Find a monic quartic polynomial over $\mathbb{Z}$ whose Galois group is $V_4$
Specifically, $V_4$ here means the subgroup $\bigl\langle (1\,2)(3\,4), (1\,3)(2\,4) \bigr\rangle$ in $S_4$.
My attempt is as following: suppose that $f$ is such a polynomial, that $x_1,x_2,x_3,x_4$ are its roots, and that $G=V_4$ is its Galois group. WLOG, we can also assume that $x_1+x_2+x_3+x_4=0$. Now define $y_{ij}=x_i+x_j$ for any $i\neq j$, so $Gy_{ij}=\{y_{ij},-y_{ij}\}$. Thus, $G$ fixes $y^2_{ij}$, i.e. $y^2_{ij}\in \mathbb{Q}$.
So we can compute the resolvent cubic $$ g = (x-y^2_{12})(x-y^2_{13})(x-y^2_{14}) = x^3+2a_2^{\,}x^2+(a^2_2-4a_0^{\,})-a_1^2 $$ where $f=x^4+a^2x^2+a_1x+a_0$. Hence, we can express $a_2,a_1,a_0$ using $y^2_{12},y^2_{13},y^2_{14}$ as $a_1 = y_{12}y_{13}y_{13}$, $a_2 = \smash{-\frac 1 2(y^2_{12}+y^2_{13}+y^2_{14})}$ and $a_0 = \smash{\frac{1}{4} (a_2^2-y^2_{12}y^2_{13}-y^2_{12}y^2_{14}-y^2_{13}y^2_{14})}$.
So conversely, we can pick $y_{12},y_{13},y_{14}\in \mathbb{R}$ such that $a_0,a_1,a_2$ are all integers, so the $f$ induced would have a Galois group $\leq V_4$.
But subgroups of $V_4$ are $\{e\}$, $\bigl\{(1\,2)(3\,4)\}, \{(1\,3)(2\,4)\}, \{(1\,4)(2\,3) \bigr\}$. But in either case, $f$ will be reducible and split either into linear factors or two quadratic factors. So we only need the $f$ induced to be irreducible. However, I struggle to pick such $y_{ij}$. Is there any trick or general method to find such $y_{ij}$?
