For a definition of symmetric polynomials, see Definition 1, here.
Δ(x1, ..., xn)2 is a symmetric polynomial
(1) Let P = ∏ (xi - xj)
(2) If any xi = xj, then P = 0. [that is, if there is a multiple root]
(3) Assume that there are no multiple roots.
(4) If we swap any two roots, then the result is either P or -P, then the result is to permute the ordering of each of the differences and to change the signs of some.
(5) Ordering doesn't change the product so the only the change that occurs is the sign of the product. That is, the result is P or -P depending upon which parameters get swapped.
(6) So it is clear that ∏ (xi - xj) is not symmetric.
(7) If we permutate the values of [∏ (xi - xj)]2, it is clear that the result is always P2 = P2 = (-P)2
Using Waring's Method (see Theorem 4, here), we know that [∏ (xi - xj)]2 can be expressed as a function of the elementary symmetric polynomials (for review, see here) so that we have:
Definition 1: The Discriminant Δ
Then the Discriminant D is:
D(s1, ..., sn) = Δ(x1, ..., xn)2
where s1, ..., sn are the elementary symmetric polynomials.
Example 1: Discriminant of a generic polynomial of degree 2
D(s1,s2) = s12 - 4s2
First, we carry out the multiplication:
Δ(x1,x2)2 = (x1 - x2)2 = x12 + x22 - 2x1x2
Then, we show it as a function of the elementary symmetric polynomials:
x12 + x22 - 2x1x2 = (x1 + x2)2 - 4x1x2=s12 - 4s2
Example 2: Discriminant of a generic polynomial of degree 3
D(s1,s2,s3) = s12s22 + 18s1s2s3 - 27s32 -4s13s3 - 4s23
We note that:
Δ(x1,x2,x3) = (x1 - x2)(x1 - x3)(x2 - x3)
We can simplify this by restating Δ(x1,x2,x3) as:
Δ(x1,x2,x3) = A - B
A = x12x2 + x22x3 + x32x1
B = x1x22 + x2x32 + x3x12
Δ(x1,x2,x3)2 = (A - B)2 = A2 + B2 - 2AB = (A + B)2 - 4AB
Now, we note that A+B and AB are symmetric polynomials and using Waring's method (see Theorem 4, here), we have:
A + B = ∑ x12x2 = s1s2 - 3s3
AB = ∑ x14x2x3 + ∑ x13x23 + 3x12x22x32 = s13s3 + s23 - 6s1s2s3 + 9s32
Now, we can combine these results to get the discriminant:
(A + B)2 - 4AB = ( s1s2 - 3s3)2 - 4( s13s3 + s23 - 6s1s2s3 + 9s32) =
= s12s22 + 18s1s2s3 - 27s32 -4s13s3 - 4s23
Example 3: Discriminant of x3 + px + q
D(s1,s2,s3) = -27q2 - 4p3
First, we note that the values of the elementary symmetric polynomials can be derived from the coefficients of a polynomial (see Theorem 1, here) so that:
s1 = 0
s2 = p
s3 = -q
s12s22 + 18s1s2s3 - 27s32 -4s13s3 - 4s23= 0 + 0 - 27(-q)2 - 0 - 4p3 = -27q2 - 4p3
Let P ∈ R[X] be a monic polynomial with real coefficients, which splits into a product of linear factors over C such that:
P = (x - u1)*...*(x - un)
for some u1, ..., un ∈ C.
Let d ∈ R be the discriminant of P
The equality d=0 holds if and only if P has a root of multiplicity at least 2 in C
If all the roots of P are real, then d ≥ 0. If n=2 or n=3 and not all the roots are real, then d ≤ 0.
(1) d = ∏ (ui - uj)2 where 1 ≤ i is less than j ≤ n
(2) If P has a root of multiplicity at least 2, then d = 0 since we have a case where ui = uj
(3) If all the roots are real, then d ≥ 0 since any real number squared is greater or equal to 0 and product of nonnegative numbers is greater or equal to 0.
(4) Assume n =2
(5) d = (u1 - u2)2 [see Definition 1 above]
(6) If u1 is not real, then u2 = u1 [see Theorem 5, here]
(7) Let u1 = a + bi
(8) Let u2 = a - bi
(9) (u1 - u2)2 = (a + bi - [a - bi])2 = (2bi)2 = -4b2 = -abs(4*b2)
(10) So that d ≤ 0.
(11) Assume that n = 3
(12) Then, d = (u1 - u2)2(u1 - u3)2(u2 - u3)2
(13) Assume that not all three roots are real. So, we can assume that u1 is not real.
(14) Then, it follows that its conjugate is also a root. So we can assume that u2 is not real and u1 = u2
(15) We know that u3 is then real. [see Theorem 3, here]
(16) So there exists real numbers a,b,c such that:
u1 = a + bi
u2 = a - bi
u3 = c
And we have:
(u1 - u2)(u1-u3)(u2 - u3) = [a+bi - (a - bi)][a+bi - c][a-bi - c] = (2bi)([a-c]+bi)([a-c]-bi)
Now, we know that:
([a - c] + bi)([a - c] - bi) = [a - c][a - c] - bi[a - c] + bi[a - c] - [bi][bi] =[a - c]2 + b2
Combining this with the above we get:
(2bi)[(a - c)2 + b2] = i[(2b)(a - c)2 + 2b2]
Now, it is clear that (2b)(a - c)2 + 2b2 is a real number since a,b,c are real and we can set s = (2b)(a - c)2 + 2b2 where s is a real number.
So d = (is)2 = -(s2) = -abs(s2)
(17) So, d ≤ 0.
x3 + px +q = 0
has three distinct real solutions if and only if (p/3)3 + (q/2)2 is less than 0.
(1) By Exercise 3 above, the discriminant of x3 + px +q is d = -27q2 - 4p3
We further note that:
d = -27q2 - 4p3 = -2233[(p/3)3 + (q/2)2]
(2) Now if (p/3)3 + (q/2)2 is less than 0, it follows that d ≥ 0.
(3) So, using Theorem 2 above, we are done.
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001