$\mathbb{Q}(\zeta)$ contains a unique subfield $K$ of degree $10$ over $\mathbb{Q}$?

Question

Let $\zeta$ be a $151$th root of unity, $L = \mathbb{Q}(\zeta)$. How do I see that the cyclotomic field $L$ contains a unique subfield $K$ of degree $10$ over $\mathbb{Q}$? Can we conclude that $\mathcal{O}_K$ has more than $3$ generators over $\mathbb{Z}$?

Answer 1

$L/\mathbb Q$ is a Galois extension with cyclic Galois group. Thus, there is exactly one subgroup for each divisor of the order of the Galois group. Therefore, there is exactly one intermediate field of each divisor of the degree of $L/\mathbb Q$. Since this degree is $\phi(151)=150$, $10$ is allowed.