Find $a \in \mathbb C$ such that $\mathbb Q (\zeta_5)= \mathbb Q(\sqrt 5, a)$ and $\left[ \mathbb Q(a): \mathbb Q \right]=2 $.

Question

Let $\zeta_5= e^{\frac{2}{5} \pi i}= \dfrac{1}{4}(\sqrt 5 -1)+ i \left( \sqrt{\dfrac{5}{8}+ \dfrac{\sqrt 5}{8}} \right)$ be the principal fifth root of the unity. I know that $\left[\mathbb Q (\zeta_5): \mathbb Q \right]=4$ because the minimal polynomial is the cyclotomic polynomial $\Phi_5(x)=x^4+x^3+x^2+x+1$. I know also that $\mathbb Q (\sqrt 5)$ is a subextension of $\mathbb Q(\zeta_5)$ and $\left[ \mathbb Q (\sqrt 5): \mathbb Q \right]=2$.
I'm struggling to find $a \in \mathbb C$ such that $\mathbb Q (\zeta_5)= \mathbb Q(\sqrt 5, a)$ and $\left[ \mathbb Q(a): \mathbb Q \right]=2 $.

Answer 1

That number $a$ does not exist. Note that the Galois group of the fifth-cyclotomic extension is isomorphic to $\mathbb{Z}_4$, who has only one subgroup of order 2. Hence the only subfield of $\mathbb{Q}(\zeta)$ is $\mathbb{Q}(\sqrt5)$.