# Fundamental Theorem of Symmetric Polynomials

• Dec 2nd 2008, 09:37 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.

• Dec 2nd 2008, 10:49 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.

Let us work with the field $\displaystyle \mathbb{C}$, 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 $\displaystyle f(x_1,x_2) = x_1^2 + x_2^2$, this is a symmetric polynomial (of two variables) because $\displaystyle f(x_1,x_2)=f(x_2,x_1)$. The elementary symmetric polynomials (of two variables) are $\displaystyle s_1 = x_1+x_2$ and $\displaystyle s_2 = x_1x_2$. We see that $\displaystyle x_1^2 + x_2^2 = x_1^2 + 2x_1x_2 + x_2^2 - 2x_1x_2$. Therefore, $\displaystyle f(x_1,x_2) = s_1^2 - 2s_2$.

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

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

*)Theorem: Let $\displaystyle K$ be a splitting field of a non-constant polynomial $\displaystyle f(x) \in F[x]$ then $\displaystyle [K:F]\leq n!$.
• Dec 2nd 2008, 12:23 PM
andreas
Thank you! (Bow)