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Thread: Cyclotomic Fields & Bases....

  1. #1
    AAM
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    Cyclotomic Fields & Bases....

    Hi everyone! :-)

    Suppose Zeta is any primitive 5th root of unity.

    Then we know that Q(Zeta) has a basis {1,Zeta,Zeta^2,Zeta^3} because Zeta is algebraic over Q.

    But how can I show that {Zeta,Zeta^-1,Zeta^2,Zeta^-2} is a basis? :-s

    Also, how does the Galois Group of Q(Zeta) act on this new basis?

    Furthermore, how can you prove the existence of a unique subfield E such that [Q(Zeta):E]=2, & find all a,b,c,d such that aZeta+bZeta^-1+cZeta^2+dZeta^-2 is in E but not Q? (apparently you can then show that E = Q(Sqrt(5)) !! :-o )

    I have absolutely no idea where to begin! :-s

    Any help on ANY of these questions would be amazing! :-)

    Thank you! x
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  2. #2
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    Quote Originally Posted by AAM View Post
    Hi everyone! :-)
    You like to smile a lot, .

    Suppose Zeta is any primitive 5th root of unity.

    Then we know that Q(Zeta) has a basis {1,Zeta,Zeta^2,Zeta^3} because Zeta is algebraic over Q.

    But how can I show that {Zeta,Zeta^-1,Zeta^2,Zeta^-2} is a basis? :-s
    Remember that $\displaystyle \zeta^{-1} = \zeta^4$ and $\displaystyle \zeta^{-2} = \zeta^3$.

    Also, how does the Galois Group of Q(Zeta) act on this new basis?
    The Galois group, $\displaystyle G=\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ is isomorphic to the cyclic group $\displaystyle \mathbb{Z}_4$. Let $\displaystyle \sigma : \mathbb{Q}(\zeta) \to \mathbb{Q}(\zeta)$ be the automorphism defined by $\displaystyle \sigma (\zeta) = \zeta^2$. Then $\displaystyle G = \left< \sigma \right>$. Since you know $\displaystyle \sigma (\zeta) = \zeta^2$ you can determine what $\displaystyle \sigma (\zeta^k)$ is for $\displaystyle k=2,3,4$.

    Furthermore, how can you prove the existence of a unique subfield E such that [Q(Zeta):E]=2, & find all a,b,c,d such that aZeta+bZeta^-1+cZeta^2+dZeta^-2 is in E but not Q? (apparently you can then show that E = Q(Sqrt(5)) !! :-o )
    $\displaystyle \zeta - \zeta^2 - \zeta^3 + \zeta^4= \sqrt{5}$
    Therefore, $\displaystyle E = \mathbb{Q}(\sqrt{5})$ is a subfield of $\displaystyle \mathbb{Q}(\zeta)$.
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