Finding an algebraic number $z \in \mathbb{C}$ with Galois group over $\mathbb{Q}(\sqrt{5})$ equal to $\mathbb{Z}/7\mathbb{Z}$

Question

Let $K = \mathbb{Q}(\sqrt{5})$. I'm trying to find an algebraic number $z \in \mathbb{C}$ such that $K(z)/K \cong \mathbb{Z}/7\mathbb{Z}$. How to go about this? The only thing I could think of as useful is the fact that $\mathbb{Q} \subset \mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\zeta_5)$ is a tower of field extensions.

Answer 1

How about this? Take $\mathbb{Q}(\zeta_{29})$. Its Galois group over $\mathbb{Q}$ is cyclic of order 28, so it has a unique subgroup of order $4$; let $L$ be the fixed field of this subgroup.

Now $L$ is linearly disjoint from $K$, so $LK$ (as an extension of $K$) has Galois group $\mathbb{Z}/7\mathbb{Z}$ as desired.