Prove that $\xi$ , the primitive 12th root of unity, is a zero of the polynomial $t^4-t^2+1$ and the other zeros are $\xi^5, \xi^7, \xi^{11}$

Question

I started with the fact that $\xi$ is a root of $t^{12}-1$. Then I factored $t^{12}-1$ so that $t^4-2t^2+1$ is a factor.

$$t^{12}-1=(t-1)(t^2+t+1)(t+1)(t^2-t+1)(t^2+1)(t^4-2t^2+1)$$

However, I still can't verify that $\xi$ is a root of that particular factor. Any suggestions?

Answer 1

Take it systematically, how did you factor it ? The first step was $$x^{12}-1=(x^{6}-1)(x^{6}+1)$$ now $\zeta$ is not a root of $x^{6}-1$ since it is a primitive 12th root. Thus it is a root of $x^{6}+1$. The next step is $$x^{6}+1=(x^2+1)(x^4-x^2+1)$$ Again $\zeta$ is not a root of $x^2+1$, thus it must be a root of $x^4-x^2+1$.

Answer 2

The primitive $12$-th roots of unity are$^{(\color{red}{*})}$ the roots of $x^{12}-1$ that are not roots of $x^6-1$ or $x^4-1$, i.e. the complex numbers of the form $\exp\left(\frac{2\pi i k}{12}\right)$ with $k\in\{1,5,7,11\}$, that is the set of natural numbers $n\in[1,12]$ such that $\gcd(n,12)=1$. Their minimal polynomial is the $12$-th cyclotomic polynomial $$ \Phi_{12}(x) = \color{red}{\frac{(x^{12}-1)(x^2-1)}{(x^6-1)(x^4-1)}}=\frac{x^6+1}{x^2+1}=x^4-x^2+1=\Phi_6(x^2)=\Phi_3(-x^2). $$