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

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


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

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?

  • 1
    $\begingroup$ The main purpose of discriminants is to detect multiple roots and distuinguishing some cases concerning the number of real roots a polynomial has. $\endgroup$
    – Peter
    Jun 19, 2020 at 19:26
  • 3
    $\begingroup$ The Galois group of a polynomial is a permutation group acting on the roots of this polynomial. The discriminant is defined in precisely such a way that it is invariant under all possible such permutations. The discriminant is (perhaps not obviously) always an element of the field the polynomial is defined over. Consider the square root of the discriminant. When a permutation acts on this, it switches sign according to the permutations signum. Galois theory then allows us to deduce that the Galois group is a subgroup of the alternating group iff the discriminant is a square. $\endgroup$
    – Thorgott
    Jun 19, 2020 at 19:54
  • $\begingroup$ The Wikipedia article gives a good hint: the discriminant must be a multiple of the Vandermonde matrix formed by the roots, so that it cancels whenever there is a multiple root. $\endgroup$
    – user65203
    Jun 23, 2020 at 7:36

1 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).

  • 1
    $\begingroup$ I think this is a very wise answer: the resultant has a definite structural purpose, after all... $\endgroup$ Jun 19, 2020 at 21:41
  • $\begingroup$ See the important reference I just gave. $\endgroup$
    – Jean Marie
    Jun 23, 2020 at 6:45

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