# Rational expressibility of polynomials which are symmetric under the permutation of all variables except one fixed variable

I am trying to prove the following statement of early group theory / classic galois theory:

Let $$\sigma_1,\ldots,\sigma_n$$ be the elementary symmetric polynomials of $$x_1,\ldots,x_n$$.

If $$f$$ is a rational function of $$x_1,\ldots,x_n$$ that is symmetric under all permutations of the $$x_i$$’s that fix $$x_1$$, then it is expressible as a rational function of $$\sigma_1,\ldots,\sigma_n$$ and $$x_1$$.

The theorem is stated in the paper The fundamental theorem on symmetric polynomials: History’s first whiff of galois theory as theorem 7. But no proof is given.

In the paper Galois for 21st-Century Readers, footnote 4, I found the hint that every symmetric polynomial in the roots $$x_2,\ldots,x_n$$ can be expressed rationally in terms of $$x_1$$.

This is explained by showing that the elementary symmetric polynomials of $$x_2,\ldots,x_n$$ can be written as polynomials of $$x_1$$. However I do not completely understand, why this is the case.

I get that if you multiply out the left side of the equation $$(x-x_2)\cdot\ldots\cdot(x-x_n)=\frac{f(x)}{(x-x_1)}$$ you get a polynomial where the coefficients are exactly the elementary symmetric polynomials of $$x_2,\ldots,x_n$$. But I do not understand why the coefficients of the polynomial on the right side can be expressed as polynomials of $$x_1$$.

What does the polynomial reminder theorem have to do with this? Yes, $$(x-x_1)$$ divides $$f(x)$$ without reminder. $$f(x_1)=0$$. So what?

Let $$\tau_1, \tau_2,\dots,\tau_{n-1}$$ be the elementary symmetric polynomials in $$x_2,x_3,\dots,x_n$$. It is straightforward to show that $$\sigma_k=\tau_k + x_1 \cdot \tau_{k-1}$$ for $$k=1,2,\dots,n-1$$ (with $$\tau_0=1\,$$).
It follows that each $$\tau_k$$ can be written as a polynomial $$t_k$$ in $$x_1$$ and $$\sigma_1, \sigma_2,\dots,\sigma_{n-1}$$:
\begin{align} \tau_1 &= \sigma_1 - x_1\,\tau_0 = \sigma_1 - x_1 = t_1(x_1, \sigma_1) \\ \tau_2 &= \sigma_2 - x_1\,\tau_1= \sigma_2 - x_1\,t_1(x_1,\sigma_1) = t_2(x_1, \sigma_1, \sigma_2) \\ \dots \\ \tau_{n-1} &= \sigma_{n-1} - x_1 \, \tau_{n-2} = \sigma_{n-1} - x_1 \, t_{n-2}(x_1, \sigma_1,\dots,\sigma_{n-2}) = t_{n-1}(x_1, \sigma_1,\dots,\sigma_{n-1}) \end{align}
A polynomial that is symmetric under permutations of the $$x_i$$’s that fix $$x_1$$ can then be written as:
\begin{align} f(x) &= \sum_{j=0}^n \;p_j(\tau_1, \tau_2, \dots,\tau_{n-1})\,x_1^j \\ &= \sum_{j=0}^n \;p_j\left(t_1(x_1, \sigma_1), t_2(x_1, \sigma_1,\sigma_2), \dots,t_{n-1}(x_1, \sigma_1,\sigma_2,\dots,\sigma_{n-1})\right)\,x_1^j \\ &= p(x_1, \sigma_1,\sigma_2,\dots,\sigma_{n-1}) \end{align}