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Math Help - Galois extension over Q

  1. #1
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    Galois extension over Q

    Why is \mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\sqrt[n]{a}) a Galois extension over \mathbb{Q}??
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  2. #2
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    Quote Originally Posted by Stiger View Post
    Why is \mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\sqrt[n]{a}) a Galois extension over \mathbb{Q}??
    Intersection of subfields (in a larger field) is a subfield. Therefore, \mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\sqrt[n]{a}) is a subfield (of the algebraic numbers, for example). Now \mathbb{Q}(\zeta)/\mathbb{Q} is a cyclotomic extension and therefore \text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) is an abelian group. Since all subgroups of this group are normal it follows that if \mathbb{Q}\subseteq L\subseteq \mathbb{Q}(\zeta_n) then L/\mathbb{Q} is a Galois extension by the fundamental theorem. Since \mathbb{Q}\subseteq \mathbb{Q}(\zeta)\cap \mathbb{Q}(\sqrt[n]{a})\subseteq \mathbb{Q}(\zeta) the rest follows.
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  3. #3
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    in general if \{K_i \}_{i \in I} is a family of (finite) Galois extensions of F, then \bigcap_{i \in I}K_i is also Galois over F. just recall that a (finite) extension is Galois iff it's separable and normal.

    Edit: what was i thinking?!! lol ... this, although true, but doesn't apply to Stiger's problem! thanks ThePerfectHacker for pointing that out.
    Last edited by NonCommAlg; December 22nd 2008 at 03:22 PM.
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    in general if \{K_i \}_{i \in I} is a family of (finite) Galois extensions of F, then \bigcap_{i \in I}K_i is also Galois over F. just recall that a (finite) extension is Galois iff it's separable and normal.
    But \mathbb{Q}(\sqrt[n]{a})/\mathbb{Q} is not Galois in general.
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