A trivial observation, which was not pointed out so far: the *non*-normalized discriminant is 'better' in that it comes 'closer' to being a 'homomorphism' $P[x]\rightarrow P$, though it 'still' is not such a homomorphism, it is merely 'multiplicative up to a multiplicative computable polynomial-in-the-coefficients constant'.

I think one could characterize the factor $([x^{\mathrm{deg}(f)}](f))^{2\deg(f)-2}$, that the OP is asking the 'essence' of, as *the unique factor which gives the 'most efficient' multiplicativity-identity*, though I have never seen this and don't have time to try to make this observation rigorous. I expect this to be widely known. It seems some sort of answer to the OP's question for the 'essence' of the normalizing factor, and perhaps even for the OP's question about the '$2m-2$'.

Here is the trivial observation.

Let $\mathrm{disc}$ denote the (standard) discriminant, in the sense of the (corrected) OP.

Let $\mathrm{d}f$ denote the degree of a polynomial $f$.

Let $\mathrm{ndisc}(f) := \frac{1}{a_{\mathrm{d}f}^{2\mathrm{d}f-2}}\cdot \mathrm{disc}(f)$ denote the normalized discriminant.^{1} .

( $\mathrm{ndisc}$ is defined since a leading coefficient by definition is non-zero )

Then, for all polynomials $f_0,f_1$ of degree $\geq 1$ we have the useful and *normalization-factor-free* identity

$\mathrm{disc}(f_0\cdot f_1) = \mathrm{resultant}(f_0,f_1)\cdot \mathrm{disc}(f_0)\cdot \mathrm{disc}(f_1)$. ${}\qquad$ (multiplicative.up.to.a.polynomial.factor)

For the 'normalized discriminant' $\mathrm{ndisc}$, i.e. the one *without* the factor that the OP is asking about, we however find a less efficient identity.

Using that $\mathrm{d}( f_0\cdot f_1) = \mathrm{d}f_0 + \mathrm{d}f_1$ and that the leading coefficient of $f_0\cdot f_1$ is $a_{\mathrm{d}f_0} \cdot a_{\mathrm{d}f_1}$, we find the less efficient identity

$\small\text{$(a_{\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1})^{2\cdot(\mathrm{d}f_0+\mathrm{d}f_1)-2}\cdot \mathrm{ndisc}(f_0\cdot f_1) = \mathrm{resultant}(f_0,f_1)\cdot a_{\mathrm{d}f_0}^{2\mathrm{d}f_0-2}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1-2}\cdot\mathrm{ndisc}(f_0)\cdot \mathrm{ndisc}(f_1)$ }$

equivalently

$\small\text{$\mathrm{ndisc}(f_0\cdot f_1) = \frac{\mathrm{resultant}(f_0,f_1)}{a_{\mathrm{d}f_0}^{2\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1}} \cdot \mathrm{ndisc}(f_0)\cdot \mathrm{ndisc}(f_1)$ }$ ${}\qquad$ (multiplicative.up.to.a.nonpolynomial.factor).

So from this point of view, the 'normalized discriminant' suddenly looks less normalized than the OP's discriminant.

The only mathematical difference between (multiplicative.up.to.a.polynomial.factor) and (multiplicative.up.to.a.nonpolynomial.factor) I can think of is:

- if $a_{\mathrm{d}f_0}^{2\mathrm{d}f_0}\cdot a_{\mathrm{d}f_1}^{2\mathrm{d}f_1}$ does not divide resultant($f_0,f_1$), where 'divides' refers to the ring generated by the coefficients,
*then* the 'multiplicative constant' in (multiplicative.up.to.a.nonpolynomial.factor) is not a polynomial in the coefficients, while in (multiplicative.up.to.a.polynomial.factor) it is.

This *perhaps* might make it possible to characterize the factor the OP is asking about by an algebraic property.

^{1} _{ Which one could argue is the 'simplest' $\mathrm{auxiliarypolynomial}(f)$ that can be used to 'encode' the logical statement 'there exists a splitting field of $f$ in which there exists a multiple root of $f$ ' in terms of *an equation over the signature of field theory', in the form '$\mathrm{auxiliarypolynomial}(f)=0$'. Though of course 'simplicity-criteria' like 'smallest number of steps to write down the auxiliar polynomial' tend to have little mathematical meaning, and the discriminant is an easy example of this. Slightly more complicated expressions are often simpler to work with. }

normalized discriminant, which simply has the factor in question left out, from the so-calledstandard discriminant, which has the factor $f_0^{2m-2}$. $\endgroup$ – Peter Heinig Sep 6 '17 at 18:40eery, not matter how it happened, since I do not remember changing the indexes of the sums, Ionlyconsciously changed the indexation of the product, which the OP made range overallpairs of distinct indices, which is (fields being free of zero-divisors) notwrong, yet highly nonstandard. Anyway, I will correct it, yet only in two hours from now, since I will now have to go offline. $\endgroup$ – Peter Heinig Sep 7 '17 at 6:06