$\zeta_5$ has degree $4$ over $\mathbb{Q}$ and $\zeta_7$ has degree $6$ over $\mathbb{Q}$

Question

To give some background, I am attempting to solve the problem from Artin's algebra textbook which was already asked in this link: Showing $\zeta_5 \notin \mathbb{Q}(\zeta_7)$

Now, I would like to prove the following two claims which are not obvious to me. If I can prove these, I will be able to show that $\zeta_5\notin \mathbb{Q(\zeta_7)}$ as needed. I suspect that both claims will have similar proofs.

Claim 1: $\zeta_5$ has degree $4$ over $\mathbb{Q}$

Claim 2: $\zeta_7$ has degree $6$ over $\mathbb{Q}$

Attempt to prove Claim 1

Let $α=\zeta_5$. I would like to find the monic irreducible polynomial for $α$, but unfortunately I do not know what it is yet. I see $α=cos(\frac {2π}5)+isin(\frac {2π}5)$ and $α^4=cos(\frac {4π}5)+isin(\frac {4π}5)$. Please help me to find the irreducible polynomial, and (ideally) to show that it is irreducible.

Answer 1

Hint: The minimal polynomial of $\zeta_n$ is the cyclotomic polynomial $\Phi_n$. When $p$ is prime, $\Phi_p(x)=x^{p-1}+\cdots+x+1$. It is irreducible by Eisenstein's criterion applied to $\Phi_p(x+1)$.

Answer 2

For $\zeta= \zeta_5$ we have $\zeta^4 + \zeta^3 + \zeta^2 + \zeta+1=0$. It follows that $\eta=\zeta+\frac{1}{\zeta}$ is a root of $y^2+ y -1$, an irreducible polynomial (even mod $2$). Therefore, $\eta$ is of degree $2$. Moreover, $\zeta$ is a root of $x^2 - \eta x + 1$. $\zeta$ cannot lie in $\mathbb{Q}(\eta)$, since $\eta$ is real, while $\zeta$ is not. Therefore, $[\mathbb{Q}(\zeta): \mathbb{Q}(\eta)]=2$.

For $\zeta= \zeta_7$ we have a similar argument, with $\eta$ a root of the irreducible $y^3 + y^2 - 2 y - 1$.