Cyclotomic field

**ZetaX** · April 13th 2009, 04:12 AM

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 $2^{n-2}.$

**ThePerfectHacker** · April 13th 2009, 09:50 AM

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.

**ZetaX** · April 13th 2009, 02:35 PM

How do you know that G is isomorphic to $\mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2<br />$ ?

**ThePerfectHacker** · April 14th 2009, 10:22 AM

Originally Posted by ZetaX

How do you know that G is isomorphic to $\mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2<br />$ ?

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

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