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

$$f(x_1,...x_n) = \frac{p(x_1,...x_n)}{q(x_1,...x_n)}$$

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

$$f(x_1,...,x_n) = f(x_{\sigma(1)},...,x_{\sigma(n)})$$

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

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 $h = x_1 + 2x_2 + ... + nx_n$. In other words, show that $h$ generates $F$ as a field extension of $S$.

Attempt at Solution:

  • 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 $S_n$.

  • Using $h$ and the 1st symmetric function $s_1 = x_1 + x_2 + \ldots + x_n$, we see that $h - s_1 = x_2 + 2x_3 + \ldots (n-1)x_n \in S(h)$.

  • Can't seem to find a good way to use the other symmetric functions $s_2,\ldots, s_n$.

Any help would be greatly appreciated. Thank you.

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1 Answer 1

up vote 1 down vote accepted

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

In more detail :

Let $P$ be the minimal polynomial of $h$ over $S$ and let $\sigma$ be in $S_n$, so that $\sigma(h) = x_{i_1} + 2 x_{i_2} + \ldots + n x_{i_n}$. Since the coefficients of $P$ are in $S$, $\sigma(P) = P$, so $0 = \sigma(0) = \sigma(P(h)) = \sigma(P)(\sigma(h)) = P(\sigma(h))$, thus $\sigma(h)$ is also a root of $P$.

Since all the $\sigma(h)$ are pairwise distinct, $P$ has degree at least $n!$, thus the extension $S(h)$ over $S$ is at least of degree $n!$

But $S(h) \subset F$, and $F$ is also of degree $n!$ over $S$, thus those two fields are equal.

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Fantastic mercio, thank you very much. To think I spent quite a bit of time manipulating symmetric polynomials! –  Conan Wong Aug 17 '12 at 8:39
    
@mercio You say according to GT; could you provide a source? I suppose what you say at least follows from results I have in my various GT books but I haven't seen it formulated as clearly as you did above. I need it for [math.stackexchange.com/questions/444973/… question)! :) –  Erik Vesterlund Jul 18 '13 at 16:06
    
@ErikVesterlund : the extension $S \subset F$ is Galois (with Galois group $S_n$, acting on the $x_i$), so there is a Galois correspondence between subgroups of $S_n$ and intermediate fields. $S(h)$ is such a field, so it has a corresponding subgroup $H$ of $S_n$ such that $S(h) = F^H$. That correspondance should be in every book about galois theory. See en.wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory –  mercio Jul 18 '13 at 16:53
    
@mercio Indeed you are right, d'oh... I mentioned you in my progress here: math.stackexchange.com/questions/444973/… –  Erik Vesterlund Jul 18 '13 at 17:42

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