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



Answered question


A rational function f in n variables is a ratio of 2 polynomials,
where q is not identically 0. The function is called symmetric if
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 hs1=x2+2x3+(n1)xnϵS(h).
3.Can't seem to find a good way to use the other symmetric functions s2,,sn.

Answer & Explanation



Beginner2022-02-16Added 5 answers

According to Galois theory, since SS(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.


Beginner2022-02-17Added 9 answers

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.

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