362 APPENDIX A. CLASSIFICATION OF REAL NUMBERS

Example A.2.4

(x− x1)(x− x2)(x− x3)

= x3− x2 (x1 + x2 + x3)+ x(x1x2 + x1x3 + x2x3)− x1x2x3.

Thus the symmetric polynomials are x1 + x2 + x3,x1x2 + x1x3 + x2x3, and x1x2x3.

Note that it follows from the above definition that

αksk (x1,x2, · · · ,xn) = sk (αx1, · · · ,αxn)

Then the following result is the fundamental theorem in the subject. It is the symmetricpolynomial theorem. This is a very remarkable theorem.

Theorem A.2.5 Let g(x1,x2, · · · ,xn) be a symmetric polynomial. Then this sym-metric polynomial g(x1,x2, · · · ,xn) equals a polynomial in the elementary symmetric poly-nomials.

g(x1,x2, · · · ,xn) = ∑k

aksk11 · · ·s

knn

and the ak in the commutative ring are unique.

Proof: The proof is by induction on the number of variables. If n = 1, it is obviouslytrue because s1 = x1 and g(x1) can only be a polynomial in x1. Suppose the theorem is truefor n−1 variables and g(x1,x2, · · · ,xn) has degree d. Thus in the sum for the polynomial,|k| ≤ d. By induction, there is a polynomial

Q(s̃1, · · · , s̃n−1) = ∑|k|≤p

aks̃k11 · · · s̃

kn−1n−1 = g(x1,x2, · · · ,xn−1,0) (1.1)

where s̃k is a symmetric polynomial for the variables {x1,x2, · · · ,xn−1} . Now let

p(x1,x2, · · · ,xn)≡ g(x1,x2, · · · ,xn)−Q(s1, · · · ,sn−1) (1.2)

Thus p(x1,x2, · · · ,xn) is a symmetric polynomial because each s j is symmetric and g isgiven to be symmetric. Notice how s̃k was replaced with sk.

If xn is set equal to 0, the right side reduces to 0 because sk (x1,x2, · · · ,xn−1,0) =s̃k (x1,x2, · · · ,xn−1). This follows from the definition of these symmetric polynomials.Indeed, the coefficient of xn−k in (x− x1)(x− x2) · · ·(x− xn−1)(x−0) is the same as thecoefficient of x(n−1)−k in (x− x1)(x− x2) · · ·(x− xn−1) . Thus, the right side of 1.2 reducesto g(x1,x2, · · · ,xn−1,0)−Q(s̃1, · · · , s̃n−1) = 0 from 1.1 when xn = 0.

Thus xn divides p(x1,x2, · · · ,xn) so every term in p(x1,x2, · · · ,xn) has a factor of xn.The same must be true with x j since otherwise, the symmetric polynomial p(x1,x2, · · · ,xn)would change if you switched x j and xn. Hence there exists a symmetric polynomialg1 (x1,x2, · · · ,xn) such that

sng1 (x1,x2, · · · ,xn) = g(x1,x2, · · · ,xn)−Q(s1, · · · ,sn−1)

Recall sn = x1x2 · · ·xn. Thus

g(x1,x2, · · · ,xn) = sng1 (x1,x2, · · · ,xn)+Q(s1, · · · ,sn−1)