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Math Help - Primitive root of unity problem

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    Primitive root of unity problem

    Show that  \sqrt {2} \in \mathbb {Q} [ \omega ] , where  \omega is a primitive 8th root of unity.

    Proof. Now  \omega = e^ {2 \pi i /8} = cos (2 \pi /8 ) + i sin (2 \pi /8 ) . So I have  \omega ^{8} = 1
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    Quote Originally Posted by tttcomrader View Post
    Show that  \sqrt {2} \in \mathbb {Q} [ \omega ] , where  \omega is a primitive 8th root of unity.

    Proof. Now  \omega = e^ {2 \pi i /8} = cos (2 \pi /8 ) + i sin (2 \pi /8 ) . So I have  \omega ^{8} = 1
    If I remember correctly \mathbb{Q}[ \omega ] would be any number of the form
    \sum_{n = 0}^7 \frac{p_n}{q_n}\omega _n
    where p_n and q_n \neq 0 are integers for all n = 0 to 7, and the \omega _n the the eighth roots of 1.

    I can construct
    \sqrt{2} = 2 \left ( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right ) + 2 \left ( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right )

    so \sqrt{2} \in \mathbb{Q}[ \omega ].

    -Dan
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    Quote Originally Posted by topsquark View Post
    If I remember correctly \mathbb{Q}[ \omega ] would be any number of the form
    \sum_{n = 0}^7 \frac{p_n}{q_n}\omega _n
    where p_n and q_n \neq 0 are integers for all n = 0 to 7, and the \omega _n the the eighth roots of 1.
    Maybe you meant, \sum_{n=0}^7 a_n \omega^n rather than \omega_n? (And a_n are rational).
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Maybe you meant, \sum_{n=0}^7 a_n \omega^n rather than \omega_n? (And a_n are rational).
    Actually I was using \{ \omega _n \} as the set of all eighth roots of unity, but yours is the more "approprate" notation if I recall.

    (You know, it's really hard to type "eighth?")

    -Dan
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    Quote Originally Posted by tttcomrader View Post
    Show that  \sqrt {2} \in \mathbb {Q} [ \omega ] , where  \omega is a primitive 8th root of unity.

    Proof. Now  \omega = e^ {2 \pi i /8} = cos (2 \pi /8 ) + i sin (2 \pi /8 ) . So I have  \omega ^{8} = 1
    Well, actually, there are 4 primitive 8th roots of unity ( e^{2 n \pi i /8} \text { for } n=1,3,5,7), so presumably your proof should work for any one of them.

    Let \omega be a primitive 8th root of unity. Note that

    0 = \omega^8 - 1 = (\omega^4 + 1) (\omega^2 + 1) (\omega + 1) (\omega - 1)

    so \omega^4+1 = 0.

    (If we had any other of the factors = 0, say \omega^2+1, then \omega would be an 8th root of unity but not a primitive root. Or, to put it another way, \omega^4+1 is the 8th cyclotomic polynomial.)

    Since  \mathbb {Q} [ \omega ] is a field, z =\omega + \omega^{-1} \in \mathbb {Q} [ \omega ] .
    And
    z^2 = \omega^2+2+\omega^{-2}=\frac{\omega^4+2 \omega^2+1}{\omega^2}=\frac{2 \omega^2}{\omega^2}=2
    so z = \pm \sqrt{2}, hence \sqrt{2} \in \mathbb {Q} [ \omega ] .
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