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.



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> 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?
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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 $3$.



Construction of a symmetric polynomial in the roots that acts like the discriminant
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 $f$.


