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Math Help - Fundamental Theorem of Symmetric Polynomials

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by andreas View Post
    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!.
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    Thank you!
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