TITLE: Prove that $f = q_1 + Gq_2$ for some $q_1, q_2 \in \mathbb{k}_{sym}(x_1,\dots,x_n)$
QUESTION [5 upvotes]: Suppose the orbit of the function $f \in \mathbb{k}(x_1,\dots,x_n)$ under the action of $\{\phi_\sigma\mid\sigma \in \mathfrak{S}_n\}$ has length $2$. Prove that $f = q_1 + Gq_2$ for some $q_1, q_2 \in \mathbb{k}_{\text{sym}}(x_1,\dots,x_n)$, where $G = W_n$ in case of $\mathrm{char}(\mathbb{k}) \neq 2$ and $G=F$ in case of  $\mathrm{char}(\mathbb{k}) = 2$.
$$F(x_1, \dots, x_n) = \sum_{\sigma \in \mathfrak{A}_n} \prod_{i=1}^nx^{i-1}_{\sigma(i)}$$
$$W_n(x_1, \dots, x_n)= \prod_{1 \leq i < j \leq n}(x_i - x_j)$$
Honestly, i'm not really good at Galois theory, so i would be glad to hear any suggestions.

REPLY [3 votes]: The key to this exercise is the bit of knowledge that the alternating group $A_n$ is the only subgroup of index two in the full symmetric group $S_n$. For $n\ge5$ this follows from the simplicity of $A_n$, and for $n<5$ it can be checked case-by-case. Leaving it to you to fill in the details to those claims. Anyway, if $f$ has an orbit of size two, we can conclude that the stabilizer of $f$ must be equal to $A_n$.
Let $K$ be the fixed field of $S_n$, and let $L$ be the fixed field of $A_n$. 
Assume first the characteristic is $\neq2$. Show that

$[L:K]=2$.
If $R$ is any polynomial such that $R\in L\setminus K$, then $\{1,R\}$ is a $K$-basis for $L$.
Show that $W\in L\setminus K$.
Figure out why steps 2. and 3. do it.

When characteristic is equal to two, then the above argument does not work, because $W\in K$ in that case. But you can use $F$ in its place in item 3.