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Let $f(x)=x^n+a_1x^{n-1}+\dots+a_n$ be a monic polynomial with coefficients in a field $K$. Its discriminant is defined like this: $$\Delta(f)=\prod_{i<j}(x_i-x_j)^2$$ Now we define the Newton's power sums polynomials: $p_k(x_1,\dots,x_n)=x_1^k+\dots+x_n^k$.

It is claimed the following equality: \begin{equation} \prod_{i<j}(x_i-x_j)^2= \begin{vmatrix} p_0 & p_1 & \dots & p_{n-1} \\ p_1 & p_2 & \dots & p_n \\ \vdots & \vdots & \ddots & \vdots \\ p_{n-1} & p_n & \dots & p_{2n-2} \end{vmatrix}. \end{equation}

I tried to show that by induction and writing the determinant using minor formula but I still have terms I don't know. Either way, it seems quite cumbersome to work with and I'm sure there is a better way to do this. Any advice?

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  • The $x_i$ are evidently roots to $f$ so it is cleanest to assume $\mathbb K$ is algebraically closed. See the 'addendum' here: https://math.stackexchange.com/questions/4460264/if-a-has-distinct-eigenvalues-then-all-matrices-close-enough-to-a-also-have/ [this assumes you know the determinant for a Vandermonde matrix] – user8675309 May 03 '23 at 15:41
  • The square root of the discriminant is the determinant of the Vandermonde matrix, and the product of the Vandermonde matrix with its transpose is the matrix of power sums – math54321 May 03 '23 at 17:23

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