A rational function $f$ in $n$ variables is a ratio of $2$ polynomials,
$f(x_1,...x_n) = \frac{p(x_1,...x_n)}{q(x_1,...x_n)}$
where $q$ is not identically $0$. The function is called symmetric if
$f(x_1,...,x_n) = f(x_{\sigma(1)},...,x_{\sigma(n)})$
for any permutation $\sigma$ of $\{1,\ldots,n\}$.
Let $F$ denote the field of rational functions and $S$ denote the subfield of symmetric rational functions. Suppose the coefficients of polynomials are all real numbers.
Show that $F = S(h)$, where $h = x_1 + 2x_2 + ... + nx_n$. In other words, show that $h$ generates $F$ as a field extension of $S$.
Attempt at Solution:
Can't seem to get very far with this one. I know that $F$ is a finite extension of $S$ of degree $n!$ and the Galois group of the extension is $S_n$.
Using $h$ and the 1st symmetric function $s_1 = x_1 + x_2 + \ldots + x_n$, we see that $h - s_1 = x_2 + 2x_3 + \ldots (n-1)x_n \in S(h)$.
Can't seem to find a good way to use the other symmetric functions $s_2,\ldots, s_n$.
Any help would be greatly appreciated. Thank you.