10.1 The Symmetric Polynomial Theorem
First here is a definition of polynomials in many variables which have coefficients in a commutative ring. A
commutative ring would be a field except you don’t know that every nonzero element has a multiplicative
inverse. If you like, let these coefficients be in a field. It is still interesting. A good example of
a commutative ring is the integers. In particular, every field is a commutative ring. Thus, a
commutative ring satisfies the following axioms. They are just the field axioms with one omission
mentioned above. You don’t have x−1 if x≠0. We will assume that the ring has 1, the multiplicative
Axiom 10.1.1 Here are the axioms for a commutative ring.
- x + y = y + x, (commutative law for addition)
- There exists 0 such that x + 0 = x for all x, (additive identity).
- For each x ∈ F, there exists −x ∈ F such that x + = 0
, (existence of additive inverse).
z = x +
,(associative law for addition).
- xy = yx,(commutative law for multiplication). You could write this as x × y = y × x.
z = x
,(associative law for multiplication).
- There exists 1 such that 1x = x for all x,(multiplicative identity).
- x =
xy + xz.(distributive law).
Next is a definition of what is meant by a polynomial.
Definition 10.1.2 Let k ≡
where each ki is a nonnegative integer. Let
Polynomials of degree p in the variables x1,x2,
,xn are expressions of the form
where each ak is in a commutative ring. If all ak = 0, the polynomial has no degree. Such a polynomial is
said to be symmetric if whenever σ is a permutation of
An example of a symmetric polynomial is
Another one is
Definition 10.1.3 The elementary symmetric polynomial sk
,n is the coefficient
kxn−k in the following polynomial.
Then the following result is the fundamental theorem in the subject. It is the symmetric polynomial
theorem. It says that these elementary symmetric polynomials raised to powers are a lot like a basis for the
symmetric polynomials. This is a really remarkable result.
Theorem 10.1.4 Let g
be a symmetric polynomial. Then g
polynomial in the elementary symmetric polynomials.
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 obviously true because
s1 = x1 and g
can only be
for some m
. Suppose the theorem is true for n −
By induction, there are unique ak such that
where si′ is the corresponding symmetric polynomial which pertains to x1,x2,
This follows from the definition of these symmetric polynomials. Indeed, the coefficient of xn−k
is the same as the coefficient of x
Thus the only terms which survive in q are those in which kn > 0. That is, all terms have a xnkn,kn > 0.
If a term in the sum has kj = 0, you could switch xn and xj and get a contradiction. Thus all ki≠0. It
and it follows that h
is symmetric of degree no more than
d − n
is the degree of
and is uniquely determined. Thus, if
is symmetric of degree
where h has degree no more than d−n. Now apply the same argument to h
repeatedly obtaining a sequence of symmetric polynomials
of strictly decreasing degree, obtaining
expressions of the form
Eventually hm must be a constant or zero. By induction, each step in the argument yields uniqueness and
so, the final sum of combinations of elementary symmetric functions is uniquely determined.
Note that if you have
then by definition, it is the sum of terms like g
If you replace x
sum over all i,
you would get ∑
which would also be a symmetric
Here is a very interesting result which I saw claimed in a paper by Steinberg and Redheffer on
Lindermannn’s theorem which follows from the above theorem.
Theorem 10.1.5 Let α1,
,αn be roots of the polynomial equation
where each ai is an integer. Then any symmetric polynomial in the quantities anα1,
integer coefficients is also an integer. Also any symmetric polynomial in the quantities α1,
rational coefficients is a rational number.
Proof: Let f
be the symmetric polynomial. Thus
From Theorem 10.1.4 it follows there are integers ak1
kn such that
where the pi are elementary symmetric polynomials defined as the coefficients of
Earlier we had them ± these coefficients. Thus
Now the given polynomial p
is of the form
Thus, equating coefficients, anpk
. Multiply both sides by ank−1.
an integer. Therefore,
and each pk
is an integer. Thus
is an integer as claimed. From this, it
is obvious that
is rational. Indeed, from
Now multiply both sides by anan2an3
an integer. Then
with the right side an integer. Thus f
is rational. If the
had rational coefficients, then mf
would have integer coefficients for a suitable m
and so mf
would be rational which yields