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Linear Algebra, Theory And Applications, 2012a

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446 FIELDS AND FIELD EXTENSIONS<br />

Then the following result is the fundamental theorem in the subject. It is the symmetric<br />

polynomial theorem. It says that these elementary symmetric polynomials are a lot like a<br />

basis for the symmetric polynomials.<br />

Theorem F.1.3 Let g (x 1 ,x 2 , ··· ,x n ) be a symmetric polynomial. Then g (x 1 ,x 2 , ··· ,x n )<br />

equals a polynomial in the elementary symmetric functions.<br />

g (x 1 ,x 2 , ··· ,x n )= ∑ k<br />

a k s k 1<br />

1 ···sk n<br />

n<br />

and the a k are unique.<br />

Proof: If n =1, it is obviously true because s 1 = x 1 . Suppose the theorem is true for<br />

n − 1andg (x 1 ,x 2 , ··· ,x n ) has degree d. Let<br />

By induction, there are unique a k such that<br />

g ′ (x 1 ,x 2 , ··· ,x n−1 ) ≡ g (x 1 ,x 2 , ··· ,x n−1 , 0)<br />

g ′ (x 1 ,x 2 , ··· ,x n−1 )= ∑ k<br />

a k s ′k 1<br />

1 ···s ′k n−1<br />

n−1<br />

where s ′ i is the corresponding symmetric polynomial which pertains to x 1,x 2 , ··· ,x n−1 .<br />

Note that<br />

s k (x 1 ,x 2 , ··· ,x n−1 , 0) = s ′ k (x 1 ,x 2 , ··· ,x n−1 )<br />

Now consider<br />

g (x 1 ,x 2 , ··· ,x n ) − ∑ k<br />

a k s k 1<br />

1 ···sk n−1<br />

n−1 ≡ q (x 1,x 2 , ··· ,x n )<br />

is a symmetric polynomial and it equals 0 when x n equals 0. Since it is symmetric, it is also<br />

0 whenever x i = 0. Therefore,<br />

q (x 1 ,x 2 , ··· ,x n )=s n h (x 1 ,x 2 , ··· ,x n )<br />

and it follows that h (x 1 ,x 2 , ··· ,x n ) is symmetric of degree no more than d − n and is<br />

uniquely determined. Thus, if g (x 1 ,x 2 , ··· ,x n ) is symmetric of degree d,<br />

g (x 1 ,x 2 , ··· ,x n )= ∑ k<br />

a k s k 1<br />

1 ···sk n−1<br />

n−1 + s nh (x 1 ,x 2 , ··· ,x n )<br />

where h hasdegreenomorethand − n. Now apply the same argument to h (x 1 ,x 2 , ··· ,x n )<br />

and continue, repeatedly obtaining a sequence of symmetric polynomials h i , of strictly decreasing<br />

degree, obtaining expressions of the form<br />

g (x 1 ,x 2 , ··· ,x n )= ∑ k<br />

b k s k 1<br />

1 ···sk n−1<br />

n−1 skn n + s n h m (x 1 ,x 2 , ··· ,x n )<br />

Eventually h m must be a constant or zero. By induction, each step in the argument yields<br />

uniqueness and so, the final sum of combinations of elementary symmetric functions is<br />

uniquely determined. <br />

Here is a very interesting result which I saw claimed in a paper by Steinberg and Redheffer<br />

on Lindemannn’s theorem which follows from the above corollary.

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