The content in today's blog is taken from Jean-Pierre Tignol's

Galois' Theory of Algebraic Equations.

For a definition of symmetric polynomials, see Definition 1,

here.

Lemma 1:Let

Then:

Î”(x_{1}, ..., x_{n})^{2} is a

symmetric polynomialProof:

(1) Let

P = ∏ (x_{i} - x_{j})(2) If any

x_{i} = x_{j}, 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

∏ (x_{i} - x_{j}) is not symmetric.

(7) If we permutate the values of

[∏ (x_{i} - x_{j})]^{2}, it is clear that the result is always

P^{2} = P^{2} = (-P)^{2}QED

Using Waring's Method (see Theorem 4,

here), we know that

[∏ (x_{i} - x_{j})]^{2 }can be expressed as a function of the

elementary symmetric polynomials (for review, see

here) so that we have:

Definition 1: The Discriminant Î”Let

Then the

Discriminant D is:

D(s_{1}, ..., s_{n}) = Î”(x_{1}, ..., x_{n})^{2}where

s_{1}, ..., s_{n} are the

elementary symmetric polynomials.

Example 1: Discriminant of a generic polynomial of degree 2

D(s_{1},s_{2}) = s_{1}^{2} - 4s_{2}First, we carry out the multiplication:

Î”(x_{1},x_{2})^{2} = (x_{1} - x_{2})^{2} = x_{1}^{2} + x_{2}^{2} - 2x_{1}x_{2}Then, we show it as a function of the

elementary symmetric polynomials:

x_{1}^{2} + x_{2}^{2} - 2x_{1}x_{2} = (x_{1} + x_{2})^{2} - 4x_{1}x_{2}=s_{1}^{2} - 4s_{2}Example 2: Discriminant of a generic polynomial of degree 3

D(s_{1},s_{2},s_{3}) = s_{1}^{2}s_{2}^{2} + 18s_{1}s_{2}s_{3} - 27s_{3}^{2} -4s_{1}^{3}s_{3} - 4s_{2}^{3}We note that:

Î”(x_{1},x_{2},x_{3}) = (x_{1} - x_{2})(x_{1} - x_{3})(x_{2} - x_{3})We can simplify this by restating

Î”(x_{1},x_{2},x_{3}) as:

Î”(x_{1},x_{2},x_{3}) = A - Bwhere:

A = x_{1}^{2}x_{2} + x_{2}^{2}x_{3} + x_{3}^{2}x_{1}and

B = x_{1}x_{2}^{2} + x_{2}x_{3}^{2} + x_{3}x_{1}^{2}So that:

Î”(x_{1},x_{2},x_{3})^{2} = (A - B)^{2} = A^{2} + B^{2} - 2AB = (A + B)^{2} - 4ABNow, we note that

A+B and

AB are

symmetric polynomials and using Waring's method (see Theorem 4,

here), we have:

A + B = ∑ x_{1}^{2}x_{2} = s_{1}s_{2} - 3s_{3}AB = ∑ x_{1}^{4}x_{2}x_{3} + ∑ x_{1}^{3}x_{2}^{3} + 3x_{1}^{2}x_{2}^{2}x_{3}^{2} = s_{1}^{3}s_{3} + s_{2}^{3} - 6s_{1}s_{2}s_{3} + 9s_{3}^{2}Now, we can combine these results to get the discriminant:

(A + B)^{2} - 4AB = ( s_{1}s_{2} - 3s_{3})^{2} - 4( s_{1}^{3}s_{3} + s_{2}^{3} - 6s_{1}s_{2}s_{3} + 9s_{3}^{2}) == s_{1}^{2}s_{2}^{2} + 18s_{1}s_{2}s_{3} - 27s_{3}^{2} -4s_{1}^{3}s_{3} - 4s_{2}^{3}Example 3: Discriminant of

x^{3} + px + qD(s_{1},s_{2},s_{3}) = -27q^{2} - 4p^{3}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:

s_{1} = 0s_{2} = ps_{3} = -qSo that:

s_{1}^{2}s_{2}^{2} + 18s_{1}s_{2}s_{3} - 27s_{3}^{2} -4s_{1}^{3}s_{3} - 4s_{2}^{3}= 0 + 0 - 27(-q)^{2} - 0 - 4p^{3} = -27q^{2} - 4p^{3}Theorem 2: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 - u_{1})*...*(x - u_{n})for some

u_{1}, ..., u_{n} ∈ C.

Let

d ∈ R be the

discriminant of

PThe equality

d=0 holds if and only if

P has a root of multiplicity at least 2 in

CIf 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.

Proof:

(1)

d = ∏ (u_{i} - u_{j})^{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

u_{i} = u_{j}(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 = (u_{1} - u_{2})^{2} [see Definition 1 above]

(6) If

u_{1} is not real, then

u_{2} = u_{1} [see Theorem 5,

here]

(7) Let

u_{1} = a + bi(8) Let

u_{2} = a - bi(9)

(u_{1} - u_{2})^{2} = (a + bi - [a - bi])^{2} = (2bi)^{2} = -4b^{2} = -abs(4*b^{2})(10) So that

d ≤ 0.

(11) Assume that

n = 3 (12) Then,

d = (u_{1} - u_{2})^{2}(u_{1} - u_{3})^{2}(u_{2} - u_{3})^{2}(13) Assume that not all three roots are real. So, we can assume that

u_{1} is not real.

(14) Then, it follows that its conjugate is also a root. So we can assume that

u_{2} is not real and

u_{1} = u

_{2}(15) We know that

u_{3} is then real. [see Theorem 3,

here]

(16) So there exists real numbers

a,b,c such that:

u_{1} = a + biu_{2} = a - biu_{3} = cAnd we have:

(u_{1} - u_{2})(u_{1}-u_{3})(u_{2} - u_{3}) = [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} + b^{2}Combining this with the above we get:

(2bi)[(a - c)^{2} + b^{2}] = i[(2b)(a - c)^{2} + 2b^{2}]Now, it is clear that

(2b)(a - c)^{2} + 2b^{2} is a real number since

a,b,c are real and we can set

s = (2b)(a - c)^{2} + 2b^{2} where

s is a real number.

So

d = (is)^{2} = -(s^{2}) = -abs(s^{2})(17)

So, d ≤ 0.QED

Corollary 2.1:x^{3} + px +q = 0has three distinct real solutions if and only if

(p/3)^{3} + (q/2)^{2} is less than

0.

Proof:

(1) By Exercise 3 above, the discriminant of x

^{3} + px +q is

d = -27q^{2} - 4p^{3}We further note that:

d = -27q^{2} - 4p^{3 }= -2^{2}3^{3}[(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.

QED

References