 Solution

All roots are $\pm \sqrt{2}$ and $\pm \sqrt{3}$, so the splitting field is $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. We have $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$.

To show that adjoining $\sqrt{3}$ doubles the degree, suppose $\sqrt{3}=a+b\sqrt{2}$ with rational $a,b$. Squaring and using irrationality of $\sqrt{2}$ gives $ab=0$. If $b=0$, then $a^2=3$; if $a=0$, then $b^2=3/2$. Neither is a rational square, as the prime exponents in a rational square are even. Thus $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. Its quadratic polynomial remains irreducible over that field, and the tower law givesA basis is $1,\sqrt{2},\sqrt{3},\sqrt{6}$.