## Monday, September 28, 2009

### Galois' Memoir: Lemma 2 (Galois Resolvent)

The following is taken from the translation of Galois' Memoir by Harold M. Edwards found in his book Galois Theory. The proof itself is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Definition 1: Galois Resolvent Function

For any equation f(x) with distinct roots, the Galois Resolvent Function is a function g(x1, ..., xn) of the roots that no matter how the roots are permuted on the function, no two of the values are equal.

Definition 2: Galois Resolvent

The Galois Resolvent is a value of the Galois Resolvent Function where the roots of the equation f(x) are passed in as parameters.

Lemma 2: Galois Resolvent Function Exists

Given any equation f(x) with distinct roots a,b,c,... one can always form a function V of the roots such that no two of the values one obtains by permuting the roots in this function are equal.

For example, one can take:

V = Aa + Bb + Cc + ...

A, B, C, ... being suitably chosen whole numbers.

Proof:

(1) Let the n distinct roots of f(x) be denoted a,b,c, ...

(2) Since these roots are distinct, the discriminant (a - b)2(a - c)2(b - c)2*... = D is not zero [For review of the discriminant, see here].

(3) What needs to be shown is that n integers A,B,C, ... can be chosen so that the n! numbers AS(a) + BS(b) + CS(c) + ... + are distinct where S ranges over all n! permutations of the roots a,b,c,...

[for details on why count(n permutations) = n!, see Corollary 1.1, here]

(4) Let P be the product of the squares of the differences of these n! numbers that is:

P = ∏ (S,T) [ A(S(a) - T(a)) + B(S(b) - T(b)) + ... ]2

where the product is all over n!(n! - 1)/2 pairs (unordered) of permutations S and T in which S≠ T.

Note: The purpose of this equation is to verify that all n! numbers are distinct. P is the product of all possible differences between any two permutations.

We know that there are n! possible permutations (see step #3 above).

We pick one of these permutations (1 out of n!) and call it S. Then, we pick a second one (1 out of n! - 1) and call it T. Since ordering doesn't matter and there are two ways to pick the same combination, we only need to deal with n!*(n!-1)/2 comparisons between S and T.

(5) To complete the proof, we only need to show that we can pick A,B,C, ... etc. such that P is nonzero.

If any of the permutations are not distinct, then the difference between S and T will be 0. If any of the differences are 0, then P = 0. So if P ≠ 0, it follows that we have found values for A,B,C,... such that all permutations are distinct.

(6) Let A, B, C, ..., be regarded at first as variables.

(7) Then P is a polynomial in variables A, B, C,... whose coefficients are polynomials in the roots a,b,c,...

(8) P is symmetric in the roots (this follows directly from the definition of P and the definition of symmetric polynomials, see Definition 1, here).

(9) Since P is symmetric, the coefficients are symmetric in the roots so using Waring's Method (see Theorem 4, here), P is a polynomial in A, B, C, ... with the coefficients symmetric in roots.

(10) So, we can determine the coefficients of P based on the elementary symmetric polynomials using the roots [see Theorem 1, here].

(11) We can therefore assume that the coefficients are known since we are assuming that the roots are known.

(12) Since P is a product of nonzero polynomials, it is nonzero. [see Theorem, here]

(13) Therefore once can assign integer values to A,B,C,... in such a way as to make P ≠ 0 [see Theorem, here].

QED

References