# Primitive 16th root of unity

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.

$$1=\zeta^{16}=(\zeta^{5})^2$$, 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.
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 $$\zeta^4$$.