9.1. THE SYMMETRIC POLYNOMIAL THEOREM 197
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) (9.2)
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 of 9.2 reduces to 0 because for k ≤ n−1,
sk (x1,x2, · · · ,xn−1,0) = s̃k (x1,x2, · · · ,xn−1)
This follows from the definition of these symmetric polynomials or their description inDefinition 9.1.3. Indeed, the coefficient of xn−k in
(x− x1)(x− x2) · · ·(x− xn−1)(x−0)
is the same as the coefficient of x(n−1)−k in (x− x1)(x− x2) · · ·(x− xn−1) . Thus, the rightside of 9.2 reduces to g(x1,x2, · · · ,xn−1,0)−Q(s̃1, · · · , s̃n−1) = 0 from 9.1 when xn = 0.
Thus xn divides p(x1,x2, · · · ,xn) . In other words, each term in p(x1,x2, · · · ,xn) has afactor of xn. The same must be true with x j since otherwise, the symmetric polynomialp(x1,x2, · · · ,xn) would change if you switched x j and xn. Hence there exists a symmetricpolynomial g1 (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) .
Now if g1 is not constant, do for g1 what was just done for g. Obtain
g(x1,x2, · · · ,xn) = sn (sng2 (x1,x2, · · · ,xn)+Q2 (s1, · · · ,sn−1))
+Q(s1, · · · ,sn−1)
Continue this way, obtaining a sequence of gk till the process stops with some gm beinga constant. This must happen because the degree of gk becomes strictly smaller witheach iteration. This yields a polynomial in the elementary symmetric polynomials for{x1,x2, · · · ,xn}. ■
Example 9.1.5 Let g(x,y) = x3 + y3. It is clear that g(x,y) = g(y,x) so g is a symmetricpolynomial. Write as a polynomial in the elementary functions.
The above proof tells how to do this. First note that x3 = s̃31 where s1 is the symmetric
polynomial associated with the single variable x. Thus p(x,y)= x3+y3−s31 where this s1 is
x+y. Then p(x,y) = x3 +y3− (x+ y)3 = −3x2y−3xy2 and this equals (−xy)(3x+3y) =−3s2s1. Thus −3s1s2 = x3 + y3− s3
1 and so g(x,y) = s31−3s1s2.
You can see that if you have a symmetric polynomial in more variables, you could usea process of reducing one variable at a time in g(x1, ...,xn−1,0) to eventually obtain thisfunction as a polynomial in the symmetric polynomials in variables {x1, ...,xn−1}.
Note that if you have ∏mi=1 (x− xi) then by definition, it is the sum of terms like
g(x1, · · · ,xm)xm−k.