 Solution

The roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. The first adjunction has degree $2$. To see that the second does too, suppose $\sqrt{3}=u+v\sqrt{2}$ with $u,v\in \mathbb{Q}$. Squaring and comparing rational and irrational parts gives $uv=0$ and $u^2+2v^2=3$. If $v=0$, then $u^2=3$; if $u=0$, then $v^2=3/2$. Neither is a rational square, as follows by comparing prime exponents in numerator and denominator. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. The tower law now yieldsA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.