Solution
= Solution
Over $K=\mathbb Q$, take
$$
\boxed{P(X)=X^4-10X^2+1.}
$$
Its roots are $\pm\sqrt2\pm\sqrt3$. For $\alpha=\sqrt2+\sqrt3$, its inverse is $\sqrt3-\sqrt2$, so both square roots belong to $\mathbb Q(\alpha)$. Since $\mathbb Q(\sqrt2,\sqrt3)$ 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.