We are going to be dealing with polynomial
s in n
x1, x2, ..., xn
. Such a polynomial
is called symmetric
if any rearrangement of the variables
leaves the polynomial unchanged.
Formally, a polynomial f(x1,...,xn)
is symmetric iff
for any permutation s in the nth Symmetric group
If n=3, then
If n=1, so there is just one variable, then every polynomial is symmetric.
If n=2, then
- x1 + x2 is symmetric.
- x1x2 is symmetric.
- x1 + (x2)2 is not. (Because
when we swap over x1 and x2 for this
polynomial we get x2 + (x1)2.)
- x1 + x2 is not symmetric because
if we consider the permutation that sends 1 -> 2 -> 3
and apply it to our polynomial we get x2 + x3.
- x1 + x2 + x3 is symmetric
though and so is (x1)2 + (x2)2 + (x3)2.
The elementary symmetric polynomials are defined as follows:
First we take the sum of all the variables
s1=Sum(1<= i <=n) xi.
Thus, s1=x1+x2 + ... + xn.
Next we take the sum of all products of pairs of distinct variables
s2=Sum(1<= i < j <=n) xixj.
- Then we take the sum of all products of three distinct variables
s3=Sum(1<= i < j < k <=n) xixjxk.
Finally we take the product of all of the variables
It is not too hard to see that s1, ..., sn
are symmetric polynomials.
In the case n=2 we get
For n=3 we get
- s1=x1 + x2 + x3.
- s2=x1x2 + x1x3 + x2x3.
It turns out that any symmetric polynomial can be constructed out of these elementary
Theorem Let f(x1,...,xn) be a symmetric
polynomial over a field then there exists a polynomial
g(x1,...,xn) such that
Here is a proof of the fundamental theorem on symmetric polynomials.