Let $f(x) = x^m+\sum\_{j=0}^{m-1}f\_{m-j}x^j\in P[x]$ be a **monic** polynomial over a field $P$ and let $f(x) = (x-\alpha\_1)\cdot\ldots\cdot(x-\alpha\_m)$ be a factorization of $f$ over an extension field $Q$ of $P$.


Then it is quite natural to consider a value (called discriminant) $$\prod\_{1\leq i<j\leq m}(\alpha\_i-\alpha\_j)^2,$$
whose being zero or not determines whether or not the polynomial has any multiple roots (in any extension field of $P$, not only $Q$).


But, when $f$ is *not* monic, say $f = f\_0 \cdot x^m+\sum\_{j=0}^{m-1}f\_{m-j}x^j $, a usual definition of the discriminant reads 
$$
f\_0^{2m-2}\cdot \prod\_{1\leq i<j\leq m}(\alpha\_i-\alpha\_j)^2.$$


So, my question is: 


**What is the essence of the factor** $f\_0^{2m-2}$?



What is the essence of the constant factor in the standard definitions of the discriminant?
As Robert said, if you want everything to work in $\mathbb Z[f\_0,\ldots,f\_m]$, you need that factor. I'll also mention that your polynomial indexing is messed up, you probably meant the sum to go from $j=0$ to $j=m-1$, not $j=1$ to $j=m$.


In any case, things become clearer if you study the theory of *resultants* and ask when two polynomials have a common root. If they're not monic, you really want to view them as homogeneous polynomials that may have a common root "at infinity" if their leading coefficients both vanish. The discriminant of $f(x)$ is, essentially, the resultant of $f(x)$ and $f'(x)$.


