Finding the irreducible polynomial of $\zeta_6, \zeta_8, \zeta_9$ over the field $\mathbb{Q}(\zeta_3).$ How can I get started on finding the irreducible polynomial of $\zeta_6, \zeta_8, \zeta_9$ over the field over $\mathbb{Q}(\zeta_3)?$ Should I construct field extensions and then use the degrees of the extensions? I have ideas over the field $\mathbb{Q}$, but $\mathbb{Q}(\zeta_3)?$ is throwing me off.
 A: You certainly recall that $[\mathbb Q(\zeta_n):\mathbb Q]=\phi(n)$ and know the corresponding irreducible polynomials over $\mathbb Q$: $$X^2-X+1,\qquad X^4+1,\qquad X^6+X^3+1.$$
Over $\mathbb Q(\zeta_3)$ (that is, in $\mathbb Q(\zeta_3)[X]$) these might becaome reducible and factor nontrivially. Indeed, we expect this for two of the cases since $\zeta_3=\zeta_6^2=\zeta_9^3$ shows that $\mathbb Q(\zeta_3)\subseteq\mathbb Q(\zeta_{6})$, resp. $\subseteq \mathbb Q(\zeta_9)$, from which we conclude $[\mathbb Q(\zeta_6):\mathbb Q(\zeta_3)]=\frac{[\mathbb Q(\zeta_6):\mathbb Q]}{[\mathbb Q(\zeta_3):\mathbb Q]}=1$ and  $[\mathbb Q(\zeta_9):\mathbb Q(\zeta_3)]=\frac{[\mathbb Q(\zeta_9):\mathbb Q]}{[\mathbb Q(\zeta_3):\mathbb Q]}=3$. Can you see the linear / cubic polynomial obeyed by $\zeta_6$ / $\zeta_9$ if we allow coefficients $\in\mathbb Q(\zeta_3)$? Hint: Look at the geometric situation.
What about $\zeta_8$? While it seems "obvious" that in this case $\zeta_3\notin \mathbb Q(\zeta_8)$, we should rely on someting more conclusive. 
We note that $\mathbb Q(\zeta_3,\zeta_8)=\mathbb Q(\zeta_{24})$ and therefore $[\mathbb Q(\zeta_8,\zeta_3):\mathbb Q(\zeta_3)]=\frac{[\mathbb Q(\zeta_{24}):\mathbb Q]}{[\mathbb Q(\zeta_3):\mathbb Q}=4$. Hence the irreducible polynomial is of degree 4 and hence is already the $X^4+1$ we found above.
