Exercise 2.2.8

In this exercise, you will prove that if φ F ( x 1 , , x n ) is symmetric, then φ is a rational function in σ 1 , , σ n with coefficients in F . To begin the proof, we know that φ = A B , where A and B are in F [ x 1 , , x n ] . Note that A and B need not be symmetric, only their quotient φ = A B is. Let

C = σ S n { e } σ B ,

where we are using the notation of Exercise 7.

(a)
Use Exercise 7 to show that BC is a symmetric polynomial.
(b)
Then use the symmetry of φ = A B to show that AC is a symmetric polynomial.
(c)
Use φ = ( AC ) ( BC ) and theorem 2.2.2 to conclude that φ is a rational function in the elementary symmetric polynomials with coefficients in F .

Answers

Proof. Let φ = A B F ( x 1 , , x n ) a symmetric rational function:

σ S n , σ φ = σ A σ B = φ = A B .

(a)
Let C = σ S n { e } σ B .

Then

BC = σ S n σ B .

By Exercise 2.2.7, BC is then a symmetric polynomial.

(b)
Note that the rules (2.31) for polynomials extend to rational functions. In particular, if φ = A B , ψ = A 1 B 1 F ( x 1 , , x n ) , and σ S n , σ ( φψ ) = ( σ φ ) ( σ ψ ) .

Indeed,

( σ φ ) ( σ ψ ) = σ A σ B σ A 1 σ B 1 = σ ( A A 1 ) σ ( B B 1 ) = σ ( φψ ) .

Using this property, for all σ S n , from AC = φBC , we obtain

σ ( AC ) = ( σ φ ) ( σ ( BC ) ) = φBC = AC .

So AC is a symmetric polynomial.

(c)
So φ = AC BC is the quotient of two symmetric polynomials, thus there exists h , k F [ x 1 , , x n ] such that φ = AC BC = h ( σ 1 , , σ n ) k ( σ 1 , , σ n ) = ( h k ) ( σ 1 , , σ n ) .

φ F ( σ 1 , , σ n ) is a rational function in the elementary symmetric polynomials with coefficients in F .

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2022-07-19 00:00
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