I have to solve the following problem:
Find $\gamma$ such that $\mathbb{Q}(\gamma)=\mathbb{Q}(\sqrt{-5},\sqrt{2})$
I am a bit confused about how to do it but as $\mathbb{Q}(\sqrt{-5},\sqrt{2})$ has degree $4$,then $\mathbb{Q}(\gamma)$ must also have degree $4$. I think this means that we must be able to write:
$$a+b\gamma + c \gamma^2 + d\gamma^3=\sqrt{-5}\\a'+b'\gamma + c' \gamma^2 + d'\gamma^3=\sqrt{2}$$
And find a suitable $\gamma$, I suspect $\gamma=\sqrt{-5}\sqrt{2}$ but I still can't solve it for $a,b,c,d$.
Consider $\gamma = \sqrt{2} + \sqrt{-5}$. By rationalizing the denominator, it follows that $$\frac{7}{\gamma} = \sqrt{2} - \sqrt{-5}$$ and hence the field $\Bbb Q(\gamma)$ contains $\sqrt{2} - \sqrt{-5}$. Now $$\sqrt{2} = \frac{1}{2}\left(\gamma + \frac{7}{\gamma}\right) \in \Bbb Q(\gamma)$$ and so $\Bbb Q(\gamma)$ contains $\gamma - \sqrt{2}$, or $\sqrt{-5}$. Therefore $\Bbb Q(\gamma) \supset \Bbb Q(\sqrt{2}, \sqrt{-5})$. As $\gamma \in \Bbb Q(\sqrt{2}, \sqrt{-5})$, the reverse containment $\Bbb Q(\gamma) \subset \Bbb Q(\sqrt{2}, \sqrt{-5})$ follows.