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.


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.

  • $\begingroup$ Fantastic mercio, thank you very much. To think I spent quite a bit of time manipulating symmetric polynomials! $\endgroup$ – Conan Wong Aug 17 '12 at 8:39
  • $\begingroup$ @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)! :) $\endgroup$ – Erik Vesterlund Jul 18 '13 at 16:06
  • 1
    $\begingroup$ @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 $\endgroup$ – mercio Jul 18 '13 at 16:53
  • 1
    $\begingroup$ @mercio Indeed you are right, d'oh... I mentioned you in my progress here: math.stackexchange.com/questions/444973/… $\endgroup$ – Erik Vesterlund Jul 18 '13 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.