerfectHacker
Quote:

Originally Posted by ZetaX
How do you know that G is isomorphic to $\displaystyle \mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2$?

It happens to be a known result from number theory that $\displaystyle \mathbb{Z}_{2^n}^{\times}$ is isomorphic to that group for $\displaystyle n>2$.