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Let $k$ be an algebraically closed field, and let $p(x)\in k[x]$. We can think of a general polynomial $p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ as a polynomial $q(x,a_n,a_{n-1},...,a_0)\in k[x,a_n,a_{n-1},...,a_0]$ where $a_i$ here are just formal coefficients, i.e "variables".

Let $\text{Disc}(p(x))=\prod_{i\neq j\text{ and }\lambda_i \text{ are all the roots of }p(x)}(\lambda_i-\lambda_j)$ be the discriminant, is it true that $\text{Disc}(p(x))=q(a_n,a_{n-1},...,a_0)\in k[x,a_n,a_{n-1},...,a_0]$? i.e a polynomial in the coefficients of $p(x)$?

I tried to prove it but I don't know how to involve the roots in a polynomial way.

Thanks in advance.

user26857
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Or Shahar
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1 Answers1

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The discriminant, as per your definition, is a symmetric polynomial in the $\lambda_{i}$. Now the $a_{i}/a_{n}$, for $0 \le i < n$, are, up to a sign, the elementary symmetric functions in the $\lambda_{i}$. It follows that the discriminant, as per your definition, is a polynomial in the $a_{i}/a_{n}$.

For instance, when $n = 2$, you have $$ \lambda_{1} + \lambda_{2} = - \frac{a_{1}}{a_{2}}, \qquad \lambda_{1} \lambda_{2} = \frac{a_{0}}{a_{2}}, $$ and \begin{align} (\lambda_{1} - \lambda_{2})^{2} &= \lambda_{1}^{2} + \lambda_{2}^{2} - 2 \lambda_{1} \lambda_{2} \\&= \lambda_{1}^{2} + \lambda_{2}^{2} + 2 \lambda_{1} \lambda_{2} - 4 \lambda_{1} \lambda_{2} \\&= (\lambda_{1} + \lambda_{2})^{2} - 4 \lambda_{1} \lambda_{2} \\&= \left(-\frac{a_{1}}{a_{2}}\right)^{2} - 4 \frac{a_{0}}{a_{2}} \\&= \frac{a_{1}^{2} - 4 a_{0} a_{2}}{a_{2}^{2}} . \end{align}

Note in fact that the usual definition of the discriminant is $$ (-1)^{n(n-1)/2} a_n^{2n-2} \prod_{i \neq j} (\lambda_i-\lambda_j), $$ and this is indeed a polynomial in the $a_{i}$.

Andreas Caranti
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