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Math Help - primitive n-th root

  1. #1
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    primitive n-th root

    There was an answered question in the morning but it is erased. Could you help me on that question.

    "Let \zeta be a primitive n-th root of unity over \mathbb{Q}, where n>2 and let \alpha=\zeta+{\zeta}^{-1}. Prove that \alpha is algebraic over \mathbb{Q} of degree \frac{\varphi(n)}{2}. "
    Last edited by superman123; April 25th 2010 at 07:55 PM.
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    MHF Contributor Bruno J.'s Avatar
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    Consider the polynomial f(x)=(x-\zeta)(x-\zeta^{-1})=x-\alpha x + 1. Its roots are \zeta, \zeta^{-1} and therefore \zeta is of degree 2 over K=\mathbb{Q}[\alpha]. What do you conclude?
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    MHF Contributor Bruno J.'s Avatar
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    Here's another proof. Let L=\mathbb{Q}(\zeta), K=\mathbb{Q}(\alpha). Let G = \mbox{Gal }L/\mathbb{Q}. It's easy to see that K is the fixed field of the subgroup \{1, \sigma\} \lhd  G where \sigma : \zeta \mapsto \zeta^{-1} (prove this!). Therefore [K:\mathbb{Q}] = |G/\{1, \sigma\}| = \phi(n)/2.
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    Quote Originally Posted by Bruno J. View Post
    Consider the polynomial f(x)=(x-\zeta)(x-\zeta^{-1})=x-\alpha x + 1. Its roots are \zeta, \zeta^{-1} and therefore \zeta is of degree 2 over K=\mathbb{Q}[\alpha]. What do you conclude?
    By  f(x)=x^2-\alpha x + 1 do we have Gal(\mathbb{Q}(\zeta)/K)
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    I am trying to understand splitting fields. Are these fields same?

    \mathbb{Q}(i,\sqrt{1+\sqrt{2}})
    \mathbb{Q}(\sqrt{1+\sqrt{2}},\sqrt{1-\sqrt{2}})
    \mathbb{Q}(i,\sqrt[4]{2})
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by superman123 View Post
    By  f(x)=x^2-\alpha x + 1 do we have Gal(\mathbb{Q}(\zeta)/K)
    That is not a question. " Gal(\mathbb{Q}(\zeta)/K)" is not a statement.
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  7. #7
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    Quote Originally Posted by Bruno J. View Post
    Consider the polynomial f(x)=(x-\zeta)(x-\zeta^{-1})=x-\alpha x + 1. Its roots are \zeta, \zeta^{-1} and therefore \zeta is of degree 2 over K=\mathbb{Q}[\alpha]. What do you conclude?
    [\mathbb{Q}(\zeta):\mathbb{Q}[\alpha]~]~[\mathbb{Q}[\alpha]:\mathbb{Q}]=[\mathbb{Q}(\zeta):\mathbb{Q}]
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