abstract algebra - The degree of the field extension $mathbb{Q}(sqrt{5},w): mathbb{Q}$

Compute the degree of the field extension $$\mathbb{Q}(\sqrt{5},w): \mathbb{Q} ,$$ where $ w = e^{2\pi i / 3}$.

I consider the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt{5})(w)$ now, $\mathbb{Q}: \mathbb{Q}(\sqrt{5})$ has degree $2$, so I am trying to find the degree of $\mathbb{Q}(\sqrt{5}) : \mathbb{Q}(\sqrt{5})(w)$ it is $\leq 2$ since $w $ satisfies $w^2+w+1 = 0$. I am trying to show that it is exactly $2$ - I know that $ w \notin \mathbb{Q}(\sqrt{5})$ but I don't see how I can justify it is exactly $2$ from here.

Answer

Since $[\Bbb Q(\sqrt 5)(w) : \Bbb Q(\sqrt 5)]$ is $\leq 2$ (as $w^2 + w + 1 = 0$), it is either $1$ or $2$, and it is $1$ iff $w \in \sqrt{5}$. Since $\Bbb Q(\sqrt 5) \leq \Bbb R$ but $w \not \in \Bbb R$, we must have $[\Bbb Q(\sqrt 5)(w) : \Bbb Q(\sqrt 5)] = 2$.