# Minimal polynomial of $\eta_4$ over $\mathbb{Q}(\eta_2)$.

I know that for each divisor $$d \mid 17-1=16$$ there is a unique subfield $$K_d ⊆ \mathbb{Q}(\zeta_{17})$$ with $$[K_d : \mathbb{Q}] = d$$. Moreover, for $$H_d ≤ (ℤ/17ℤ)^*$$ the subgroup of index $$d$$, and $$\eta_d := \sum\limits_{k ∈ H_d}\zeta_{17}^k$$ the Gaussian period of degree $$d$$, we have $$K_d = \mathbb{Q}(\eta_d).$$ I have already found a closed formula (which I believe to be correct) for the minimal polynomial of $$\eta_d$$ over $$\mathbb{Q}$$, namely, for $$\sigma_k ∈ \text{Gal}(\mathbb{Q}(\zeta_{17})/\mathbb{Q})$$ given by $$\zeta_{17} ↦ \zeta_{17}^k$$: $$f^{\eta_d}_{\mathbb{Q}} = \prod_{kH_d ∈ (\mathbb{Z}/17\mathbb{Z})^*/H_d} (X - \sigma_k(\eta_d)).$$ Now my gut strongly tells me that to find the minimal polynomial of $$\eta_4$$ over $$K_2$$, we need simply replace the whole unit group by $$H_2$$: $$f^{\eta_4}_{\mathbb{Q}(\eta_2)} = \prod_{kH_4 ∈ H_2/H_4} (X - \sigma_k(\eta_4))$$ but I'm not entirely sure this is true. This is in line with the fact that the latter polynomial must divide the former, and we indeed get fewer factors in our product. Am I on the right track here?

The Galois group $$\rm Gal(\mathbf Q(\zeta_{17})/\mathbf Q)$$ is isomorphic to $$(\mathbf Z/17\mathbf Z)^\times$$.
Since $$3$$ is a generator of $$(\mathbf Z/17\mathbf Z)^\times$$, $$H_4=\langle 3^4\rangle =\{1,-1,4,-4\}$$ is the unique subgroup of index $$4$$, so $$K_4=\mathbf Q(\eta_4)=\mathbf Q(\zeta+\zeta^{-1}+\zeta^4+\zeta^{-4})$$.
Furthermore, $$H_2=\langle 3^2\rangle=\{-8,-4,-2,-1,1,2,4,8\}$$ is the unique subgroup of index $$2$$, so $$K_2=\mathbf Q(\eta_2)=\mathbf Q(\zeta^{-8}+\zeta^{-4}+\zeta^{-2}+\zeta^{-1}+\zeta+\zeta^2+\zeta^4+\zeta^8)$$.
Since $$[K_4:K_2]=2$$, the minimal polynomial $$f^{\eta_4}_{K_2}$$ is of degree 2.
Check that the automorphism $$\sigma_2:\zeta_{17}\mapsto \zeta_{17}^2$$ fixes $$\eta_4$$. Then you can conclude that $$f^{\eta_4}_{K_2}=(X-\eta_4)(X-\sigma_2(\eta_4))$$.