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Math Help - Cyclotomic field

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