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Thread: Cyclotomic field

  1. #1
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    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 $\displaystyle 2^{n-2}.$
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  2. #2
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    Quote Originally Posted by ZetaX View Post
    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.
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  3. #3
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    How do you know that G is isomorphic to $\displaystyle \mathbb{Z}_{2^{n-2}}\times \mathbb{Z}_2
    $?
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  4. #4
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    Quote Originally Posted by ZetaX View Post
    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 $\displaystyle n>2$.
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