Galois group of an irreducible even quartic ($X^4+b{X}^2+c$)

Let $g(X)=X^4+b{X}^2+c\in \mathbb{Q}[X]$ be irreducible, let $\alpha$ be a root, and put $\delta =\sqrt{c}$. Its roots areso its splitting field is $K=\mathbb{Q}(\alpha ,\delta )$. If $\delta \notin \mathbb{Q}(\alpha )$, then $[K:\mathbb{Q}]=8$ and the Galois group is $D_8$. If $\delta \in \mathbb{Q}(\alpha )$, then $[K:\mathbb{Q}]=4$ and the group is $V_4$ when $\delta \in \mathbb{Q}$, and $C_4$ otherwise.

For irreducible $X^4+b{X}^2+c\in \mathbb{Q}[X]$, the Galois group is $C_2\times C_2$ exactly when $c$ is a square in $\mathbb{Q}$. In the transitive Klein-four action on the roots $\alpha ,-\alpha ,\beta ,-\beta$, all three nonidentity double transpositions fix $\alpha \beta =\sqrt{c}$, forcing it into the fixed field $\mathbb{Q}$.