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I wonder if anybody has seen the following natural polynomial. Given a monic univariate polynomial $P(z)$ of degree $N$, denote its roots by $z_1,..., z_N$. Now form a new polynomial $Q(z)$ of degree $N(N-1)/2$ which I call the discriminantal polynomial of $P(z)$. The set of roots of $Q(z)$ are all possible squares of differences $(z_i-z_j)^2$ with $i\neq j$. Observe that the constant term of $Q(z)$ is the usual discriminant of $P(z)$ and other coefficients are also polynomials in the coefficients of $P(z)$.

I wonder if this polynomial has already appeared in the literature as well as if there is a reasonable way to calculate it (for example, as the characteristic polynomial of some appropriate matrix). 

                Boris Shapiro, Stockholm
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    $\begingroup$ I'm not sure this is MO level. The resultant in $y$ of $P(x+y)$ and $P(y)$ has roots $z_i-z_j$. If $H$ denotes this polynomial, then $H(x)H(-x)=Q(x^2)$ for some polynomial $Q$ which is your $Q$, up to a power of $x$. $\endgroup$
    – Felipe Voloch
    Jun 17, 2014 at 21:08
  • $\begingroup$ @FelipeVoloch As you've noted, there's an elementary characterization which gives a way to calculate it. But I think that since he asked the question "Has this polynomial already appeared in the literature?", it is fine for MO. $\endgroup$
    – Joe Silverman
    Jun 18, 2014 at 0:35
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    $\begingroup$ @JoeSilverman I am not so sure about that. "Has blank appeared in the literature?" can get pretty vague. Also, you can never answer negatively for sure. $\endgroup$
    – Felipe Voloch
    Jun 18, 2014 at 1:10
  • $\begingroup$ Dear Felipe and Joe, thank you for your kind comments. Since I am professional, I am almost embarrassed that I have not found this easy recipe myself. I need this polynomial for a rather advanced project on multivalued abelian differential on Riemann surfaces. Yours Boris $\endgroup$
    – user53092
    Jun 19, 2014 at 7:33
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    $\begingroup$ Somewhat related is the Bezoutant $$ \Delta_k(x):=\det(B_k(x)) =\sum_{i_1<i_2<\dots<i_k}(x_{i_1}-x_{i_2})^2\dots(x_{i_1}-x_{i_k})^2 \dots(x_{i_{k-1}}-x_{i_k})^2 $$ see 3.1 in <A Href="mat.univie.ac.at/~michor/roots.pdf">this paper</A> which contains many references. $\endgroup$
    – Peter Michor
    Jun 23, 2014 at 19:59

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