The discriminant $\Delta(P)$ of a monic polynomial $P(x)=x^n + a_{n-1} x^{n-1} + \dotsb + a_0$ of degree $n$, when expanded (using elementary symmetric polynomials), is a symmetric polynomial of degree $n(n − 1)$ in the roots ($x_i$). This can be obtained by the definition of the discriminant
$$ \Delta(P)=\prod_{i<j}(x_i -x_j)^2.\tag 1\label{1}$$
When $\Delta=0$, $P$ has a double root.
For an arbitrary degree $n$, is it possible to construct a symmetric polynomial of degree $<n(n − 1)$ in the roots (I'd call it a false discriminant $\Delta_{\text{false}}(P)$), which is equal to $0$ if and only if the polynomial has a double root? Or is $n(n − 1)$ the minimal degree?
As a naïve example, for degree $2$, the discriminant, when expanded in the roots, is simply $x_1^2-2x_1x_2+x_1^2$ and there's no symmetric polynomial in the roots of degree $1$ that is $0$ iff $x_1=x_2$. For degree $3$, it's not so clear. Its expansion in the roots is
$$x_1^4x_2^2 – 2x_1^3x_2^3 + x_1^2x_2^4 – 2x_1^4x_2x_3 + 2x_1^3x_2^2x_3 + 2x_1^2x_2^3x_3 – 2x_1x_2^4x_3 + x_1^4x_3^2 + 2x_1^3x_2x_3^2 – 6x_1^2x_2^2x_3^2 + 2x_1x_2^3x_3^2 + x_2^4x_3^2 – 2x_1^3x_3^3 + 2x_1^2x_2x_3^3 + 2x_1x_2^2x_3^3 – 2x_2^3x_3^3 + x_1^2x_3^4 – 2x_1x_2x_3^4 + x_2^2x_3^4$$ and I wasn't able to find a symmetric polynomial in the roots of degree less than $6$, $\Delta_{\text{false}}(P)$, that satisfies $\Delta_{\text{false}}(P)=0$ iff $P$ has a double root ($x_i=x_j$ for some $i\ne j$). I wonder if it even exists for degree $3$.
Best Answer
In characteristic not equal to $2$, the discriminant is optimal. In characteristic $2$, the polynomial $\prod_{i<j} (x_i+x_j)$ works and has degree $\binom{n}{2}$.
Proof: Let $f(x_1, x_2, \ldots, x_n)$ be a nonzero symmetric polynomial of the sort that you describe. Then $f$ must vanish whenever $x_i = x_j$, so $f$ is divisible by $x_i - x_j$. Furthermore, $\tfrac{f(x_1, x_2, \ldots, x_n)}{x_i - x_j}$ must be antisymmetric in $x_i$ and $x_j$ so, assuming the characteristic is not two, $\tfrac{f(x_1, x_2, \ldots, x_n)}{x_i - x_j}$ must be divisible by $x_i - x_j$ as well, so $(x_i-x_j)^2$ divides $f$. Since all the $(x_i - x_j)^2$ are pairwise coprime, we deduce that $\Delta$ divides $f$.