TITLE: About the construction of resolvents in Galois theory (over $\mathbb{Q}$ in $\mathbb{C}$)
QUESTION [5 upvotes]: I have to say that my question is quite long and I apologize for this. The  main idea is that I would like to show how to construct resolvents for any transitive
subgroup of the permutation group to some students. I present here the whole construction but I think there is a problem at the end and I would be happy if someone could help me to fix it.
I am looking for a way to compute the Galois group of an irreducible monic polynomial. That is, given an integer $n$ and a transitive subgroup of $\mathfrak{S}_n$. Find a way to answer (with a little computation) for each $P\in\mathbb{Q}[X]$ do we have $Gal(P,\mathbb{Q})\subseteq $ some conjugate of the group $G$. 
A resolvent $R_G$ for $G$ is then a polynomial in $n+1$ variables :
$$R_G\in\mathbb{Q}[A_1,...,A_n,X] $$
Such that if :
$$P(Y)=Y^n+\sum_{k=1}^na_kY^{n-k} $$
We have $Gal(P,\mathbb{Q})\subseteq $ some conjugate of the group $G$ if and only if $R_G(a_1,...,a_n,X)$ has a root in $\mathbb{Q}$.
I claim that we all know such an example of resolvent for $G=\mathfrak{A}_n$ :
$$R_{\mathfrak{A}_n}(X)=X^2-Discr(P) $$
Now, imitating the idea for $G=\mathfrak{A}_n$ would go as follows :
Given some $G$ a transitive subgroup of $\mathfrak{S}_n$, first find :
$$U\in\mathbb{Z}[Y_1,...,Y_n]\text{ such that } Stab_{\mathfrak{S}_n}(U)=G  $$
Then using a system $(g_i)$ of representative for $\mathfrak{S}_n/G$ (its cardinal being $d$) :
$$S_G(Y_1,...,Y_n,X):=\prod_{i=1}^d(X-[g_i.U](Y_1,...,Y_n)) $$
Finally :
$$S_G(Y_1,...,Y_n,X)=X^d+\sum_{r=1}^dV_r(Y_1,...,Y_n)X^{d-r} $$
From the first expression of $S_G$ we actually see that it is fixed by every permutation of $Y_1,..,Y_n$ so that each of the $V_r$ is a symmetric polynomial in the variables $Y_1,...,Y_n$.  If we set :
$$A_k=(-1)^k\sum_{1\leq l_1<...<l_k\leq n}Y_{l_1}...Y_{l_k} $$
By the Newton's theorem about symmetric polynomials we finally get that there exists $R_G$ :
$$S_G(Y_1,...,Y_n,X)=R_G(A_1,...,A_n,X) $$
Now I would like to proove that such a $R_G$ is a resolvent for $G$. 
Take :
$$P(Y)=(Y-y_1)...(Y-y_n)=Y^n+\sum_{k=1}^na_kY^{n-k}$$
Suppose that $Gal(P,\mathbb{Q})\subseteq $ some conjugate of $G$, let's say $g_kGg_k^{-1}$ then  we have that for all $\sigma\in Gal(P,\mathbb{Q})$ $g_k^{-1}\sigma g_k$ is in $Stab(U)$ (here I identify the decomposition field of $P$automorphism with the permutation induced on the roots of my polynomial $P$) then we have :
$$\sigma([g_k.U](y_1,...,y_n))=g_k(g_k^{-1}\sigma g_k).U(y_1,...,y_n)=g_k.U(y_1,...,y_n) $$
So we see that the complex number $[g_k.U](y_1,...,y_n)$ is fixed by any automorphism of the decomposition field so it must be rational. This gives the first part.
Now, here comes my problem when you reverse the process you want to  do the following implication (recalling that an automorphism of the decomposition field is at the same time a permutation of the roots and the correspondence is faithfull) because a root of $R_G(a_1,...,a_n,X)$ is supposed to be rational we get for some $k$ :
$$[g_k.U](y_1,...,y_n)\in \mathbb{Q} $$
We see that this means, in particular that $\sigma\in Stab([g_k.U](y_1,...,y_n))$ and I would like to deduce from that $\sigma\in Stab(g_k.U(Y_1,...,Y_n))$. I cannot figure if this works or not in my setting (that is the $(y_1,...,y_n)$ must be the roots of an irreducible polynomial).

REPLY [3 votes]: The proof you are looking for is theorem 5 in
MR0327712 (48 #6054) 
Stauduhar, Richard P.
The determination of Galois groups. 
Math. Comp. 27 (1973), 981–996
The extra ingredient you need is for the resolvent not having repeated roots, that is permutations of the variables would imply a change of the roots.