## Saturday, October 03, 2009

### A polynomial invariant on all but one variable

The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Lemma:

Let g be a polynomial in n indeterminates x1, ..., xn over some field K.

Let g be invariant under every permutation of x2, ..., xn

Then:

g can be written as a polynomial in x1 and the elementary symmetric polynomials s1, ..., sn-1 in x1, ..., xn

Proof:

(1) We can view g as a polynomial in x2, ..., xn with coefficients in K[x1].

(2) Using Waring's Method [see Theorem 4, here], we know that g can be written as a polynomial in the elementary symmetric polynomials s'1, ..., s'n-1 in x2, ..., xn with coefficients in K[x1]

(3) Therefore, there exists a polynomial g' such that:

g(x1, ..., xn) = g'(x1, s'1, ..., s'n-1)

where:

s'1 = x2 + ... + xn

s'2 = x2x3 + ... + xn-1xn

...

s'n-1 = x2*...*xn

(4) To complete the proof, we need to show that s'1, s'2, ..., s'n-1 can be restated in terms of s1, s2, ..., sn where:

s1 = x1 + ... + xn

s2 = x1x2 + ... + xn-1xn

...

sn = x1*...*xn

(5) We know that for any given polynomial [see Theorem 1, here]:

(X - x1)*...*(X - xn) = Xn - s1Xn-1 + ... + (-1)nsn

(6) Now, we can use the same principle to get:

(X - x2)*...*(X - xn) = Xn-1 - s'1Xn-2 + ... + (-1)n-1s'n-1

(7) Multiplying the above equation by (X - x1) gives us:

(X - x1)*...*(X - xn) = (X - x1)Xn-1 - (X - x1)s'1Xn-2 + ... + (X - x1)(-1)n-1s'n-1 =

Xn - (x1+s'1)Xn-1 + (x1s'1 + s'2)Xn-2 - (x1s'2 + s'3)Xn-3 + ... + (-1)n(x1s'n-1)

(8) Combining step #5 and step #7 gives us:

s1 = x1 + s'1

so that:

s'1 = s1 - x1

s2 = x1s'1 + s'2

so that:

s'2 = s2 - x1s'1 = s2 - x1(s1 - x1) = s2 - x1s1 + x12

and so on...

(9) Since we can subtitute all values s'i in terms of x1 and s1, ..., sn, we can use the equation in step #3 to get:

g(x1, ..., xn) = g'(x1, s'1, ..., s'n-1) = g'(x1, s1 - x1, s2 - s1x1 + x12, ... )

QED

References
• Jean-Pierre Tignol, , World Scientific, 2001