# A generalization of the discriminant of a polynomial

Let $\mathbb{K}$ be a field and let $f \in \mathbb{K}[x]$ be a monic polynomial of degree $n$. Suppose that $\alpha_1, \ldots, \alpha_n$ are all the roots of $f$ (in some algebraic closure of $\mathbb{K}$) and define $$\Delta_f(y) = \prod_{\substack{1 \leq i , j \leq n \\ i \neq j}} (y - (\alpha_i - \alpha_j)).$$ Clearly, $\Delta_f(0)$ is the discriminant of $f$, so $\Delta_f(y)$ can be considered as a generalization of the notion of discriminant of a polynomial. It is also easy to see that $\Delta_f(y) \in \mathbb{K}[y]$ and that $\Delta_f(y)$ satisfies the identity $$\operatorname{Res}_x(f(x), f(x + y)) = y^n \Delta_f(y),$$ where $\operatorname{Res}_x$ is the resultant respect to $x$. My questions are:

(1) Does $\Delta_f(y)$ have a name? Is there any article/book where $\Delta_f(y)$ is introduced and its properties are studied?

(2) Assume that $\mathbb{K}$ has characteristic 0. If $n = 2$ then $\Delta_f(y) = y^2 - \Delta_f(0)$ and it follows that $\Delta_f(y)$ is irreducible in $\mathbb{K}[y]$ if and only if $f(x)$ is irreducible in $\mathbb{K}[x]$. What can we say about the irreducibility of $\Delta_f(y)$ for $n \geq 3$?

Thanks for any suggestion/reference.

• I think I've seen the resultant formulation come up in a slick proof that the difference of algebraic numbers is algebraic, without using Galois Theory or messing with symmetric functions. – Gerry Myerson Feb 16 '17 at 10:55

## 1 Answer

As to question (2): Suppose that $\Delta_f(y)$ is separable. Then $\Delta_f(y)$ is irreducible if and only if the Galois group of $f$ acts transitively on the set of differences $\alpha_i-\alpha_j$, $i\ne j$, which is equivalent to transitivity on the set of pairs $(\alpha_i,\alpha_j)$. This, however, is equivalent to the Galois group of $f$ acting doubly transitively on the roots $\alpha_i$ of $f$.