# Conjugates of elements of given field

Let $$\zeta=\zeta_{77}$$. For each of the following, find all conjugates of the given element over the given field.

i) of $$\zeta^{44}$$ over $$\mathbb{Q}$$.

ii) of $$\zeta$$ over $$\mathbb{Q}(\zeta^{44})$$.

Proof:

i) We see $$\zeta^{44}$$ is the primitive $$7$$th root of unity. Hence $$[\mathbb{Q}(\zeta^{44}):\mathbb{Q}]=\varphi(7)=6$$ and hence $$\text{irr}(\zeta^{44},\mathbb{Q},x)=\Phi_7(x)=x^6+\dots+x+1$$ and hence the conjugates of $$\zeta^{44}$$ over $$\{\mathbb{Q}$$ are $$\zeta^{44i}: 1\leq i<7\}$$.

ii) Since $$\zeta^{44}$$ is the primitive $$7$$th root of unity and taking into account that $$[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\varphi(n)$$ we see that $$[\mathbb{Q}(\zeta):\mathbb{Q}(\zeta^{44})]=10$$ and it follows that $$\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}(\zeta^{44}))\cong \mathbb{Z}_{10}$$. Suppose its Galois group is generated by $$\sigma$$ then $$\sigma:\zeta \mapsto \zeta^i$$ where $$(i,77)=1$$. Since $$o(\sigma)=10$$ then $$i^{10}\equiv 1 \pmod {77}$$. Also since $$\sigma: \zeta^{44}\mapsto \zeta^{44}$$ then it follows that $$i\equiv 1 \pmod {7}$$ and then $$i\in \{1,8,15,29,36,43,50,57,64,71\}$$.

That's all what I was able to do in this part of the problem? Is there any other way to solve this problem or how to complete my solution without messy computations?

Would be very grateful for answer!