Havlishkq

2022-02-15

A rational function f in n variables is a ratio of 2 polynomials,
$f\left({x}_{1},\dots {x}_{n}\right)=\frac{p\left({x}_{1},\dots {x}_{n}\right)}{q\left({x}_{1},\dots {x}_{n}\right)}$
where q is not identically 0. The function is called symmetric if
$f\left({x}_{1},\dots {x}_{n}\right)=f\left({x}_{\sigma \left(1\right)},\dots {x}_{\sigma \left(n\right)}\right)$
for any permutation $\sigma$ of $\left\{1,\dots ,n\right\}$.
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 $h={x}_{1}+2{x}^{2}+\dots +n{x}_{n}$. In other words, show that h generates F as a field extension of S.
Attempt at Solution:
1.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 ${S}_{n}$.
2.Using h and the 1st symmetric function ${x}_{1}={x}_{1}+{x}_{2}+\dots +{x}_{n}$, we see that $h-{s}_{1}={x}_{2}+2{x}^{3}+\dots \left(n-1\right){x}_{n}ϵS\left(h\right)$.
3.Can't seem to find a good way to use the other symmetric functions ${s}_{2},\dots ,{s}_{n}$.

### Answer & Explanation

razlikaml42

According to Galois theory, since $S\subset S\left(h\right)\subset F,S\left(h\right)$ is ${F}^{H}$, the field of elements of F fixed by some subgroup H of ${S}^{H}$. Since h is only fixed by the identity automorphism, $H=\left\{id\right\}$, and S(h)=FZSK.

Pregazzix2a

In more detail:
Let P be the minimal polynomial of h over S and let $\sigma$ be in ${S}^{H}$, so that $\sigma \left(h\right)={x}_{{i}_{1}}+2{x}_{{i}_{1}}+\dots +n{x}_{{i}_{1}}$. Since the coefficients of P are in S, $\sigma \left(P\right)=P$, so $0=\sigma \left(0\right)=\sigma \left(P\left(h\right)\right)=\sigma \left(P\right)\left(\sigma \left(h\right)\right)=P\left(\sigma \left(h\right)\right)$, thus $\sigma \left(h\right)$ is also a root of P.
Since all the $\sigma \left(h\right)$ are pairwise distinct, P has degree at least n!, thus the extension S(h) over S is at least of degree n!
But $S\left(h\right)\subset F$, and F is also of degree n! over S, thus those two fields are equal.

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