A question about the field of rational functions

Question

Let $G$ be a finite group with $n$ elements. Let $f$ be an element from $\mathbb{Q}(x_1,\dots,x_n)^G$. Let $e_1,\dots,e_n$ be the elementary symmetric functions on $x_1,\dots,x_n$. Is it true, that there exist polynomials $p,q$ from $\mathbb{Q}[x_1,\dots,x_n]$ such that $f=p(e_1,\dots,e_n)/q(e_1,\dots,e_n)$ ?
If so, what is the proof?

Answer 1

This is not true. If you take $G=S_n$ with the standard action on $x_1,...,x_n$, then the fundamental theorem of symmetric polynomial states that $\mathbb{Q}(x_1,...,x_n)^{S_n}=\mathbb{Q}(e_1,...,e_n)$, and moreover $\mathbb{Q}(x_1,...,x_n)/\mathbb{Q}(x_1,...,x_n)^{S_n}$ is Galois of degree $n!$. If you take a smaller group $G\lneq S_n$ that acts on these $n$ variables, then $\mathbb{Q}(x_1,...,x_n)/\mathbb{Q}(x_1,...,x_n)^{G}$ is still Galois, but of degree $|G|<|S_n|$ so that $\mathbb{Q}(x_1,...,x_n)^{G}$ properly contains $\mathbb{Q}(x_1,...,x_n)^{S_n}=\mathbb{Q}(e_1,...,e_n)$.