A rational function f in n variables is a ratio of 2 polynomials,f(x1,…xn)=p(x1,…xn)q(x1,…xn)where q is not identically 0. The function is called symmetric iff(x1,…xn)=f(xσ(1),…xσ(n))for any permutation σ of {1,…,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=x1+2x2+…+nxn. 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 Sn.2.Using h and the 1st symmetric function x1=x1+x2+…+xn, we see that h−s1=x2+2x3+…(n−1)xnϵS(h).3.Can't seem to find a good way to use the other symmetric functions s2,…,sn.

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A rational function f in n variables is a ratio of 2 polynomials,f(x1,…xn)=p(x1,…xn)q(x1,…xn)where q is not identically 0. The function is called symmetric iff(x1,…xn)=f(xσ(1),…xσ(n))for any permutation σ of {1,…,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=x1+2x2+…+nxn. In other words, show that h generates F as a field extension of S.Attempt at Solution:1.Can&#039;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 Sn.2.Using h and the 1st symmetric function x1=x1+x2+…+xn, we see that h−s1=x2+2x3+…(n−1)xnϵS(h).3.Can&#039;t seem to find a good way to use the other symmetric functions s2,…,sn.

razlikaml42 · Accepted Answer

According to Galois theory, since S⊂S(h)⊂F,S(h) is FH, the field of elements of F fixed by some subgroup H of SH. Since h is only fixed by the identity automorphism, H={id}, and S(h)=FZSK.

Pregazzix2a · Accepted Answer

In more detail:  Let P be the minimal polynomial of h over S and let σ be in SH, so that σ(h)=xi1+2xi1+…+nxi1. Since the coefficients of P are in S, σ(P)=P, so 0=σ(0)=σ(P(h))=σ(P)(σ(h))=P(σ(h)), thus σ(h) is also a root of P.  Since all the σ(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)⊂F, and F is also of degree n! over S, thus those two fields are equal.

A rational function f in n variables is a ratio of 2 polynomials, f(x_1,...x_n)=\frac{p(x_1

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