For $k$ between $1$ and $10$ inclusive, let $\lambda_k=\big|1+e^{\frac{k}{11}2\pi i}+e^{\frac{3k}{11}2\pi i}\big|$. What is the degree of the extension ${\mathbb Q}(\lambda_1,\ldots,\lambda_{10})$ over $\mathbb Q$ ?
My thoughts : any individual $\lambda_k$ has minimal polynomial $\lambda^{10} - 12\lambda^8 + 51\lambda^6 - 96\lambda^4 + 80\lambda^2 - 23$. With the help of a computer I learnt that the degree of ${\mathbb Q}(\lambda_1,\lambda_2,\lambda_{3})$ over $\mathbb Q$ is 40, but I was unable to push the computations further.
$\lambda_i^2$ is an algebraic integer of norm $23$ in $\Bbb Z[2\cos(2\pi/11)]$, and so in this ring, the principal ideal $(23)$ splits in the product of principal ideals $\prod_{i=1}^5 (\lambda_i^2)$.
Therefore, the $\lambda_i^2$ are distinct primes and no nontrivial product combination of them can be a square in $\Bbb Z[2\cos(2\pi/11)]$.
This shows that $\Bbb Q((\lambda_i)_{i=1\ldots 10}) = \Bbb Q((\sqrt{\lambda_i^2})_{i=1 \ldots 5})$ has degree $2^5$ over $\Bbb Q(2\cos(2\pi/11))$, and so it has degree $32 \times 5 = 160$ over $\Bbb Q$.