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Let $n \geq 2$ be an integer, and let $f(x) = \prod\limits_{k = 1}^n(x - \alpha_k)$ be a monic irreducible polynomial in $\mathbb Z[x]$, with the property that $f(-\alpha_k) \neq 0$ for any $k = 1, 2, \ldots, n$.

Is there anything meaningful that we can say about $\operatorname{Res}(f(x), f(-x))$, the resultant of $f(x)$ and $f(-x)$?

To rephrase, what can be said about the value of the product $\prod\limits_{k = 1}^nf(-\alpha_k)$? Perhaps, it can be expressed somehow through $n$ and the discriminant of $f(x)$?

One thing that I can note is that $f(-\alpha_1), \ldots, f(-\alpha_k)$ are algebraic conjugates, which means that their product is equal to the norm of $f(-\alpha_1)$. Thus, up to a sign, the product $\prod\limits_{k = 1}^nf(-\alpha_k)$ is equal to the constant coefficient of the minimal polynomial of $f(-\alpha_1)$. But what is this constant coefficient is a mystery.

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    $\begingroup$ Just $n$ and the discriminant of $f$ are definitely not enough to determine the resultant you want - there will be polynomials of same degree and same discriminant but different resultants. $\endgroup$
    – Wojowu
    Jan 29, 2021 at 0:52
  • $\begingroup$ @Wojowu, I see, that makes sense. $\endgroup$
    – Anton
    Jan 29, 2021 at 1:16
  • $\begingroup$ Up to sign it's $\prod_{i=1}^n \prod_{j=1}^n (\alpha_i + \alpha_j)$. $\endgroup$
    – Will Sawin
    Jan 29, 2021 at 1:22
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    $\begingroup$ @Anton That follows from the product formula $\prod_{i,j=1}^n (\alpha_i + \alpha_j)$. The $i=j$ factors give $2^n \prod_i \alpha_i = (-2)^n a_n$, and each of the $i\neq j$ factors appears twice. (Their product, which you call $b_n$, is $\pm$ the resultant w.r.t. $\beta$ of the linear and constant coefficients of the remainder of $f(x) \bmod x^2-\beta$.) $\endgroup$ Jan 29, 2021 at 4:20
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    $\begingroup$ @NoamD.Elkies, thanks! I think your observation regarding the connection between $b_n$ and the remainder of $f(x)$ mod $x^2 - \beta$ is actually what I was looking for. $\endgroup$
    – Anton
    Jan 29, 2021 at 12:47

2 Answers 2

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We have $\mathrm{Res}(f(x),f(-x))=2^n a_n P(\alpha)^2$, where $P(\alpha)=\prod_{1\leq i<j\leq n}(\alpha_i+\alpha_j)$. By e.g. the case $d=2$ of Exercise 7.30 in Enumerative Combinatorics, vol. 2, we have $P(\alpha)=s_{n-1,n-2,\dots,1}(\alpha)$, where $s_{n-1,n-2,\dots,1}$ is a Schur function. By the dual Jacobi-Trudi identity (Corollary 7.16.2 of the above reference), we get $$ \mathrm{Res}(f(x),f(-x)) = 2^n a_n \left( \det[a_{n-2i+j}]_{i,j=1}^{n-1}\right)^2, $$ where we set $a_0=1$ and $a_{-k}=0$ for $k>0$. For instance, when $n=3$ the determinant is $$ \left| \begin{array}{cc} a_2 & a_3\\ 1 & a_1\end{array} \right| =a_2a_1-a_3. $$

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See the discussion above.

Also, if $f(x) = x^n + \dotsb + a_{n-1}x + a_n$, then

  • for $n = 2$, $\operatorname{Res}(f(x), f(-x)) = 2^2a_2a_1^2$

  • for $n = 3$, $\operatorname{Res}(f(x), f(-x)) = 2^3a_3(a_3 - a_1a_2)^2$

  • for $n = 4$, $\operatorname{Res}(f(x), f(-x)) = 2^4a_4(a_4a_1^2 - a_1a_2a_3 + a_3^2)^2$.

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  • $\begingroup$ I edited in a link to @RichardStanley's answer, which I assume is what you meant by "above". $\endgroup$
    – LSpice
    Jan 29, 2021 at 20:18
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    $\begingroup$ @LSpice Since Richard Stanley’s answer was posted 11 hours after Anton’s, that’s clearly impossible. The “discussion above” must have referred to the comment thread under the question. $\endgroup$ Jan 29, 2021 at 20:34
  • $\begingroup$ @EmilJeřábek, agreed, thanks; I didn't notice the time stamps. Hopefully @‍Anton will edit in the appropriate links themselves (I am reluctant to guess again). Reminder about how to link to comments. ‍ $\endgroup$
    – LSpice
    Jan 29, 2021 at 21:09

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