Rational functions

In the set of ordered couples of 𝕂[x1, ..., xn] x 𝕂[x1, ..., xn]\{0} we define the following equivalence relation

(f(x1,..., xn), g(x1,..., xn)) ρ (f*(x1,..., xn), g*(x1,..., xn))
⇐⇒
f(x1,..., xn) ⋅ g*(x1,..., xn) = f*(x1,..., xn) ⋅ g(x1,..., xn)

which leads to the following definition

5.11.1 Definition. A rational function in the n indeterminates x1,..., xn, is defined as an element of the quotient set

𝕂[x1, ..., xn] x 𝕂[x1, ..., xn] \ {0}/ ρ

and indicated as

f(x1,..., xn) / g(x1,..., xn),   g(x1,..., xn) ≠ 0

As previuosly done on the field of rational numbers, we introduce on 𝕂[x1, ..., xn] the two following operations

f/g + h/k := (fk + gh)/(k+g)   (addition)

f/g ⋅ h/k := (fh)/(gk)   (multiplication)

hence the set 𝕂[x1, ..., xn] becomes a field which is the field of quotients of the ring 𝕂[x1, ..., xn] (compare with the rationals).

5.11.2 Definition. A rational symmetric function is a rational function invariable with respect to any permutation of its indeterminates.

This definition is well-defined since does not depend on the particular represenetative of the class.

We can exten the fundamental theorem on symmetric functions to rational functions

5.11.3 Theorem. (Fundamental theorem of rational symmetric functions). Every symmetric rational function in the indeterminates x1, ..., xn with coefficients in the field 𝕂 can be represented as rational function in terms of elementary symmetric polynomials σ1, σ2, ..., σn with coefficients in 𝕂.

Prove. Let

f ( x 1 , x 2 , … , x n ) g ( x 1 , x 2 , … , x n ) , g ( x 1 , x 2 , … , x n ) ≠ 0

a rational symmetric function in x1,..., xn.

If we are able to represent f/g as an equivalent rational function p/q we are done, since it would be sufficient then to express the numerator and denominator in terms of elementary symmetric functions using the fundamental theorem of symmetric polynomials.

Let g1, ..., gn!−1 be the n!−1 various distinct polynomials (other than g) obtained from g by permutation of the indeterminates. The product g g1gn!−1 is then symmetric since any permutation of the indeterminates merely permutes the factors. By multiplying both numerator of f/g by this product of function, we have

f⋅g1⋅gn!−1 / g⋅g1⋅gn!−1

the denominator is clearly a symmetric polynomial. Since this rational function is equivalent to the initial one itfollows that the polynomial fg1gn!−1 is symmetric too. □

The set SĖƒ of all symmetric rational functions with coefficients in the field 𝕂 in the indeterminates x1,..., xn is a field contained in 𝕂[x1, ..., xn]. We just proved that

SĖƒ = 𝕂[x1, ..., xn]

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