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

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