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.
According to Galois theory, since $S \subset S(h) \subset F$, $S(h)$ is $F^H$, the field of elements of $F$ fixed by some subgroup $H$ of $S_n$. Since $h$ is only fixed by the identity automorphism, $H = \{id \}$, and $S(h) =F$.
In more detail :
Let $P$ be the minimal polynomial of $h$ over $S$ and let $\sigma$ be in $S_n$, so that $\sigma(h) = x_{i_1} + 2 x_{i_2} + \ldots + n x_{i_n}$. Since the coefficients of $P$ are in $S$, $\sigma(P) = P$, so $0 = \sigma(0) = \sigma(P(h)) = \sigma(P)(\sigma(h)) = P(\sigma(h))$, thus $\sigma(h)$ is also a root of $P$.
Since all the $\sigma(h)$ are pairwise distinct, $P$ has degree at least $n!$, thus the extension $S(h)$ over $S$ is at least of degree $n!$
But $S(h) \subset F$, and $F$ is also of degree $n!$ over $S$, thus those two fields are equal.