Let $f(X)=X^6-X^3-1 \in \mathbb{Q}[X]$ and $\rho=\sqrt[3]{\frac{1+\sqrt{5}}{2}}$.
Given that $f$ is irreducible over the rationals and $w$ the cube root of unity($w^3=1$,$w \neq 1$) i proved that $L=\mathbb{Q}[w,\rho]$ is splitting field of $f$ over $\mathbb{Q}$ and all the roots of $f$ are $\rho,-w/ \rho, -w^2/ \rho,w \rho,w^2 \rho,-1/ \rho$
I also proved the existence of isomorphisms $\sigma$ and $\tau$ such that $\sigma(\rho)=-w/ \rho ,\sigma(w)=w^2$ and $\tau(\rho)=\rho,\tau(w)=w^2$ and the galois group of $L/ \mathbb{Q}$ is isomorphic with the dihedral group of $D_6$.
I've been asked to make a table with the values to $\rho ,w$ of each automorphism-element in the galois group.
But i find a problem in calculating $\sigma^4(\rho)$ (thus and $\sigma^5(\rho)$) because every element in galois group has to permute theo roots of $f$ and $\sigma^4(\rho)$ does not give a root of $f$ .
Can someone help me with this please?
Thank you in advance!
Yes, it does:
$$\sigma\rho:=-\frac\omega\rho\implies \sigma^2\rho=\sigma\left(-\frac\omega\rho\right)=-\frac{\sigma\omega}{\sigma\rho}=-\frac{\omega^2}{-\frac\omega\rho}=\omega\rho\implies$$
$$\sigma^3\rho=\sigma\left(\omega\rho\right)=\omega^2\left(-\frac\omega\rho\right)=-\frac{\omega^3}\rho=-\frac1\rho\implies$$
$$\color{red}{\sigma^4\rho}=\sigma\left(-\frac1\rho\right)=-\frac1{-\frac\omega\rho}=\frac\rho\omega=\color{red}{\omega^2\rho}$$