## Compute the degree of a given extension of the rationals

Compute the degrees of $\mathbb{Q}(\sqrt{3+4i} + \sqrt{3-4i})$ and $\mathbb{Q}(\sqrt{1+\sqrt{-3}} + \sqrt{1-\sqrt{-3}})$ over $\mathbb{Q}$.

Let $\zeta = \sqrt{3+4i} + \sqrt{3-4i}$. Evidently, $\zeta^2 = 16$. (WolframAlpha agrees.) That is, $\zeta$ is a root of $p(x) = x^2 - 16 = (x+4)(x-4)$. Thus the minimal polynomial of $\zeta$ over $\mathbb{Q}$ has degree 1, and the extension $\mathbb{Q}(\zeta)$ has degree 1 over $\mathbb{Q}$.

Similarly, let $\eta = \sqrt{1+\sqrt{-3}} + \sqrt{1-\sqrt{-3}}$. Evidently, $\eta^2 = 6$ (WolframAlpha agrees), so that $\eta$ is a root of $q(x) = x^2-6$. $q$ is irreducible by Eisenstein’s criterion, and so is the minimal polynomial of $\eta$ over $\mathbb{Q}$. The degree of $\mathbb{Q}(\eta)$ over $\mathbb{Q}$ is thus 2.