Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 16I / b / Solution

Put $a=\sqrt{p}$, $b=\sqrt{-q}$ and $\alpha =a+b$. Since $a^2-b^2=p+q$, the identity $(a+b)(a-b)=p+q$ givesThus $\mathbb{Q}(\alpha )=\mathbb{Q}(\sqrt{p},\sqrt{-q})$. The latter has degree $4$: $p$ is not a square and $\mathbb{Q}(\sqrt{p})$ is real, whereas $\sqrt{-q}$ is not real. Squaring ${\alpha }^2=p-q+2\sqrt{-pq}$ gives its monic degree-$4$ minimal polynomial