# Fundamental Theorem of Symmetric Polynomials

• Dec 2nd 2008, 10:37 AM
andreas
Fundamental Theorem of Symmetric Polynomials
I do not quite understand this theorem.Please ,show me how to convert symmetric polynomial to elementary symmetric.

Show me nontrivial examples please!
• Dec 2nd 2008, 11:49 AM
ThePerfectHacker
Quote:

Originally Posted by andreas
I do not quite understand this theorem.Please ,show me how to convert symmetric polynomial to elementary symmetric.

Show me nontrivial examples please!

Let us work with the field $\mathbb{C}$, it ultimately does not matter because it will work for any field,
I just want to use a specific field to make it clearer to you.

Consider the polynomial $f(x_1,x_2) = x_1^2 + x_2^2$, this is a symmetric polynomial (of two variables) because $f(x_1,x_2)=f(x_2,x_1)$. The elementary symmetric polynomials (of two variables) are $s_1 = x_1+x_2$ and $s_2 = x_1x_2$. We see that $x_1^2 + x_2^2 = x_1^2 + 2x_1x_2 + x_2^2 - 2x_1x_2$. Therefore, $f(x_1,x_2) = s_1^2 - 2s_2$.

In general let $f(x_1,...,x_n)$ be a polynomial in $n$ variables. We say that $f$ is symmetric iff $f(x_{\sigma(1)}, ... , x_{\sigma(n)}) = f(x_1,...,x_n)$ where $\sigma$ is any permutation of $\{1,2,...,n\}$ i.e. $\sigma \in S_n$. For example, the polynomial $f(x_1,...,x_n) = x_1^2 + ... + x_n^2$ is obviously a symmetric polynomial because no matter how we permute those summands we still have the same polynomial. The polynomials $s_1 = x_1 + ... + x_n$, $s_2 = x_1x_2 + x_1x_3 + ... + x_{n-1}x_n$, ... , $s_n = x_1x_2...x_n$ are referred to the elementary symmetric polynomials for $n$ variables. There is a theorem that says that if $f$ is a symmetric polynomial then we can write it terms of elementary symmetric polynomials $s_1,...,s_n$. This is what we want to prove.

Let $K = \mathbb{C}(x_1,...,x_n)$ be the field of rational functions in $n$ variables i.e. the field of $\frac{f(x_1,...,x_n)}{g(x_1,...,x_n)}$ (where $f,g$ are any polynomials they need not be symmetric). This is a field under addition and multiplication of polynomials. Now for any $\sigma \in S_n$ we can define $\sigma \left( \frac{f(x_1,...,x_n)}{g(x_1,...,x_n)} \right) = \frac{f(x_{\sigma(1)},...,x_{\sigma(n)})}{g(x_{\si gma(1)},...,x_{\sigma(n)})}$. It can be show (in a straightforward manner) that $\sigma$ is an automorphism of $K$, so $\sigma \in \text{Aut} (K)$. We can identity $S_n$ therefore as a subgroup of $\text{Aut}(K)$. Now we will define $F = K^{S_n}$ i.e. $F$ is the fixed field under the automorphism subgroup $S_n$. Then $K/F$ is a Galois extension and $\text{Gal}(K/F) = S_n$. Notice that $F$ is the field of all symmetric rational functions. Let $s_1,...,s_n$ be the elementary symmetric polynomials, we wish to show that $\mathbb{C}(s_1,...,s_n) = F$, this will show that any symmetric rational function can be expressed in terms of elementary symmetric polynomials alone, exactly what we are trying to prove. Let $p(t) = t^n - s_1t^{n-1}+s_2t^{n-2} - ... + (-1)^n s_n \in \mathbb{C}(s_1,...,s_n)[t]$. But notice that $p(t) = (t - x_1)(t-x-2)...(t-x_n)$ (just expand RHS to see this). Therefore, $K$ is a splitting field of $p(t)$ over $\mathbb{C}(s_1,...,s_n)$. This forces, $[K: \mathbb{C}(s_1,...,s_n)] \leq n!$*. But $[K:F] = |S_n| = n!$, this implies that $[K:\mathbb{C}(s_1,...,s_n)] \leq n!$ since $\mathbb{C}(s_1,...,s_n)\subseteq F$. This forces, $[K:\mathbb{C}(s_1,...,s_n)] = n!$ and so $\mathbb{C}(s_1,...,s_n) = F$. And this completes the proof.

*)Theorem: Let $K$ be a splitting field of a non-constant polynomial $f(x) \in F[x]$ then $[K:F]\leq n!$.
• Dec 2nd 2008, 01:23 PM
andreas
Thank you! (Bow)