# Thread: Primitive root of unity problem

1. ## 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 $\omega ^{8} = 1$

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

3. Originally Posted by topsquark
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).

4. Originally Posted by ThePerfectHacker
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

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