Havlishkq

2022-02-15

A rational function f in n variables is a ratio of 2 polynomials,

$f({x}_{1},\dots {x}_{n})=\frac{p({x}_{1},\dots {x}_{n})}{q({x}_{1},\dots {x}_{n})}$

where q is not identically 0. The function is called symmetric if

$f({x}_{1},\dots {x}_{n})=f({x}_{\sigma \left(1\right)},\dots {x}_{\sigma \left(n\right)})$

for any permutation$\sigma$ of $\{1,\dots ,n\}$ .

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$h={x}_{1}+2{x}^{2}+\dots +n{x}_{n}$ . 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 $h-{s}_{1}={x}_{2}+2{x}^{3}+\dots (n-1){x}_{n}\u03f5S\left(h\right)$ .

3.Can't seem to find a good way to use the other symmetric functions$s}_{2},\dots ,{s}_{n$ .

where q is not identically 0. The function is called symmetric if

for any permutation

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

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

2.Using h and the 1st symmetric function

3.Can't seem to find a good way to use the other symmetric functions

razlikaml42

Beginner2022-02-16Added 5 answers

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

Beginner2022-02-17Added 9 answers

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.

Let P be the minimal polynomial of h over S and let

Since all the

But