 Solution

Over $K=\mathbb{Q}$, takeIts roots are $\pm \sqrt{2}\pm \sqrt{3}$. For $\alpha =\sqrt{2}+\sqrt{3}$, its inverse is $\sqrt{3}-\sqrt{2}$, so both square roots belong to $\mathbb{Q}(\alpha )$. Since $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree four, $P$ is irreducible and this field is its splitting field. Independently changing the signs of the two square roots gives its Galois group $C_2\times C_2=V_4$, which is abelian and satisfies the hypothesis.