For a course on Galois theory, we proved the fundamental theorem of symmetric polynomials, which states that every symmetric polynomial can be uniquely written as a polynomial in the elementairy symmetric polynomials. As an exercise we were asked to prove that every symmetric rational function also is a rational function in the elementary symmetric polynomials.For this I have the following (partial) proof. Let K be a field, let fϵK(T1,…,Tn) be a symmetric function. We can write f=gh with g,h co'. Let σϵSn, then σ(g)σ(h)=σ(gh)=gh, so σ(g)h=σ(h)g. Since g and h are co', it follows that g∣σ(g), and since the total degrees of the polynomials are the same it follows there exists some kϵeK× such that σ(g)=kg. Similarly for h.Of course I would like to conclude from this that we must have σ(g)=g since σ doesn't change the coefficients, but I realised the argument is more subtle than that, since this is not necessarily true. For example, with K=FF7 and n=3 we have(321)⋅(X1X2+2X2X3+4X3X1)=2⋅(X1X2+2X2X3+4X3X1).So, does anyone know how to proceed? Ofcourse, once you have σ(g)=g and σ(h)=h for all σϵSn the result follows from the fundamental theorem of symmetric polynomials.As a sidenote: I know there's a proof using the fundamental theorem of Galois theory, but I would like to finish a proof using this approach.

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For a course on Galois theory, we proved the fundamental theorem of symmetric polynomials, which states that every symmetric polynomial can be uniquely written as a polynomial in the elementairy symmetric polynomials. As an exercise we were asked to prove that every symmetric rational function also is a rational function in the elementary symmetric polynomials.For this I have the following (partial) proof. Let K be a field, let fϵK(T1,…,Tn) be a symmetric function. We can write f=gh with g,h co&#039;. Let σϵSn, then σ(g)σ(h)=σ(gh)=gh, so σ(g)h=σ(h)g. Since g and h are co&#039;, it follows that g∣σ(g), and since the total degrees of the polynomials are the same it follows there exists some kϵeK× such that σ(g)=kg. Similarly for h.Of course I would like to conclude from this that we must have σ(g)=g since σ doesn&#039;t change the coefficients, but I realised the argument is more subtle than that, since this is not necessarily true. For example, with K=FF7 and n=3 we have(321)⋅(X1X2+2X2X3+4X3X1)=2⋅(X1X2+2X2X3+4X3X1).So, does anyone know how to proceed? Ofcourse, once you have σ(g)=g and σ(h)=h for all σϵSn the result follows from the fundamental theorem of symmetric polynomials.As a sidenote: I know there&#039;s a proof using the fundamental theorem of Galois theory, but I would like to finish a proof using this approach.

Kathryn Duggan · Accepted Answer

Let fϵQ(T1,…,Tn) be a symmetric rational function, then we can write f=gh with g,hϵZ[T1,…,Tn].If h is a symmetric polynomial, then g=hf is symmetric too. By the fundamental theorem of symmetric polynomials, f is a rational function in the elementary symmetric polynomials.If h is not a symmetric polynomial, consider h′=∏σϵSn{e}σh. Then hh&#039; is a symmetric polynomial by construction. By writing f=gh=gh′hh′, we reduce ourselves to the previous case.

For a course on Galois theory, we proved the fundamental theorem of symmetric polynomials, which sta

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