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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- Dec 2nd 2008, 09:37 AMandreasFundamental 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, 10:49 AMThePerfectHacker
Let us work with the field , 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 , this is a symmetric polynomial (of two variables) because . The elementary symmetric polynomials (of two variables) are and . We see that . Therefore, .

In general let be a polynomial in variables. We say that is__symmetric__iff where is any permutation of i.e. . For example, the polynomial is obviously a symmetric polynomial because no matter how we permute those summands we still have the same polynomial. The polynomials , , ... , are referred to the__elementary symmetric polynomials__for variables. There is a theorem that says that if is a symmetric polynomial then we can write it terms of elementary symmetric polynomials . This is what we want to prove.

Let be the field of rational functions in variables i.e. the field of (where are any polynomials they need not be symmetric). This is a field under addition and multiplication of polynomials. Now for any we can define . It can be show (in a straightforward manner) that is an automorphism of , so . We can identity therefore as a subgroup of . Now we will define i.e. is the fixed field under the automorphism subgroup . Then is a Galois extension and . Notice that is the field of all symmetric rational functions. Let be the elementary symmetric polynomials, we wish to show that , 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 . But notice that (just expand RHS to see this). Therefore, is a splitting field of over . This forces, *. But , this implies that since . This forces, and so . And this completes the proof.

*)**Theorem**: Let be a splitting field of a non-constant polynomial then . - Dec 2nd 2008, 12:23 PMandreas
Thank you! (Bow)