Prove that the discriminant of a degree $n$ polynomial f is a polynomial of degree $2n - 2$ in the coefficients of $f$

Question

The discriminant of some some degree $n$ polynomial $$ f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 $$ is given by $$ \Delta_f = a_n^{2n-2} \prod\limits_{i < j} (\alpha_i - \alpha_j)^2 $$ where $\alpha_0, \dots, \alpha_n$ are the roots of $f$.

I have seen it stated that the discriminant is always a homogenous polynomial of degree $2n - 2$ in the coefficients of $f$, $a_n, \dots, a_0$, but I have yet to see this proven.

One can easily calculate the discriminant for some low degrees and see that this holds, but what approach could I use to show that it holds in general?

Answer 1

The discriminant is the resultant of the polynomial and its derivative. This gives a square matrix in the coefficients of size $n+(n-1)=2n-1$. I suspect that the final reduction comes from the fact that $a_0$ will be common to all terms and it is cancelled.

Answer 2

The discriminant of a univariate polynomial $f$ is the resultant of $f$ and of its derivative $f'$.

So let's take a step back to talk about resultants. Let $f$, $g$ be two univariate polynomials of degree $d$ and $e$ respectively. Then the resultant is a polynomial in the coefficients of $f$ and $g$ that arises as the determinant of the Sylvester matrix. See the wikipedia article on resultant. The Sylvester matrix has size $d+e$.

In particular if $f$ is univariate of degree $n$, its derivative has degree $n-1$, so the resultant of $f$ and $f'$ has degree $2n-1$. However, you can see that you can factor out $a_n$ from the Sylvester matrix (depending on how you write it, the first row or the first column will be a multiple of $a_n$). This tells you that the resultant of $f$ and $f'$ is $a_n \cdot \Delta$ where $\Delta$ is some polynomial in the coefficients of $f$. That $\Delta$ is the discriminant.