I've been trying to solve the following question, but I wasn't able to.
Given $p\in\Bbb Q[x]$ an irreducible quartic polynomial, and $a,b,c,d$ its roots, consider $K=\Bbb Q(a,b,c,d)$ its splitting field, and $G=G(K/\Bbb Q)\subseteq S_4$ its Galois group. Prove that $G\cap V$, with $V=\{\textit{Id},(12)(34),(13)(24),(14)(23)\}$ the Klein group, has $L=\Bbb Q(ab+cd,ac+bd,ad+bc)$ as fixed field.
I see that each element of $V$ fixes one of the three generators of $L$, but that's not enough, since every generator must be fixed. I also know that we can get rid of one of the generators using Vieta's formulas, but there are still two and it's not clear $V$ fixes both of them. I've read something related to this problem about a cubic resolvent, but I've never used that before, so it's unlikely it's necessary for the solution.
Any advice? Thanks in advance.
It follows immediately from the definition of $V$ that it fixes $\alpha= ab+cd, \beta= ac+bd, \gamma= ad+bc$. Conversely, it is easily seen (by tedious checking)that any permutation of the roots $a,b,c,d$ fixing $\alpha, \beta, \gamma$ belongs to $V$. Hence $L=\mathbf Q(\alpha, \beta, \gamma)$ is the fixed field of $G\cap V$, and $G/G\cap V$ is the Galois group of $L/\mathbf Q$.
NB. 1) The "resolvent cubic" of your irreducible quartic $P(X)$ is $(X-\alpha)(X-\beta)(X-\gamma)$. The name comes from the fact that the degree $n$ of the splitting field of the resolvent determines completely $G$, unless when $n=2$, in which case a supplementary computation is necessary. 2) The base field need not be $\mathbf Q$, it could be any field of characterisic $\neq 2$.