How to show that $\sqrt{5}$ is not in $Q(\sqrt{2},\sqrt{3})$ My strategy is that suppose $\sqrt{5}$ is in $Q(\sqrt{2},\sqrt{3})$, then it should be of the form $a+b\sqrt{2} +c\sqrt{3}+d\sqrt{6}$. After some tedious computation I can get a contradiction. Is there any simpler way to prove this?

One way is to note that $\Bbb Q(\sqrt 5)$ is a subextension of your extension of degree $2$ over $\Bbb Q$. But one can find out exactly all subextensions of $\Bbb Q(\sqrt 2,\sqrt 3)$ of degree $2$ over $\Bbb Q$, and these are $\Bbb Q(\sqrt 2),\Bbb Q(\sqrt 3),\Bbb Q(\sqrt 6)$ (these correspond to the subgroups generated (separately!) by $(1,0),(0,1)$ and $(1,1)$ in $C_2\times C_2$. You're now reduced to showing that $\sqrt 5$ lies in none of those, and this is easier. 
