I'm struggling with a question I've got as I can't seem to wrap my head around Galois theory much.
i)Prove that $\zeta^{2}$ is a primitive 8th root of unity and $\zeta^{4}$ is a primitive 4th root of unity. Deduce then that $\zeta$ is a constructible number, where $\zeta$ is primitive 16th root of unity.
I've tried different methods for this now and I'm just confusing myself more unfortunately. We had a question earlier which I was trying to work off since they seemed similar but now I'm not sure. The question was
- If $\zeta$ is the primitive 10th root of unity then show $-\zeta$ is a 5th root of unity.
They answered this by saying:
$1=\zeta^{16}=(\zeta^{5})^2$, thus $\zeta^{5}$ is one of $\pm$1. Since $\zeta$ is a primitive 10th root of unity then $\zeta^{5}=1$. Thus: $(-\zeta)^5=(-1)^5\zeta^{5}=(-1)(-1)=1$ (eqn.2), which shows its a 5th root and the fact that $-\zeta \neq 1$ we get that $-\zeta$ is primitive 5th root of unty.
I have been trying to apply this to my question as its the only similar example weve been given but i cant get my eqn.2 to equal 1.
Am I going about this in the worst way/ just a wrong way? Any help would be appreciated since Galois theory is very tricky for me.
Best Answer
You can prove this directly from definitions. Obviously $\zeta^2$ is an $8$th root of unity and $\zeta^4$ is a $4$th root of unity, so all you need to show is that they're primitive. Suppose they weren't. That would mean, in the one case, that $\zeta^2$ is a $4$th root of unity, which would mean that $\zeta$ is an $8$th root of unity, which would mean that $\zeta$ is not a primitive $16$th root of unity, contrdicting our choice of $\zeta$. An analogous argument works for $\zeta^4$.