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Math Help - Galois closure

  1. #1
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    Galois closure

    Let L be the Galois closure of K=\mathbb{Q}(\sqrt[n]{a}), where a \in \mathbb{Q}, a>0 and suppose [K:\mathbb{Q}]=n.
    Prove that [L:\mathbb{Q}]=n \varphi(n) or \frac{1}{2}n\varphi(n)
    ( \varphi Euler's function)
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  2. #2
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    Quote Originally Posted by Byun View Post
    Let L be the Galois closure of K=\mathbb{Q}(\sqrt[n]{a}), where a \in \mathbb{Q}, a>0 and suppose [K:\mathbb{Q}]=n.
    Prove that [L:\mathbb{Q}]=n \varphi(n) or \frac{1}{2}n\varphi(n)
    ( \varphi Euler's function)
    i haven't thought about the details yet but what is clear to me is that the Galois closure of K would be the splitting field of the polynoimal x^n - a, because [K:\mathbb{Q}]=n, which means we've

    assumed that x^n - a is irreducible over \mathbb{Q}. so if we let \xi = \exp(2\pi i/n), then L=\mathbb{Q}(\xi, \sqrt[n]{a}). thus [L:\mathbb{Q}]=[\mathbb{Q}(\xi, \sqrt[n]{a}): \mathbb{Q}(\xi)] [\mathbb{Q}(\xi):\mathbb{Q}]=\varphi(n)[\mathbb{Q}(\xi, \sqrt[n]{a}): \mathbb{Q}(\xi)]. so what you need to prove

    is that the only possible values of [\mathbb{Q}(\xi)(\sqrt[n]{a}): \mathbb{Q}(\xi)] are n and \frac{n}{2}.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    is that the only possible values of [\mathbb{Q}(\xi)(\sqrt[n]{a}): \mathbb{Q}(\xi)] are n and \frac{n}{2}.
    Yes, but show do you prove this?

    Notice by the Natural Irrationalities theorem: Let K = \mathbb{Q}(\sqrt[n]{a}) and E = \mathbb{Q}(\xi) and so [KE:E] = [K:E\cap K].

    Therefore, [ \mathbb{Q}(\sqrt[n]{a},\xi) : \mathbb{Q}(\sqrt[n]{a})] = [\mathbb{Q}(\sqrt[n]{a}): \mathbb{Q}(\sqrt[n]{a})\cap \mathbb{Q}(\xi)].

    But, \mathbb{Q}(\sqrt[n]{a})\cap \mathbb{Q}(\xi) is a subfield of both \mathbb{Q}(\xi) and \mathbb{Q}(\sqrt[n]{a}). But subfields of \mathbb{Q}(\sqrt[n]{a}) have form \mathbb{Q}(\sqrt[m]{a}), m|n. Thus, somehow the only possibilities for this degree are 1\text{ or }2. But there does not seem to be direct way to show this. This there some kind of theorem about real subfields of a cyclotomic extension?
    ---

    The other thing I see is that \mathbb{Q}(\xi)(\sqrt[n]{a})/ \mathbb{Q}(\xi) is an n-Kummer extension. Therefore, the degree is equal to the order of the subgroup \left< a(F^{\times})^n \right> in F^{\times}/(F^{\times})^n where F=\mathbb{Q}(\xi). But this seems to be really messy.
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  4. #4
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    here's a nice problem which will also prove that [\mathbb{Q}(\xi, \sqrt[n]{a}): \mathbb{Q}(\xi)]=n \ \text{or} \ \frac{n}{2}:

    let \xi_k=\xi^k, \ 1 \leq k \leq n, \ n > 1. clearly \sum_{k=1}^n \xi_k = 0. now prove that if \sum_{j=1}^m \xi_{k_j}=0, where 1 \leq k_1 < k_2 < \cdots < k_m \leq n, then m=n \ \text{or} \ \frac{n}{2}.
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