Discriminant as a polynomial 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.
 A: 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 $a_{i}$.
