Given a monic polynomial $p(t) = t^n + … + c_1 t + c_0$ with integer (or rational) coefficients and with roots $a_1, \dots a_n$, we can compute its discriminant, which is defined to be $\prod_{i< j}(a_i – a_j)^2$.
In my case, I have a polynomial which is the characteristic polynomial of some invertible matrix $T$. It is palindromic — i.e., $c_{n-i} = c_i$ for all $0 \leq i \leq n$ — so the roots come in inverse pairs $a$ and $\frac{1}{a}$. There are no repeated roots, so the discriminant is non-zero.
My question is: is there any way of knowing which primes divide this discriminant, i.e. from the coefficients of the polynomial or from the matrix $T$?
Best Answer
I disagree with the definition of the discriminant as the resultant of $P$ and $P'$. When $P$ is a polynomial with integer coefficients, then a prime $q$ should divide the discriminant of $P$ if and only if the reduction of $P$ modulo $q$ has a multiple root (possibly at infinity, when the degree decreases by at least 2 under reduction). But now consider $P=2X^2+ 3X+1$. The resultant of $P$ and $P'$ is $-2$, and the reduction of $P$ modulo 2 has no multiple root. In this case, the well known discriminant $b^2-4ac$ is actually 1. The correct relation between the discriminant and the resultant for a polynomial $P(t)=a_nt^n+\cdots+a_1t+a_0$ is $\mathrm{disc}(P)= (-1)^{n(n-1)/2}\mathrm{res}(P,P')/a_n$.