$\mathbb{Q}(\sqrt2,\sqrt3)=\mathbb{Q}(\lambda\sqrt2 + \mu\sqrt3)$

Question

I am trying to prove that $\mathbb{Q}(\sqrt2,\sqrt3)=\mathbb{Q}(\lambda\sqrt2+ \mu\sqrt3)$ for all $\lambda, \mu \in \mathbb{Q}*$. I understand that $\mathbb{Q}(\sqrt2,\sqrt3)=\mathbb{Q}(\sqrt2+\sqrt3)$, but I am not sure what to do with the $\lambda, \mu$ .

Answer 1

Let $\lambda,\mu\in\Bbb{Q}^{\times}$ and let $\alpha:=\lambda\sqrt{2}+\mu\sqrt{3}$. Then $\alpha\in\Bbb{Q}(\sqrt{2},\sqrt{3})$ so the degree $[\Bbb{Q}(\alpha):\Bbb{Q}]$ divides $[\Bbb{Q}(\sqrt{2},\sqrt{3}):\Bbb{Q}]=4$. If $[\Bbb{Q}(\alpha):\Bbb{Q}]\neq4$ then there exist $u,v\in\Bbb{Q}$ such that $$\alpha^2+u\alpha+v=0,\tag{1}$$ where $$\alpha^2=2\lambda^2+3\mu^2+2\lambda\mu\sqrt{6}.$$ Then comparing the coefficients of $\sqrt{6}$ in $(1)$ shows that $\lambda\mu=0$, a contradiction.

This shows that $[\Bbb{Q}(\alpha):\Bbb{Q}]=4$ and hence that $\Bbb{Q}(\alpha)=\Bbb{Q}(\sqrt{2},\sqrt{3})$.