Symmetric rational function is a rational function in the elementary symmetric polynomials.

Question

For a course on Galois theory, we proved the fundamental theorem of symmetric polynomials, which states that every symmetric polynomial can be uniquely written as a polynomial in the elementairy symmetric polynomials. As an exercise we were asked to prove that every symmetric rational function also is a rational function in the elementary symmetric polynomials.

For this I have the following (partial) proof. Let $K$ be a field, let $f \in K(T_1, \dots, T_n)$ be a symmetric function. We can write $f = g/h$ with $g, h$ coprime. Let $\sigma \in S_n$, then $\sigma(g)/\sigma(h) = \sigma(g/h) = g/h$, so $\sigma(g)h = \sigma(h)g$. Since $g$ and $h$ are coprime, it follows that $g \mid \sigma(g)$, and since the total degrees of the polynomials are the same it follows there exists some $k \in K^\times$ such that $\sigma(g) = kg$. Similarly for $h$.

Ofcourse I would like to conclude from this that we must have $\sigma(g) = g$ since $\sigma$ doesn't change the coefficients, but I realised the argument is more subtle than that, since this is not necessarily true. For example, with $K = \mathbb{F}_7$ and $n = 3$ we have $$ (3 2 1) \cdot (X_1 X_2 + 2 X_2 X_3 + 4 X_3 X_1) = 2 \cdot (X_1 X_2 + 2 X_2 X_3 + 4 X_3 X_1). $$ So, does anyone know how to proceed? Ofcourse, once you have $\sigma(g) = g$ and $\sigma(h) = h$ for all $\sigma \in S_n$ the result follows from the fundamental theorem of symmetric polynomials.

As a sidenote: I know there's a proof using the fundamental theorem of Galois theory, but I would like to finish a proof using this approach.

Answer 1

For $\sigma \in S_n$, let $\lambda_\sigma \in K^\times$ be such that $\sigma(g) = \lambda_\sigma g$ and $\sigma(h) = \lambda_\sigma h$. Then $\sigma \mapsto \lambda_\sigma$ is a group homomorphism $\lambda : S_n \to K^\times$. (When $n \geq 6$, there are at most two such homomorphisms.)

We want to show that $\lambda$ is trivial. Suppose not, and let $(i, j)$ be a transposition that does not lie in the kernel of $\lambda$. We will arrive at a contradiction with the fact that $g$ and $h$ are coprime. Because $\lambda_{(i, j)} \neq 1$, we obtain that $g(T_1, \ldots, \widehat T_j, T_i, \ldots, T_n) = 0$. (We substituted $T_i$ for $T_j$.) When $g$ is considered as a polynomial over $K((T_k)_{k \neq j})$, this means that $T_i$ is a root of $g$. Thus $g$ is divisible by $(T_i - T_j)$. Similarly, $h$ is divisible by $(T_i - T_j)$. But by Gauss's lemma, $g$ and $h$ are coprime in $K((T_k)_{k \neq j})[T_j]$, a contradiction.

Answer 2

Let $f\in \mathbf{Q}(T_1,\ldots,T_n)$ be a symmetric rational function, then we can write $f=\frac{g}{h}$ with $g,h\in \mathbf{Z}[T_1,\ldots,T_n]$.

If $h$ is a symmetric polynomial, then $g=hf$ is symmetric too. By the fundamental theorem of symmetric polynomials, $f$ is a rational function in the elementary symmetric polynomials.

If $h$ is not a symmetric polynomial, consider $h'=\prod_{\sigma\in S_n\setminus \{e\}} \sigma h$. Then $hh'$ is a symmetric polynomial by construction. By writing $f=\frac{g}{h}=\frac{gh'}{hh'}$, we reduce ourselves to the previous case.