Why the sign in the definition of the discriminant? Consider the split monic $f=\prod_{i=1}^n(x-x_i)\in \mathbb Z[x_1 ,\dots ,x_n,x]$. Its discriminant is usually defined as $$(-1)^{n(n-1)/2}\prod_{i=1}^nf^\prime(x_i)=\prod_{1\leq i<j\leq n}(x_i-x_j)^2.$$
What is the reason for taking this definition as opposed to $\prod_{i=1}^nf^\prime(x_i)$? The product of the derivatives at the roots "feels" to me more canonical than the product on the RHS.
 A: The reason is that the formula on the right side should be considered more fundamental, not the formula on the left, when seeking a symmetric expression in the roots. Don't use a product of anything "at" the roots, but a symmetric expression in the roots that vanishes if any pair of roots are equal. That explains the factors $(x_i-x_j)^2$. Do you consider the simplest polynomial with a double root at 0 to be $x^2$ or $-x^2$? The product on the left (without the sign) is also interesting and has a name: it is called the resultant of $f(x)$ and $f'(x)$.
A: The definition of the resultant $\text{res}(f,g)$ of two (monic) polynomials $f,g\in {\mathbb Z}[x]$ as the determinant of the corresponding Sylvester matrix, readily implies
that if $\alpha_1,\ldots,\alpha_n$ are the roots of $f(x)$ (say, in ${\mathbb C}$) then
$$\text{res}(f,g)=\prod_{i=1}^ng(\alpha_i).$$
Now, if you define the discriminant of the (monic) polynomial $f(x)$ as
$$\Delta(f(x))=\text{res}(f(x),f'(x))=\prod_{i=1}^nf'(\alpha_i),$$
writing $f(x)=(x-\alpha_1)\cdots (x-\alpha_n)$ it follows that
$$f'(\alpha_j)=\prod_{i\neq j}^n(\alpha_j-\alpha_i).$$
In this equality, once you order the indices, for each pair $i<j$ there are two factors $\pm(\alpha_i-\alpha_j)$ and thus
\begin{align*}
\Delta(f(x))&=\prod_{i=1}^nf'(\alpha_i)=\prod_{i\neq j}^n(\alpha_j-\alpha_i)\\
&=(-1)^{n(n-1)/2}\prod_{i< j}^n(\alpha_j-\alpha_i)^2
\end{align*}
where the sign is because there are precisely $n(n-1)/2$ pairs $\pm(\alpha_i-\alpha_j)$.
