What is the underlying reason behind the definition of the discriminant as an expression of the roots?

Question

Background:

The discriminant of a polynomial $A(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ can be expressed in terms its roots as

$$\text{Disc}(A)=a_{n}^{2n-2}\prod_{i<j}(r_i -r_j)^2\tag 1$$

so that for a quadratic $ax^2 + bx +c,$ the discriminant would predictably be

$$a^2\left( \frac{-b-\sqrt{b^2-4a}}{2a} - \frac{b-\sqrt{b^2-4a}}{2a} \right)^2=b^2-4a$$

The generalized form in $(1)$ I thing may be motivated by symmetric functions of the roots of a polynomial, $x_1, x_2, \dots, x_n,$ such as

$$\begin{align} S_1 &= x_1+x_2+\cdots+x_n=\sum x_i\\ S_2 &= x_1 x_2 + \cdots+ x_{n-1}x_n=\sum_{i<j}x_ix_j\\ S_3 &= x_1x_2x_3 +\cdots + x_{n-2} x_{n-1} x_n =\sum_{i<j<k} x_i x_j x_k\\ S_n &= \prod x_i \end{align}$$

or

$$\begin{align} \sigma_1 &= S_1= x_1+x_2+\cdots+x_n=\sum x_i\\ \sigma_2 &= S_1^2 - 2 S_2= x_1^2+x_2^2+\cdots+x_n^2=\sum x_i^2\\ \sigma_3 &= S_1^3 - 3 S_1 S_2 + 3S_3= x_1^3+x_2^3+\cdots+x_n^3=\sum x_i^3\\ \end{align}$$

and Newton's recursive formulae

$$\begin{align} \sigma_1 &= S_1\\ \sigma_2 &= S_1 \sigma_1 - 2 S_2\\ \sigma_3 &= S_1 \sigma_2 - S_2 \sigma_1 + 3 S_3\\ \sigma_4 &= S_1 \sigma_3 - S_2 \sigma_2 + S_3 \sigma_1 - 4 S_4\\ \end{align}$$

But is this even true? And if so, what is the link?

For instance for a monic polynomial of degree $2$

$$\begin{align} (x_1 - x_2)^2 &= (x_1 + x_2)^2 - 4 x_1 x_2\\ &= S_1^2 - 4 S_2 \end{align}$$

but what is the significance of that? Symmetry to achieve what purpose?

Answer 1

I am uncertain whether what I give here answers your question, but I think it is by "enlarging the picture" that one gets a better understanding of a concept like this one.

The discriminant of a polynomial is the particular case $Res(f,f')$ of the concept of resultant $Res(f,g)$ of two monic polynomials $f$ and $g$ ("monic" meaning that their dominant coefficients are $1$) [with $Res(f,g)=0$ expressing that $f$ and $g$ have a common root : here $Res(f,f')=0$ expresses that $f$ and $f'$ have a common root, which is necessarily a double root of $f$ ; therefore, it shouldn't come as a surprise that the factors have the form $(r_i-r_j)$].

A very interesting property of $Res(f,g)$ is that it is the product $f(\beta_1)\cdots f(\beta_n)$ of the values of the first polynomial at at the roots $\beta_k$ of the second one [In fact, as $Res(g,f)=Res(f,g)$, it is also equal to the product $g(\alpha_1)\cdots g(\alpha_m)$ of $g$ computed at the roots of $\alpha_k$ of $f$].

In particular, the discriminant is the product of the values of $f$ evaluated at the roots of its derivative, otherwise said the product of the ordinates of the local extrema of $f$. See the way I have used this property in an answer I recently gave here ; please note that I use there a (third !) way to compute the discriminant using a certain determinant.

For all this, see the excellent book of Gelfand et al. "Discriminants, Resultants and multidimensional determinants (advise : begin p. 397).