Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?
How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?