The Discriminant

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:

Δ(x₁, ..., x_n)² is a symmetric polynomial

Proof:

(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 P² = P² = (-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 Δ

Let



Then the Discriminant D is:

D(s₁, ..., s_n) = Δ(x₁, ..., x_n)²

where s₁, ..., s_n 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

where:

A = x₁²x₂ + x₂²x₃ + x₃²x₁

and

B = x₁x₂² + x₂x₃² + x₃x₁²

So that:

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

So that:

s₁²s₂² + 18s₁s₂s₃ - 27s₃² -4s₁³s₃ - 4s₂³= 0 + 0 - 27(-q)² - 0 - 4p³ = -27q² - 4p³


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 - u₁)*...*(x - u_n)

for some u₁, ..., u_n ∈ 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.

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 = (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.

QED

Corollary 2.1:

x³ + px +q = 0

has 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 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.

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001