<s>[INST] Prove that Q($\sqrt{2}$, $\sqrt{3}$) is a field Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \{a+b\sqrt{2} +c\sqrt{3} +d\sqrt{6}\ |\ a,b,c,d \in \mathbb{Q}\}$ is a field.
I am doing the subfield test, but having trouble in showing how to express the inverse in such a form. Anyone can help?
 [/INST] HINT: $$(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6})=(a+d\sqrt{6})^2-(b\sqrt{2}+c\sqrt{3})^2:=r+s\sqrt{6}$$ where $r,s\in\mathbb{Q}$.   Then $$(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6})(r-s\sqrt{6})=r^2-6s^2=\alpha$$ and $\alpha$ is rational. Hence you have got the inverse.
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