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Thread: Primitive root of unity problem

  1. #1
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    Primitive root of unity problem

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

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

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

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

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

    -Dan
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    Quote Originally Posted by topsquark View Post
    If I remember correctly $\displaystyle \mathbb{Q}[ \omega ]$ would be any number of the form
    $\displaystyle \sum_{n = 0}^7 \frac{p_n}{q_n}\omega _n$
    where $\displaystyle p_n$ and $\displaystyle q_n \neq 0$ are integers for all n = 0 to 7, and the $\displaystyle \omega _n$ the the eighth roots of 1.
    Maybe you meant, $\displaystyle \sum_{n=0}^7 a_n \omega^n$ rather than $\displaystyle \omega_n$? (And $\displaystyle a_n$ 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, $\displaystyle \sum_{n=0}^7 a_n \omega^n$ rather than $\displaystyle \omega_n$? (And $\displaystyle a_n$ are rational).
    Actually I was using $\displaystyle \{ \omega _n \}$ 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 $\displaystyle \sqrt {2} \in \mathbb {Q} [ \omega ] $, where $\displaystyle \omega $ is a primitive 8th root of unity.

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

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

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

    so $\displaystyle \omega^4+1 = 0$.

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

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