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.
Definition F.1.1Letk ≡
(k1,k2,⋅⋅⋅,kn)
where each k_{i}is a nonnegative integer. Let
|k | ≡ ∑ ki
i
Polynomials of degree p in the variables x_{1},x_{2},
⋅⋅⋅
,x_{n}are expressions of the form
∑
g(x1,x2,⋅⋅⋅,xn) = akxk11⋅⋅⋅xknn
|k|≤p
where each a_{k}is ina commutative ring. If all a_{k} = 0, the polynomial has no degree. Such apolynomial is said to be symmetric if whenever σ is a permutation of
Then the following result is the fundamental theorem in the subject. It is the symmetric
polynomial theorem. It says that these elementary symmetric polynomials are a lot like a basis for
the symmetric polynomials.
Theorem F.1.3Let g
(x1,x2,⋅⋅⋅,xn)
be a symmetric polynomial. Then g
(x1,x2,⋅⋅⋅,xn )
equalsa polynomial in the elementary symmetric functions.
∑ k k
g(x1,x2,⋅⋅⋅,xn) = aks11⋅⋅⋅snn
k
and the a_{k}are unique.
Proof:If n = 1, it is obviously true because s_{1} = x_{1}. Suppose the theorem is true for n − 1
and g
(x1,x2,⋅⋅⋅,xn)
has degree d. Let
g′(x1,x2,⋅⋅⋅,xn−1) ≡ g(x1,x2,⋅⋅⋅,xn−1,0)
By induction, there are unique a_{k} such that
′ ∑ ′k1 ′kn−1
g (x1,x2,⋅⋅⋅,xn−1) = aks1 ⋅⋅⋅sn−1
k
where s_{i}^{′} is the corresponding symmetric polynomial which pertains to x_{1},x_{2},
⋅⋅⋅
,x_{n−1}. Note
that
′
sk(x1,x2,⋅⋅⋅,xn −1,0) = sk(x1,x2,⋅⋅⋅,xn −1)
Now consider
∑ k1 kn−1
g(x1,x2,⋅⋅⋅,xn)− aks1 ⋅⋅⋅sn−1 ≡ q (x1,x2,⋅⋅⋅,xn)
k
is a symmetric polynomial and it equals 0 when x_{n} equals 0. Since it is symmetric, it is also 0
whenever x_{i} = 0. Therefore,
q (x1,x2,⋅⋅⋅,xn) = snh(x1,x2,⋅⋅⋅,xn)
and it follows that h
(x ,x ,⋅⋅⋅,x )
1 2 n
is symmetric of degree no more than d − n and is uniquely
determined. Thus, if g
(x ,x ,⋅⋅⋅,x )
1 2 n
is symmetric of degree d,
∑
g(x1,x2,⋅⋅⋅,xn) = aksk11⋅⋅⋅sknn−−11 + snh (x1,x2,⋅⋅⋅,xn )
k
where h has degree no more than d − n. Now apply the same argument to h
(x1,x2,⋅⋅⋅,xn )
and
continue, repeatedly obtaining a sequence of symmetric polynomials h_{i}, of strictly decreasing
degree, obtaining expressions of the form
∑ k1 kn−1 kn
g(x1,x2,⋅⋅⋅,xn) = bks1 ⋅⋅⋅sn−1 sn + snhm (x1,x2,⋅⋅⋅,xn )
k
Eventually h_{m} 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. ■
Here is a very interesting result which I saw claimed in a paper by Steinberg and Redheffer on
Lindemannn’s theorem which follows from the above theorem.
Theorem F.1.4Let α_{1},
⋅⋅⋅
,α_{n}be roots of the polynomial equation
p (x ) ≡ anxn + an−1xn− 1 + ⋅⋅⋅+ a1x + a0 = 0
where each a_{i}is an integer. Then any symmetric polynomial in the quantities a_{n}α_{1},
⋅⋅⋅
,a_{n}α_{n}having integer coefficients is also an integer. Also any symmetric polynomial in the quantitiesα_{1},
⋅⋅⋅
,α_{n}having rational coefficients is a rational number.
Proof: Let f
(x1,⋅⋅⋅,xn)
be the symmetric polynomial. Thus
f (x1,⋅⋅⋅,xn) ∈ ℤ [x1⋅⋅⋅xn], the polynomials having integer coefficients
From Theorem F.1.3 it follows there are integers a_{k1}
⋅⋅⋅
k_{n} such that
∑ k1 kn
f (x1,⋅⋅⋅,xn) = ak1⋅⋅⋅knp1 ⋅⋅⋅pn
k1+⋅⋅⋅+kn≤m
where the p_{i} are the elementary symmetric polynomials defined as the coefficients of
n∏
(x − xj)
j=1
Earlier we had them ± these coefficients. Thus
f (an∑α1,⋅⋅⋅,anαn)
= ak1⋅⋅⋅knpk11(anα1,⋅⋅⋅,anαn )⋅⋅⋅pknn(anα1,⋅⋅⋅,anαn )
k1+⋅⋅⋅+kn=d
Now the given polynomial p
(x)
is of the form
( )
∏n ∑n n−k
an (x− αj) ≡ an pk(α1,⋅⋅⋅,αn )x
j=1 k=0
= a xn + a xn−1 + ⋅⋅⋅+ a x+ a
n n−1 1 0
Thus, equating coefficients, a_{n}p_{k}
(α1,⋅⋅⋅,αn)
= a_{n−k}. Multiply both sides by a_{n}^{k−1}.
Thus