Splitting field of resolvent equals that of $f$ Lemma: Let $\Psi \in k[X_1,...,X_n]=:B$ be s.t. $stab_{S_n}(\Psi)=H \subset S_n$, $S_n/H=\{ \Psi, t_2 \Psi,..., t_e\Psi \}$, $\Delta_\Psi$ the discriminant of $L_\Psi:=\prod_{i=1}^e (X-t_i \Psi)$, and let $B_H$ be the integral closure of $k[\sigma_1,...,\sigma_n]$ in $k(X_1,...,X_n)^H$ ($\sigma_i$ are the elementary symm. polynomials). Then $B_H \subset (1/ \Delta_\Psi)k[\sigma_1,...,\sigma_n][\Psi]$.
Theorem: Let $k$ be an infinite field and $A$ a subring of $k$ s.t. $k=Frac(A)$. Let $\Psi$ be as in the lemma and s.t. $L$ evaluated at the roots $(\rho_1,...,\rho_n)$ of $f$ (separable of degree $n \geq 5$; denote said evaluation by $L_f$) is separable, and assume that $H \notin \{S_n,A_n\}$. Then the splitting field of $L_f$ equals the splitting field $E$ of $f$.
Proof: Let $\Psi_1,...,\Psi_e$ be the conjugates of $\Psi$ over $k(\sigma_1,...,\sigma_n)$ with $\Psi_1=\Psi$. Put $\tilde{\Psi}_i = \Psi_i(\rho_1,...,\rho_n)$. Choose $a_1,...,a_e \in A$ s.t. the $e!$ elements $\lambda_s = \sum_{i=1}^e a_i \tilde{\Psi}_{s(i)}$ are distinct ($s \in S_e$). The natural action of $S_n = Gal(k(X_1,...,X_n)/k(\sigma_1,...,\sigma_n))$ is transitive (1). As $n \geq 5$, the only normal proper subgroups of $S_n$ are $A_n$ and the trivial group. As $e\geq 3$, the above action is faithful (2), which implies that $k(\sigma_1,...,\sigma_n)[\Psi_1,...,\Psi_e]=k(X_1,...,X_n)$ (3).
Let $\Gamma_n$ be the image of $S_n$ in $S_e$ resulting from the above representation. The $n!$ elements $\sum_{i=1}^e a_i \Psi_{s(i)}$ ($s \in \Gamma_n$) of $k(X_1,...,X_n)$ are obviously distinct; therefore $\Phi :=\sum_{i=1}^e a_i \Psi_i$ is a primitive element of $k(X_1,...,X_n)/k(\sigma_1,...,\sigma_n)$ (4). Apply the Lemma with $H=\{id\}$ and with the primitive invariant $\Phi$ of $H$. Here $B_H = B$; thus $B \subset (1/\Delta_\Phi)k[\sigma_1,...,\sigma_n][\Phi]$, where $\Delta_\Phi$ is the $X$-discriminant of $\prod_{s \in \Gamma_n} (X-\sum_{i=1}^e a_i \Psi_{s(i)})$. From $k[\sigma_1,...,\sigma_n][\Phi] \subset k[\sigma_1,...,\sigma_n][\Psi_1,...,\Psi_e]$ we infer
$B \subset (1/\Delta_\Phi)k[\sigma_1,...,\sigma_n][\Psi_1,...,\Psi_e]$, i.e. $\Delta_\Phi B \subset k[\sigma_1,...,\sigma_n][\Psi_1,...,\Psi_e]$. (Eq. 1)
Set $\tilde{\Delta}_\Phi=\Delta_\Phi(\rho_1,...,\rho_n)=(-1)^{n!(n!-1)/2} \prod_{s,t\in \Gamma_n, s\neq t} (\lambda_s-\lambda_t)$. Clearly, by the choice of $\Phi$, we have $\tilde{\Delta}_\Phi \neq 0$. Specializing in Eq. 1, we see that $E = \tilde{\Delta}_\Phi k[\rho_1,...,\rho_n] \subset k[\tilde{\Psi}_1,...,\tilde{\Psi}_e] \subset E$; hence $E=k[\tilde{\Psi}_1,...,\tilde{\Psi}_e].$ (end of proof)
I have many questions but let's begin with the numbered ones:
1) I suppose they mean the action $S_n$ on $\lambda_s$. Doesn't this just come down to claiming that for any pair $\lambda_s,\lambda_t$ there exists $\tau \in S_n$ s.t $\tau \lambda_s = \lambda_t$, or basically just $\tau s = t$? Wouldn't $\tau = ts^{-1} \in S_e$ suffice?
2) This follows from the terms of $\lambda_s$ being distinct, right?
3) Two-fold question: Why does that follow from the preceding text, or rather, what should I read to understand why that follows? (something in Lang, perhaps) Second, we have $k(X_1,...,X_n)^H = k(\sigma_1,...,\sigma_n)(\Psi)$, but why do all the conjugates generate (?) $k(X_1,...,X_n)$? (I should probably know that.)
4) Why? Key material needed to understand this?
Progress: This is thanks to the user @merco's answer in this thread, regarding my question (4); Let $K:=k(X_1,...,X_n)$. Then $K^{S_n}(\Phi) \subset K$ whence $K^{S_n}(\Phi)=K^H$ for some $H \subset S_n$. Since the stabilizer in $S_n$ of $\Phi$ is the trivial group, $K^{S_n}(\Phi)=K^H = K^{\langle id \rangle}=K$. I shall have to investigate the final assertion more closely but it "seems reasonable". This insight should also give me some insight wrt question (3), as well as making the rest of the proof more understandable/"contextual".
Progress 2: I haven't gone over this rigorously, these are just a few thoughts. Assuming that the "natural action" is that of $S_n$ on $T:=\{\Psi_1,...,\Psi_e\}$, the kernel of that action is normal. The kernel is a normal subgroup of $S_n$, hence the trivial, alternating or symmetric group. I think the connecting piece I don't fully understand (how it comes into the picture, I mean) is the fact that the orbits of $T$ partition $T$, these orbits number at least three, whence the kernel must be trivial. I will look more closely into this.
Progress 3: @JackSchmidt solved question (2) for me in another post I made; since the kernel is a normal subgroup of $S_n$ so it's either $A_n$ or trivial, but since $e \geq 3$ it must be trivial. From this explanation, I can now understand @mercio's answer to (3), and so this entire question is answered.
 A: 1) They are talking about the action of $S_n$ on the $\Psi_i$: for any pair $i,j$ :
$\sigma t_i \Psi = t_j \Psi \iff t_j^{-1}\sigma t_i \Psi = \Psi \iff t_j^{-1}\sigma t_i \in H \iff \sigma \in t_j H t_i^{-1}.$ And this set is non-empty.
2) Since $n \ge 5$, the kernel $K$ of the action is either trivial, $A_n$, or $S_n$.
The action is transitive and $e \ge 3$ so there is at least $3$ different actions (one taking $\Psi_1$ ot $\Psi_i$, for $i=1,2,3$), so the kernel can't be $A_n$ nor $S_n$. Hence it is trivial, so the action is faithful.
3) is a consequence of the fundamental theorem of Galois theory. $k(\sigma_i) \subset k(\sigma_i)(\Psi_j) \subset k(X_i)$. Let $H = \{ \sigma \in S_n/ \forall x \in k(\sigma_i)(\Psi_j), \sigma(x) = x\}$. Then, $k(\sigma_i)(\Psi_j) = k(X_i)^H$. If $\sigma \in H$, then $\sigma(\Psi_j) = \Psi_j$, forall $j$, and so $H = K$. But we have just seen that $K = \{id\}$, and so $k(\sigma_i)(\Psi_j) = k(X_i)^K = k(X_i)$.
Finally, $k(\sigma_i)[\Psi_j] = k(\sigma_i)(\Psi_j)$, because $1/ \Psi_j = (\prod_{k \neq j} \Psi_k )/ \prod \Psi_k$, and $\prod \Psi_k \in k(\sigma_i)$.
4) is again a consequence of the fundamental theorem of Galois theory. $k(\sigma_i) \subset k(\sigma_i)(\Phi) \subset k(\sigma_i)(\Psi_j) = k(X_i)$. Let $H = \{ \sigma \in S_n/ \forall x \in k(\sigma_i)(\Phi), \sigma(x) = x\}$. If $\sigma \in H$ then $\sigma(\Phi) = \Phi$. But we have just chosen $\Phi$ such that the $\sigma(\Phi)$ are $n!$ distinct polynomials. Hence $\sigma = id$, and so $H = \{id\}$ and $k(\sigma_i)(\Phi) = k(X_i)$. Finally, we apply the lemma to learn that $\Delta_\Phi k[X_i] \subset k[\sigma_i][\Phi] \subset k[\sigma_i][\Psi_j]$
Finally we specialise $X_i$ in $\rho_i$, and since $E = k(\rho_i) = k[\rho_i]$, we get $E = \tilde \Delta_\Phi E = \tilde \Delta_\Phi k[\rho_i] \subset k[\tilde \Phi] \subset k[\tilde{\Psi_j}] \subset k[\rho_i] = E$. Hence $ E = k[\tilde \Psi_j]$
