Exercise 4.2.5

Proof.

(a)

The instruction Sage ’factor’ gives the decomposition

$x^{24}-1=(x^8-x^4+1)(x^4-x^2+1)(x^4+1)(x^2+x+1)(x^2-x+1)(x^2+1)(x+1)(x-1)$

(b)

The Sage instructions

     zeta = exp(2*i*pi/24)
     (x^8 - x^4 + 1).subs(x=zeta).expand()

return the value 0.

Thus ${\zeta }_{24}=e^{2\mathit{i\pi }}∕24$ is a root of $x^8-x^4+1$, irreducible over $\mathbb{Q}$ by (a).

$x^8-x^4+1$ is so the minimal polynomial over $\mathbb{Q}$ of ${\zeta }_{24}$.

Verification: ${\zeta }_{24}^8-{\zeta }_{24}^4+1=e^{2\mathit{i\pi }∕3}-e^{\mathit{i\pi }∕3}+1=\omega +{\omega }^2+1=0$.

Note: If we know the cyclotomic polynomials, since $3$ is prime:

$$\begin{array}{llll}\hfill {\Phi }_3(x)& =x^2+x+1,& \hfill & \\ \hfill {\Phi }_6(x)& ={\Phi }_3(-x)=x^2-x+1,& \hfill & \\ \hfill {\Phi }_{24}(x)& ={\Phi }_{\mathrm{rad}(24)}(x^{\frac{24}{\mathrm{rad}(24)}})={\Phi }_6(x^4)=x^8-x^4+1,& \hfill & \end{array}$$

( $24=3\times 2^3,\mathrm{rad}(24)=3\times 2=6$).

${\Phi }_{24}$ is the minimal polynomial of ${\zeta }_{24}$ over $\mathbb{Q}$. The decomposition in (a) is the decomposition

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